# Find Equivalence Classes for the relation $S$ on the set $\mathbb{R}^*$ is definied as $aSb \iff ab > 0$

I am trying to understand how to determine the equivalence classes. I have this question in the book

The relation $$S$$ on the set $$\mathbb{R}^*$$ is definied as $$aSb \iff ab > 0$$

The answer in the book is $$ = \mathbb{R}^+$$ and $$[-1] = \mathbb{R}^-$$.

I don't even know where to start with this problem. I know that $$ab > 0$$ so then $$a \cdot b$$ always have to be positive and I know that equivalence class means disjoint categories but why are $$$$ and $$[-1]$$ the equivalence classes?

• "I know that ab>0 so then a⋅b always have to be positive". – user247327 Mar 29 at 12:59

First, for all $$a,b \in \mathbb{R}^-$$, $$ab > 0$$. So all the elements in $$\mathbb{R}^-$$ are in the same equivalence class.
Secondly, for all $$a,b \in \mathbb{R}^+$$, $$ab > 0$$. So all the elements in $$\mathbb{R}^+$$ are in the same equivalence class.
Finally, if $$a \in \mathbb{R}^-$$ and $$b \in \mathbb{R}^+$$, $$ab < 0$$. So $$a$$ and $$b$$ are not in the same equivalence class.
This shows you that the two equivalence classes (containing $$\mathbb{R}^-$$ and $$\mathbb{R}^+$$) are disjoint, so you have indeed two equivalent classes : $$\mathbb{R}^-$$ and $$\mathbb{R}^+$$.
• I thought $$ and $[-1]$ where the equivalent classes? – Sam Mar 29 at 13:02
• Yes, but $$ denotes the equivalence class in which $1$ belongs (i.e. $\mathbb{R}^+$), and $[-1]$ denotes the equivalence class in which $-1$ belongs (i.e. $\mathbb{R}^-$). – TheSilverDoe Mar 29 at 13:04
Numbers $$a$$ and $$b$$ are equivalent with respect to relation $$R$$ if their product is positive. If you take $$a>0$$ what numbers are in relation with $$a$$? All $$b>0$$. Similarly for $$a<0$$, $$b$$ is in relation with $$a$$ if $$ab>0$$, so $$b$$ must be negative. $$R$$ partitions $$\mathbb{R}^*$$ into two nonempty, disjoint sets which add up to the whole $$\mathbb{R}^*$$, so the two equivalence classes are $$\mathbb{R}_+$$ and $$\mathbb{R}_-$$. Any positive number is in relation with $$1$$, so we can write $$_R=\mathbb{R}_+$$, and similarly, we can denote $$\mathbb{R}_-=[-1]_R$$.