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Prove that for for all $p,q\in \mathbb{Z}$, $q>0$ we have: $$ \left| \sqrt{2}-\frac{p}{q} \right| > \frac{1}{3q^2}. $$

To be honest, I do not know where to start - any help would be appreciated.

Jyrki Lahtonen
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Mat Dyl
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2 Answers2

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You can assume that $p>0$ and $q>1$, and $\sqrt 2 + p/q ≤ 3$, otherwise this is easy: if $\sqrt 2 + p/q > 3$ then $\sqrt 2-p/q < 2\sqrt 2 - 3<0$, so $$\left|\sqrt2 - \frac{p}{q}\right| > 3-2\sqrt 2 > 1/12 ≥ 1/(3q^2)$$

The highest power of $2$ dividing $2q^2$ is odd, while the highest power of $2$ dividing $p^2$ is even. Then, $p^2$ and $2q^2$ must be distinct integers, thus $|2 q^2 - p^2| \geq 1$. Then

$$\left|\sqrt2 - \frac{p}{q}\right| = \frac{|2p^2-q^2|}{q^2(\sqrt{2}+p/q)} \ge \frac{1}{q^2(\sqrt2 + p / q)} \ge \frac{1}{3q^2},$$

as desired.

Watson
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    $|2q^2-p^2|\ge 1$ can also by explained by the irrationality of $\sqrt{2}$ (irrationality gives it nonzero, the difference between 2 integers is at least 1). – tong_nor Feb 14 '16 at 20:49
  • @Watson Not really sure why the inequality is trivial if $\sqrt{2}+p/q>3$. – egreg Feb 14 '16 at 20:52
  • @egreg : I hope this is clearer now. You're right, I should have written a little bit more about this part. – Watson Feb 14 '16 at 20:59
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$\sqrt{2}$ is an algebraic number of order $2$ for it is irrational and satisfies $f(x)=x^2-2=0$. For $x\in [\sqrt{2}-1,\sqrt{2}+1]$, $\frac{1}{4} <f'(x)=2x<5$. Thus, by the mean value theorem $$ |f(\sqrt{2})-f(x)|=|x^2-2|< 5|\sqrt{2}-x|$$

Claim: for any integers $p$, $q$ ($q\geq1$) $$ \begin{align} \big|\sqrt{2}-\frac{p}{q}\big|>\frac{1}{5q^2}\tag{1}\label{one} \end{align} $$

This clearly holds true if $|\sqrt{2}-\tfrac{p}{q}|> 1$. If $|\sqrt{2}-\tfrac{p}{q}|\leq 1$, then by $\eqref{one}$ $$ \big|f(p/q)|=\big|\tfrac{p^2}{q^2}-2|< 5|\sqrt{2}-\tfrac{p}{q}| $$ and so $$ q^2|f(p/q)|<5q^2|\sqrt{2}-\tfrac{p}{q}| $$ Clearly $q^2|f(p/q)|=|p^2-2q^2|$ is a nonnegative integer. Since the equation $f(x)=x^2-2=0$ has no rational solution, $q^2|f(p/q)|\geq1$. Putting things together, $$ \big|\sqrt{2}-\tfrac{p}{q}\big|>\frac{1}{5q^2} $$ for all $p,q\in\mathbb{Z}$ with $q\geq1$.

Comment: This solution I outlined above is an adaptation of standard technique used by Liouville to prove the following result:

Theorem: For any real algebraic number of degree $n>1$, there exists a positive integers $M$ such that $$ \Big|z-\frac{p}{q}\Big|>\frac{1}{Mq^n} $$ for all integers $p$ and $q$, $q\geq1$.

Mittens
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