Show that if $x, y, \sqrt{x} + \sqrt{y}∈ \mathbb Q$ then $\sqrt{x}, \sqrt{y} ∈ \mathbb Q$

Question

Show that if $x, y, \sqrt{x} + \sqrt{y}∈ \mathbb{Q}$ then $\sqrt{x}, \sqrt{y} ∈ \mathbb Q$.

If $r=\sqrt{x} + \sqrt{y} ∈ \mathbb{Q}$, then so is $$r^2 = x + y + 2\sqrt{xy},$$ which implies that $$\sqrt{xy} = \frac{1}{2}(r^2 - x - y) ∈ \mathbb{Q}.$$

From $\sqrt{xy} ∈ \mathbb{Q}$ we have that $\sqrt{x}, \sqrt{y} ∈ \mathbb{Q}$. To show this is true we suppose the opposite.

Wlog if $\sqrt{y} \not∈ \mathbb{Q}$ and $\sqrt{x} ∈ \mathbb{Q}$, then we would have $\sqrt{xy} \not∈ \mathbb{Q}$.

Assume that $\sqrt{xy} ∈ \mathbb{Q}$ when $\sqrt{y} ∈ \mathbb{Q}$.

So we have $$\sqrt{xy} = \sqrt{x}\sqrt{y} = \frac{p}{q}\sqrt{y},$$ where $\dfrac{p}{q} = \sqrt{x}$ and $(p,q)=1$, which is irrational. Contradiction.

I don't know if this is right. Can somebody please help me with this problem and tell me what should I do?

Answer 1

$(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)=x-y$ this implies that $\sqrt x-\sqrt y\in\mathbb{Q}$ as a quotient of two rational.

$2\sqrt x=(\sqrt x+\sqrt y)+(\sqrt x-\sqrt y)\in\mathbb{Q}$ implies that $\sqrt x\in\mathbb{Q}$.