Prove that $(a,b)R(c,d) \longleftrightarrow ad=bc $ is equivalence relation on $A=R^2-\{(0,0)\}$

Question

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!

score 5 · Accepted Answer · edited Jul 31 '13 at 17:30

5

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 $(a,b)R(e,f)$

edited Jul 31 '13 at 17:30

Thomas Andrews

177,126

answered Jul 31 '13 at 17:08

How I can show the equivalence classes in geometric way? and what is the partition set? – Ofir Attia Jul 31 '13 at 17:25
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@OfirAttia Two points are equivalent if the line through them passes through $0$. – Thomas Andrews Jul 31 '13 at 17:32
but $R$ not include $(0,0)$ – Ofir Attia Jul 31 '13 at 17:41

score 2 · Answer 2 · answered Jul 31 '13 at 17:18

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?

score 1 · Answer 3 · answered Jul 31 '13 at 17:08

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?

pppqqq · Answer 4 · 2013-07-31T18:41:53.330

0

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.

edited Jul 31 '13 at 18:41

answered Jul 31 '13 at 17:07

pppqqq

2,150

1

You need to take care. The definition also allows for $b=0$. – Mark Bennet Jul 31 '13 at 17:12
Thank you. I read the $-{ (0,0) }$ condition to hurrily. – pppqqq Jul 31 '13 at 18:10

score 0 · Answer 5 · answered Jul 31 '13 at 17:16

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 $(a,b)R(e,f).$

Prove that $(a,b)R(c,d) \longleftrightarrow ad=bc $ is equivalence relation on $A=R^2-\{(0,0)\}$

5 Answers5