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Let $R$ be a relation defined on $S = \mathbb{Z} \times \mathbb{Z}^*$ by $(a,b)R(c,d)$ iff $ad = bc$ for $(a,b), \; (c,d)\; \in S$. Show that $R$ is an equivalence relation and describe equivalence classes with respect to $R$ on $S$.

To show that $R$ is an equivalence relation, we have to show that it is symmetric, transitive, and reflexive right?

For Reflexivity, by the definition of $R$, we have $$ ab = ba$$ so consequently, we have $\forall (a,b) \in \mathbb{Z} \times \mathbb{Z}^*, \; (a,b)R(a,b)$. I.e. the relation is reflexive.

To show that $R$ is symmetric, let $(a,b), (c,d) \in \mathbb{Z} \times \mathbb{Z}^*$ and assume that $(a,b)R(c,d)$. Then by the definition of $R$, we have $$ad = bc$$. Since $$ad = bc \Leftrightarrow cb = ad$$. Therefore, by the definition of $R$, we have $$(c,d)R(a,b)$$. Thus, we showed that $$\forall (a,b), (c,d) \in \mathbb{Z}\times \mathbb{Z}^*, \; \left [ (a,b)R(c,d) \implies (c,d)R(a,b)\right ]$$

To show that $R$ is transitive, let $(a,b), (c,d), (e,f), \in \mathbb{Z}\times \mathbb{Z}^*$. Then by the definition of $R$, we have $$(a,b)R(c,d) \Leftrightarrow ad = bc \;, \; (c,d)R(e,f) \Leftrightarrow cf = ed $$ Therefore, we have $$(ad = bc ) \wedge (cf = ed) \text{ so,}$$ $$adcf = bced $$ $$af = eb$$. Hence, by the definition of $R$, we have $$(a,b)R(e,f)$$ Therefore, we have $$\forall (a,b), (c,d), (e,f) \in \mathbb{Z} \times \mathbb{Z}^*, \; \left [(a,b)R(c,d) \wedge (c,d)R(e,f) \implies (a,b)R(e,f) \right ] $$

My question is, is the the first part of the question done correctly? Also, I am unsure what to do about the last part of the question of describing the equivalence class of $S$.

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up vote 3 down vote accepted

If $(a,b)R(c,d)$ iff $ad=bc$, we can then say that $$\frac{a}{b}=\frac{c}{d}$$ Thus equivalence classes are just the rational numbers whose values are the same. For example, $\frac{1}{2}, \frac{3}{6}, \frac{2048}{4096}$ are all members of the class $\left[\frac{1}{2}\right]=\left\{\frac{k}{2k}:k\in\mathbb{N}\right\}$ since $$1\cdot 6=2\cdot 3$$ $$1\cdot 4096=2\cdot 2048$$ etc... Thus the classes are $\left[\frac{a}{b}\right]$, where $a,b\in \mathbb{Z}, \gcd(a,b)=1, b\neq 0$

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Ah okay. Did I do the first part of the question correctly? – Ozera May 7 '14 at 21:27
Yeah, it look pretty good. just change your symmetric argument to $cb=da$, since it is this that yields $(c,d)R(a,b)$, not $cb=ad$. THe idea is there, just nitpicking so you have it right.... – Eleven-Eleven May 7 '14 at 21:40
Ah, okay thanks! – Ozera May 7 '14 at 23:54

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