Relation on set, how is it transitive?

Question

Show that the relation R on the set:

$$ S={\{(a,b) : a,b \in Z, a,b \ne 0}\} $$ defined by $$ (a_1,b_1)R(a_2,b_2) <- -> a_1\times b_2 = a_2 \times b_1 $$ Show that it is an equivalence relation. (reflexive, symmetrical, transitive)

I see that it is: reflexive. Every pair is related to itself, e.g. (3,3)R(3,3).

I see that it is symmetrical. For every pair, e.g. (3,3)R(4,4), (4,4)R(3,3) $$ a_1 \times b_2 = a_2\times b_1 \textrm{ you also have } a_2 \times b_1 = a_1 \times b_2 $$

But how is it transitive? And why is it not an equivalence relation if a, b can be 0?

Answer 1

Let's suppose $(a_1,b_1)R(a_2,b_2)$ and $(a_2,b_2)R(a_3,b_3)$. So we have $a_1\times b_2 = a_2 \times b_1$ and $a_2\times b_3 = a_3 \times b_2$.

Else: $a_1b_3=a_1b_2*b_3/b_2=a_2b_1*b_3/b_2=a_2b_3*b_1/b_2=a_3b_2*b_1/b_2=a_3b_1$, this is what we wanted to prove.

It is no longer a equivalence relation if you let $a$ and $b$ be egal to $0$ because it is no longer transitive. Indeed let's see this counter exemple:

we have $(1,2)R(0,0)$ because $1*0=0*2$. For the same reason, we have $(0,0)R(2,1)$. But $1*1\neq 2*2$ so $(1,2)$ and $(2,1)$ are not in relation.

Answer 2

Suppose you know $(a_1,b_1)R(a_2,b_2)$ and $(a_2,b_2)R(a_3,b_3)$. You need to show $(a_1,b_1)R(a_3,b_3)$.

So rewrite what this actually means. You know $a_1b_2=a_2b_1$ and $a_2b_3=a_3b_2$. You want to show $a_1b_3=a_3b_1$.

The task now is to use those first two questions to obtain the third equation. Note that the third equation does not contain $a_2$ or $b_2$. So the hint is try to eliminate either $a_2$ or $b_2$ in the first two equations and see what happens.

If you do this you'll see that the elimination involves division, and so things need to be nonzero for the process to work. This should help guide you toward a counterexample when $0$ is involved.