I have a question that goes as follows: Let d be a positive integer. Define the relation Rho on the integers Z as follows: for all m,n element of the integers.
m rho n if and only if d|(m-n) Prove that rho is an equivalence relation. Then list its equivalence classes.
Now the first d that comes to mind is 1, so I proved it was an equivalence relation as follows:
Reflexive: m rho m <=> d|m-m
Symmetric: m rho n <=> d|m-n => d|n-m with n-m = -(m-n) <=> n rho m
Transitive: k is an element of integers: m rho n and n rho k => d|m-n & d|n-k => d|(m-n) + (n-k) => d|m-k => m rho k
I am unsure if this is a sufficient proof or if my logic holds. However I can't think of any equivalence classes for this as d can vary. If d was 7 for example I would think equivalence classes would be 1 = {...,-13,-6,1,8,15,...}, 2 = {...,-12,-5,2,9,15,...} etc...
Does this relation have equivalence classes and I am missing something or?
There is a different relation for each $d$. What is being asked is "for all $d$, is the corresponding relation an equivalence relation?"
As for your proof, it is correct, but you may want to be clearer with some of the steps, depending on how familiar the intended audience is with divisibility.