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I am trying to figure out how to prove that an equivalence relation of the relation x~y defined on Z <=> x-y is a multiple of 3.

My attempt was:

1) Reflexive: x = x => x ~ x

2) Symmetric: x ~ y => x - y => - y + x => y ~ x

3) Transitive: ... <--- how do I make use of the fact that x-y is a multiple of 3?

Thanks for help!

user138913
  • 339
  • "x = x => x ~ x" What do you mean? This proves nothing. "x - y = - y + x => (x,y) ~ (y,x)" What does this mean? (x,y) ~ (y,x) is not even defined since ~ is a relation between numbers, not pairs of numbers. – Did May 11 '14 at 17:10
  • Can be useful if you specify what numbers you are using : natural, integer, rational... – Mauro ALLEGRANZA May 11 '14 at 17:13
  • @MauroALLEGRANZA it would appear pretty clear that this is modular arithmetic on the naturals. Surely if someone was doing otherwise they would specify for thats a bit more advanced. – DanZimm May 11 '14 at 17:15
  • @MauroALLEGRANZA I don't exactly understand the question, but it is clear if someone was dealing with an equivalence relation like this on a set other than the integers they would say so. – DanZimm May 11 '14 at 17:19
  • On the line that begins "Symmetric", you have a statement implying a number, which in turn implies another number, which implies a statement. Implication makes sense only between two statements, so none of the implications on that line make sense. – Andreas Blass May 11 '14 at 22:16

1 Answers1

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You have not proven anything.

$x-x=0$ is a multiple of $3$, so it is reflexive.

If $x-y=3k$ for some integer $k$, then $y-x=-3k$ is a multiple of $3$, so it is symmetric.

If $x-y=3m$ for some integer $m$ and $y-z=3n$ for some integer $n$, then $x-z=3(m+n)$ is a multiple of $3$, so it is transitive.