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

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

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

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

the chain rule and mathematical induction imply

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

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

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

Since the coefficients of x⁰, x¹, ..., x^n − 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,

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

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:

$frac12\; int\_0^pi\; f(x)sin(x),dx=F(0)!$

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

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

By the fundamental theorem of calculus

$frac12\; int\_0^pi\; f(x)sin(x),dx=\; frac12\; bigl(F\text{'}(x)sin\; x\; -\; F(x)cos\; xbigr)Big|\_\{x=0\}^\{x=pi\}.!$

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

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

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

$frac12\; int\_0^pi\; f(x)sin(x),dxle\; frac\{b^n\}\{n!\}Bigl(fracpi2Bigr)^\{2n+1\}!$

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).

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

$begin\{align\}$

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

$I\_n\; =\; int\_\{-1\}^1\; (1\; -\; x^2)^n\; cos(alpha\; x),dx.$

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

$J\_n\; =\; 2n(2n\; -\; 1)J\_\{n-1\}\; -\; 4n(n\; -\; 1)alpha^2\; J\_\{n-2\}.$

Also

$J\_0\; =\; 2sinalpha,quad\; J\_1\; =\; -4alphacosalpha\; +\; 4sinalpha.,$

Hence for all n,

$J\_n\; =\; alpha^\{2n+1\}I\_n\; =\; n!(P\_nsinalpha\; +\; Q\_ncosalpha),$

where P_n, Q_n 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

$frac\{b^\{2n+1\}\}\{n!\}\; I\_n\; =\; P\_n\; a^\{2n+1\}.$

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

and

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

Hence for sufficiently large n

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

References

• Ivan Niven, "A simple proof that π is irrational", Bull. Amer. Math. Soc., Vol. 53, No. 6, p. 509, (1947) Online
• Harold Jeffreys, Scientific Inference, 3rd edition, Cambridge University Press, 1973, page 268.