Putnam 2011, Problem B-1

Let $h$ and $k$ be positive integers. Prove that for every $\varepsilon > 0$, there are positive integers $m$ and $n$ such that
$$\varepsilon < |h\sqrt{m} - k\sqrt{n}| < 2\varepsilon.$$

Solution:
Since the rational numbers are dense in the reals, we can find positive integers \(a,b\) such that  
\[
\frac{3\varepsilon}{hk} < \frac{b}{a} < \frac{4\varepsilon}{hk}.
\]  
By multiplying \(a\) and \(b\) by a suitably large positive integer, we can also ensure that \(3a^2 > b\). We then have  
\[
\frac{\varepsilon}{hk} < \frac{b}{3a} < \frac{b}{\sqrt{a^2 + b} + a} = \sqrt{a^2 + b} - a
\]  
and  
\[
\sqrt{a^2 + b} - a = \frac{b}{\sqrt{a^2 + b} + a} \leq \frac{b}{2a} < 2 \frac{\varepsilon}{hk}.
\]  
We may then take \(m = k^2(a^2 + b)\), \(n = h^2 a^2\).