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.

  • $\begingroup$ When you divide an integer by $\;10\;$ , what are all the possible residues you can get? $\endgroup$ – Timbuc Apr 10 '15 at 22:48
  • $\begingroup$ 0-9. Is this just like with mods but screwing me up because it's using div instead? $\endgroup$ – Bob Apr 10 '15 at 22:49
  • $\begingroup$ Well, I thought $\;div\;$ is exactly the same as modulo...it isn't?? $\endgroup$ – Timbuc Apr 10 '15 at 22:50
  • $\begingroup$ 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. $\endgroup$ – Bob Apr 10 '15 at 22:51
  • $\begingroup$ 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 $\;22\,div\,10=2\;$ ...? $\endgroup$ – 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 $[10q]_\approx = \{10q , 10q+1, 10q+2, \dots, 10q+9\}$


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