I'm working on a proof that shows for any $a \in \mathbb Q$ such that $a < 2$, there exists some $b \in \mathbb Q$ such that $a < b^2 < 2$. It was suggested that as part of the proof, I should consider the series defined by $a_1=1, a_{n+1}=\frac{2a_n+2}{a_n+2}$ and prove by induction that $2 - a_n^2 \le \frac{1}{n}$. I can then use this fact to help pick some $b^2$ larger than $a$. I have a decent idea on what to do with the result, but I seem to be struggling with the induction. Here is what I have so far:
First, we consider our base inductive case. Let $n_0=1$. We have $$2-a_1^2=2-(1)^2=2-1=1 \le1$$ so our hypothesis holds. Now, assume that it holds for some $k > n_0$. We want to show that $2-a_{k+1}^2 \le \frac{1}{k+1}$. We obtain the following: \begin{align} \\2-a_{k+1}^2 & = 2-\left( \frac{2a_k+2}{a_k+2} \right) ^2 \\ & = 2 - \frac{(2a_k+2)^2}{(a_k+2)^2} \\ & = 2 - \frac{4a_k^2+8a_k+4}{(a_k+2)^2} \\ & = \frac{2(a_k+2)^2-(4a_k^2+8a_k+4)}{(a_k+2)^2} \\ & = \frac{2a_k^2+8a_k+8-4a_k^2-8a_k-4}{(a_k+2)^2} \\ & = \frac{-2a_k^2+4}{(a_k+2)^2} \end{align}
It isn't totally clear to me how to proceed from here. From our initial induction hypothesis, we have $2-a_k^2 \le \frac{1}{k}$, so $-a_k^2 \le \frac{1}{k} -2$. Returning to where we left off, we can see:
\begin{align} \\ \frac{-2a_k^2+4}{(a_k+2)^2} & = \frac{2(-a_k^2+2)}{(a_k+2)^2} \\ & \le \frac{2\left((\frac{1}{k}-2)+2\right)}{(a_k+2)^2} \\ & = \frac{2}{k(a_k+2)^2} \end{align}
Unfortunately, I don't really see how this can help me reach what I'm trying to show. Any suggestions on where to go from here (or where I went wrong previously) would be more than welcome. I'm not sure if this is relevant, but I'd like to reiterate that this is going to be used as a lemma in a proof regarding the rationals and not the reals, so I'd like to avoid taking the square root of anything that may not result in a rational number.
You are essentially done at this point. All that's left to note is that $\,a_k \ge 0\,$, and therefore:
$$ \\2-a_{k+1}^2 \,=\, \frac{2}{(a_k+2)^2} \cdot (2-a_k^2) \;\le\; \frac{2}{2^2} \cdot \frac{1}{k} \;\le\; \frac{1}{k+1} $$