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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.

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    $\begingroup$ Don't continue that way. Binomial expansion of $(1-\sqrt{2})^n$ yields the result comparatively easily. $\endgroup$ – Daniel Fischer Sep 4 '13 at 10:14
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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.

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For every $n\in \mathbb{N}$, you can find two integers $a_n$ and $b_n$ such that $(1-\sqrt{2})^n=a_n+\sqrt{2}b_n$ (to prove that use the binomial theorem). After that, you only need to show that for every $n$, $b_n\neq 0$.

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Using the binomial theorem you can show

$$(1-\sqrt2)^n=\text{something rational}-\sqrt 2\sum_{i \text{ odd}}\binom ni2^{\frac{i-1}{2}}$$

Can you finish?

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