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 $\sqrt{7}+\sqrt{3}={a\over b}$ rational number say
then ${4b\over a}={4\over \sqrt{7}+\sqrt{3}}=4 \cdot {\sqrt{7}-\sqrt{3}\over (\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}=(\sqrt{7}-\sqrt{3})$
so $\sqrt{7}+\sqrt{3}+\sqrt{7}-\sqrt{3}={a\over b}+{4b\over a}$
so $\sqrt{7}={a\over 2b}+{2b\over a}$
so we are getting $\text{ an irrational number }\sqrt{7}=\text{ some rational number}(\Rightarrow\Leftarrow)$
