Bound on difference between irrational square root and rational number I'm trying to show that if $\alpha$ is the square root of a non-perfect square $d$, then for any rational $p/q$ there's a positive real $x$ such that $|\alpha - p/q| > x/q^2$. I'm having trouble starting on the problem, could someone give a hint? Thanks

Let $ d $ be a positive integer which isn't a square, and consider a rational $ \frac{p}{q} $ ($ p, q $ integers, $ q > 0 $).
Looking at $ \left| \sqrt{d} - \frac{p}{q} \right| $, it is $ \dfrac{1}{q} |q\sqrt{d} - p | $ $ = \dfrac{1}{q} \dfrac{|q^2 d - p^2|}{|q\sqrt{d}+p|} $. As $ \sqrt{d} $ is irrational, numerator $ |q^2 d - p^2 | $ is non-zero (and an integer), making $ | q^2 d - p^2 | \geq 1 $.
So $ \left| \sqrt{d} - \dfrac{p}{q} \right| \geq \dfrac{1}{q^2} \dfrac{1}{\left| \sqrt{d} + \frac{p}{q} \right|}$. Notice denominator $$\left| \sqrt{d} + \frac{p}{q} \right|= \left| 2\sqrt{d} -\sqrt{d} + \frac{p}{q} \right|\leq 2\sqrt{d} + \left| -\sqrt{d} + \frac{p}{q} \right|,$$ giving us
$$  \left| \sqrt{d} - \dfrac{p}{q} \right| \geq \dfrac{1}{q^2} \left(\dfrac{1}{2\sqrt{d} + \left| \sqrt{d} - \frac{p}{q} \right|} \right).$$
Writing $ t := \left| \sqrt{d} - \dfrac{p}{q} \right| $, this is $ t \geq \dfrac{1}{q^2 (2\sqrt{d} + t)} $. So $ q^2 t^2 + q^2 2\sqrt{d} t - 1 \geq 0 $.
Quadratic on LHS has roots $ \dfrac{-q \sqrt{d} \pm \sqrt{ q^2 d + 1 }}{q} $, so the inequality becomes $$ \left(t-\frac{-q\sqrt{d} - \sqrt{q^2 d+1}}{q} \right) \left( t - \frac{-q\sqrt{d}+\sqrt{q^2 d+1}}{q} \right) \geq 0.$$
The first factor is anyways $ > 0 $, giving $ t \geq \dfrac{-q\sqrt{d} + \sqrt{q^2 d + 1}}{q} $. Now RHS $$ \dfrac{-q\sqrt{d} + \sqrt{q^2 d + 1}}{q} = \dfrac{1}{q(q\sqrt{ d} + \sqrt{q^2 d + 1})}=\dfrac{1}{q^2\left(\sqrt{d} + \sqrt{d + \dfrac{1}{q^2}}\right)}\geq  \dfrac{1}{q^2(\sqrt{d} + \sqrt{d+1})}= \dfrac{\sqrt{d+1}-\sqrt{d}}{q^2}$$

So finally, we have
$$ \left| \sqrt{d} - \dfrac{p}{q} \right| \geq \dfrac{\sqrt{d+1}-\sqrt{d}}{q^2} $$
for any integers $ p, q $ with $ q > 0 $.
