Prove that R is an equivalence relation

Question

Let R be a relation on $\mathbb{Z} × ( \mathbb{Z} \setminus \{0\} )$ where $$( a, b ) R ( c, d ) \Leftrightarrow ad = bc.$$ Prove that R is an equivalence relation. Describe the distinct equivalence classes of R in terms of rational numbers and justify your answer.

I copy and pasted this question from the textbook, sorry if the formatting is weird. Anyways, I don't understand the $\mathbb{Z} \setminus \{0\}$ part of the question. If you're willing to solve the problem, can you somewhat explain your answer? Thanks I'm very new to this stuff.

Answer 1

You elements are pairs $(a,b)$ where $b \ne 0$.

$(a,b)R(c,d)$ means that $ad = bc$

TO be an equivalence relationship the must be:

1) Reflexive:

For all $(a,b)$ we must have $(a,b) = R(a,b)$

So we must prove that $ab = ab$ for all $(a,b)$.

2) Symetric:

If $(a,b) R (c,d)$ it must be true that $(c,d) R (a,b)$

So we must prove that if $ad = cb$ then $cb = ad$.

3) Transitive:

If $(a,b) R (cd)$ and $(c,d)R (e,f)$ it must be true that $(a,b)R(e,f)$.

So we must prove that if $ad=cb$ and $cf=ed$ then $af=eb$.

Answer 2

It is very important to keep definitions in mind when you're solving such questions. For a relation to be an equivalence class it must be:

1) Symmetric: $\forall (a,b), (c,d) \in \mathbb{Z}\times {\mathbb{Z}\ -{0}}$

if $(a,b)R(c,d) \implies ad=bc \implies bc=ad \implies (c,d)R(a,b)$

2) Reflexive: $\forall (a,b) \in \mathbb{Z}\times {\mathbb{Z}\ -{0}}$

$ab= ab \implies (a,b)R(a,b) \implies$ R is reflexive

3) Transitive: consider $(a,b), (c,d), (e,f) \in \mathbb{Z}\times {\mathbb{Z}\ -{0}}$

If $(a,b)R(c,d) and (c,d)R(e,f) \implies ad=bc and cf=de \implies c=de/f \implies ad=b(de/f) \implies fad=bde \implies fa = be \implies af=be \implies (a,b)R(e,f)$

Then, R is an equivalence relation.

Try proving the other part on your own. Remember, definitions are very important!