Is this proposition "$\sqrt2$ + $\sqrt3$ is an irrational number." true or false?

Question

I know $\sqrt2$ and $\sqrt3$ are irrational numbers. However, how can I prove that $\sqrt2$ + $\sqrt3$ is an irrational number? Thank you.

Answer 1

If it were rational, its square would also be rational, so $2+3+2\sqrt{6}$ would be rational and hence $\sqrt{6}$ would be rational.

Answer 2

$$\sqrt2+\sqrt3=\frac pq\implies 2=\left(\frac pq-\sqrt 3\right)^2\implies\frac{3+\frac{p^2}{q^2}-2}{2 \frac pq}=\sqrt 3$$

Can this be?

Answer 3

If $u=\sqrt 2+\sqrt 3$, we have \begin{align*}u-\sqrt 2=\sqrt 3&\implies u^2+2-2\sqrt 2u=3\iff u^2-1=2\sqrt 2 u\\ &\implies (u^2-1)^2=8u^2 \iff u^4-10u^2+1=0 \end{align*} This is a polynomial with integer coefficients. By the rational roots theorem, if $u=\frac pq$ is a rational root of this polynomial, necessarily $u=\pm 1$. As one can check, none is a root. So $u$ is irrational.