$\sqrt{a},~\sqrt{b}$ is irrational but $\sqrt{a}-\sqrt{b}\in\Bbb Q$, then they are equal

Question

Let $a,~b\in\Bbb Q$ and suppose $\sqrt{a},~\sqrt{b}$ is irrational and $\sqrt{a}-\sqrt{b}\in\Bbb Q$. I want to prove that $\sqrt{a}-\sqrt{b}=0$; that is, $\sqrt{a}=\sqrt{b}$. It seems straightforward by observing some practical numbers. However I found it hard to write a formal proof.

I have tried squaring them, but gained nothing. And I have also searched on this site but nothing was found.

Answer 1

$\sqrt{a} - \sqrt{b} = r$

$\sqrt{a} = r + \sqrt{b}$

$a = (r + \sqrt{b})^2 = r^2 + 2 r \sqrt{b} +b$

If $r \neq 0$, then :

$\frac{a - b - r^2}{2r} = \sqrt{b}$.

The left side is rational.

Answer 2

Observe that $(\sqrt a-\sqrt b)(\sqrt a +\sqrt b)=a-b\in \Bbb Q$, if $\sqrt a\ne\sqrt b$ then $\sqrt a+\sqrt b=\frac {a-b}{\sqrt a-\sqrt b}\in\Bbb Q$. But so $\sqrt a=\frac 12(\sqrt a-\sqrt b+ \sqrt a+\sqrt b)\in\Bbb Q$, a contradiction.

We conclude $\sqrt a=\sqrt b$

Answer 3

This is a hint

Suppose $\sqrt a-\sqrt b\in\mathbb{Q}$. Then $(\sqrt a-\sqrt b)^2\in\mathbb{Q}$ and so $$ \sqrt{ab}=\frac{a+b}2\in\mathbb{Q} $$ with $ab\in\mathbb{Q}$.

What are the $q\in\mathbb{Q}$ such that $\sqrt q\in\mathbb{Q}$, say in terms of their prime factorization?