elementary number theory - Prove that $6 - sqrt{2}$ is Irrational by contradiction

What is a Proof by Contradiction, and how to prove by contradiction that $6 - \sqrt{2}$ is an irrational number?

Answer

A proof by contradiction is assuming something then building on it and finding that it leads to contradiction, concluding that the assumed statement is false.

Assume $x=6-\sqrt{2}=\frac{p}{q}$

$x^2=8-12\sqrt{2}=\frac{p^2}{q^2}$

$8q^2=p^2+12\sqrt{2}q^2$

$8q^2-p^2=12\sqrt{2}q^2$

$8q^2-p^2$ is rational, and $q^2$ is rational, thus $12\sqrt{2}$ is rational.

However, we know that that is not true, and thus, $6-\sqrt{2}$ is irrational.