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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.

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    $\begingroup$ 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
    Commented Oct 18, 2020 at 1:53
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    $\begingroup$ (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
    Commented Oct 18, 2020 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? $\endgroup$
    – user831952
    Commented Oct 18, 2020 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
    Commented Oct 19, 2020 at 4:27

1 Answer 1

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  • $\sim$ 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, $(a,b) \sim (e,f)$.
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  • $\begingroup$ 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? $\endgroup$
    – user831952
    Commented Oct 18, 2020 at 14:02
  • $\begingroup$ How would one go about the next part? $\endgroup$
    – user831952
    Commented Oct 18, 2020 at 20:20

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