<s>[INST] Show that the number $a_{n}=\sqrt{2+\sqrt{2+\sqrt{2+...}}}$, is irrational for every positive integer n. Define $a_{n}=\sqrt{2+\sqrt{2+\sqrt{2+...}}},\quad n=1,2,...$ where the radical contains $n$ twos.
Provide a recursive definition of $a_n$ and then show that the number $a_n$ is irrational for every positive integer n.
This is what I have so far-
$$a_0=\sqrt{2}, \quad a_{n+1}=\sqrt{2+a_n}$$
Assume $a_{n+1}$ is rational. Then
\begin{align*}
a_{n+1}&=\frac{b}{c} \quad grd(b,c)=1\\
(a_{n+1})^2&=\frac{b^2}{c^2}\\
2+a_n&=\frac{b^2}{c^2}\\
(2+a_n)c^2&=b^2 \quad =>(2+a_n)\big|b^2\\
\end{align*}
Is this correct so far? I am not sure what the next step is or if I am even on the right path?
 [/INST] Using induction we suppose that $a_n$ is irrational. Suppose now that $a_{n+1}$ is rational. Then $a^2_{n+1}=2+a_n$, therefore $a_n$ is rational, contradiction.
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