Why is $\sum_{i=1}^n a$ always irrational if $n>0$ and $a$ is irrational?

Question

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.

Answer 1

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...

Answer 2

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

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

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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.