5
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ 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. $\endgroup$
    – Zaragosa
    Commented Oct 19, 2021 at 6:16

2 Answers 2

7
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$
    – 5xum
    Commented Oct 19, 2021 at 6:38
4
$\begingroup$

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

$\endgroup$

You must log in to answer this question.