Question:Let $X=\mathbb{Z}\times(\mathbb{Z}\setminus\{0\})$. Define a relation $\sim$ on $X$ by declaring that $(a, b)\sim(c, d)$ if and only if $ad = bc$
Prove that the relation $\sim$ is an equivalence relation.
My attempt
$ab = ba \Leftrightarrow (a, b) \sim (a, b)$. When I focus on left side of the biconditional, the relation $\sim$ is symmetric, but when I see the right side, the relation is reflexive. I can't clearly say the relation.
$ab= ab \Leftrightarrow (a, a) \sim (b, b)$. I don't know how to define this relation when I look at the right side. But when I look at the left side, it's reflexive relation.
For a transitive relation, I'm not clearly sure if I have to write "$ab=bd=cd$". But what should follow after "$ab=bd=cd$"?
Can you complete the proof?
[Edit] Now I know how to prove it.
(i) $ab = ba \Leftrightarrow (a, b) \sim (a, b)$
Therefore,~ is reflexive
(ii) $[ad = bc \Leftrightarrow (a, b) \sim (c, d)]$
$\implies$ [cb=da $\Leftrightarrow$ (c, d) $\sim$ (a, d)]
Therefore, ~ is symmetric
(iii) $[ad = bc \Leftrightarrow (a, b) \sim (c, d)]$
$\land$ [cf=de $\Leftrightarrow$ (c, d) $\sim$ (e, f)]
$\implies$ [ $\dfrac {ad} {de}$= $\dfrac {bc}{cf}$ $\Leftrightarrow$ af=be $\Leftrightarrow$ (a, b) $\sim$ (e, f)]
Therefore ~ is transitive
By (i), (ii), (iii), the relation ~ is an equivalence relation.
[FYI]
A relation $R$ from $A$ to $B$ is a subset of the Cartesian product $A\times B$. It is customary to write $aRb$ for $(a, b)\in R$. The symbol $aRb$ is read "$a$ is $R$-related to $b$.
$A$ relation on a set $A$ is a relation from $A$ to $A$.
$R$ is reflexive if and only if $\forall x\in X, xRx$
$R$ is symmetric if and only if $xRy\Rightarrow yRx$
$R$ is transitive if and only if $xRy\land yRz\Rightarrow xRz$.
$R$ is an equivalence relation if and only if R is reflexive, symmetric, and transitive.
Best Answer
Since multiplication is commutative in $\Bbb Z,$ then $ab=ba.$ Hence, as you noted, the relation is reflexive.
To prove symmetry, start by supposing that $(a,b)\sim(c,d),$ meaning that $a,b,c,d$ are integers with $b,d\ne0,$ and $ad=bc.$ You must then prove that $(c,d)\sim(a,b),$ which you will do by showing that $cb=da.$
To prove transitivity, start by assuming that $(a,b)\sim(c,d)$ and $(c,d)\sim(x,y).$ This means that $a,b,c,d,x,y$ are integers with $b,d,y\neq0,$ such that $ad=bc$ and $cy=dx.$ You must use these assumptions to prove that $(a,b)\sim(x,y),$ which you will do by proving that $ay=bx.$
Added: As a side note, your second paragraph--the one about $(a,a)\sim(b,b)$--isn't in any way pertinent to the proof.