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.
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
and denote by
the alternating sum of f and its first n even derivatives.
Claim 1: F(0) = F(π)
for all the derivatives, in particular
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
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,
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).
Proof: Since f (2n + 2) is the zero polynomial, we have
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
and 0 ≤ sin x ≤ 1 for 0 ≤ x ≤ π, we have
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).
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.
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
then this becomes
Hence for all n,
Take α = (1/2)π, and suppose if possible that
where a and b are integers. Then
The right side is an integer. But 0 < In < 2 since
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.