I'm asking this question because I was unable to find an answer elsewhere as most questions are about the summation of different irrational numbers, which is not what this question is about. Here, I'm interested in demonstrating that the result of the summation of the same irrational number is always irrational: $\sum_{i=1}^n a$, where $n$ is a non-negative integer $>0$ and $a$ is an irrational constant.

  • $\begingroup$ Note: $a$ factors out of that expression so, sure. But...is your question clear? There is no $i$ anywhere in your summand so your expression is just $na\times \sum_{i=0}^n1$. But I expect you meant something else. $\endgroup$
    – lulu
    Sep 24, 2018 at 16:04
  • $\begingroup$ Good eye. I'll correct this mistake and replace $n$ by $i$ in the equation. Also, $i$ should start at 1 and not 0. I'll correct that as well. $\endgroup$
    – m_power
    Sep 24, 2018 at 16:44
  • 1
    $\begingroup$ If you're going to write $\sum_{i=1}^n [\cdots]$, then $n$ had better be not just rational but an integer (probably a non-negative one, at that). $\endgroup$ Sep 24, 2018 at 17:16
  • 17
    $\begingroup$ Why do you have to write "summation of the same number"? Why not just call it multiplication by an integer? $\endgroup$
    – user253751
    Sep 25, 2018 at 5:57
  • $\begingroup$ You write that $n$ is "a non-negative integer". That allows for it to be zero, but in that case how do you intend $\sum_{i=1}^n a$ to work? Do you mean it to be the sum of no terms, in which case the sum is zero? As fleablood pointed out, the answer to your question changes depending on whether you include this case. $\endgroup$
    – LarsH
    Sep 25, 2018 at 14:39

2 Answers 2


It is trivially so. If you sum the irrational number $x$ $n$ times ($n$ being an integer "of course"), you end up with $nx$.

If $nx=\frac ab$, with $(a,b)$ integers, then $x=\frac{a}{nb}$, thus is rational, which is not true...

  • 6
    $\begingroup$ I love it when "trivial" things have to be proven in more than one line. $\endgroup$
    – mastov
    Sep 25, 2018 at 16:47
  • 9
    $\begingroup$ Trivial in mathematics often means "follows immediately from the definitions". I think this qualifies, but trivial is often in the eye of the beholder. $\endgroup$ Sep 25, 2018 at 16:50
  • 3
    $\begingroup$ As @EricWilson points out, the word "trivial" has taken on an almost technical meaning in mathematics which is not really the same as that used in common speech. This makes it somewhat useful in mathematical writing for experts, but in my opinion very off-putting when responding to new mathematicians or the general public. Just adding this here in case the response appeared to have a derogatory tone to the OP. It was probably not intended that way. $\endgroup$ Sep 25, 2018 at 22:02
  • 3
    $\begingroup$ Hello to all. Indeed my use of "trivial" was not intended to be derogatory at all. I was just pointing out that with a relatively simple (at least once you see it) process, you can prove it, with no advanced concepts (does not mean it is deriving stricly from definition). $\endgroup$
    – Martigan
    Sep 26, 2018 at 7:37
  • 2
    $\begingroup$ @MilanStojanovic Hello, you can't "sum" something an irrational number of times. You can multiply something with an irrational number, but in that context (as shown by the OP question), it is clearly we are talking about summing the number with itself a certain amount of integer times. $\endgroup$
    – Martigan
    Sep 26, 2018 at 7:38

$\sum k_ia = a(\sum k_i)$ and $\sum k_i$ is rational.

And if $\sum k_i \ne 0$ then a (non-zero) rational times an irrational is irrational.


I guess it's somewhat important to point out that if the the sum is $0$ such as $a + 2a - 3a$ or $k_1a + k_2 a + ..... + k_n a$ where $k_1 + k_2 + .... + k_n = 0$, then the statement is trivially false.

But that is the only exception.

  • 5
    $\begingroup$ .. unless that rational factor were $=0$ $\endgroup$ Sep 24, 2018 at 16:07
  • 16
    $\begingroup$ Yeah, I was drinking my coffee and eating a bagel and think "oh ###, unless $0$! Better get back and fix it before anyone notices!" .... oh, well. $\endgroup$
    – fleablood
    Sep 24, 2018 at 16:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .