<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] This is from here:
Prove that $\sqrt{2}+\sqrt{3}$ is irrational.
More generally,
suppose
$r
=\sqrt{a}+\sqrt{b}
$
is rational,
where
$a$ and $b$
are positive integers.
Then
$r(\sqrt{a}-\sqrt{b})
=a-b
$
so
$\sqrt{a}-\sqrt{b}
=\dfrac{a-b}{r}
$
is also rational.
Adding and subtracting these,
$\sqrt{a}$ and $\sqrt{b}$
are rational.
Therefore,
if either or both
of
$\sqrt{a}$ and $\sqrt{b}$
are irrational,
then
$\sqrt{a}+\sqrt{b}$
is irrational
(and similarly for
$\sqrt{a}-\sqrt{b}$).
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