Solutions to Exercises

Appendix C Solutions to Exercises

1 Sets
1.4 Exercises

1.4.1.

Solution.

\begin{align*} A_1 \amp = \set{1} \\ A_2 \amp = \set{-1,0,1} \\ A_3 \amp = \set{ 3, 6, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216}\\ A_4 \amp = \set{\ldots, -12, -7, -2, 3, 8, 13, \ldots}\\ A_5 \amp = \set{2,3}\\ A_6 \amp = \set{7,11,13,17,19}\\ A_7 \amp = \set{-2,-1,0,1,2,3,4,5,6,7,8 } \end{align*}

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1.4.2.

Solution.

$\displaystyle A=\set{5n\st n\in\mathbb{N}}$
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$\displaystyle B=\set{n\in\mathbb N\st 10\leq n\leq 100}$
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$\displaystyle C=\set{n^2-1\st n\in\mathbb N}$
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$\displaystyle D=\set{\frac{m}{m^2+1} \st m \in \mathbb Z}$
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$\displaystyle E=\set{n\in\mathbb{N}\st n=2^{(2^k)} \text{ for some nonnegative integer } k}$
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$\displaystyle F=\set{2^a 3^b \st (a,b \in \mathbb Z), (a,b \geq 0), \text{ and }(a+b\neq 0)}$
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1.4.3.

Solution.

We can write $A$ in set builder notation in the following two equivalent ways:

\begin{equation*} A=\set{m\st m\in\mathbb{Z}, \; 0\leq m\leq 100, \; m\text{ is even}} = \set{2k\st k\in\mathbb{Z}, \; 0\leq k\leq 50} \end{equation*}

We could also describe it in words by saying that $A$ is the set of all even non-negative integers that are at most 100.

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We can write $B$ in set builder notation as follows:

\begin{equation*} B = \set{3^n \st n\in\mathbb{N}} \end{equation*}

Also, $B$ is the set of all positive integer powers of three.

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The set $C$ is given in set builder notation. Listing out its elements,

\begin{equation*} C=\set{-3, -2, -1, 0, 1, 2, 3} \end{equation*}

Also, $C$ is the set of all integers that are at most three in absolute value.

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Listing out the elements of $D\text{,}$ we have

\begin{equation*} D =\set{\dots,-11, -7, -3, 1, 5, 9, 13, \dots} \end{equation*}

Since the elements of $D$ “go off” to both positive and negative infinity, we have ellipsis at the beginning and end of the set. We can describe the set in words by saying that $D$ is the set of all integers that are one more than a (possibly negative) multiple of four. Alternatively, we could describe $D$ as the set of all integers that have a remainder of one when divided by four.

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The set $E$ is described in words. The set $E$ can be given by listing elements, and using set builder notation, as follows:

\begin{equation*} E = \set{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},\dots}=\set{\frac{1}{n}\st n\in\mathbb{N}} \end{equation*}

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The set $F$ can be given by listing elements, and using set builder notation, as follows:

\begin{equation*} F = \set{\dots,-18, -13, -8, -3, 2, 7, 12, 17, 22,\dots}=\set{5k+2 \st k\in\mathbb{Z}} \end{equation*}

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1.4.4.

Solution.

The set is ill-defined because we haven’t written out enough elements to determine the set’s “rule”. We could, for example, interpret this set as the set of all positive even integers,

\begin{equation*} \set{2, 4, 6, 8,\dots}, \end{equation*}

or the set of all positive integer powers of two,

\begin{equation*} \set{2, 4, 8, 16,\dots}. \end{equation*}

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1.4.5.

Solution.

We have not specified what the variable $k$ is, so this set is ambiguous. For example, one person might assume that $k$ is a constant, while another person assumes $k\in \mathbb{R}\text{.}$ Neither of these gives the desired set.
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We have used the letter $N$ rather than the symbol $\mathbb{N}\text{,}$ which indicates the natural numbers.
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We use the letter $c$ to represent this set. It is more conventional to use capital letters to denote sets.
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We have used the letter $k$ to write the set elements $2k+1\text{,}$ and then we have stated that another variable $n$ is a natural number. Because of this, the variable $k$ is not defined, and the set is ambiguous.
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We have used the Greek letter epsilon ($\oldepsilon$) rather than the “element of” symbol $\in\text{.}$
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We have used capital letters to denote elements of the set. It is more conventional to use lowercase letters for set elements when possible.
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We have used the Greek letter epsilon ($\varepsilon$) rather than the “element of” symbol $\in\text{.}$ Note that this is the same issue as part (e), but is included twice because of the two representations of epsilon.
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This is an issue you would encounter if you are typing up math using a program such as LaTeX. We have written this set using normal text, rather than in math mode. In LaTeX, one uses dollar signs (e.g. “$x$”) or backslash with parentheses (e.g. “$x$”)around their math. This produces italicized math characters such as “$x$” instead of “x.” Notice that in the set, none of the letters are italicized, and the word “in” is used, rather than the symbol “$\in$”. This is very sloppy, and makes it more difficult for the reader to follow. When typing, you should always be careful about mixing up text and mathematics. Make sure you type characters deliberately to avoid confusion.
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1.4.6.

Solution.

The goal of this question is to highlight that the empty set is not the same as the zero set, nor is it same as the set containing the empty set. When in doubt, refer the empty bag analogy provided in Section 1.2.

False. The empty set contains no elements, but the set $\set{0}$ contains $1$ element, namely the number 0.
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False. This is for the same reason as above. $\{\es \}$ is a set with $1$ element; it contains the empty set.
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True. There are $0$ elements in the empty set.
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False. The set $\{\es\}$ is a set, containing a set with no elements (this is like a bag containing another empty bag), whereas the set $\{\{\es\}\}$ is a set containing a set containing a set with no elements (like a bag containing a bag containing a third empty bag).
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True. These are both sets containing a set with no elements.
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1.4.7.

Solution.

The only integer that belongs to $A$ is $-1\text{,}$ so all of the numbers listed are not elements of $A\text{.}$
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The numbers $2$ and $8$ do not belong to $B$ because their squares do not exceed 100; $-12$ is not an element of $B$ because it is negative.
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The elements of $C$ are the set containing $2\text{,}$ the set containing $8\text{,}$ and the set containing $-12\text{.}$ $C$ itself does not actually have the elements $2\text{,}$ $8\text{,}$ and $-12\text{.}$
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The number $2$ does not belong to $D$ because it is not a multiple of $4\text{,}$ $8$ does not belong to $D$ because it is not an odd multiple of $4\text{,}$ and $-12$ does not belong to $D$ because it is not a positive multiple of $4\text{.}$
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1.4.8.

Solution.

$\mathbb{Z} \neq\{ a: a \in \mathbb{N} \text{ or } -a\in\mathbb{N}\}\text{.}$ This is because $0\in \mathbb{Z}\text{,}$ but $0 \not\in \{ a: a \in \mathbb{N} \text{ or } -a\in\mathbb{N}\}\text{.}$
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$\{1,2,2,3,3,3,2,2,1\}=\{1,1,1,1,1,2,2,2,2,2,3,3,3,3,3\}\text{.}$ The number of occurrences of an element of a set does not matter. Both of these sets are equal to the set $\{1,2,3\}\text{.}$
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$\{d : d \text{ is a day of the week with 40 hours}\}=\{w: w \text{ is a week with 6 days}\}\text{.}$ Since there are no days with $40$ hours and no weeks with $6$ days, both of these are equal to the empty set.
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$\{p: p \text{ is prime, } p\lt 42\}\neq\{1,2,3,5,7,11,13,17,19,23,29,31,37,41\}\text{.}$ The number $1$ is not a prime, thus there is an element in the set on the right hand side that is not present in the left hand side.
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1.4.9.

Solution.

The sets $B\text{,}$ $C\text{,}$ $E\text{,}$ and $F$ equal $S\text{.}$ The sets $A$ and $D$ do not.

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$A\neq S$ because $1\not\in A$ but $1\in S\text{.}$
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$B=S$ since the elements of $B$ are just the reciprocal of the absolute value of non-zero integers. Hence $-1,1$ both give $1/1\text{,}$ $-2,+2$ both give $1/2$ and so on.
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$C=S$ since the elements of $C$ can be reduced to the form $1/n\text{,}$ $n\in\mathbb{N}\text{.}$ Indeed, $1/n=2/k$ when $k=2n\text{,}$ and in the definition of $C\text{,}$ we require that $k$ is even.
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$D\neq S$ because $3/2\in D$ but $3/2\not\in S\text{.}$ Since $D$ is the set of rational numbers, there are many elements in $D$ that are not in $S\text{.}$
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$E=S$ because we compensate for writing its elements as $\frac{1}{n-1}$ by restricting to $n\in\mathbb{N}$ that are greater than one. Then $n=2$ gives $1/1\text{,}$ $n=3$ gives $1/2\text{,}$ and so on.
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$F=S$ since the requirement that $m\in\mathbb{Z}, \; m>0 $ is the same as saying $m\in\mathbb{N}\text{.}$
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2 A little logic
2.7 Exercises

2.7.1.

Solution.

This is a statement, and it is an implication. Since the hypothesis of the implication is true, but the conclusion is false, the full statement is false.
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This is a statement. Since the hypothesis of the implication is false, the full statement is automatically true.
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This is an open sentence, since it is true or false depending on the function $f\text{.}$
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This is a statement, and a conjunction of the statement “13 is prime” and the statement “6 is prime.” Since the first is true, but the second is false, the conjunction is false.
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This is a statement, and a disjunction of the statement “13 is prime” and the statement “6 is prime.” Since the first is true, but the second is false, the disjunction is true.
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This is an open sentence, since it is true or false depending on the circle.
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2.7.2.

Solution.

True. A weekend consists of the days Saturday and Sunday, hence, if it is Saturday, we can conclude that it is the weekend.
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False. Since a weekend consists of Saturday and Sunday, knowing that it is a weekend is not enough to conclude that it is Saturday.
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True. The hypothesis is false, so we automatically conclude that the implication is true.
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2.7.3.

Solution.

False. Recall that we can also read this statement as “whenever $x$ is an even number, it is also twice a natural number .” Notice that the set in the conclusion is the set of positive even numbers, so taking a negative even number, such as $x=-2\text{,}$ we see that there is no $n\in\mathbb N$ such that $x= 2n\text{.}$
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False. The statement can also be read as “every prime number is odd,” but $2$ is an even prime number, so the statement must be false.
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False. This statement can be read as “every multiple of $3$ is a multiple of $6\text{.}$” The number $3$ is a multiple of $3$ (in particular, $3=3\cdot 1$), but it is not a multiple of $6\text{.}$
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True. This statement can be read as “every multiple of $6$ is a multiple of $3\text{.}$” Here is a brief proof of our claim:
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Proof.
Suppose that $x\in \{6k : k\in \mathbb{Z}\}\text{.}$ By definition, we may write $x = 6k$ for some $k\in \mathbb{Z}\text{.}$ Factoring the $6$ yields $x = 3(2k)\text{.}$ Since $2k$ is an integer, we have shown that we can write $x$ as a multiple of $3\text{.}$ Hence we may conclude that $x\in \{3k : k\in \mathbb{Z}\}\text{,}$ as desired.
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We will do many more proofs like this in Chapter 3.
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2.7.4.

Solution.

False. The statement “$3$ is even” is false, so the conjunction is false.
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True. The statement “$3$ is prime” is true, so the disjunction is true.
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True. Both of the statements “$x^2 > x$ when $x>1$” and “$18$ is composite” are true, so the conjunction is true.
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True. Both of the statements “$x^2 > x$ when $x>1$” and “$18$ is composite” are true, so the disjunction is also true.
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2.7.5.

Solution.

This sentence can be written as $P\land Q$ for

\begin{align*} P: \amp \quad 8 \text{ is even },\\ Q: \amp \quad 5 \text{ is prime}. \end{align*}

This is a statement.

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This sentence can be written as $(P\land Q)\implies R$ for

\begin{align*} P: \amp \quad n \text{ is a multiple of } 4,\\ Q: \amp \quad n \text{ is a multple of } 6,\\ R: \amp \quad n \text{ is a multiple of } 24. \end{align*}

Even though $P\text{,}$ $Q$ and $R$ are open sentences, this sentence is a conditional statement.

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This sentence can be written as $\neg P\implies \big(Q\land (\neg R)\big)$ for

\begin{align*} P: \amp \quad n \text{ is a multiple of } 10,\\ Q: \amp \quad n \text{ is a multiple of } 2,\\ R: \amp \quad n \text{ is a multiple of } 5. \end{align*}

This is a statement.

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This expression can be written as $P\land Q$ for

\begin{align*} P: \amp \quad x\geq 3,\\ Q: \amp \quad x\leq 6. \end{align*}

This is an open sentence.

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This sentence can be written as $R\implies (P\lor Q)$ for

\begin{align*} P: \amp \quad x \text{ is less than } -2,\\ Q: \amp \quad x \text{ is greater than } 2,\\ R: \amp \quad x^2 \text{ is greater than } 4. \end{align*}

This is a statement.

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This sentence can be written as $P\implies (Q\implies R)$ for

\begin{align*} P: \amp \quad f(x) \text{ is differentiable everywhere},\\ Q: \amp \quad x \text{ is a local maximum of } f(x),\\ R: \amp \quad f'(x)=0. \end{align*}

This is a statement.

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2.7.6.

Solution.

We list a few possible English statements:

- If $x$ is a real number, then $x^2$ is also a real number and $x^2$ is non-negative.
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- If $x$ is a real number, then $x^2$ is a non-negative real number.
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- The square of a real number is non-negative and real.
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- $4$ is an element of the set of positive, even numbers.
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- We can write $4=2\ell$ for some $\ell\in \mathbb{N}\text{.}$
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- $4$ is a positive even number.
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- For a natural number $x\text{,}$ it is not the case that $x^2=0\text{.}$
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- When $x$ is a natural number, $x^2$ is never $0\text{.}$
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- The square of a natural number is non-zero.
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- If $x$ is an integer, then $x$ is an even number or $x$ is an odd number.
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- If $x$ is an integer, then it is either even or odd.
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- Every integer is even or odd.
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2.7.7.

Solution.

We can refer to the truth table for the implication, except this time the hypothesis is $\neg P$ rather than $P\text{.}$
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$P$ $Q$ $\neg P$ $(\neg P)\implies Q$

T T F T

T F F T

F T T T

F F T F

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We can use the truth table from part (a) to help figure out the new truth table.
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$P$ $Q$ $(\neg P)\implies Q$ $P\land Q$ $(P\land Q)\lor \big((\neg P)\implies Q \big)$

T T T T T

T F T F T

F T T F T

F F F F F

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From the truth table, we see that $P\land (\neg P)$ is always false.
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$P$ $\neg P$ $P\land (\neg P) $

T F F

F T F

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From the truth table, we see that $P\lor (\neg P)$ is always true.
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$P$ $\neg P$ $P\lor (\neg P) $

T F T

F T T

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From the truth table, we see that $(P\implies Q) \iff (Q\implies P) $ is true only when $P$ and $Q$ have the same truth value.
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$P$ $Q$ $P\implies Q$ $Q\implies P$ $(P\implies Q) \iff (Q\implies P) $

T T T T T

T F F T F

F T T F F

F F T T T

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2.7.8.

Solution.

We start with the truth table for $P\implies Q$ given in Definition 2.4.1. The next column (fourth column of the table) gives the truth table for $\neg(P\implies Q)\text{.}$ We compare this column to the last column of the table, which is the truth table for $P\land\neg Q\text{.}$

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$P$	$Q$	$P\implies Q$	$\neg(P\implies Q)$	$P\land\neg Q$
T	T	T	F	F
T	F	F	T	T
F	T	T	F	F
F	F	T	F	F

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2.7.9.

Solution.

Recall the truth table of the implication:

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$P$	$Q$	$P\implies Q$
T	T	T
T	F	F
F	T	T
F	F	T

We cannot determine whether or not it was raining on Monday. The conclusion of the implication ($Q=$ “I brought an umbrella to work”) is true, which means that the hypothesis ($P=$ “it’s raining”) may be true or false, while the implication remains true. See the first and third rows of the truth table.
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It was not raining on Tuesday. Because the conclusion of the implication if false, the hypothesis must also be false. If the hypothesis were true, then the implication would be false. See the second and fourth rows of the truth table.
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It was raining on Wednesday. Because the hypothesis of the implication ($P=$ “I am late for work”) is true, the conclusion ($Q=$ “it’s raining”) must also be true. If the conclusion were false, then the implication would also be false. See the first and second rows of the truth table.
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We cannot determine whether or not it was raining on Thursday. The hypothesis of the implication is false, which means that the conclusion may be true or false, while the implication remains true. See the third and fourth rows of the truth table.
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2.7.10.

Solution.

We cannot conclude anything. We have not satisfied the hypothesis of the conditional statement.
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We conclude that sailors should take warning by modus ponens.
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We cannot conclude anything. If we did, we would be “affirming the consequent,” and this is a false deduction. There are many other things that could cause sailors to take warning, such as pirates or a group of hungry sharks.
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We cannot conclude anything. If we did, we would be “denying the antecedent,” and this is a false deduction.
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We conclude that it is not true that (the sky is red and it is morning) by modus tollens.
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2.7.11.

Solution.

If $n$ is not a multiple of $24\text{,}$ then it is not a multiple of $4$ or $6\text{.}$
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Let’s start by rewriting the original statement to make it a bit more clear:
- If $n$ is not a multiple of $10\text{,}$ then (it is a multiple of $2$) and (it is not a multiple of $5$).
  
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This means it easier to find the contrapositive:
- If $n$ is not a multiple of $2$ or it is a multiple of $5\text{,}$ then $n$ is a multiple of 10.
  
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First we rearrange the original statement to be in a format where we can more easily identify the hypothesis and the conclusion:
- If $x$ is a real number whose square is greater than $4\text{,}$ then $x$ is less than $-2$ or greater than $2\text{.}$
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We should also make sure we are careful with the inequalities. In the given statement these are strict, $x \lt -2\text{,}$ $x \gt 2\text{,}$ so when we form the contrapositive they will be inclusive, $x \geq -2\text{,}$ $x \leq 2\text{.}$ So the contrapositive is
- If $x$ is between $-2$ and $2$ (inclusive), then the square of $x$ is less than or equal to $4\text{.}$
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$(x^2\not \in \mathbb{R}) \lor (x^2 \lt 0) \implies x\not \in \mathbb{R}\text{.}$
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$x^2=0\implies x\not\in \mathbb{N}\text{.}$
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$\displaystyle x \not\in \set{6k:k\in \mathbb{Z}} \implies x\not\in \set{3k:k\in \mathbb{Z}}.$
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2.7.12.

Solution.

Let $P=$ “$m^2$ is even,” $Q=$ “$m$ is even,” and $R=$ “$m^2$ is divisible by 4.”

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The statement “If $m$ is odd, then $m^2$ is odd” converts to $(\neg Q)\implies (\neg P).$ Its contrapositive is $\neg(\neg P)\implies \neg(\neg Q)\text{,}$ which is the same as $P\implies Q\text{.}$ In words, the contrapositive is “If $m^2$ is even, then $m$ is even.” Since an implication and its contrapositive have the same truth table, and $(\neg Q)\implies (\neg P)$ is true, $P\implies Q$ is also true.

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The statement “If $m$ is even, then $m^2$ is divisible by 4” converts to $Q\implies R\text{.}$ Its contrapositive is $(\neg R)\implies (\neg Q)\text{,}$ or “If $m^2$ is not divisible by 4, then $m$ is odd.”

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We have the four distinct implications, $P\implies Q\text{,}$ $(\neg Q)\implies (\neg P)\text{,}$ $Q\implies R\text{,}$ and $(\neg R)\implies (\neg Q)\text{.}$

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Chaining the implications $P\implies Q$ and $Q\implies R\text{,}$ we end up with the implication $P\implies R\text{,}$ or

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“If $m^2$ is even, then $m^2$ is divisible by 4.”
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Chaining the implications $(\neg R)\implies (\neg Q)$ and $(\neg Q)\implies (\neg P)\text{,}$ we end up with the implication $(\neg R)\implies (\neg P)\text{,}$ or

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“If $m^2$ is not divisible by 4, then $m^2$ is odd.”
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Notice that this implication is the contrapositive of the other implication we formed.

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2.7.13.

Solution.

Let $P=$ “$p$ is prime” and $Q=$ “$\sqrt{p}$ is irrational.”

The contrapositive of $P\implies Q$ is $\neg Q\implies \neg P\text{,}$ or “If $\sqrt{p}$ is rational, then $p$ is not prime.”

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The converse of $P\implies Q$ is $Q\implies P\text{,}$ or “If $\sqrt{p}$ is irrational, then $p$ is prime.”

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The inverse of $P\implies Q$ is $\neg P \implies \neg Q\text{,}$ “If $p$ is not prime, then $\sqrt{p}$ is rational.”

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Since the truth tables of the implication and its contrapositive are the same, the contrapositive is also a true statement. Since the truth tables of the converse and inverse differ from that of the original implication, we cannot determine whether or not the converse and inverse are true statements from the information provided (although they are indeed false, as evidenced by the fact that $\sqrt{6}$ is irrational; this is something you could prove using the tools developed later on in this book).

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2.7.14.

Solution.

$P$ is true and $R$ is false. The truth value of $Q$ cannot be determined from the given information.

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2.7.15.

Solution.

$P$ is true, $Q$ is true, and $R$ is false.

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2.7.16.

Solution.

$P$ is true, $Q$ is false, and $R$ is false.

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3 Direct proofs
3.5 Exercises

3.5.1.

Solution.

Proof.

Assume $n$ is even. Then we know $n=2k$ for some $k\in\mathbb{Z}\text{.}$ Hence, $n^2+3n+5=(2k)^2+3(2k)+5=2(2k^2+3k+2)+1\text{.}$ Since $2k^2+3k+2\in\mathbb{Z}$ for $k\in\mathbb{Z}\text{,}$ we see $n^2+3n+5$ is odd.

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3.5.2.

Solution.

Proof.

Assume $n,m$ are odd integers. Then we know $n=2k+1$ and $m=2\ell+1$ for some $k,\ell\in\mathbb{Z}\text{.}$ Thus, $nm=(2k+1)(2\ell+1)=2(2k\ell+k+\ell)+1\text{.}$ Now since $k,\ell \in \mathbb{Z}$ we know that $2k\ell+k+\ell\in\mathbb{Z}\text{,}$ and so $nm$ is odd.

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3.5.3.

Solution.

The sum of two odd numbers is even.
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Proof.

Let $m,n$ be two odd numbers. Then there are some $k,\ell\in\mathbb{Z}$ such that $m=2k+1$ and $n=2\ell+1\text{,}$ and so

\begin{equation*} m+n = (2k+1)+(2\ell+1) =2(k+\ell+1). \end{equation*}

Since $k,\ell\in\mathbb{Z}\text{,}$ we also have that $k+\ell+1$ is an integer, and thus $m+n$ is even.

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The sum of two even numbers is even.
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Proof.

Let $m,n$ be two even numbers. Then there are some $k,\ell\in\mathbb{Z}$ such that $m=2k$ and $n=2\ell\text{,}$ and so

\begin{equation*} m+n = (2k)+(2\ell) =2(k+\ell). \end{equation*}

Since $k,\ell\in\mathbb{Z}\text{,}$ we also have that $k+\ell$ is an integer, and thus $m+n$ is even.

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The sum of an even number and an odd number is odd.
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Proof.

Let $m$ be an even number and $n$ be an odd number. Then there are some $k,\ell\in\mathbb{Z}$ such that $m=2k$ and $n=2\ell+1\text{,}$ and so

\begin{equation*} m+n = (2k)+(2\ell+1) =2(k+\ell)+1. \end{equation*}

Since $k,\ell\in\mathbb{Z}\text{,}$ we also have that $k+\ell$ is an integer, and thus $m+n$ is odd.

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The product of two even numbers is even.
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Proof.

Let $m,n$ be two even numbers. Then there are some $k,\ell\in\mathbb{Z}$ such that $m=2k$ and $n=2\ell\text{,}$ and so

\begin{equation*} mn = (2k)(2\ell) =2(2k\ell). \end{equation*}

Since $k,\ell\in\mathbb{Z}\text{,}$ we also have that $2k\ell$ is an integer, and thus $mn$ is even.

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The product of an even number and an odd number is even.
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Proof.

Let $m$ be an even number and $n$ be an odd number. Then there are some $k,\ell\in\mathbb{Z}$ such that $m=2k$ and $n=2\ell+1\text{,}$ and so

\begin{equation*} mn = (2k)(2\ell+1) =2( 2k\ell+k). \end{equation*}

Since $k,\ell\in\mathbb{Z}\text{,}$ we also have that $2k\ell+k)$ is an integer, and thus $mn$ is even.

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3.5.4.

Solution.

We have accidentally proven the converse rather than the desired statement. We started by assuming the conclusion, and ended up showing that the hypothesis is true. A correct proof can be given by using cases or proving the contrapositive Chapter 5.

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3.5.5.

Solution.

The main issue with proof as written is that the variables are undefined. The variables $a\text{,}$ $b\text{,}$ $k\text{,}$ and $\ell$ came out of nowhere! Presumably, $a$ and $b$ are odd integers, and $k$ and $\ell$ are integers, but the proof didn’t define them as such. Moreover, note that saying $k+\ell+1\in \mathbb{Z}$ does not guarantee that $k,\ell\in\mathbb{Z}\text{;}$ for example, taking $k=\ell=1/2$ shows why.

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So while the ideas of the proof are correct, it is, at best, very sloppy. You should not assume that the reader will “just get it” - you need to explain what “it” is!

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Let’s rewrite the proof with these issues resolved:

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Proof.

Let $a$ and $b$ be odd integers. Then there are $k,\ell\in\mathbb{Z}$ such that $a=2k+1$ and $b=2\ell+1\text{.}$ Hence

\begin{equation*} a+b = (2k+1)+(2\ell+1)=2(k+\ell+1). \end{equation*}

3.5.9.

Solution.

Proof.

Let $n\in\mathbb{Z}$ and assume that $3\mid (n-4)\text{.}$ Then we see that $(n-4)=3k$ for some $k\in\mathbb{Z}\text{.}$ Thus, $n=3k+4\text{,}$ which implies $n^2-1= (3k+4)^2-1=(9k^2+24k+16)-1= 3(3k^2+8k+5)\text{.}$ Therefore, since $k\in\mathbb{Z}\text{,}$ we know that $3k^2+8k+5\in\mathbb{Z}\text{,}$ and so we see $3\mid (n^2-1)$ as required.

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3.5.10.

Solution.

The issue with this proof is that we used the variable $k$ in two different situations. In general, the integer multiple that $b$ is of $a$ would be different than the integer multiple that $c$ is of $b\text{.}$ While the logic of the proof is mostly correct, we end up with the statement $c=k^2a$ (that $c$ is a square multiple of $a$), which is incorrect. For example, if $a$ is positive while $c$ is negative, then we definitely can’t find an integer $k$ satisfying $c=k^2a\text{.}$

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Let’s rewrite the proof with these issues resolved:

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Proof.

Assume $a\text{,}$ $b\text{,}$ and $c$ are integers such that $a\mid b$ and $b\mid c\text{.}$ Since $a$ divides $b\text{,}$ we have that $b=ka$ for some $k\in\mathbb{Z}\text{.}$ Moreover, since $b$ divides $c\text{,}$ we have that $c=\ell b$ for some $\ell\in\mathbb{Z}\text{.}$ But then

\begin{equation*} c=k(\ell a)=(k\ell)a, \end{equation*}

Since $k$ and $\ell$ are integers, $k\ell\in\mathbb{Z}\text{,}$ and it follows that $a$ divides $c\text{.}$

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3.5.11.

Solution.

First, this solution is quite terse, and it could benefit from more explanation. Explaining the steps taken in each line could make the error more clear:

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Assume that $x=y\text{.}$ Then

\begin{align*} x^2 \amp = xy \amp \text{multiplying by } x\\ \Rightarrow x^2-y^2 \amp = xy - y^2 \amp \text{Subtracting } y^2\\ \Rightarrow (x+y)(x-y) \amp = y(x-y) \amp \text{Factoring}\\ \Rightarrow x+y \amp = y \amp \text{Dividing by } (x-y)\\ \Rightarrow 2y \amp = y \amp \text{Using } x=y \end{align*}

Letting $x=y=1\text{,}$ we have shown that $2=1\text{.}$

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The above proof is now more clear to read, and we notice that in the fourth line, we divide by $x-y\text{.}$ We should always take care when we divide; we cannot divide by zero. Recall that we previously assumed that $x=y\text{,}$ thus we are dividing by $0$ in the fourth line. This is our logical error, and we can be reassured that $2\neq 1\text{.}$

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In this proof, a careless division of an equation caused all the problems. Similar problems can arise when we work with inequalities. For example, multiplication or division of an inequality by a negative number will change the sign of the inequality, so we must take extra care to check if quantities are negative or not.

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3.5.12.

Solution.

Proof.

Start by letting $x \in \mathbb{R}$ and write $n = \lfloor x \rfloor\text{.}$ By definition of the floor function $n$ is an integer.

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Now we prove each implication in turn.

Assume that $n = x\text{.}$ By definition of the floor function $n$ is an integer, so $x$ is an integer as required.
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Now assume that $x \in \mathbb{Z}\text{.}$ We prove the result by showing that $n \leq x$ and $n \geq x\text{.}$
- Then, since the floor function returns the greatest integer less than or equal to $x\text{,}$ we know that $n \leq x\text{.}$
  
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- By definition of the floor function, we know that for any integer $k$ satisfying $k \leq x\text{,}$ we must have that $k \leq n\text{,}$ since $n$ is the greatest such integer. Since $x$ is an integer and $x \leq x\text{,}$ we know that $x \leq n$
  
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Because $x \leq n$ and $x \geq n\text{,}$ we know $x = n = \lfloor x \rfloor\text{.}$

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What we are really doing here is proving a “biconditional statement”. We will see much more on this in the next couple of chapters.

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3.5.13.

Solution.

Proof.

Assume that $a, b$ are integer roots, then, by definition, there are natural numbers $k,\ell$ and integers $m,n$ such that $a^k=n$ and $b^\ell=m\text{.}$ Therefore we see that $(ab)^{k\ell}=n^\ell m^k\text{.}$ Since $k\ell\in\mathbb{N}$ and $n^\ell m^k\in\mathbb{Z}\text{,}$ we see that $ab$ is also an integer root.

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3.5.14.

Solution.

The issue with the proof is that it is written in the wrong order — we start with the conclusion of the implication, and end up with the hypothesis of the implication. Essentially, the proof is backwards. Often, when doing scratchwork for proofs involving inequalities, you will end up with a backwards proof. However, when you’re writing up the proof formally, always remember to start off with the hypothesis.

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Here’s a correct proof of the statement:

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Solution.

Proof.

Assume that $0 \lt y \lt x \text{.}$ This means that $x-y\gt 0\text{.}$ Then, since $x,y\gt 0\text{,}$ we can factor the expression $x-y$ and get

\begin{equation*} (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) \gt 0 \end{equation*}

and since the square root function is nonnegative, and $x,y\gt 0\text{,}$ we also know that $\sqrt{y} + \sqrt{x} \gt 0\text{.}$ So dividing both sides of the inequality by $\sqrt{x}+\sqrt{y}$ gives

\begin{equation*} \sqrt{x}-\sqrt{y} \gt 0 \end{equation*}

Hence we conclude that $\sqrt{x}\gt \sqrt{y}$ as required.

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It is a good extension of this exercise to think how you might extend this to the slightly more general result:

\begin{equation*} x \geq y \geq 0 \implies \sqrt{x} \geq \sqrt{y}. \end{equation*}

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3.5.17.

Solution.

There are a few issues with this proof. First of all, the flow of logic in this proof is backwards. The computation presented in the question could be used as scratchwork, but is not a proof. A proof should start with a basic fact (here: the square of any nonzero real number is positive), and work towards the desired conclusion. Furthermore, this solution lacks explanation. A good proof should explain the steps taken to reach the conclusion. Our proofs should not only be correct but they should also be easy on the reader. Often, we forget to include extra explanation because our brains are filling in those extra details as we write. Think about the reader - they don’t have the same brain as you to fill in those gaps! How can you make it easier for them to understand?

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Proof.

Let $a,b\in \mathbb{R}$ and $0 \lt a \lt b\text{.}$ Then $a-b \neq 0\text{,}$ and we know that $(a-b)^2 \gt 0\text{.}$ We expand and complete the square to get $(a+b)^2\text{,}$

\begin{align*} 0 \amp \lt (a-b)^2\\ \amp = a^2 -2ab +b^2\\ \amp = a^2-2ab+b^2 + 4ab-4ab\\ \amp =(a^2+2ab+b^2)-4ab\\ \amp =(a+b)^2-4ab \end{align*}

Therefore $4ab \lt (a+b)^2\text{.}$

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We would now like to take square roots. From Exercise 3.5.16, the function $f(x) = \sqrt{x}$ is a positive, increasing function when $x \gt 0$ (these two conditions are necessary to ensure that the inequality will not change when we take the square root of each side). Now, taking square roots, we see

\begin{equation*} 2\sqrt{ab} \lt (a+b). \end{equation*}

Dividing by two yields $\sqrt{ab} \lt \frac{a+b}{2}\text{,}$ as desired.

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3.5.18.

Solution.

Proof.

Let $x,y\in\mathbb{R}$ such that $x,y\geq0\text{.}$ Since the square root function is never negative, $\sqrt{x}\sqrt{y}\geq0\text{.}$ Then multiplying by $2$ and adding $x+y$ to both sides, we have

\begin{equation*} x + y \leq x+2\sqrt{x}\sqrt{y}+y. \end{equation*}

Since $x,y\geq0\text{,}$ we have that $(\sqrt{x})^2=x$ and $(\sqrt{y})^2=y\text{,}$ and so we may factor the right-hand side of the above inequality to obtain

\begin{equation*} x+y \leq (\sqrt{x}+\sqrt{y})^2. \end{equation*}

Since $x+y\geq0\text{,}$ we then have

\begin{equation*} \sqrt{x+y} \leq \sqrt{(\sqrt{x}+\sqrt{y})^2} \end{equation*}

But

\begin{equation*} \sqrt{(\sqrt{x}+\sqrt{y})^2}=\sqrt{x}+\sqrt{y}, \end{equation*}

since $\sqrt{x}+\sqrt{y}\geq0\text{.}$ Therefore

\begin{equation*} \sqrt{x+y} \leq \sqrt{x}+\sqrt{y}, \end{equation*}

as required.

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4 More logic
4.3 Exercises

4.3.1.

Solution.

$({{\neg}} P)\lor Q$ and $P\Rightarrow Q\text{.}$
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$P$ $Q$ $({\neg} P)$ $({\neg} P) \lor Q$ $P \implies Q$

T T F T T

T F F F F

F T T T T

F F T T T

We see that the last two columns are the same. Therefore, these are logically equivalent.
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$P\Leftrightarrow Q$ and $({{\neg}} P)\Leftrightarrow ({{\neg}} Q)\text{.}$
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$P$ $Q$ ${\neg} P$ ${\neg} Q$ $P \Leftrightarrow Q$ $({\neg} P) \Leftrightarrow ({\neg} Q)$

T T F F T T

T F F T F F

F T T F F F

F F T T T T

We see that the last two columns are the same. Therefore, these are logically equivalent.
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$P\Rightarrow (Q\lor R)$ and $P\Rightarrow (({{\neg}} Q)\Rightarrow R)\text{.}$

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$P$	$Q$	$R$	$Q\lor R$	$({\neg} Q)\Rightarrow R$	$P \Rightarrow (Q\lor R)$	$P \Rightarrow (({\neg} Q)\Rightarrow R)$
T	T	T	T	T	T	T
T	T	F	T	T	T	T
T	F	T	T	T	T	T
F	T	T	T	T	T	T
T	F	F	F	F	F	F
F	T	F	T	T	T	T
F	F	T	T	T	T	T
F	F	F	F	F	T	T

We see that the last two columns are the same. Therefore, these are logically equivalent.

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$(P\lor Q)\Rightarrow R$ and $(P\Rightarrow R)\land(Q\Rightarrow R)\text{.}$

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$P$	$Q$	$R$	$P\lor Q$	$P\Rightarrow R$	$Q \Rightarrow R$	$(P\lor Q) \Rightarrow R$	$(P \Rightarrow R)\land(Q\Rightarrow R)$
T	T	T	T	T	T	T	T
T	T	F	T	F	F	F	F
T	F	T	T	T	T	T	T
F	T	T	T	T	T	T	T
T	F	F	T	F	F	F	F
F	T	F	T	T	F	F	F
F	F	T	F	T	T	T	T
F	F	F	F	T	T	T	T

We see that the last two columns are the same. Therefore, these are logically equivalent.

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$P\Rightarrow (Q\lor R)$ and $(Q\land R)\Rightarrow P\text{.}$

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$P$	$Q$	$R$	$Q\lor R$	$Q\land R$	$P\Rightarrow(Q\lor R)$	$(Q \land R)\Rightarrow P$
T	T	T	T	T	T	T
T	T	F	T	F	T	T
T	F	T	T	F	T	T
F	T	T	T	T	T	F
T	F	F	F	F	F	T
F	T	F	T	F	T	T
F	F	T	T	F	T	T
F	F	F	F	F	T	T

We see that the last two columns are different. Therefore, these are not logically equivalent. Indeed, we see that when $P$ is false and $Q, R$ are true (i.e. the fourth row), we see that $P\Rightarrow(Q\lor R)$ is true, whereas $(Q \land R)\Rightarrow P$ is false.

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Even though questions like this are very easy to show and may feel quite unnecessary, they play an important role in understanding what logical equivalences we can think of when we are proving statements.

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For example, part (c) tells us that if we have a conditional statement where the conclusion is an `or’-statement, we can indeed turn that into a double implication statement which may be easier to prove.

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Similarly part (d) is going to play an important role in proving statements using cases in the upcoming chapter.

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So, it is important to think about which statements are logically equivalent and how we can use that to our advantage when we want to prove different statements.

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4.3.2.

Solution.

In this question, first we are going to write the sentences in symbolic logic, then negate them using the Theorem 4.2.3 and finally we will rewrite these negations in English.

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This sentence can be written as $P\land Q$ for

\begin{align*} P: \amp \quad 8 \text{ is even}\\ Q: \amp \quad 5 \text{ is prime}. \end{align*}

So, its negation is:

\begin{equation*} {\neg} (P\land Q)\equiv ({\neg} P)\lor ({\neg} Q) \end{equation*}

Here, we used DeMorgan’s law. Then, the negation can be rewritten as:

\begin{gather*} \text{8 is not even or 5 is not prime}. \end{gather*}

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This sentence can be written as $(P\land Q)\implies R$ for

\begin{align*} P: \amp \quad n \text{ is a multiple of } 4,\\ Q: \amp \quad n \text{ is a multple of } 6,\\ R: \amp \quad n \text{ is a multiple of } 24. \end{align*}

Then its negation is:

\begin{equation*} {\neg}\left( (P\land Q)\Rightarrow R\right)\equiv {\neg}\left( {\neg} (P\land Q) \lor R \right)\equiv (P\land Q)\land {\neg} R. \end{equation*}

Here, we used the equivalence of implication, DeMorgan’s law and double negation. Then, the negation can be rewritten as:

\begin{gather*} n \text{ is a multiple of 4 and 6, but it is not a multiple of 24.} \end{gather*}

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This sentence can be written as $(\neg P)\implies \big(Q\land (\neg R)\big)$ for

\begin{align*} P: \amp \quad n \text{ is a multiple of } 10,\\ Q: \amp \quad n \text{ is a multiple of } 2,\\ R: \amp \quad n \text{ is a multiple of } 5. \end{align*}

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Then, we see that the negation is:

\begin{align*} \neg\!\left(\neg P\implies \big(Q\land (\neg R)\big)\right) \amp \equiv\ \neg\!\left(\neg(\neg P)\lor \big(Q\land(\neg R)\big)\right)\\ \amp \equiv (\neg P)\land \neg\big(Q\land (\neg R)\big)\\ \amp \equiv (\neg P)\land \big((\neg Q) \lor R\big)\\ \amp \equiv \big((\neg P)\land (\neg Q)\big)\lor \big( (\neg P)\land R \big). \end{align*}

Here, first we used the equivalence of implication. Then DeMorgan’s law and double negation, and finally distribution law. This means that we can rewrite the negation as:

\begin{align*} \amp n \text{ is not a multiple of 10 and } n \text{ is not a multiple of 2, or }\\ \amp n \text{ is not a multiple of 10 and } n \text{ is a multiple of 5.} \end{align*}

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This expression can be written as $P\land Q$ for

\begin{align*} P: \amp \quad x \geq 3,\\ Q: \amp \quad x\leq 6. \end{align*}

Then its negation is:

\begin{equation*} {\neg} (P\land Q)\equiv ({\neg} P)\lor ({\neg} Q). \end{equation*}

Here we used DeMorgan’s law. Therefore, the negation can be rewritten as:

\begin{gather*} x \lt 3 \text{ or } x \gt 6. \end{gather*}

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This sentence can be written as $R\implies (P\lor Q)$ for

\begin{align*} P: \amp \quad x \text{ is less than -2},\\ Q: \amp \quad x \text{ greater than 2},\\ R: \amp \quad x^2 \text{ is greater than 4}.\text{.} \end{align*}

Then its negation is:

\begin{equation*} {\neg} (R\Rightarrow ( Q\lor P))\equiv {\neg} \left({\neg} R\lor ( Q\lor P)\right)\equiv R\land (({\neg} Q)\land ({\neg} P)). \end{equation*}

Here, first we used the equivalence of implication. Then DeMorgan’s law and double negation. Thus, the negation can be rewritten as:

\begin{align*} \amp \text{The square of a real number } x \text{ is greater than 4 and }\\ \amp x \text{ is greater than or equal to -2, and less than or equal to 2}. \end{align*}

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This sentence can be written as $P\implies (Q\implies R)$ for

\begin{align*} P: \amp \quad f \text{ is differentiable everywhere},\\ Q: \amp \quad x \text{ is a local maximum of } f,\\ R: \amp \quad f'(x)=0. \end{align*}

Here, first we used the equivalence of implication. Then DeMorgan’s law and double negation. Then its negation is:

\begin{equation*} {\neg} (P\Rightarrow ( Q\Rightarrow R))\equiv {\neg} \left({\neg} P\lor ({\neg} Q\lor R)\right)\equiv P\land (Q\land ({\neg} R)). \end{equation*}

Here, first we used the equivalence of implication twice. Then DeMorgan’s law and double negation. Hence, the negation can be rewritten as:

\begin{align*} \amp \text{A function } f \text{ is differentiable everywhere and } x\in\mathbb{R} \text{ is a local maximum of } f,\\ \amp \text{ but } f'(x)\neq 0. \end{align*}

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4.3.3.

