How to prove an equivalence relation with more than 2 variables? Let $R$ be a relation of positive integers
$$((a,b),(c,d)) \in R \iff ac = bd.$$
Prove that $R$ is an equivalence relation.
So I need to prove that this relation is reflexive , transitive and symmetric.
How i do it ??
I never encountered equivalence problems with more than 2 variables before.
 A: $R$ as you define it is not an equivalence relation.
Some of your uncertainty should go away once you specify the set $A$ on which $R$ is a relation. As you've indicated,
$$
\text{$R$ is a relation on $A=\Bbb N_+ \times \Bbb N_+$}.
$$
It's still the same notion of equivalence relation. Only in this case, there's a twist: the things being $x$ and $y$ that can be $R$-related have some structure — they're pairs $p = (a,b), q=(c,d)$. That is,
$$
R\subseteq A\times A =(\Bbb N_+ \times \Bbb N_+) \times (\Bbb N_+ \times \Bbb N_+).
$$
Here's how you the definition of $R$ looks without giving names $a,b,c,d$ to the coordinates:
$$
(p,q)\in R\iff first(p) \cdot first(q) = second(p) \cdot second(q) 
$$
For the record, $p = (a,b)$ if and only if $first(p)=a,second(p)=b$.
When reasoning about relations, you can reduce clutter by using "infix notation" — $p R q := (p,q)\in R$.
Hints
(Reflexive) If $p = (a,b)\in A=\Bbb N^2$, then do we have $pRp$? i.e. $(p,p)\in R$? Expand the definition: $pRp$ means $(a,b)R(a,b)$, which means $a^2 = b^2$. Does this hold for every pair of positive integers $p=(a,b)$? Is it true of $p=(1,2)$? So is $R$ reflexive?
Just as an exercise,
(Symmetric) If $pRq$, then do we have $qRp$? Let $p=(a,b), q=(c,d)$. Suppose that $pRq$. What does that mean? Well, it means $(a,b)R(c,d)$, or $ac=bd$. Now, what does $qRp$ mean, in terms of the coordinates?
(Transitive) If $pRqRs$, show that $pRs$. That is, if $(a,b)R(c,d)$ and $(c,d)R(e,f)$, show that $(a,b)R(e,f)$.
By the way, the relation
$$
(a,b)S(c,d)\iff ad = cb
$$
is an equivalence relation (on $\Bbb N_+ \times \Bbb N_+$ and on $\Bbb Z \times \Bbb N_+$). Essentially, it's the equivalence relation used to define fractions: $(a,b)S(c,d)$ holds iff $\frac a b = \frac c d$.
