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
 A: 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 $\text{ an irrational number }\sqrt{7}=\text{ some rational number}(\Rightarrow\Leftarrow)$
A: 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
A: @Doorknob: I think he means that he took the square: $(\sqrt{3} + \sqrt{7})^2 = 10 + 2\sqrt{21}$.
If you want to prove that $\sqrt{21}$ is rational/irrational, do as follows. Suppose $a,b \in \mathbb{N}$ are such that $\frac{a}{b} = \sqrt{21}$ and that $\gcd(|a|,|b|) = 1$. I.e., $\frac{a}{b}$ can not be simplified. Then, we must have that $\frac{a^2}{b^2} = 21$. 
Or: $a^2 = b^2\cdot 21$. 
Every prime factor of both $a^2$ and $b^2$ must occur an even amount of times in their respective factorisation. We see, however, that in this case, it will not happen, due to the extra factors 3 and 7 on the right hand side of the equation. This yields a contradiction, thus $\sqrt{21}$ is irrational. 
