Sum of two irrational radicals is irrational? If $a,b,m$ and $n$ are positive integers such that $\sqrt[m]{a}$ and $\sqrt[n]{b}$ are
irrational numbers, how can we prove that the sum $\sqrt[m]{a}+\sqrt[n]{b}$ is also
irrational?
 A: Let $a'$, $b'$, $m'$ and $n'$ be positive integers such that $\sqrt[m']{a'}$ and $\sqrt[n']{b'}$ are irrational. Let $a$, $b$, $m$ and $n$ be the minimal positive integers such that $\sqrt[m]{a}=\sqrt[m']{a'}$ and $\sqrt[n]{b}=\sqrt[n']{b'}$, so that their minimal polynomials are $f_a=X^m-a$ and $f_b=X^n-b$, respectively. Note that $m,n>1$ as $\sqrt[m]{a}$ and $\sqrt[n]{b}$ are irrational.
Suppose $\sqrt[m]{a}+\sqrt[n]{b}=q\in\Bbb{Q}$. Then $\sqrt[m]{a}$ and $\sqrt[n]{b}$ are roots of $f_b(q-X)$ and $f_a(q-X)$, respectively, which shows that $f_a$ divides $f_b(q-X)$ and $f_b$ divides $f_a(q-X)$, respectively. In particular we see that $m\leq n$ and $n\leq m$, so $n=m$, and hence $f_a=cf_b(q-X)$ for some nonzero constant $c\in\Bbb{Q}$. Then
$$X^m-a=f_a=cf_b(q-X)=c(q-X)^n-cb=c(q-X)^m-cb,$$
which immediately shows that $q=0$ because $m,n>1$. It follows that $c=\pm1$ and $a=cb$. Because $a$ and $b$ is positive it follows that $c=1$ and so
$$\sqrt[m]{a}+\sqrt[n]{b}=2\sqrt[m]{a}=2\sqrt[n]{b},$$
which is irrational because $\sqrt[n]{b}$ is, by assumption.
A: One way to show that a sum of square roots is an irrational number(let me consider this case which includes the idea for tackling the general problem- although whether it can be put into practice is not clear) is to notice that you can write a sum of square roots as a single square root.
Consider $\sqrt2$ + $\sqrt5$. You want to find a number $z$ such that $z=(\sqrt2+\sqrt5)^2$ So $\sqrt z=\sqrt{(2+2\sqrt2\sqrt5+5)}$.  That is $\sqrt{(7+2\sqrt{10 })}$ ,which is a square root of an irrational number. And a square root of an irrational number is always irrational.
