Prove that the sqrt(2) is irrational
To do this we will assume sqrt(2) is rational, a fraction, which means: 21/2 = m/n; m,n belong to integers. Also, m/n is an irreductible fraction, meaning m and n have no common divisors.
21/2 = 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.
