# 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.

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?