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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?

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What are your thoughts on the problem so far? – Servaes Aug 29 '13 at 13:44
You actually only need that at least one of them is irrational – user2566092 Aug 29 '13 at 13:46
Related to… – user26857 Sep 20 '13 at 21:18

Let positive integers $a'$, $b'$, $m'$ and $n'$ be given. Let $a$, $b$, $m$ and $n$ be minimal such that $\sqrt[m]{a}=\sqrt[m']{a'}$ and $\sqrt[n]{b}=\sqrt[n']{b'}$. Then the minimal polynomials of $\sqrt[m]{a}$ and $\sqrt[n]{b}$ are $f_a(X)=X^m-a$ and $f_b(X)=X^n-b$, respectively.

Suppose $\sqrt[m]{a}+\sqrt[n]{b}=q\in\Bbb{Q}$. Then $f_a(X)=f_b(q-X)$, and hence $X^m-a=(q-X)^n-b$. In particular $m=n$, $q=0$ and $a=b$. Then $\sqrt[m]{a}+\sqrt[n]{b}=2\sqrt[m]{a}=2\sqrt[n]{b}$ is rational, and hence both $\sqrt[m']{a'}$ and $\sqrt[n']{b'}$ are rational, a contradiction.

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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.

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To say that one "considers a case for simplicity" is to say that the general case follows from the specific one. This does not seem to be so with the question at hand. – Did Sep 20 '13 at 19:13
I was convinced that I could generalize the result at the time of writing, but admittedly I didn´t look too deeply into it. – Adam Sep 20 '13 at 19:44
"let me consider this case which includes the idea for tackling the general problem"... How do you know? – Did Sep 20 '13 at 20:42
It's also not the case that $(\sqrt{a}+\sqrt{b})^2$ is always irrational. (Consider the case where $a=20$ and $b=5$, for instance) – Steven Stadnicki Sep 20 '13 at 21:59
@Steven Stadnicky: I see your point, but I still wonder if sqrt(5)+sqrt(20) cannot be written as a single square using a different approach? – Adam Oct 2 '13 at 13:14

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