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

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...


$\sum k_ia = a(\sum k_i)$ 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.

Tags:

Summation

Irrational Numbers

Rationality Testing

Related

Query on a Solution to the Problem: $\gcd(5a+2,7a+3)=1$ for all integer $a$. Geometric meaning of the quantity $|a|^2 |b|^2 |c|^2 - (a \cdot b)(b \cdot c)(c \cdot a)$ for non-coplanar vectors $a$, $b$, $c$ Why isn't this approach in solving $x^2+x+1=0$ valid? How many $5$-digit numbers with distinct digits are divisible by 3 (and generalize) How to easily create a polynomial function that gives a desired output? Is the set $\mathcal{A}$ of all matrices whose trace is $0$ nowhere dense in $\mathbb{M}_n(\mathbb{R}),n \ge 2$? Trying to arrange sums in neat way Verify a Matrix Proof Given $A = A^2$ "categorical" proof of a seemingly symmetric statement about Noetherian/Artinian modules Can the empty set serve as universe of a semigroup (i.e. set equipped with associative binary operation)? (Combinatorial?) Proof of the identity $\sum_{k=1}^n \frac {(-1)^k}{k\,(k+1)}\binom nk = \frac 12 + \frac 13 + \dots + \frac 1{n+1}$? 4-adic numbers and zero divisors

Recent Posts