Showing that there exists infinitely many rational numbers $\frac{p}{q}$ satisfying $|x - p/q|\leq \frac{1}{q^{2+t}}$ for irrational $x$ and $t>0$

Question

Let $t>0$ be a fixed parameter and $x\in\mathbb{R}$ be an irrational number. I am trying to show that there exists infinitely many rational numbers $\frac{p}{q}$ such that $\left|x - \frac{p}{q}\right|\leq \frac{1}{q^{2 + t}}$. I was told that this claim follows almost immediately from contined fractions. Unfortunately I haven't really worked with continued fractions so I am clueless on how I should prove the result. The tricky part seems to be getting the $t$-dependence right to the upperbound.

An alternative approach would be to maybe use a proof by contradiction and the density of rational numbers, although I don't know what you really can say about the denominator of the rational numbers close enough to $x$.

Answer 1

Mathworker21 and Conrad gave examples showing that the assertion is false. In fact it is false for almost every $x$ according to Lebesgue measure (Of course they know this well, one example suffices to refute a general assertion.) This follows from Khinchin's theorem on metric Diophantine approximation [1], see also [2], [3]. As a special case of this theorem, we find that for every $\epsilon>0$ and almost every real $x$, there are only finitely many rational numbers $p/q$ such that $|x-p/q|< q^{-2-\epsilon}$. The statement in the previous line just uses the easy direction of Khinchin's theorem, which follows immediately from the first Borel-Cantelli Lemma.

[1] A. Khintchine, "Zur metrischen Theorie der diophantischen Approximationen" Math. Z. , 24 (1926) pp. 706–714

[2]https://en.wikipedia.org/wiki/Diophantine_approximation#Khinchin's_theorem_on_metric_Diophantine_approximation_and_extensions

[3] https://carmamaths.org/pdf/retreat2013/retreat2013-mumtaz_hussain.pdf

Answer 2

Let us consider the number set in the form of $\;\mathbf{z=\dfrac m{3q^n}},\;$ where $$\mathbf{n\in\mathbb N},\quad \mathbf{n>2+t},\quad m\in \mathbb Z.$$

Since the interval between the neighboring points of the building set is less than half of the given interval, then exists two or more points of the set, which belongs to the given interval. Let the left of obtained point is $\;\dfrac l{3q^n},\;$ then the next one is $\;\dfrac {l+1}{3q^n}.\;$

I.e., obtained a pair of the inner rational points of the given interval.

Calculating the center of the obtained interval, we can get two new intervals. $$\left(\dfrac{l}{3q^n},\dfrac {2l+1}{6q^n}\right), \left(\dfrac{2l+1}{6q^n}, \dfrac{l+1}{3q^n}\right).$$

Repeating pointed dichotomi procedure for each of the obtained intervals, easily to get arbitrary many rational numbers, which belongs to the given interval.

$\color{green}{\textbf{Appendium.}}$

Continued fractions defined by the formulas $$x=[q_0;q_1,q_2,\dots]=q_0+\cfrac1{q_1+\cfrac1{q_2+\dots}},$$ where $$q_0=\lfloor x\rfloor,\quad r_0 = x - q_0,\quad q_{s+1} =\genfrac\lfloor\rfloor{}01{r_s},\quad r_{s+1} = \dfrac1{r_s}-q_{s+1},\quad s=0,1\dots$$

Approximations of the number $x$ of the form $\dfrac{P_s}{Q_s}$ can be calculated using the formulas: $$P_0=1,\quad Q_0=0,\quad P_1=q_0,\quad Q_1=1,$$ $$P_{s+1}=q_{s+1}P_s+P_{s-1},\quad Q_{s+1}=q_{s+1}Q_s+Q_{s-1},\quad s=1,2,\dots$$

According to the logic of continued fractions (compact approximations of the number $x$), this process should not be infinite. Therefore, calculations should be terminated not only when the fractional part is zero, but also when the specified accuracy is reached.