Measuring Irrationality à la 3Blue1Brown vs Liouville So, I was watching this 3B1B video on measure theory and music (as you do) and was struck by how Grant uses epsilon to measure "how irrational" a real number is.
He takes all the rationals $[0, 1]$ and assigns them progressively smaller ranges to cover as follows:

Where each range is centered on its corresponding rational and has a width corresponding to the nth number in the sequence shown above (where $\epsilon \in [0, 1]$ obviously). He then claims that these ranges won't cover $\frac {\sqrt 2}{2}$ when $\epsilon = 0.3$

So, my main question is: how did he determine that? Does $\frac {\sqrt2}{2}$ have a continuing fraction representation which shows it "hiding" in the gaps between these intervals? How does he know that some large denominator fraction doesn't happen to be an extraordinarily good approximation?
Beyond that, I was wondering if we could use this idea to measure "how irrational" a real number is. To formalize this slightly, I'll define $g(x)$ as the smallest $\epsilon$ you must have in order to cover a given real number $x$ with the intervals shown above. So, $q \in \Bbb Q \implies g(q)=0$ and the video asserts that $g(\frac {\sqrt 2}{2}) > 0.3$ (and also that $g(2^{\frac {7}{12}}-1) < 0.01$ which can be shown easily)
Has this irrationality measure (or something like it) been named or studied? What $x$s give values close to 0? Values of 1? What is this function's relationship to the Liouville-Roth Irrationality Measure? etc
Any musings welcome really. Just having some fun here 
 A: I have recently got interested in continued fractions, so I'll give it a try!

*

*CONTINUED FRACTIONS.

It is a basic property of continued fractions that if $h_n/k_n$ is the sequence of convergent to $x\in\mathbb R\setminus\mathbb Q$, then
$$
\frac 1{k_n(k_n+k_{n+1})}<\left|x-\frac{h_n}{k_n}\right|<\frac 1{k_nk_{n+1}}.
$$
It follows that for all $h,k$ coprime we have
$$
\left|x-\frac hk\right|\geq\left|x-\frac{h_n}{k_n}\right|
$$
for all $k\leq k_n$.
Now take $x=\sqrt{2}/2$, which is known to have the continued fraction $[0,1,2,2,2,...]$
$$
\frac{\sqrt{2}}2=\frac 1{1+\frac 1{2+\frac 1{2+\cdots}}}.
$$
It is possible to show by the recursion $k_{n+1}=2k_n+k_{n-1}$ that
$$
k_n=\frac{(1+\sqrt 2)^n+(1-\sqrt 2)^n}2.
$$
(Yes, it is always an integer!) Incidentally, $k_n$ is well approximated as
$$
k_n\sim\frac {(1+\sqrt 2)^n}2.
$$


*THE FUNCTION $g(x)$.

Now, what is the meaning of $g(x)$ from the video. It strongly depends on the enumeration of the rationals (in the open unit interval) you choose. Let us call this enumeration $N(h/k)\in\mathbb N:=\{1,2,3,\dots\}$.
Clearly the inequality $g(x)>\epsilon$ is ensured by the inequality
$$
\left|x-\frac hk\right|>\epsilon 2^{-N(h/k)},\ \ \forall h,k\mbox{ coprime integers,}
$$
and more precisely, $g(x)$ could be defined as the sup of $\epsilon$'s making this inequality true for all $h,k$ coprime integers.
The natural choice made in the video is
$$
N(1/2)=1,\quad N(1/3)=2,\quad N(2/3)=3,\quad N(1/4)=4,\quad N(3/4)=5,\dots
$$
hence $N(h/k)\geq k$ for all $k\geq 4$ (absolutely not sharp estimate, but enough for what we need later on).
(Such $N$ is explicit essentially in terms of Euler totient function, but it does not matter here.)
Thus $g(x)>\epsilon$ is ensured by the inequality
$$
\left|x-\frac hk\right|>\epsilon 2^{-k},\ \ \forall h,k\mbox{ coprime},\ k\geq 4
$$
and by taking care separately of the cases with $k\leq 3$.


*THE INEQUALITY $g(\sqrt 2/2)>0.3$.

Suppose that we can prove the following inequality
$$
k_n(k_n+k_{n+1})<\frac {2^{k_{n-1}}}\epsilon\qquad(\star)
$$
for all $n$ sufficiently large, say $n\geq n_0$. Then by the properties of continued fractions above we get that for $k>k_{n_0}$ we have
$$
\left|x-\frac hk\right|>\left|x-\frac{h_M}{k_M}\right|>\frac\epsilon{2^{k_{M-1}}}>\frac\epsilon{2^k}.
$$
where here $M\geq 1$ is defined by $k_{M-1}<k\leq k_M$.
In this case, the inequality $(\star)$ is clearly true at least for $n$ sufficiently large (use that $k_n\sim (1+\sqrt 2)^n/2$). Thus we only have a finite number of cases to examine.
More specifically, we look at the first numerical values
$$
\begin{array}{c|cccccccc}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \cdots
\\ \hline
k_n(k_n+k_{n+1}) &4&30&168&986&5740&33462&195024&\cdots
\end{array}
$$
and
$$
\begin{array}{c|cccccccc}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \cdots
\\ \hline
\frac{2^{k_{n-1}}}\epsilon & 6.67 & 6.67 & 26.67 & 426.67 & 436907. & 7.33\times 10^{12} & 2.11\times 10^{30}&\cdots
\end{array}
$$
So one needs to consider separately that also the neighborhoods around the "first" rationals (in the enumeration $N$) do not contain $x=\sqrt 2/2$, but I will omit this check.
