Section 3.7 Optional — Rational and irrational numbers
In this optional section we shall use series techniques to look a little at rationality and irrationality of real numbers. We shall see the following results.
- A real number is rational (i.e. a ratio of two integers) if and only if its decimal expansion is eventually periodic. “Eventually periodic” means that, if we denote the
decimal place by then there are two positive integers and such that whenever So the part of the decimal expansion after the decimal point looks like and so that the decimal expansion ends with an infinite string of zeros. is irrational. is irrational.
Subsection 3.7.1 Decimal expansions of rational numbers
We start by showing that a real number is rational if and only if its decimal expansion is eventually periodic. We need only consider the expansions of numbers If a number is negative then we can just multiply it by and not change the expansion. Similarly if the number is larger than then we can just subtract off the integer part of the number and leave the expansion unchanged.
Subsection 3.7.2 Eventually periodic implies rational
Let us assume that a number has a decimal expansion that is eventually periodic. Hence we can write
This is a ratio of integers, so is a rational number.
Subsection 3.7.3 Rational implies eventually periodic
Let be rational with where and are positive integers. We wish to show that ’s decimal expansion is eventually periodic. Start by looking at the last formula we derived in the “eventually periodic implies rational” subsection. If we can express the denominator in the form with and integers, we will be in business because From this we can generate the desired decimal expansion by running the argument of the last subsection backwards. So we want to find integers such that To do so consider the powers of up to
To do so, start with and repeatedly subtract from it until the remainder drops strictly below The ’s can take at most different values, namely and we now have ’s, namely So we must be able to find two powers of 10 which give the same remainder . That is there must be so that Hence
1
This is an application of the pigeon hole principle — the very simple but surprisingly useful idea that if you have items which you have to put in boxes, and if then at least one box must contain more than one item.
and we have
Next divide the numerator by and compute the remainder. That is, write with Notice that as otherwise That is, has at most digits and has at most digits. This, finally, gives us
which gives the required eventually periodic expansion.
Subsection 3.7.4 Irrationality of
We will give 2 proofs that the number is irrational, the first due to Fourier (1768–1830) and the second due to Pennisi (1918–2010). Both are proofs by contradiction — we first assume that is rational and then show that this implies a contradiction. In both cases we reach the contradiction by showing that a given quantity (related to the series expression for ) must be both a positive integer and also strictly less than 1.
2
Proof by contradiction is a standard and very powerful method of proof in mathematics. It relies on the law of the excluded middle which states that any given mathematical statement is either true or false. Because of this, if we can show that the statement being false implies something contradictory — like or — then we can conclude that must be true. The interested reader can certainly find many examples (and a far more detailed explanation) using their favourite search engine.
Subsubsection 3.7.4.1 Proof 1
This proof is due to Fourier. Let us assume that the number is rational so we can write it as
Now multiply both sides by to get
The left-hand side of this expression is an integer. We complete the proof by showing that the right-hand side cannot be an integer (and hence that we have a contradiction).
The first sum, is finite sum of integers:
Consequently must be an integer. Notice that we simplified the ratio of factorials using the fact that when we have
Now we turn to the second sum. Since it is a sum of strictly positive terms we must have
We complete the proof by showing that To do this we bound each term from above:
Indeed the inequality is strict except when Hence we have that
And since is a positive integer, we have shown that
and thus cannot be an integer.
Subsubsection 3.7.4.2 Proof 2
This proof is due to Pennisi (1953). Let us (again) assume that the number is rational. Hence it can be written as
where are positive integers. This means that we can write
Using the Maclaurin series for we have
Before we do anything else, we multiply both sides by — this might seem a little strange at this point, but the reason will become clear as we proceed through the proof. The expression is now
The left-hand side of the expression is an integer. We again complete the proof by showing that the right-hand side cannot be an integer.
We split the series on the right-hand side into two pieces:
Every term in the sum is an integer. To see this we simplify the ratio of factorials as we did in the previous proof:
Let us now examine the series Again clean up the ratio of factorials:
Hence is an alternating series of decreasing terms and by the alternating series test (Theorem 3.3.14) it converges. Further, it must converge to a number between its first and second partial sums (see the discussion before Theorem 3.3.14). Hence the right-hand side lies between
Since is a positive integer the above tells us that converges to a real number strictly greater than and strictly less than 1. Hence it cannot be an integer.
This gives us a contradiction and hence cannot be rational.
Subsection 3.7.5 Irrationality of
This proof is due to Niven (1946) and doesn’t require any mathematics beyond the level of this course. Much like the proofs above we will start by assuming that is rational and then reach a contradiction. Again this contradiction will be that a given quantity must be an integer but at the same time must lie strictly between and
Assume that is a rational number and so can be written as with positive integers. Now let be a positive integer and define the polynomial
It is certainly not immediately obvious why and how Niven chose this polynomial, but you will see that it has been very carefully crafted to make the proof work. In particular we will show — under our assumption that is rational — that, if is really big, then
Subsubsection 3.7.5.1 Bounding the integral
Consider again the polynomial
Notice that
We could work out a more precise upper bound, but this one is sufficient for the analysis that follows. Hence
3
You got lots of practice finding the maximum and minimum values of continuous functions on closed intervals when you took calculus last term.
for all Using this inequality we bound
We will later show that, if is really big, then We’ll first show, starting now, that is an integer.
Subsubsection 3.7.5.2 Integration by parts
In order to show that the value of this integral is an integer we will use integration by parts. You have already practiced using integration by parts to integrate quantities like
and this integral isn’t much different. For the moment let us just use the fact that is a polynomial of degree Using integration by parts with and gives us
Use integration by parts again with
Use integration by parts yet again, with
And now we can see the pattern; we get alternating signs, and then derivatives multiplied by sines and cosines:
This terminates at the derivative since is a polynomial of degree We can check this computation by differentiating the terms on the right-hand side:
and similarly
When we add these two expressions together all the terms cancel except as required.
Subsubsection 3.7.5.3 The derivatives are integers
Now take the derivative and set Note that, if then for all and, if then is some number times which evaluates to zero when we set So
If then this is zero since If this is an integer because is an integer and is an integer. If then is again an integer. Thus all the derivatives of evaluated at are integers.
But what about the derivatives at To see this, we can make use of a handy symmetry. Notice that
You can confirm this by just grinding through the algebra:
Using this symmetry (and the chain rule) we see that
and if we keep differentiating
Setting in this tells us that
So because all the derivatives at are integers, we know that all the derivatives at are also integers.
Subsubsection 3.7.5.4 Putting it together
Based on our assumption that is rational, we have shown that the integral
satisfies
and also that is an integer.
We are, however, free to choose to be any positive integer we want. If we take to be very large — in particular much much larger than — then will be much much larger than (we showed this in Example 3.6.8), and consequently
Which means that the integral cannot be an integer. This gives the required contradiction, showing that is irrational.