Prove that the number $\sqrt{3}+\sqrt{7}$ is irrational.

(N/A) Let us assume,to the contrary,that $\sqrt{3}+\sqrt{7}$ is a rational number. Let $\sqrt{3}+\sqrt{7} = r$,where $r$ is a rational number.
Squaring both sides,we get: $(\sqrt{3}+\sqrt{7})^2 = r^2$.
$3 + 7 + 2\sqrt{21} = r^2$.
$10 + 2\sqrt{21} = r^2$.
$2\sqrt{21} = r^2 - 10$.
$\sqrt{21} = \frac{r^2 - 10}{2}$.
Since $r$ is a rational number,$r^2$ is also a rational number. Therefore,$\frac{r^2 - 10}{2}$ is a rational number.
This implies that $\sqrt{21}$ is a rational number.
However,we know that $\sqrt{21}$ is an irrational number because $21$ is not a perfect square.
This is a contradiction to our assumption that $\sqrt{3}+\sqrt{7}$ is rational.
Therefore,our assumption is false,and $\sqrt{3}+\sqrt{7}$ must be an irrational number.