Proof: by Ivan Niven (19151999),
Bull.Amer.Math.Soc. 53 (1947),509):
Assume = a/b with positive integers a and b. We will deduce a contradiction. For an integer n>0 define
f(x) = x^{n} (ab x)^{n}/n!

It satisfies f(x) = f(x) and the inequality
0 < f(x) <^{n} a^{n}/n!

for all x in [0,]. For all j, smaller or equal to n, the jth derivative of f
is zero for x=0 and x=. For j larger or equal to n, the jth derivative of f
is an integer at 0 and . The function
F(x) = f(x)  f^{(2)}(x) + f^{(4)}(x)  ... + (1)^{n} f^{2n}(x)

now has the property that F(0) and F() are integers and F + F'' = f. Therefore,
(F'(x) sin(x)  F(x) cos(x))' = f sin(x)

By the fundamental theorem of calculus, f(x) sin(x) dx is an integer.
By the inequality however, this integral is strictly between 0 and 1 for large n. QED.

