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.
Best Answer
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 $\sqrt{a}-\sqrt{b}$).