0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $[10q]_\approx = \{10q , 10q+1, 10q+2, \dots, 10q+9\}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.