<s>[INST] Prove $\sqrt{3} + \sqrt{5}$ is irrational If it is assumed that $\sqrt{3}$ is known to be irrational (not the case for $\sqrt{5}$), then prove that $\sqrt{3}+\sqrt{5}$ is irrational.
My approach:
Assume that $\sqrt{3}+\sqrt{5}$ is rational. Then there exist coprime integers $p$ and $q$ so that $\frac{p}{q}$ is rational and $\frac{p}{q}=\sqrt{3}+\sqrt{5}$. Thus $(\sqrt{3}+\sqrt{5})^2=2(4+\sqrt{15})=\frac{p^2}{q^2}$, which implies that $p^2$ is even, so $p$ is also even. Let $p:=2m$ for some integer $m$, then $p^2=4m^2$. Thus $q^2(4+\sqrt{15})=2m^2$, which implies that $q^2$ is even, and so $q$ is even. But this contradicts that $p$ and $q$ are coprime, and we arrive at a contradiction.

My way of proving this does not use the fact that $\sqrt{3}$ is
  irrational. Please let me know if my proof is correct, and how to use
  the above mentioned fact.

 [/INST] If $\sqrt 3 + \sqrt 5 = p/q; p,q \in \mathbb Z $ then...
$\sqrt 5 = p/q - \sqrt 3$
$5 = (p/q)^2 - 2 (p/q)\sqrt 3 +3$
$2 (p/q)\sqrt 3 = (p/q)^2 - 2$
$\sqrt 3 = (p/2q) - (q/p) $ which is rational.  Which is a contradiction as you had already proven that.
.....
However if you hadn't...
This is not the easiest but 
If $\sqrt 3 + \sqrt 5 = p/q $
$3 + 2\sqrt {15} + 5 = p^2/q^2$
$\sqrt {15} = p^2/2q^2 - 4 = a/b $
Where $a = p^2- 8q^2; b=2q^2$.
Let $a= 3^j5^k c; 3^l5^m d $ where $c,d $ don't have any factors of $3$ or $5$.  ($j,k,l,m $ may be $0$)
$15 =\frac {3^{2j}5^{2k}c}{3^{2l}5^{2m}d}$
$3^{2l+1}5^{2m+1}c = 3^{2j}5^{2k} d$
But the LHS had an odd number of factors of 3 and 5, while the RHS has even (or 0) number of factors of 3 and 5 so they can't be equal.
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