# Proof: $$\sqrt{2}$$ is an irrational number

I suppose that you know definition of rational number. Irrational number is real number but is not rational number.

Here, I prove that ( sqrt{2} ) is a irrational number.

## Proof

Suppose ( sqrt{2} ) a rational number, then it can be represented with coprime natural numbers , ( m, n ) , as the following

[ sqrt{2} = frac{m}{n} ]

Now,

[ 2n^2 = m^2 . ]

Left hand ( 2 n^2 ) is even, so ( m^2 ) is also even. Therefore ( m ) can be represented as ( m = 2k ) , with the natural number ( k ) , and

begin{array}[crcl] & & 2 n ^2 & = & ( 2k )^2 \ & & = & 4k^2 \ Leftrightarrow & n^2 & = & 2k^2 end{array}

Therefore ( n ) is also even, but it is against the supposition that ( m, n ) are coprime. Above all, ( sqrt{2} ) is a irrational number.