Showing $\mathbb{Q}(\sqrt{a},\sqrt{b})= \mathbb{Q}(\sqrt{a}+\sqrt{b})$

Question

For $a, b\in\mathbb{Q}$ I have to show that $\mathbb{Q}(\sqrt{a},\sqrt{b})= \mathbb{Q}(\sqrt{a}+\sqrt{b})$.

Unfortunately I dont even know where to start. We never used these brackets $(-)$ before, so I dont know how these sets are actually defined.

I would be grateful for the definitions (:D) and any kind of help or advice! Thank you!

Answer 1

Hint: ${1\over{\sqrt a+\sqrt b}}$ $={{\sqrt a-\sqrt b}\over {a-b}}$ implies that $\sqrt a-\sqrt b\in\mathbb{Q}(\sqrt a+\sqrt b)$.

Answer 2

If $a=b$ it's obvious.

Now,let $a\neq b$.

$$\sqrt{ab}=\frac{(\sqrt{a}+\sqrt{b})^2-a-b}{2}\in\mathbb Q(\sqrt{a}+\sqrt{b}).$$ Thus, $$(\sqrt{a}+\sqrt{b})\sqrt{ab}=a\sqrt{b}+b\sqrt{a}=a(\sqrt{a}+\sqrt{b})+(b-a)\sqrt{a}\in\mathbb Q(\sqrt{a}+\sqrt{b}).$$ Can you end it now?