Answer to Question #134203 in Discrete Mathematics for Promise Omiponle

Question

Prove that there is a positive integer that equals the sum of the positive integersnot exceeding it.

Answer 1

Statement that need to be proved: there is a positive integer that equals the sum of the positive integers not exceeding it. In quantifiers this can be written as

$\displaystyle \exists \, n \in N\; (n = \sum_{i=0}^n i )$

We know that for finite sum $\displaystyle \sum_{i=0}^n i = \frac{n(n+1)}{2}$

So, $\displaystyle n = \frac{n(n+1)}{2}$

$n^2+n =2n$

$n^2-n=n(n-1)=0$

This has two solutions: $n=0$ and $n=1$. $n=0$ doesn't satisfy the condition of being positive integer (by definition positive is >0, 0 is not greater than 0). Therefore, only $n=1$ makes the statement true. We have showed that indeed there is a positive integer that equals the sum of the positive integers not exceeding it, and this positive integer is $n=1.$