I saw this one in an analysis text. The author didn't say where this inequality comes from or where it is used. I'd like to know more about it if anyone has information on it. Looks like some kind of number theoretic inequality since $\displaystyle \sqrt {p}$ with $\displaystyle p$ prime is always irrational.

Let $\displaystyle a \in \mathbb{N}$ be such that $\displaystyle \sqrt{a}$ is irrational. We want to prove there is some $\displaystyle c > 0$ such that for all integers $\displaystyle p, q$ with $\displaystyle q > 0$, we have $\displaystyle |q \sqrt{a}-p| > \frac{c}{q}$.

I have no clue on how to prove this one. The hint was to rationalize. So $\displaystyle (q \sqrt{a}-p)(-q\sqrt{a}-p)=-q^2a+p^2.$ Then $\displaystyle |p^2-q^2a| \geq 0$ but supposing $\displaystyle |p^2-q^2a| = 0$, we have $\displaystyle |p-q\sqrt{a}||p+q\sqrt{a}| = 0 $ and $\displaystyle |p-q\sqrt{a}| = 0 $ implies $\displaystyle q\sqrt{a}$ is an integer which is a contradiction. Therefore $\displaystyle |p^2-q^2a| > 0$ and $\displaystyle |p^2-q^2a|$ is an integer. So $\displaystyle |p^2-q^2a| \geq 1$.