Solution.

Using the biconditional equivalence, we have

\begin{equation*} (P\iff Q) \equiv ((P\implies Q)\land (Q\implies P)). \end{equation*}

Then using the contrapositive equivalence, we have

\begin{equation*} ((P\implies Q)\land (Q\implies P)) \equiv \big(((\neg Q)\implies (\neg P))\land ((\neg P)\implies (\neg Q))\big) \end{equation*}

and by commutativity, this is logically equivalent to

\begin{equation*} ((\neg P)\implies (\neg Q)) \land ((\neg Q)\implies (\neg P)). \end{equation*}

Finally, by the biconditional equivalence, this is logically equivalent to

\begin{equation*} (\neg P) \iff (\neg Q) \end{equation*}

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By the implication equivalence and the double negation equivalence, we have

\begin{equation*} (P\implies(\neg Q\implies R))\equiv \big(P\implies((\neg(\neg Q))\lor R)\big)\equiv (P\implies(Q\lor R)). \end{equation*}

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By the implication equivalence and DeMorgan’s law,

\begin{equation*} ((P\lor Q)\implies R)\equiv ((\neg(P\lor Q))\lor R) \equiv \big(((\neg P)\land (\neg Q))\lor R\big). \end{equation*}

Then using commutativity and then the distribution law, this is logically equivalent to

\begin{equation*} \big(R\lor ((\neg P)\land (\neg Q))\big)\equiv \big((R\lor (\neg P))\land (R\lor (\neg Q))\big) \end{equation*}

Then by commutativity and then the implication equivalence, this is logically equivalent to

\begin{equation*} \big(((\neg P)\lor R)\land ((\neg Q) \lor R)\big) \equiv ((P\implies R)\land(Q\implies R)). \end{equation*}

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5 More proofs
5.5 Exercises

5.5.1.

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5.5.7.

Solution.

Proof.

We proceed by contrapositive. Without loss of generality, assume that $b \gt a\text{.}$ Assume that $a$ and $b$ are consecutive, so that $b=a+1\text{.}$ We compute,

\begin{equation*} a+b = a + (a+1) = 2a+1. \end{equation*}

Since $a\in \mathbb{Z}\text{,}$ we see that $a+b$ fits the definition of an odd integer, as desired.

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5.5.8.

Solution.

Proof.

Assume $n$ is an even integer. Then we see $n=2a$ for some $a\in\mathbb Z\text{.}$ Since $a\in\mathbb Z$ we see that $a$ is either even or odd.

Case 1: Assume $a$ is even. So in this case, we see that $a=2m$ for some $m\in\mathbb Z\text{.}$ Thus, $n=2a=2(2m)=4m$ for some integer $m\text{.}$
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Case 2: Now assume that $a$ is odd. We see that $a=2t+1$ for some $t\in\mathbb Z\text{.}$Thus, $n=2a=2(2t+1)=4t+2$ for some integer $t\text{.}$
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Therefore, if $n$ is an even integer than $n=4k$ or $n=4k+2$ for some integer $k\text{.}$

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5.5.9.

Solution.

Proof.

Let $n\in\mathbb{Z}\text{.}$ We prove each implication in turn.

$(\Rightarrow)\text{:}$ Assume that $2\mid (n^4-7)\text{.}$ Then we see that $n^4-7=2k$ for some $k\in\mathbb{Z}\text{.}$ Thus, we see that $n^4=2(k+3)+1\text{.}$ Since $(k+3)\in\mathbb{Z}\text{,}$ we see that $n^4$ is odd. Since the square of an even number is even and $n^4=n^2n^2\text{,}$ we see that $n^2$ is odd. Similarly, we see that $n$ is odd. Thus $n=2m+1$ for some $m\in\mathbb{Z}\text{.}$ Hence, $n^2+3=(4m^2+4m+1)+3=4(m^2+m+1)$ and since $(m^2+m+1)\in\mathbb{Z}\text{,}$ we see that $4\mid (n^2+3)\text{.}$
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$(\Leftarrow)\text{:}$ Assume that $4\mid (n^2+3)\text{.}$ Then we see that $n^2+3=4k$ for some $k\in\mathbb{Z}\text{.}$ Thus, we see that $n^2=2(2k-2)+1\text{.}$ Since $(2k-2)\in\mathbb{Z}\text{,}$ we see that $n^2$ is odd. Moreover, since the product of two odd numbers is odd, we see that $n^4=n^2n^2$ is odd. Thus $n^4=2m+1$ for some $m\in\mathbb{Z}\text{.}$ Hence, $n^4-7=2m-6=2(m-3)$ and since $(m-3)\in\mathbb{Z}\text{,}$ we see that $2\mid (n^4-7)\text{.}$
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Since both implications are true, the biconditional holds.

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5.5.10.

Solution.

Proof.

This is a biconditional statement. Hence, we need to prove both implications:

\begin{gather*} (3\mid 5a)\implies(3\mid a) \qquad \text{ and } (3\mid a)\implies (3\mid 5a). \end{gather*}

$(\Leftarrow)\text{:}$ Assume that $3\mid a\text{.}$ Then we see that $a=3k$ for some $k\in\mathbb Z\text{.}$ Hence, $5a=5(3k)=3(5k)\text{.}$ Since $5k\in\mathbb Z\text{,}$ we get $3\mid 5a\text{.}$
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$(\Rightarrow)\text{:}$ Assume that $3\mid 5a\text{.}$ This implies $5a=3m$ for some $m\in\mathbb Z\text{.}$ Therefore we see that by adding $a$ to both sides, we get $6a=3m+a\text{.}$ Thus, $a=6a-3m=3(2a-m)\text{.}$ Since $(2a-m)\in\mathbb Z\text{,}$ we see $3\mid a\text{.}$
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Therefore the result follows.

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Notice that we have indicated to the reader which implication we are proving by starting with $(\Leftarrow)$ or $(\Rightarrow)\text{.}$ This is a very simple useful way to help tell the reader what is going on. Always make life easier for your reader.

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5.5.11.

Solution.

Proof.

Let $n\in\mathbb{Z}\text{.}$ We consider the cases that $n$ is odd and $n$ is even. First suppose that $n$ is even, so that $n=2k$ for some $k\in\mathbb{N}\text{.}$ Then

\begin{equation*} n^2+2n=(2k)^2+2(2k)=4k^2+4k=4(k^2+k) \end{equation*}

and so

\begin{equation*} (n^2-1)(n^2+2n)= 4(k^2+k)(n^2-1). \end{equation*}

Since $k,n\in\mathbb{Z}\text{,}$ $(k^2+k)(n^2-1)\in\mathbb{Z}\text{,}$ and thus $(n^2-1)(n^2+2n)$ is divisible by $4\text{.}$

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Next suppose that $n$ is odd, so that $n=2k+1$ for some $k\in\mathbb{N}\text{.}$ Then

\begin{equation*} n^2-1=(2k+1)^2-1=4k^2+4k+1-1=4(k^2+k) \end{equation*}

and so

\begin{equation*} (n^2-1)(n^2+2n)= 4(k^2+k)(n^2+2n). \end{equation*}

Since $k,n\in\mathbb{Z}\text{,}$ $(k^2+k)(n^2+2n)\in\mathbb{Z}\text{,}$ and thus $(n^2-1)(n^2+2n)$ is divisible by $4\text{.}$

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We have proved the statement in all cases, and so the statement holds for all $n\in\mathbb{N}\text{.}$

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5.5.12.

Solution.

Proof.

We proceed by contrapositive. We see that thee contrapositive of the statement is: If $x$ and $y$ have the same parity, then $x+y$ is even. Now assume that $x$ and $y$ are both even or both odd. Then we have 2 cases:

Case 1: Assume that both $x$ and $y$ are even. Then $x=2n$ and $y=2m$ for some $n,m\in \mathbb{Z}\text{.}$ Therefore,

\begin{equation*} x+y = 2n+2m =2(n+m). \end{equation*}

Since $n, m \in \mathbb{Z}$ and the sum of integers is also an integer, we see that $n+m\in \mathbb{Z}\text{,}$ so that $x+y$ fits the definition of an even number, as required.

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Case 2: Assume that $x$ and $y$ are both odd. Then $x=2j+1$ and $y=2k+1$ for some $j,k\in \mathbb{Z}\text{.}$ Then

\begin{equation*} x+y = 2j+1+2k+1 = 2(j+k+1). \end{equation*}

Since $j,k,1 \in \mathbb{Z}$ and the sum of integers is an integer, $j+k+1\in \mathbb{Z}\text{,}$ and we see that $x+y$ is even, as desired.

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5.5.13.

Solution.

Proof.

We prove the contrapositive. Let $n\not\equiv 1\mod 3\text{.}$ Then we see that we have $2$ cases: $n=3k$ or $n=3k+2$ for some $k\in\mathbb Z\text{.}$

Case 1: $n=3k$ for some $k\in\mathbb Z\text{.}$ In this case, we see that

\begin{equation*} n^2+4n+1=(3k)^2+4(3k)+1=9k^2+12k+1=3(3k^2+4k)+1. \end{equation*}

Hence, since $3k^2+4k\in\mathbb Z\text{,}$ we see that $3\nmid (n^2+4n+1)\text{.}$

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Case 2: $n=3k+2$ for some $k\in\mathbb Z\text{.}$ In this case, we see that

\begin{equation*} n^2+4n+1=(3k+2)^2+4(3k+2)+1=9k^2+12k+4+12k+8+1=3(3k^2+8k+4)+1. \end{equation*}

Since $3k^2+8k+4\in\mathbb Z\text{,}$ this implies $3\nmid (n^2+4n+1)\text{.}$

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Hence the contrapositive holds, and therefore, we see that if $3\mid (n^2+4n+1)\text{,}$ then $n\equiv 1\mod 3\text{.}$

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5.5.14.

Solution.

Proof.

Assume that $5\nmid m\text{.}$ By the division algorithm, there are four cases for $m\text{:}$ $m=5k+1\text{,}$ $m=5k+2\text{,}$ $m=5k+3\text{,}$ or $m=5k+4$ for some $k\in\mathbb Z\text{.}$

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Case 1: $m=5k+1$ for some $k\in\mathbb Z\text{.}$ In this case, we have $m^2=25k^2+10k+1\text{.}$ Thus, we see $m^2-1=5(5k^2+2k)\text{.}$ Since $(5k^2+2k)\in\mathbb Z\text{,}$ we see $5\mid (m^2-1)\text{,}$ that is $m^2\equiv 1 \mod{5}\text{.}$
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Case 2: $m=5k+2$ for some $k\in\mathbb Z\text{.}$ In this case, we have $m^2=25k^2+20k+4\text{.}$ Thus, we see $m^2+1=5(5k^2+4k+1)\text{.}$ Since $(5k^2+4k+1)\in\mathbb Z\text{,}$ we see $5\mid (m^2+1)\text{,}$ that is $m^2\equiv -1 \mod{5}\text{.}$
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Case 3: $m=5k+3$ for some $k\in\mathbb Z\text{.}$ In this case, we have $m^2=25k^2+30k+9\text{.}$ Thus, we see $m^2+1=5(5k^2+6k+2)\text{.}$ Since $(5k^2+6k+2)\in\mathbb Z\text{,}$ we see $5\mid (m^2+1)\text{,}$ that is $m^2\equiv -1 \mod{5}\text{.}$
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Case 4: $m=5k+4$ for some $k\in\mathbb Z\text{.}$ In this case, we have $m^2=25k^2+40k+16\text{.}$ Thus, we see $m^2-1=5(5k^2+8k+3)\text{.}$ Since $(5k^2+8k+3)\in\mathbb Z\text{,}$ we see $5\mid (m^2-1)\text{,}$ that is $m^2\equiv 1 \mod{5}\text{.}$
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Therefore, if $5\nmid m\text{,}$ then $m^2\equiv 1 \mod{5}$ or $m^2\equiv -1 \mod{5}\text{.}$

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5.5.15.

Solution.

Proof.

Assume that $3\nmid q\text{.}$ Then we have two possible cases, $q\equiv 1 \mod 3$ or $q\equiv 2 \mod 3\text{.}$

Case 1: $q\equiv 1 \mod 3.$ Then by definition, $q=3k+1$ for some $k\in \mathbb{Z}\text{.}$ Squaring $q\text{,}$ we see that

\begin{equation*} q^2 = (3k+1)^2 = 9k^2+6k+1 = 3(3k^2+2k) + 1. \end{equation*}

Because $3k^2+2k\in \mathbb{Z}\text{,}$ we conclude that $q^2\equiv 1 \mod 3 \text{.}$

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Case 2: $q\equiv 2 \mod 3.$ Then by definition, $q=3k+2$ for some $k\in \mathbb{Z}\text{.}$ Squaring $q\text{,}$ we see that

\begin{equation*} q^2 = (3k+2)^2 = 9k^2+12k+4 = (9k^2+12k+3)+1 = 3(3k^2+4k+1)+1. \end{equation*}

Because $3k^2+4k+1\in \mathbb{Z}\text{,}$ we conclude that $q^2\equiv 1 \mod 3 \text{.}$

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Since we have verified both possible cases, we deduce that if $3\nmid q\text{,}$ then $q^2 \equiv 1 \mod 3\text{.}$

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5.5.16.

Solution.

We see that this is a conditional statement, so we can assume the hypothesis and try to show the conclusion. This means that we assume that $n\in\mathbb Z$ and show that $n^3+(n+1)^3+(n+2)^3$ is divisible by $9\text{.}$ Since the hypothesis is very broad, this suggests that we use proof by cases; and since the result is about divisibility by $9\text{,}$ this suggests that we use $9$ cases using division algorithm. Oof! That sounds ugly.

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Instead, let’s try to rewrite what we want to show and see if it can be simplified. Notice that

\begin{align*} n^3+(n+1)^3+(n+2)^3 \amp =n^3+(n^3+3n^2+3n+1)+(n^3+6n^2+12n+8)\\ \amp =3n^3+9n^2+15n+9\\ \amp =3(n^3+5n)+9(n^2+1)\\ \amp =3n(n^2+5)+9(n^2+1). \end{align*}

This means that all we need to do is show that $3n(n^2+5)+9(n^2+1)=9k$ for some $k\in\mathbb Z\text{.}$ In fact since one of the terms on the LHS is already a multiple of $9\text{,}$ we only need to show that $3n(n^2+5)=9t$ for some $t\in\mathbb Z\text{,}$ i.e. $3\mid n(n^2+5)\text{.}$ Even though this still requires cases, we now only need 3. Let’s see how we can make use of this in the proof.

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Proof.

Let $n\in\mathbb Z\text{.}$ Then, we have

\begin{align*} n^3+(n+1)^3+(n+2)^3 \amp =n^3+(n^3+3n^2+3n+1)+(n^3+6n^2+12n+8)\\ \amp =3n^3+9n^2+15n+9\\ \amp =3(n^3+5n)+9(n^2+1)\\ \amp =3n(n^2+5)+9(n^2+1). \end{align*}

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We see that $9\mid 9(n^2+1)\text{.}$ Hence, to show $9\mid \big(n^3+(n+1)^3+(n+2)^3\big)\text{,}$ it is enough to show $9\mid 3n(n^2+5)\text{,}$ or equivalently $3\mid n(n^2+5)\text{.}$ To prove that we use cases.

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Case 1: $n\equiv 0 \mod 3$ : In this case, we see that $n(n^2+5)\equiv 0\mod 3\text{,}$ which implies that $3\mid n(n^2+5)\text{.}$
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Case 2: $n\equiv 1 \mod 3$ : In this case, we see that $n^2\equiv 1\mod 3\text{,}$ which implies that $(n^2+5)\equiv 0\mod 3\text{.}$ Thus, $3\mid n(n^2+5)\text{.}$
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Case 3: $n\equiv 2 \mod 3$ : In this case, we see that $n^2\equiv 4\equiv 1\mod 3\text{,}$ which implies that $(n^2+5)\equiv 0\mod 3\text{.}$ Thus, $3\mid n(n^2+5)\text{.}$
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Therefore, we see that $3\mid n(n^2+5)$ for all $n\text{,}$ and thus, for every integer $n\geq 0\text{,}$ the sum $n^3+(n+1)^3+(n+2)^3$ is divisible by $9\text{.}$

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In our proof here we have proved a small additional result to help us prove the main result. We could have, instead, made that small result a separate lemma and proved it separately. Both approaches are common practice.

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5.5.17.

Solution.

Proof.

Let $a\in\mathbb{Z}\text{.}$ By Euclidean division, the number $a$ can be written uniquely as $a=5k+r$ with $k\in \mathbb{Z}$ and $r \in \set{0,1,2,3,4}\text{.}$ This means that

\begin{equation*} a \equiv r \pmod5 \qquad \text{ with } r \in \set{0,1,2,3,4}. \end{equation*}

This means that we can consider five cases — one for each value of $r\text{:}$

$a\equiv 0\pmod 5\text{:}$ In this case, we see that $a^5\equiv 0^5\equiv 0\equiv a \pmod 5\text{.}$
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$a\equiv 1\pmod 5\text{:}$ In this case, we see that $a^5\equiv 1^5\equiv 1\equiv a \pmod 5\text{.}$
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$a\equiv 2\pmod 5\text{:}$ In this case, we see that $a^5\equiv 2^5\equiv 32\equiv 2\equiv a \pmod 5\text{.}$
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$a\equiv 3\pmod 5\text{:}$ In this case, we see that $a^5\equiv 3^5\equiv 243\equiv 3 \equiv a \pmod 5\text{.}$
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$a\equiv 4\pmod 5\text{:}$ In this case, we see that $a^5\equiv 4^5\equiv 1024 \equiv 4\equiv a \pmod 5\text{.}$
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Adding $2$ to everything, we end up with

\begin{equation*} 1 \lt x \lt 3 \end{equation*}

which implies that $|x| \lt 3\text{.}$

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Furthermore, note that

\begin{equation*} |2x^2-3x-2|=|(2x+1)(x-2)|=|2x+1|\cdot|x-2|. \end{equation*}

Using the inequality $|x-2| \lt 1\text{,}$ we therefore have

\begin{equation*} |2x^2-3x-2| \lt |2x+1|\cdot1. \end{equation*}

However, using the triangle inequality, and the bound $|x| \lt 3$ we already established, we have

\begin{equation*} |2x+1|\leq |2x|+1=2|x|+1 \lt 2\cdot3+1=7. \end{equation*}

Thus

\begin{equation*} |2x^2-3x-2| \lt |2x+1|\cdot1 \lt 7\cdot 1=7. \end{equation*}

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5.5.21.

Solution.

Proof.

Assume that $x,y\in \mathbb{R}\text{.}$ Recall the triangle inequality, which states: For any $a,b\in \mathbb{R}\text{,}$ $|a+b|\leq |a|+|b|\text{.}$ We set $a = x-y$ and $b = y\text{.}$ Since $x,y\in \mathbb{R}\text{,}$ $x-y \in \mathbb{R}\text{,}$ so $a$ and $b$ satisfy the hypothesis of the triangle inequality. Plugging our values of $a$ and $b$ into the triangle inequality yields,

\begin{equation*} |x-y+y|\leq |x-y|+|y|. \end{equation*}

Subtracting $|y|$ from both sides gives

\begin{equation*} |x|-|y|\leq |x-y|. \tag{1} \end{equation*}

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We use the triangle inequality again. This time setting $a=x$ and $b= y-x\text{.}$ Again, since $x$ and $y-x$ are real numbers, the hypothesis of the triangle inequality is satisfied, so we obtain

\begin{equation*} |x+y-x|\leq |x|+|y-x|. \end{equation*}

Subtracting $|x|$ from each side, we see $|y|-|x|\leq |y-x|\text{.}$ Dividing both sides by $-1$ yields

\begin{equation*} |x|-|y| \geq -|y-x| =-|x-y| \tag{2}. \end{equation*}

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Combining equations (1) and (2), we obtain the desired result

\begin{equation*} -|x-y|\leq |x|-|y| \leq |x-y| \end{equation*}

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5.5.22.

Solution.

Proof.

Let $x,y\in(0,\infty)$ and suppose that $x\leq y\text{.}$ Since both $x$ and $y$ and positive, we can divide this equation by $xy$ to obtain

\begin{equation*} f(y)=\frac{1}{y}\leq \frac{1}{x}=f(x). \end{equation*}

Therefore $f$ is decreasing.

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We could have instead restricted $f$ to the domain $(a,\infty)$ for any $a \gt 0\text{,}$ or to the domain $(-\infty,b)$ for any $b \lt 0\text{,}$ and also obtained a decreasing function. So even though the function is decreasing on $(-\infty,0)$ and also on $(0,\infty)\text{,}$ it is not decreasing; in order to show that a function is decreasing, we have to look at the function and its domain as a whole.

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6 Quantifiers
6.6 Exercises

6.6.1.

Solution.

Proof.

Let $n\in\mathbb{Z}\text{.}$ Notice that $n^3-n = n(n-1)(n+1)\text{,}$ so it is sufficient to show that $3\mid n(n-1)(n+1)\text{.}$ By division algorithm we see that we have $3$ cases: $n=3k\text{,}$ $n=3k+1$ or $n=3k+2$ for some $k\in\mathbb Z\text{.}$

Case 1: $n=3k\text{.}$ In this case we have $n^3-n=n(n-1)(n+1)=3k(3k-1)(3k+1)\text{.}$ Then, since $k(3k-1)(3k+1)\in\mathbb Z$ we can conclude $3\mid (n^3-n)\text{.}$
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Case 2: $n=3k+1\text{.}$ Now we have $n^3-n=n(n-1)(n+1)=(3k+1)(3k)(3k+2)\text{.}$ Then, since $k(3k+1)(3k+2)\in\mathbb Z$ we get $3\mid (n^3-n)\text{.}$
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Case 3: $n=3k+2\text{.}$ Finally $n^3-n=n(n-1)(n+1)=(3k+2)(3k+1)(3k+3)=3(3k+2)(3k+1)(k+1)\text{.}$ Then, since $(3k+2)(3k+1)(k+1)\in\mathbb Z$ we have $3\mid (n^3-n)\text{.}$
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Hence, for all $n\in\mathbb Z\text{,}$ $3\mid (n^3-n)\text{.}$

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6.6.2.

Solution.

Proof.

Let $n,k\in\mathbb{Z}\text{.}$ Assume $k\mid (2n+1)$ and $k\mid(4n^2+1)\text{.}$ This implies that $2n+1=ka$ and $4n^2+1=kb$ for some $a, b\in\mathbb{Z}\text{.}$ Combining these equations gives

\begin{align*} k^2a^2 \amp =(2n+1)^2 = 4n^2+4n+1\\ \amp = kb+4n = kb+2(ka-1) \amp \text{ since }2n=ka-1. \end{align*}

Hence, if we group all the $k$ terms together, we get

\begin{gather*} 2 = kb+2ka-k^2a^2 = k(b+2a-ka^2). \end{gather*}

But this tells us that $k \mid 2\text{.}$ The only divisors of $2$ are $\pm 1, \pm 2\text{.}$

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However, since $k \mid 2n+1\text{,}$ we know that $k$ must be odd (if it were even, then $2n+1$ would have to be even). Therefore $k$ must be an odd divisor of $2\text{,}$ that is $k= \pm 1\text{.}$

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6.6.3.

Solution.

Proof.

We are going to use proof by contrapositive. We see that the expression $\forall n\in\mathbb Z\text{,}$ comes before the conditional statement starts, and hence is not a part of it. Thus, we see that the contrapositive of the statement is:

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$\forall n\in\mathbb Z\text{,}$ if $n\equiv 3\pmod 4\text{,}$ then $\forall a,b\in\mathbb Z\text{,}$ we have $a^2+b^2\neq n\text{.}$
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Now, let $n\in\mathbb Z$ and assume that $n\equiv 3 \pmod 4\text{.}$ To prove the statement, we are going to use $3$ cases: $a$ and $b$ are both even, $a$ and $b$ are both odd, and only one of $a$ or $b$ is even.

Case 1: $a$ and $b$ are both even: In this case, we know that $a=2k$ and $b=2\ell$ for some $k,\ell\in\mathbb{Z}\text{.}$ Hence, we see that $a^2=4k^2$ and $b^2=4\ell^2\text{.}$ Thus, $a^2+b^2=4k^2+4\ell^2=4(k^2+\ell^2)\text{.}$ Since $k^2+\ell^2\in\mathbb Z$ we see $4\mid (a^2+b^2)\text{.}$ Therefore $(a^2+b^2)\equiv 0\pmod 4\text{.}$
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Case 2: $a$ and $b$ are both odd: Then, we know that $a=2k+1$ and $b=2\ell+1$ for some $k,\ell\in\mathbb{Z}\text{.}$ In this case, we see that $a^2=4k^2+4k+1$ and $b^2=4\ell^2+4\ell+1\text{.}$ Thus, $a^2+b^2=4k^2+4k+1+4\ell^2+4\ell+1=4(k^2+k+\ell^2+\ell)+2\text{.}$ Since $k^2+k+\ell^2+\ell\in\mathbb Z$ we see $4\mid (a^2+b^2-2)\text{.}$ Therefore $(a^2+b^2)\equiv 2\pmod 4\text{.}$
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Case 3: only one of $a$ or $b$ is even: For this case, WLOG, we can assume that $a$ is odd and $b$ is even. Then, we know that $a=2k+1$ and $b=2\ell$ for some $k,\ell\in\mathbb{Z}\text{.}$ This implies that $a^2=4k^24k+1$ and $b^2=4\ell^2\text{.}$ Thus, $a^2+b^2=4k^2+4k+1+4\ell^2=4(k^2+k+\ell^2)+1\text{.}$ Since $k^2+k+\ell^2+\in\mathbb Z$ we see $4\mid (a^2+b^2-1)\text{.}$ Therefore $(a^2+b^2)\equiv 1\pmod 4\text{.}$
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Hence, there does not exist $a,b\in\mathbb{Z}\text{,}$ where $(a^2+b^2)\equiv 3\pmod 4\text{.}$ Therefore there does not exist $a,b\in\mathbb{Z}\text{,}$ where $a^2+b^2= n\text{.}$

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6.6.4.

Solution.

Proof.

We are going to use proof by contrapositive. Let $a,b\in\mathbb{Z}\text{.}$ Assume that $3\nmid a$ or $3\nmid b\text{.}$ WLOG we can assume $3\nmid a\text{.}$ Then we see that we have 2 cases $a=3k+1$ or $a=3k+2$ for some $k\in\mathbb Z\text{.}$

Case 1: Let $a=3k+1$ for some $k\in\mathbb Z\text{,}$ then we see that $a^2=9k^2+6k+1=3(3k^2+2k)+1\text{.}$
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Case 2: Now let $a=3k+2$ for some $k\in\mathbb Z\text{.}$ Then $a^2=9k^2+12k+4=3(3k^2+4k+1)+1\text{.}$
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Thus, we see that since $3\nmid a\text{,}$ we have $a^2=3m+1$ for some $m\in\mathbb Z\text{.}$

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Now, either $3 \mid b$ or $3 \nmid b\text{.}$

When $3\mid b\text{,}$ $b = 3\ell$ for some $\ell \in \mathbb{Z}\text{,}$ and so $a^2+b^2 = 3m + 9\ell^2 +1 = 3(m+3\ell^2)+1\text{.}$
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On the other hand, when $3\nmid b\text{,}$ then, by the reasoning above, we get $b^2=3\ell+1$ for some $\ell\in\mathbb Z\text{.}$ And so $a^2+b^2=3(m+\ell)+2\text{.}$
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In both cases, we see that $3\nmid (a^2+b^2)$ as required.

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6.6.5.

Solution.

Proof.

Let $n=4\text{,}$ $a=2\text{,}$ and $b=2\text{.}$ Since $4\mid 4\text{,}$

\begin{equation*} n\mid a\cdot b \end{equation*}

but $4\nmid 2$ and so $n\nmid a$ and $n\nmid b\text{.}$ Therefore we see that the statement, if $n\mid ab\text{,}$ then $n\mid a$ or $n\mid b\text{,}$ does not hold for all $n,a,b,\in\mathbb{Z}\text{.}$

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6.6.6.

Solution.

Proof.

Let $f:\mathbb{R}\to\mathbb{R}$ be defined by $f(x)=x^2\text{,}$ and let $g:\mathbb{R}\to\mathbb{R}$ be defined by $g(x)=x-x^2\text{.}$ Then for any $x\in\mathbb{R}\text{,}$

\begin{equation*} f(x)=x^2=(-x)^2=f(-x) \end{equation*}

so $f$ is even, not odd. Moreover, $(f+g)(x)=x^2+(x-x^2)=x\text{.}$ Then

\begin{equation*} (f+g)(x)=x=-(f+g)(-x) \end{equation*}

and so $f+g$ is odd. Therefore just because the sum of two functions is an odd function, it does not follow that those two functions are odd functions.

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6.6.7.

Solution.

True. We just need to find one example of $x,y$ so that $x+y=3\text{.}$ So take $x=0,y=3\text{.}$
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False. We show the negation is true; the negation is $\forall x \in \mathbb{Z}, \exists y\in \mathbb{Z} \text{ s.t. } x+y \neq 3\text{.}$ We must be careful to choose $x$ before we choose $y\text{.}$ So, let $x$ be any integer, and then set $y=-x\text{.}$ Then $x+y = 0 \neq 3\text{.}$ Since the negation is true, the original is false.
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True. Let $x$ be any integer and then pick $y=3-x\text{.}$ Then $x+y=3\text{.}$
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False. We show the negation is true; the negation is $\exists x\in \mathbb{Z} \text{ s.t. } \exists y\in \mathbb{Z} \text{ s.t. } x+y \neq 3\text{.}$ So we only need one example of $x,y$ so that $x+y \neq 3\text{.}$ Pick $x=3,y=7$ then $x+y \neq 3\text{.}$ Since the negation is true, the original is false.
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6.6.8.

Solution.

True. We only need an example of $x,y$ so that $x^2 \lt y\text{.}$ Take $x=0,y=1\text{.}$
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False. We show the negation is true; the negation is $\forall x\in\mathbb{R}, \exists y\in\mathbb{R} \text{ s.t. } x^2 \geq y\text{.}$ Let $x$ be any real number, then we know $x^2 \geq 0\text{,}$ so set $y=0\text{.}$ Then we must have $x^2 \geq 0 = y$ as required. Since negation is true, original is false.
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True. Let $x$ be any real number and then pick $y=x^2+1$ (remember we must pick $y$ after we pick $x\text{.}$ Then we have $x^2 \lt x^2+1 = y$ as required.
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False. We show the negation is true; the negation is $\exists x\in\mathbb{R}\text{ s.t. } \exists y\in\mathbb{R} \text{ s.t. } x^2 \geq y\text{.}$ So we only need an example of $x,y\text{.}$ Pick $x=1,y=0$ then $x^2 = 1 \gt 0 = y\text{.}$ Since the negation is true, the original is false.
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6.6.9.

Solution.

The fact that 2 is the only even prime really helps to solve this problem.

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Disproof: This statement is false. The negation of the statement is

\begin{gather*} \exists x\in P \text{ s.t. } \exists y\in P \text{ s.t. } x+y \not\in P. \end{gather*}

So we to prove the negation is true we simply need an example of $x,y\text{.}$ That example is then a counterexample to the original. Take $x=y=3\in P\text{.}$ Then we see that $x+y=6\notin P\text{.}$ Notice that there are many such counter examples; one can take any pair of odd primes.

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Disproof: This statement is false. The negation of the statement is

\begin{gather*} \exists x\in P \text{ s.t. } \forall y\in P \text{ s.t. } x+y \not\in P. \end{gather*}

Take $x=7 \in P\text{,}$ and let $y \in P\text{.}$ Either $y$ is even or $y$ is odd.
- When $y$ is even, it must be 2, but then $x+y=9 \not\in P\text{.}$
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- When $y$ is odd, $x+y \gt 2$ is even, and so not prime.
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In either case the negation is true, and so the original statement is false.

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Disproof: This statement is false. The negation of the statement is

\begin{gather*} \forall x\in P, \exists y\in P \text{ s.t. } x+y\notin P. \end{gather*}

We can use a similar proof to the previous statement. Let $x \in P\text{,}$ then either $x=2$ or $x$ is odd.
- Case 1: $x=2$: In this case, we can pick $y=7\in P\text{,}$ and get $x+y=9\notin P\text{.}$
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- Case 2: $x$ is odd: In this case, we can take $y=x\text{.}$ This implies that $x+y \gt 2$ is even and thus, $x+y\notin P\text{.}$
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Therefore the negation of the statement is true, and hence, the original statement is false.

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Proof: This statement is true. As an example, we can take any $x=2\in P\text{,}$ and $y=3\in P\text{.}$ Then we see that $x+y=5\in P\text{.}$
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6.6.10.

Solution.

$\forall a\in A\text{,}$ $\exists b\in B$ such that $a+b\in C\text{.}$
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This statement is true.
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Let $a\in A\text{.}$ Then we see that $a=3k$ for some $k\in\mathbb{Z}\text{.}$ Thus, since $k=2n$ or $k=2n+1$ for some $n\in\mathbb{Z}\text{,}$ we see that $a=6n$ or $a=6n+3$ for some $n\in\mathbb{Z}\text{.}$ Therefore if we pick $b=1\in B\text{,}$ we get, $a+b=6n+1$ or $a+b=6n+4\text{.}$ Hence, $6\nmid (a+b)\text{.}$
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$\exists a\in A$ such that $\forall b\in B\text{,}$ $a+b\in C\text{.}$
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This statement is false.
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It helps to write out the negation: $\forall a \in A, \exists b\in B \text{ such that } a+b\notin C\text{.}$ Let $a$ be any element of $A\text{,}$ then, as we saw in part (a), we know that $a=6n$ or $a=6n+3$ for some $n\in\mathbb{Z}\text{.}$
- If $a=6n\text{,}$ then we can pick $b=6\in B\text{,}$ and we get $a+b=6n+6=6(n+1)\text{.}$ Hence, $6\mid (a+b)\text{,}$ that is, $(a+b)\notin C\text{.}$
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- Similarly, if $a=6n+3\text{,}$ we can pick $b=3\in B\text{,}$ and get $a+b=6n+6=6(n+1)\text{.}$ Hence, $6\mid (a+b)\text{,}$ that is, $(a+b)\notin C\text{.}$
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So in either case, we can pick $b \in B$ so that $a+b \notin C\text{.}$ Since the negation is true, the original statement is false.

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$\forall a\in A\text{,}$ $\forall b\in B\text{,}$ $a+b\in C\text{.}$
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This statement is false.
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Again, it helps to write out the negation: $\exists a \in A \text{ such that } \exists b\in B\text{ such that } a+b \notin C \text{.}$ So it suffices to find an example of $a,b$ so that $a+b \notin C\text{.}$ We can take $a=3\in A$ and $b=3\in B\text{;}$ and get $a+b=6\text{.}$ This means $6\mid (a+b)\text{.}$ Thus, $(a+b)\notin C\text{.}$ Since the negation is true, the original statement is false.
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$\exists a\in A$ and $\exists b\in B$ such that $a+b\in C\text{.}$
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This statement is true.
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Since both quantifiers are “there exists”, we only need an example. We can take $a=0\in A$ and $b=1\in B\text{,}$ and get $a+b=1$ and hence, $6\nmid (a+b)\text{.}$ Thus, $(a+b)\in C\text{.}$
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6.6.11.

Solution.

True. We only need an example, so pick $x=y=1\text{.}$ Then $xy = 1 \gt 0$ and $x+y=2 \gt 0\text{.}$ Since both the hypothesis and conclusion are true, the implication is true.
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True. Note that when the hypothesis is false, the implication is always true. By picking $x=0\text{,}$ then no matter what $y$ is chosen, $xy=0\text{,}$ making the hypothesis false and the implication true.
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True. We can use a similar “trick” to the previous statement. Let $x$ be any real number, and then pick $y=0\text{.}$ Then $xy=0\text{,}$ making the hypothesis false and so the implcation holds.
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False. The negation is $\exists x\in\mathbb{R} \text{ s.t. } \exists y\in\mathbb{R} \text{ s.t. } (xy \gt 0) \land (x+y \leq 0) \text{,}$ so we only need an example of $x,y\text{.}$ Pick $x=y=-1\text{.}$ Then $xy=1 \gt 0$ and $x+y = -2 \leq 0\text{.}$ Since the negation is true, the original is false.
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True. Let $x=1\text{,}$ and then let $y \in \mathbb{R}\text{.}$ When $y \lt 0\text{,}$ the hypothesis is false, since $xy=y\lt 0\text{,}$ so the implication is true. While if $y \geq 0\text{,}$ the hypothesis is true, and the conclusion is true, since $xy=y \geq 0$ and $x+y = 1+y \geq 1 \gt 0\text{.}$ Thus the statement holds.
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True. Let $x$ be any real number and then set $y=1\text{.}$ When $x \lt 0\text{,}$ the hypothesis is false since $xy=x \lt 0\text{,}$ and thus the implication is true. On the other hand, when $x \geq 0\text{,}$ the hypothesis and conclusion are both true since $xy=x \geq 0$ and $x+y=x+1 \geq 1 \gt 0\text{.}$ So the statement is true.
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6.6.12.

Solution.

True. Let $y,z$ be any real numbers, and then pick $x=z-y$ (which is a real). Then $x+y = z-y+y = z$ as required.
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False. The negation of the statement is $\forall x\in \mathbb{R}, \exists y \in \mathbb{R} \text{ s.t. } \exists z \in \mathbb{R} \text{ s.t. } x+y \neq z\text{.}$ So let $x$ be any real number and then set $y=-x$ and $z=1\text{.}$ Then $x+y = x-y = 0 \neq 1 = z\text{.}$ Since the negation is true, the original is false.
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True. Recall that when the hypothesis is false the implication is true. So let $x$ be any real number and set $y=1, z=0\text{.}$ Then $z = 0 \lt 1 =y\text{,}$ so the hypothesis is false, making the implication true.
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False. The negation is $\exists x \in \mathbb{R} \text{ s.t. } \forall y \in \mathbb{R}, \exists z \in \mathbb{R} \text{ s.t. } (z \gt y) \land (z \leq x+y)\text{.}$ So pick $x=2$ and then let $y$ be any real number. Now set $z=y+1\text{,}$ then $z =y+1 \gt y$ and $z = y+1 \leq y+2 = x+y\text{.}$ Since the negation is true, the original is false.
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6.6.13.

Solution.

Notice that to prove either (a) or (b) to be true, we must show that they work for any functions that satisfy the hypotheses. But to show they are false, it is sufficient to find counter-examples.

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Disproof: This statement is false. Consider the function $f(x)=\sin(x)\text{.}$ We show that this is type A but not type B.
- To see that $f$ is type A, let $x$ be any real number. Since sine is a periodic function with period $2\pi$ and that $\sin(\pi/2)=1\text{,}$ we know that there is some point $y \in (x,x+2\pi)$ so that $f(y)=1\text{.}$ Hence sine is a type A function.
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- To see that $f$ is not type B, first consider the negation of the definition. A function $g$ is not type B when
  
  \begin{equation*} \forall x \in \mathbb{R}, \exists y \in \mathbb{R} \text{ such that } (y \geq x) \land (|g(y)|\lt 1). \end{equation*}
  
  So let $x$ be any real number, and then (using similar reasoning to the above), we know that there will be $y \in (x, x+2\pi)$ so that $f(y)=\sin(y)=0$ (just pick the smallest integer multiple of $\pi$ that is larger than $x$). Thus sine is not type B.
  
