# $\sqrt{6}+\sqrt{3}$ is not rational proof

I want to prove $$\sqrt{6}+\sqrt{3}$$ is not rational; here is my attempt:

Assume for the sake of contradiction that $$\sqrt{6}+\sqrt{3}$$ is rational. Then $$(\sqrt{6}+\sqrt{3})^2$$ must also be rational. Since $$(\sqrt{6}+\sqrt{3})^2=9+2\sqrt{6}\sqrt{3}=9+2\sqrt{2}\sqrt{3}\sqrt{3}=9+6\sqrt2,$$

we see $$9+6\sqrt2$$ must be rational. But, since $$\sqrt{2}$$ is not rational we have a contradiction and hence $$\sqrt{6}+\sqrt{3}$$ is not rational.

Is it correct?

• It seems like a good solution to me, since you suppose that $\sqrt{6}+\sqrt{3}$ is rational then $\sqrt{2}=\frac{(\sqrt{6}+\sqrt{3})^2-9}{6}$ is also rational. Which is a contradiction. But it is only valid assuming that you have already proved that $\sqrt{2}$ is irrational. Commented Oct 19, 2021 at 6:16

Looks more or less ok, I would just add one more step.

You correctly prove that if $$\sqrt 6+\sqrt 3$$ is rational, then $$9+6\sqrt2$$ is also rational.

I would now add one more step and say that this also means that because $$\sqrt{2} = \frac{9+6\sqrt{2} - 9}{6}$$, that therefore, $$\sqrt{2}$$ is also rational, and only then move on to the conclusion.

It's a minor thing, but I would rather be a little too clear than skip an important step. Also, you can then completely explicitly write that

$$\sqrt{2} = \frac{(\sqrt{6} + \sqrt{3})^2 - 9}{6}$$

and make the connection between the two numbers even more explicit.

• Can the user that downvoted this answer please explain why they think the answer deserves a downvote? Usually, downvotes are reserved for answers that are wrong, so what in the answer is wrong?
– 5xum
Commented Oct 19, 2021 at 6:38

You can also appeal to the rational root theorem, \begin{align*} x &= \sqrt{3} + \sqrt{6}\\ x^2 &= 9 + 6\sqrt{2}\\ x^4-18x^2 &+9=0 \end{align*} So if $$x = p/q$$ with $$p,q$$ coprime then $$p |9, q|1$$ or $$p = 1,3,9$$ and $$q=1,-1$$, but none of the options qualify as $$x$$ is not an integer