prove that root 2+root 5 is irrational

Question

parthkohli · Answer

Let $\sqrt{2} + \sqrt{5}$ be rational. Then$$\sqrt{2} + \sqrt{5} = \dfrac{p}{q}\tag{p,q integers and q not 0}$$$$\Rightarrow 2 + 5 + \sqrt{10} = \dfrac{p^2}{q^2}$$$$\Rightarrow \sqrt{10} = \dfrac{p^2}{q^2} - 7$$$$\Rightarrow \sqrt{10} = \dfrac{p^2 - 7q^2}{q^2}$$Now $p^2 - 7q^2$ and $q^2$ are integers. But $\sqrt{10}$ cannot be written as a ratio of integers since it is irrational. This is a contradiction! And so $\sqrt{2} + \sqrt{5}$ is irrational.

preetha · Answer

Parth, that is pretty impressive.

anonymous · Answer

Here's another solution. Note that:$$(\sqrt{2}+\sqrt{5})^2=7+\sqrt{10}$$and $$(\sqrt{2}+\sqrt{5})^4=(7+\sqrt{10})^2=59+14\sqrt{10}$$Therefore its clear that if:$$x=\sqrt{2}+\sqrt{5}$$then$$x^4-14x^2+39=0$$Using the rational roots theorem, the only possible rational solutions to that equation are:$$\pm 1, \pm 3, \pm 13, \pm 39$$However, plugging those numbers in shows that none of them are solutions. Hence if x is a solution to that polynomial, it is irrational.

parthkohli · Answer

@Preetha Thanks, I just used conventional methods. Joe's solution is out of the box IMHO!