Let $a\in \mathbb{Z}^{+}$ such that $\alpha=\sqrt{a}$ is irrational. Show that there is a positive number $c$ such that for every $p,q\in \mathbb{Z}^{+}$, we have, $$|q\alpha-p|>\frac{c}{q}.$$
My attempt:
Notice that, $$|q\alpha-p|=|q\sqrt{a}-p|=\left|\dfrac{(q\sqrt{a}-p)(q\sqrt{a}+p)}{q\sqrt{a}+p}\right|=\dfrac{|q^2a-p^2|}{q\sqrt{a}+p}.$$ It's clear that $|q^2a-p^2|\geq 1$, because $\sqrt{a}$ is irrational. So, we have, $$|q\alpha-p|\geq \frac{1}{q\sqrt{a}+p}.$$ How can we use this lower bound in order to find $c$? I would appreciate any help. Thanks!
Let's continue working on the last inequality $|q \alpha - p| \geq \frac{1}{q \sqrt{a} + p}$ that you have established. We want a further lower bound, but without the $p$. This entails enforcing some bound of the form $g(q) \geq p$.
First, notice that we only really care about $p/q$ with $(p/q)^2 \approx a$ or equivalently $p \approx \sqrt{a} q$. Thus, we do not lose generality by only considering pairs $(p,q)$ obeying $p < q\sqrt{10a + 3}$, because pairs $(p,q)$ with $p/q \geq \sqrt{10a+3}$ would be ludicrously bad approximations to $\sqrt{a}$ anyway. Hence we have the estimate $$|q \alpha - p| \geq \frac{1}{q \sqrt{a} + p} > \frac{1}{q \sqrt{a} + q\sqrt{10a + 3}} = \frac{C}{q}$$
where $C=\frac{1}{\sqrt{a} + \sqrt{10a+3}}$. (This is of course far from tight).