Prove that square root of 6 is irrational by contradiction .

Assume that $\sqrt{(}6)$ is rational.
Then $\sqrt{(}6)=\frac{p}{q}$ where p and q are coprime integers.
${\sqrt{6}}^2=6=\frac{p^2}{q^2}$
$p^2=6q^2$
therefore $p^2$ is an even number since an even number multiplied by any other integer is also an even number. If $p^2$ is even then p must also be even since if p were odd, an odd number multiplied by an odd number would also be odd.
So we can replace p with 2k where k is an integer.
$(2k{)}^2=6q^{2}4k^2=6q^{2}2k^2=3q^2$
Now we see that $3q^2$ is even. For $3q^2$ to be even, $q^2$ must be even since 3 is odd and an odd times an even number is even. And by the same argument above, if $q^2$ is even then q is even.
So both p and q are even which means both are divisible by 2. But that means they are not coprime, contradicting our assumption so sqrt(6) is not rational.