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.


If $\sqrt3+\sqrt5=x$ is rational, then $5=(x-\sqrt3)^2=x^2-2\sqrt3x+3$, and
$$\sqrt3=\frac{x^2-2}{2x}$$
But then $\sqrt3$ is also rational. However, we know that $\sqrt3$ is irrational, so $x$ is also irrational.

And we know $\sqrt3$ is irrational by the usual argument: is $\sqrt3=p/q$, with coprime integers $p$ and $q$, then $p^2=3q^2$. Hence $3$ divides $p$, but then $9$ divides $p^2$ and $3$ divides $q$, so $p$ and $q$ are not coprime, contradiction.
