0
$\begingroup$

Let $Q$ be the following subset of $\mathbb{Z}\times \mathbb{Z}$:

$Q=\left \{ (a,b)\in \mathbb{Z}\times \mathbb{Z}: b\neq 0 \right \}$

Define the relation $\sim $ on $Q$ as

$(a,b)\sim (c,d)\Leftrightarrow ad=bc$

Proof that $\sim$ is an equivalence relation, and specify $[(2,3)]$ and more generally the equivalence class $[(a,b)]$. Try to give an explanation of $Q/\sim $

I know that an equivalence relation is reflexive, symmetric, and transitive. I am not sure on how to approach such a proof and then to specify the values.

$\endgroup$
4
  • 1
    $\begingroup$ The fact that this relation is indeed an equivalence follows straight from the definition: you merely need to verify that it is indeed reflexive, symmetric and transitive, with the first two properties being quite obvious and the third one requiring just a little bit of work. What is important to realise is that this is the construction by which one obtains the field $\mathbb{Q}$ from the ring $\mathbb{Z}$. The class of any pair $(m, n)$ represents nothing else than the fraction $\frac{m}{n}$. (to be cont.) $\endgroup$ – ΑΘΩ Oct 18 '20 at 1:53
  • 1
    $\begingroup$ (cont.) In general, if $k=(m; n)$ is the greatest common divisor of $m$ and $n$ and if $m=kr, n=rs$, then the class of $(m, n)$ will be given by $\{(qr, qs)\}_{q \in \mathbb{Z^{\times}}}$. $\endgroup$ – ΑΘΩ Oct 18 '20 at 1:55
  • $\begingroup$ I see, I was having difficulties on how to formulate and write it up even though I know all the concepts. But I see it is quite simple! Thanks for the explanation! Now, to the tricky part; how would i specify the two classes? $\endgroup$ – mathstudent23 Oct 18 '20 at 14:04
  • $\begingroup$ What exactly do you mean by "specifying" the two classes? If you are referring to describing them explicitly, then the way to go is via the description I mentioned above, which requires some notions of elementary arithmetic in order to be proved. $\endgroup$ – ΑΘΩ Oct 19 '20 at 4:27
1
$\begingroup$
  • $\sim$ is reflexive: For all $(a,b) \in Q$ we have $ab = ba$, that is, $(a,b) \sim (a,b)$.
  • $\sim$ is symmetric: For all $(a,b),(c,d) \in Q$ such that $(a,b) \sim (c,d)$ we have $ad=bc$ and then $cb = da$, that is, $(c,d) \sim (a,b)$.
  • $\sim$ is transitive: For all $(a,b),(c,d),(e,f) \in Q$ such that $(a,b) \sim (c,d)$ and $(c,d) \sim (e,f)$ we have $ad=bc$ and $cf = de$. Thus, $$(af)\color{red}{d} = (ad)f = (bc)f = b(cf) = b(de) = (be)\color{red}{d}$$ and since $d\neq0$, it follows that $af = be$, that is, $(a,b) \sim (e,f)$.
$\endgroup$
2
  • $\begingroup$ I see, I was having difficulties on how to formulate and write it up even though I know all the concepts. But I see it is quite simple! Thanks for the explanation! Now, to the tricky part; how would i specify the two classes? $\endgroup$ – mathstudent23 Oct 18 '20 at 14:02
  • $\begingroup$ How would one go about the next part? $\endgroup$ – mathstudent23 Oct 18 '20 at 20:20

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.