Proof that π is irrational

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Although the mathematical constant known as π (pi) has been studied since ancient times, and so has the concept of irrational number, it was not until the 18th century that π was proved to be irrational.

In the 20th century, proofs were found that require no prerequisite knowledge beyond integral calculus. One of those, due to Ivan Niven, is widely known. A somewhat earlier similar proof is by Mary Cartwright. She set it as an examination problem but did not publish it. Cartwright's proof is reproduced in Jeffreys, in an appendix.

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[edit] Niven's proof

The proof uses the characterization of π as the smallest positive zero of the sine function. As in many proofs of irrationality, the argument proceeds by reductio ad absurdum.

Preparation: Suppose that π is rational, i.e. π = a / b for some integers a and b ≠ 0, which may be taken without loss of generality to be positive. Given any positive integer n, we define the polynomial function

f(x) = {x^n(a - bx)^n \over n!},\quad x\in\mathbb{R}\!

and denote by

F(x) = f(x) + \cdots + (-1)^j f^{(2j)}(x) + \cdots + (-1)^n f^{(2n)}(x). \!

the alternating sum of f and its first n even derivatives.

Claim 1: F(0) = F(π)

Proof: Since

f(x)=b^n{x^n(\pi - x)^n \over n!}=f(\pi-x),\quad x\in\mathbb{R}\!

the chain rule and mathematical induction imply

f^{(j)}(x) = (-1)^j f^{(j)}(\pi - x),\quad x\in\mathbb{R}\!

for all the derivatives, in particular

f^{(2j)}(0)=f^{(2j)}(\pi)\!

for j = 1, 2, ...,n and Claim 1 follows from the definition of F.

Claim 2: F(0) is an integer.

Proof: Using the binomial formula to expand (a – bx)n and the index transformation j = k + n, we get the representation

f(x)={1\over n!}\sum_{j=n}^{2n}{n \choose j-n}a^{2n-j}(-b)^{j-n}x^{j}.\!

Since the coefficients of x0, x1, ..., xn − 1 are zero and the degree of the polynomial f is at most 2n, we have f (j)(0) = 0 for j < n and j > 2n. Furthermore,

f^{(j)}(0)={j!\over n!}{n \choose j-n}a^{2n-j}(-b)^{j-n}\quad\mbox{for } n\le j\le 2n\!

Since j ≥ n, the fraction of the two factorials is an integer. The same holds for the binomial coefficient, as can be seen from its combinatorical interpretation or Pascal's triangle. Hence f and every derivative of f at 0 is an integer and so is F(0).

Claim 3:

\frac12 \int_0^\pi f(x)\sin(x)\,dx=F(0)\!

Proof: Since f (2n + 2) is the zero polynomial, we have

F'' + F = f.\,

The derivatives of the sine and cosine function are given by (sin x)' = cos x and (cos x)' = −sin x, hence the product rule implies

(F'\cdot\sin - F\cdot\cos)' = f\cdot\sin\!

By the fundamental theorem of calculus

\frac12 \int_0^\pi f(x)\sin(x)\,dx= \frac12 \bigl(F'(x)\sin x - F(x)\cos x\bigr)\Big|_{x=0}^{x=\pi}.\!

Since sin 0 = sin π = 0 and cos 0 = –cos π = 1 (here we use the abovementioned characterization of π as a zero of the sine function), Claim 3 follows from Claim 1.

Conclusion: Since f(x) > 0 and sin x > 0 for 0 < x < π (because π is the smallest positive zero of the sine function), Claims 2 and 3 show that F(0) is a positive integer. Since

x(\pi -x) = \Bigl(\frac\pi2\Bigr)^2-\Bigl(x-\frac\pi2\Bigr)^2\le\Bigl(\frac\pi2\Bigr)^2,\quad x\in\mathbb{R}\!

and 0 ≤ sin x ≤ 1 for 0 ≤ x ≤ π, we have

\frac12 \int_0^\pi f(x)\sin(x)\,dx\le \frac{b^n}{n!}\Bigl(\frac\pi2\Bigr)^{2n+1}\!

which is smaller than 1 for large n, hence F(0) < 1 by Claim 3 for these n. This is impossible for the positive integer F(0).

[edit] Analysis of Niven's proof

The above proof is a polished version, which is kept as simple as possible concerning the prerequisites, of an analysis of the formula

\begin{align} \frac12\int_0^\pi f(x)\sin(x)\,dx &=\frac12\sum_{j=0}^n (-1)^j \bigl(f^{(2j)}(\pi)+f^{(2j)}(0)\bigr)\\ &\qquad+\frac{(-1)^{n+1}}2\int_0^\pi f^{(2n+2)}(x)\sin(x)\,dx, \end{align}

which is obtained by 2n + 2 partial integrations. Claim 3 essentially establishes this formula, where the use of F hides the iterated partial integrations. The last integral vanishes because f (2n + 2) is the zero polynomial. Claims 1 and 2 show that the remaining sum is an integer.

[edit] Cartwright's proof

Jeffreys, page 268, says:

The following was set as an example in the Mathematics Preliminary Examination at Cambridge in 1945 by Dame Mary Cartwright, but she has not traced its origin.

Consider the integrals

I_n = \int_{-1}^1 (1 - x^2)^n \cos(\alpha x)\,dx.

Two integrations by parts give the recurrence relation

\alpha^2 I_n = 2n(2n - 1) I_{n-1} - 4n(n-1)I_{n-2},\quad n \ge 2.

If

J_n = \alpha^{2n+1}I_n,\,

then this becomes

Jn = 2n(2n − 1)Jn − 1 − 4n(n − 1)α2Jn − 2.

Also

J_0 = 2\sin\alpha,\quad J_1 = -4\alpha\cos\alpha + 4\sin\alpha.\,

Hence for all n,

Jn = α2n + 1In = n!(Pnsinα + Qncosα),

where Pn, Qn are polynomials in α of degree ≤ 2n, and with integral coefficients depending on n.

Take α = (1/2)π, and suppose if possible that

\frac{1}{2}\pi = \frac{b}{a}

where a and b are integers. Then

\frac{b^{2n+1}}{n!} I_n = P_n \alpha^{2n+1}.

The right side is an integer. But 0 < In < 2 since

0 < (1 - x^2)^n\cos\left(\frac{1}{2}\pi x\right) < 1\text{ for }-1 < x < 1

and

\frac{b^{2n+1}}{n!} \to 0\text{ as }n \to \infty.

Hence for sufficiently large n

0 < \frac{b^{2n+1}I_n}{n!} < 1,

that is, we could find an integer between 0 and 1. That is the contradiction that follows from the assumption that π is rational.

[edit] References