Recent content by nitenglo

So I redid it again...here it goes. Reflexive: let n∈ℤ, then for any n∈ℤ n-n=0 is a multiple of 10. ∴ n~n Symmetric: Let m, n∈ℤ and m~n. Then m-n is a multiple of 10. That is m-n ≡ 10k, k∈ℤ and n-m ≡ 10(-k), -k∈ℤ. ∴ n-m is a multiple of 10 and n~m. Transitive: Let m, n, p∈ℤ and m~n, n~p...

Sorry I forgot to post this earlier...this is what I have came up with so far... Reflexive: m-n = m=n mod 10. ∈ ℤ and is a divisible by 10 (same as being a multiple) Therefore, m~n Symmetric: m~n = m-n = n mod 10. ∈ ℤ and is a divisible by 10. Then -(m-n)= n-m = n=m mod 10. Therefore, n~m...

Thank you fourier...can you take a look at the work I did for proving m~n and see if it is correct.

I edited the problem.

Prove that the following is an equivalence relation on the indicated set. Then describe the partition associated with the equivalence relation. 1. In Z, let m~n iff m-n is a multiple of 10.2. The attempt at a solution Reflexive: m-n = 0 0 ∈ Z, and 0 is a multiple of every number...