Originally Posted by

**Bingk** Um, I think melese's method works for $\displaystyle \sqrt{3} + \sqrt{12}$

$\displaystyle \sqrt{3} + \sqrt{12} = r$ where $\displaystyle r$ is rational.

$\displaystyle r^2 = (\sqrt{3} + \sqrt{12})^2$

$\displaystyle r^2 = 3+12+2 \sqrt{36}$

$\displaystyle r^2 = 3+12+12$

$\displaystyle r^2 = 27$

$\displaystyle r = 3 \sqrt{3}$

$\displaystyle r_a = \sqrt{3}$, where $\displaystyle r_a$ is rational

After that, it's the regular way of proving an irrational is not rational ...

Also, be careful when you say that $\displaystyle a,b\in\mathbb{Q}$, because if $\displaystyle a$ and $\displaystyle b$ could be equal to $\displaystyle c^2$ and $\displaystyle d^2$ respectively, where $\displaystyle c,d\in\mathbb{Q}$, respectively ... then $\displaystyle \sqrt{a} + \sqrt{b} = c + d$.