A question about $ax = b$ I am studying inequality and come across the following statement. I don't understand it and want to believe the book must have made mistakes. I am going to copy what the book says here exactly.

A linear inequality with one variable is in the form: $ax>b$. When $a>0$, the solution is $x>b/a$, When $a<0$, the solution is $x<b/a$. When $a=0$, if $b<0$, there are infinite number of solutions; if $b>0$, there are no solutions.  


Your book is correct.
When $a>0$, multiply both sides of the inequality by $\frac{1}{a}$. Since this is a positive number, the direction of the inequality doesn't change:
\begin{align*}
ax&>b
\\\frac{1}{a}\cdot ax &> \frac{1}{a} \cdot b\\
x&>\frac{b}{a}
\end{align*}
When $a<0$, again multiply both sides by $\frac{1}{a}$. Now the direction of the inequality reverses since $\frac{1}{a}$ is negative.
\begin{align*}
ax&>b
\\\frac{1}{a}\cdot ax &< \frac{1}{a} \cdot b\\
x&<\frac{b}{a}
\end{align*}
When $a=0$, we cannot multiply by $\frac{1}{a}$. The inequality becomes $0\cdot x>b$, i.e., $0>b$. If $b>0$, there is a contradiction - $b$ cannot be both positive and negative. However, when $b<0$, every value of $x$ satisfies the equation $0 \cdot x >b$.