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Proof. We show that any function that is type B is also type A. So assume that the function $h$ is type B. Hence, there is some $t \in \mathbb{R}$ so that when $y \geq t$ then $|h(y)| \geq 1\text{.}$ Now, let $x \in \mathbb{R}\text{,}$ and then set $y = \max\{x,t\}\text{.}$ Since $h$ is type B and $y \geq t\text{,}$ it follows that $|h(y)| \geq 1\text{.}$ Thus $h$ is type A.
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6.6.14.

Solution.

$\exists x\in \mathbb Z$ such that $x \gt 84$ and $x\equiv 75 \pmod{84}\text{.}$
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The negation of this sentence is:
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“$\forall x\in \mathbb Z$ such that $x \gt 84\text{,}$ $x\not\equiv 75 \pmod{84}$”.
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Or, equivalently,
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“$\forall x\in\mathbb{Z}\text{,}$ $x \leq 84$ or $x \not\equiv 75 \pmod{84}$”.
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We see that the original statement is true. For an example, we can take $x=159$ (since $159 = 84\times1+75$).
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$\exists x, y\in \mathbb Z$ such that if $1\geq x^2\geq y^2\text{,}$ then $x\geq y\text{.}$
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The negation of this sentence is:
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“$\forall x,y\in \mathbb Z\text{,}$ $1\geq x^2\geq y^2$ and $y \gt x$”.
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We see that the original statement is true. For an example, we can take $x=y=0\text{.}$
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$\forall z\in\mathbb N\text{,}$ $\exists x,y\in\mathbb Z$ such that $z=x^2+y^2\text{.}$
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The negation of this sentence is:
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“$\exists z\in \mathbb N\text{,}$ $\forall x,y\in\mathbb{Z}\text{,}$ $z\neq x^2+y^2$”.
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We see that the negation of the statement is true. For an example, we can take $z=3\text{.}$ Then, there doesn’t exist $x,y\in\mathbb{Z}$ such that $x^2+y^2=3\text{.}$ Since the negation is true, the original statement is false.
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If we wanted to give a more formal solution to this problem, we could use the fact that the only way we can have $x^2 \leq 3$ is when $x \in \set{-1,0,1}\text{.}$ Thus, in order for $x^2+y^2 \leq 3$ we have only 9 possibilities to check, namely $x,y \in \set{-1,0,1}\text{,}$ and none of them work.
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$\exists a\in \mathbb R$ such that $a \gt 0$ and $\forall x\in\mathbb R\text{,}$ if $x\geq a\text{,}$ then $2^{-x} \lt \dfrac{1}{100}\text{.}$
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The negation of this sentence is:
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“$\forall a\in \mathbb R$ such that $a \gt 0\text{,}$ $\exists x\in\mathbb R\text{,}$ $x\geq a$ and $2^{-x}\geq \dfrac{1}{100}$”.
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Or
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“$\forall a\in \mathbb R\text{,}$ $a\leq0\text{;}$ or $\exists x\in\mathbb R\text{,}$ $x\geq a$ and $2^{-x}\geq \dfrac{1}{100}$”.
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We see that the original statement is true. For an example, we can take $a=7\text{.}$ Then we see that $\dfrac{1}{2^7}=\dfrac{1}{128} \lt \dfrac{1}{100}\text{,}$ and whenever $x\geq 7\text{,}$ we get $\dfrac{1}{2^x}\leq\dfrac{1}{2^7} \lt \dfrac{1}{100}\text{.}$
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Notice that we are implicitly using the fact that the function $2^x$ is an increasing function of $x\text{,}$ and so $2^{-x}$ is a decreasing function of $x\text{.}$ This is not hard to prove. If we assume that $a \lt b\text{,}$ then $\Delta = 2^b-2^a = 2^a \left(2^{b-a}-1\right)\text{.}$ And since $b-a \gt 0\text{,}$ we know that $2^{b-a} \gt 1\text{.}$ Thus $\Delta \gt 0\text{,}$ and so $2^b \gt 2^a\text{.}$
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$\forall n\in\mathbb R\text{,}$ $n$ is even if and only if $n^2$ is even.
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The negation of this sentence is:
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“$\exists n\in \mathbb R$ such that $n$ is even but $n^2$ is not even; or $n^2$ is even but $n$ is not even.”
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We see that the negation of the statement is true. For an example, we can take $n=\sqrt{2}\text{.}$ Then, we see $n^2=2$ is even whereas $n$ is not even since it is not an integer.
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6.6.15.

Solution.

Proof: Let $a,b\in\mathbb{N}\text{.}$ Assume that $b \lt a\text{.}$ Thus, since $b^2\geq 0\text{,}$ we get $b-b^2\leq b \lt a\text{.}$
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Disproof: Consider the negation:

\begin{equation*} \exists p\in\mathbb N \text{ such that } \exists q\in\mathbb{N} \text{ such that } \sqrt{\dfrac{p}{q}}\in\mathbb N \text{ and }\left( \sqrt{p}\notin\mathbb N \text{ or } \sqrt{q}\not\in\mathbb N \right). \end{equation*}

So we can prove that this is true using an example --- that is a counter-example to the original statement. Take $p=8$ and $q=2\text{.}$ Then we see that $\sqrt{\dfrac{p}{q}}=\sqrt{4}=2\in\mathbb N\text{,}$ whereas $\sqrt{p}=\sqrt{8}\notin\mathbb N$ and $\sqrt{q}=\sqrt{2}\notin\mathbb N\text{.}$

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Note that We haven’t proved that $\sqrt{2}, \sqrt{8}$ are not integers, but we do so later in the book 11. For the moment, it is safe to assume this.
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Proof: Let $a,b\in\mathbb{R}\text{,}$ then we can pick $c=a$ and $d=b$ and get $ab=cd$ and moreover, since $a=c$ and $b=d\text{,}$ we see that the statement is true.
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Disproof: Consider the negation:

\begin{equation*} \exists a,b\in\mathbb{N} \text{ such that } \left(\exists x,y\in\mathbb{Z} \text{ and } \exists k\in\mathbb{N} \text{ such that } ax+by=k \right) \text{ and } (k \nmid a \lor k \nmid b). \end{equation*}

Since all of these are existential quantifiers we can prove the negation is true by an example --- this is just a counter-example to the original statement.

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Take $a=2$ and $b=3$ so that both are prime numbers. At the same time, taking $x=y=1$ and $k=5$ and get $ax+by=k\text{,}$ but $5 \nmid 2$ and $5 \nmid 3\text{.}$
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Similarly, one can take $a=b=1\text{,}$ and $x=y=1$ with $k=2\text{.}$This gives $ax+by=k\text{,}$ but $2 \nmid 1$ and $2 \nmid 1\text{.}$
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6.6.16.

Solution.

Let $\epsilon \gt 0$ be given. Take $\delta=\epsilon/6\text{.}$ Let $x\in\mathbb{R}$ such that $0 \lt |x-0| \lt \delta\text{.}$ Since $|\sin(\theta)|\leq1$ for all $\theta\in\mathbb{R}\text{,}$ we have

\begin{equation*} |f(x)-0|=\left|6x\sin\left(\frac{1}{x}\right)\right|=|6x|\left|\sin\left(\frac{1}{x}\right)\right|\leq |6x|. \end{equation*}

Let $\epsilon \gt 0$ be given. Let $N=\left\lceil \frac{5}{\epsilon}\right\rceil\text{.}$ Notice that when $n \geq N\text{,}$ we have that

\begin{equation*} \frac{5}{n} \lt \frac{5}{N} \leq \epsilon. \end{equation*}

Since $n \geq 1\text{,}$ we know that

\begin{equation*} \frac{1}{n} \gt \frac{1}{n^2} \geq \frac{1}{n^3} \end{equation*}

Hence

\begin{equation*} \left|1-\frac{2}{n^2}-\frac{3}{n^3}-1\right| =\frac{2}{n^2}+\frac{3}{n^3}\leq\frac{5}{n}\lt\epsilon. \end{equation*}

Thus the sequence converges to 1.

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We shouldn’t feel bad that we made a more complicated proof first; It is a normal part of doing mathematics. Our first proof of a result is often “bettered” by our second proof.

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6.6.22.

Solution.

The symbolic statement

\begin{equation*} \forall M\gt0, \exists N\in\mathbb{N} \text{ such that } \forall n\in\mathbb{N}, n\geq N\implies s_n\geq M \end{equation*}

means that for all $M\gt0\text{,}$ there is some $N\in\mathbb{N}$ so that for all $n\in\mathbb{N}\text{,}$ if $n\geq N\text{,}$ then we have $s_n\geq M\text{.}$ So, $s_n$ can be made arbitrarily large (larger than $M\text{,}$ where $M$ is given), by taking $n$ large enough (larger than $N\text{,}$ where the choice of $N$ depends on $M$).

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The negation of the statement is

\begin{equation*} \exists M\gt0 \text{ such that } \forall N\in\mathbb{N}, \exists n\in \mathbb{N} \text{ such that } n\geq N \text{ and } s_n\lt M. \end{equation*}

This means that there is some $M\gt 0$ so that no matter how large we take $N\text{,}$ there’s some $n\geq N$ with $s_n\lt M\text{.}$ This doesn’t necessarily mean that all the terms are bounded by some fixed number, but rather, that we can always find some `small’ value.

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Proof of (c).

Let $M\gt 0\text{.}$ Take $N=M^2\text{.}$ Let $n\in\mathbb{N}\text{,}$ and suppose $n\geq N\text{.}$ Then $n\geq M^2\text{,}$ and so $\sqrt{n}\geq M\text{.}$ Therefore

\begin{equation*} \displaystyle\lim_{n\to\infty}\sqrt{n}=+\infty. \end{equation*}

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Proof of (d).

Take $M=1\text{,}$ and let $N\in \mathbb{N}\text{.}$ If $N$ is odd, then

\begin{equation*} (-1)^N\sqrt{N}\lt0\lt M. \end{equation*}

If $N$ is even, then $N+1$ is odd, and

\begin{equation*} (-1)^{N+1}\sqrt{N+1}\lt0\lt M. \end{equation*}

Thus in any case, there is some $n\geq N$ with $(-1)^n\sqrt{n}\lt M\text{.}$ Therefore

\begin{equation*} \displaystyle\lim_{n\to\infty}(-1)^n\sqrt{n}\neq+\infty. \end{equation*}

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Proof of (e).

Let $M\gt 0\text{.}$ Take $N=\max\{M,101\}\text{.}$ Let $n\in\mathbb{N}\text{,}$ and suppose $n\geq N\text{.}$ Since $n\geq N\geq 101\text{,}$ we have $n-100\geq 1\text{.}$ Thus

Solution.

We saw in part (d) of Exercise 6.6.22 that $\lim_{n\to\infty}(-1)^n\sqrt{n}\neq+\infty\text{,}$ and $((-1)^n\sqrt{n})_{n\in\mathbb{N}}$ is an unbounded sequence. Indeed, given $M\geq 0\text{,}$ we choose $N$ be an integer strictly greater than $M^2\text{.}$ Using the ceiling function, we set $N = \lceil M^2 \rceil +1\text{.}$ Then $N\gt M^2$ implies $|a_N|=\sqrt{N}\gt M\text{.}$
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We need to find a sequence that increases, and does not go to infinity. Take $a_n=-1/n\text{.}$ Then $a_{n+1}\geq a_n$ for all $n\in\mathbb{N}\text{.}$ Moreover, given any $M\geq 0\text{,}$ we have that $a_n\lt M$ for all $n\text{.}$ This implies that $\displaystyle\lim_{n\to\infty}a_n\neq+\infty\text{.}$
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Proof of (c).

Suppose that $\displaystyle\lim_{n\to\infty}a_n\neq+\infty\text{,}$ and that $a_n$ is increasing. We show that $(a_n)_{n\in\mathbb{N}}$ is bounded. Since $\displaystyle\lim_{n\to\infty}a_n\neq+\infty\text{,}$ there is some $C\gt0$ such that for any $n\in\mathbb{N}\text{,}$ there’s some $m\geq n$ with $a_m\lt C\text{.}$ Fix $n\in \mathbb{N}\text{,}$ and let $m\geq n$ be such that $a_m\lt C\text{.}$ Since $(a_n)_{n\in\mathbb{N}} $ is increasing and $m\geq n\text{,}$ we have

\begin{equation*} a_n\leq a_m \lt C. \end{equation*}

But as $(a_n)_{n\in\mathbb{N}} $ is increasing, we also know that $a_1\leq a_n\text{.}$ Let $M=\max\{|a_1|,|C|\}\text{.}$ Note that

\begin{equation*} -|a_1|\leq a_1\leq a_n \lt C \leq |C| \end{equation*}

But by choice of $M\text{,}$ $|C|\leq M\text{,}$ and $-M\leq -|a_1|\text{.}$ Therefore

\begin{equation*} -M \leq a_n \leq M, \end{equation*}

and so $|a_n|\leq M\text{.}$ This holds for all $n\text{,}$ and so $a_n$ is bounded. Thus if $\displaystyle\lim_{n\to\infty}a_n\neq+\infty\text{,}$ either $(a_n)_{n\in\mathbb{N}} $ is not increasing or $(a_n)_{n\in\mathbb{N}} $ is bounded.

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6.6.25.

Solution.

Proof of (a).

We are going to show that $D$ is a distance, by showing that it satisfies triangle inequality. Since the function $D$ is defined piecewise, we see that we would need to prove this statement using cases.

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Let $x,y,z \in\mathbb R\text{.}$ Now either $x=z$ or $x \neq z\text{,}$ so we have two cases to check.

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Case 1: If $x = z$ then $D(x,z) = 0\text{.}$ Since $D(x,y),D(y,z) \geq 0 = D(x,z)\text{,}$ we know that $D(x,y) \leq D(x,y) + D(y,z)\text{.}$
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Case 2: If $x \neq z$ then $D(x,z) = 1\text{.}$ So we must show that $D(x,y) + D(y,z) \geq 1\text{.}$ Now either $x = y$ or $x\neq y\text{;}$ this gives us two sub-cases to check.
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- Case 2a: If $x = y$ then we must have $y \neq z$ since $x \neq z\text{.}$ But then $D(x,y) + D(y,z) = 0+1=1 = D(x,z)\text{.}$
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- Case 2b: If $x \neq y$ then $D(x,y) = 1\text{.}$ Now we don’t know if $y = z$ or $y \neq z\text{,}$ but it doesn’t matter, we know that $D(y,z) \geq 0\text{.}$ Putting this together we have that $D(x,y)+D(y,z) = 1 + D(y,z) \geq 1 = D(x,z)\text{.}$
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So in both sub-cases, $D(x,z) \leq D(x,y) + D(y,z)$ as required.

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So $D$ satisfies the triangle inequality and therefore it is a distance.

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Proof.

Assume that $x_n \rightarrow L\text{.}$ Let $\epsilon = 1/2\text{.}$ Then, by definition there is some $N \in\mathbb N$ s.t. for all $n \gt N$ we must have $D(x_n,L)\lt 1/2\text{.}$ By definition of $D(x,y)$ it follows that we must have $D(x_n,L) = 0$ and so $x_n = L\text{.}$ Hence when $x_n \neq L$ we have $n \lt N\text{.}$ Thus, the set $\set{ n\in\mathbb{N}: x_n\neq L}$ is finite.

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7 Induction
7.3 Exercises

7.3.1.

Solution.

We have seen this statement before where we used cases to prove it. In this question, we are asked to use induction instead. This tells us that we can prove statements like this in more than one way.

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Proof.

We are going to use mathematical induction.

Base Case: We see that for $n=1\text{,}$ the statement is “$3\mid (1^3-1)=0$”. Hence, we see that this statement is true for $n=1\text{.}$
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Inductive Step: Let $k\geq 1\text{,}$ and assume that the statement is true for $n=k\text{,}$ that is, $3\mid (k^3-k)\text{.}$ Then we see that $k^3-k=3m$ for some $m\in\mathbb{Z}\text{.}$ Thus,
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\begin{equation*} (k+1)^3-(k+1)=k^3+3k^2+3k+1-k-1=(k^3-k)+3(k^2+k)=3(m+k^2+k), \end{equation*}

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and since $(m+k^2+k)\in\mathbb{Z}\text{,}$ we see $3\mid \big((k+1)^3-(k+1)\big)\text{.}$
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Therefore the statement is true for $n=k+1\text{,}$ and hence, by induction, it is true for all $n\in\mathbb{N}\text{.}$

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7.3.2.

Solution.

Proof.

We prove the result by induction.

Base case: when $n=2$ we have $n! = 2 \cdot 1 = 2$ and $n^n = 2^2 = 4\text{.}$ Since $2 \leq 4\text{,}$ the result holds when $n=2\text{.}$
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Inductive step: Assume the result holds when $n=k\geq 2\text{.}$ Then we have

\begin{align*} k! \amp \leq k^k \amp \text{multiply both sides by } k+1 \gt 0\\ (k+1)! \amp \leq (k+1) \cdot k^k \amp \text{since } k \lt k+1\\ \amp \leq (k+1) \cdot (k+1)^k\\ \amp = (k+1)^{k+1} \end{align*}

That is $(k+1)! \leq (k+1)^{k+1}\text{,}$ so the result holds when $n=k+1\text{.}$

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Since the base case and inductive step both hold, the result follows by induction.

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7.3.3.

Solution.

Proof.

We are going to use mathematical induction.

Base Case: We see that for $n=7\text{,}$ the stament is “$7!\gt 3^7$”. We also know that $7!=5040$ and $3^7=2187\text{.}$ Thus, the statement is true for $n=7$
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Inductive Step: Let $k\geq 7\text{,}$ and assume that the statement is true for $n=k\text{,}$ that is, $k!\gt 3^k\text{.}$ Then we see that
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\begin{equation*} (k+1)!=(k+1)k!\gt (k+1)3^k \gt (3)3^k=3^{k+1}, \end{equation*}

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since $(k+1)\gt 3\text{.}$
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Therefore the statement is true for $n=k+1\text{,}$ and hence, by induction, it is true for all $n\in\mathbb{N}\text{.}$

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We should also note that the inductive step doesn’t really require the assumption $n\geq 7\text{.}$ For the inductive step to be true, all we need is $n\geq 3\text{.}$ But, as we have seen in our scratchwork, the statement isn’t true for $n\lt 7\text{,}$ so the corresponding base cases would be false. This means that we can only prove this statement for $n\geq 7\text{.}$

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7.3.4.

Solution.

The question wants us to use mathematical induction. However, we also see that this is not a standard induction question in the sense that for all $n\in\mathbb N\text{,}$ we actually want to prove an existential statement, that is, the existence of $x,y\text{,}$ and $z$ satisfying $x^2+y^2=z^{2n+1}\text{.}$ Let’s see how we can do that in the proof.

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Proof.

We are going to use mathematical induction.

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Base Case: We see that for $n=1\text{,}$ the statement is “$\exists x,y,z\in \mathbb Z$ such that $x\geq 2\text{,}$ $y\geq 2\text{,}$ and $z\geq 2$ and satisfy $x^2+y^2=z^{3}$”. We see that for $x=2\text{,}$ $y=2\text{,}$ and $z=2\text{,}$ this statement is true.
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Inductive Step: Let $k\geq 1\text{,}$ and assume that the statement is true for $n=k\text{,}$ that is, $\exists x,y,z\in \mathbb Z$ such that $x\geq 2\text{,}$ $y\geq 2\text{,}$ and $z\geq 2$ and satisfy $x^2+y^2=z^{2k+1}\text{.}$ Then we see that if we multiply the equation by $z^2\text{,}$ we get $(xz)^2+(yz)^2=z^{2k+3}=z^{2(k+1)+1}\text{.}$ Moreover, we see that $(xz),(yz)\geq 2$ since $x,z,y\geq 2\text{.}$
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Therefore the statement is true for $n=k+1\text{,}$ and hence, by induction, it is true for all $n\in\mathbb{N}\text{.}$

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7.3.5.

Solution.

Proof.

Let $n\in\mathbb{N}\text{.}$ We proceed by induction on $n\text{.}$ When $n=1\text{,}$ we have

\begin{equation*} 3^{2\cdot 1}-1 =9-1=8, \end{equation*}

Let $n\in\mathbb{N}\text{.}$ We proceed by induction on $n\text{.}$ When $n=0\text{,}$ we have

\begin{equation*} 2^{2n+1}+3^{2n+1} =2+3=5, \end{equation*}

which is divisible by $5\text{.}$ Now suppose that the result holds for $n=k\text{,}$ so

\begin{equation*} 5\mid \left( 2^{2k+1}+3^{2k+1} \right). \end{equation*}

Then there is some $\ell\in\mathbb{Z}$ such that

\begin{equation*} 2^{2k+1}+3^{2k+1} = 5\ell. \end{equation*}

Note that

\begin{equation*} 2^{2(k+1)+1}+3^{2(k+1)+1} =4\cdot2^{2k+1}+9\cdot 3^{2k+1}=4(2^{2k+1}+3^{2k+1})+5\cdot 3^{2k+1}. \end{equation*}

Then by the inductive hypothesis

\begin{equation*} 2^{2(k+1)+1}+3^{2(k+1)+1} =4\cdot 5\ell+5\cdot 3^{2k+1} =5(4\ell+3^{2k+1}) \end{equation*}

and so $5$ divides $2^{2(k+1)+1}+3^{2(k+1)+1}\text{.}$ Therefore, by induction, $5$ divides $2^{2n+1}+3^{2n+1}$ for all $n\in\mathbb{N}\text{.}$

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7.3.9.

Solution.

Proof.

Let $n\in\mathbb{N}\text{.}$ Suppose $n=2\text{,}$ and that $x_1,x_2\in\mathbb{Q}\text{.}$ Then there are $a_1,a_2,b_1,b_2\in\mathbb{Z}$ with $b_1\neq0$ and $b_2\neq0$ so that $x_1=a_1/b_1$ and $x_2=a_2/b_2\text{.}$ Then

\begin{equation*} x_1+x_2 = \frac{b_2a_1+b_1a_2}{b_1b_2} \end{equation*}

where $b_2a_1+b_1a_2,b_1b_2\in\mathbb{Z}$ with $b_1b_2\neq0\text{.}$ Thus $x_1+x_2\in\mathbb{Q}\text{.}$

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Now assume the result holds for $n=k$ for some $k\geq 2\text{,}$ and suppose that $x_1,x_2,\dots,x_{k+1}\in\mathbb{Q}\text{.}$ By the inductive hypothesis, $x_1+x_2+\dots+x_k\in\mathbb{Q}\text{.}$ Moreover, by the base case, the sum of two rationals is rational, and therefore

\begin{equation*} x_1+x_2+\dots+x_{k+1}=(x_1+x_2+\dots+x_k)+x_{k+1}\in\mathbb{Q}. \end{equation*}

By induction, the result holds for any $n\geq2\text{.}$

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Note that the proof by induction only gives the result for a sum of finitely many terms. We cannot conclude that the sum of an infinite number of rational numbers is rational; indeed that statement is false. To see this, suppose that $x_0=1\text{,}$ $x_1=4\text{,}$ $x_3=1\text{,}$ and in general, $x_n$ is the $n^{th}$ digit of $\sqrt{2}\text{.}$ Then $x_n 10^{-n} \in\mathbb{Q}$ for all $n=0,1,2,3,\dots\text{.}$ But

\begin{equation*} \sqrt{2}=\sum_{n=0}^\infty x_n 10^{-n} \not\in\mathbb{Q}. \end{equation*}

We’ll take the statement that $\sqrt{2}=\sum_{n=0}^\infty x_n 10^{-n}$ to be a fact, but this does require a proof, since we’re dealing with an infinite number of terms. Really, we’d need to prove that

\begin{equation*} \sqrt{2}=\lim_{N\to\infty}\sum_{n=0}^Nx_n 10^{-n} \end{equation*}

which is beyond the scope of what we are trying to do in this chapter.

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7.3.10.

Solution.

Proof.

We are going to use mathematical induction.

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Base Case: We see that for $n=1\text{,}$ the statement is “$\displaystyle{\sum\limits_{k=1}^1 k^3=1=\left( \sum\limits_{k=1}^1 k \right)^2}$”. Thus, the statement is true for $n=1\text{.}$
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Inductive Step: Let $m\geq 1$ and assume that the statement is true for $n=m\text{,}$ that is, assume that $\displaystyle{\sum\limits_{k=1}^m k^3=\left( \sum\limits_{k=1}^m k \right)^2}\text{.}$ Then,

\begin{equation*} \sum\limits_{k=1}^{m+1} k^3=\sum\limits_{k=1}^m k^3+(m+1)^3=\left( \sum\limits_{k=1}^m k \right)^2+(m+1)^3. \end{equation*}

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Moreover, we know that $\displaystyle{\sum\limits_{k=1}^m k}=\dfrac{m(m+1)}{2}\text{.}$ Hence,

\begin{equation*} \sum\limits_{k=1}^{m+1} k^3=\frac{m^2(m+1)^2}{4}+(m+1)^3= \frac{(m+1)^2(m^2+4m+1)}{4}=\frac{(m+1)^2(m+2)^2}{4}= \left( \sum\limits_{k=1}^{m+1} k \right)^2. \end{equation*}

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Therefore the statement is true for $n=m+1\text{,}$ and hence, by induction, it is true for all $n\in\mathbb{N}\text{.}$

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7.3.11.

Solution.

Proof.

We are going to prove this statement using induction.

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Base case: We see that $\sum\limits_{j=1}^1 j^3=1\gt\frac{1}{4}=\frac{1}{4}n^4\text{.}$ Thus, the statement is true for $n=1\text{.}$
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Inductive step: Assume that the statement is true for $n=k$ for some $k\geq 1\text{,}$ that is,
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$\sum\limits_{j=1}^k j^3\gt\frac{1}{4}k^4\text{.}$ Then, we see,
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\begin{align*} \sum\limits_{j=1}^{k+1} j^3 \amp =\sum\limits_{j=1}^k j^3+(k+1)^3 \\ \amp \gt\frac{1}{4}k^4+(k+1)^3 \amp \text{by assumption}\\ \amp =\frac{1}{4}k^4+k^3+3k^2+3k+1\\ \amp =\frac{1}{4}(k+1)^4+\left(\frac 32k^2+2k+ \frac{3}{4}\right)\\ \amp \gt\frac{1}{4}(k+1)^4, \end{align*}

since $\frac{3}{2}k^2+2k+\frac{3}{4} \gt 0$ for all $k\in\mathbb N\text{.}$ This means that the statement is true for $n=k+1\text{.}$

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Therefore, by mathematical induction, we see that the statement is true for all $n\in\mathbb N\text{.}$

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7.3.12.

Solution.

Proof.

Let $r\in\mathbb{R}\text{,}$ $r\neq 1\text{.}$ We proceed with induction on $n\text{.}$ For the base case, let $n=0\text{.}$ Then

\begin{equation*} \sum_{i=0}^nr^i=r^0=1=\frac{1-r}{1-r}=\frac{1-r^{n+1}}{1-r}. \end{equation*}

Now suppose that

\begin{equation*} \sum_{i=0}^kr^i = \frac{1-r^{k+1}}{1-r} \end{equation*}

for $n=k\text{,}$ where $k\in\mathbb{Z}$ and $k\geq0\text{.}$ Then

\begin{equation*} \sum_{i=0}^{k+1}r^i = \left(\sum_{i=0}^kr^i \right)+r^{k+1} =\frac{1-r^{k+1}}{1-r}+r^{k+1}, \end{equation*}

by the inductive hypothesis. Some arithmetic gives

\begin{equation*} \frac{1-r^{k+1}}{1-r}+r^{k+1}=\frac{1-r^{k+1}+r^{k+1}(1-r)}{1-r}=\frac{1-r^{k+2}}{1-r}. \end{equation*}

Putting everything together, we have

\begin{equation*} \sum_{i=0}^{k+1}r^i = \frac{1-r^{k+2}}{1-r}=\frac{1-r^{(k+1)+1}}{1-r} \end{equation*}

and so the result holds for $n=k+1\text{.}$ Then, by induction, the result holds for any $n\in\set{0,1,2,\dots}\text{.}$

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7.3.13.

Solution.

Claim:

\begin{equation*} f^{(k)} (x)= \frac{n!}{(n-k)!}x^{n-k} \end{equation*}

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Proof.
We proceed by induction on $k\text{.}$ As a base case, take $k=1\text{.}$ Computing the derivative of $f\text{,}$ we see $f'(x) = nx^{n-1} =\frac{n!}{(n-1)!}x^{n-1}\text{,}$ as desired.
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Now assume that the derivative holds for an arbitrary $k\in \mathbb{N}\text{:}$

\begin{equation*} f^{(k)} (x)= \frac{n!}{(n-k)!}x^{n-k}. \end{equation*}

We would like to show that the claim holds for $k+1\text{.}$ We compute $f^{(k+1)(x)}$ by taking one derivative of $f^{(k)}(x)\text{.}$

\begin{align*} f^{(k+1)} (x) \amp = \frac{d}{dx} f^{(k)}(x) \end{align*}

By the inductive hypothesis,

\begin{align*} \amp = \frac{d}{dx} \frac{n!}{(n-k)!}x^{n-k}. \end{align*}

We compute the derivative

\begin{align*} \amp = \frac{n!}{(n-k)!} (n-k) x^{(n-k)-1}. \end{align*}

Finally, we rearrange into the desired form

\begin{align*} \amp = \frac{n!}{(n-(k+1))!} x^{n-(k+1)}. \end{align*}

By the inductive principle, our claim is true.

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Claim:

\begin{equation*} g^{(k)}(x) =\frac{(-1)^k (n+(k-1))!}{(n-1)!}x^{-n-k} \end{equation*}

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Proof.
We proceed by induction on $k\text{.}$ As a base case, take $k=1\text{.}$ Taking derivatives, we see that $g'(x) = -nx^{-n-1} = \frac{(-1)n!}{(n-1)!}x^{-n-1}\text{.}$ Since this is the desired form, the base case is true.
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Now, we assume that the derivative holds for an arbitrary $k\in \mathbb{N}\text{:}$

\begin{equation*} g^{(k)}(x) =\frac{(-1)^k (n+(k-1))!}{(n-1)!}x^{-n-k}. \end{equation*}

We would like to show that the result holds for $k+1\text{.}$ We compute

\begin{align*} g^{(k+1)(x)} \amp = \frac{d}{dx}g^{(k)}(x). \end{align*}

By the inductive hypothesis:

\begin{align*} \amp =\frac{d}{dx} \frac{(-1)^k (n+(k-1))!}{(n-1)!}x^{-n-k}. \end{align*}

We compute the derivative

\begin{align*} \amp = \frac{(-1)^k (n+(k-1))!}{(n-1)!} (-n-k) x^{-n-k-1}, \end{align*}

and rearrange to obtain the desired form

\begin{align*} \amp =\frac{(-1)^{k+1} (n+k)!}{(n-1)!} x^{-n-(k+1)}. \end{align*}

By induction, our claim holds.

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Claim:

\begin{equation*} h^{(k)}(x) = (2k-1)!! (9-2x)^{-(2k+1)/2} \end{equation*}

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Proof.
We prove our claim by induction on $k\text{.}$ As a base case, let $k=1\text{.}$ By the power rule and the chain rule, we see that $h'(x) = \frac{-1}{2}(9-2x)^{-3/2}(-2) = (9-2x)^{-3/2}\text{.}$ Hence, our base case holds.
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Now, we assume that the derivative holds for an arbitrary $k\in \mathbb{N}\text{:}$

\begin{equation*} h^{(k)}(x) = (2k-1)!! (9-2x)^{-(2k+1)/2} \end{equation*}

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We would like to show that the result holds for $k+1\text{.}$ We compute,

\begin{align*} h^{(k+1)}(x) \amp = \frac{d}{dx}h^{(k)}(x). \end{align*}

By the inductive hypothesis,

\begin{align*} \amp = \frac{d}{dx}(2k-1)!! (9-2x)^{-(2k+1)/2}. \end{align*}

We take the derivative

\begin{align*} \amp =(2k-1)!! \frac{-(2k+1)}{2}(-2 ) (9-2x)^{-(2k+1)/2 -1} \end{align*}

Finally, we rearrange

\begin{align*} \amp = (2k+1)!! (9-2x)^{-(2k+1)/2 -1} \end{align*}

This is the desired form, so our claim holds by the principle of induction.

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7.3.14.

Solution.

Proof.

As a base case, we take $n=1\text{.}$ Then we see that

\begin{equation*} \sum_{k=1}^1 k^2 = 1^2 = \frac{1(2)(3)}{6}. \end{equation*}

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We assume that the statement is true for $n = j \in \mathbb{N}\text{,}$ that is

\begin{equation*} \sum_{k=1}^j k^2 = \frac{j(j+1)(2j+1)}{6}, \end{equation*}

and show that the statement is true for $n=j+1\text{.}$

\begin{align*} \sum_{k=1}^{j+1} k^2 \amp = (j+1)^2 + \sum_{k=1}^{j} k^2 \end{align*}

By the inductive hypothesis,

\begin{align*} \amp = (j+1)^2 + \frac{j(j+1)(2j+1)}{6}. \end{align*}

Factoring out $j+1$ and giving both terms a common denominator yields

\begin{align*} \amp =\frac{(j+1)(2j^2+7j+6)}{6} \end{align*}

We now factor the numerator to get

\begin{align*} \amp = \frac{(j+1)(j+2)(2j+3)}{6} \end{align*}

Finally, we rearrange into the desired form

\begin{align*} \amp = \frac{(j+1)((j+1)+1)(2(j+1)+1)}{6}. \end{align*}

And so the statement holds for $n=j+1\text{,}$ as required.

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Since the base case and inductive step are true, by the principle of induction, the statement holds true for each $n\in \mathbb{N}\text{.}$

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7.3.15.

Solution.

Proof.

As a base case, we take $n=1\text{.}$ The base case is true since $\frac{1}{2} \lt 1\text{.}$

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Now assume that the statement is true for an arbitrary value $n=j\in \mathbb{N}\text{:}$

\begin{equation*} \sum_{k=1}^j \frac{1}{2^k} = \frac{1}{2}+\frac{1}{4} + \cdots + \frac{1}{2^j} \lt 1. \end{equation*}

Now we consider the desired sum

\begin{align*} \sum_{k=1}^{j+1} \frac{1}{2^k} \amp = \frac{1}{2}+\frac{1}{4} + \cdots + \frac{1}{2^j} + \frac{1}{2^{j+1}} \end{align*}

We factor $\frac{1}{2}$ from all but the first term to obtain

\begin{align*} \amp = \frac{1}{2} + \frac{1}{2} \left(\frac{1}{2}+\frac{1}{4} + \cdots + \frac{1}{2^j}\right) \end{align*}

By the inductive hypothesis, $\frac{1}{2}+\frac{1}{4} + \cdots + \frac{1}{2^j} \lt 1\text{,}$ therefore

\begin{align*} \amp \lt \frac{1}{2} + \frac{1}{2}(1)\\ \amp = 1. \end{align*}

So the inductive hypothesis is true.

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By the principle of mathematical induction, we have our desired result.

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7.3.16.

Solution.

When we prove a statement, say $P(n)\text{,}$ by induction, we prove the statement for all $n\in\mathbb{N}\text{.}$ Even though $n$ can be as large as we like, it must always be a finite number. We can’t actually set $n$ to infinity.

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The proposed proof actually shows that if $A\subseteq\mathbb{N}$ and $|A|=n$ for some $n\in\mathbb{N}\text{,}$ then $A$ has a maximum element. So, we have shown that if $A$ is a non-empty subset of $\mathbb{N}$ with a finite number of elements, then $A$ has a maximum element. This is not true if $A$ has infinitely many elements. For example, if $A=\mathbb{N}\text{,}$ then $A$ does not have a maximum.

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7.3.17.

Solution.

Proof of (a).

Let $n\in\mathbb{Z}\text{,}$ $n\geq0\text{.}$ We proceed by induction on $n\text{.}$ First assume $n=0\text{.}$ Then

\begin{equation*} \lim_{x\to\infty}x^ne^{-x}=\lim_{x\to\infty}e^{-x}=0, \end{equation*}

and so the result holds. Next suppose the result holds for $n=k\text{,}$ so that

\begin{equation*} \lim_{x\to\infty}x^ke^{-x}=0. \end{equation*}

By l’Hôpital’s rule,

\begin{equation*} \lim_{x\to\infty}x^{k+1}e^{-x}=\lim_{x\to\infty}\frac{x^{k+1}}{e^{x}}=\lim_{x\to\infty}\frac{(k+1)x^{k}}{e^{x}}. \end{equation*}

Pulling out the constant, and applying the inductive hypothesis, we have

\begin{equation*} \lim_{x\to\infty}\frac{x^{k+1}}{e^{x}}=(k+1)\lim_{x\to\infty}\frac{x^{k}}{e^{x}}=(k+1)\cdot0=0. \end{equation*}

Therefore the result holds for $n=k+1\text{,}$ and by induction, for any $n\text{.}$

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Proof of (b).

Let $n\in\mathbb{Z}\text{,}$ $n\geq0\text{.}$ We proceed by induction on $n\text{.}$ First assume $n=0\text{.}$ Then

\begin{equation*} \int_0^\infty x^ne^{-x}\mathrm{d}x =\int_0^\infty e^{-x}\mathrm{d}x = \lim_{t\to\infty}(-e^{-x})\Big]_0^t = 1=0!. \end{equation*}

Next suppose that the result holds for $n=k\text{;}$ that is,

\begin{equation*} k! = \int_0^\infty x^ke^{-x}\mathrm{d}x. \end{equation*}

Then, by integration by parts,

\begin{align*} \int_0^\infty x^{k+1}e^{-x}\mathrm{d}x \amp =(-x^{k+1}e^{-x})\Big]_0^\infty +(k+1)\int_0^\infty x^ke^{-x}\mathrm{d}x\\ \amp =\lim_{t\to\infty} \left(-t^{k+1}e^{-t}\right)+0^{k+1}e^{-0} +(k+1)\int_0^\infty x^ke^{-x}\mathrm{d}x\\ \amp = 0 +(k+1)\int_0^\infty x^ke^{-x}\mathrm{d}x \end{align*}

where we have used the fact that

\begin{equation*} \lim_{x \to \infty} x^k e^{-x} = 0. \end{equation*}

which we proved in (a). Now using the induction hypothesis, we therefore have

\begin{equation*} \int_0^\infty x^{k+1}e^{-x}\mathrm{d}x=(k+1)\int_0^\infty x^ke^{-x}\mathrm{d}x=(k+1)k!=(k+1)!. \end{equation*}

Thus, by induction, the result holds for all $n\text{.}$

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7.3.18.

Solution.

This is a generalization of the triangle inequality to arbitrary sets of finitely many real numbers. This proof is a standard way of generalizing statements that are stated for only two elements to arbitrary number of elements. This will also come up in later chapters when we discuss the union and intersection of multiple sets.

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Proof.

We are going to prove this statement using induction.

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Base case: We see that for $n=1$ we have the statement $a_1\leq a_1\text{,}$ which is true for all real numbers. Moreover for $n=2$ the statement is:
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“Let $a_1, a_2$ be real numbers. Then we have $\left| \sum_{k=1}^2 a_k \right|=|a_1+a_2|\leq |a_1|+|a_2| =\sum_{k=1}^2|a_k|\text{.}$”
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This is the triangle inequality (see Theorem 5.4.6) which we proved in Section 5.4.
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🔗
Inductive step: Assume that this statement is true for $m=n\geq 2\text{.}$ This means that whenever $a_1, a_2, \ldots a_n\in\mathbb R\text{,}$ then we have

\begin{equation*} \left| \sum_{k=1}^n a_k \right|\leq\sum_{k=1}^n|a_k|. \end{equation*}

Now assume that we have some set of $n+1$ real numbers, $b_1, b_2, \ldots b_n, b_{n+1}\in\mathbb R\text{.}$ Then we see

\begin{align*} \left| \sum_{k=1}^{n+1} b_k\right| \amp =\left| \left(\sum_{k=1}^n b_k\right)+b_{n+1}\right|\\ \amp \leq \left| \left(\sum_{k=1}^n b_k\right)\right|+|b_{n+1}| \amp \text{by Triangle Inequality}\\ \amp \leq \sum_{k=1}^n|b_k|+|b_{n+1}| \amp \text{by induction hypothesis}\\ \amp \leq \sum_{k=1}^{n+1}|b_k|. \end{align*}

Hence the statement is true for $m=n+1\text{.}$

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Therefore, by induction, the statement is true for all $n\in\mathbb N\text{.}$

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7.3.19.

Solution.

Proof.

