I am trying to prove $1-\sqrt{2}$ is an irrational number.

Question

I tried to prove that $1-\sqrt{2}$ is rational:
$1-\sqrt{2} = p/q$

$(1-\sqrt2)^2 = (p/q)^2$

$1^2 - 2\sqrt2 +2 = p^2/q^2$

$3-2\sqrt2 = p^2/q^2$

$3-p^2/q^2 = 2\sqrt2$

I know that $\sqrt2$ is irrational, but how should I proceed from here?

Answer 1

Note that $(1-\sqrt{2})^{2}\not=1^{2}-2$ as you stated. If there is a rational $p/q$ such that $1-\sqrt{2}=p/q$, then $1-p/q=\sqrt{2}$. Note that $1-p/q=\frac{q-p}{q}$ is rational. But what do you know about $\sqrt{2}$?

Answer 2

Setting $$1-\sqrt{2}=\frac{p}{q}$$ squaring we get $$\left(\frac{p}{q}\right)^2-2\frac{p}{q}=1$$ Can you finish?