Given a rational number $x$ and $x^2 < 2$, is there a general way to find another rational number $y$ that such that $x^2Suppose I have a rational number $a$ and $a^2 < 2$. Can I find another rational number $B$ such that $a^2<B^2<2$?
Based on the answer to this question, I thought of doing the following:

$$ a^2 < 2 \implies a < \frac{2}{a}\\
\text{Let}\hspace{1cm} B=\frac{a+\frac{2}{a}}{2}=\frac{a^2+2}{2a}
$$

$B$ is greater than $a$ because:

$$ \begin{array} {aa} B>a & \implies \frac{a^2+2}{2a}>a \\
& \implies a^2 + 2 > 2a^2 \\
& \implies 2 > a^2 \\
& \implies a^2 < 2
\end{array}$$

If $B^2$ is less than $2$, then $B^2-2<0$, but:

$$\begin{array} {aa} B^2-2 < 0 & \implies \left( \frac{a^2+2}{2a} \right)^2 - 2 < 0 \\
& \implies \frac{a^4+4a^2+4}{4a} - \frac{8a^2}{4a^2} < 0 \\
& \implies \frac{(a^2-2)^2}{(2a)^2} < 0
\end{array}$$

Which is a contradiction since the left hand side of the inequality will be positive for all values of $a$.
But I think we must be able to find such a $B$ since based on my understanding of this answer, we can find a another rational number whose distance from $a$ is less than the distance between $a$ and $\sqrt{2}$
Therefore, I have 2 questions to ask:


*

*Why does this approach work in the case of $a^2>2$ but not when $a^2<2$?

*How should I approach these kind of questions since it seems that there are a few ways to construct a $B$ that satisfies a given set of restrictions? For example, see here (the proof is immediately before the section "13. The Completeness Axiom".

The substitution $$a \leftarrow \frac{a^2+2}{2a} $$ is exactly what you get from Newton's method for the function $f(a)=a^2-2$.  This will result in iterations that are greater than $\sqrt{2}$ since $f$ is an increasing convex function on $(0, \infty)$.  I will sketch two alternatives, both based on Newton's method, to get iterations less than $\sqrt{2}$.
First idea.  Use some other function than $f(a) = a^2-2$ with Newton's method.  For example, if $0 \leq p < \sqrt{2}$ then
$$
\frac{1}{\sqrt{2}+p} = \frac{\sqrt{2}-p}{2-p^2}
$$
and so $\sqrt{2}$ is a root of the function $$f(a) = \frac{2-p^2}{a+p}-a+p = \frac{2-a^2}{a+p}.$$
This function is convex and decreasing on $(0, \infty)$.  Applying Newton's method to this function results in the iteration
$$
a \leftarrow \frac{p a^2+4 a+2 p}{a^2+2 p a+2}
$$
which produces an increasing sequence that converges to $\sqrt{2}$ from below as required. For example for $p=0$ you get
$$
a \leftarrow \frac{4a}{a^2 + 2}
$$
which is exactly the harmonic mean of $a$ and $\frac{2}{a}$ as suggested by haruspex.  For $p=1$ and $p=\frac{7}{5}$ you get
$$
a \leftarrow \frac{a^2+4a+2}{a^2+2a+2} \textrm{ and } a \leftarrow \frac{7 a^2+20 a+14}{5 a^2+14 a+10}
$$
respectively, to give just two other examples.  Substituting $a=\frac{7}{5}$ in the latter results in $\frac{1393}{985}$ which is less than $4\times 10^{-7}$ below $\sqrt{2}$.
Second idea.  For this idea I will assume that $a\geq 1$.  As you already found out we have $$a < \sqrt{2} < \frac{a^2+2}{2a}.$$
Therefore we could try to take some weighted average of $a$ and $\frac{a^2+2}{2a}$ that ends up below $\sqrt{2}$.  So we're looking for some factor $\lambda \in (0,1)$ such that
$$
\lambda a + (1-\lambda)\frac{a^2+2}{2a} < \sqrt{2}
$$
for all $a \in [1, \sqrt{2})$.  It is not difficult to show that this is the case exactly if $\lambda \in (3 - 2\sqrt{2}, 1)$.  We can take $\lambda = \frac{1}{5}$ or $\lambda = \frac{5}{29}$ for example to get the iterations
$$
a \leftarrow \frac{3 a^2+4}{5a} \textrm{ and } a \leftarrow \frac{17 a^2 + 24}{29a}
$$
respectively.
