# Prove that $\sqrt{2}+\sqrt{4}$ is irrational [duplicate]

As the question says, how can I prove that $$\sqrt{2}+\sqrt{4}$$ is irrational?

I have tried setting it to be equal to $$a$$, and $$\sqrt{4}$$ equal to $$a^2$$, but I haven't gotten anywhere. The solution does not have to use the above. Any help is appreciated. (I know this is a duplicate, but I haven't seen any answers that I have gotten a full solution from)

• One application of Fermat's Last Theorem is to show that $\sqrt{2}$ is irrational. :) – user844292 Nov 2 '20 at 4:55
• That application is discussed here – J. W. Tanner Nov 2 '20 at 22:39
• Not only is this a duplicate. It is a duplicate of a duplicate. Shame on the high rep users who don't search (took me less than 5 seconds on Approach0). – Jyrki Lahtonen Nov 11 '20 at 10:58

Let $$\alpha=\sqrt2+\sqrt4$$.
Then $$\alpha^3=2+4+3\sqrt{32}+3\sqrt{16}=6+6\sqrt4+6\sqrt2=6+6\alpha.$$
Apply the rational root theorem to $$x^3-6x-6$$.
Since, $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ and $$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(a-b)^2=0$$ for $$a=b=c$$ only and $$\sqrt2\neq\sqrt4,$$ we obtain: $$\sqrt2+\sqrt4-x=0$$ is equivalent to $$2+4-x^3+3\sqrt2\cdot\sqrt4\cdot x=0$$ or $$x^3-6x-6=0.$$ But the polynomial $$x^3-6x-6$$ is irreducible by Eisenstein (just take $$p=2$$, for example): https://en.wikipedia.org/wiki/Eisenstein%27s_criterion
Thus, the polynomial $$x^3-6x-6$$ has no rational roots.