Let $n$ be defined as the number of $1$’s between $100$ and $7\text{.}$ For example, in the $n=0$ case, the number is $1007$ and in the $n=4$ case, the number we are working with is $10011117\text{.}$ We proceed by induction on $n\text{.}$

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For our base case, take $n=0\text{.}$ That is, we wish to show that $1007$ is divisible by $53\text{.}$ Since $1007 = 19\cdot 53\text{,}$ our base case is true.

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We assume that the statement holds true for $n=k$ and wish to prove it for the $n=k+1$ case. Notice that we can write

\begin{equation*} 100\underbrace{1\dots 1}_{\substack{{k+1} \\ \text{times}}}7 = 100\underbrace{1\dots 1}_{\substack{{k} \\ \text{times}}}7 + 901\underbrace{0\dots 0}_{\substack{{n} \\ \text{times}}}. \end{equation*}

Since $901 = 17 \cdot 53\text{,}$ we see that $901\underbrace{0\dots 0}_{k} = 17 \cdot 53 \cdot 10^k\text{.}$ Therefore, $53\mid 901\underbrace{0\dots 0}_{k}\text{.}$ By the inductive hypothesis, $53 \mid 100\underbrace{1\dots 1}_{k}7\text{.}$ Finally, applying the hint tells us that $53$ divides $100\underbrace{1\dots 1}_{k+1}7\text{,}$ as required.

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Since the base case and inductive hypothesis are both true, the result follows.

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7.3.20.

Solution.

Proof.

Let $n \in \mathbb{N}\text{.}$ We prove, using strong induction, that for all $\ell \in \mathbb{N}$

\begin{equation*} F_{n+\ell} = F_{\ell} \cdot F_{n+1} + F_{\ell-1} \cdot F_{n}. \end{equation*}

and then use this to prove the result.

Base case: When $\ell=1$ we have

\begin{equation*} F_{n+1} = F_1 \cdot F_{n+1} + F_0 \cdot F_{n} \end{equation*}

and since $F_0=0, =F_1=1\text{,}$ this follows.

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🔗
Inductive step: Now assume that
\begin{equation*} F_{n+k} = F_{k} \cdot F_{n+1} + F_{k-1} \cdot F_{n}. \end{equation*}
holds for all $k \leq \ell\text{.}$ Then
\begin{align*} F_{n+\ell+1} \amp = F_{n+\ell} + F_{n+\ell-1}\\ \amp = (F_{\ell} \cdot F_{n+1} + F_{\ell-1} \cdot F_{n}) + (F_{\ell-1} \cdot F_{n} + F_{\ell-2} \cdot F_{n-1})\\ \amp = (F_{\ell}+F_{\ell-1} ) \cdot F_{n} + (F_{\ell-1}+F_{\ell-2}) \cdot F_{n-1}\\ \amp = F_{\ell+1} \cdot F_{n} + F_{\ell} \cdot F_{n-1} \end{align*}
as required.

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Thus

\begin{equation*} F_{n+\ell} = F_{\ell} \cdot F_{n+1} + F_{\ell-1} \cdot F_{n}. \end{equation*}

holds for all $\ell \in\mathbb{N}\text{.}$

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Now we prove the main result by induction.

Base case: Since $F_q \mid F_q$ the case case is true.

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Inductive step: Assume that

\begin{equation*} F_q \mid F_{qn} \end{equation*}

and consider $F_{qn+q}\text{.}$ By the inductive result we proved above (with $n \mapsto qn$ and $\ell \mapsto q$) we know that

\begin{equation*} F_{qn+q} = F_{q+1} \cdot F_{qn} + F_q \cdot F_{qn-1} \end{equation*}

Now since $F_q \mid F_{qn}$ by assumption, this shows that $F_q \mid F_{qn+q}$ as required.

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Thus $F_q \mid F_{qn}$ for all $n \in \mathbb{N}\text{.}$

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7.3.21.

Solution.

We prove each result in turn. First, Pascal’s identity:

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Proof.

Let $n,r$ be as given, then

\begin{align*} \binom{n}{r} + \binom{n}{r-1} \amp = \frac{n!}{(n-r)! r!} + \frac{n!}{(n-r+1)! (r-1)!}\\ \amp = \frac{n! (n-r+1)}{(n-r+1)! r!} + \frac{n! \cdot r}{(n-r+1)! r!}\\ \amp = \frac{n!}{(n-r+1)! r!} \left(n-r+1 + r \right)\\ \amp = \frac{n! (n+1)}{(n+1-r)! r!}\\ \amp = \frac{(n+1)!}{( (n+1)-r)! r!}\\ \amp = \binom{n+1}{r} \end{align*}

as required.

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Next, the Binomial Theorem:

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Proof.

Let $a,b$ be as stated, and then we prove the result by induction on $n\text{.}$

When $n=1$ we have that

\begin{equation*} (a+b)^1 = a+b = \binom{1}{0} a + \binom{1}{1} b \end{equation*}

and so the base case holds.

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🔗
Now assume that the result holds for some $n = \ell \in \mathbb{N}$ and we show that it holds for $n = \ell+1\text{.}$ Start with

\begin{equation*} (a+b)^\ell = \sum_{k=0}^\ell \binom{\ell}{k} a^{\ell-k} b^k. \end{equation*}

Multiplying both sides by $(a+b)$ gives:

\begin{align*} (a+b)^{\ell+1} \amp = (a+b)\sum_{k=0}^\ell \binom{\ell}{k} a^{\ell-k} b^k\\ \amp = \sum_{k=0}^\ell \binom{\ell}{k} a^{\ell-k} b^k (a+b)\\ \amp = \sum_{k=0}^\ell \binom{\ell}{k} a^{\ell-k+1} b^k + \sum_{k=0}^\ell \binom{\ell}{k} a^{\ell-k} b^{k+1} \end{align*}

Extract out the very first and very last terms

\begin{align*} \amp = \binom{\ell}{0} a^{\ell+1} + \sum_{k=1}^\ell \binom{\ell}{k} a^{\ell-k+1} b^k + \sum_{k=0}^{\ell-1} \binom{\ell}{k} a^{\ell-k} b^{k+1} + \binom{\ell}{\ell} a^0 b^{\ell+1} \end{align*}

re-index the second sum to make it start from 0 by replacing $k$ with $k-1$

\begin{align*} \amp = a^{\ell+1} + \sum_{k=1}^{\ell} \binom{\ell}{k} a^{\ell-k+1} b^{k} + \sum_{k=1}^{\ell} \binom{\ell}{k-1} a^{\ell-k+1} b^{k} + a^0 b^{\ell+1} \end{align*}

If we group-together terms with similar powers of $a,b\text{,}$ we get

\begin{align*} \amp = a^{\ell+1} + \sum_{k=1}^{\ell} \left[ \binom{\ell}{k} + \binom{\ell}{k-1} \right] a^{\ell-k+1} b^{k+1} + b^{\ell+1} \end{align*}

then, Pascal’s identity gives

\begin{align*} \amp = a^{\ell+1} + \sum_{k=1}^{\ell} \binom{\ell+1}{k} a^{\ell+1-k} b^{k} + b^{\ell+1} \end{align*}

Finally, since $\binom{\ell+1}{0}=\binom{\ell+1}{\ell+1}=1 \text{,}$ we have

\begin{align*} \amp = \binom{\ell+1}{0} a^{\ell+1} b^0 + \sum_{k=1}^{\ell} \binom{\ell+1}{k} a^{\ell+1-k} b^{k} + \binom{\ell+1}{\ell+1} a^0 b^{\ell+1}\\ \amp = \sum_{k=0}^{\ell+1} \binom{\ell+1}{k} a^{\ell+1-k} b^{k} \end{align*}

as required.

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Since the base-case and inductive step hold, the result follows by induction.
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7.3.22.

Solution.

Proof.

We proceed by induction on $k\text{.}$

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As a base case, let $k=1\text{.}$ We show that

\begin{equation*} \mathcal{L}\{f'\}(s) = s\mathcal{L}\{f\}(s) - f(0) \end{equation*}

using integration by parts. Using the definition of $\mathcal{L}\{f'\}(x)\text{,}$ we compute

\begin{align*} \mathcal{L}\{f'\}(s) \amp = \int_0^\infty f'(x)e^{-sx} \mathrm{d}x \end{align*}

Integrating by parts, we see,

\begin{align*} \amp = f(x)e^{-sx}\bigg|_{0}^\infty - \int_0^\infty f(x) (-se^{-sx}) \mathrm{d}x\\ \amp = \lim_{t\to \infty} f(t)e^{-st} - f(0) + s\int_0^\infty f(x) e^{-sx} \mathrm{d}x \end{align*}

By assumption 2, the first term is $0\text{,}$ so we have

\begin{align*} \amp =-f(0) + s\mathcal{L}\{f\}(s) \end{align*}

This proves the base case.

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Now for some $k\in \mathbb{N}\text{,}$ we assume that

\begin{equation*} \mathcal{L}\left\{f^{(k)}\right\}(s) = s^k\mathcal{L}\{f\}(x) - \sum_{i=0}^{k-1} s^{k-1-i}f^{(i)}(0) \end{equation*}

We wish to show that the result holds for $k+1\text{.}$ The computation is quite similar to the base case above.

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We use integration by parts to compute

\begin{align*} \mathcal{L}\left\{f^{(k+1)}\right\}(s) \amp = \int_0^{\infty} f^{(k+1)}(x)e^{-sx}\mathrm{d}x\\ \amp = f^{(k)}(x)e^{-sx} \bigg|_{0}^{\infty} - \int_0^\infty f^{(k)}(x)(-se^{-sx})\mathrm{d}x\\ \amp = \lim_{t\to \infty} f^{(k)}(t)e^{-st} - f^{(k)}(0) + s\mathcal{L}\{f^{(k)}\}(s) \end{align*}

By assumption 3. the first term is $0\text{,}$ so we have

\begin{align*} \amp = - f^{(k)}(0) + s\mathcal{L}\{f^{(k)}\}(s). \end{align*}

By the inductive hypothesis,

\begin{align*} \amp = - f^{(k)}(0) + s\left( s^k\mathcal{L}\{f\}(x) - \sum_{i=0}^{k-1} s^{k-1-i}f^{(i)}(0)\right) \end{align*}

Expand the bracketed terms

\begin{align*} \amp = - f^{(k)}(0) + s^{k+1}\mathcal{L}\{f\}(x) - \sum_{i=0}^{k-1} s^{k-i}f^{(i)}(0) \end{align*}

Finally, we absorb $f^{(k)}(0)$ into the sum

\begin{align*} \amp =s^{k+1}\mathcal{L}\{f\}(x) - \sum_{i=0}^{k} s^{k-i}f^{(i)}(0) \end{align*}

So the inductive step holds.

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Since both the base case and inductive step are true, the result follows by induction.

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7.3.23.

Solution.

We offer two proofs of the statement. We first use mathematical induction with $n=0,\, n=1$ as the base case.

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Proof.

Let $\alpha$ be as stated and then we proceed by strong induction.

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As a base case, we take both $n=0$ and $n=1\text{.}$ When $n=0\text{,}$ we have $\alpha^0 + \frac{1}{\alpha^0} = 2\in \mathbb{Z}\text{,}$ and the $n=1$ case follows directly by assumption.

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For our inductive hypothesis, we assume that for an arbitrary $n=k\in \mathbb{N}\text{,}$

\begin{equation*} \alpha^{k-1} +\frac{1}{\alpha^{k-1}}\in \mathbb{Z} \qquad \text{and} \qquad \alpha^k+\frac{1}{\alpha^k}\in \mathbb{Z} \end{equation*}

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We wish to show that the statement holds true for $n=k+1\text{.}$

\begin{equation*} \alpha^{k+1} + \frac{1}{\alpha^{k+1}} = \left(\alpha^k+\frac{1}{\alpha^k}\right)\left(\alpha + \frac{1}{\alpha}\right) -\left(\alpha^{k-1}+\frac{1}{\alpha^{k-1}}\right) \end{equation*}

By assumption, $\alpha + \frac{1}{\alpha}\in \mathbb{Z}\text{.}$ Since the product and difference of integers is an integer, we conclude that $\alpha^{k+1} + \frac{1}{\alpha^{k+1}} \in \mathbb{Z}\text{.}$

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Since both the base case and inductive step are true, the result follows.

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We now appeal to the principle of strong mathematical induction for an alternate proof.

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Proof.

We proceed by strong induction. As a base case, we take $n=0\text{.}$ Then we have $\alpha^0 + \frac{1}{\alpha^0} = 2\in \mathbb{Z}\text{.}$

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Fix $n\in \mathbb{N}\text{.}$ For our inductive hypothesis, we assume that for any $m\in \mathbb{N}\cup \{0\}$ such that $m\leq n\text{,}$

\begin{equation*} \alpha^{m} +\frac{1}{\alpha^{m}}\in \mathbb{Z} \end{equation*}

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We wish to show that the statement holds true for $n+1\text{.}$

\begin{equation*} \alpha^{n+1} + \frac{1}{\alpha^{n+1}} = \left(\alpha^n+\frac{1}{\alpha^n}\right)\left(\alpha + \frac{1}{\alpha}\right) -\left(\alpha^{n-1}+\frac{1}{\alpha^{n-1}}\right) \end{equation*}

By assumption, $\alpha + \frac{1}{\alpha}\in \mathbb{Z}\text{.}$ By the inductive hypothesis,

\begin{equation*} \alpha^n+\frac{1}{\alpha^n}\in \mathbb{Z} \hspace{10pt} \text{and} \hspace{10pt} \alpha^{n-1}+\frac{1}{\alpha^{n-1}} \in \mathbb{Z}. \end{equation*}

Since the product and difference of integers is an integer, we conclude that

\begin{equation*} \alpha^{n+1} + \frac{1}{\alpha^{n+1}} \in \mathbb{Z}. \end{equation*}

By the principle of strong mathematical induction, the result is proven.

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7.3.24.

Solution.

Proof.

We are going to use strong induction on $a\text{.}$

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Base case: We see that this statement is true for $n=1$ since we can take $m=0$ and $3\nmid 1\text{.}$
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🔗
Inductive step: Assume that the statement is true for all $1\leq \ell \leq k\text{.}$
🔗
Then for $n=k+1$ we have two cases.
- Case 1: $3\nmid n\text{.}$ In this case, we can take $m=0, a=n\text{,}$ that is $n = 3^0 n\text{.}$ Thus, the statement is true.
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- Case 2: $3\mid n\text{.}$ Then, we know that $n=3b$ for some $b\in\mathbb N\text{.}$ Moreover, we see that $b \lt k+1\text{,}$ and so $b \leq k\text{.}$ Thus, by inductive hypothesis, we see that
  
  \begin{equation*} b=3^{m} a, \end{equation*}
  
  for some $m,a \in\mathbb N$ with $3 \nmid a \text{.}$ Hence
  
  \begin{equation*} n = 3b = 3^{m+1}a \end{equation*}
  
  And since $3 \nmid a\text{,}$ the statement is true for $n=k + 1\text{.}$
  
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  🔗
So the inductive hypothesis holds for all $n \in \mathbb{N}\text{.}$

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🔗

Therefore the statement is true for all $n\in\mathbb N\text{.}$

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7.3.25.

Solution.

Claim: You can buy any souvenir you want!

🔗

Proof.

We proceed by strong induction. As a base case, we take $p=8\text{,}$ $p=9\text{,}$ and $p=10\text{.}$ Since $p=3+5\text{,}$ we can pay for an $\$8$ item with one $3$ dollar bill and one $5$ dollar bill. Similarly, $9= 3 \cdot 3 \text{,}$ so we can use three $\$3$ bills to purchase a $\$9$ item, and $10 = 2\cdot 5\text{,}$ so we can use two $\$5$ bills to purchase a $\$10$ item.

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Let $n \in \mathbb{N}$ with $n \gt 10\text{.}$ As our inductive hypothesis, we assume that we can pay for items with price $p=8, \, p=9, \dots, p=n$ using only $3$ and $5$ dollar bills. We show that we can purchase an $\$(n+1)$ item.

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By the inductive hypothesis, we can purchase an $\$(n-2)$ souvenir. That is, there exist natural numbers $a, b$ such that $3a+5b = n-2\text{.}$ Adding $3$ to each side, we see $3(a+1) + 5b = n+1\text{.}$ Therefore, we can purchase an $\$(n+1)$ souvenir using $(a+1)$ $\$3$ bills and $b$ $\$5$ bills.

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By the principle of strong mathematical induction, the result follows.

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7.3.26.

Solution.

Proof.

Let $n\in\mathbb{N}\text{.}$ We proceed with strong induction on $n\text{.}$ The result holds for $n=0$ and $n=1\text{,}$ since

\begin{equation*} a_0=2=(-2)^0+3^0, \quad a_1=1=(-2)^1+3^1. \end{equation*}

Now suppose that $k\geq 1$ and for all $n\leq k\text{,}$ we have $a_n=(-2)^n+3^n\text{.}$ Then by the recurrence for the sequence, and the inductive hypothesis, we have

\begin{equation*} a_{k+1}=a_k+6a_{k-1}=(-2)^k+3^k+6\left((-2)^{k-1}+3^{k-1}\right) \end{equation*}

We can rewrite this as

\begin{equation*} a_{k+1}=(-2)^k+3^k-3(-2)^{k}+2\cdot 3^{k}=-2\cdot (-2)^k+3\cdot 3^k=(-2)^{k+1}+3^{k+1} \end{equation*}

Therefore the result holds for $n=k+1\text{,}$ and so, by strong induction, all $n\in\mathbb{N}\cup\{0\}\text{.}$

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7.3.27.

Solution.

Proof.

We are going to use strong induction on $n\text{.}$

🔗

Base case: We see that the statement is true for $n=1$ since we can take $m=0$ and $c_0=1\text{.}$
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🔗
Inductive step: Assume that the statement is true for all $1 \leq \ell \leq k$ Then for $n=k+1$ we have two cases.
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- Case 1: $n$ is even. So $n=2a$ for some $a \in \mathbb{N}\text{.}$ Since $a \lt n\text{,}$ we know $a \leq k$ and so the induction hypothesis implies that there exists $m\in\mathbb Z\text{,}$ $m\geq 0$ and constants $c_0,c_1,c_2,\dots, c_m\in\set{0,1} $ such that
  
  \begin{equation*} a = c_m\cdot 2^m+c_{m-1}\cdot 2^{m-1}+\cdots+c_1\cdot 2+c_0. \end{equation*}
  
  This implies that
  
  \begin{equation*} n=2a = c_m\cdot 2^{m+1}+c_{m-1}\cdot 2^{m}+\cdots+c_1\cdot 2^2+c_0 \cdot 2, \end{equation*}
  
  Hence, the statement is true for $n=k+1\text{.}$
  
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- Case 2: $n$ is odd. So, $n=2b+1$ for some $b \in \mathbb{N}\text{.}$ Again, since $b \lt n\text{,}$ we know that $b \leq k$ and so the induction hypothesis tells us that there exists $m\in\mathbb Z\text{,}$ $m\geq 0$ and constants $c_0,c_1,c_2,\dots, c_m\in\set{0,1} $ such that
  
  \begin{equation*} b = c_m\cdot 2^m+c_{m-1}\cdot 2^{m-1}+\cdots+c_1\cdot 2+c_0. \end{equation*}
  
  This, then, implies
  
  \begin{equation*} 2b = c_m\cdot 2^{m+1}+c_{m-1}\cdot 2^{m}+\cdots+c_1\cdot 2^2 +c_0\cdot 2. \end{equation*}
  
  and so
  
  \begin{equation*} n=2b+1 = c_m\cdot 2^{m+1}+c_{m-1}\cdot 2^{m}+\cdots+c_1\cdot 2^2 +c_0\cdot 2+1. \end{equation*}
  
  Hence, the statement is true for $k=n+1\text{.}$
  
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Since both cases are true, the inductive step holds.

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🔗

Therefore the statement is true for all $n\in\mathbb N\text{.}$

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7.3.28.

Solution.

Proof.

We prove the result by strong induction.

When $n=1$ we have $T(1)=1$ and since $1 \log_2 1 +1 = 1\text{,}$ the result holds.
🔗

🔗
Now assume that the result holds for all $1 \leq k \leq \ell$ and we wish to show that it holds for $k= \ell+1\text{.}$ Since the recurrence is slightly different for even or odd $k=\ell+1\text{,}$ we split into two cases.
- Assume that $k=\ell+1 = 2j\text{,}$ then
  
  \begin{align*} T(k) \amp = 2T(j) + k \\ \amp \leq 2\left( j \log_2(j) + j \right) + k \\ \amp = 2j \log_2(j) + 2j + k \\ \amp = k \log_2\left(\frac{2j}{2}\right) + 2k \\ \amp = k \left( \log_2(k) - \log_2(2) \right) +2k \\ \amp = k\log_2(k) + k, \end{align*}
  
  since $\log_2(2) = 1\text{.}$
  
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- Now assume that $k=\ell+1 = 2j+1\text{,}$ then
  
  \begin{align*} T(k) \amp = 2T(j) + k \\ \amp \leq 2\left( j \log_2(j) + j \right) + k \\ \amp = 2j \log_2(j) + 2j + k \\ \amp = k \log_2\left(\frac{2j}{2}\right) + 2k-1 \\ \amp = k\log_2(k) + k-1 \lt k\log_2(k)+k \end{align*}
  
  where again we have used $\log_2(2) = 1\text{.}$
  
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In both cases the inductive step holds.

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🔗

Since both the base case and inductive step are true, the result follows by induction.

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7.3.29.

Solution.

Proof.

We are going to use mathematical induction.

Base Case: We see that for $n=2\text{,}$ we have only one way of splitting them, by splitting them into two piles of one stone each. Thus, our number becomes $1\times 1=1$ and we also have $\dfrac{2(2-1)}{2}=1\text{.}$ Hence, the statement is true for $n=2\text{.}$
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🔗
Inductive Step: Let $m\geq 2\text{,}$ and assume that the statement is true for all $k\leq m\text{.}$ Now, assume we have $m+1$ stones. Then for the first splitting, we have two cases.
- Case 1: Splitting into two piles of $1$ and $m$ stones: In this case, we have our first number to be $m\times 1=m\text{.}$ Moreover, from the inductive hypothesis, we see that if we keep splitting the pile of $m$ stones we get the number $\dfrac{m(m-1)}{2}\text{.}$ Thus, our final number is $m+\dfrac{m(m-1)}{2}=\dfrac{(m+1)m}{2}\text{.}$
  🔗
  
  Therefore, in this case, the statement is true for $n=m+1\text{.}$
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- Case 2: Splitting into two piles of $p$ and $q$ stones, $p,q \gt 1$: In this case, we have our first number to be $pq\text{.}$ Moreover, since $p,q \gt 1\text{,}$ we have $p,q \lt m\text{.}$ Thus, by inductive hypothesis, the number we should get by splitting the two piles into smaller piles are $\dfrac{p(p-1)}{2}$ and $\dfrac{q(q-1)}{2}\text{.}$ We also know that $q=(m+1)-p\text{,}$ and hence, our final number is
  
  \begin{equation*} \dfrac{p(p-1)}{2}+\dfrac{q(q-1)}{2}+pq= \dfrac{(p^2-p+q+2-q)}{2}+pq=\dfrac{p^2+q^2-(p+q)+2pq}{2}, \end{equation*}
  
  which implies
  
  \begin{equation*} \dfrac{(p+q)^2-(p+q)}{2}=\dfrac{(m+1)^2-(m+1)}{2}=\dfrac{(m+1)m}{2}. \end{equation*}
  
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  Therefore, in this case, the statement is true for $n=m+1\text{.}$
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Since both cases hold, the inductive step is true.

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Hence, we conclude that the statement is true for all $n\in\mathbb{N}\text{.}$

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8 Return to sets
8.6 Exercises

8.6.1.

Solution.

$A_2- A_3=\set{ 6, 9, 12, 24, 27, 36, 54, 72, 216 }\text{.}$
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$A_5\cap A_6=\emptyset\text{.}$
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$\pow{A_1}=\set{ \emptyset, \set{-1},\set{0},\set{1},\set{-1,1},\set{-1,0},\set{0,1}, \set{-1,0,1} }\text{.}$ Notice that since $|A_1|=3$ we have $|\pow{A_1}| = 2^3 = 8\text{.}$
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$\pow{\pow{A_1-\set{-1} }}\text{.}$ We see that $A_1-\set{-1}=\set{0,1}\text{.}$ Then we get,
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$\pow{A_1-\set{-1}})=\set{\emptyset, \set{0},\set{1},\set{0,1} }\text{.}$ Thus we see,
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\begin{align*} \pow{\pow{A_1-\set{-1} }} \amp =\Big\{ \emptyset, \set{\emptyset}, \set{\set{0}},\set{\set{1}},\set{\set{0,1}},\\ \amp \quad \set{\emptyset, \set{0}}, \set{\emptyset,\set{1}}, \set{\emptyset,\set{0,1}} \set{\set{0},\set{1}}, \set{\set{0},\set{0,1}} \set{\set{1},\set{0,1}},\\ \amp \quad \set{\emptyset,\set{0},\set{1}}, \set{\emptyset, \set{0},\set{0,1}}, \set{\emptyset,\set{1},\set{0,1}}, \set{\set{0},\set{1},\set{0,1}}, \\ \quad \amp \qquad \set{\emptyset, \set{0},\set{1},\set{0,1}} \Big\} . \end{align*}

Since $|\pow{A_1-\set{-1} }| = 4\text{,}$ the power set of that has $2^4 = 16$ elements. We have arranged our answer here so that the single subset with 0 elements is followed by the four subsets with 1 element, then the 6 with two elements, the four with 3 elements and finally the single subset with 4 elements. Of course, the order doesn’t matter, as long as all 16 are there.

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$A_3\cap A_4=\set{3}\text{,}$
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$A_2-A_7=\emptyset\text{.}$
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$A_5\cup A_2=\set{ 3, 7, 6, 9, 11, 12, 13, 17 18, 19, 24, 27, 36, 54, 72, 108, 216}\text{.}$
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$A_2\cap A_7=\set{ 3, 6, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216}=A_2\text{.}$
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$A_5\cap \overline{A_2}\text{,}$ given the universal set $U=\mathbb R\text{.}$ We have $A_5\cap\overline{A_2}=\set{ 7,11,13,17,19 }=A_5\text{.}$
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Verify whether $(A_4\times B)\cap(B\times A_4)=A_4\times A_4$ and $(A_4\times B)\cup(B\times A_4)=B\times B\text{.}$
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We see

\begin{equation*} A_4\times B=\set{ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6) }, \end{equation*}

and

\begin{equation*} B\times A_4=\set{ (1,2),(2,2),(3,2),(4,2),(5, 2), (6, 2),(1, 3),(2,3),(3,3),(4,3),(5,3),(6,3) }. \end{equation*}

Then,

\begin{equation*} (A_4\times B)\cap (B\times A_4)=\set{ (2,2),(2,3),(3,2),(3,3) }=A_4\times A_4. \end{equation*}

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We also see,

\begin{align*} (A_4 \amp \times B)\cup (B\times A_4)\\ \amp =\{ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),\\ \amp \qquad (1,2),(4,2),(5, 2), (6, 2),(1, 3),(4,3),(5,3),(6,3) \}\neq B\times B, \end{align*}

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since $(1,1)\in B\times B$ but $(1,1)\not\in (A_4\times B)\cup (B\times A_4)\text{.}$
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8.6.2.

Solution.

Proof.

We see that $F\cap G=\set{(a,b)\in\mathbb R^2\colon (a,b)\in F \text{ and } (a,b)\in G}$

\begin{align*} F\cap G= \amp \set{(a,b)\in\mathbb R^2\colon (a,b)\in F \text{ and } (a,b)\in G}\\ = \amp \set{(a,b)\colon a,b\in\mathbb R, b=a^2-3a+2 \text{ and } b=a+2}\\ = \amp \set{(a,b)\colon a,b\in\mathbb R, b=a^2-3a+2=a+2}\\ = \amp \set{(a,b)\colon a,b\in\mathbb R, a^2-4a=0 \text{ and } b=a+2}\\ = \amp \set{(a,b)\colon a,b\in\mathbb R, (a=0 \text{ and } b=a+2=2), \text{ or } (a=4 \text{ and } b=a+2=6) }\\ = \amp \set{(0,2),(4,6)} \end{align*}

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Proof.

We prove that

\begin{equation*} F \cap G \subseteq \set{(0,2), (4,6)} \qquad \text{and} \qquad \set{(0,2), (4,6)} \subseteq F\cap G \end{equation*}

Let $(x,y) \in F\cap G\text{.}$ So we know that $(x,y) \in F$ and $(x,y) \in G\text{.}$ Since $(x,y) \in F$ we know that $x,y \in \mathbb{R}$ so that $y=x^2-3x+2\text{.}$ Then since $x,y \in G\text{,}$ we know that $y=x+2\text{.}$ Combining these two equations we see that we must have $x+2=x^2-3x+2$ or, equivalently $x^2-4x=0\text{.}$ This, in turn, implies that $x=0,4\text{.}$ And thus $(x,y) = (0,2)$ or $(x,y)=(4,6)\text{.}$ In both cases $(x,y) \in \set{(0,2), (4,6)} $ as required.
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Now let $(x,y) \in \set{(0,2), (4,6)}\text{.}$ Then either $(x,y) = (0,2)$ or $(x,y) = (4,6)\text{.}$
- Let $(x,y)=(0,2)\text{.}$ Then since $2=0^2-3\cdot 0+2\text{,}$ it follows that $(x,y) \in F\text{.}$ Then, since $2=0+2\text{,}$ we know that $(x,y) \in G\text{.}$
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- Very similarly, let $(x,y)=(4,6)\text{.}$ Then since $6 = 4^2-3\cdot 4 + 2\text{,}$ we know that $(x,y) \in F\text{.}$ And since $6=4+2$ we also know that $(x,y) \in G\text{.}$
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So in both cases, we have that $(x,y) \in F$ and $(x,y) \in G\text{,}$ and thus $(x,y) \in F \cap G$ as required.

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Since both inclusions hold, the sets are equal.

Solution.

Proof.

Let $m,n\in\mathbb Z\text{.}$ Assume that $s\in \{x \in\mathbb{Z} : mn\mid x\}\text{.}$ Then we see that $mn\mid s$ and thus, $s=mnk$ for some $k\in\mathbb Z\text{.}$

Now since $s = m(nk)$ and $nk \in \mathbb{Z}\text{,}$ we know that $m \mid s\text{.}$
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Similarly, since $s=n(mk)$ and $mk \in \mathbb{Z}\text{,}$ we also know that $n \mid s\text{.}$
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Consequently $s\in \{x \in\mathbb{Z} : n\mid x\}$ and $s\in \{x \in\mathbb{Z} : m\mid x\}\text{.}$ Since $s$ is an element of both of those sets, it is an element of their intersection. Thus

\begin{equation*} \{x \in\mathbb{Z} : mn\mid x\}\subseteq \{x \in\mathbb{Z} : m\mid x\}\cap\{x \in\mathbb{Z} : n\mid x\} \end{equation*}

as required.

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8.6.7.

Solution.

Claim: The statement is true.
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Proof.
By the definition of the Cartesian product, an element of $A\times \emptyset$ takes the form $(a,b)$ where $a\in A$ and $b\in \emptyset\text{.}$ But by the definition of the empty set, no such $b$ exists. Therefore the ordered pair $(a,b)$ cannot exist, and we see that there are no elements in $A\times \emptyset\text{.}$ Therefore, $A\times \emptyset = \emptyset\text{.}$ Hence, $A\times \emptyset \subseteq A\text{.}$
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Claim: The statement is false for general sets $A\text{.}$ It is, however, true when A$=\emptyset\text{.}$
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Proof.
We saw in part (a), that $A\times \emptyset = \emptyset\text{.}$ Therefore $A \subseteq A\times \emptyset = \emptyset$ if and only if $A = \emptyset\text{.}$ Thus, $A\times \emptyset = A$ if and only if $A = \emptyset\text{.}$
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8.6.8.

Solution.

Proof of (a).

We require two cases.

Case 1: $C= \emptyset\text{.}$ By the definition of the Cartesian product, an element of $A\times C$ takes the form $(a,c)$ where $a\in A$ and $c\in C = \emptyset\text{.}$ But by the definition of the empty set, no such $c$ exists. Therefore the ordered pair $(a,c)$ cannot exist, and we see that there are no elements in $A\times C\text{.}$ Therefore, $A\times C = \emptyset\text{.}$ We can similarly argue that $B\times C = \emptyset\text{.}$ We conclude that

\begin{equation*} \emptyset = A\times C \subseteq B\times C = \emptyset. \end{equation*}

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Case 2: $C\neq \emptyset\text{.}$ Suppose that $x\in A\times C\text{.}$ Then by the definition of the Cartesian product, $x = (a,c)$ for some elements $a\in A$ and $c\in C\text{.}$ Since $A\subseteq B\text{,}$ we also have $a\in B\text{.}$ Then by the definition of the Cartesian product, $x = (a,c) \in B\times C\text{.}$ We conclude that $A\times C \subseteq B\times C.$
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In order for (b) to be true, we claim that we require that $C\neq \emptyset\text{.}$

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Proof of (b).

First, notice that if $C = \emptyset\text{,}$ then by Case 1 of the proof of part (a), we would have $A\times C = \emptyset = B\times C\text{.}$ Thus, we must require that $C\neq \emptyset\text{.}$

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Now, we repeat the argument of Case 2 from part (a).

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Suppose that $x\in A\times C\text{.}$ Then by the definition of the Cartesian product, $x = (a,c)$ for some elements $a\in A$ and $c\in C\text{.}$ Since $A\subset B\text{,}$ we also have $a\in B\text{.}$ Then by the definition of the Cartesian product, $x = (a,c) \in B\times C\text{.}$ Therefore $A\times C \subseteq B\times C.$

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Let $A$ be a finite set. We will prove this statement using induction on the size of $A\text{,}$ where $|A|=n\text{,}$ for $n\in\mathbb N\cup \set{0}\text{.}$

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Base case: We see that if $|A|=0\text{,}$ then $A=\emptyset\text{.}$ Therefore $\pow{A}=\set{\emptyset}\text{.}$ This means that $|\pow{A}|=1=2^0\text{.}$ Hence, the statement holds for $n=0\text{.}$
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Moreover, we see that if $|A|=1\text{,}$ then $A=\set{a}$ for some $a\text{.}$ Thus, we see that $\pow{A}=\set{\emptyset, \set{a}}\text{.}$ Thus, $|\pow{A}|=2=2^1\text{.}$ This means that the statement is true for $n=1$ as well.
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Inductive step: Assume that the statement is true for $k=n\text{,}$ that is, if $|A|=n\text{,}$ then we have $|\pow A|=2^n\text{.}$
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Now, let $B$ be a set with $|B|=n+1\text{.}$ Then, we see $B=\set{b_1, b_2, b_3, \ldots, b_{n+1}}\text{.}$ Therefore, we see that we can write

\begin{equation*} B=\set{b_1, b_2, b_3, \ldots, b_{n}}\cup\set{b_{n+1}}=\tilde{B}\cup\set{b_{n+1}}, \end{equation*}

where $|\tilde{B}|=n\text{.}$

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Therefore we know that $|\pow{\tilde{B}}|=2^n\text{.}$
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Now we can write $\pow{B}$ as a union of all subsets that contain the element $b_{n+1}\text{,}$ call $\mathcal P\text{,}$ and subsets that don’t contain the element $b_{n+1}\text{,}$ call $\mathcal Q\text{.}$ Moreover, we see that the set of all subsets that don’t contain $b_{n+1}$ is the set of all subsets of $\tilde{B}\text{,}$ i.e. $\mathcal P=\pow{\tilde{B}}\text{.}$
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We also see that a subset, $D\text{,}$ of $B$ contains $b_{n+1}$ if and only if $D-\set{b_{n+1}}$ is a subset of $\tilde{B}\text{,}$ that is

\begin{equation*} |\mathcal Q|=|\mathcal P|=|\pow{\tilde{B}}|. \end{equation*}

Thus, since $\mathcal P\cap\mathcal Q=\emptyset\text{,}$ we get

\begin{align*} |\pow{B}| \amp =|\mathcal P|+|\mathcal Q|\\ \amp = |\pow{\tilde{B}}|+|\mathcal Q|\\ \amp = |\pow{\tilde{B}}|+|\pow{\tilde{B}}|=2 \cdot |\pow{\tilde{B}}|\\ \amp =2 \cdot 2^n=2^{n+1}. \end{align*}

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Hence, the statement is true for $k=n+1\text{.}$
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Therefore, by mathematical induction, we conclude that if $A$ is a finite set with $|A|=n\text{.}$ Prove that $|\pow A|=2^n\text{.}$

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8.6.12.

Solution.

We give a disproof of both statements.

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Disproof.

Both statements are false. We disprove each in turn.

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Let $A = \set{1}$ and $B = \set{2}\text{.}$ Then $\emptyset$ is an element of $\pow{A}, \pow{B}$ and $\pow{A-B}\text{.}$ But this means that $\emptyset \in \pow{A-B}$ but $\emptyset \not\in \pow{A}-\pow{B}\text{.}$ Hence the first inclusion does not hold.
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Now let $A = \set{1,2}, B = \set{2}\text{.}$ Then

\begin{equation*} \pow{A} - \pow{B} = \set{\set{1}, \set{1,2}} \quad\text{but}\quad \pow{A-B} = \set{\emptyset, \set{1}}. \end{equation*}

Hence $\set{1,2} \in LHS$ but $\set{1,2} \not\in RHS\text{.}$ So the second inclusion does not hold.

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8.6.13.

Solution.

Proof of (a).

We show that each side is included in the other.

Suppose $x\in \cup_{n=3}^\infty (1/n,1-1/n)\text{.}$ Then $x\in (1/n,1-1/n)$ for some $n\in\mathbb{N}\text{.}$ Since $0\lt 1/n$ and $1-1/n\lt 1\text{,}$ we also have $x\in (0,1)\text{.}$ Therefore the LHS is a subset of $(0,1)\text{.}$
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Suppose $y\in (0,1)\text{.}$ Let $N_1=\left\lceil\frac{1}{y} \right\rceil+1$ and $N_1=\left\lceil\frac{1}{1-y} \right\rceil+1\text{.}$ Take $N=\max\{N_1,N_2\}\text{.}$ Then

\begin{equation*} \frac{1}{1-y}\lt N_2\leq N. \end{equation*}

Multiplying the inequality through by $1-y\text{,}$ which is positive as $y \lt 1\text{,}$ and dividing by $N\text{,}$ we have $1/N\lt 1-y\text{,}$ from which we may obtain $y\lt 1-1/N\text{.}$ Moreover, $N\geq N_1$ implies that

\begin{equation*} \frac{1}{N}\leq \frac{1}{N_1}\lt y. \end{equation*}

Thus $y\in (1/N,1-1/N)$ and so $y$ is an element of the LHS. This gives us the reverse inclusion as well.

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Since both inclusions hold, we can conclude

\begin{equation*} \bigcup_{n=3}^\infty \left(\frac{1}{n},1-\frac{1}{n}\right) = (0,1) \end{equation*}

as required.

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Proof of (b).

We prove each side is included in the other.

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Suppose $x\in[0,1]\text{.}$ Then for any $n\in\mathbb{N}\text{,}$

\begin{equation*} -\frac{1}{n}\lt 0 \leq x\leq 1 \lt 1+\frac{1}{n}. \end{equation*}

So $x\in \cap_{n=1}^\infty (-1/n,1+1/n)\text{,}$ and therefore $[0,1]$ is a subset of the LHS.

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Conversely, suppose $y\not\in[0,1]\text{.}$ If $y\lt0\text{,}$ then $N= \left\lceil\frac{1}{|y|}\right\rceil+1$ is a natural number, and $y\lt -1/N\text{.}$ But then $y\not\in (-1/N,1+1/N)\text{.}$ Therefore $y$ is not an element of the LHS.
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If $y\gt 1\text{,}$ then $N= \left\lceil\frac{1}{y-1}\right\rceil+1$ is a natural number, and we’d have $N\gt 1/(y-1)\text{.}$ Multiplying through by $y-1\text{,}$ which is positive as $y \gt 1\text{,}$ and dividing by $N\text{,}$ we’d obtain $y-1\gt 1/N\text{.}$ But then $y\gt 1+1/N\text{,}$ and so $y\not\in (-1/N,1+1/N)\text{.}$ So in either case we have that $y$ is not an element of the LHS. Thus, by contrapositive, we have

\begin{equation*} \bigcap_{n=1}^\infty (-1/n,1+1/n)\subseteq[0,1]. \end{equation*}

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As we have proved the reverse inclusion as well, the two sets are equal.

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8.6.14.

Solution.

We claim that

\begin{equation*} \bigcup_{n\in\mathbb{N}}[-n,n] = \mathbb{R}. \end{equation*}

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Proof.

