Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $S$.

share|improve this question

1 Answer 1

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 $a,b\in \mathbb{Z}, \gcd(a,b)=1, b\neq 0$

share|improve this answer
    
Ah okay. Did I do the first part of the question correctly? –  Ozera May 7 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 at 21:40
    
Ah, okay thanks! –  Ozera May 7 at 23:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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