Definitions

# Proof that π is irrational

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.

## 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

$f\left(x\right) = \left\{x^n\left(a - bx\right)^n over n!\right\},quad xinmathbb\left\{R\right\}!$

and denote by

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

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

Claim 1: F(0) = F(π)

Proof: Since

$f\left(x\right)=b^n\left\{x^n\left(pi - x\right)^n over n!\right\}=f\left(pi-x\right),quad xinmathbb\left\{R\right\}!$

the chain rule and mathematical induction imply

$f^\left\{\left(j\right)\right\}\left(x\right) = \left(-1\right)^j f^\left\{\left(j\right)\right\}\left(pi - x\right),quad xinmathbb\left\{R\right\}!$

for all the derivatives, in particular

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

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

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

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,

$f^\left\{\left(j\right)\right\}\left(0\right)=\left\{j!over n!\right\}\left\{n choose j-n\right\}a^\left\{2n-j\right\}\left(-b\right)^\left\{j-n\right\}quadmbox\left\{for \right\} nle jle 2n!$

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:

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

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

$F\text{'}\text{'} + F = f.,$

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

$\left(F\text{'}cdotsin - Fcdotcos\right)\text{'} = fcdotsin!$

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

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

$x\left(pi -x\right) = Bigl\left(fracpi2Bigr\right)^2-Bigl\left(x-fracpi2Bigr\right)^2leBigl\left(fracpi2Bigr\right)^2,quad xinmathbb\left\{R\right\}!$

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

$frac12 int_0^pi f\left(x\right)sin\left(x\right),dxle frac\left\{b^n\right\}\left\{n!\right\}Bigl\left(fracpi2Bigr\right)^\left\{2n+1\right\}!$

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).

### 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\left\{align\right\}
frac12int_0^pi f(x)sin(x),dx &=frac12sum_{j=0}^n (-1)^j bigl(f^{(2j)}(pi)+f^{(2j)}(0)bigr) &qquad+frac{(-1)^{n+1}}2int_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.

## 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

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

Two integrations by parts give the recurrence relation

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

If

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

then this becomes

$J_n = 2n\left(2n - 1\right)J_\left\{n-1\right\} - 4n\left(n - 1\right)alpha^2 J_\left\{n-2\right\}.$

Also

$J_0 = 2sinalpha,quad J_1 = -4alphacosalpha + 4sinalpha.,$

Hence for all n,

$J_n = alpha^\left\{2n+1\right\}I_n = n!\left(P_nsinalpha + Q_ncosalpha\right),$

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

$frac\left\{1\right\}\left\{2\right\}pi = frac\left\{b\right\}\left\{a\right\}$

where a and b are integers. Then

$frac\left\{b^\left\{2n+1\right\}\right\}\left\{n!\right\} I_n = P_n a^\left\{2n+1\right\}.$

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

$0 < \left(1 - x^2\right)^ncosleft\left(frac\left\{1\right\}\left\{2\right\}pi xright\right) < 1text\left\{ for \right\}-1 < x < 1$

and

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

Hence for sufficiently large n

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

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

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