For any $n\in\mathbb{N}\text{,}$ we have $[-n,n]\subseteq\mathbb{R}\text{,}$ and so

\begin{equation*} \bigcup_{n\in\mathbb{N}}[-n,n] \subseteq \mathbb{R}. \end{equation*}

We need to show that the reverse inclusion holds. To this end, let $x\in \mathbb{R}\text{.}$ If $x=0\text{,}$

\begin{equation*} 0 \in [-1,1]\subseteq \bigcup_{n\in\mathbb{N}}[-n,n]. \end{equation*}

If $x\neq0\text{,}$ let $N=\lceil |x| \rceil\text{.}$ Recall that $f(y)=\lceil y\rceil$ is the ceiling function, defined to be the smallest integer $m$ such that $y\leq m\text{.}$ We have $N\in\mathbb{N}\text{,}$ and $|x|\leq N\text{.}$ Hence

\begin{equation*} x\in [-N,N]\subseteq \bigcup_{n\in\mathbb{N}}[-n,n] \end{equation*}

for this case as well. Thus

\begin{equation*} \mathbb{R}\subseteq \bigcup_{n\in\mathbb{N}}[-n,n] \end{equation*}

holds as well, establishing that the sets are equal.

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For a fixed $r\in\mathbb{R}\text{,}$ $r\gt 0\text{,}$ the set

\begin{equation*} B_r = \{(x,y):x^2+y^2\lt r\} \end{equation*}

represents the inside of a disk in $\mathbb{R}^2$ with radius $r\text{.}$ We claim that

\begin{equation*} B_r = \{(x,y)\in\mathbb{R}\times\mathbb{R}: x^2+y^2\lt r\}= \mathbb{R}\times\mathbb{R}. \end{equation*}

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Proof.

For any $r\in \mathbb{R}\text{,}$ $r\gt 0\text{,}$ we have by definition that

\begin{equation*} B_r = \{(x,y)\in \mathbb{R}\times\mathbb{R}:x^2+y^2\lt r\}\subseteq \mathbb{R}\times\mathbb{R}. \end{equation*}

Therefore

\begin{equation*} \bigcup_{r\in\mathbb{R}, r\gt0} B_r \subseteq\mathbb{R}\times\mathbb{R}. \end{equation*}

We need to show that the reverse inclusion holds. To this end, let $(x,y)\in \mathbb{R}\times \mathbb{R}\text{.}$ Let $r=x^2+y^2+1\text{.}$ Then $r \gt 0\text{,}$ and $x^2+y^2\lt r\text{.}$ Hence

\begin{equation*} (x,y)\in B_r\subseteq \bigcup_{r\in\mathbb{R}, r\gt0} B_r. \end{equation*}

As $(x,y)$ was arbitrary, we have

\begin{equation*} \mathbb{R}\times \mathbb{R}\subseteq \bigcup_{r\in\mathbb{R}, r\gt0} B_r \end{equation*}

and so the sets are in fact equal.

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\begin{equation*} 2 \lt a+a \lt 2y = a+3 \lt 3+3 \end{equation*}

Then $y\in (1,3)$ but $y \gt a\text{,}$ so $a$ is not an upper bound of $(1,3)\text{.}$ Thus $3$ is the supremum of $(1,3)\text{,}$ as claimed.

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Now let $x\in S\text{.}$ Then $x=2-1/n$ for some $n\in\mathbb{N}\text{.}$ But then

\begin{equation*} \frac{1}{n+1}\lt \frac{1}{n}, \end{equation*}

and so

\begin{equation*} x=2-\frac{1}{n}\lt 2-\frac{1}{n+1}=y. \end{equation*}

Since $y \in S$ and $y \gt x\text{,}$ $x$ is not the maximum of $S\text{.}$ As $x$ was an arbitrary element of $S\text{,}$ the set has no maximum.

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Proof of (e).

Let $a\in\mathbb{R}\text{.}$ We show that $a$ is not an upper bound of the set. Let $k=\left\lceil a/\pi\right\rceil+1\text{.}$ Then

\begin{equation*} \pi k \gt \pi\cdot \frac{a}{\pi}=a. \end{equation*}

Since $\pi k$ is an element of $\set{x\in \mathbb{R}:\cos(2x)=1}\text{,}$ $a$ is not an upper bound of the set. Since $a$ was arbitrary, the set has no upper bound, and hence no maximum or supremum.

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After completing this question, you may be wondering when a set has a supremum or a maximum. Every non-empty subset of the real numbers that has an upper bound also has a supremum; this property is called completeness of the real numbers. This may fail if the set has no upper bound, as we saw in part (e).

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However, even if a non-empty subset of the real numbers has an upper bound, it may not have a maximum. For example, in part (b), we saw that $(1,3)$ has an upper bound (for example, 3), but no maximum.

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Note that rational numbers do not share this property; they are not complete. For example, the set

\begin{align*} A \amp = \set{r \in \mathbb{Q} \;:\; 0 \lt r \leq \sqrt{2} } \end{align*}

is the set of positive rational numbers that are at most $\sqrt{2}\text{.}$ This set has a supremum in the real numbers; it is $\sqrt{2}\text{.}$ We will show in Chapter 11 that $\sqrt{2} \not \in \mathbb{Q}\text{.}$ We could consider $A$ to be a subset of the rational numbers, rather than the real numbers. Then $A$ does not have a maximum nor a supremum in the rational numbers, even though it has an upper bound in the rational numbers (for example, $2$). With some more work, one can show that for any $\frac{a}{b}$ in this set, you can construct another $\frac{c}{d}$ in the set which is larger. In fact, we’ll make that an exercise in Chapter 11. This statement implies that $\sqrt{2}$ is the set’s supremum in the real numbers, and that the set has no supremum in the rational numbers.

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8.6.16.

Solution.

Proof of (a).

Let $S,T\subset\mathbb{R}\text{,}$ and $s=\sup(S)\text{,}$ $t=\sup(T)\text{.}$ Suppose $a=\max\{s,t\}\text{.}$ Then for any $x \in S \cup T$ we know that $x \in S$ or $x \in T\text{.}$

If $x\in S\text{,}$ then $x\leq s\leq a\text{.}$
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Similarly, when $x\in T\text{,}$ then $x\leq t\leq a\text{.}$
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Hence $a$ is an upper bound for $S\cup T\text{.}$

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Now let $b$ be any upper bound for $S\cup T\text{.}$ By definition, this means that for all $x\in S\cup T\text{,}$ $x\leq b\text{.}$ Since all elements of $S$ and $T$ are also elements of $S \cup T\text{,}$ it follows that $b$ is an upper bound for $S$ and $T\text{.}$ Now, since $s$ and $t$ are the least upper bounds of $S$ and $T$ respectively, we must have $s\leq b$ and $t\leq b\text{.}$ Hence $a=\max\{s,t\}\leq b\text{.}$ Thus $a$ is smaller than any upper bound of $S \cup T\text{,}$ and so is the least upper bound of $S\cup T\text{,}$ as required.

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For (b), one cannot determine the supremum of the intersection $S \cap T$ from $s,t\text{.}$ One requires more detailed information about the sets. See the scratchwork above.

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Proof of (c).

Let $S,T\subset\mathbb{R}\text{,}$ and $s=\sup(S)\text{,}$ $t=\sup(T)\text{.}$ Let $a=s+t\text{.}$

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Then $a$ is an upper bound for $S\cup T\text{;}$ indeed, for any $x\in S+ T\text{,}$ we may write $x=y+z$ for some $y\in S\text{,}$ $z\in T\text{.}$ But by definition of the supremum, $y\leq s$ and $z\leq t\text{.}$ Thus $x=y+z\leq s+t=a\text{.}$

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Next, suppose that $b \lt a\text{.}$ Then $\epsilon=\frac{a-b}{2}$ is a positive real number. Hence $s-\epsilon$ and $t-\epsilon$ are not upper bounds for $S$ and $T$ respectively, by definition of the supremum. Consequently, there are some $c\in S\text{,}$ $d\in T$ such that $c \gt s-\epsilon$ and $d \gt t-\epsilon\text{.}$ But then

\begin{equation*} c+d \gt (s-\epsilon) +(t-\epsilon)=a-2\epsilon=a-(a-b)=b, \end{equation*}

and since $c+d\in S+T\text{,}$ we see that $b$ is not an upper bound for $S+T\text{.}$

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Thus $a$ is the least upper bound of $S+T\text{,}$ as required.

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8.6.17.

Solution.

Proof.

Let $\{a_n\}_{n\in\in\mathbb{N}}$ be a sequence such that $a_{n+1}\geq a_n$ for all $n\in \mathbb{N}\text{,}$ and such that

\begin{equation*} a = \sup\{a_n:n\in\mathbb{N}\} \end{equation*}

exists as a real number. Let $\epsilon\gt0$ be given. Then $a-\epsilon\lt a\text{,}$ so $a-\epsilon$ is not an upper bound of the set $\{a_n:n\in\mathbb{N}\}\text{.}$ Consequently, there is some $N\in\mathbb{N}$ such that $a-\epsilon \gt a_N\text{.}$ But then for any $n\geq N\text{,}$ we have

\begin{equation*} a-\epsilon \lt a_N\leq a_n. \end{equation*}

Moreover, since $a$ is an upper bound of the set $\{a_n:n\in\mathbb{N}\}\text{,}$ we have

\begin{equation*} a_n\leq a\lt a+\epsilon \end{equation*}

for all $n\in\mathbb{N}\text{,}$ and particularly, all $n\geq N\text{.}$ Adding $a$ to both sides of these inequalities, and putting these two inequalities together, we have

\begin{equation*} -\epsilon\lt a_n-a\lt \epsilon \end{equation*}

for all $n\geq N\text{.}$ Equivalently, $|a_n-a|\lt\epsilon$ for all $n\geq N\text{.}$

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9 Relations
9.7 Exercises

9.7.1.

Solution.

The relation is

\begin{align*} R=\big\{ \amp (2,1),(2,3),(2,5),(3,1),(3,2),(3,4),(3,5),(4,1),(4,2),(4,3),(4,5),\\ \amp (4,6),(5,1),(5,2),(5,3),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5) \big\}. \end{align*}

The relation is not reflexive: $(1,1)\not\in R\text{.}$ The relation is not symmetric: $(2,1)\in R\text{,}$ but $(1,2)\not\in R\text{.}$ The relation is not transitive: $(2,3)\in R$ and $(3,2)\in R\text{,}$ but $(2,2)\not\in R\text{.}$

We must show that $R$ is reflexive, symmetric, and transitive.

Reflexive:For any $a\in\mathbb{Z}\text{,}$ we have $(2a-5a)=3(-a)\text{,}$ which implies $3\mid (2a-5a)\text{.}$ Thus $a\rel a\text{.}$
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Symmetric: Let $a,b\in\mathbb{Z}$ and assume $a\rel b\text{.}$ Then we see $3\mid (2a-5b)\text{,}$ and so $2a-5b=3k$ for some $k\in\mathbb{Z}\text{.}$ Then

\begin{equation*} 2b-5a = (5b-2a) - 3b-3a = -3k -3b-3a = 3(-k-b-a). \end{equation*}

Since $(-b-a-k)\in\mathbb{Z}$ we see that $3\mid (2b-5a)\text{.}$ Therefore $R$ is symmetric.

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Transitive: Let $a,b,c\in\mathbb{Z}$ and assume $a\rel b$ and $b\rel c\text{.}$ Then we see $3\mid (2a-5b)$ and $3\mid (2b-5c)\text{,}$ so that $2a-5b=3k$ and $2b-5c=3\ell$ for some $k,\ell\in\mathbb{Z}\text{.}$ Then

\begin{equation*} 2a-5c = (2a-5b)+3b+(2b-5c) = 3(k+b+\ell) \end{equation*}

Since $(k+b+\ell)\in\mathbb{Z}$ we see that $3\mid (2a-5c)\text{.}$ Therefore $R$ is transitive.

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Thus $R$ is an equivalence relation.

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9.7.7.

Solution.

Proof.

Let us prove that $\rel$ is an equivalence relation.

Reflexivity: Let $A\in \mathcal P (E)\text{.}$ Then $(q\in A)\vee (q\in \overline{A})$ which we can rewrite as $(q\in A\cap A)\vee (q\in \overline{A}\cap \overline{A})\text{.}$ Hence, $A\rel A\text{.}$
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Symmetry: The symmetry is immediate from the symmetry of the intersection of sets.
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Transitivity: Let $A,B,C \in \mathcal P(E)$ and assume that $A\rel B$ and $B\rel C$ so that

\begin{gather*} \left((q\in A\cap B)\vee (q\in \overline{A}\cap \overline{B})\right) \land \left((q\in B\cap C)\vee (q\in \overline{B}\cap \overline{C})\right). \end{gather*}

Now we can study 4 cases in turn:
- Case 1: $(q\in A\cap B)\land (q\in B\cap C)\text{.}$ Then $q\in A$ and $q\in C$ so $q\in A\cap C\text{,}$ which implies $A \rel C.$
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- Case 2: $(q\in A\cap B) \land (q\in \overline{B}\cap \overline{C})\text{,}$ which entails that $q\in B\cap \overline{B}$ so this case never happens.
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- Case 3: $(q\in \overline{A}\cap \overline{B}) \land (q\in B\cap C)\text{.}$ This case does not happen for the same reason as above.
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- Case 4: $\left(q\in \overline{A}\cap \overline{B})\right) \land (q\in \overline{B}\cap \overline{C})\text{.}$ From there $q\in \overline{A} \cap \overline{C}$ and so $A \rel C\text{.}$
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Therefore we see that the relation is an equivalence relation.
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9.7.8.

Solution.

Proof.

To prove that this is an equivalence relation, we need to show that $R$ is reflexive, symmetric, and transitive.

$R$ is reflexive: Let $a\in\mathbb Z\text{.}$ Notice that $7a^2 - 2a^2 = 5a^2$ and $5\mid 5a^2\text{.}$ Thus, $a\rel a\text{,}$ and so $R$ is reflexive.
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$R$ is symmetric: Let $a,b \in\mathbb Z$ such that $a\rel b\text{.}$ This means that $7a^2\equiv 2b^2\mod{5}\text{,}$ that is $5\mid (7a^2-2b^2)\text{,}$ and so $7a^2-2b^2 = 5\ell$ for some $\ell \in \mathbb{Z}\text{.}$ Then notice that we can write

\begin{equation*} 7b^2-2a^2 = 5a^2+5b^2 - (7a^2-2b^2) = 5(a^2+b^2-\ell) \end{equation*}

Thus $5\mid (7b^2-2a^2)\text{,}$ which implies $7b^2\equiv 2a^2\mod{5}$ Thus, $b\rel a$ and so $R$ is symmetric.

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$R$ is transitive: Let $a,b,c \in\mathbb Z$ such that $a\rel b$ and $b\rel c\text{.}$ This means that $7a^2\equiv 2b^2\mod{5}\text{,}$ and $7b^2\equiv 2c^2\mod{5}\text{.}$ So for some $k,\ell \in \mathbb{Z}$ we have

\begin{align*} 7a^2 -2b^2 \amp = 5k \amp \text{and} \amp \amp 7b^2-2c^2 \amp = 5\ell \end{align*}

Then

\begin{align*} 7a^2-2c^2 \amp = (7a^2-2b^2)+(7b^2-2c^2)-5b^2 = 5(k+\ell-b^2) \end{align*}

Hence, $5\mid (7a^2-2c^2)\text{,}$ which implies $7a^2\equiv 2c^2\mod{5}$ Thus, $a\rel c\text{,}$ and so $R$ is symmetric.

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Now let’s consider the equivalence classes of $R\text{.}$ Let $a,b\in\mathbb{Z}$ and suppose that $a \rel b\text{.}$ We know that $7a^2 \equiv 2b^2 \mod{5}\text{.}$ It makes sense to consider the integers modulo 5, and by the division algorithm there are unique integers $q_a,q_a, r_a, r_b$ with $0\leq r_a, r_b \lt 5$ so that

\begin{align*} a \amp = 5q_a + r_a \amp b = 5q_b + r_b \end{align*}

Then

\begin{align*} 7a^2 \amp = 7(5q_a+r_a)^2 = 5(35q_a^2 + 14q_ar_a+r_a^2) + 2ra^2\\ \amp \equiv 2r_a^2 \mod{5} \amp \text{ and similarly}\\ 2b^2 \amp \equiv 2r_b^2 \mod{5} \end{align*}

Hence

\begin{gather*} a \rel b \qquad \iff \qquad 2r_a^2 \equiv 2r_b^2 \end{gather*}

So, for example, to determine all the numbers equivalent to $a=0\text{,}$ which has $r_a=0\text{,}$ we need to find all the remainders that give $2r_b^2 \equiv 0 \mod{5}\text{.}$ Rather than do the specific case, we compute $2r^2 \mod{5}$ for $r=0,1,2,3,4$ and then compare the results:

\begin{align*} 2\cdot0^2 \amp \equiv0\mod{5}, \amp 2\cdot 1^2 \amp \equiv 2\mod{5} \amp 2\cdot2^2 \amp =8\equiv3\mod{5},\\ 2\cdot3^2 \amp = 18\equiv3\mod{5} \amp 2\cdot4^2 \amp =32\equiv2\mod{5} \end{align*}

Hence the integers equivalent to 0 must have $r=0\text{.}$ That is

\begin{equation*} [0] = \set{5 n \st n \in \mathbb{Z} } \end{equation*}

The integers equivalent to $1$ must have $r=1,4\text{,}$ since they both result in $2r^2 \equiv 2 \mod{5}\text{.}$ That is

\begin{equation*} [1] = \set{5n+1 \st n \in \mathbb{Z}} \cup \set{5n+4 \st n \in \mathbb{Z}} \end{equation*}

Notice that since $4 \equiv -1 \mod{5}$ we can also write this as

\begin{equation*} [1] = \set{5n \pm 1 \st n \in \mathbb{Z}} \end{equation*}

The integers equivalent to $2$ must have $r=2,3$ since they both result in $2r^2 \equiv 3 \mod{5}\text{.}$ And since $3 \equiv -2 \mod{5}$ we can write

\begin{equation*} [2] = \set{5n \pm 2 \st n \in \mathbb{Z}} \end{equation*}

Since every integer has remainder $0,1,2,3,4$ modulo 5, the above are all the equivalence classes.

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9.7.9.

Solution.

Proof.

We need to check that the relation is reflexive, symmetric and transitive.

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Reflexivity: Let $f\in P\text{.}$ Then, since $f-f=0$ and $0\in\mathbb R\text{,}$ we see that $fTf\text{.}$ Hence $T$ is reflexive.

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Symmetry: Let $f,g\in P$ and assume that $fTg\text{.}$ Then, by definition, we see that $f-g=c$ for some $c\in \mathbb R\text{.}$ Hence, we see that $g-f=-c\text{,}$ and since $-c\in \mathbb R\text{,}$ we have $gTf\text{,}$ that is, $T$ is symmetric.

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Transitive:. Let $f,g,h\in P$ and assume that $fTg$ and $gTh\text{.}$ This means $f-g=c$ for some $c\in\mathbb R$ and $g-h=d$ for some $d\in\mathbb{R}\text{.}$ Hence, we see that

\begin{equation*} f-h=(f-g)+(g-h)=c+d. \end{equation*}

Thus, since $c+d\in\mathbb R\text{,}$ we see $fTh\text{.}$ Therefore $T$ is transitive.

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Hence the relation $T$ is an equivalence relation.

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9.7.10.

Solution.

This statement is false.
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Proof.

As a counterexample, take $A=\set{1,2,3}$ and the relations

\begin{equation*} R=\set{(1,1),(2,2),(3,3),(1,2),(2,1)} \text{ and } S=\set{(1,1),(2,2),(3,3),(1,3),(3,1)}. \end{equation*}

Note that $R$ and $S$ are equivalence relations. However,

\begin{equation*} R\cup S=\set{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)} \end{equation*}

is not an equivalence relation since it is not transitive; $(2,1),(1,3)\in R\cup S\text{,}$ but $(2,3)\notin R\cup S\text{.}$

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This statement is true.
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Proof.
We need to show that $R\cap S$ is reflexive, symmetric, and transitive.
- Reflexive: Let $a\in A\text{.}$ Since $R$ and $S$ are reflexive, $(a,a)\in R$ and $(a,a)\in S\text{.}$ But then $(a,a)\in R\cap S\text{,}$ so $R\cap S$ is reflexive.
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- Symmetric: Suppose $(a,b)\in R\cap S\text{.}$ Then $(a,b)\in R$ and $(a,b)\in S$ by definition of the intersection. Since $R$ and $S$ are symmetric, $(b,a)\in R$ and $(b,a)\in S\text{.}$ Thus $(b,a)\in R\cap S\text{,}$ and so $R\cap S$ is symmetric.
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- Transitive: Suppose $(a,b)\in R\cap S$ and $(b,c)\in R\cap S\text{.}$ Then $(a,b),(b,c)\in R$ and $(a,b),(b,c)\in S\text{,}$ by definition of the intersection. Since $R$ and $S$ are transitive, $(a,c)\in R$ and $(a,c)\in S\text{.}$ Hence $(a,c)\in R\cap S\text{,}$ and $R\cap S$ is transitive.
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9.7.11.

Solution.

Take $A=\set{1,2,3}$ and take the relation $R=\set{(1,1),(1,2),(2,1),(2,2)}\text{.}$ Then $R$ is a symmetric and transitive relation, but it’s not reflexive, since $3\in A$ but $(3,3)\not\in R\text{.}$
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Proof of (b).

Assume that for all $a \in A\text{,}$ there exists $b \in A$ such that $a \rel b\text{.}$ Let $x\in A$ be arbitrary. Then, by the assumption, there is some $y\in\ A$ such that $x\rel y\text{.}$ Since $R$ is symmetric, we see that $y\rel x\text{.}$ Moreover, since $R$ is transitive and $x\rel y$ and $y \rel x\text{,}$ we see $x\rel x\text{.}$ As $x$ was arbitrary, $R$ is reflexive.

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9.7.12.

Solution.

This statement is false.
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Proof.
Let $a\in A\text{.}$ Then, we see that if $R$ is reflexive, then $(a,a)\in R\text{.}$ Thus, $(a,a)\notin \overline{R}\text{.}$ Therefore $\overline{R}$ is not reflexive.
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One can make a more explicit counter-example by considering the set $A= \set{1}$ and $R = \set{(1,1)}\text{.}$ Then $\bar{R}=\emptyset\text{.}$ While $\bar{R}$ is a relation it is not reflexive since $(1,1) \not\in \bar{R}\text{.}$
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This statement is true.
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Proof.
Let $a,b\in A\text{.}$ Assume $R$ is symmetric and assume that $(a,b)\in \overline{R}\text{.}$ By definition, we see that $(a,b)\notin R\text{.}$ Since $R$ is symmetric we know that if $(b,a)\in R\text{,}$ then $(a,b)\in R\text{.}$ The contrapositive of this statement implies that if $(a,b)\notin R\text{,}$ then $(b,a)\notin R\text{.}$ Hence, $(b,a)\in\overline{R}\text{.}$ Therefore $\overline{R}$ is symmetric.
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Notice that we are using modus-tollens in this proof. We know that $(a,b) \in R \implies (b,a) \in R\text{,}$ and that $(b,a) \not\in R\text{,}$ so we conclude $(a,b) \not \in R\text{.}$
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This statement is false.
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Proof.
For a counterexample, we can take $A=\set{1,2}$ and $R=\set{(1,1),(2,2)}\text{.}$ Notice that $R$ is the “equality” relation, and it is transitive. Then the complementary relation $\overline{R}=\set{(1,2),(2,1)}\text{.}$ Since $(1,2)\in\overline{R}$ and $(2,1)\in\overline{R}$ but $(1,1)\not\in \overline{R}\text{,}$ we see that $\overline{R}$ is not transitive.
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9.7.13.

Solution.

Proof of (a).

We prove each implication in turn.

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First, suppose that $R$ is sparse, and observe that since $n\mid 0$ for all $n \in \mathbb{N}\text{,}$ $(a,a) \in \mathbb{R}$ for all $a \in \mathbb{Z}\text{.}$ Then since $R$ is sparse, $(a,a+1) \notin R$ and hence $n \nmid (a - a - 1) = -1\text{.}$ In particular, $n \neq 1\text{,}$ as desired.

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For the second implication we will prove the contrapositive. Suppose that $R$ is not sparse. Then we have two cases: either there exist $a,b \in \mathbb{Z}$ so that $(a,b)$ and $(a,b+1)$ are in $R\text{,}$ or there exist $a,b \in \mathbb{Z}$ so that $(a,b)$ and $(a+1,b)$ are in $R\text{.}$

In the first case, we have that $n|(a-b)$ and $n|(a-(b+1))\text{,}$ and hence $n$ divides their difference. That is, $n|1\text{.}$ Since $n \in \mathbb{N}\text{,}$ this implies $n =1\text{.}$
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The second case is very similar. We have that $n|(a-b)$ and $n|(a+1-b)\text{,}$ and hence $n|1\text{.}$ In particular $n =1\text{.}$
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Note that these are the modular equivalence relations.

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The statement in (b) is false.

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Disproof of (b).

Consider $R = \mathbb{Z} \times \mathbb{Z}\text{.}$ This is an equivalence relation: all three conditions are trivially satisfied since $R$ contains all pairs of integers. However, both $(0,0)$ and $(0,1)$ are elements of $R\text{,}$ so $R$ is not sparse.

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Note that the counter-example in this disproof is just the case $n =1$ in the previous part of the question.

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9.7.14.

Solution.

Proof.

Let $A,P,Q,R$ be as given in the statement of the problem. Then to show that $R$ is a partition of $A\text{,}$ we need to show that

\begin{gather*} \bigcup_{X\in R}X=A \end{gather*}

and

\begin{gather*} (U_1, U_2\in R, U_1\neq U_2)\implies (U_1\cap U_2=\emptyset). \end{gather*}

We prove each of these in turn.

To prove the set equality we need to show that each side is included in the other. One of these inclusions is straight-forward, since if $X\in R\text{,}$ then, by definition of $R\text{,}$ $X\subseteq A\text{.}$ Therefore, we see that $\bigcup_{X\in R}X\subseteq A\text{.}$
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Thus, we just need to show that $A\subseteq \bigcup_{X\in R}X\text{.}$ Let $x\in A\text{.}$ Since $P$ and $Q$ are partitions of $A\text{,}$ there exists $S\in P$ and $T\in Q$ such that $x\in S$ and $x\in T\text{.}$ This entails that $x\in S\cap T$ and $S\cap T\in R\text{.}$ Therefore $x\in \bigcup_{X\in R}X\text{.}$
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To show the second property of partitions let $U_1,U_2\in R\text{.}$ By definition $U_1=S_1\cap T_1$ for some $S_1\in P$ and $T_1\in Q\text{.}$ Similarly, $U_2 = S_2 \cap T_2$ for some $S_2 \in P$ and $T_2 \in Q\text{.}$
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We can prove the contrapositive of the implication. Assume that $U_1 \cap U_2 \neq \emptyset\text{.}$ Hence there is some $x \in U_1 \cap U_2\text{.}$ Since $x \in U_1$ this implies that $x \in S_1 \cap T_1\text{,}$ and since $x \in U_2$ we know that $x \in S_2 \cap T_2\text{.}$ Thus $x$ is in all of $S_1, S_2, T_1, T_2\text{.}$
- Since $x \in S_1$ and $x \in S_2\text{,}$ it follows that $x \in S_1 \cap S_2\text{.}$ Thus, since $S_1, S_2$ are parts of a partition, we have $S_1 = S_2\text{.}$
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- Similarly, since $x \in T_1$ and $x \in T_2\text{,}$ we know that $T_1 = T_2\text{.}$
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Thus we must have $U_1 = S_1 \cap T_1 = S_2 \cap T_2 = U_2\text{.}$ Hence when $U_1 \cap U_2 \neq \emptyset$ it follows that $U_1 = U_2\text{.}$

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In the end, $R$ is a partition of $A\text{.}$

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9.7.15.

Solution.

We first prove that $R$ is an equivalence relation.

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Proof.

We have to show that $R$ is reflexive, symmetric and transitive.

(reflexive) We have $[x]_n\rel [x]_n$ since $[x]_n[1]_n=[x]_n$ for all $[x]_n \in \mathbb{Z}_n\text{.}$
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(symmetric) Suppose $[x]_n\rel [y]_n\text{.}$ Then by definition of $R\text{,}$ we know that $[x]_n[u]_n=[y]_n$ for some invertible $[u]_n\text{.}$ Thus, we know that $[u]_n$ has an inverse in $\mathbb Z_n\text{,}$ and call $[v]_n \in \mathbb{Z}_n$ to be a inverse of $[u]_n\text{,}$ i.e. $[u]_n[v]_n=[1]_n\text{.}$ Then since $[y]_n=[x]_n[u]_n\text{,}$ we know that $[y]_n[v]_n = [x]_n[u]_n[v]_n = [x]_n[1]_n = [x]_n$ and thus $[y]_n\rel [x]_n\text{.}$
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(transitive) Suppose $[x]_n\rel [y]_n$ and $[y]_n\rel [z]_n\text{,}$ that is $[x]_n[u]_n=[y]_n$ and $[y]_n[v]_n=[z]_n$ with $[u]_n,[v]_n$ both admitting multiplicative inverse.
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Then we have $[x]_n[u]_n[v]_n=[y]_n[v]_n=[z]_n\text{.}$ Denoting the multiplicative inverses of $[u]_n$ and $[v]_n$ by $[\bar{u}]_n$ and $[\bar{v}]_n\text{,}$ respectively, we see that

\begin{equation*} [u]_n[v]_n[\bar{v}]_n[\bar{u}]_n = [u]_n [1]_n [\bar{u}]_n = [u]_n [\bar{u}]_n = [1]_n. \end{equation*}

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This means that $[u]_n[v]_n$ admits a multiplicative inverse.
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Therefore, we have $[x]_n\rel [z]_n\text{.}$
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For the second part, we first note that the set of elements in $\mathbb{Z}_{6}$ with a multiplicative inverses are $U = \{[1]_{6}, [5]_{6} \}\text{.}$ We can see this by looking at the multiplication table for $\mathbb Z_6\text{:}$

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\begin{equation*} \begin{array}{|c||*{7}{c|}} \hline \amp [0]_6 \amp [1]_6 \amp [2]_6 \amp [3]_6 \amp [4]_6 \amp [5]_6 \\ \hline\hline [0]_6 \amp [0]_6 \amp [0]_6 \amp [0]_6 \amp [0]_6 \amp [0]_6 \amp [0]_6 \\ \hline [1]_6 \amp [0]_6 \amp [1]_6 \amp [2]_6 \amp [3]_6 \amp [4]_6 \amp [5]_6 \\ \hline [2]_6 \amp [0]_6 \amp [2]_6 \amp [4]_6 \amp [0]_6 \amp [2]_6 \amp [4]_6 \\ \hline [3]_6 \amp [0]_6 \amp [3]_6 \amp [0]_6 \amp [3]_6 \amp [0]_6 \amp [3]_6 \\ \hline [4]_6 \amp [0]_6 \amp [4]_6 \amp [2]_6 \amp [0]_6 \amp [4]_6 \amp [2]_6 \\ \hline [5]_6 \amp [0]_6 \amp [5]_6 \amp [4]_6 \amp [3]_6 \amp [2]_6 \amp [1]_6 \\ \hline \end{array} \end{equation*}

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Notice that the only products that are equal to $[1]_ 6$ are $[1]_6[1]_6=[1]_6$ and $[5]_6[5]_6=[1]_6\text{.}$

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Thus we may list the equivalence classes defined by $R\text{:}$

We see that

\begin{align*} [[0]_{6}] \amp = \set{[y]_6\in\mathbb Z_6\colon [y]_6=[0]_6[u]_6 \text{ for some invertible } [u]_6\in\mathbb Z_6}\\ \amp =\{[0]_{6}u: u \in U \} = \{ [0]_{6} \}. \end{align*}

Then similarly,

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$[[1]_{6}] = \{[1]_{6}u: u \in U \} = \{[1]_{6}, [5]_{6}\}\text{.}$
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$[[2]_{6}] = \{[2]_{6}u: u \in U \} = \{ [2]_{6}, [4]_{6}\}\text{.}$
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$[[3]_{6}] = \{[3]_{6}u: u \in U \} = \{ [3]_{6}\}\text{.}$
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Since these equivalence classes contain all the elements in $\mathbb Z_6\text{,}$ we see that there are no other equivalence classes.

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9.7.16.

Solution.

Proof of (a).

Let $n\in\mathbb{Z}\text{.}$ Assume that $3\mid n$ and $8\mid n\text{.}$ This means that $n=3k$ and $n=8l$ for some $k,l\in\mathbb{Z}\text{.}$ Thus, we see $8n=24k$ and $3n=24l\text{.}$ Moreover, since $\gcd(3,8)=1\text{,}$ by Bézout’s identity, we see that $\exists x,y\in\mathbb{Z}$ such that $3x+8y=1$ (we can take x=3, y=-1). Using the $x,y$ in the Bézout’s identity, we get $3xn=24xl$ and $8yn=24yk\text{.}$ Thus, adding two equations together, we get $(3x+8y)n=n=24(xl+yk)\text{.}$ Since $(xl+yk)\in\mathbb{Z}\text{,}$ we see $24\mid n$ as desired.

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We use the result of (a) to prove (b).

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Proof of (b).

Assume $p$ is prime. We see that proving $p^2\equiv 1 \pmod {24}$ is equivalent to showing $24 \mid (p^2- 1)$ and thus by part (a), it is enough to show $8\mid (p^2-1)$ and $3\mid (p^2-1)\text{.}$

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Proof that $3\mid (p^2-1)$: We know, since $p $ is prime, that $3\nmid p\text{.}$ Then we see that we have 2 cases $p\equiv 1\pmod 3$ or $p\equiv 2 \pmod 3\text{.}$

Case 1: $p\equiv 1\pmod 3\text{:}$ Then we see that $p^2\equiv 1\pmod 3\text{,}$ that is, $3\mid (p^2-1)\text{.}$
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Case 2: $p\equiv 2\pmod 3\text{:}$ Then we see that $p^2\equiv 4\equiv 1\pmod 3\text{,}$ which again proves $3\mid (p^2-1)\text{.}$
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Proof that $8\mid (p^2-1)$: Since $p$ is prime and $p\geq 5$ we know that $p$ is odd. Thus, $p=2k+1$ for some $k\in\mathbb{Z}\text{.}$ This means $p^2-1=4k(k+1)\text{.}$ Moreover, we see that for $k$ we have two cases: $k=2n$ or $k=2n+1$ for some $n\in\mathbb{Z}\text{.}$

Case 1: $k=2n$: In this case we see that $p^2-1=4k(k+1)=8n(k+1)\text{;}$ and since $n(k+1)\in\mathbb{Z}\text{,}$ we see that $8\mid (p^2-1)\text{.}$
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Case 2: $k=2n+1$: In this case we see that $p^2-1=4k(k+1)=4(2n+1)(2n+2)=8(n+1)(2n+1)\text{;}$ and since $(n+1)(2n+1)\in\mathbb{Z}\text{,}$ we see that $8\mid (p^2-1)\text{.}$
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Therefore, $3\mid (p^2-1) $ and $8\mid (p^2-1)$ which implies $24\mid (p^2-1)\text{.}$

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9.7.17.

Solution.

Proof of (a).

Let $p$ be a prime number, and suppose that $n\in\mathbb{Z}$ is such that $n\not\equiv0\mod{p}\text{.}$ Then $p\nmid (n-0)\text{,}$ so $p\nmid n\text{.}$ Since $p$ is prime, its only divisors are $1$ and $p\text{,}$ and as $p$ isn’t a divisor of $n\text{,}$ we have $\gcd(n,p)=1\text{.}$ Then, by Bézout’s identity, there are some $k,\ell\in\mathbb{Z}$ so that $nk+\ell p=1\text{.}$ Rearranging, we have $\ell p=1-nk\text{,}$ and so $p\mid (1-nk)\text{.}$ Then, by definition, $nk\equiv 1\mod{p}\text{.}$

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For (b), take $p=4$ and set $n=2\text{.}$ Then for any $k\text{,}$ we know that $nk$ is even, and so $nk-1$ is odd. Since $nk-1$ is odd it is not divisible by 4. Thus $nk \not\equiv 1 \mod{4}\text{.}$

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9.7.18.

Solution.

Proof.

Let $a,b,d\in\mathbb Z$ such that $d\mid ab\text{.}$ By Bézout’s lemma, there are some $x,y\in \mathbb Z$ such that $\gcd(a,d)=xa+yd\text{.}$ Multiplying the equation by $b\text{,}$ we have $b\gcd(a,d)=xab+ydb\text{.}$ Since $d\mid ab\text{,}$ there is some $k\in \mathbb Z$ so that $ab=kd\text{.}$ Hence

\begin{equation*} b\gcd(a,d)=xab+ydb=xkd+ydb=d(xk+yb). \end{equation*}

Now $\gcd(a,d)\mid d\text{,}$ and so there is some $\ell\in\mathbb Z$ so that $\ell\gcd(a,d)= d\text{.}$ But then

\begin{equation*} b\gcd(a,d)=\ell\gcd(a,d)(xk+yb) \end{equation*}

Pascal’s identity tells us that every binomial coefficient $\binom{n}{r}$ is either a boundary term $\binom{n}{n}=1, \binom{n}{0}=1 $ or can be written as the sum of two earlier terms $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}\text{.}$ We can use this as the basis of an induction argument.

Base case: When $n=0$ we have $\binom{n}{n}=1 \text{.}$
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Inductive step: Assume that $\binom{k}{r} \in \mathbb{Z} $ for all $0 \leq r \leq k\text{.}$ Then consider $\binom{k+1}{s} \text{.}$ If $s = 0, k+1 $ then, by the boundary conditions, we know that the coefficient is an integer. Otherwise, we can write $\binom{k+1}{s} = \binom{k}{s-1}+\binom{k}{s}$ which is, by hypothesis, the sum of two integers, and so is itself an integer.
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Thus, by mathematical induction, the binomial coefficients are integers.

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Proof of (b).

Let $p$ be prime and let $0 \lt k \lt p$ be an integer. Then the binomial coefficient

\begin{equation*} \binom{p}{k} = \frac{p!}{(p-k)! k!}. \end{equation*}

The numerator contains a factor of $p\text{.}$ However, since $0 \lt k \lt p$ the denominator is a product of positive integers that are strictly smaller than $p\text{.}$ Thus there is no term in the denominator that can cancel the prime $p$ in the numerator. Thus the binomial coefficient is an integer that is divisible by $p\text{.}$

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Proof of (c).

Let $x,y$ and $p$ be as given. Then, by the Binomial Theorem

\begin{equation*} (x+y)^p = x^p + y^p + \sum_{k=1}^{p-1} \binom{p}{k} x^{p-k} y^k \end{equation*}

Hence it suffices to show that

\begin{equation*} \binom{p}{k} \equiv 0 \pmod p \qquad \text{when } 0 \lt k \lt p \end{equation*}

This is precisely what we proved above. From this, the result follows since every term in the expansion of $(x+y)^p $

\begin{equation*} \binom{p}{k} x^{p-k} y^k \end{equation*}

is a multiple of $p\text{,}$ excepting $x^p$ and $y^p\text{.}$

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10 Functions
10.8 Exercises

10.8.1.

Solution.

The set is a function from $\mathbb{R}^2$ to $\mathbb{R}^3\text{.}$

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Proof.

We see that for every pair $(x,y)\in \mathbb R^2\text{,}$ there is a unique ordered pair $((x, y), (5y, 4x, x + y))\in\mathbb R^2\times \mathbb R^3\text{.}$ Therefore, $\theta$ defines a function from $\mathbb R^2$ to $\mathbb R^3\text{.}$ Moreover, this implies that the domain of $\theta$ is $\mathbb R^2\text{.}$

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We know that the range of $\theta$ is the set $Y\subseteq \mathbb R^3$ satisfying:

\begin{align*} Y \amp =\set{(a,b,c)\in\mathbb R^3: (a,b,c)=(5y, 4x, x + y)\text{ for some } (x,y)\in\mathbb R^2}\\ \amp =\set{(a,b,c)\in\mathbb R^3: a=5y, b=4x, c=x+y \text{ for some } (x,y)\in\mathbb R^2} \end{align*}

Writing $c$ in terms of $a$ and $b\text{,}$

\begin{align*} \amp =\set{(a,b,c)\in\mathbb R^3: a=5y, b=4x, c=a/5+b/4 \text{ for some } (x,y)\in\mathbb R^2}\\ \amp =\set{ (a,b,a/5+b/4)\in\mathbb R^3: a=5y, b=4x \text{ for some } (x,y)\in\mathbb R^2} \end{align*}

Using the fact that $\{(5y, 4x) : (x,y)\in \mathbb{R}^2\} = \mathbb{R}^2\text{,}$

\begin{align*} \amp =\set{ (a,b,a/5+b/4)\in\mathbb R^3: a,b\in\mathbb R}. \end{align*}

Hence the range of the function is the plane $20z-4a-5b=0$ in $\mathbb{R}^3\text{.}$

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10.8.2.

Solution.

