Prove that $(\mathbb{R}\setminus \{0\},\sim):= ab>0$ is transitive.

I have the following problem:

A relation $\sim$ on $\mathbb{R}\setminus\{0\}$ is defined by $a\sim b$ if $ab>0$. Show that $\sim$ is an equivalence relation and identify the equivalence classes.

I've been able to easily demonstrate that $\sim$ is both reflexive and symmetric, but I'm not sure how to approach demonstrating that it is transitive.

If $ab>0$, $bc>0$, then

$$ab\cdot bc >0 \implies ab^2c>0$$

Since $ab>0$, $b\neq 0$ so $b^2>0$. From the previous inequality, we have that $ac>0$.

• Ah. For whatever reason I had not considered multiplying them together. I had thought of dividing them out, but was getting nowhere that way. – agent154 Sep 20 '15 at 18:25

Hint: $ac = \frac{(ab)(bc)}{b^2}$.

• Shows there is no need for cases +1 – rschwieb Sep 20 '15 at 18:21

Let a~b i.e.ab>0 and b~c i.e. bc>0 Now ab>0 implys a and b have same sign i.e. either both are positive or both are negative. Similarly b and c have same sign. Thus,ultimately we get that a and c have same sign (because both have same sign as that of b), thus ac>0 and hence a~c.

Let $a\sim b$ and $b\sim c$. This means $ab>0$ and $bc>0$. If $a,b>0$, then $c>0$, hence $bc>0$. If $a<0$ and $b<0$, then $c<0$, hence also $bc>0$. So in both cases we have $b\sim c$.

First, if $ab>0$ then $ab^{-1}>0$, because if $ab>0$, then $ab^{-1}=(ab)b^{-2}>0$ (given that $\frac{1}{b^2}>0$). Now, suppose $a~b$ and $b~c$, so $ab>0$ and $bc>0$. Then $ac=(ab)(b^{-1}c)>0$.