<s>[INST] 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
 [/INST] @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. 
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