# Cardinal of the set $A=\Big\{x=\frac pq\in\mathbb{Q} :|\sqrt 2-x|<\frac1{q^3}\Big\}$

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.

• 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 $\mu\leq 2$. Aug 29, 2021 at 22:29
• Check this, it might help to reduce from the numbers to check "manually". Aug 29, 2021 at 22:36

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.

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

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

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

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 $$x=\frac{3}{2}$$ ($$0< \frac{3}{2} - \sqrt{2} < \frac{1}{10} < \frac{1}{8}$$).

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