Today is 3/14/15 — Super Pi Day — so was I telling my 7-year-old son all about the number this afternoon. When I told him that keeps on going forever and ever he asked “How do you know that?” Although I don’t know a proof that I could explain to a 7-year-old, I wanted to record the following proof which uses only basic calculus. It is essentially Niven’s famous proof, as generalized by Alan Parks, but I have tried to write it in a way which is more motivated than the usual treatments. As a bonus, the proof also shows that e is irrational.
Let be a positive real number. Our goal is to assume that is rational and use this assumption to construct a continuous function and a sequence of polynomials for with the following properties:
(1) and all are positive on the interval .
(2) There is a sequence of real numbers with as such that for each , for all
(3) The integral is an integer for all .
To see why this gives a contradiction, note that (1) and (3) imply that for all , but if is the maximum value of on then (2) implies that
which is less than 1 for sufficiently large.
Let be the set of all continuous functions on with the property that both and are integers. Note that is closed under addition, subtraction, and multiplication (it is what mathematicians call a ring).
We define to be the set of all in whose derivative belongs to for all . The product rule for derivatives shows that is closed under addition, subtraction, and multiplication, and by definition it is closed under taking derivatives.
Key Observation #1: If and , then for all
Proof: The proof is by induction on . The case is trivial. Assume that . We have by assumption, and by the chain rule we have . Since and is closed under multiplication, the result follows. Q.E.D.
We define to be the set of all in with a antiderivative in for all More formally, consists of all functions in such that for each , there exists in whose derivative is . Note that by definition we in fact have .
Key Observation #2: If and is a polynomial, then is an integer.
Proof: The proof is by induction on the degree of . If then is constant, and the result follows easily from the fundamental theorem of calculus. Assume the result is true for polynomials of degree . Letting be an antiderivative of in and applying integration by parts, we obtain
which is an integer by the inductive hypothesis. Q.E.D.
We can now prove our main theorem, which as we will see implies that both and are irrational.
Main Theorem: Suppose there is a function with for . Then is irrational.
Proof: Assume for the sake of contradiction that with positive integers. Then the function belongs to and satisfies . (The derivative of is and its second derivative is .) It follows from Key Observation #1 that belongs to for all By Key Observation #2, is an integer for all . On the other hand, both and all are positive on . Thus conditions (1) and (3) above are satisfied. To see that (2) is satisfied as well, let be the maximum value of on . Then , which goes to zero as . As discussed above, conditions (1),(2),(3) together now lead to a contradiction. Q.E.D.
Corollary 1: is irrational.
Proof: Take and . Q.E.D.
Corollary 2: is irrational.
Proof: Assume with positive integers and take and . Q.E.D.
1. As mentioned in the introduction, this post is a reorganization and repackaging of the proof given in this article by Alan E. Parks, which originally appeared in the American Math Monthly. I first came across Parks’s proof in the book “Biscuits of Number Theory”, edited by Arthur Benjamin and Ezra Brown.
2. As Parks notes in his article, the main theorem easily implies the following generalizations of Corollary 1 and Corollary 2:
- If and if and are both rational, then is irrational.
- If is a positive rational number then is irrational.
3. The strategy of proof here is a very natural one. Every irrationality proof that I know of for or involves reaching a contradiction by constructing a positive integer less than 1. And the argument used to bound the integral in (3) appears frequently (albeit in a different context) in analysis, especially elementary complex analysis. And the Key Observations, while clever, are simple consequences of the product rule for derivatives, the chain rule, and the fundamental theorem of calculus. So unlike other irrationality proofs that I’ve seen, I find this argument quite motivated. It is also accessible to anyone who has studied calculus and mathematical induction. And finally, it’s nice that the same argument proves irrationality of and at the same time.
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I know that this is a slick way to prove that π is irrational. However, I’ve always been fond of the original concept behind Lambert’s proof, the idea being that tan(r) is irrational for r rational. This follows from the continued fraction expansion tan(r) = r/(3-r^2/(5-r^2/(7-r^2… .
The point here is that if r is rational, the the partial convergents of this continued fraction give rational numbers which converge too quickly to tan(r) for tan(r) to be irrational. Of course the proof requires some care, both to prove convergence of the continued fractions and to show the limit is irrational. But I like this approach because of the analogy with other irrationality proofs using continued fractions. A complete argument is given here: http://www.maths.unsw.edu.au/~angell/5535/chapter7.ps
Yes, Lambert’s continued fraction argument is also great! (And a similar analysis with tanh instead of tan proves that e is irrational, though of course this is a much easier fact than the irrationality of pi.)