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Show that for all $n\in \mathbb{N}$ the number $(\sqrt{2}-1)^n$ is irrational.

I do not get the idea of the proof at all, any help appreaciated.

edit: I am also thinking whether it will be possible to show $(\sqrt{2}-1)^n=\sqrt{m+1}-\sqrt{m}$

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  • $\begingroup$ What was the proof that you did not get? $\endgroup$ – Pieter21 Oct 9 '14 at 15:36
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Since $$ (\sqrt{2}-1)^n=\frac{1}{(\sqrt{2}+1)^n} $$ it is enougth to prove that $(\sqrt{2}+1)^n$ is irrrational. We have after direct calculation that $$ (\sqrt{2}+1)^n=a+b \sqrt{2}, $$ for some natural $a,b,$ $b\neq 0.$ Suppose now that $(\sqrt{2}+1)^n$ is rational. It follows that $\sqrt{2}$ is also rational since $$ \sqrt{2}=\frac{1}{b}((\sqrt{2}+1)^n-a). $$ We obtain a contradiction so $(\sqrt{2}-1)^n$ is an irrational number.

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  • $\begingroup$ well, you are right, but since $\sqrt{2}-1=\frac{1}{\sqrt{2}+1}}$ it is enouth that $\sqrt{2}+1}$ is rational. Then $b \neq 0.$ $\endgroup$ – Leox Oct 9 '14 at 15:35
  • $\begingroup$ Can't we just show that (\sqrt{2}-1)^n-\sqrt{m}? $\endgroup$ – mortar Oct 9 '14 at 15:39
  • $\begingroup$ I have explained it in main text $\endgroup$ – Leox Oct 9 '14 at 15:51
  • $\begingroup$ Proving that $b\ne0$ is exactly the problem. You can't say $b\ne0$ without proving it. $\endgroup$ – egreg Oct 9 '14 at 16:30
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    $\begingroup$ I dont see any problem. If $\sqrt{2}>0, 1>0$ then $a>0$ and $b>0$, it is easy to prove. $\endgroup$ – Leox Oct 9 '14 at 17:56
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Note that if we set $a=1-\sqrt 2$ and $b=1+\sqrt 2$ we have $a+b=2$ and $ab=-1$, so that $a,b$ are roots of the quadratic $x^2-2x-1=0$, and $(ab)^n=(-1)^n$.

Now note that $a^2-2a-1=0$ can be multiplied by $a^n$ to obtain the equation $a^{n+2}=2a^{n+1}+a^n$. If we add the equivalent equation for $b^{n+2}$ and set $u_n=a^n+b^n$ we find that $$u_{n+2}=2u_{n+1}+u_n$$

Since $u_0=u_1=2$ the $u_n$ are increasing positive (even) integers and $a^n$ and $b^n$ are roots of $x^2-u_nx+(-1)^n=0$.

Now apply the rational root theorem.

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$(\sqrt{2}-1)^n$ can be written as $p(n) + q(n)\sqrt{2}$, with both $p(n)$ and $q(n)$ integers, with $q(n)\neq0$.

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  • $\begingroup$ $q(n)\ne0$ is exactly what's to be proved. Induction should suffice. $\endgroup$ – egreg Oct 9 '14 at 16:31
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Let a vector $[x\ y]^T$, $x,y \in \mathbb{Q}$ represent a number of the form $x + \sqrt{2}y$. Multiplying by $(\sqrt{2} - 1)$ then corresponds to applying the matrix $$ A = \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix}. $$

Multiplying by $(\sqrt{2} - 1)^n$ corresponds to applying $A^n$.

Suppose $(\sqrt{2} - 1)^n = r \in \mathbb{Q}$ for some $n \in \mathbb{Z}^+$. That would mean that $A^n = rI$. We show that it is impossible.

$A$, considered as a real matrix, has two distinct eigenvalues: $\pm\sqrt{2} - 1$. Thus, $A^n$ also has two distinct eigenvalues: $(\pm\sqrt{2} - 1)^n$. But then it can't be of the form $rI$.

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