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.

 A: 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
$\sqrt{a}-\sqrt{b}$).
A: Let $x=\sqrt 3 + \sqrt 5$. Then:
$$(x - \sqrt 3)^2 = 5$$
$$x^2 - 2 \sqrt 3 x + 3 = 5$$
$$x^2 - 2 = 2 \sqrt 3 x$$
$$(x^2 - 2)^2 = 12 x^2$$
$$x^4 -16 x^2 + 4 = 0$$
By the rational root theorem, if the last equation had rational roots then they would be integer divisors of $4$. But none of $\pm1, \pm2,\pm4$ are roots, so $x$ must be irrational.
A: 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.
A: Hint:

If $\sqrt{15}=p/q$ is rational and $\sqrt3$ is irrational then $\sqrt5=\frac{\sqrt{15}}{\sqrt3}=\frac{p}{q\sqrt3}$ so
  $$5=\sqrt5^2=\frac{p^2}{3q^2}.$$
  Hence $3$ appears an odd number of times in the prime factorisation of $5,$ a contradiction.


Hope this helps.

P.S. I don't know what I was thinking in putting the old Hint I here.
A: 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.
