# Question about a number theory result

## Main Question or Discussion Point

Suppose that $n,j \in \mathbb{N}$, $j \in [0, n-1]$, and $n~|~2j$. Why is it the case that $j = 0$ or $2j = n$? This is used in a proof of something else, but I am getting tripped up on this part. I know it has to do with the fact that $j \in [0, n-1]$. Is it because $n$ can't ever divide 2 or j separately and the quantity 2j is never greater than or equal to 2n, then if it divides anything it must divide 0 or $n$?

$0 \leq 2j < 2n$. To be a multiple of n it has to be either 0 or n, everything else is not in that range.
Instead of using interval notation from analysis, in number theory we denote $[n] := \{ k\leq n\}\subset\mathbb N$. The interval notation is also used in lattice theory, which might confuse the reader into thinking you are viewing the interval $[0,n-1]$ in the lattice $(\mathbb N, \mid)$.
Suppose $j > 0$. We have $kn = 2j$. Assuming $k>1$ immediately puts you out of range since $2j\leq 2(n-1)< kn$. Obviously $k=0$ is out of question so there is only one option.