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

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.

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More important, $c\gt0$. – Gerry Myerson Nov 14 '12 at 6:12
Ah. Thanks for pointing out that. I will correct them. – wikizero Nov 14 '12 at 6:12
For what it's worth, you can easily reduce it to the case where $a$ and $c$ are both integers by multiplying by the squares of the denominators. – Jonas Meyer Nov 14 '12 at 6:25
Assume a and c are integers. You need to find a square number between $k^2a$ and $k^2c$ for some $k$. Let $f(x)$ be the number of perfect squares $\leq x$. Then we need to show that $f(k^2 c) -f(k^2 a)>1$ for some $k$ (it would not suffice to show $f(k^2 b) -f(k^2 a)=1$ because c may be a square). This will require an estimate on the growth of $f(k^2x)$ as a function of $k$. – Jeff Tolliver Nov 14 '12 at 6:29
Thanks for commenting my question.//Then, I think that it suffice to prove $f(k^2 c)−f(k^2 a)>1$ in case that c=a+1. Is it true? – wikizero Nov 14 '12 at 6:54

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.

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It seems $+\frac{a}2$ should read $-\frac{c}2$. (And with the same idea as in your proof, one could start from $2\sqrt{a}n+1\geqslant(c-a)n^2$, deduce that $(a+1)n+n\geqslant(c-a)n^2$, hence $n\leqslant(a+2)/(c-a)$, et voilà!) – Did Nov 14 '12 at 7:05
Thanks everyone for answering my question. Have a good day. – wikizero Nov 14 '12 at 7:06
@did: I did throw away the wrong term, didn’t I? Thanks. – Brian M. Scott Nov 14 '12 at 7:09
@wikizero: You’re welcome. – Brian M. Scott Nov 14 '12 at 7:09