When is $\sqrt{a}-\sqrt{b}$ rational?

Question

Here's what I have so far:

$$\sqrt{a}-\sqrt{b} = m/n \implies a^2 + b^2 -2\sqrt{ab} = (m/n)^2$$ Therefore, $\sqrt{ab}$ must be rational, $ab$ must be a perfect square. Also,

$$\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right) = a - b$$

Therefore, in order for $\sqrt{a}-\sqrt{b}$ to be rational, $\sqrt{a}+\sqrt{b}$ must also be rational for all $a$ and $b$ such that $a \neq b$.

Is this a good enough answer?

$\sqrt{\pi^2}-\sqrt{(\pi-1)^2}$ is rational... What restrictions do you have on $a$ and $b$? — Ian Miller, Nov 07 '16 at 12:26
$\sqrt{a}-\sqrt{b}\in\mathbb{Q}\implies(\sqrt{a}\in\mathbb{Q}\wedge\sqrt{b}\in\mathbb{Q})\vee(\sqrt{a}\not\in\mathbb{Q}\wedge\sqrt{b}\not\in\mathbb{Q})$ — barak manos, Nov 07 '16 at 12:30
Of course, it doesn't help with the other direction, which is what your question is all about... — barak manos, Nov 07 '16 at 12:31

Arthur · Answer 1 · 2016-11-07T12:37:53.760

1

If you require $a,b$ to be rational and unequal, then your last part is not quite finished: since $\sqrt a-\sqrt b$ and $\sqrt a+\sqrt b$ are both rational, their sum, $2\sqrt a$ and their difference $2\sqrt b$ must also be rational.

edited Nov 07 '16 at 12:37

answered Nov 07 '16 at 12:31

Arthur

199,419

When is $\sqrt{a}-\sqrt{b}$ rational?

1 Answers1