Determine if R is an equivalence relation I've got this question: Consider the relationship $R$ between ordered pairs of natural numbers such that $(a, b)$ is related to $(c, d)$ (denoted by $(a, b) R (c, d)$) if and only if $ad = bc$.
Discuss whether $R$ is an equivalence relation. 
I'm pretty new to set theory and am wondering if someone can explain how you would go about evaluating if $R$ is an equivalence relation (I really want to understand how to work out the answer, not just the answer, thanks!)

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*Check reflexivity: Is it the case that for all $(a, b)\in \mathbb
   N\times \mathbb N$, it is true that $(a, b) R (a, b)$? That is, is it
true that for all such $(a, b)$, $ab = ba$?

*Check symmetry:  Is it the case that for all $(a, b), (c, d) \in \mathbb N\times \mathbb N,$ that If $(a, b) R (c, d)$, then $(c, d) R (a, b)$? This means that if $ad = bc,$ is it true that $cb = da$?

*Check transitivity: Is it the case that for all $(a, b), (c, d), (e, f) \in \mathbb N\times \mathbb N,$ that If $(a, b) R (c, d)$ and $(c, d) R (e, f)$, then it follows that $(a, b) R (e, f)?$ We need to show that $$(ad = bc \text{ and } cf = de) \implies af = be$$ This is slightly (very slightly) tricky, but just a little algebra gives the desired result:  Suppose we know that $(a, b) R (c, d)$ and $(c, d) R (e, f)$. Then by the definition of $R$, we know that $ad = bc$ and $cf = de$. Then $adcf = bcde$, and so canceling the factors $c, d$ on each side of the equation gives us $af = be$, as desired, since this means $(a, b) R (e, f)$. Hence transitivity holds.
If all three properties hold for $R$ (if you can answer yes to all of the above questions), then $R$ is an equivalence relation.
