Given $x,y\in\mathbb{Q}$ such that $x<y$, $\exists r\in\mathbb{Q} : x<r^2<y$

Question

Given two postive rational numbers $x$ and $y$ such that $x<y$, how do you prove that there exists a rational number $r$ such that $x<r^2<y$. I am aware that since $\mathbb{Q}$ is dense in $\mathbb{R}$, and $x$ & $y$ are positive, there exists a rational number $r$ such that $\sqrt{x}<r<\sqrt{y}$. But I would like a proof which gives such a rational number, while just staying within the context of the rational number system.

Answer 1

Assuming that $y>x\geq0$, then $x < r^2 <y $ a valid answer for $r^2 = \frac{x+y}{2}$ but this means that $r = \sqrt{\frac{x+y}{2}}$ , and this means that $r$ not necessarily a rational number.

To fix this, let $r_0 = \frac{x+y}{2}$ and $r_{n+1}= \frac{1}{2} (r_0 + \frac{x+y}{2r_n})$ (Newton's method for computing the square root).

And stop when $x<r_n^2 <y$ , we are now sure that $r_n \in \mathbb{Q}$ ,since $\mathbb{Q}$ is closed under addition,multiplication.

For example : $x=\frac{1}{3} $ and $y=\frac{1}{2}$ thus

$r_0 = \frac{5}{12}$ and $r_0^2 \leq \frac{1}{3}$ so it does not work,

Next $r_1 = \frac{17}{24}$ and $r_1^2 \geq \frac{1}{2}$ so it does not work,

Next $r_2 = \frac{529}{816}$ and $\frac{1}{3} < r_2^2 < \frac{1}{2}$ and we are done.

Note : Newton's method for approximating square roots converges very fast, yet there are many better method, so you will not reach very big $n$ to find a good $r_n$.