Show $\mathbb{Q}(2+\sqrt{5})=\mathbb{Q}(\sqrt{5})$

Question

I'm doing a problem that's similar to this and wasn't sure if I have the right idea. Here it is:

$\mathbb{Q}(2+\sqrt{5}) = \{{ a_0+a_1(2+\sqrt{5}) | a_0,a_1 \in \mathbb{Q} }\}$ $=\{a_0+2a_1+a_1\sqrt{5} | a_0,a_1 \in \mathbb{Q} \}$

But $a_0+2a_1$ is just some rational, so we get:

$\mathbb{Q}(\sqrt{5}) = \{{ b_0+b_1\sqrt{5} | b_0,b_1 \in \mathbb{Q} }\}$.

I have a feeling that this isn't enough. Any help is appreciated!

P.S. Sorry about the title, I'm typing this from my phone.

Answer 1

You can solve $\sqrt{5} = a + b(2+\sqrt{5})$. You can see that $a = -2 \in \mathbb{Q}, b = 1 \in \mathbb{Q}$. So $\sqrt{5} \in \mathbb{Q}(2+\sqrt{5})$. This suffices to prove the statement...