How to prove the existence of $b$ in $Q$ such that $aI would like to prove the existence of $b \in \mathbb Q$ such that $a<b^2<c$ for any given $a,c \in \mathbb Q$ with $a,c>0$ 
I want to use the statement above to prove a statement in a link
I thought that '$b$' must be exist. But, in my opinion, $\sqrt{}$ can't not be used because $\sqrt{a}$ or $\sqrt{c}$  may not exist in $\mathbb Q$ for some $a$ and $c$. 
I couldn't find a clue to prove the statement before the real number is constructed from $\mathbb Q$. Would you help me to prove that? 
Thanks all for replying and pointing out errors.
 A: Suppose that there is no such $b$. Then for each $n\in\Bbb Z^+$ there is a $k_n\in\Bbb N$ such that $$\frac{k_n^2}{n^2}\le a<c\le\frac{(k_n+1)^2}{n^2}=\frac{k_n^2}{n^2}+\frac{2k_n+1}{n^2}\;.$$ For each $n\in\Bbb Z^+$ we then have $$\frac{2k_n+1}{n^2}\ge c-a$$ and hence $$k_n\ge\frac{(c-a)n^2-1}2$$
and 
$$\begin{align*}
\frac{k_n^2}{n^2}&\ge\frac{\left((c-a)n^2-1\right)^2}{4n^2}\\
&=\frac{(c-a)^2n^2}4-\frac{c-a}2+\frac1{4n^2}\\
&\ge\frac{(c-a)^2n^2}4-\frac{c}2\;.
\end{align*}$$
But $a\ge\dfrac{k_n^2}{n^2}$, so we have $$a\ge\frac{(c-a)^2n^2}4-\frac{c}2$$ and hence 
$$n^2\le\frac{2(2a+c)}{(c-a)^2}$$
for all $n\in\Bbb Z^+$, contradicting the Archimedean property.
A: Here's a constructive answer based on Newton-Raphson approximation. The construction is guaranteed to work by the Archimedean property (AP), which I'll explicitly state wherever it is needed. It should be clear that this gives an actual simple program to find the desired answer.
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$

First we can assume that $1 < a$, because we can (by AP) multiply the desired equation by $k^2$ for some sufficiently large natural $k$ such that $1 < k^2·a$. Then $a < a^2$.
Let $f(b) = \lfrac{b^2+a}{2b}$ for every rational $b > 0$. Note that $f$ maps rationals to rationals.
Given any rationals $a,b>0$ such that $a < b^2$:
  $f(b)^2 > a$ since $(b^2-a)^2 > 0$.
  $f(b)^2 = \Big( \lfrac{b}{2}+\lfrac{a}{2b} \Big)^2 = \lfrac{b^2}{4}+\lfrac{a}{2}+\lfrac{a^2}{4b^2} < \lfrac{b^2}{4}+\lfrac{a}{2}+\lfrac{a}{4} = \lfrac{3a+b^2}{4}$.
  Thus $0 < f(b)^2-a < \lfrac14(b^2-a)$.
  Thus $0 < f^n(b)^2-a < 4^{-n}(b^2-a)$ for every natural $n > 0$ (by induction).
Therefore $a < f^n(a)^2 < c$ for every natural $n > 0$ such that $4^{-n} < \lfrac{c-a}{a^2-a}$.
We only need one such natural $n$, which exists (by AP) because $\lfrac{c-a}{a^2-a} > 0$.

For simple proofs of the claims involving AP, note that $k^2 \ge k$ for every natural $k$ (by induction), and $4^n \ge n$ for every natural $n$ (by induction), so it suffices to find a natural number $c$ such that $\lfrac1c < \lfrac1a$ and $\lfrac1c < \lfrac{c-a}{a^2-a}$ respectively, which exists by AP.
