# 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?

• 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
• – user26857 Sep 20 '13 at 21:18

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.

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.

• 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