# Sum of irrational numbers, a basic algebra problem [duplicate]

Let $x_1,\dots,x_n$ be positive rational numbers. If $\sqrt[l_1]{x_1},\dots,\sqrt[l_n]{x_n}$ are all irrational numbers (where $l_1,l_2,\dotsc,l_n\in\Bbb N^*$), does it follow that $$\sqrt[l_1]{x_1}+ \dotsb + \sqrt[l_n]{x_n}$$ is an irrational number, too?

• What do you mean by $\mathbb N^*$? Commented Jul 10, 2013 at 14:19
• @tomasz I gess it means strictly positive integers Commented Jul 10, 2013 at 14:20
• I mean integers no smaller than $0$. What does it matter? I'm not the one who asked the question, am I? On the other hand, if I wrote $\mathbb N^*$, I would probably have meant some saturated extension of natural numbers, or the natural numbers within a saturated extension of reals. Commented Jul 10, 2013 at 14:58
• I asked OP what he meant by a symbol whose meaning I'm not sure about. Why would I need to explain anything? Maybe that's what he meant, but it still doesn't hurt to ask (or clarify where there's confusion...). Commented Jul 10, 2013 at 15:02
• See the papers by Mordell and Siegel that I cite in this answer (the radicands need to be multiplicatibvely independents). Commented Nov 30, 2022 at 19:37

This is and old question and the answer is yes.

Mordell was interested in a particular case of this problem in his paper On the linear independence of algebraic numbers.

Later on, Ursell in the paper The degrees of radical extensions solves this problem in the particular case $$x_i\in\mathbb N^*$$ and $$(x_i,x_j)=1$$ for $$i\neq j$$.

In fact, the problem can be easily reduced to the case where $$x_i\in\mathbb N^*$$ and $$l_1=\cdots=l_n$$. This is in fact a problem proposed/solved by Preda Mihailescu long time go. Later on Toma Albu worked on this problem from different point of view and proved this as Theorem 2.2 in this survey paper. Basically he proved the theorem in an older paper where he acknowledges that his results have appeared in many other sources, such as Karpilovski's book Field Theory.

Anyway, the proof given by Mihailescu is the most "elementary" and I suggest you to try to read it (well, maybe using Google translate).

• The link to a paper (or whatever that is) by Mihailescu is dead. Could you provide a reference as to what you were linking to? Commented Feb 3, 2016 at 14:09
• Mission 835 of page(23:19) gdz.sub.uni-goettingen.de/id/PPN378850199_0036?tify={%22pages%22:[23],%22panX%22:0.41,%22panY%22:0.503,%22view%22:%22info%22,%22zoom%22:0.496} Commented Jan 5, 2019 at 22:48

Too long for a comment.

Your problem is actually equivalent to the following:

Problem: If $a_1,..,a_k$ are positive integers, and $m$ is a positive integer, so that $\sqrt[m]{a_i} \notin \mathbb Q$ then

$$\sum \sqrt[m]{a_i} \notin \mathbb Q \,$$

You can do the reduction to this problem in two steps:

Step 1: Let $m=lcm (l_1,..,l_n)$. Then $\frac{1}{l_i}=\frac{k_i}{m}$. Let $y_i=x_i^{k_i}$.

Then, you know that $y_i$ are positive rational numbers and $x_i^\frac{1}{l_i}=\sqrt[m]{y_i}$.

Step 2: Let $y_i=\frac{b_i}{c_i}$ and let $d= lcm (c_1,..,c_n)$. Then you can write

$$y_i=\frac{a_i}{d^m}$$

and you get that problem.

I am pretty sure that I saw some Theorems in Galois Theory which imply stronger versions of the above problem.

Added Theorem 2.9 in This Paper is exactly the question asked, it provides some reference to the History of the problem and a proof.

• This is true at least if $a_i$ are coprime and $m=2$, and the proof is quite a bit more involved than one could think at first glance; elementary methods work for $n=3$, but I think they fail as soon as $n=4$. That is, if I recall correctly from my abstract algebra classes. :) Commented Jul 10, 2013 at 14:24
• @tomasz I actually found a reference, which proves the asked question ;) If I remember right, the issue with the relatively primess is that otherwise the radicals could cancel, which is not the case because of the positivity..... And I remember a very nice elementary proof for $m=2$, which cannot be extended:) Commented Jul 10, 2013 at 14:28