Claim: $\phi$ is a function for all $a,b \in \mathbb{N}$ satisfying $b \mid 6$ and $b \mid a\text{.}$

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Proof.

For $\phi$ to define a function $\forall x\in \mathbb Z\text{,}$ there has to be a unique point $y\in\mathbb Z$ such that $(x,y)\in\phi\text{.}$ Let $x\in\mathbb Z\text{.}$ Then, by definition of $\phi\text{,}$ we see that $y=\dfrac{6-ax}{b}\text{,}$ which is uniquely defined for any $x\in\mathbb{Z}\text{.}$ Then, we see that for $\phi$ to be a function, $\dfrac{6-ax}{b}$ must be an integer.

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Since this has to be true for any $x\in \mathbb Z\text{,}$ we see that it has to be true for $x=0\text{,}$ so $b\mid 6\text{.}$ Thus, $b\in\set{1,2,3,6}\text{.}$

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This also means that for $\phi$ for be a function $\dfrac{ax}{b}$ must also be an integer, that is $b \mid ax\text{.}$ Setting $x=1$ tells us that $a$ must be a multiple of $b$ as claimed.

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Now assume that $b \mid 6$ and $b \mid a\text{,}$ so that $6 = b k, a = b \ell$ for some $k,\ell \in \mathbb{Z}\text{.}$ Notice that $k \neq 0\text{.}$ Then the condition that $ax+by=6$ becomes

\begin{equation*} \ell b x + b y = b k \end{equation*}

and dividing through by $k$ we have

\begin{equation*} \ell x + y = k \end{equation*}

or $y=k-\ell x\text{.}$ Then for any $x \in \mathbb{Z}$ we know that $y \in \mathbb{Z}$ and so $\phi$ is a function.

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10.8.3.

Solution.

Proof.

Let $f$ be defined as above. To show that

\begin{gather*} f(\mathbb{R}) = [-1,1] \end{gather*}

we prove each side is included in the other.

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$f(\mathbb{R}) \subseteq [-1,1] \text{:}$ Let $y\in f(\mathbb R)\text{.}$ Then, by definition, we know that there is a $x\in\mathbb R\text{,}$ such that $y=f(x)=\dfrac{2x}{1+x^2}\text{.}$
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Now, for all $x \in \mathbb{R}$ we know that

\begin{align*} (1+2x+x^2) = (1+x)^2 \amp \geq 0 \amp \text{and} \amp \amp (1-2x+x^2) = (1-x)^2 \amp \geq 0 \end{align*}

We can rearrange this to give

\begin{align*} 1+x^2 \amp \geq -2x \amp \text{and} \amp \amp 1+x^2 \amp \geq 2x \end{align*}

Combining these gives us

\begin{align*} -(1+x^2) \amp \leq 2x \leq 1+x^2 \end{align*}

Since $1+x^2 \geq 1$ we can divide both sides of this inequality by $1+x^2$ to get

\begin{align*} -1 \amp \leq \frac{2x}{1+x^2} \leq 1 \end{align*}

which is precisely the result we require. Thus $f(\mathbb{R}) \subseteq [-1,1]\text{.}$

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$[-1,1]\subseteq f(\mathbb R)\text{:}$ Let $y\in[-1,1]\text{.}$ We need to show that $f \in f(\mathbb R)\text{.}$ That is, we need to find $x\in\mathbb R\text{,}$ such that $f(x)=\dfrac{2x}{1+x^2}=y\text{.}$ Now, either $y=0$ or $y \neq 0\text{.}$
- When $y=0\text{,}$ set $x=0$ and then note that $f(0)=0$ as required.
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- When $y \neq 0$ set
  
  \begin{align*} x \amp = \frac{1 + \sqrt{1-y^2}}{y}. \end{align*}
  
  Since $y \neq 0$ and $y \in [-1,1]\text{,}$ the above is a real number. Then we verify that
  
  \begin{align*} f(x) \amp = \dfrac{ \frac{2+2\sqrt{1-y^2}}{y} }{\left(\frac{1+\sqrt{1-y^2}}{y}\right)^2 +1} \amp \text{multiply by }y^2\\ \amp = \dfrac{2y + 2y\sqrt{1-y^2}}{\left(1+\sqrt{1-y^2} \right)^2+y^2} \amp \text{expand}\\ \amp = \dfrac{2y + 2y\sqrt{1-y^2}}{\left(1+2\sqrt{1-y^2} +1-y^2\right)+y^2} \\ \amp = \dfrac{2y + 2y\sqrt{1-y^2}}{2+2\sqrt{1-y^2}} \\ \amp = y \end{align*}
  
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Thus $y \in f(\mathbb{R})$ and so $[-1,1]\subseteq f(\mathbb R)\text{.}$

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Thus, we see that $f(\mathbb R)=[-1,1]\text{.}$

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10.8.4.

Solution.

We see that the function can be written as $f :\{1,2,3,4,5,6,7\}\rightarrow \{0,1,2,3,4,5,6,7,8,9\} $ defined by

\begin{align*} f(1) \amp =3 \amp f(2) \amp = 8 \amp\\ f(3) \amp = 3 \amp f(4) \amp = 1\\ f(5) \amp = 2 \amp f(6) \amp = 4\\ f(7) \amp = 6. \end{align*}

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Therefore, we compute

\begin{align*} f\big(\{1,2,3\}\big) \amp =\set{3,8},\\ f\big(\{4,5,6,7\}\big) \amp =\set{1,2,4,6},\\ f(\emptyset) \amp =\emptyset\\ f^{-1}\big(\{0,5,9\}\big) \amp =\emptyset, \text{ and}\\ f^{-1}\big(\{0,3,5,9\}\big) \amp =\set{1,3}. \end{align*}

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We begin by writing the function $g$ in a few ways. Below, we have written the quadratic in standard form, factored form, and vertex form, respectively.

\begin{equation*} g(x) = 4x^2-x-3 = (4x+3)(x-1) = \frac{1}{16}(8x-1)^2 - \frac{49}{16}. \end{equation*}

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From the vertex form of $g(x)\text{,}$ we see that the function is symmetric around $x=\frac{1}{8}\text{.}$ Then plugging in $\frac{1}{8}$ to $g(x)\text{,}$ we find

\begin{equation*} g\left(\set{\frac{1}{8}}\right) = \set{-\frac{49}{16}}. \end{equation*}

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Moreover, from the factored form of $g(x)\text{,}$ we see that

\begin{equation*} g^{-1}(\set{0}) = \set{-\frac{3}{4}, 1}. \end{equation*}

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In particular, the previous two computations tell us that $g(x)$ has a minimum when $x = \frac{1}{8}\text{.}$ Therefore, $g(x)$ is decreasing on the interval $\left(-\infty, \frac{1}{8}\right)$ and increasing on $\left(\frac{1}{8}, \infty\right)\text{.}$ With this in mind, we compute

\begin{align*} g((-1, 0)\cup [3, 4]) \amp = g((-1, 0)) \cup g([3,4])\\ \amp = \left(g(0), g(-1)\right) \cup [g(3), g(4)]\\ \amp = (-3, 2) \cup [30, 57]. \end{align*}

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Finally, since $g(x)$ has a minimum when $x = \frac{1}{8}\text{,}$ and $g\left(\frac{1}{8}\right) = -\frac{49}{16}\text{,}$ we find

\begin{equation*} g^{-1}([-10, -5)) = \emptyset. \end{equation*}

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First, we notice that $h(t)$ is $1$-periodic, that is the function satisfies $h(t)=h(t+1)\text{.}$ We use this fact to compute

\begin{align*} h(\mathbb{Z}) \amp = h(\set{0}) = \set{0} \text{ and}\\ h\left(\set{\frac{1}{4}, \frac{7}{2}, \frac{19}{4}, 22} \right) \amp = h\left(\set{\frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1} \right)\\ \amp = \set{h\left(\frac{1}{4}\right), h\left(\frac{1}{2}\right), h\left(\frac{3}{4}\right), h(1)} = \set{-1, 0, 1}. \end{align*}

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To find the desired preimages, we first find the preimage of $h$ for $t\in [0,1)\text{,}$ and then use $1$-periodicity to find the preimage for $t\in \mathbb{R}\text{.}$
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For $t\in [0,1)\text{,}$ $h^{-1}(\set{1}) = \set{\frac{1}{4}}\text{.}$ Therefore for $t\in \mathbb{R}\text{,}$

\begin{equation*} h^{-1}(\set{1}) = \set{\frac{1}{4}+k: k\in \mathbb{Z}}. \end{equation*}

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Finally, for $t\in [0,1)\text{,}$ $h^{-1}([0,1)) = \left[0, \frac{1}{4}\right) \cup \left(\frac{1}{4}, \frac{1}{2}\right]\text{.}$ Therefore for $t\in \mathbb{R}\text{,}$

\begin{equation*} h^{-1}([0,1)) = \bigcup_{k\in \mathbb{Z}}\left(\left[k,k+ \frac{1}{4}\right) \cup \left(k+\frac{1}{4}, k+\frac{1}{2}\right]\right). \end{equation*}

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10.8.5.

Solution.

Proof.

We have to prove that $LHS \subseteq RHS$ and $RHS \subseteq LHS\text{.}$

Let $x \in LHS\text{.}$ Then, by definition of the preimage, we know that $f(x) \in E-F\text{.}$ Since $x \in E\text{,}$ we know that $x \in f^{-1}(E)\text{.}$ Similarly since we know (via modus tollens 2.5.2) that $f(x) \not \in f(F)$ implies $x \not\in f^{-1}(F) \text{.}$ Hence $x \in f^{-1}(E)-f^{-1}(F) = RHS\text{.}$
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Now let $x \in RHS\text{,}$ so that $x \in f^{-1}(E)$ and $x \not\in f^{-1}(F)\text{.}$ Since $x\in f^{-1}(E)\text{,}$ the definition of the preimage implies that $f(x) \in E\text{.}$ Similarly (again via modus tollens 2.5.2) since $x \not\in f^{-1}(F)\text{,}$ it follows that $f(x) \not\in F\text{.}$ So $f(x)\in E-F$ and the definition of preimage implies that $x\in f^{-1}(E-F) = LHS\text{.}$
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We now conclude: $f^{-1}(E)-f^{-1}(F)= f^{-1}(E-F)$

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10.8.6.

Solution.

Proof.

We will show that this function is neither injective nor surjective. First, observe that, by completing the square, we can rewrite the function as

\begin{equation*} f(x)=\left(x+\frac{a}{2} \right)^2+(b-\frac{a^2}{4}). \end{equation*}

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$f$ is not injective. Since $f$ can be rewritten as above we see that

\begin{align*} f\left(-1-\frac a2\right) \amp =\left(-1-\frac a2+\frac{a}{2}\right)^2+\left(b-\frac{a^2}{4}\right)\\ \amp =1+\left(b-\frac{a^2}{4}\right) \end{align*}

and also

\begin{align*} f\left(1-\frac a2\right) \amp =\left(1-\frac a2+\frac{a}{2}\right)^2+\left(b-\frac{a^2}{4}\right)\\ \amp =1+\left(b-\frac{a^2}{4}\right) = f\left(-1-\frac a2\right) \end{align*}

Since $-1-\frac a2\neq 1-\frac a2\text{,}$ we see that $f$ is not injective.

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$f$ is not surjective. Since $f(x)=\left(x+\frac{a}{2}\right)^2+\left(b-\frac{a^2}{4}\right)$ and since $\left(x+\frac{a}{2}\right)^2\geq 0$ for all $x\in\mathbb R\text{,}$ we can conclude that $f(x)\geq b-\frac{a^2}{4}$ for all $x\in\mathbb R\text{.}$ So if we take $y=b-a^2-1 \lt b-a^2\leq b-\frac{a^2}{4} \leq f(x)$ for all $x \in \mathbb{R}\text{.}$ Thus there is no $x\in\mathbb R$ such that $f(x)=y\text{,}$ and hence $f$ is not surjective.
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10.8.7.

Solution.

Proof.

First we see that the function defined by $g(n)=n$ satisfies the properties above.

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Then, let us prove that any function $f$ that satisfies the above has to be equal to $g\text{.}$ Let us prove this using strong induction on the property $P(n): f(n)=n\text{.}$

Base case: $n=1\text{.}$ Since $f(1)\leq 1$ and $f(1)\in \mathbb N\text{,}$ then $f(1)=1\text{.}$
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Induction step: assume that $f(k)=k$ for all $k\le n\text{.}$ By assumption, $f(n+1)\in \{1,2,3,\dots ,n+1\}\text{.}$ We cannot have $f(n+1)=k$ for any $k\leq n$ by injectivity of $f\text{,}$ because we already have $f(k)=k$ and $k\neq n+1\text{.}$ Hence the only possible value is $f(n+1)=n+1\text{.}$
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In the end, we conclude by strong induction that for all $n\in \mathbb N$ we have $f(n)=n\text{.}$

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10.8.8.

Solution.

Proof.

Let $f:[3,\infty)\rightarrow [5,\infty)\text{,}$ defined by $f(x)=x^2-6x+14\text{.}$ We want to show that $f$ is a bijective function. To do that, we need to show that $f$ is both injective and surjective.

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Injective: Let $a,b\in [3,\infty)$ and $f(a)=f(b)\text{.}$ This means that $a^2-6a+14=b^2-6b+14\text{.}$ Cancelling $14$ from both sides and collecting terms together, we see $a^2-b^2-6(a-b)=(a+b-6)(a-b)=0\text{.}$ Then, we see that $a+b-6=0$ or $a-b=0\text{.}$ If $a-b=0$ then we know that $a=b\text{.}$ On the other hand, when $a+b=6\text{,}$ since $a,b\geq 3\text{,}$ we see that $a=3=b\text{.}$ Thus, we see that in all the cases, $a=b\text{.}$
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Hence, $f$ is injective.
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Surjective: Let $y\in [5,\infty)\text{.}$ Then set $x=3+\sqrt{y-5}\text{.}$ Since $y \geq 5\text{,}$ we know that $\sqrt{y-5}$ is a non-negative real number, and thus $x \geq 3$ lies in the domain of the function. We now verify that $f(x)=y\text{:}$

\begin{align*} f(x) \amp = (3+\sqrt{y-5})^2 -6(3+\sqrt{y-5}) +14\\ \amp = 9 + 6\sqrt{y-5} + (y-5) - 18 - 6\sqrt{y-5} + 14\\ \amp = y \end{align*}

as required. Hence $f$ is surjective.

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Therefore $f$ is a bijective function.

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10.8.9.

Solution.

Proof of cardinality of $F$.

We see that for $f:A\rightarrow \set{0,1}$ to be a function, every element in $A$ has to go to a unique element in $\set{0,1}\text{.}$ Then we see that for every element $a_i\text{,}$ $i\in\set{1,2,\cdots, n}\text{,}$ there are two different options for the value of $f(a_i)\text{,}$ $f(a_i)=0$ or $f(a_i)=1\text{.}$ Thus, there are $2^n$ different ways we can construct a function from $A$ to $\set{0,1}\text{.}$ Therefore, $|F|=2^n\text{.}$

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Proof that $g$ is injective and surjective.

Now, let $g:F \rightarrow \mathcal{P}(A)$ be defined as $g(f)=\{ a\in A: f(a)=1 \}\text{.}$ We prove show that this function is both surjective and injective.

Surjective: Let $B\in\mathcal P(A)\text{.}$ Then, we see that $B\subseteq A\text{.}$ Thus, we can define $f_B: A\rightarrow \set{0,1}\text{,}$ defined as $f_B(x)=1$ if $x\in B$ and $f_B(x)=0$ if $x\notin B\text{.}$ Then we see that $f_B$ is well-defined and moreover, we have $g(f_B)=B\text{.}$ Therefore, $g$ is surjective.
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Injective: Let $f_1,f_2\in F$ and assume that $g(f_1)=g(f_2)\text{.}$ This implies,

\begin{equation*} \set{a\in A: f_1(a)=1}=\set{a\in A: f_2(a)=1}. \end{equation*}

Call this set $B\text{.}$ To show that $f_1=f_2\text{,}$ we show that the functions agree at all inputs. Let $x \in A\text{,}$ then either $x \in B$ or $x \not\in B\text{.}$ When $x=\in B\text{,}$ we have, by definition of $B\text{,}$ $f_1(x)=1=f_2(x)\text{.}$ Then when $x \not\in B$ we have $f_1(x)=0=f_2(x)\text{.}$Therefore $\forall x\in A\text{,}$ we have $f_1(x)=f_2(x)\text{,}$ which implies $f_1=f_2\text{.}$ Therefore $g$ is injective.

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Notice that we can also prove the injectiveness of $g$ as follows:

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Alternative proof that $g$ is injective.

Let $f_1, f_2 \in F$ so that $f_1 \neq f_2\text{.}$ This means that there is some $x \in A$ so that $f_1(x) \neq f_2(x)\text{.}$ Now, either $f_1(x)=0\text{,}$ or $f_1(x)= 1\text{.}$

Assume $f_1(x) = 1$ and so $f_2(x)=0\text{.}$ Then $x \in g(f_1)$ but $x \not\in g(f_2)\text{,}$ and so $g(f_1) \neq g(f_2)\text{.}$
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Similarly, when $f_1(x)=0$ we know $f_2(x)=1\text{.}$ Hence $x \not\in g(f_1)$ but $x \in g(f_2)\text{,}$ and again $g(f_1) \neq g(f_2)$
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Thus when $f_1 \neq f_2$ we know that $g(f_1) \neq g(f_2)$ and thus $g$ is injective.

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10.8.10.

Solution.

Claim: The function is injective, but not surjective.

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Proof.

To prove that the function is injective, let $m,n \in \mathbb{Z}$ so that $f(n)=f(m)\text{.}$ Then $(2n+1,n+2) = (2m+1,m+2)\text{.}$ This gives us two equations

\begin{equation*} 2n+1=2m+1 \qquad \text{and} \qquad n+2=m+2 \end{equation*}

Both of these give $n=m$ as required.

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Now, to show that the function is not surjective, we find a point in $(y_1, y_2) \in \mathbb{Z}\times \mathbb{Z}$ which is not the image of any $x\in \mathbb{Z}\text{.}$ Consider $(2,0)\in\mathbb Z\times \mathbb Z\text{.}$ We see that $(2n+1, n+2)\neq (2,0)$ for any $n\in\mathbb Z$ since $2$ is even and $2n+1$ is always odd, that is $2\neq 2n+1$ for any $n\in\mathbb Z\text{.}$

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Notice that we can also prove that the function is an injection as follows:

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Alternative proof of injectiveness.

Let $n,m\in\mathbb Z$ and assume $n\neq m\text{.}$ Then we see that $n+2\neq m+2\text{.}$ Hence, we see $(2n+1, n+2)\neq (2m+1, m+2)\text{,}$ that is $f(n)\neq f(m)\text{.}$ Therefore the function is injective.

Let $y\in f(A-B)\text{.}$ Then, by definition, there exists $x\in A-B$ such that $f(x)=y\text{.}$ Then, we see that $y\in f(A)\text{.}$ Now we must show that $y \not\in f(B)\text{.}$ Let $z$ be any element of $B\text{.}$ Then we must have that $x \neq z\text{,}$ and so, since $f$ is injective, we know that $f(x) \neq f(z)\text{.}$ This means that $y$ cannot be the image of any $z \in B\text{,}$ and so $y \not\in f(B)\text{.}$ Hence $y \in f(A) - f(B)\text{,}$ and so $f(A-B) \subseteq f(A)-f(B)$ as required.
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We note that the contrapositive of this argument is illuminating. If $y \in f(B)$ then there is some $z \in B$ so that $f(z)=y\text{.}$ Now, since $y=f(x)$ this means $f(x) = f(z)\text{.}$ But since $x \not\in B$ and $z \in B\text{,}$ we know $x\neq z$ and thus $f$ is not injective.
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Now, let $y\in f(A)-f(B)\text{.}$ This means that $y \in f(A)$ and so there is some $x \in A$ so that $y=f(x)\text{.}$ At the same time, $y\not\in f(B)\text{,}$ which means that $y\neq f(z)$ for any $z\in B\text{.}$ Hence it cannot be the case that $x \in B\text{.}$ Thus $x \in A-B$ and so $y = f(x) \in f(A-B)\text{,}$ and therefore $f(A)-f(B)\subseteq f(A-B)$ as required.
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Notice that the proof that $f(A)-f(B)\subseteq f(A-B)$ does not actually require that $f$ is surjective. It holds for all functions. Also notice that when we prove first inclusion we are actually proving that

\begin{align*} f \text{ is injective} \amp \implies (LHS \subseteq RHS) \end{align*}

So we could prove the contrapositive of this instead. Namely

\begin{align*} (LHS \not\subseteq RHS) \amp \implies f \text{ is not injective}. \end{align*}

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Contrapositive proof.

Assume that $f(A-B) \not\subseteq f(A)-f(B)\text{.}$ Hence there is some $y \in f(A-B)$ so that $y \not\in f(A)-f(B)\text{.}$ So we know that there is some $x \in A-B$ with $f(x)=y\text{.}$ Notice that since $x \in A\text{,}$ we know that $y=f(x) \in f(A)\text{.}$

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Now since $y \not\in f(A)-f(B)$ and $y \in f(A)$ it must be the case that $y \in f(B)\text{.}$ However, this means that there is some $z \in B$ so that $f(z)=y\text{.}$ This implies that $f(z)=y=f(x)$ and at the same time $x \neq z$ and thus $f$ is not injective.

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10.8.13.

Solution.

Proof.

We will show that $f$ is injective and then that it is surjective.

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To prove that $f$ is injective, let $(a,b),(c,d)\in\mathbb N \times \mathbb{N}$ and assume that $f(a,b)=f(c,d)\text{.}$ Without loss of generality, let us further assume that $a\geq c\text{;}$ the case that $a \lt c$ is very similar.

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Using the definition of the function, we see that $2^{a-1}(2b-1)=2^{c-1}(2d-1)\text{,}$ which implies

\begin{equation*} 2^{a-c}(2b-1)=(2d-1). \end{equation*}

Notice that the right hand side of this expression, $(2d-1)\text{,}$ is odd. On the other hand, the left hand side of the expression has a factor of $2^{a-c}$ and $a-c\geq 0\text{.}$ Therefore, we must have $a-c=0$ (so that $2^{a-c}=1$), otherwise the left hand side will be even. Thus, $a=c\text{.}$ Hence, we get $(2b-1)=(2d-1)\text{,}$ that is, $b=d\text{.}$ Hence $(a,b)=(c,d)\text{,}$ and we see that $f$ is injective.

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To prove that $f$ is surjective, let $n \in\mathbb N\text{.}$ Then, we know that $n $ has a unique prime factorisation. We can write the factorisation as $n=2^{a_1}3^{a_2}5^{a_3}7^{a_4}\cdots p_m^{a_m}$ for some $m\in\mathbb N$ where $a_i\in\mathbb N\cup \set{0}$ for all $i\in\set{1,2,3,\cdots, m}\text{.}$

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Since the product of odd numbers is odd, we see that $3^{a_2}5^{a_3}7^{a_4}\cdots p_m^{a_m}$ is odd. Thus, there exists $\ell\in\mathbb{Z} $ such that $3^{a_2}5^{a_3}7^{a_4}\cdots p_m^{a_m}=2 \ell-1\text{.}$ Note that since that product of primes is positive and odd, we know that $\ell \geq 1$ and so $\ell \in \mathbb{N}\text{.}$ Similarly, since $a_1 \geq 0\text{,}$ we know that $k = a_1+1 \in \mathbb{N} $ and so we can write $n=2^{k-1} (2\ell-1)=f(k,\ell) $ with $k,\ell \in \mathbb{N}\text{.}$ Therefore $f$ is surjective.

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10.8.14.

Solution.

Proof.

Let $A, B$ be nonempty sets and assume that there is a bijection $f:A\rightarrow B\text{.}$ Then, since $f$ is a function, we can define a new function $g:\mathcal{P}(A)\rightarrow \mathcal{P}(B)$ as follows:

\begin{equation*} g(X)=\set{ f(x) : x\in X } \in \mathcal P(B). \end{equation*}

That is, our new function $g$ takes each element of a subset to its image under $f\text{.}$ We need to check that $g$ is injective and surjective.

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Surjective: Let $Y\in\mathcal P(B)\text{.}$ Then we know that $Y\subseteq B\text{.}$ We also know that $f$ is surjective. Then, we see that $\forall y\in Y$ there is an $x\in A$ such that $f(x)=y\text{.}$ Consider the set $X=\set{x\in A: f(x)\in Y}\text{.}$ We would like to show that $g(X) = Y\text{.}$
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First, let $b \in g(X)\text{.}$ Then by the definition of $g\text{,}$ $b = f(a)$ for some $a\in X\text{.}$ In particular, $X$ is the set of $x\in A$ such that $f(x) \in Y\text{,}$ so $a \in X$ implies $a = f(b) \in Y\text{.}$ Thus, $g(X) \subseteq Y\text{.}$
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Now, let $b \in Y\text{.}$ Since $f$ is surjective, there exists $a \in A$ such that $f(a)=b\text{.}$ Thus, by the definition of $X$ we have $a \in X\text{,}$ and by the definition of $g\text{,}$ $f(a) = b \in g(X)\text{.}$
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Therefore we see $g(X)=Y\text{.}$ Hence $g$ is surjective.
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Injective: We proceed by contrapositive. Let $Z, W \in \mathcal{P}(A)\text{.}$ We will show that if $g(Z) = g(W)$ then $Z=W\text{.}$ Assume $g(Z)=g(W)\text{.}$
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Let $z \in Z\text{.}$ By the definition of $g\text{,}$ $f(z) \in g(Z) = g(W)\text{.}$ Thus, there is some $w\in W$ so that $f(w) = f(z)\text{.}$ By the injectivity of $f\text{,}$ $w=z\text{,}$ so $z\in W\text{,}$ and we see $Z \subseteq W\text{.}$
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Now, let $w\in W\text{.}$ By the definition of $g\text{,}$ $f(w) \in g(W) = g(Z)\text{.}$ Thus, there is some $z\in Z$ so that $f(w) = f(z)\text{.}$ By the injectivity of $f\text{,}$ $w=z\text{,}$ so $w\in Z\text{,}$ and we see $W \subseteq Z\text{.}$ Therefore, $Z=W\text{,}$ and we see that $g$ is injective.
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We conclude that $g$ is a bijection.
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10.8.15.

Solution.

Proof.

Let $\rel$ be the relation on $\mathbb R^2$ defined as

\begin{equation*} (x,y)\rel (s,t) \text{ if } x^2+y^2=s^2+t^2. \end{equation*}

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We need to show that this relation is reflexive, symmetric and transitive.
- {Reflexivity:} Let $(x,y)\in\mathbb R^2\text{.}$ Then, we see that since $x^2+y^2=x^2+y^2$ , we get $(x,y)\rel(x,y)\text{.}$ Hence $\rel$ is reflexive.
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- {Symmetry:} Let $(x,y),(s,t)\in\mathbb R^2$ and assume that $(x,y)\rel (s,t)\text{.}$ This means $x^2+y^2=s^2+t^2\text{.}$ Then, we see that $s^2+t^2=x^2+y^2\text{.}$ Hence, we see that $(s,t)\rel (x,y)\text{,}$ that is, $\rel$ is symmetric.
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- {Transitive:}. Let $(x,y),(s,t),(a,b)\in\mathbb R^2$ and assume that $(x,y)\rel (s,t)$ and $(s,t\rel (a,b))\text{.}$ This means $x^2+y^2=s^2+t^2$ and $s^2+t^2=a^2+b^2$ . Then, we see that $x^2+y^2=a^2+b^2\text{.}$ Hence, we see that $(x,y)\rel (a,b)\text{,}$ that is, $\rel$ is transitive.
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Hence the relation $\rel$ is an equivalence relation.

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Now, let $\mathcal S$ be the set of equivalence classes of the relation $\rel$ defined in part (a) and define $f:\mathcal S\to (0,\infty)\text{,}$ defined by $f([(x,y)])=\sqrt{x^2+y^2}.$
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- Show $f$ is a function: Let $[(x,y)] $ be an equivalence class. Then since $x,y \in \mathbb{R}$ we know that $x^2+y^2 \geq 0$ and so $\sqrt{x^2+y^2} \geq 0\text{.}$ Hence $f$ is defined everywhere on its domain. Now let $[(x,y)]=[(s,t)]\text{.}$ This means that $x^2+y^2=s^2+t^2\text{.}$ Thus, we get
  
  \begin{equation*} f\big([(x,y)]\big)=\sqrt{x^2+y^2}=\sqrt{s^2+t^2}=f\big([(s,t)]\big). \end{equation*}
  
  Therefore, $f:\mathcal s\to (0,\infty)$ defines a function.
  
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- Show $f$ is bijective: We need to show that $f$ is injective and surjective.
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  - Injective: Let $(x,y), (s,t)\in\mathbb R^2\text{,}$ and assume that
    
    \begin{equation*} f\big([(x,y)]\big)=f\big([(s,t)]\big). \end{equation*}
    
    This means that $\sqrt{x^2+y^2}=\sqrt{s^2+t^2}\text{,}$ which in turn implies that
    
    \begin{equation*} x^2+y^2=s^2+t^2. \end{equation*}
    
    Thus, we get $(x,y)\rel(s,t)\text{,}$ that is,
    
    \begin{equation*} [(x,y)]=[(s,t)]. \end{equation*}
    
    Hence $f$ is injective.
    
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  - Surjective: Let $z\in[0,\infty)\text{.}$ Then, we can take $[(x,y)]=[(0,z)]\in\mathcal S\text{,}$ and see that
    
    \begin{equation*} f\big([(x,y)]\big)=f\big([(0,z)]\big)=\sqrt{z^2}=z. \end{equation*}
    
    Hence, $f$ is surjective.
    
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  Therefore, we see that $f:\mathcal S\to [0,\infty)$ is bijective.
  
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10.8.16.

Solution.

Proof of (a).

Let $[z]_n \in \mathbb{Z}_n\text{,}$ then we know that $z \in [z]_n\text{.}$ By Euclidean division we know that $z = qn + r$ with $q \in \mathbb{Z}$ and $r \in \set{0,1,\dots,n-1}\text{.}$ Hence $f([x]_n) \in \set{0,1,\dots,n-1}$ as required.

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Now let $x,y\in\mathbb{Z}\text{,}$ and suppose that $[x]_n=[y]_n\text{.}$ Since $x,y \in \mathbb{Z}\text{,}$ by Euclidean division, there are integers $q_x,q_y$ and $r_x,r_y \in \set{0,1,\dots,n-1}$

\begin{equation*} x=q_xn+r_x \qquad \text{and} \qquad y=q_yn+r_y. \end{equation*}

So, by our definition of $f\text{,}$ $f([x]_n) = r_x$ and $f([y_n])=r_y\text{.}$

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However, since $[x]_n=[y]_n$ we need to show that $f([x]_n)=f([y]_n)\text{.}$ That is, we need to show that $r_x = r_y\text{.}$ Since $[x]_n = [y]_n\text{,}$ we must have that $x$ and $y$ are congruent modulo $n\text{,}$ so $n=x-y\text{.}$ That is, there is some $k\in\mathbb{Z}$ so that $(x-y)=kn\text{.}$

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Hence the equation $(x-y)=kn$ gives

\begin{equation*} q_xn+r_x-(q_yn+r_y) =kn \end{equation*}

and rearranging, we have

\begin{equation*} r_x-r_y=n(k+q_y-q_x). \end{equation*}

Thus $n\mid r_x-r_y\text{.}$ Notice that since $0 \leq r_x,r_y \leq n-1\text{,}$

\begin{equation*} -(n-1)\leq r_x-r_y\leq n-1. \end{equation*}

Equivalently, we can say $|r_x-r_y|\lt n\text{.}$ Since $n \mid (r_x-r_y)$ and $|r_x-r_y| \lt n\text{,}$ it must be the case that $r_x-r_y=0\text{.}$ That is, $r_x=r_y\text{,}$ and hence $f([x]_n)=f([y]_n)\text{.}$

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Proof of (b).

We prove that $f$ is both injective and surjective, and so, bijective.

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First we prove that $f$ is injective. Suppose that for $[x]_n,[y]_n\in\mathbb{Z}_n\text{,}$ we have $f([x]_n)=f([y]_n)\text{.}$ Then there is some $r\in\set{0,1,\dots,n-1}$ so that $f([x]_n)=f([y]_n)=r\text{.}$ By definition of $f\text{,}$ we know that $r$ is the remainder when $x$ or $y$ is divided by $n\text{.}$ By Euclidean division, there are integers $p$ and $q$ so that $x=pn+r$ and $y=qn+r\text{.}$ Thus

\begin{equation*} x-y=(pn+r)-(qn+r)=(p-q)n \end{equation*}

and since $p-q\in\mathbb{Z}\text{,}$ we have that $n$ divides $x-y\text{.}$ Therefore $x$ and $y$ are congruent mod $n\text{,}$ and so $[x]_n=[y]_n\text{.}$ Hence $f$ is injective.

Let $f,g:\mathbb{R}\to \mathbb{R}$ be two strictly increasing functions. Fix $x_1, x_2 \in \mathbb{R}$ with $x_1 \lt x_2\text{.}$ Then because $g$ is strictly increasing,

\begin{equation*} z_1=g(x_1) \lt g(x_2)=z_2. \end{equation*}

Now, using the fact that $f$ is strictly increasing and that $z_1\lt z_2\text{,}$ we see that

\begin{equation*} (f\circ g)(x_1) = f(g(x_1)) = f(z_1) \lt f(z_2) = f(g(x_2)) = (f\circ g)(x_2). \end{equation*}

Therefore $f\circ g$ is a strictly increasing function.

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Proof of (b).

Let $f,g:\mathbb{R}\to \mathbb{R}$ be two strictly decreasing functions. Fix $x_1, x_2 \in \mathbb{R}$ with $x_1 \lt x_2\text{.}$ Then because $g$ is strictly decreasing,

\begin{equation*} z_1 = g(x_1) \gt g(x_2)= z_2. \end{equation*}

Now, using the fact that $f$ is strictly decreasing and $z_2 \lt z_1\text{,}$ we see that

\begin{equation*} (f\circ g)(x_2) = f(g(x_2)) = f(z_2) \gt f(z_1) = f(g(x_1)) = (f\circ g)(x_1) \end{equation*}

Therefore $f\circ g$ is a strictly increasing function.

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10.8.20.

Solution.

Claim: This statement is false.
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Proof.

Consider $f(x)=x^2$ and $g(x)=h(x) = x\text{.}$ Then

\begin{align*} f\circ (g+h)(x) \amp = f((g+h)(x))\\ \amp = f(g(x) +h(x)) \end{align*}

Using the choice of $g(x)$ and $h(x)\text{,}$

\begin{align*} \amp = f(2x) \end{align*}

Using the choice of $f(x)\text{,}$

\begin{align*} \amp = 4x^2 \end{align*}

On the other hand, we compute

\begin{align*} (f\circ g + f\circ h)(x) \amp = f(g(x)) + f(h(x)) \end{align*}

Using the definitions of $g(x)$ and $h(x)\text{,}$

\begin{align*} \amp =f(x) + f(x) \end{align*}

Using the definitions of $f(x)\text{,}$

\begin{align*} \amp =2x^2 \end{align*}

Consider $x=1\text{.}$ Since

\begin{equation*} f\circ (g+h)(1) = (4)1^2 = 4 \neq 2 = (2)1^2 = (f\circ g + f\circ h)(1), \end{equation*}

we have found a value of $x$ such that the functions are not equal. Therefore the statement is false.

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Claim: The statement is true.
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Proof.

Fix $x\in \mathbb{R}\text{.}$ We compute, using the definition of function composition

\begin{align*} (g+h)\circ f(x) \amp = (g+h)(f(x)). \end{align*}

Using the definition of function addition, we obtain

\begin{align*} \amp =g(f(x)) + h(f(x)) . \end{align*}

We now use the definition of function composition to rewrite the statement as

\begin{align*} \amp = \left(g\circ f\right)(x) + \left(h\circ f\right)(x). \end{align*}

Finally, we use the definition of function addition to obtain

\begin{align*} \amp = (g\circ f + h\circ f)(x) . \end{align*}

Since $x$ was chosen arbitrarily, we have that

\begin{equation*} (g+h)\circ f(x) = (g\circ f + h\circ f)(x) \end{equation*}

for every $x\text{.}$ We conclude that

\begin{equation*} (g+h)\circ f = g\circ f + h\circ f \end{equation*}

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Claim: The statement is true.
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Proof.

Fix $x\in \mathbb{R}\text{.}$ First, we use the definition of function composition to see that

\begin{align*} \left(\frac{1}{f}\circ g \right)(x) \amp = \left(\frac{1}{f}\right)(g(x)). \end{align*}

We regard the numerator as the constant function $h(x)=1\text{.}$ This lets us rewrite the expression as

\begin{align*} \amp = \left(\frac{h}{f}\right)(g(x)) \end{align*}

By the definition of function division, we have

\begin{align*} \amp = \frac{h(g(x))}{f(g(x))} \end{align*}

Since $h$ is a constant function, we have $h(g(x)) = 1 = h(x)\text{,}$ thus

\begin{align*} \amp = \frac{h(x)}{f(g(x))} \end{align*}

By the definition of function composition, we have

\begin{align*} \amp =\frac{h(x)}{(f\circ g)(x)} \end{align*}

Using the definition of function division yields

\begin{align*} \amp =\left(\frac{h}{f\circ g}\right)(x) \end{align*}

Finally, we use the definition of $h$ to conclude,

\begin{align*} \amp =\left(\frac{1}{f\circ g}\right)(x). \end{align*}

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Since $x$ was chosen arbitrarily, we see that

\begin{equation*} \left(\frac{1}{f}\circ g\right) (x) = \left(\frac{1}{f\circ g}\right)(x) \end{equation*}

for every $x\in \mathbb{R}\text{.}$ We conclude that

\begin{equation*} \frac{1}{f}\circ g = \frac{1}{f\circ g}. \end{equation*}

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Claim: The statement is false.
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Proof.

Consider $f(x)=x+2$ and $g(x)=x^2\text{.}$ We compute

\begin{align*} \frac{1}{f\circ g}(x) \amp = \frac{1}{f(g(x))} \end{align*}

Using the definition of $g(x)\text{,}$

\begin{align*} \amp = \frac{1}{f(x^2)} \end{align*}

Using the definition of $f(x)\text{,}$

\begin{align*} \amp = \frac{1}{x^2+2}. \end{align*}

On the other hand,

\begin{align*} \left(f\circ \frac{1}{g}\right) (x) \amp = f\left(\frac{1}{g(x)}\right) \end{align*}

Using the definition of $g(x)\text{,}$

\begin{align*} \amp = f\left(\frac{1}{x^2} \right) \end{align*}

Using the definition of $f(x)\text{,}$

\begin{align*} \amp = \frac{1}{x^2}+2. \end{align*}

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Consider $x=1\text{.}$ Since

\begin{equation*} \frac{1}{f\circ g}(1) = \frac{1}{1^2+2} = \frac{1}{3} \neq 3 = \frac{1}{1^2}+2 = \left(f\circ \frac{1}{g}\right) (x), \end{equation*}

we have found a value of $x$ such that the functions are not equal. Therefore the statement is false.

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And on the other hand,

\begin{equation*} (g\circ f(x) )= g(f(x)) = g(x+1) = x+1. \end{equation*}

Thus, $f\circ g = g\circ f\text{,}$ as desired.

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Proof of (b).

Claim: $g$ can be any function such that $g(c)=c\text{.}$

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Notice that $(f\circ g)(x) = f(g(x))=c\text{,}$ since $f$ is a constant function. Then in order for $(f\circ g)(x) = (g \circ f)(x)$ for every $x\in \mathbb{R},$

\begin{equation*} c= (g\circ f)(x) = g(f(x))=g(c). \end{equation*}

Hence, we require that $g(c)=c\text{.}$ There are no other conditions on the function $g\text{.}$

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Proof of (c).

Since the statement

\begin{equation*} (f\circ g)(x) = (g\circ f)(x) \tag{1} \end{equation*}

must be true for any $x\in \mathbb{R}\text{,}$ and any function $g:\mathbb{R}\to \mathbb{R}\text{,}$ we begin by examining the restrictions placed on $f$ when $g$ is a constant function.

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Fix $c\in \mathbb{R}$ and define the function $g_c:\mathbb{R}\to \mathbb{R}$ by $g_c(x) = c\text{.}$ By assumption, the statement $(f\circ g_c)(x) = (g_c\circ f)(x)$ must hold. Applying part (b), we see that we must have $f(c)=c\text{.}$

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Moreover, since the choice of $c\in \mathbb{R}$ was arbitrary and $(f\circ g_c)(x) = (g_c\circ f)(x)$ must hold for every $c\in \mathbb{R}\text{,}$ we know that $f(c)=c$ for every $c\in \mathbb{R}\text{.}$ This tells us that the function $f(x)=x$ is the only function which could possibly satisfy (1).

