# Help Representing Equivalence Classes

In the set $\mathbb{Z}$, we define two integers $x$ and $y$ to be equivalent ($x ≈ y$) if and only if $x \operatorname{div} 10 = y \operatorname{div}10$. How would one select a representative from each equivalence class?

I understand why it's an equivalent relation (it's reflexive, symmetric, and transitive), but I can't figure out what the different equivalence classes would look like.

• When you divide an integer by $\;10\;$ , what are all the possible residues you can get? – Timbuc Apr 10 '15 at 22:48
• 0-9. Is this just like with mods but screwing me up because it's using div instead? – Bob Apr 10 '15 at 22:49
• Well, I thought $\;div\;$ is exactly the same as modulo...it isn't?? – Timbuc Apr 10 '15 at 22:50
• No, div is the quotient part. So a = qd + r. The quotient is written q = a div d, and the remainder is written r = a mod d. That's why I'm having trouble representing the classes. – Bob Apr 10 '15 at 22:51
• But then something else must be said, like $\;0\le r<10\;$ or stuff, otherwise the relation isn't well defined: $\;22=1\cdot 10+12=2\cdot10+2\;$ . I'm guessing it must be what I wrote above, right? So $\;22\,div\,10=2\;$ ...? – Timbuc Apr 10 '15 at 22:56

First note that $$a \approx b$$ iff $$f(a) = f(b)$$ with $$\DeclareMathOperator{div}{div} f:\mathbb{Z} \to \mathbb{Z}, x \mapsto x \div 10$$
Now, given $$y\in \mathbb{Z}$$, we have $$f(x) = y$$ with $$x = y \cdot 10$$. So $$f$$ is surjective and $$10\mathbb{Z}$$ is a system of representatives.
Now, let $$x\in \mathbb{Z}$$ be arbitrary. Then we can write $$x = q\cdot 10 + r$$ with $$q, r \in \mathbb{Z}, 0\leq r < 10$$. Now, since $$f(x) = q = f(q \cdot 10)$$ we have $$x \approx q \cdot 10 = x-r$$
So, to summarize: $$\mathbb{Z}/_\approx = 10 \mathbb{Z}$$ and $$[10q]_\approx = \{10q , 10q+1, 10q+2, \dots, 10q+9\}$$