Irrational number

For the proving that π is an irrational number, please refer to the following web page. It is quite difficult.
http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational
Below is the proving that √2 is a rational number.
The proposition is proved by assuming the contrary and showing that doing so leads to a contradiction (hence the proposition must be true).

Assume that √2 is a rational number. This assumption implies that there exist integers m and n with n ≠ 0 such that m/n = √2.
Then √2 can also be written as an irreducible fraction m/n (the fraction is shortened as much as possible). This means that m and n are coprime integers, i.e., they have no common factor greater than 1.
From m/n = √2 it follows that m = n√2, and so m2 = (n√2)2 = 2n2.
So m2 is an even number, because it is equal to 2n2, which is even.
It follows that m itself is even (since only even numbers have even squares).
Because m is even, there exists an integer k satisfying m = 2k.
We may therefore substitute 2k for m in the last equation of (3), thereby obtaining the equation (2k)2 = 2n2, which is equivalent to 4k2 = 2n2 and may be simplified to 2k2 = n2.
Because 2k2 is even, it now follows that n2 is also even, which means that n is even (recall that only even numbers have even squares).
Then, by (5) and (8), m and n are both even, which contradicts the property stated in (2) that m/n is irreducible.
Since we have found a contradiction, the initial assumption (1) that √2 is a rational number is false; that is to say, √2 is irrational.

This proof can be generalized to show that any root of any natural number is either a natural number or irrational.