Proof that $(a, b) \mathrel{R} (c, d)$ iff $ad = bc$ is an equivalence relation

Question

Let $X = \{(a,b) \mid a,b \in \Bbb Z; b \ne 0\}$. We define $(a,b)\mathrel R (c,d)$ iff $ad = bc$. Prove that $R$ is an equivalence relation on the set $X$. Which known set do the equivalence classes of the relation form?

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Any help on solving this please?

Answer 1

You need to show that the given relation is

• reflexive
  
  For all $(a, b) \in X$, $ab = ba$. This is clearly true. Hence R is reflexive.

• symmetric
  
  For all $(a, b), (c, d) \in X$, suppose $(a, b) R(c, d).$ Then $ad = bc $ if and only if $cb = da$ if and only if $(c, d) R (a,b)$. Therefore, R is symmetric.

• transitive
  
  Now, see what you can do with the following: Take arbitrary $(a, b), (c, d), (e, f)$ and assume $(a, b)R (c, d)$, and $(c, d) R (e, f)$. So $$ad = bc,\quad \text{and} \quad cf = de$$ Now, we use a little algebra to show that this implies $af = be$, and hence $(a, b) R (e, f)$.
  
  $ad = bc \iff adf = bcf = b(cf).$ Also, we have $cf = de$. So $adf = b(de) \iff af = be$, as desired!

The relation has all three properties, and hence, is by definition, an equivalence relation.

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Hint: $$(a, b) R (c, d) \;\text{ if and only if }\;\frac{a}{b} = \frac{c}{d}\; \text{ if and only if } \;ad = bc, \;\;(b\neq 0, d\neq 0)$$

Answer 2

To show this is an equivalence relation, you need to show, for pairs $(a,b),(c,d), (e,f)$:

i)$(a,b)R(a,b)$, for any pair $(a,b)$ ii) If $(a,b)R(c,d)$ , then $(c,d)R(a,b)$ iii)If $(a,b)R(c,d)$ and $(c,d)R(e,f)$ , then $(a,b)R(e,f)$.

Can you see how to it?

Let's set up i): We want to show that, for any pair $(a,b)$ , i.e., for any choice of integers $a,b$ , we have $(a,b)R(a,b)$. This means that, according to the definition of $R$, we have:

$(a,b)R(a,b)$ iff $ab=ba$ . Can you do the rest?

Answer 3

Consider that $$(a,b)R(c,d)\quad \text{ iff }\quad ad = bc\quad \text{ iff }\quad\frac{a}{b}=\frac{c}{d}$$ and since $=$ is an equivalence relation, $R$ is an equivalence one, and $\mathbb{Q}$ is an example of such classes.