Partition and Equivalence Relation: In $\mathbb{Z}$, let $m\sim n$ iff $m - n$ is a multiple of 10

Question

Prof. Pinter's "A Book of Abstract Algebra" presents this exercise:

Prove that each of the following is an equivalence relation on the indicated set. Then, describe the partition associated with that equivalence relation.

In $\mathbb{Z}$, let $m\sim n$ iff $m - n$ is a multiple of 10.

This partition seems intuitive, but I'm not sure how to prove it. I can show different examples. But, I'm not sure how to form the equivalence classes to start the proof.

Can you please give me a hint?

Answer 1

Given any natural number $n$ when we do the divition by $10$ we have the posibilities $n\cong 0,1,2,3,4,5,6,7,8,9 \bmod 10$ the posibles residues so this are the 10 sets of the partition : $$\lbrace 0,10,20,30,\ldots \rbrace$$ $$\lbrace 1,11,21,31,\ldots \rbrace$$ $$\lbrace 2,12,22,32,\ldots \rbrace$$ $$\lbrace 3,13,23,33,\ldots \rbrace$$ $$\lbrace 4,14,24,34,\ldots \rbrace$$ $$\vdots$$ $$\lbrace 8,18,28,38,\ldots \rbrace$$ $$\lbrace 9,19,29,39,\ldots \rbrace$$

Answer 2

First you prove its an equivalence relation. Reflexive: for all $ n \in \mathbb{Z}: n \sim n$ because $n-n = 0 = 0\times 10$ Symmetric: for all $m, n \in \mathbb{Z}: m \sim n \to m - n = 10k \to n - m =10(-k) \to n \sim m$. Transitivity: for all $m, n, p \in \mathbb{Z}: m \sim n, n \sim p \to m - n = 10k, n - p = 10l \to m - p = 10(k+l) \to m \sim p$.

Thus $\sim$ is an equivalence relation. Now for $a \in \mathbb{Z}$, consider $[a]$. Using the Division Algorithm: $a = 10q+r$, thus $[a] = [r], 0 \leq r \leq 9$. Thus we have $\mathbb{Z}/{\sim} = \{[r]: 0 \leq r \leq 9\}$