Lower Bound on |x-y√D|

Stub explanation

We know that $\sqrt{2}$ is irrational. But "how irrational" is it?

One fruitful way to "measure irrationality" is to look at the distances between $\sqrt{2}$ and various rational numbers. More precisely, we can ask

How far is $\sqrt{2}$ away from a given rational number $x/y$? The rationals are dense in the reals, so some fractions $x/y$ will be arbitrarily close to the real number $\sqrt{2}$: approximation gets easier as the denominator $y$ increases. But can we get a lower bound on the distance $|\frac{x}{y} - \sqrt{2}|$ in terms of the denominator $y$?

(It will be more natural to try to lower bound $|x - y\sqrt{2}|$ instead, and then divide everything by $y$ at the end.)

Yes! Let $x/y$ be any rational with numerator $x$ and denominator $y\ne0$. Since $\sqrt{2}$ is irrational, $x/y \ne \sqrt{2}$, or more naturally, $x^2/y^2 \ne 2$, or even more naturally, $x^2 \ne 2y^2$.

But $x,y$ are integers, so $x^2$ and $2y^2$ are unequal integers, which means they differ by at least $1$: $$|x^2 - 2y^2| \ge 1.$$ Let $N = x^2 - 2y^2$, so $|N| \ge 1$. By difference of squares, this factors as $$|x - y\sqrt{2}|\cdot|x + y\sqrt{2}| = |N| \ge 1.$$

We want a lower bound on $|x - y\sqrt{2}|$ (basically), so it would suffice to have an upper bound on $|x+y\sqrt{2}|$. From the triangle inequality, $|x+y\sqrt{2}| \le |x|+|y|\sqrt{2}$ (with equality when $x,y$ have the same sign—which is always the case when $x/y$ is close enough to the positive number $\sqrt{2}$). So $$|x - y\sqrt{2}| = \frac{|N|}{|x+y\sqrt{2}|} \ge \frac{|N|}{|x|+|y|\sqrt{2}} \ge \frac{1}{|x|+|y|\sqrt{2}}.$$ So we wanted a lower bound in terms of $|y|$, but right now we only have a lower bound in terms of $|x|$ and $|y|$. That's not good enough if $|x|$ is a lot larger than $|y|$, but let's rephrase our original question:

In the case that $|x - y\sqrt{2}|$ is small, how small can it be? In other words, our job is easier when $|x - y\sqrt{2}|$ is not small, because we don't have to prove anything then!

So here's one way to finish:

If $|x|$ and $|y|\sqrt{2}$ are close (say within $\epsilon = 0.1$ of each other), then $|x - y\sqrt{2}| \ge \frac{1}{|x|+|y|\sqrt{2}} \ge \frac{1}{2\sqrt{2}|y| - \epsilon}$. Otherwise, if $|x|$ and $|y|\sqrt{2}$ are not within $\epsilon$ of each other, then certainly $x$ and $y\sqrt{2}$ must be more than $\epsilon$ apart, so $|x - y\sqrt{2}| \ge \epsilon$.

Always one of the two cases holds, so we conclude that for any integers $x,y$ with $y\ne 0$, we have the lower bound $$|x - y\sqrt{2}| \ge \min(\frac{1}{2\sqrt{2}|y| - \epsilon},\epsilon) \ge \frac{C}{|y|} = C|y|^{-1}.$$ for some positive constant $C>0$ (we're using rough asymptotic approximations here, to bound a complicated function in $|y|$ below by a simple power function in $|y|$, which captures the most important information).

Finally, dividing by $|y|$, gives $|\frac{x}{y} - \sqrt{2}| \ge C|y|^{-2}$. By the way, equality "basically holds" whenever $N = x^2 - 2y^2$ is small enough (intuitively, whenever our lower bound $|N|\ge1$ above is "close to holding"), including when $x^2 - 2y^2 = \pm 1$, a Pell-type equation discussed here.