Prove that the sqrt(2) is irrational

To do this we will assume sqrt(2) is rational, a fraction, which means: 2^1/2 = m/n; m,n belong to integers. Also, m/n is an irreductible fraction, meaning m and n have no common divisors.

2^1/2 = m/n <=> 2 = m²/n² <=>  2n²=m²;  this means m² is even (divisible by 2), which implies m is even (can be proven).

Hence, m can be rewritten as: m = 2k. Thus: 2n²=(2k)² <=> 2n²=4k² <=> n²=2k²; the same logic is applied as above: n² is even (divisible by 2), which implies n is even.

We have now arrived at a contradiction: m/n was supposed to be an irreductable fraction, however both m and n can be divided by 2.

Ergo, by contradiction, we can conclude that srqt(2) cannot be a rational number, hence, srqt(2) is irrational.