Proving that $\sqrt{2}+\sqrt{3}$ is irrational [duplicate]

Solution 1:

$(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)=3-2=1$ is rational, so if the first factor were rational, so would be the second, and their difference $2\sqrt2$. But surely you know that the latter isn't rational.

Solution 2:

Hint $\ \sqrt2 +\sqrt3 = r\in\Bbb Q\,\overset{\large\rm square}\Rightarrow\ 5+2\sqrt 6 = r^2 \,\Rightarrow\ \sqrt 6 = (r^2-5)/2 \in \Bbb Q,\,$ contradiction,

since $\,\sqrt{6} = a/b\,\Rightarrow\, 6b^2 = a^2\,$ has an odd number of $2$'s on the LHS, but an even number on RHS (by uniqueness of prime factorizations).

There are also many other ways, e.g. using the Rational Root Test on its minimal polynomial, or using Bezout, etc see prior posts here.

Solution 3:

Let $x=\sqrt2+\sqrt3$. Then this is a solution to the equation $0=x^4-10x^2+1$. By rational root theorem, this equation has no rational solution. Therefore, $x$ must be irrational.