# Equivalence Relation, and finding the subset that defines the relation.

Let \$xsim y\$ means \$xy > 0\$. Prove this is an equivalence relation. Find the subset of \$mathbb{R} times mathbb{R}\$ which defines the relation. Find the equivalence classes.

\$xsim y Leftrightarrow xy > 0\$ is an equivalence relation since it is:

1) Reflexive: \$xsim x\$ since \$xtimes x > 0\$, that is \$x^{2} > 0\$, which is trivially true.

2) Symmetric: If \$xsim y\$ such that \$xy > 0\$ and \$yx > 0\$, we know that \$xy > 0\$ will be positive so either \$x > 0\$ and \$y > 0\$ or \$x < 0\$ and \$y < 0\$ since \$xy > 0\$ will be positive so will \$yx > 0\$.

3) Transitive: \$xsim y\$ and \$ysim z\$ implies \$xsim z\$: \$xy > 0\$ and \$yz > 0\$ where \$z\$ is an integer. Since, \$(xy)(yz) > 0\$ then \$xy^{2}z > 0\$ since \$y^{2}\$ is strictly positive for any \$y\$, \$xz > 0\$.

I am not sure if this is right, and I am confused about finding the subset and the equivalence classes, any suggestions is greatly appreciated.

## Answer

Mostly right, which means wrong.

The Transitive proof is correct.

The symmetric proof is correct, but cluttered. You just have to say that: as multiplication of reals is commutative, then \$xy>0\$ implies \$yx>0\$.

However, the reflexive proof is wrong.   It ignores the case of \$x=0\$.   \$0sim 0\$ is `false` because \$0times 0 not> 0\$.

Thus the relation is not an equivalence relation for \$Bbb R\$. It is only an equivalence relation for \$Bbb Rsetminus{0}\$.

In such case, the equivalence classes of \$xin Bbb Rsetminus{0}\$ are:

\$\$[x] mathop{:=}{yinBbb Rsetminus {0}: xy>0}\$\$

Can you now identify both of them?