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It remains to check that (1) holds for the function $f(x)=x\text{.}$ Working with the left hand side, we see

\begin{equation*} (f\circ g)(x) = f(g(x)) = g(x) \end{equation*}

Where $f(g(x)) = g(x)$ by the definition of $f\text{.}$ On the other hand,

\begin{equation*} (g\circ f)(x) = g(f(x)) = g(x) \end{equation*}

Where we obtain the second equality using the fact that $f(x)=x\text{.}$ Therefore,

\begin{equation*} (f\circ g)(x) = g(x) = (g\circ f)(x), \end{equation*}

and we conclude that $f(x)=x$ is the only solution to (1).

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10.8.24.

Solution.

Proof of (a).

Assume $f\circ f$ is bijective. Let $x,y\in A$ and assume that $f(x)=f(y)\text{.}$ Then, since $f$ is a function, we get $(f\circ f)(x)=(f\circ f)(y)\text{.}$ Moreover, since $f\circ f$ is injective we see $x=y\text{.}$ Therefore $f$ is injective.

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Moreover, if $b\in A\text{,}$ then we see that since $f\circ f$ is surjective, there exists $a\in A$ such that $(f\circ f)(a)=b\text{.}$ Thus, we see that $f(f(a))=b\text{,}$ which means that for $c=f(a)\in A\text{,}$ we have $f(c)=b\text{.}$ Therefore $f$ is surjective.

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Hence, $f$ is bijective.

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Proof of (b).

We see that if $f(x)=\log\left( \dfrac{e^x+1}{e^x-1} \right)\text{,}$ then

\begin{align*} f\circ f(x) \amp =f(f(x)) \end{align*}

Using the definition of $f$ for the “inner” function,

\begin{align*} \amp =f\left( \log\left( \dfrac{e^x+1}{e^x-1} \right) \right) \end{align*}

Using the definition of $f$ for the “outer” function,

\begin{align*} \amp =\log\left( \dfrac{e^{\log\left( \dfrac{e^x+1}{e^x-1} \right)}+1}{e^{\log\left( \dfrac{e^x+1}{e^x-1} \right)}-1} \right) \end{align*}

We now use the fact that $e$ and $\log$ are inverse functions to obtain,

\begin{align*} \amp =\log\left( \dfrac{\left(\dfrac{e^x+1}{e^x-1}\right)+1}{ \left(\dfrac{e^x+1}{e^x-1}\right)-1} \right) \end{align*}

Simplifying, we see

\begin{align*} \amp = \log \left(\dfrac{ \left(\dfrac{2e^x}{e^x-1} \right)}{ \left(\dfrac{2}{e^x-1}\right) }\right)\\ \amp =\log(e^x)\\ \amp =x. \end{align*}

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Therefore, we see that $f\circ f$ is the identity function, $i_{(0,\infty)}\text{.}$ And then by part (a), since $i_{(0,\infty)}$ is bijective, $f$ is bijective.

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10.8.25.

Solution.

Proof.

Let $f : \mathbb R-\set{-2}\rightarrow\mathbb R=\set{1}\text{,}$ be defined by $f(x)=\dfrac{x+1}{x+2}\text{.}$ Then we want to show that $f$ is both injetive and surjective.

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Injective: Assume that $a,z\in\mathbb \mathbb R-\set{-2}\text{,}$ and $f(x)=f(z)\text{.}$ Then

\begin{align*} \dfrac{x+1}{x+2} \amp =\dfrac{z+1}{z+2} \amp \text{since }x,z \neq -2\\ (z+2)(x+1) \amp = (x+2)(z+1) \amp \text{expand}\\ xz+z+2x+2 \amp = xz+x+2x+2 \amp \text{cancel}\\ x \amp = z \end{align*}

Hence $x=z$ and so $f$ is injective.

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Surjective: Let $y\in\mathbb R-\set{1}\text{.}$ Then set $x= -2+\frac{1}{1-y}\text{.}$ Since $y \neq 1$ we know that $x \in \mathbb{R}\text{.}$ And since $\frac{1}{1-y} \neq 0\text{,}$ we know that $x \neq -2\text{.}$ Hence $x$ lies in the domain of the function. Then we verify that $f(x)=y$ as follows:

\begin{align*} f(x) \amp = \frac{x+1}{x+2}\\ \amp = \dfrac{-1 +\frac{1}{1-y}}{\frac{1}{1-y}}\\ \amp = \frac{y-1+1}{1} = y \end{align*}

as required. Hence, $f$ is surjective.

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Therefore $f$ is bijective.

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Since it is bijective it has an inverse, denoted $f^{-1}\text{,}$ such that $f^{-1}:\mathbb R-\set{1}\rightarrow \mathbb R-\set{-2}\text{.}$ Our proof of surjectiveness shows that for $x=\dfrac{2y-1}{1-y}\text{,}$ we have $f(x)=f\left(\dfrac{2y-1}{1-y}\right)=y\text{,}$ which in turn implies that $f^{-1}(y)=\dfrac{2y-1}{1-y}\text{.}$

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10.8.26.

Solution.

Proof.

Let $g:\mathbb Z \rightarrow \mathbb Z$ be defined as

\begin{equation*} g(n) = \begin{cases} 3-n \amp \text{if } n \text{ is odd } \\ n-7 \amp \text{if } n \text{ is even } \end{cases} \end{equation*}

We prove that $g$ is a left and right inverse of $f\text{,}$ which implies, via Lemma 10.6.3 and Lemma 10.6.4, that $f$ is bijective and that $g$ is its inverse.

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$g$ is the left inverse of $f$: Let $n\in\mathbb Z$ and consider $g(f(n))\text{.}$ Then $n$ is either even or odd.
- $n$ is even: In this case, $g(f(n))=g(-n+3)=-(-n+3)+3=n$ since $-n+3$ is odd.
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- $n$ is odd: Then, we see that $g(f(n))=g(n+7)=(n+7)-7=n$ since $n+7$ is even.
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Hence $g\circ f=i_\mathbb Z\text{,}$ where $i_\mathbb Z$ is the identity function on $\mathbb Z\text{,}$ and so $g$ is a left-inverse of $f\text{.}$

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$g$ is the right inverse of $f$: Let $n\in\mathbb Z$ and consider $f(g(n))\text{.}$ Then, again, we have two cases, $n$ is either even or odd;
- $n$ is even: Then, we have $f(g(n))=f(n-7)=(n-7)+7=n$ since $n-7$ is odd.
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- $n$ is odd : Finally, in this case we have $f(g(n))=f(-n+3)=-(-n+3)+3=n$ since $-n+3$ is even.
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Hence $f\circ g=i_\mathbb Z$ and so $g$ is a right-inverse of $f\text{.}$

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Therefore $f$ is bijective and $g$ is the inverse of $f$ .

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10.8.27.

Solution.

Proof of (a).

Let $B$ a set and assume that $g:A\to A$ satisfies the condition $g\circ g\circ g=i_A\text{.}$ We need to show that $g$ is injective and surjective.

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Injective: Let $x,y\in A$ so that $g(x)=g(y)\text{.}$ Since $g$ is a function, this implies that $g( g(x) ) = g( g(y) )\text{.}$ Applying $g$ again gives $g( g( g(x) ) ) = g( g( g(y)))\text{,}$ that is $(g \circ g \circ g)(x) = (g\circ g\circ g)(y)\text{.}$ But, since $g\circ g\circ g=i_A\text{,}$ the above implies that $i_A(x) = i_A(y)$ and so $x=y\text{.}$ Therefore $g$ is injective.
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Surjective: Let $b\in A\text{.}$ The,n since $g\circ g\circ g=i_A\text{,}$ we get $g\circ g\circ g(b)=b\text{.}$ Therefore, if we set $a = (g\circ g)(b) \in A\text{,}$ we have $g(a)=b\text{.}$ Hence, $g$ is surjective.
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Therefore, we see that $g$ is bijective.

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Proof of (b).

This requires some careful algebraic manipulations. Start by computing and simplifying $(f\circ f)(x)$

\begin{align*} f(f(x)) \amp = 1 - \dfrac{1}{1-\frac{1}{x}}\\ \amp = 1 - \frac{x}{x-1} = \frac{x-1-x}{x-1}\\ \amp = \frac{-1}{x-1} = \frac{1}{1-x} \end{align*}

Then use this to compute $(f\circ f \circ f)(x)\text{:}$

\begin{align*} (f\circ f \circ f)(x) = f( (f\circ f)(x) ) \amp = 1 - \dfrac{1}{\frac{1}{x-1}}\\ \amp = 1 - \frac{x-1}{1} = x \end{align*}

as required. Thus $f\circ f \circ f = i_A$ as required.

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Proof of (c).

The result from (a), implies that $f$ is bijective. We can compute the inverse in two ways. First, observe that since $f\circ f \circ f$ is the identity function, it follows that $\forall x\in A\text{,}$ we have $(f \circ f)(x) =f^{-1}(x)= \frac{1}{1-x}$ as calculated in part (b).

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We could also solve for $f^{-1}$ directly. Namely, if $f^{-1}(x)=y\text{,}$ then we have

\begin{align*} x \amp = f(y) = 1 - \frac{1}{y}\\ xy \amp = y-1\\ xy-y \amp = -1 \end{align*}

from which we can conclude that

\begin{equation*} y = f^{-1}(x) = \frac{1}{1-x}. \end{equation*}

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11 Proof by contradiction
11.3 Exercises

11.3.1.

Solution.

Proof.

Assume for a contradiction that such an integer $a$ exists. Then the first congruence implies that $a=6k+2$ for some integer $k\text{.}$ Similarly, the second congruence implies that $a=9\ell+7$ for some integer $\ell\text{.}$ Thus we have

\begin{equation*} 6k+2 = 9\ell+7. \end{equation*}

We can rewrite this as

\begin{equation*} 6k-9\ell=3(2k-3\ell) = 5. \end{equation*}

Therefore, since $2k+3\ell\in\mathbb Z\text{,}$ we conclude that $3 \mid 5\text{,}$ which is a contradiction.

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Hence, there is no integer $a$ so that $a \equiv 2 \mod 6$ and $a \equiv 7 \mod 9\text{.}$

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11.3.2.

Solution.

Proof.

Let $a,b,c\in\mathbb Z$ be such that $a^2+b^2=c^2\text{.}$ Assume for a contradiction that and $a$ and $b$ are both odd. Then, there are $k,\ell\in\mathbb Z$ such that $a=2k+1$ and $b=2\ell+1\text{.}$ Then we see that

\begin{equation*} c^2=a^2+b^2=(2k+1)^2+(2\ell+1)^2=4k^2+4k+4\ell^2+4\ell+2=2(2k^2+2\ell^2+2k+2\ell+1). \end{equation*}

Since $(2k^2+2\ell^2+2k+2\ell+1)\in\mathbb Z\text{,}$ we see that $c^2$ is even, which implies $c$ is even. Thus, $c=2m$ for some $m\in\mathbb Z\text{.}$ Substituting this into the equation

\begin{equation*} 4k^2+4k+4\ell^2+4\ell+2=c^2 \end{equation*}

we obtain

\begin{equation*} 4k^2+4k+4\ell^2+4\ell+2=4m^2. \end{equation*}

But then we may write

\begin{equation*} 2=4m^2-( 4k^2+4k+4\ell^2+4\ell)=4(m^2-k^2-k-\ell^2-\ell) \end{equation*}

and as $m^2-k^2-k-\ell^2-\ell\in\mathbb Z\text{,}$ we can conclude that $4\mid 2\text{,}$ a contradiction. Hence our original assumption was false, and so $a$ or $b$ must be even.

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11.3.3.

Solution.

Proof.

Suppose for a contradiction that there are integers $x$ and $y$ satisfying $5y^2-4x^2=7\text{.}$ Then we see that $5y^2-4x^2\equiv 7 \pmod 4\text{.}$ This implies

\begin{align*} 1y^2-0x^2 \amp \equiv 3\pmod 4\\ y^2 \amp \equiv 3 \pmod 4. \end{align*}

However, we have $2$ cases for $y\text{,}$ $y$ is even or $y$ is odd.

$y$ is even: In this case, we see that $y=2k$ for some $k\in\mathbb Z\text{.}$ Therefore $y^2=4k^2\equiv 0\pmod 4\text{,}$ which contradicts with $y^2\equiv 3 \mod 4\text{.}$
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$y$ is odd: Then, we see that $y=2k+1$ for some $k\in\mathbb Z\text{.}$ Thus, we get $y^2=4k^2+4k+1\equiv 1 \pmod 4\text{,}$ which, again, contradicts with $y^2\equiv 3 \mod 4\text{.}$
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Hence we can conclude that there doesn’t exist $x,y\in\mathbb Z$ satisfying the equation $5y^2-4x^2=7\text{.}$

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It is worth noting that this approach to showing the non-existence of integer solutions to equations can work very effectively (as it does above). However, consider what happens in the above if we choose a “bad” modulus. Consider the equation modulo 3:

\begin{align*} 5y^2 - 4x^2 \amp \equiv 7 \pmod 3\\ 2y^2 + 2y^2 \amp \equiv 1 \pmod 3 \end{align*}

This equation does have a solution $x,y = 1 \pmod 3$ since

\begin{gather*} 2 \cdot 1 + 2 \cdot 1 = 4 \equiv 1 \pmod 3. \end{gather*}

This does not mean the equation has a solution over the integers. Let us be careful with the flow of logic:

If the equation has a solution, $x=a,y=b$ over the integers, then
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those numbers satisfy $5b^2-4a^2 = 7 \text{,}$ so then
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taking that equation modulo $n$ gives the equation $5b^2 -4a^2 \equiv 7 \pmod n$
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That chain of reasoning has to be true for all moduli $n\text{.}$ So when we find a modulus with no solutions, we can use modus tollens 2.5.2 to tell us that the equation has no solution. Unfortunately, when we do find a solution for a particular modulus, we cannot infer that there has to be a solution over the integers; that would be an example of affirming the consequent 2.5.3.

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11.3.6.

Solution.

Proof of (a).

Assume, for a contradiction, that $r\in\mathbb{Q}$ is the smallest positive rational number. Then there are $a,b\in\mathbb{Z}\text{,}$ $b\neq0\text{,}$ such that $r=a/b\text{.}$ Thus

\begin{equation*} \frac{r}{2} = \frac{a}{2b} \end{equation*}

and as $a,2b\in\mathbb{Z}$ with $2b\neq0\text{,}$ we have that $r/2$ is rational. But as $r\gt 0\text{,}$

\begin{equation*} 0\lt \frac{r}{2}\lt r, \end{equation*}

so $r/2$ is a positive rational number smaller than $r\text{.}$ This contradicts our assumption on $r\text{,}$ and so there is no smallest positive rational number.

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Proof of (b).

Assume, for a contradiction, that $r\in\mathbb{R}$ is the smallest positive irrational number. We claim that $r/2$ is also irrational. If this were not the case, then there would be $a,b\in\mathbb{Z}$ with $b\neq0$ such that $r/2=a/b\text{.}$ But then

\begin{equation*} r=\frac{2a}{b} \end{equation*}

and as $2a,b\in\mathbb{Z}$ with $b\neq0\text{,}$ we have that $r$ is rational. This is a contradiction, and so $r/2$ must be irrational. Since $r\gt0\text{,}$

\begin{equation*} 0\lt \frac{r}{2}\lt r, \end{equation*}

so $r/2$ is a positive irrational number smaller than $r\text{.}$ This contradicts our assumption on $r\text{,}$ and so there is no smallest positive irrational number.

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11.3.7.

Solution.

Proof of (a).

Assume, to the contrary that $\sqrt{6}$ is rational. Then we can write $\sqrt{6} = \frac{a}{b}$ with $a \in \mathbb{Z}, b \in \mathbb{N}$ and $\gcd(a,b)=1\text{.}$ Squaring both sides gives us $6 = \frac{a^2}{b^2} $ and so $a^2 = 6b^2\text{.}$ Thus $6 \mid a^2\text{.}$ This means that $a^2$ is even, and so $a$ is even.

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Since $a$ is even we can write $a=2k$ for some $k \in \mathbb{Z}\text{.}$ But then

\begin{equation*} a^2 =4k^2 = 6b^2 \end{equation*}

This implies that $2k^2 = 3b^2\text{.}$ This, in turn, implies that $3b^2$ is even, since $3$ is odd, we must have that $b^2$ is even and so $b$ is even. Now we have a contradiction since $2$ divides both $a$ and $b$ but we assumed that $\gcd(a,b)=1\text{.}$

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Hence $\sqrt{6}$ is irrational.

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Notice that we did not actually need, as part of the contradiction, to show that $6 \mid a \implies 6 \mid b\text{.}$ We only needed that $\gcd(a,b) \neq 1$ and it was sufficient to show that both were even.

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We give two proofs of (b).

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Therefore, $\sqrt[3]{25}$ is irrational.

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11.3.9.

Solution.

Proof.

Assume for a contradiction that $k$ is a positive integer, $\sqrt{k}$ is not an integer, but that $\sqrt k$ is rational. Then we see that $\sqrt k=\dfrac{a}{b}$ for some $a\in\mathbb Z$ and $b\in\mathbb N\text{,}$ where $\gcd(a,b)=1\text{.}$ This gives us two equations:

By rearranging the expression for $\sqrt{k}$ we have

\begin{equation*} kb^2 = a^2. \end{equation*}

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At the same time, since $\gcd(a,b)=1\text{,}$ Bézout’s identity implies that there exists integers $x,y$ so that

\begin{equation*} ax +by =1. \end{equation*}

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We can link these two equations together by multiplying the second by $a$ and then substituting in the first gives

\begin{equation*} kb^2x + aby = a. \end{equation*}

Factoring this then gives

\begin{equation*} b (kbx + ay) = a \end{equation*}

and since $kbx + ay \in \mathbb{Z}\text{,}$ this implies that $b \mid a\text{.}$

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We know that $\gcd(a,b)=1$ and so if $b \mid a\text{,}$ we must have that $b=1\text{.}$ This, in turn, implies that we can write $\sqrt{k} = \frac{a}{b} = a \in \mathbb{Z}\text{.}$ This contradicts our assumption that $k$ is not an integer.

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Therefore the statement is true.

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11.3.10.

Solution.

Proof.

Let $r,x\in\mathbb{R}$ with $r\neq0\text{.}$ Suppose that $r$ is rational and $x$ is irrational. Assume, toward a contradiction, that $rx\in\mathbb{Q}\text{.}$ Then there are integers $p,q\text{,}$ with $q$ non-zero, such that $rx=p/q\text{.}$ As $r\neq0$ and rational, there are non-zero integers $m,n$ such that $r=m/n\text{.}$ Then

\begin{equation*} \frac{p}{q}=rx=\frac{m}{n}x, \end{equation*}

and since $m\neq0$ we have

\begin{equation*} x=\frac{pn}{qm}. \end{equation*}

Assume for a contradiction that $x$ is a rational number. This means that we can write $x=\dfrac{m}{n}\text{,}$ where $m \in\mathbb Z\text{,}$ $n\in\mathbb N\text{,}$ where $\gcd(m,n)=1\text{.}$ Then, plugging this into the equation, we get

\begin{equation*} \dfrac{m^7}{n^7}+5\dfrac{m^2}{n^2}-3=0. \end{equation*}

Then multiplying both sides by $n^7$ we get

\begin{equation*} m^7+5m^2n^5-3n^7=0. \end{equation*}

Since, by assumption, this is satisfied by $m,n\text{,}$ it must also be satisfied modulo 2. It becomes:

\begin{equation*} m^7 + m^2n^5 + n^7 \equiv 0 \pmod 2 \end{equation*}

There are four possibilities depending on the parity of $m,n\text{:}$ both even, both odd, or different parities,

both even: This implies that $2 \mid m$ and $2 \mid n$ which contradicts our assumption that $\gcd(m,n)=1\text{.}$ Thus this cannot happen.
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both odd: in this case $m \equiv 1 \pmod 2, n \equiv 1 \pmod 2\text{,}$ but then

\begin{equation*} m^7 + m^2n^5 + n^7 \equiv 3 \equiv 1 \pmod 2 \end{equation*}

But this cannot happen since $m^7 + m^2n^5 + n^7 \equiv 0 \pmod 2\text{,}$ and $1 \not\equiv 0 \pmod 2\text{.}$

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$m$ even and $n$ odd: In this case $m \equiv 0 \pmod 2$ and $n \equiv 1 \pmod 2$ and so

\begin{equation*} m^7 + m^2n^5 + n^7 \equiv 1 \pmod 2 \end{equation*}

Again, this cannot happen since $m^7 + m^2n^5 + n^7 \equiv 0 \pmod 2\text{.}$

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$m$ odd and $n$ even: Now $m \equiv 1 \pmod 2$ and $n \equiv 0 \pmod 2$ and so

\begin{equation*} m^7 + m^2n^5 + n^7 \equiv 1 \pmod 2 \end{equation*}

This cannot happen since $m^7 + m^2n^5 + n^7 \equiv 0 \pmod 2\text{.}$

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This gives us a contradiction since there is no integer solution to this equation modulo 2.

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Let $b\in\mathbb{N}\text{,}$ $b\neq1\text{,}$ be odd. Assume, towards contradiction, that $\log_2(b)$ is rational. Then there are integers $m,n$ with $n\neq 0$ such that

\begin{equation*} \log_2(b)=\frac{m}{n}. \end{equation*}

Moreover, since $\log_2(b)\gt0\text{,}$ we may assume that $m$ and $n$ are positive. This equation implies that $2^{m/n}=b\text{,}$ and so $2^m=b^n\text{.}$ Since $m \gt 0\text{,}$ we have that $2\mid b^n\text{.}$

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However, we claim that $b^k$ is odd for all $k\in\mathbb{N}\text{,}$ and so for $k=n\text{.}$ Indeed, the statement holds for $k=1\text{,}$ since $b$ is odd. Assume that $b^k$ is odd. Then $b^{k+1}=b^k\cdot b$ is the product of two odd numbers, and so odd itself. By induction, $b^k$ is odd for all $k\in\mathbb{N}\text{.}$

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Thus we have shown that simultaneously $2\mid b^n$ and that $b^n$ is odd, a contradiction. Thus $\log_2(b)$ is irrational.

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Thus $\log_2(b)$ with $b$ odd is rational when $b=1$ and irrational for $b \gt 1\text{.}$

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Finally, consider any natural number $n = 2^a b$ with $b$ odd. Then if $b=1\text{,}$ so $n=2^a$ then we know that $\log_2(n)=a$ is rational. While if $b$ is odd and greater than $1\text{,}$ then $\log_2(n) = a + \log b\text{,}$ where $\log_2(b)$ is irrational. Since the sum of a rational and irrational is irrational, we know that $\log_2(n)$ is irrational.

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11.3.14.

Solution.

Proof.

Assume, to the contrary, that this set does have a maxium. Then we can write that maximum as $\frac{a}{b} \in \mathbb{Q}\text{.}$ Since $\sqrt{2} \not\in \mathbb{Q}\text{,}$ we must have that

\begin{equation*} \sqrt{2} - \frac{a}{b} = \delta \gt 0. \end{equation*}

Now let $n = \left\lceil \frac{1}{\delta} \right\rceil\text{.}$ Then

\begin{equation*} 0 \lt \frac{1}{n} \lt \delta = \sqrt{2}-\frac{a}{b} \end{equation*}

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We prove the claim using two cases.

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Case 1: $x=y\text{.}$

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If $x=y\text{,}$ then $\frac{x}{y} = \frac{y}{x} = 1\text{.}$ Therefore

\begin{equation*} \frac{x}{y}+ \frac{y}{x} = 1+1 = 2. \end{equation*}

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Case 2: $x\neq y\text{.}$

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By either of the previous parts, we have

\begin{equation*} \frac{x}{y}+ \frac{y}{x} \gt 2. \end{equation*}

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Combining cases 1 and 2, we conclude that for all $x, y\in \mathbb{R}$ with $x, y \gt 0$

\begin{equation*} \frac{x}{y}+\frac{y}{x} \geq 2. \end{equation*}

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11.3.18.

Solution.

Proof of (a).

By way of contradiction, suppose there exist $a, b\in \mathbb{R}$ with $a,b \gt 0$ such that

\begin{equation*} \frac{2}{a} + \frac{2}{b} = \frac{4}{a+b}. \end{equation*}

We rearrange the expression, multiplying both sides by $ab$ to achieve

\begin{equation*} 2b+2a = \frac{4ab}{a+b}. \end{equation*}

Multiplying both sides by $a+b\text{,}$ we have

\begin{equation*} 4ab+2b^2 + 2a^2 = 4ab. \end{equation*}

Subtracting $4ab$ from each side and dividing by $2$ yields,

\begin{equation*} a^2+b^2 = 0. \end{equation*}

This gives a contradiction since $a, b\in\mathbb{R}\setminus\{0\}$ implies $a^2+b^2 \gt 0\text{.}$

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Proof of (b).

Recall that the square of a positive real number is positive. Therefore, $a^2 \gt 0$ and $b^2 \gt 0\text{,}$ and we must have

\begin{equation*} 2a^2+2b^2 \gt 0. \end{equation*}

We add $4ab$ to each side of our expression

\begin{equation*} 2a^2+2b^2+4ab \gt 4ab \end{equation*}

and factor the left hand side to obtain

\begin{equation*} (2a+2b)(a+b) \gt 4ab. \end{equation*}

Since $a,b \gt 0\text{,}$ $a+b \gt 0$ and we may divide both sides by $a+b$ to achieve

\begin{equation*} 2a+2b \gt \frac{4ab}{a+b} \end{equation*}

Finally, since $a, b\gt 0\text{,}$ we may divide both sides by $ab$ to obtain

\begin{equation*} \frac{2}{b}+\frac{2}{a} \gt \frac{4}{a+b}. \end{equation*}

In particular,

\begin{equation*} \frac{2}{b}+\frac{2}{a} \neq \frac{4}{a+b}. \end{equation*}

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11.3.19.

Solution.

Proof.

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous, bijective function.

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By way of contradiction, assume that $f$ is neither strictly increasing, nor strictly decreasing.

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Then there exists $x_1, x_2, x_3\in \mathbb{R}$ with $x_1 \lt x_2 \lt x_3$ such that

$f(x_1) \leq f(x_2)$ and $f(x_3)\leq f(x_2)$ (the function increases and then decreases). Or,
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$f(x_1)\geq f(x_2)$ and $f(x_3) \geq f(x_2)$ (the function decreases and then increases).
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First, we consider case (1). In this case, the function increases and then decreases, but we do not have information about how $f(x_1)$ compares to $f(x_3)\text{.}$ This gives us two further sub-cases.

Suppose $f(x_1)\leq f(x_3)\text{.}$ We combine this with the inequalities from (1) to find

\begin{equation*} f(x_1)\leq f(x_3)\leq f(x_2). \end{equation*}

We can regard $f(x_3)$ as the value $c$ in the statement of the Intermediate Value Theorem. Since $f$ is continuous, by the Intermediate Value Theorem, there exists $x_0\in [x_1, x_2]$ such that $f(x_0) = f(x_3)\text{.}$ Because $x_0 \leq x_2 \lt x_3\text{,}$ $x_0 \neq x_3\text{.}$ This contradicts the assumption that $f$ is injective.

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Suppose $f(x_3)\leq f(x_1)\text{.}$ Combining this with the inequalities from (1) yields

\begin{equation*} f(x_3)\leq f(x_1)\leq f(x_2). \end{equation*}

We can now regard $f(x_1)$ as the value $c$ in the statement of the Intermediate Value Theorem. Since $f$ is continuous, the Intermediate Value Theorem tells us that there exists $x_0\in [x_2, x_3]$ such that $f(x_0) = f(x_1)\text{.}$ Because $x_1 \lt x_2 \leq x_0\text{,}$ $x_1 \neq x_0$ and we find that $f$ is not injective.

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The argument is similar for case (2). We break into two further sub-cases which compares the value of $f(x_1)$ with $f(x_3)\text{.}$

Suppose $f(x_1)\leq f(x_3)\text{.}$ Then the inequalities from (2) tell us

\begin{equation*} f(x_2)\leq f(x_1)\leq f(x_3). \end{equation*}

We regard $f(x_1)$ as the value $c$ in the statement of the Intermediate Value Theorem. Since $f$ is continuous, the Intermediate Value Theorem says there exists $x_0\in [x_2, x_3]$ such that $f(x_0) = f(x_1)\text{.}$ Notice that $x_1 \lt x_2 \leq x_0\text{,}$ so $x_1 \neq x_0\text{.}$ We conclude that $f$ is not injective.

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Finally, suppose $f(x_3)\leq f(x_1)\text{.}$ $f$ is continuous, and from (2) we know that

\begin{equation*} f(x_2)\leq f(x_3)\leq f(x_1). \end{equation*}

We regard $f(x_3)$ as the value $c$ in the statement of the Intermediate Value Theorem. By the Intermediate Value Theorem, there exists $x_0\in [x_1, x_2]$ such that $f(x_0) = f(x_3)\text{.}$ Because $x_3 \gt x_2 \geq x_0\text{,}$ we see that $x_3\neq x_0\text{,}$ so $f$ is not injective.

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In each of the four cases, we reach the conclusion that $f$ is not injective, contradicting our assumption that $f$ is bijective.

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We conclude that $f$ must be strictly increasing or strictly decreasing

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12 Cardinality
12.7 Exercises

12.7.1.

Solution.

We can list the elements of this set as $\{0,1,2,3,\dots\}\text{.}$ The function corresponding to this list is $f:\mathbb{N}\rightarrow\mathbb{N}\cup\{0\}$ defined by $f(n)=n-1\text{.}$ This function is well-defined, since $n-1\in\mathbb{N}\cup\{0\}$ whenever $n\in\mathbb{N}\text{.}$ We need to show that this function is a bijection.
- Surjective: Let $m\in\mathbb{N}\cup\{0\}\text{.}$ Then $m+1\in\mathbb{N}\text{,}$ and $f(m+1)=m\text{.}$
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- Injective: If $f(m)=f(n)\text{,}$ then $m-1=n-1\text{.}$ Hence $m=n\text{.}$
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Alternatively, we could show that $f$ is a bijection by showing that it has an inverse function. Indeed, its inverse is given by $g:\mathbb{N}\cup\{0\}\to\mathbb{N}$ with $g(n)=n+1\text{.}$

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Let $f:\mathbb{N}\rightarrow\{5,6,7,8,\dots\}$ be defined by $f(n)=n+4\text{.}$ This function is well-defined, since $n+4\in\{5,6,7,8,\dots\}$ whenever $n\in\mathbb{N}\text{.}$ We need to show that this function is a bijection.
- Surjective: Let $m\in\{5,6,7,8,\dots\}\text{.}$ Then $m-4\in\mathbb{N}\text{,}$ and
  
  \begin{equation*} f(m-4)=(m-4)+4=m. \end{equation*}
  
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- Injective: If $f(m)=f(n)\text{,}$ then $m+4=n+4\text{.}$ Hence $m=n\text{.}$
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Let $f:\mathbb{N}\rightarrow\set{1,3,3^2,3^3,\dots}$ be given by $f(n)=3^{n-1}\text{.}$ We need to show that this function is a bijection.
- Surjective: Let $a\in\set{1,3,3^2,3^3,\dots}\text{.}$ Then $a=3^k$ for some $k\in\{0,1,2,\dots\}\text{.}$ But then $k=n-1$ for some $n\in\mathbb{N}\text{,}$ and $f(n)=3^{n-1}=3^k=a\text{.}$
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- Injective: If $f(m)=f(n)\text{,}$ then $3^m=3^n\text{.}$ By taking logarithm with base $3$ of each side, we have $m=n\text{.}$
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We can construct a bijection in a similar fashion as in Theorem 12.2.3, where we showed that $\mathbb{Z}$ is denumerable. We can alternate between the positive and negative integers, listing out the elements of $\mathbb{Z}\setminus\{0\}$ as

\begin{equation*} 1, -1, 2, -2, 3, -3,\dots \end{equation*}

So, the odd natural numbers will be set to the positive integers, and the even natural numbers will be sent to the negative integers. Define the function $f:\mathbb{N}\rightarrow\mathbb{Z}\setminus\{0\}$ by

\begin{equation*} f(n)=\left\{\begin{array}{ll} \frac{n+1}{2} \amp \text{ if $n$ is odd} \\ -\frac{n}{2} \amp \text{ if $n$ is even}. \end{array}\right. \end{equation*}

We show that $f$ is a bijection:
- Surjective: Let $m\in\mathbb{Z}\setminus\{0\}\text{.}$ First suppose that $m\geq1\text{.}$ Then $2m-1$ is an integer, and $2m-1\geq1\text{,}$ so $2m-1\in\mathbb{N}\text{.}$ Moreover, $2m-1$ is odd, and hence
  
  \begin{equation*} f(2m-1)=\frac{(2m-1)+1}{2}=\frac{2m}{2}=m. \end{equation*}
  
  Next suppose that $m\leq -1\text{.}$ Then $-2m\in\mathbb{N}\text{,}$ and $-2m$ is even. Hence
  
  \begin{equation*} f(-2m)=-\frac{-2m}{2}=m. \end{equation*}
  
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- Injective: Suppose $f(m)=f(n)\text{.}$ First note that this implies both $m$ and $n$ are even, or both $m$ and $n$ are odd. Indeed, if this were not the case, than the sign of $f(m)$ and $f(n)$ would be opposite, meaning they could not be equal. If both $m$ and $n$ are even, then
  
  \begin{equation*} \frac{m+1}{2}=f(m)=f(n)=\frac{n+1}{2} \end{equation*}
  
  implying $m=n\text{.}$ If both $m$ and $n$ are odd, then
  
  \begin{equation*} -\frac{m}{2}=f(m)=f(n)=-\frac{n}{2} \end{equation*}
  
  and so $m=n\text{.}$
  
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12.7.2.

Solution.

Claim: It is true that $|A|=|\mathbb{N}|\text{.}$

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Proof.

We define a function $f:\mathbb{N}\rightarrow A$ by $f(k)=(k,k\pi)\text{.}$

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Suppose $f(n) = f(\ell)$ for some $n, \ell \in \mathbb{N}\text{.}$ Then $(n, n\pi) = (\ell, \ell\pi)\text{,}$ and in particular, the first entries in the pair must match, so $n=\ell\text{.}$ Thus, $f$ is injective.

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Now, consider $a\in A\text{.}$ By the definition of $A\text{,}$ this element takes the form $a = (m, \pi m)$ for some $m\in \mathbb{N}\text{.}$ Therefore $f(m) = a\text{,}$ and we see that $f$ is surjective.

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We conclude that $|A|=|\mathbb{N}|\text{.}$

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12.7.3.

Solution.

Proof.

We show that the function

\begin{equation*} f(m,n) = 2^{m-1}(2n-1), \end{equation*}

is a bijection, and so show that $\mathbb N\times\mathbb N$ is denumerable.

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We first show that the function is injective. So assume that $f(m_1,n_1)=f(m_2,n_2)\text{,}$ then $2^{m_1-1}(2n_1-1)=2^{m_2-1}(2n_2-1)\text{.}$ WLOG, assume $m_1\geq m_2\text{.}$ Then $2^{m_1-m_2} (2n_1-1) = 2n_2-1\text{.}$ If $m_1\not=m_2\text{,}$ then the left side is even but this contradicts the right side is odd. Thus, $m_1=m_2\text{.}$ Then $2n_1-1=2n_2-1\text{,}$ hence $n_1=n_2\text{.}$ We conclude that $f$ is injective.

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Now we show that $f$ is also surjective. Let $k\in \mathbb N\text{.}$ Then we can write $k$ as a nonnegative power of 2 times an odd number, namely, $k = 2^n (2q-1)$ where $n\geq 0$ is an integer and $q\in\mathbb N\text{.}$ To see that this expression for $k$ is unique, assume $k = 2^{n_1}(2q_1-1)=2^{n_2}(2q_2-1)\text{.}$ WLOG, assume $n_1\geq n_2\text{.}$ Then $2^{n_1-n_2}(2q_1-1)=2q_2-1\text{.}$ If $n_1 \gt n_2$ then the left side is even but the right side is odd, a contradiction. Thus $n_1=n_2$ and it follows $q_1=q_2\text{.}$ Now we have $n+1,q\in\mathbb N$ $f(n+1,q) = 2^{n}(2q-1)\text{.}$ So $f$ is surjective.

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Hence, since $f$ is a bijection, we see that $\mathbb N\times\mathbb N$ is denumerable.

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12.7.4.

Solution.

Proof.

Here we can use the fact that the set of prime numbers is countably infinite. Let $P$ be the set of prime numbers. We have seen that $P$ is countably infinite. Then we can list the elements of $P$ as $\{ p_1, p_2, p_3, \ldots \}$ and $\forall i\in\mathbb{N}$ we can define the sets

\begin{equation*} A_i=\{ p_i^n: n\in\mathbb{N} \}. \end{equation*}

We see by definition that $|A_i|=|\mathbb{N}|\text{.}$

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Moreover consider the set:

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$A_0=\{1\}\cup\{n\in\mathbb{N}: \exists p_i,p_j\in P, \text{ s.t. } p_i\neq p_j, \text{ and } p_i\mid n \text{ and } p_j\mid n\}\text{.}$ Then we see that $A_0$ is also countably infinite (since, say, it contains all the multiples $6$ and still a subset of the set of natural numbers).

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We also observe that there are countable infinitely many $A_k$’s, which are all countably infinite themselves and $\bigcup\limits_{n\geq 0} A_n=\mathbb{N}\text{.}$

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Another proof.

Another solution: We know that we have a bijection, $f:\mathbb N\times\mathbb N\rightarrow\mathbb N\text{,}$ defined as $f(n,m) = 2^{n-1}(2m-1)\text{.}$ Then, $\forall n\in\mathbb N\text{,}$ we can define the set $A_n=\set{2^{n-1}z: z \text{ is odd}}\text{.}$ Then we see that $A_n$ is countably infinite for all $n\text{,}$ there are countable many of such $A_n$’s, where $A_k\cap A_l=\emptyset$ if $k\neq l\text{,}$ and $\bigcup\limits_{n\geq 0} A_n=\mathbb{N}\text{.}$

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12.7.5.

Solution.

Proof.

First we should check that $f$ is well-defined; in particular, if $n\in\mathbb{N}\text{,}$ then $f(n)\in\mathbb{Z}\text{.}$ If $n$ is odd, then $2$ divides $1-n\text{,}$ and so

\begin{equation*} f(n)=\frac{1-n}{2}\in\mathbb{Z}. \end{equation*}

If $n$ is even, then $2$ divides $n\text{,}$ and so

\begin{equation*} f(n)=\frac{n}{2}\in\mathbb{Z}. \end{equation*}

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Now we need to check that $f$ is a bijection.

Surjective: Let $m\in\mathbb{Z}\text{.}$ First suppose that $m\leq0\text{.}$ Since $m$ is an integer, $1-2m\in\mathbb{Z}\text{.}$ Moreover, since $-2m\geq 0\text{,}$ we have $1-2m\geq 1\text{.}$ Thus $1-2m\in\mathbb{N}\text{.}$ Furthermore,$1-2m=2(-m)+1$ is odd, and hence

\begin{equation*} f(1-2m)=\frac{1-(1-2m)}{2}=\frac{2m}{2}=m. \end{equation*}

Next suppose that $m\geq1\text{.}$ Then $2m\in\mathbb{N}\text{,}$ and $2m$ is even. Hence

\begin{equation*} f(2m)=\frac{2m}{2}=m. \end{equation*}

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Injective: Suppose $f(m)=f(k)\text{.}$ If $f(m)=f(k)$ is non-positive, then both $m$ and $k$ are odd, since $f(\ell)\gt0$ for even $\ell\text{.}$ In this case, we have

\begin{equation*} \frac{1-m}{2}=f(m)=f(k)=\frac{1-k}{2} \end{equation*}

implying $m=k\text{.}$

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If $f(m)=f(k)$ is positive, then both $m$ and $k$ are even, since $f(\ell)\leq0$ for odd $\ell\text{.}$ In this case,

\begin{equation*} \frac{m}{2}=f(m)=f(k)=\frac{k}{2} \end{equation*}

and so $m=k\text{.}$ We have reached this conclusion in all cases, so $f$ is injective.

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The argument for surjectivity is similar as part (e): if $x\in \set{a q:q\in\mathbb{Q}}\text{,}$ then $x=a q$ for some $q\in \mathbb{Q}\text{,}$ and $g(q)=x\text{.}$ However, we have to be careful with injectivity. Suppose $g(q)=g(r)\text{,}$ in which case $a q=ar\text{.}$ This implies that $q=r$ as long as $a\neq0\text{.}$

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Thus, we see that when $a\neq0\text{,}$ $g$ is a bijection and $\set{a q:q\in\mathbb{Q}}$ is denumerable. If $a=0\text{,}$ then $\set{a q:q\in\mathbb{Q}}=\set{0}\text{,}$ and so the set is finite, not denumerable.

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12.7.7.

