A Lower Bound for Dirichlet’s Approximation

Question

In this problem, I have successfully deduced that $\forall x\in\mathbb{R}\backslash\mathbb{Q}$, if $x$ is of the form $x=nr-[nr]$, with $n\in\mathbb{N}$ and $r\in\mathbb{R}\backslash\mathbb{Q}$ that if we choose two numbers of this form of $x$, their difference is less than $\varepsilon$ and from this we can derive, Dirichlet’s Approximation.

Issue I am Facing : It appears, rather quite challenging to find a lower bound for Dirichlet’s Approximation say for instance the following : $\left|x-\frac{p}{q}\right|\geq \frac{1}{q^{2}(1+2\sqrt{2})}$ with $p\in\mathbb{Z}$ and $q\in\mathbb{Z^{+}}$. I wish to prove this inequality

My Attempt : I thought of considering two cases, the first case is to consider $\left|x-\frac{p}{q}\right|\geq1$ and second case to be $\left|x-\frac{p}{q}\right|\leq1$ for the purpose of simplicity, moreover, assume that $x=\sqrt{2}$, if I could find the minimum of $|p^{2}-2q^{2}|$ then I would make some progress but I am unable to take any further steps

Sidenote : How is the name 'Dirichlet' correctly pronounced?

dee-ree-shlay (sometimes dih-rih-shlay) is how English speakers pronounce the name (the first is closer to the original French) — Greg Martin, Oct 04 '20 at 20:31
The inequality, you are trying to prove, depends on $x$. See this conversation. — rtybase, Oct 04 '20 at 20:41

score 0 · Accepted Answer · answered Oct 04 '20 at 20:45

Certainly it is not true in general that $|x-p/q|$ must always be larger than $1/q^2(1+\sqrt2)$, and indeed irrational numbers can have extremely good rational approximations (see Liouville numbers for example).

For any real number $x$, there exist infinitely many rational approximations $p/q$ with $|x-p/q| < 1/q^2\sqrt5$; this is Hurwitz's theorem. This is best possible in general (the golden ratio is an example).

For the specific number $x=\sqrt2$ we have much more information, coming from its periodic continued fraction expansion. The inequality $|x-p/q| \ge 1/q^2C$ holds for any $C>2\sqrt2$ with only finitely many exceptions, but there are infinitely many exceptions (every other convergent to $\sqrt2$) for the inequality $|x-p/q| \ge 1/q^22\sqrt2$ itself.

I see, therefore I must show that $|x-p/q| \ge 1/q^2(1+2\sqrt2)$, I request your assistance — , Oct 04 '20 at 22:36
See for example the chapter on continued fractions of Niven, Zuckerman, and Montgomery's book. It's a theorem that if $|x-p/q| < 1/2q^2$ then $p/q$ must be a convergent to the continued fraction of $x$, and then one can measure q^2|x-p/q|$ quite precisely. — Greg Martin, Oct 04 '20 at 22:46

A Lower Bound for Dirichlet’s Approximation

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