1. Assume that \(\sqrt{2}\) is a rational number. That means there exist two integers a and b such that a / b =  \(\sqrt{2}\) 
2. So it can be written as a fractional fraction (a fraction that can not be reduced again): a / b with a, b is two primes together and (a / b)² = 2.
3. From (2) denotes a² / b² = 2 and a² = 2.b².
4. Then a² is even because it is equal to 2.b² (obviously even)
5. From this inference, a must be an even number because a² is the even number of digits (odd numbers are square rooted as odd, even square digits have square root).
6. Since a is even, there are some k satisfactions: a = 2k.
7. Substitution (6) into (3) we have: (2k)² = 2b² <=> 4k² = 2b² <=> 2k² = b².
8. Because 2k² = b² where 2k² is even, then b² is even, this inference b is also even [reason similar to (5)].
9. From (5) and (8) we have: a and b are even numbers, which contradicts the assumption that a / b is the minimal number in (2).
From the contradiction inference:  \(\sqrt{2}\)  is a rational number that is false and must conclude. \(\sqrt{2}\)  is an irrational number.

Go to this link and see the "Proofs of irrationality" : en.wikipedia.org/wiki/Square_root_of_2