Prove that $6 - \sqrt{2}$ is Irrational by contradiction

Question

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

Answer 1

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.

Answer 2

The method of proof by contradiction is done by assuming the negation of a given proposition is true, then finding a contradiction with that assumption, hence proving the original statement. For instance, suppose $6-\sqrt{2}$ is rational, then we can say that for $a,b\in \mathbb{Z}$ we have $$6-\sqrt{2}=\frac{a}{b}$$ where $a$ and $b$ are coprime. Now, we can do some algebra and find a contradiction.

Answer 3

If $6-\sqrt{2}$ is rational, then $\sqrt{2}$ is rational. Then just prove it for $\sqrt{2}$ which is easy.