Prove that for all positive reals $k,j$ with $2k>3j$, there exists positive integers $s,t$ such that $k > |\sqrt{s}-\sqrt{t}| > j$ Prove that for all positive reals $k,j$ with $2k&gt;3j$, there exists positive integers $s,t$ such that

$k &gt; |\sqrt{s}-\sqrt{t}| &gt; j$
Let's show that there are positive integers $s, t$ such that:

$3j/2 &gt; |\sqrt{s} - \sqrt{t}| &gt; j$.

Note that $2k &gt; 3j$ iff $k &gt; 3j/2$, so that if we can find $s, t$ as above we'll be done.

Observe $\sqrt{n+1} - \sqrt{n} = 1/(\sqrt{n+1} + \sqrt{n}) \longrightarrow 0$ as $n$ tends to infinity. Pick $n$ large enough for this difference to be less than $j/2$. Then choose $a \in \mathbb{Z}^+$ such that $3j/2 &gt; a(\sqrt{n+1} - \sqrt{n}) &gt; j$. 

(This will be possible, since increasing our prospective $a$ by $1$ amounts to taking "steps" of size less than $j/2$. Thus, you don't run the risk of totally "overstepping" the $j/2$ gap between $3j/2$ and $j$.)

Pick $s = a^2 (n+1)$ and $t = a^2 n$.
Another solution:

Select a rational $b/a$ with $a,b$ positive integers such that $2k&gt;b/a&gt;3j$. By multiplying the numerator and denominator by a constant integer, if necessary, we could ensure $3a^2&gt;b$. Now we claim $s=a^2+b,t=a^2$ satisfy the condition. Proof:
$k&gt;\frac{b}{2a}&gt;\sqrt{a^2+b}-\sqrt{a^2}=\frac{b}{\sqrt{a^2+b}+a}&gt;\frac{b}{3a}&gt;j$