Showing that $R$ is an equivalence relation on $X \times X$ Let $X = \{1,2,3,..,10\}$ define a relation $R$ on $X \times X$ by $(a,b)R(c,d)$ if $ad=bc$. Show that R is an equivalence relation on $X \times X$. 
I know that the $R$ have to be reflexive (because $ad=bc$), but I don't know how to prove that $R$ is reflexive, symmetric, and transitive? thx
 A: Hint: it might help to note that your relation $$R: (a, b)\,R\,(c, d) \iff ad = bc$$ can be equivalently expressed as
$$R: (a, b)\,R\,(c,d) \iff \frac ab= \frac cd$$
That is, if the ratio of $a$ to $b$ equals the ratio of $c $ to $d$, then $(a, b) R (c, d)$.
You need to apply the definitions of reflexivity, symmetry, and transitivity: in this case
($a$) Reflexivity: Is it the case that for all $(a, b) \in X\times X$, $(a,b)\,R\,(a,b)$? 
($b$) Symmetry: Is it the case that for all $(a, b), (c, d) \in X\times X,$ if $(a, b) R (c, d),$ then $\,(c, d) R (a, b)\,?\,$  Why does $(a, b) R (c, d)$ immediately imply about whether $(c,d)\,R\,(a,b)$?
($c$) Transitivity: Is it the case that for all $(a, b), (c, d), (e, f) \in X\times X,$ that if $(a,b)\,R\,(c,d)$ and $(c,d)\,R\,(e,f)$, then it must follow that $(a,b)\,R\,(e,f)$?
If you can answer yes to each of the above (and you can), then your relation $R$ is an equivalence relation. 
A: Hint: ($1$) Reflexive: What does it mean to say $(a,b)\,R\,(a,b)$? Write it out and you'll see it immediately.
($2$) Symmetric: Using the definition for $R$, what does it mean to say $(a,b)\,R\,(c,d)$? Why does it immediately imply $(c,d)\,R\,(a,b)$?
($3$) Transitive: Again, take three elements, $(a,b)\,R\,(c,d)$ and $(c,d)\,R\,(e,f)$. Show how this implies $(a,b)\,R\,(e,f)$.
A: Hint $\ $ You have $\rm\:(a,b)\sim (b,d)  \smash[t]{\overset{\ def}{\iff}} a/b = c/d\iff f(a,b) = f(c,d)\ $ for $\rm\:f(x,y) = x/y.$
Now it is quite easy to prove that any relation of the above form is an equivalence relation.
More generally, suppose $\rm\ u\sim v\ \smash[t]{\overset{\ def}{\iff}}\, f(u) \approx f(v)\ $ for a function $\rm\,f\,$ and equivalence relation $\,\approx.\, \ $ Then the equivalence relation properties of $\,\approx\,$ transport (pullback) to $\,\sim\,$ along $\rm\,f\,$ as follows:


*

*reflexive  $\rm\quad\ f(v) \approx f(v)\:\Rightarrow\:v\sim v$

*symmetric  $\rm\,\ u\sim v\:\Rightarrow\ f(u) \approx f(v)\:\Rightarrow\:f(v)\approx f(u)\:\Rightarrow\:v\sim u$

*transitive $\rm\ \ \ u\sim v,\, v\sim w\:\Rightarrow\: f(u)\approx f(v),\,f(v)\approx f(w)\:\Rightarrow\:f(u)\approx f(w)\:\Rightarrow u\sim w$
Such relations are called (equivalence) kernels. One calls $\, \sim\,$ the $\,(\approx)\,$ kernel of $\rm\,f.$ 
A: Consider the function $f:X\times X\rightarrow \mathbb{R}$ defined by
$$
f(a,b):=\frac{a}{b}
$$
and note that
$$
(a,b)\;R\;(c,d)\quad\Leftrightarrow\quad f(a,b)=f(c,d).
$$
The fact that it is an equivalence relation on $X\times X$ follows from the following general result.
If $f:Y\longrightarrow Z$ is a function, then the relation
$$
y_1\;R_f\;y_2\quad\mbox{if}\quad f(y_1)=f(y_2)
$$
is an equivalence relation on $Y$.
The proof is very close to trivial, if not trivial.
Note that the equivalence classes are the inverse images of singletons of $Z$ by $f$, or the level sets of $f$, if you prefer.
