TITLE: Prove $1+\sqrt2$ is irrational
QUESTION [5 upvotes]: I am trying to prove $1+\sqrt 2$ is an irrational number.
I start with contradiction 
Proof:
assume that $1+\sqrt 2$ is a rational number such that $1+\sqrt 2=\frac{m}{n}$ where m and n  are some integers.
Then, $$1+\sqrt 2=\frac{m}{n}$$
$$\implies \sqrt 2 =\frac{m}{n}-1$$
$$\implies \sqrt 2=\frac{m-n}{n}$$
$$\implies \sqrt{2} n=m-n$$
$$\implies 2n^2=(m-n)^2$$ 
I get stuck at this step, can anyone give a hint or a suggestion?
Thanks!

REPLY [7 votes]: A striking way to prove such irrationality is via Euclid's gcd algorithm, which works for rationals. Namely, if $\,w=\sqrt{2}+1\in\Bbb Q\,$ then $\,w-2 =\sqrt{2}-1\in \Bbb Q\,$ therefore
$$\begin{align} d = \gcd(1,\sqrt{2}+1)\ &=\, (\sqrt{2}+1)\,\gcd(\sqrt{2}-1,\,1)\quad {\rm by}\ \ \gcd(ab,ac)\, =\, a\gcd(b,c)\\ &=\, (\sqrt{2}+1)\, \gcd(\sqrt{2}+1,\,1)\quad {\rm by}\ \ \gcd(a,b)\, =\, \gcd(a\!+\!2b,b)\\\end{align}\quad $$
hence the gcd $\,d\,$ satisfies $\, d = (\sqrt{2}+1)d,\,$ contradiction! Therefore $\,\sqrt{2}+1\not\in\Bbb  Q$