<s>[INST] If $x \in \mathbb{Q}$, $\frac{x^2}{2} < 2$ then $\exists a \in \mathbb{Q}$, $a>0$ where $\frac{x^2}{2}There is this demonstration in my calculus book that I can't understand:
I want to prove that:
If $x \in \mathbb{Q}$, $\frac{x^2}{2} < 2$ then $\exists a \in \mathbb{Q}$,  $a>0$ where $\frac{x^2}{2}<a^2<2$.
Dem:
If $\frac{x^2}{2} \leq 1$ then $a = \frac{11}{10}$ satisfies the proposition.
If $y = \frac{x^2}{2} >1$, let's take an $n \in \mathbb{N}$ where
$(1+\frac{1}{n})^2<y$ and $(1+\frac{1}{n})^2 <\frac{2}{y} $
We can find this $n$, just see that
 $(1+\frac{1}{n})^2 = 1 + \frac{2}{n} + \frac{1}{n^2} \leq 1 + \frac{3}{n} $
If we take an $n$ where $1+\frac{3}{n} < y^* \implies n>\frac{3}{y^*-1}$ with $y^* = min\{y,\frac{2}{y}\}$ then $(1+\frac{1}{n})^2 = 1 + \frac{2}{n} + \frac{1}{n^2} \leq 1 + \frac{3}{n} $.
If $u = 1+\frac{1}{n}$ where $n>\frac{3}{y^*-1}$. Then there will be a term $v$ in the progression $u^2, u^4, u^6,...,u^{2k},...$
where $\frac{x^2}{2} < v <2$
Let $m \in \mathbb{N}$ where $\frac{x^2}{2}<u^{2m}<2$ then $ a = u^k$

Here are my questions:
Question 1:
How we can prove the inequality
$(1+\frac{1}{n})^2 = 1 + \frac{2}{n} + \frac{1}{n^2} \leq 1 + \frac{3}{n} $
I tried to use induction, but it did not work for me.
Solution: $\frac{1}{n^2} \leq \frac{1}{n} \implies 1 + \frac{2}{n} + \frac{1}{n^2} \leq 1 + \frac{3}{n} $
Question 2:
Why if $u = 1+\frac{1}{n}$ where $n>\frac{3}{y^*-1}$. Then there will be a term $v$ in the progression $u^2, u^4, u^6,...,u^{2k},...$
where $\frac{x^2}{2} < v <2$.
 [/INST] For Question 2:
The idea is that $u^2,u^4,u^6,\ldots$ is a geometric progression with ratio $r$ greater than $1$ and with:
$$r=u^2= \left(1+\frac{1}{n} \right)^2 < \frac{2}{y}=\frac{2}{\frac{x^2}{2}}$$
and with initial term:
$$u^2 < y=\frac{x^2}{2}$$

To prove it let us suppose that there is not $k$ such that $u^{2k} \in (x^2/2,2)$.
As $u^2 <x^2/2$ and $(u^{2k})$ is an increasing sequence going to $+\infty$ (as the ration is $>1$) there exists $k_0$ such that:
$$u^{2k_0} \leq \frac{x^2}{2} < 2 \leq u^{2(k_0+1)}  $$
so:
$$\frac{u^{2(k_0+1)}}{u^{2k_0}} \geq \frac{2}{\frac{x^2}{2}}$$
i.e:
$$\left(1+\frac{1}{n} \right)^2 \geq \frac{2}{y}$$
which is in contradiction with the choice of $n$.
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