Prove that $(a,b)R(c,d) \longleftrightarrow ad=bc $ is equivalence relation on $A=R^2-\{(0,0)\}$ I am trying to prove that $$(a,b)R(c,d) \longleftrightarrow ad=bc $$ is equivalence relation on $$A=\mathbb{R}^2-\{(0,0)\}$$
$A$ is all points on the plane.
If I want to show that is reflexivity so I need to take $a$ and $c$, set $(a,a)\in R$ and $(c,c)\in R$ how I can show that $a^2 = c^2$? 
how to show transitivity? 
Thanks!
 A: Reflexivity:
$$(a,b)R(a,b)\iff ab=ba\,\text{(which is true)}$$
Symmetry:
$$(a,b)R(c,d)\iff ad=bc\iff cb=da\iff(c,d)R(a,b)$$
Transitivity:
if $(a,b)R(c,d)$ and $(c,d)R(e,f)$ then $ad=bc$ and $cf=de$ so by multiplying we have $adcf=bcde$ and WLG (why?) we can suppose that $cd\ne0$ so $af=be$ hence $(a,b)R(e,f)$
A: Note that the relation $R$ is reflexive iff $(a,b)R(a,b)$ for all $(a,b)\in A$. 
It is transitive iff for all triples $(a,b)$, $(c,d)$, $(e,f)$ in $A$, $(a,b)R(c,d)$ and $(c,d)R(e,f)$ implies $(a,b)R(e,f)$. Here is where you need that the pair $(0,0)$ is not an element of $A$. Can you see why?
A: It seems to me that reflexivity boils down to $(a,b)R(a,b)$, namely $ab=ab$.
Similarly, symmetry means that $(a,b)R(c,d)$ implies $(c,d)R(a,b)$, which again means that $cb=da$.
Finally, transitivity means that $(a,b)R(c,d)$ and $(c,d)R(e,f)$ implies $(a,b)R(e,f)$. But then $ad=bc$ and $cf=de$. You want to prove that $af=be$. Can you continue?
A: Reflexivity and simmetry: obvious.
Transitivity: suppose $(a,b)R(c,d)$ and $(c,d)R(e,f)$. We have two cases: $b=0$, $b\neq0$. Suppose $b=0$. Since $(0,0)\notin A$, $ad=bc$ implies $d=0$.  Now, if $(c,d)R(e,f)$, for the same argument $f=0$, and so $af=be$, or $(a,b)R(e,f)$. Now suppose $b\neq0$: we have $$(a,b)R(c,d)\iff a/b=c/d$$
and
$$(c,d)R(e,f)\iff c/d=e/f.$$
So $$a/b=c/d=e/f$$ follows that $(a,b)R(e,f)$.
If you know something about determinants, a much simpler proof is as follows:$$(a,b)R(c,d)\iff \det A=0\iff(a,b)=\lambda (c,d),$$
where $A$ is the matrix with rows $(a,b)$, $(c,d)$ and $\lambda\in \mathbb R$, $\lambda \neq 0$. Note that the last equivalence is true because we exclude $(0,0)$. The last relation is clearly transitive.
A: For reflexivity, you need to show that $\forall z \in \mathbb{R}^2\backslash\{(0,0)\}\, (z,z)\in R$. So start by assuming $z \in \mathbb{R}^2\backslash\{(0,0)\}$. So $z=(x,y)$ for some $x, y\in \mathbb{R}$ where $x\ne 0$ or $y\ne 0$. Now showing $(x,y)R(x,y)$ is easy.
For symmetry, assume $(a,b)R(c,d)$ and prove $(c,d)R(a,b)$.
For transitivity, assume $(a,b)R(c,d)$ and $(c,d)R(e,f)$ and then prove $(a,b)R(e,f).$
