proving a binary relation on $\mathbb{Z} × \mathbb{Z}$ is an equivalence relation.

Question

Define a binary relation ∼ on $\mathbb{Z}×\mathbb{Z}$ by $(a, b) ∼ (c, d)$ if and only if $ad = bc$.

Prove that ∼ is an equivalence relation.

I'm trying to prove symmetric of this relation. Do I just show that $ab=bc$ and $bc=ab$ are equivalent by commutative property? how can I prove it?

You have to show that $(a_1,a_2)\sim(b_1,b_2)\implies (b_1,b_2)\sim(a_1,a_2)$, Now $(a_1,a_2)\sim(b_1,b_2)\implies a_1b_2=a_2b_1$,We have to show $b_1a_2=b_2a_1$.Which is true by commutativity of reals. — PNDas, Nov 19 '20 at 06:53

score 1 · Answer 1 · answered Nov 19 '20 at 07:04

1

We have to show that $(a_1,a_2)\sim(b_1,b_2)\implies (b_1,b_2)\sim(a_1,a_2)$.

Now $$(a_1,a_2)\sim(b_1,b_2)\\\implies a_1b_2=a_2b_1\\\implies a_2b_1=a_1b_2\;\;\text{,since '=' is symmetric}\\\implies b_1a_2=b_2a_1\;\;\text{,since multiplication is commutative for reals}\\\implies (b_1,b_2)\sim(a_1,a_2)\,\,\square$$

answered Nov 19 '20 at 07:04

PNDas

937

so it would be same even though second pair is just $\mathbb{Z}$, not $\mathbb{Z} - 0$? – delete2020 Nov 19 '20 at 07:06
What do you mean by 'second pair is just $\mathbb{Z}'$. – PNDas Nov 19 '20 at 07:10
don't we need to think about when second pair is $0$ also? since it's $\mathbb{Z}$? – delete2020 Nov 19 '20 at 07:18
It'll not make any difference .If we write $(b_1,b_2)=(0,0)$ then the result is still valid. – PNDas Nov 19 '20 at 07:21
oh okay. then, I guess transitivity would be also same? – delete2020 Nov 19 '20 at 07:27

proving a binary relation on $\mathbb{Z} × \mathbb{Z}$ is an equivalence relation.

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