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?