Show that $\sqrt{5} \notin \mathbb{Q}(\sqrt{3})$

Question

Show that $\sqrt{5} \notin \mathbb{Q}(\sqrt{3})$.

I don't even know where to start. I can't find references to this in my textbook anywhere. I feel like the notation came out of nowhere.

Answer 1

Assume $\sqrt{5}\in \mathbb{Q}(\sqrt{3}) \Rightarrow \sqrt{5} = a+b\sqrt{3}, a,b \in \mathbb{Q}$. Square both sides and deduce that $\sqrt{3}$ is rational, which is absurd, proving the claim.

Answer 2

Try solving $(a+b\sqrt{3})^2 = 5$.

So $a^2+3b^2=5$ and $2ab=0$.

Show this isn't possible.

(It's actually not possible to find rational $a,b$ such that $a^2+3b^2=5$, even without the $2ab=0$ condition.)

Answer 3

I actually had to solve this in a test yesterday.

$(a+b\sqrt 3)^2=a+2ab\sqrt3+3b^2$ so if $(a+b\sqrt3)^2=5$ then we have $ab=0$ since otherwise $ab\sqrt3$ would be irrational and irrational plus rational is irrational, while $5$ is clearly rational. From here $ab=0\implies$ $a=0$ or $b=0$

Now you just have to check if it is possible to have $a^2=5$ and $(b\sqrt3)^2=5$

But $\sqrt{5}$ is not rational and $(b\sqrt{3})^2=3b^2=5\implies b^2=\frac{5}{3}$. We also know $\sqrt\frac{5}{3}$ is not rational.