TITLE: Prove that for every positive integer $n$ the number $(1−\sqrt 2)^n$ is not rational.
QUESTION [5 upvotes]: Prove that for every positive integer $n$ the number $(1−\sqrt 2)^n$ is not rational. 
Here is something I have, I'm not quite sure if I'm on the right track.
Proof:
Assume that $(1-\sqrt 2)^n \in \mathbb Q$, that is, it is a rational number, then we can express it as:
$(1-\sqrt 2)^n = \frac{p}{q}$ for some $p$, $q$ $\in \mathbb Z$, where $p$ and $q$ are relatively prime. I wanted to see if I can reach to a contradiction somewhere along the way.
Then $1-\sqrt 2 = \sqrt[n]{\frac{p}{q}}$. 
So $\sqrt 2 = 1-\sqrt[n]{\frac{p}{q}}$.
Since $\sqrt 2$ is an irrational number, it follows that $\sqrt[n]{\frac{p}{q}}$ must be irrational for the equation to hold. 
But $\frac{p}{q}$ is rational, 
That's where I stopped and didn't know what to continue as.
Please enlighten! Thanks in advance.

REPLY [5 votes]: We can prove by induction that $(1-\sqrt2)^n=r-s\sqrt2$, where $r$ and $s$ are positive integers.
First, the result is clear for $n=1$. Now suppose that it holds for $n=k$: say$$(1-\sqrt2)^k=r-s\sqrt2\quad\text{with}\quad r,s>0.$$Then$$(1-\sqrt2)^{k+1}=(r-s\sqrt2)(1-\sqrt2)=r+2s-(r+s)\sqrt2,$$which is of the required form to complete the inductive step.