(Injectivity) Suppose that $f, h \in S$ such that $G(f) = G(h)\text{.}$ Then we have

\begin{equation*} \set{n: f(n)=1 } = \set{n: h(n)=1 }, \end{equation*}

So the functions $f,h$ take the value $1$ at the same inputs. At any other value of $n$ the functions must be zero. Hence they take the value $1$ at the same times and the value $0$ at the same time. So $f = h\text{.}$

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(Surjectivity) Suppose that $X \in \mathcal{P}(\mathbb{N})$ and so $X \subseteq \mathbb{N}\text{.}$ Define $f: \mathbb{N} \rightarrow \set{0, 1}$ by

\begin{equation*} f(n) = \begin{cases} 1 \quad \text{if } n \in X \\ 0 \quad \text{otherwise.} \end{cases} \end{equation*}

Then we have $G(f) = X$ by definition.

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12.7.10.

Solution.

First proof.

Let $h:(0,1) \to \mathbb{R}$ be defined by

\begin{equation*} h(x)=\begin{cases} \frac{1}{x}-2 \amp \text{if } 0\lt x \leq \frac{1}{2} \\ \frac{1}{x-1}+2 \amp \text{if } \frac{1}{2}\lt x\lt 1 \end{cases} \end{equation*}

In order to prove that this is bijective, we show that it has an inverse function, namely

\begin{equation*} h^{-1}(x)=\begin{cases} \frac{1}{x+2} \amp \text{if } x\geq0 \\ \frac{1}{x-2}+1 \amp \text{if } x\lt 0 \end{cases} \end{equation*}

It is then a straightforward (and slightly tedious) computation to show that $h \circ h^{-1} = id$ and $h^{-1} \circ h = id\text{.}$

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Second proof.

Let $y:(0,1)\rightarrow\mathbb{R}$ defined by

\begin{equation*} y(x) = \frac{1}{x}-\frac{1}{1-x}=\frac{1-2x}{x(1-x)} \end{equation*}

We prove that this is a bijection.

Injective: Suppose that $y(a)=y(b)\text{,}$ with $a,b\in(0,1)\text{.}$ Then

\begin{equation*} \frac{1-2a}{a(1-a)}=\frac{1-2b}{b(1-b)} \implies (1-2a)b(1-b)=(1-2b)a(1-a) \end{equation*}

Expanding and simplifying, we have

\begin{equation*} b-b^2+2ab^2=a-a^2+2a^2b. \end{equation*}

We want to show that $a=b\text{,}$ so we’d like to rewrite this expression with a factor of $a-b\text{.}$ We write

\begin{align*} 0 \amp = a-b-(a^2-b^2)+2a^2b-2ab^2 \\ 0 \amp = a-b-(a-b)(a+b)+2ab(a-b) \\ 0 \amp = (a-b)(1-(a+b)+2ab). \end{align*}

Now, we need to show that $1-(a+b)+2ab\neq0\text{,}$ as then $a-b=0\text{.}$ To this end, we’ll show that

\begin{equation*} 1+2ab\gt a+b. \end{equation*}

We will split the argument into two cases, $a+b\leq 1$ and $a+b\gt1\text{,}$ and use the assumption that $0\lt a,b\lt 1\text{.}$ If $a+b\leq 1\text{,}$ then $a+b\lt 1+2ab\text{,}$ as $a$ and $b$ are positive. If $a+b\gt 1\text{,}$ then either $a\gt 1/2$ or $b\gt 1/2\text{.}$ Without loss of generality, assume $a\gt 1/2\text{,}$ as we may relabel if necessary. Then $2a\gt 1\text{,}$ and as $b\gt0\text{,}$ we also have $2ab\gt b\text{.}$ Therefore

\begin{equation*} 1+2ab \gt 1+b\gt a+b \end{equation*}

with the last two inequality following from $1\gt a\text{.}$ In all cases, we have $1+2ab\gt a+b\text{,}$ and so $1-(a+b)+2ab\neq0\text{.}$ As mentioned earlier, this implies that $a=b\text{,}$ and so $y$ is injective.

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Surjective: Let $d\in \mathbb{R}\text{.}$ We need to find $c\in(0,1)$ such that $y(c)=d\text{.}$ That is, we need to show that there is such a $c$ that solves the equation

\begin{equation*} d=\frac{1-2x}{x(1-x)}. \end{equation*}

This equation simplifies to

\begin{equation*} dx^2-(d+2)x+1=0. \end{equation*}

If $d=0\text{,}$ then we have the solution $x=1/2\text{.}$ If $d\neq0\text{,}$ by the quadratic formula, the equation has solutions

\begin{equation*} x=\frac{d+2\pm\sqrt{(d+2)^2-4d}}{2d}=\frac{d+2\pm\sqrt{d^2+4}}{2d}. \end{equation*}

By Exercise 3.5.16, we know that

\begin{equation*} \sqrt{d^2+2}\geq \sqrt{d^2}=|d| \end{equation*}

Suppose $d\gt0\text{.}$ Then

\begin{equation*} d+\sqrt{d^2+4}\geq d+d=2d, \end{equation*}

implying that

\begin{equation*} \frac{d+2+\sqrt{d^2+4}}{2d}\geq\frac{2d+2}{2d}\gt1, \end{equation*}

so this candidate for $c$ doesn’t work, as this solution lies outside the interval $(0,1)\text{.}$ Therefore, we aim to show that

\begin{equation*} 0 \lt\frac{d+2-\sqrt{d^2+4}}{2d}\lt1. \tag{*} \end{equation*}

We’ll split the proof of this into the cases $d\gt0$ and $d\lt0\text{.}$

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First assume $d\gt0\text{.}$ Then (*) is equivalent to the inequality

\begin{equation*} 0 \lt d+2-\sqrt{d^2+4}\lt 2d. \tag{**} \end{equation*}

By Exercise 3.5.18, we know that

\begin{equation*} \sqrt{d^2+4}\lt \sqrt{d^2}+\sqrt{4}=d+2. \end{equation*}

Therefore

\begin{equation*} d+2-\sqrt{d^2+4}\gt d+2-(d+2)= 0. \end{equation*}

Moreover, by Exercise 3.5.16, we know that $d^2+4\geq 4$ implies

\begin{equation*} \sqrt{d^2+4}\geq \sqrt{4}=2. \end{equation*}

Thus

\begin{equation*} d+2-\sqrt{d^2+4}\leq d+2-2 \lt 2d. \end{equation*}

So we have established (**), and so (*), for the case $d\gt 0\text{.}$

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Now assume $d\lt0\text{.}$ Then (*) is equivalent to the inequality

\begin{equation*} 0 \gt d+2-\sqrt{d^2+4}\gt 2d. \tag{$\dagger$} \end{equation*}

By Exercise 3.5.18, we know that

\begin{equation*} \sqrt{d^2+4}\lt \sqrt{d^2}+\sqrt{4}=|d|+2=-d+2. \end{equation*}

Therefore

\begin{equation*} d+2-\sqrt{d^2+4}\gt d+2-(-d+2)=2d. \end{equation*}

Moreover, by Exercise 3.5.16, we know that $d^2+4\geq 4$ implies

\begin{equation*} \sqrt{d^2+4}\geq \sqrt{4}=2\gt 2+d, \end{equation*}

with the last inequality holding as $d\lt 0\text{.}$ Thus

\begin{equation*} d+2-\sqrt{d^2+4}\lt d+2-(2+d)=0. \end{equation*}

So we have established ($\dagger$), and so (*), for the case $d\lt 0\text{.}$

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We have found a solution $c\in(0,1)$ to the equation $y(c)=d$ in all cases, and so $y$ is surjective.
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Third proof.

We show that both sets are equinumerous to the interval $(1,\infty)\text{.}$

First we find a bijection $f:\mathbb{R}\rightarrow(1,\infty)\text{.}$ A function that almost works is the exponential function, $2^x\text{;}$ the issue is its range is $(0,\infty)$ rather than $(1,\infty)\text{.}$ We can resolve this by taking $f(x)=2^x+1\text{.}$ Moreover, we know that $f$ has an inverse, $f^{-1}:(1,\infty)\rightarrow\mathbb{R}\text{,}$ given by $f^{-1}(x)=\log_2(x-1)\text{.}$ Then by Theorem 10.6.8, $f$ is bijective. Thus $\mathbb{R}$ and $(1,\infty)$ are equinumerous.
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Next, we want to find a bijection $g:(1,\infty)\rightarrow(0,1)\text{.}$ Take $g(x)=1/x\text{.}$ This function is well-defined, as for $x \gt 1\text{,}$ we have $0 \lt 1/x \lt \infty\text{.}$ Moreover, its inverse is given by $g^{-1}:(0,1)\rightarrow(1,\infty)\text{,}$ with $g^{-1}(x)=1/x\text{.}$ Again by Theorem 10.6.8, $g$ is bijective. Thus $(1,\infty)$ and $(0,1)$ are equinumerous.
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Combining the two statements we proved, and appealing to Theorem 12.1.8, we can conclude that $\mathbb{R}$ and $(0,1)$ are equinumerous, as desired. Indeed, the function $g(f(x))=1/(2^x+1)$ is a bijection from $\mathbb{R}$ to $(0,1)\text{.}$

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12.7.11.

Solution.

Proof of (a).

We know that the function $g:\mathbb{R}\rightarrow (0,\infty)$ defined by $g(x)=e^x$ is a bijective function. Using this function we can define a bijection $f:\mathbb{R}\rightarrow (\sqrt{2},\infty)$ as $f(x)=g(x)+\sqrt2\text{.}$

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To see that $f$ is injective, consider $x, y \in \mathbb{R}$ such that $f(x) = f(y)\text{.}$ Then by the definition of $f\text{,}$ $g(x)+\sqrt{2} = g(y)+\sqrt{2}\text{.}$ Subtracting $\sqrt{2}$ from each side, we see that we must have $g(x) = g(y)\text{.}$ By the injectivity of $g\text{,}$ we see that $x=y\text{,}$ and conclude that $f$ is injective.

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Moreover, consider $z\in (\sqrt{2}, \infty)\text{.}$ Then $z-\sqrt{2}\in (0, \infty)\text{.}$ Since $g$ is bijective, there exists a value $x\in \mathbb{R}$ such that $g(x) = z-\sqrt{2}\text{.}$ Adding $\sqrt{2}$ to each side, we see that $z = g(x) + \sqrt{2} = f(x)\text{,}$ so $f$ is surjective.

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We conclude that $|\mathbb{R}| = |(\sqrt{2}, \infty)|\text{.}$

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Proof of (b).

Let $O$ denote the set of odd integers and $E$ denote the set of even integers. Then we can define the function $f:O\rightarrow E$ as $f(x)=x+1\text{.}$

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To see that $f$ is injective, consider $x,y \in O$ such that $f(x)=f(y)\text{.}$ Then $x+1=y+1\text{.}$ Subtracting $1$ from each side, we see that $x=y\text{.}$ Also, for any even integer $y\text{,}$ we see $y-1$ is odd and $f(y-1)=y\text{,}$ which implies that $f$ is surjective. Therefore $f$ is bijective, and we see that $|O| = |E|$

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Proof of (c).

We see that $S = \{x \in\mathbb{R}: \sin x = 1\}=\{ x\in\mathbb{R}: x=\pi/2+2\pi n \text{ for some } n\in\mathbb{Z} \}\text{.}$ Thus we can define a bijection $f: \mathbb{Z}\rightarrow S\text{,}$ where $f(k)=\dfrac{\pi}{2}+2\pi k\text{.}$ Consider $i,j \in \mathbb{Z}$ such that $f(i)=f(j)\text{.}$ Then by the definition of $f\text{,}$ $\dfrac{\pi}{2}+2\pi i = \dfrac{\pi}{2}+2\pi j\text{.}$ Subtracting $\dfrac{\pi}{2}$ from each side gives $2\pi i = 2\pi j\text{.}$ Finally, dividing by $2\pi\text{,}$ we see that $i=j\text{,}$ so $f$ is injective.

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Now, consider $s\in S\text{.}$ As we noticed above, $s = \dfrac{\pi}{2} + 2\pi k$ for some $k\in \mathbb{Z}\text{.}$ Thus, $f(k)=s\text{,}$ and we see that $f$ is a surjection.

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Hence $f$ is bijective, and we conclude that $|\mathbb{Z}| = |S|\text{.}$

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Proof of (d).

Define a function $f: \mathbb{Z}\rightarrow \set{0,1} \times\mathbb{N}\text{,}$ by

\begin{equation*} f(k)= \begin{cases} (1,k+1) \amp \text{ if } k\geq 0 \\ (0,-k) \amp \text{ if } k\leq -1 \end{cases} \end{equation*}

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Then we see that for $k,m\in\mathbb{Z}\text{,}$ if $f(k)=f(m)\text{,}$ then either $k,m\leq -1$ or $k,m\geq 0\text{.}$ If $k,m\geq 0$ we have $(1,k+1)=(1,m+1)$ in which case $k=m\text{.}$ If $k,m\leq -1$ we have $(0,-k)=(0,-m)\text{,}$ in which case $k=m$ too. We also see that these two are the only two cases since if $k\geq 0$ and $m\leq -1$ or $m\geq 0$ and $k\leq -1$ we see $f(k)\neq f(m)\text{.}$

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We also see that $f$ is surjective. If $(a,n)\in(0,1)\times\mathbb{N}$ we have two cases: $a=0$ or $a=1\text{.}$ If $a=0\text{,}$ we can define $k=-n$ so that $f(-k)=(0,n)$ and if $a=1\text{,}$ we ca define $k=n-1$ so that $f(k)=(1,n)\text{.}$ We conclude that $f$ is a bijection, and therefore $|\set{0,1}\times\mathbb{N}| = |\mathbb{Z}|\text{.}$

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12.7.12.

Solution.

Proof.

Since $|A|=|B|$ and $|C|=|D|\text{,}$ there are bijections $f:A\to B$ and $g:C\to D\text{.}$ Since $A\cap C=\emptyset\text{,}$ any $x\in A\cup C\text{,}$ either $x\in A$ or $x\in C$ but not in both. Define $F:A\cup C\to B\cup D$ by

\begin{equation*} F(x) = \begin{cases} f(x) \amp \text{ if } x \in A \\ g(x) \amp \text{ if } x \in C. \end{cases} \end{equation*}

And since $A\cap C = \emptyset$ this is well defined.

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We now show $F$ is bijective. To see $F$ is injective: assume $F(x)=F(y)\text{.}$

$x,y\in A\text{.}$ So $f(x)=f(y)$ hence $x=y$ as $f$ is injective.
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$x,y\in C\text{.}$ So $g(x)=g(y)$ hence $x=y$ as $g$ is injective.
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$x\in A$ and $y\in C$ (or similarly $x\in C$ and $y\in A$), $F(x) = f(x) \in B$ and $F(y)=g(y)\in D\text{.}$ But $B\cap D=\emptyset\text{,}$ which gives us a contradiction. So this case cannot happen.
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Since we always have $x=y\text{,}$ we conclude $F$ is injective.

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To see $F$ is surjective: For any $y\in B\cup D\text{,}$ %since $B\cap D=\emptyset\text{,}$ $y\in B$ or $y\in D$ but not in both.

$y\in B\text{.}$ There is $x\in A$ s.t. $f(x)=y$ since $f$ is surjective. So $F(x)=f(x)=y\text{.}$
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$y\in D\text{.}$ There is $x\in C$ s.t. $g(x)=y$ since $g$ is surjective. So $F(x)=g(x)=y\text{.}$
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We conclude $F$ is surjective. Hence $F$ is bijective and $|A\cup C|=|B\cup D|\text{.}$

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12.7.13.

Solution.

Proof.

We can define such a bijection, $f:(0,\infty) \to (0,\infty)-\set{1}\text{,}$ as follows:

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\begin{equation*} f(x)=\left\{ \begin{array}{lll} x+1 \amp \amp x\in \mathbb N \\ x \amp \amp \text{ otherwise}. \end{array} \right. \end{equation*}

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Now, we need to show that $f$ is indeed a bijective function.

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Show $f$ is surjective: Let $y\in (0,\infty)-\set{1}= (0,1)\cup (1,\infty)\text{.}$ Then we have 2 cases.

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Case 1: $y\in \mathbb N\text{,}$ $y\neq 1\text{,}$ that is, $y\geq 2\text{.}$ Then, we see that we can take $x=y-1$ and get $f(x)=x+1=y\text{.}$

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Case 2: $y\notin \mathbb N\text{.}$ Then, we see that we can take $x=y$ and get $f(x)=x=y\text{.}$

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Show $f$ is injective: Let $x,y\in (0.\infty)\text{,}$ such that $f(x)=f(y)\text{.}$ Then we know that if $f(x)\in\mathbb N\text{,}$ then $f(y)\in\mathbb N\text{,}$ which implies that $x,y \in\mathbb N$ and since for $x,y\notin\mathbb N\text{,}$ we see that $f(x)=x\notin\mathbb N$ and $f(y)=y\notin\mathbb N\text{.}$

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Hence, we see that $x=f(x)-1=f(y)-1=y\text{.}$

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Moreover, we see that if $f(x),f(y)\notin\mathbb N\text{,}$ we see that $x=f(x)=f(y)=y\text{.}$

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Therefore $f$ is injective, which in turn implies that $f$ is bijective.

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12.7.14.

Solution.

Proof of (a).

We will prove this statement using Cantor-Schröder-Bernstein 12.5.1, defining an injection from $(0,1)$ to $(0,1)\times(0,1)\text{,}$ and another injection in the reverse direction. This is easier than trying to define an explicit bijection from one set to the other; as we will see, the fact that some numbers have two distinct decimal representations causes trouble if we tried that approach instead.

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First let’s define an injection $f:(0,1)\rightarrow (0,1)\times(0,1)\text{.}$ We can take

\begin{equation*} f(x)=(x,1/2), \end{equation*}

so that $f$ maps the interval $(0,1)$ onto a strip contained in the square $(0,1)\times(0,1)\text{.}$

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Next, we try to define an injection $g:(0,1)\times(0,1)\rightarrow (0,1)\text{.}$ This direction is trickier. In Section 12.3, we saw that any element $a\in (0,1)$ can be written as a decimal expansion,

\begin{equation*} a = 0.a_1a_2a_3\dots \quad a_i\in\{0,1,\dots,9\}. \end{equation*}

Moreover, we discussed that this representation is unique, with the exception of numbers whose digits are eventually all zero or nine, such as

\begin{equation*} 1/2 = 0.5000\dots = 0.4999\dots \end{equation*}

In these cases, we choose the representation ending in all zeros. Therefore, none of the decimal expansions will end in all repeating nines.

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Let $(a,b)\in (0,1)\times(0,1)\text{,}$ which we write as

\begin{equation*} (a,b)=(0.a_1a_2a_3\dots, 0.b_1b_2b_3\dots). \end{equation*}

We want to map this to a single element in $(0,1)\text{.}$ We’ll identify this element by its own decimal representation, which will be defined in terms of the digits $a_i$ and $b_i\text{.}$ Since there are infinitely digits in the expansions of $a$ and $b\text{,}$ we’ll need to combine their digits by going back and forth between the $a_i$ and $b_i\text{.}$ Define

\begin{equation*} g((a,b))=0.a_1b_1a_2b_2a_3b_3\dots \end{equation*}

We need to check that this map is injective.

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Injectivity will follow in most cases by the uniqueness of the decimal representation. However, we need to be careful for numbers that have two representations, like $1/2\text{.}$ In such a case, one of these representations ends in repeating nines. In order for $0.a_1b_1a_2b_2a_3b_3\dots$ to end in repeating nines, the digit expansions chosen for $a$ and $b$ would also need to end in repeating nines. But this can’t happen, since we already made the choice to not use any decimal expansions ending in all nines, opting for the expansion ending in zeros instead.

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Notice that the function $g$ isn’t a surjective. For example, the element

\begin{equation*} (0.5, 0.1119191919\dots) \end{equation*}

is not in the range of $g\text{;}$ indeed, if it were, then we would need $b=0.1999\dots\text{,}$ but for this number we use the representation $0.200\dots$ instead.

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Proof of (b).

From Exercise 12.7.10, we know that $|(0,1)|=|\mathbb{R}|\text{.}$ So there is a bijection $h:(0,1)\to\mathbb{R}\text{.}$ We can use $h$ to form a bijection $H$ from $(0,1)\times(0,1)$ to $\mathbb{R}\times\mathbb{R}\text{.}$ Indeed, let

\begin{equation*} H((x,y))=(h(x),h(y)). \end{equation*}

Since $h$ is invertible, $H$ is as well; namely,

\begin{equation*} H^{-1}((x,y))=(h^{-1}(x),h^{-1}(y)). \end{equation*}

Thus $H$ is indeed bijective. Therefore

\begin{equation*} |(0,1)\times(0,1)|=|\mathbb{R}\times\mathbb{R}|. \end{equation*}

From part (a), we know that

That is, infinite sequences of natural numbers from $0$ to $9$ which do not end in infinitely many consecutive $9$’s. We will first show that $\mathcal{B}$ is uncountable by constructing a bijection from $\mathcal{B}$ to $[0,1)\text{.}$ Define $f: \mathcal{B}\to [0,1)$ by

\begin{equation*} f((b_1, b_2, b_3, b_4, b_5, \dots)) = 0.b_1b_2b_3b_4b_5\dots. \end{equation*}

This functions takes the $i^{th}$ coordinate of $(b_1, b_2, b_3, b_4, b_5, \dots)$ and sends it to the $i^{th}$ decimal place following $0\text{.}$ For example

\begin{equation*} f((1,2,3,4,5,6, \dots))= 0.123456\dots. \end{equation*}

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To see that $f$ is injective, let $x= (x_1, x_2, x_3, \dots)$ and $y=(y_1, y_2, y_3, \dots)$ be elements of $\mathcal{B}$ and suppose that $f(x) = f(y)\text{.}$ Then

\begin{equation*} 0.x_1x_2x_3\dots = 0.y_1y_2y_3\dots. \end{equation*}

Since our sequence does not terminate in 9’s, each real number in $[0,1)$ has a unique decimal representation, and we must have

\begin{align*} x_1 \amp = y_1 \amp x_2 \amp =y_2 \amp x_3 \amp = y_3 \amp \cdots \end{align*}

Therefore $x=y$ and $f$ is injective. It remains to show that $f$ is surjective. Suppose $s\in [0,1)\text{.}$ Then $s$ takes the form

\begin{equation*} 0.s_1s_2s_3s_4s_5\dots \end{equation*}

where $s_j\in \{0, 1, \dots, 9\}\text{.}$ Since

\begin{equation*} f((s_1, s_2, s_3, s_4, s_5, \dots)) = 0.s_1s_2s_3s_4s_5\dots \end{equation*}

we have found an element of $\mathcal{B}$ mapping to $s\text{,}$ so $f$ is surjective. Therefore, $|\mathcal{B}| = |[0,1)|\text{,}$ so $\mathcal{B}$ is uncountable. Since $\mathcal{B} \subset \mathcal{A}\text{,}$ $\mathcal{A}$ is uncountable as well.

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Alternative proof.

We are going to prove this set is uncountable by showing that it has an uncountable subset. Let $\mathcal{A} = \set{(a_1, a_2, a_3, \ldots): a_i \in\mathbb{Z}}\text{,}$ and let $\mathcal{C}=\{(c_1, c_2, c_3, \ldots): c_i \in\set{0,1}\}\text{.}$ Then, we see that $\mathcal{C}\subset \mathcal{A}\text{.}$ Now, we can show that $\mathcal{C}$ is uncountable by finding a bijection from $\mathcal{C}$ to $\mathcal{P}(\mathbb{N})\text{.}$

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We can define the bijection $f$ as follows:

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\begin{align*} f:\mathcal{C} \amp \rightarrow \mathcal{P}(\mathbb{N})\\ f(c_1, c_2, c_3, \ldots) \amp =\{ k\in\mathbb{N}: c_k=1 \}. \end{align*}

for example, $f(0,1,0,1,0,0,0,0,\ldots)=\{2,4\}\text{.}$

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We see that $f$ is surjective since for any $X\subseteq\mathbb{N}\text{,}$ we can construct the sequence $(x_1, x_2, x_3, \ldots)\text{,}$ where $x_k=1$ if $k\in X$ and $0$ otherwise. Then, $f(x_1, x_2, x_3,\ldots)=X\text{.}$

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We also see that $f$ is injective since if $(a_1, a_2, a_3, \ldots)\neq (b_1, b_2, b_3, \ldots)\text{,}$ then $\exists m\in\mathbb{N}$ such that $a_m=1$ and $b_m=0\text{,}$ in which case $m\in f(a_1, a_2, a_3, \ldots)$ and $m\notin f(b_1, b_2, b_3, \ldots)\text{,}$ or $a_m=0$ and $b_m=1\text{,}$ in which case $m\notin f(a_1, a_2, a_3, \ldots)$ and $m\in f(b_1, b_2, b_3, \ldots)\text{.}$ In both cases, we see $f(a_1, a_2, a_3, \ldots)\neq f(b_1, b_2, b_3, \ldots)\text{.}$ We conclude that $f$ is a bijection. Therefore we see that $|\mathcal{C}|=| \mathcal{P}(\mathbb{N})|\text{,}$ which means $\mathcal{C}$ is uncountable. Thus, $\mathcal{A}$ is uncountable too.

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Note: We can also use a diagonal argument directly to show that the set $\{(a_1, a_2, a_3, \ldots): a_i \in\mathbb Z\}$ is uncountable.

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12.7.17.

Solution.

Proof.

For each $X\in P\text{,}$ since $X$ and $h(X)$ have the same cardinality, there exists a bijection $g_X\colon X\to h(X)\text{.}$ Define $f$ to be the union of all these $g_X\text{:}$

\begin{equation*} f = \{ (a,b)\colon \text{there exists } X\in P \text{ such that } a\in X \text{ and } b = g_X(a) \}. \end{equation*}

Of course, it is not immediately obvious that this is a function, let alone that it is a bijection. We must prove that.

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We show that $f$ is a function from $A$ to $B\text{,}$ that it is surjective, and that it is injective.

For any $(a,b)\in f\text{,}$ choose $X\in P$ such that $a\in X$ and $b=g_X(a)\text{.}$ Since $X\in P$ and $P$ is a partition of $A\text{,}$ we know that $X\subset A\text{,}$ and therefore $a\in A\text{.}$ Similarly, $g_X(a)\in h(X)$ and $h(X)\in Q\text{;}$ since $Q$ is a partition of $B\text{,}$ we see that $h(X) \subset B\text{,}$ and so $b=g_X(a)\in B\text{.}$ This shows that $f\subset A\times B\text{.}$
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Let $a\in A$ be arbitrary. Since $P$ is a partition of $A\text{,}$ there is a unique $X\in P$ with $a\in X\text{.}$ Since $g_X$ is a function from $X$ to $h(X)\text{,}$ there is a unique $b$ such that $g_X(a)=b\text{.}$ This shows that there is a unique $b\in B$ with $f(a)=b\text{,}$ and so $f$ is a function.
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Let $b\in B$ be arbitrary. Since $Q$ is a partition of $B\text{,}$ there exists $Y\in Q$ with $b\in Y\text{.}$ Since $h$ is surjective, there exists $X\in P$ with $h(X)=Y\text{,}$ and therefore $g_X$ is a function from $X$ to $Y\text{.}$ Since $g_X$ is surjective, there exists $a\in X$ such that $g_X(a)=b\text{.}$ By definition, we see that $f(a) = b\text{.}$ Therefore $f$ is surjective.
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Let $a_1,a_2\in A\text{,}$ and suppose that $f(a_1)=f(a_2)\text{.}$ Since $Q$ is a partition of $B\text{,}$ there is a unique $Y\in Q$ with $f(a_1)=f(a_2)\in Y\text{.}$ Since $h\colon P\to Q$ is injective, there is a unique $X\in P$ with $h(X) = Y\text{.}$ This $X$ corresponds to the only function $g_X$ whose codomain is $Y\text{,}$ and so we must have $a_1,a_2\in X$ and $f(a_1) = g_X(a_1)$ and $f(a_2)=g_X(a_2)\text{.}$ Since $g_X$ is injective, we conclude that $a_1=a_2\text{.}$ Therefore $f$ is injective.
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\begin{align*} f: F_n \amp \rightarrow \mathbb{N}^n,\\ f(\{ a_1, a_2, a_3, \cdots, a_n \}) \amp =(a_{k_1}, a_{k_2}, a_{k_3}, \cdots, a_{k_n}), \end{align*}

where $\{ a_1, a_2, a_3, \cdots, a_n \}=\{a_{k_1}, a_{k_2}, a_{k_3}, \cdots, a_{k_n}\}$ and $a_{k_1} \lt a_{k_2} \lt a_{k_3} \lt \cdots \lt a_{k_n}\text{.}$ We see that $f$ is a well-defined function since if the sets $\{ a_1, a_2, a_3, \cdots, a_n \}=\{ b_1, b_2, b_3, \cdots, b_n \}\text{,}$ then $(a_{k_1}, a_{k_2}, a_{k_3}, \cdots, a_{k_n})=(b_{k_1}, b_{k_2}, b_{k_3}, \cdots, b_{k_n})\text{,}$ that is, when the $a_i$’s and $b_j$’s are arranged in ascending order, the corresponding $n$-tuples are equal.

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Moreover, if $f(\{ a_1, a_2, a_3, \cdots, a_n \})=f(\{ b_1, b_2, b_3, \cdots, b_n \})\text{,}$ then

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$(a_{k_1}, a_{k_2}, a_{k_3}, \cdots, a_{k_n})=(b_{k_1}, b_{k_2}, b_{k_3}, \cdots, b_{k_n})$ and since

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$\{ a_1, a_2, a_3, \cdots, a_n \}=\{a_{k_1}, a_{k_2}, a_{k_3}, \cdots, a_{k_n}\}$ and $\{ b_1, b_2, b_3, \cdots, b_n \}=\{b_{k_1}, b_{k_2}, b_{k_3}, \cdots, b_{k_n}\}$ we see $\{ a_1, a_2, a_3, \cdots, a_n \}=\{b_{1}, b_{2}, b_{3}, \cdots, b_{n}\}\text{,}$ meaning that $f$ is injective.

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We have also seen that the finite Cartesian product of countable sets is countable (via Result 12.2.6). Thus, we see $\mathbb{N}^n$ is countable. Therefore since $f$ is an injection from $F_n$ to $\mathbb{N}^n$ and since $\mathbb{N}^n$ is countable, we see $F_n$ is countable.

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Proof of (c).

and that $B_n\cap B_m=\emptyset$ for $n\neq m\text{.}$ First we show (1). Suppose that

\begin{equation*} x\in \bigcup_{n=1}^\infty A_n. \end{equation*}

Let

\begin{equation*} N = \min\{n:x\in A_n\}. \end{equation*}

Then $N$ exists, since $\{n:x\in A_n\}$ is non-empty (as $x$ is an element of the union, so must be in some $A_n$), and a subset of the natural numbers (so it has a minimum). Then

\begin{equation*} x\not\in \bigcup_{m=1}^{N-1} A_m \end{equation*}

as $x\not\in A_m$ for $1\leq m\leq N-1$ by choice of $N\text{.}$ Combining this with the fact that $x\in A_N\text{,}$ we have

\begin{equation*} x\in B_N = A_N\setminus\left(\cup_{m=1}^{N-1} A_m\right). \end{equation*}

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Therefore

\begin{equation*} x\in B_N\subset \bigcup_{n=1}^\infty B_n. \end{equation*}

Conversely, suppose

\begin{equation*} y\in \bigcup_{n=1}^\infty B_n. \end{equation*}

Then there is some $n$ such that

\begin{equation*} y\in B_n = A_n \setminus\left(\cup_{m=1}^{n-1} A_m\right) \end{equation*}

implying that $y\in A_n\text{.}$ Thus (1) is established.

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Now we show that the sets $B_n$ are disjoint. Let $m\neq n\text{.}$ Without loss of generality, we may assume that $m\lt n\text{,}$ as we may relabel if necessary. Then

\begin{equation*} B_m\subset A_m\subset \bigcup_{k=1}^{n-1} A_k. \end{equation*}

So, when choosing elements of $A_n$ to include in $B_n\text{,}$ we leave out any elements that appear in $B_m\text{.}$ That is,

\begin{equation*} B_m\cap B_n =B_m \cap \left(A_n\setminus \left(\cup_{k=1}^{n-1} A_k \right)\right) \end{equation*}

is empty.

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Finally, for all $n\in\mathbb{N}\text{,}$ $B_n\subset A_n\text{,}$ and so $|B_n|\leq |A_n|\text{.}$ Since $A_n$ is countable, $B_n$ is countable as well.

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By part (b), $\cup_{n=1}^\infty B_n $ is countable, and hence by (1), $\bigcup_{n=1}^\infty A_n $ is countable as well.

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12.7.21.

Solution.

Proof of (a).

We will show this by constructing a bijection. Let $f: P_m\rightarrow \left(\mathbb Q^{m}\times \left(\mathbb Q \setminus \set{0}\right)\right)$ be defined by

\begin{equation*} f(a_0+a_1x+a_2x^2+\cdots+a_mx^m)=\big((a_0,a_1,a_2,\cdots,a_{m-1}), a_m\big). \end{equation*}

Consider $\left((b_0, \dots, b_{m-1}), b_m\right)\in \left(\mathbb Q^{m}\times \left(\mathbb Q \setminus \set{0}\right)\right)\text{.}$ By construction,

\begin{equation*} f(b_0+b_1x+b_2x^2+\dots+b_mx^m) = \left((b_0, \dots, b_{m-1}), b_m\right), \end{equation*}

so we see $f$ is surjective.

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Moreover, let $(a_0+a_1x+a_2x^2+\cdots+a_mx^m),(b_0+b_1x+b_2x^2+\cdots+b_mx^m)\in P_m$ such that $f(a_0+a_1x+a_2x^2+\cdots+a_mx^m)=f(b_0+b_1x+b_2x^2+\cdots+b_mx^m)\text{.}$ Then $\big((a_0,a_1,a_2,\cdots,a_{m-1}), a_m\big)=\big((b_0,b_1,b_2,\cdots,b_{m-1}), b_m\big)\text{,}$ and we see that $a_i=b_i$ for all $i\in\set{0,1,2,\ldots,m}\text{.}$ Thus, $a_0+a_1x+a_2x^2+\cdots+a_mx^m=b_0+b_1x+b_2x^2+\cdots+b_mx^m\text{,}$ which implies that the function $f$ is injective.

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Therefore $|P_m|=\left|\left(\mathbb Q^{m}\times \left(\mathbb Q \setminus \set{0}\right)\right)\right|\text{.}$ Since the Cartesian product of finitely many countable sets is countable, we have $|P_m|=\left|\left(\mathbb Q^{m}\times \left(\mathbb Q \setminus \set{0}\right)\right)\right| = |\mathbb{N}|\text{.}$

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Proof of (b).

By part (a), $|P_m| = |\mathbb{N}|\text{.}$ Notice that $P=\bigcup\limits_{m\in\mathbb N}P_m\text{.}$ By exercise Exercise 12.7.20, the union of countably many countable sets is countable. Therefore, $|P| = |\mathbb{N}|\text{.}$

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Proof of (c).

Claim: $|A|=|P|\text{.}$

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Let $m\in \mathbb{N}\text{.}$ By part (a), the polynomials of degree $m$ are countable. Therefore, they are in bijection with $\mathbb{N}\text{,}$ and we may enumerate them as $p_{m, 1}, p_{m,2}, p_{m,3}, \dots\text{.}$ That is, any polynomial in $P_m$ can be described as $p_{m,k}$ for $k\in \mathbb{N}\text{,}$ where $m$ refers to the degree of the polynomial and $k$ comes from the enumeration of $P_m\text{.}$ We define the set

\begin{equation*} R_{p_{m,k}}=\set{x\in\mathbb R: p_{m,k}(x)=0} \end{equation*}

to be the set of real roots of the polynomial $p_{m,k}\text{.}$ By the Fundamental Theorem of Algebra, we know that every polynomial of order $m$ has at most $m$ distinct real roots. Therefore for any $k\in\mathbb N\text{,}$ $|R_{p_{m,k}}|\leq m\text{,}$ in particular, $R_{p_{m,k}}$ is countable.

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Thus, we see that $A_m=\bigcup\limits_{k\in\mathbb N}R_{p_{m,k}}$ gives us the set of all possible roots of polynomials of order $m\text{.}$ Notice that $A_m$ is the countable union of countable set, and thus, by Exercise 12.7.20, is countable.

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Again, using Exercise 12.7.20, since $A=\bigcup\limits_{m\in\mathbb N} A_m\text{,}$ we see that $A$ is also countable union of countable sets, and hence is countable.

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Moreover, since any $q\in\mathbb Q$ is a zero of a polynomial with rational coefficients, e.g. $f(x)=x-q\text{,}$ we see that $\mathbb Q\subseteq A\text{.}$ This implies that $A$ is infinite. Thus, we see that $A$ is countably infinite.

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Therefore $|A|=|P|\text{.}$

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(This is a very important result. These numbers that are solutions to rational (or equivalently integer) polynomials are called algebraic numbers. As an example we see $\sqrt 2$ is irrational, but is the solution to the equation $x^2-2=0\text{.}$ Thus, since the set of algebraic numbers is countable and the set of real numbers is uncountable, we see that there has to be uncountable many real numbers that cannot be written as a solution to a polynomial with rational coefficients. We call such numbers transcendental numbers. For example, $\pi$ is a transcendental number-as one can guess, showing a number is transcendental is generally a very hard question.)

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Prev Top Next

\(P\)	\(Q\)	\(\neg P\)	\((\neg P)\implies Q\)
T	T	F	T
T	F	F	T
F	T	T	T
F	F	T	F

\(P\)	\(Q\)	\((\neg P)\implies Q\)	\(P\land Q\)	\((P\land Q)\lor \big((\neg P)\implies Q \big)\)
T	T	T	T	T
T	F	T	F	T
F	T	T	F	T
F	F	F	F	F

\(P\)	\(Q\)	\(P\implies Q\)	\(Q\implies P\)	\((P\implies Q) \iff (Q\implies P) \)
T	T	T	T	T
T	F	F	T	F
F	T	T	F	F
F	F	T	T	T

\(P\)	\(Q\)	\(P\implies Q\)	\(\neg(P\implies Q)\)	\(P\land\neg Q\)
T	T	T	F	F
T	F	F	T	T
F	T	T	F	F
F	F	T	F	F

\(P\)	\(Q\)	\(({\neg} P)\)	\(({\neg} P) \lor Q\)	\(P \implies Q\)
T	T	F	T	T
T	F	F	F	F
F	T	T	T	T
F	F	T	T	T

\(P\)	\(Q\)	\({\neg} P\)	\({\neg} Q\)	\(P \Leftrightarrow Q\)	\(({\neg} P) \Leftrightarrow ({\neg} Q)\)
T	T	F	F	T	T
T	F	F	T	F	F
F	T	T	F	F	F
F	F	T	T	T	T

\(P\)	\(Q\)	\(R\)	\(Q\lor R\)	\(({\neg} Q)\Rightarrow R\)	\(P \Rightarrow (Q\lor R)\)	\(P \Rightarrow (({\neg} Q)\Rightarrow R)\)
T	T	T	T	T	T	T
T	T	F	T	T	T	T
T	F	T	T	T	T	T
F	T	T	T	T	T	T
T	F	F	F	F	F	F
F	T	F	T	T	T	T
F	F	T	T	T	T	T
F	F	F	F	F	T	T

Appendix C Solutions to Exercises

1 Sets1.4 Exercises

1.4.1.

1.4.2.

1.4.3.

1.4.4.

1.4.5.

1.4.6.

1.4.7.

1.4.8.

1.4.9.

2 A little logic2.7 Exercises

2.7.1.

2.7.2.

2.7.3.

Proof.

2.7.4.

2.7.5.

2.7.6.

2.7.7.

2.7.8.

2.7.9.

2.7.10.

2.7.11.

2.7.12.

2.7.13.

2.7.14.

2.7.15.

2.7.16.

3 Direct proofs3.5 Exercises

3.5.1.

Proof.

3.5.2.

Proof.

3.5.3.

Proof.

Proof.

Proof.

Proof.

Proof.

3.5.4.

3.5.5.

Proof.

3.5.6.

Proof.

3.5.7.

Proof.

3.5.8.

First proof.

Alternative proof.

3.5.9.

Proof.

3.5.10.

Proof.

3.5.11.

3.5.12.

Proof.

3.5.13.

Proof.

3.5.14.

Proof.

3.5.15.

Proof.

3.5.16.

Proof.

3.5.17.

Proof.

3.5.18.

Proof.

4 More logic4.3 Exercises

4.3.1.

4.3.2.

4.3.3.

5 More proofs5.5 Exercises

5.5.1.

Proof.

5.5.2.

Proof.

5.5.3.

Proof.

1 Sets
1.4 Exercises

2 A little logic
2.7 Exercises

3 Direct proofs
3.5 Exercises

4 More logic
4.3 Exercises

5 More proofs
5.5 Exercises

6 Quantifiers
6.6 Exercises