Question about a number theory result

  • #1
1,456
44

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##?
 

Answers and Replies

  • #2
34,043
9,891
##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.
 
  • #3
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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.
 

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