Proving that $\sqrt{3} +\sqrt{7}$ is rational/irrational I took $\sqrt{3}+\sqrt{7}$ and squared it.  This resulted in a new value of $10+2\sqrt{21}$.
Now, we can say that $10$ is rational because we can divide it with $1$ and as for $2\sqrt{21}$, we divide by $2$ and get $\sqrt{21}$.
How do I prove $\sqrt{21}$ to be rational/irrational?
Thanks

Suppose that
$$\sqrt 7 +\sqrt 3=r, r\in\mathbb{Q}$$
Now we have:
$$\sqrt 7 =r-\sqrt 3/^2$$
$$7=r^2-2\sqrt 3 +3$$
$$4=r^2-2\sqrt 3$$
$$2\sqrt 3=r^2-4$$
$$\sqrt 3=\frac{r^2-4}{2}$$
This is a contradiction, because the left both is $\sqrt 3\in\mathbb{I}$, and the right both is $\frac{r^2-4}{2}\in\mathbb{Q}$. The contradiction is due to make the wrong assumption i.e. $\sqrt 7 +\sqrt 3$ is irational number
