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I am currently in abstract mathematics. I am unclear on how to make a formal equivalence relation proof. I know I must prove reflexive, transitive, and symmetric, but I am not sure the formal set up or even how to for my specific example.

I have to define the relation $\sim$ on $Z \times Z-\{0\}$ by $(a,b) \sim (c,d) \Leftrightarrow ad=bc$ and we must prove that it's an equivalence relation.

Please let me know if you can help!

manofbear
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Bridget
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    I believe you meant $ad=bc$? – Siong Thye Goh Apr 03 '17 at 20:08
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    Welcome! Please have a look at how to format mathematics on Mathematics SE and take the Tour of the site. –  Apr 03 '17 at 20:16
  • Is it true that (a,b)~ (a,b) i.e. is it always true that ab = ba? Is it true that if (a,b)~(c,d) then (c,d)~(a,b) i.e. is it alwas true that if ad=bc then is it true that cb = da always? Is it always true that if(a,b)~(c,d) and (c,d)~(e,f) then (a,b)~(e,f).i.e is it always true if ad = bc and cf=de then af = be? (That is if none of b,d or f are ever zero.) If you can prove all those things are always true it is an equivalence relation. Two should be trivial. One should be easy. – fleablood Apr 03 '17 at 20:17
  • As an aside, this is exactly the equivalence relation that is used for saying whether or not two fractions are equal. $\frac{a}{b}=\frac{c}{d}$ iff $ad=bc$. For reflexive, it is basically asking if $\frac{a}{b}=\frac{a}{b}$, for symmetric, if $\frac{a}{b}=\frac{c}{d}$ whether or not $\frac{c}{d}=\frac{a}{b}$, and for transitive if $\frac{a}{b}=\frac{c}{d}$ and $\frac{c}{d}=\frac{e}{f}$ whether or not $\frac{a}{b}=\frac{e}{f}$ – JMoravitz Apr 03 '17 at 20:27

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~ is reflexive $\iff$ $(a,b)$ ~ $(a,b)$ for all $(a,b)$ $\iff$ $ab = ba$ for all $a$ and all $b \ne 0$.

Can you prove that? (Hint: It's trivial)

~ is symmetric $\iff $ $(a,b)$ ~ $(c,d)$ means $(c,d)$~$(a,b)$ $\iff$ $ad = bc$ means $cb = da$. Can you prove that? (Hint: it's trivial)

~ is transitive $\iff$ $(a,b)$~$(c,d)$ and $(c,d)~(e,f)$ means $(a,b) $~$(e,f)$ $\iff$ $ad = bc$ and $cf = de$ means $af=be$ if $b,d,f \ne 0$. Can you prove that? (Hint: It's not trivial but it is not very hard.)

fleablood
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