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Let $A=\Big\{x=\frac pq\in\mathbb{Q} :|\sqrt 2-x|<\frac1{q^3}\Big\}$ We want to prove that $A$ is finite and find its cardinal? we can prove that it's finite by using directly Roth’s theorem which is a generalized theorem of Lionville's theorem. But, I am stuck on finding the number of its elements. Any help, and thanks in advance.

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  • $\begingroup$ There is a result by K. F. Roth, Mathematika 2, 1955, part 1, no. 3, pp. 1-20 that states that if $x$ is an irrational algebraic number and there are infinitely many $p/q$ such that $|x=\tfrac{p}{q}|\leq q^{-\mu}$, then $\mu\leq 2$. $\endgroup$
    – Mittens
    Aug 29, 2021 at 22:29
  • $\begingroup$ Check this, it might help to reduce from the numbers to check "manually". $\endgroup$
    – rtybase
    Aug 29, 2021 at 22:36

4 Answers 4

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Notice that your problem is equivalent to show that: $$-\frac{1}{q^3}<\sqrt{2}-\frac{p}{q}< \frac{1}{q^3}$$ And this is equivalent to solve the following system of inequalities over $\mathbb{Z}$: $$\sqrt{2}q^3-pq^2-1<0$$ $$\sqrt{2}q^3-pq^2+1>0$$ where $q\neq0$. From this we can split into two cases: $q>0$ or $q<0$

If $q>0$ then $q\geq1$ and so $\sqrt{2}q^3-pq^2=q^2\left(\sqrt{2}q-p\right)\geq \sqrt{2}-p$. Thus: $$\sqrt{2}-p-1\leq\sqrt{2}q^3-pq^2-1<0\longrightarrow p>\sqrt{2}-1$$ $$\sqrt{2}q^3-pq^2+1>\sqrt{2}-p+1>0 \longrightarrow p<\sqrt{2}+1$$ Therefore $\sqrt{2}+1>p>\sqrt{2}-1$ which implies $p\in\lbrace0,1,2\rbrace$. Checking the three possible values for $p$ one can easily get that $(p,q)\in\lbrace{(1,1),(2,1)\rbrace}$

In a similar way with $q<0$ one gets $(p,q)\in\lbrace{(-1,-1),(-2,-1)\rbrace}$.

Since we do not care about the specific values of $p$ and $q$,but on the quotient $\frac{p}{q} $we get that $\frac{p}{q}\in\lbrace1,2\rbrace$ and so $A=\lbrace 1,2\rbrace$.

Hope you can fill the details.

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if $|x- \sqrt 2 | < 1,$ then $x>0 $ and $ x + \sqrt 2 < 1 + 2 \sqrt 2 = 1 + \sqrt 8 $

then $$ (1 + \sqrt 8) \; |x- \sqrt 2 | > (x+ \sqrt 2 ) |x- \sqrt 2 | > \frac{|p^2 - 2 q^2|}{q^2} \geq \frac{1}{q^2} $$

and $$ |x- \sqrt 2 | > \frac{1}{ (1 + \sqrt 8) q^2} $$

IF we also demand $$ \frac{1}{q^3} > |x- \sqrt 2 | ..... \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

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If $0<|x-\sqrt 2|<1/q^3\le 1$ then $0<q$

and $0<-1+\sqrt 2<x<1+\sqrt 2<3$ so $0< x+\sqrt 2<4.$ So we have $$1\le|p^2-2q^2|=q^2\cdot |x-\sqrt 2|\cdot (x+\sqrt 2)<$$ $$<q^2\cdot q^{-3}\cdot(x+\sqrt 2)<4/q \implies$$ $$\implies 1<4/ q\implies q\in \{1,2,3\}.$$ This narrows the search to those $p/q$ such that $-1+\sqrt 2<p/q=x<1+\sqrt 2$ with $q\in \{1,2,3\}$ and $p\in\Bbb Z^+$.... I leave the rest to you.

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Note that $|2 - \frac{p^2}{q^2}| \ge \frac{1}{q^2}$, and $\sqrt{2}+\frac{p}{q}< \sqrt{2} + (\sqrt{2} + \frac{1}{q^3})$ therefore

$$\frac{1}{q^3} >|\sqrt{2} - \frac{p}{q}| = \frac{|2 - \frac{p^2}{q^2}|}{\sqrt{2}+\frac{p}{q}}> \frac{\frac{1}{q^2}}{2 \sqrt{2}+\frac{1}{q^3}} $$ and so $$2\sqrt{2}+\frac{1}{q^3}> q$$ We conclude that $q\le 2$, and so $x=1$, $x=2$, or $x=\frac{3}{2}$ ($0< \frac{3}{2} - \sqrt{2} < \frac{1}{10} < \frac{1}{8}$).

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    $\begingroup$ $q=2$ is possible. Let $x=3/2$. $\endgroup$ Aug 30, 2021 at 0:41
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    $\begingroup$ @DanielWainfleet: Thank you! Corrected that. $\endgroup$
    – orangeskid
    Aug 30, 2021 at 2:37

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