Prove that

Let us assume that $\sqrt{3}+\sqrt{4}$ is rational .Then, there exist positive co primes $a$ and $b$ such that

$\sqrt{3}+\sqrt{4}=\frac{a}{b}$

$\sqrt{4}=\frac{a}{b}-\sqrt{3}$

$(\sqrt{4})^{2}=\left(\frac{a}{b}-\sqrt{3}\right)^{2}$

$4-3=\left(\frac{a}{b}\right)^{2}-\frac{2 a \sqrt{3}}{b}$

$4-3=\left(\frac{a}{b}\right)^{2}-\frac{2 a \sqrt{3}}{b}$

$1=\left(\frac{a}{b}\right)^{2}-\frac{2 a \sqrt{3}}{b}$

$\left(\frac{a}{b}\right)^{2}-1=\frac{2 a \sqrt{3}}{b}$

$\frac{a^{2}-b^{2}}{b^{2}}=\frac{2 a \sqrt{3}}{b}$

$\left(\frac{a^{2}-b^{2}}{b^{2}}\right)\left(\frac{b}{2 a}\right)=\sqrt{3}$

$\sqrt{3}=\left(\frac{a^{2}-b^{2}}{2 a b}\right)$

Here we see that $\sqrt{3}$ is a rational number which is a contradiction as we know that $\sqrt{3}$ is an irrational number

Hence $\sqrt{3}+\sqrt{4}$ is irrational