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?
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.