# The class of equivalence.

Let $Q$ be the following set of $\mathbb{Z}\times \mathbb{Z}$ \begin{align*} Q=\{(a,b)\in \mathbb{Z}\times \mathbb{Z} | b\neq 0\} \end{align*} Define the relation $\sim$ on $Q$ as \begin{align*} (a,b)\sim (c,d)\Leftrightarrow ad=bc \end{align*} Prove that $\sim$ is a relation of equivalence, and give the class of equivalence of $[(2,3)]$. Furthermore the general order of equivalence of $[(a,b)]$. Try to give a description $Q/\sim$.

I've asked the same question (I deleted it one hour later) but it didn't look well when I didn't try to add what I tried. Here we go. I have proven that $\sim$ is a relation of equivalence by showin it's i) reflexive, ii) symmetric and iii) transitive.

i) $\forall (a,b)\in Q:(a,b) \sim (a,b)$.

$\forall (a,b)\in Q:(a, b) \sim (a, b) \Leftrightarrow ab \stackrel{\surd}= ab$

ii) $(a, b) \sim (c, d) \Rightarrow (c, d) \sim (a,b)$

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

then $ad = bc \stackrel{\surd}\Rightarrow cb = da$.

iii) $[(a, b) \sim (c,d) \wedge (c, d) \sim (e, f)] \Rightarrow (a, b) \sim (e, f)$

$(a, b)\sim (c, d) \Leftrightarrow ad = bc$, $(c, d) \sim (e, f) \Leftrightarrow cf = de$ and $(a, b) \sim (e, f) \Leftrightarrow af = be$

then $((ad = bc) \wedge (cf = de))\stackrel{\surd}\Rightarrow af = be$.

The class of eq. $[(2,3)]$ is the class, that contains the element $(2,3)$, so for every element $(a,b)$ in this class satisfies $(a,b) \sim (2,3) \Leftrightarrow 3a=2b$, that is \begin{equation} [(2,3)] = \lbrace (a,b) \in \mathbb{Z}\times\mathbb{Z} | b \neq 0 \wedge 3a = 2b \rbrace \end{equation}

So now I've shown what I tried to show. Is there something I'm missing or something I should have mentioned? I need some help to answer the rest. Thanks.

• Thanks for your comment. I edited my post. I meant the "class" of equivalence. – UnknownW Oct 29 '13 at 23:27
• I didn't read your question. As for the equivalence class $[(2,3)]_\sim$, you're correct. It's useful to note, though, how the elements of this class all correspond (in some sense) to the rational number $\frac 2 3$. – Git Gud Oct 29 '13 at 23:37

The point is that $Q/\sim$ wants to be the set of rational numbers, $(a,b)$ representing $\displaystyle\frac ab$. We also have $$ad=bc \iff \frac ab=\frac cd\,,$$ and forming the quotient set basically means to transform $\sim$ to become equality.
So, the equivalence class of $(2,3)$ is $\{\dots,(-2,-3),(2,3),(4,6),(6,9),\dots\}$. By the general order I don't know what is meant..