How to prove the existence of $b$ in $Q$ such that $a<b^2<c$ in $Q$?

Question

I 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.

Answer 1

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.