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.  

 A: 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 $0 \cdot x >b$.
A: If we have the inequality
$a x > b$
then we can only divide by $a$ on both sides, if $a \ne 0$.
The next thing to watch out is that a negative value of $a$ will flip the $>$ into a $<$, while a positive $a$ will not change the operator, if we multiply or divide both sides by $a$.
This means
$$
a x > b \wedge a > 0 \iff x > b / a
$$
then
$$
a x > b \wedge a < 0 \iff x < b / a
$$
The third case was
$$
a x > b \wedge a = 0 \quad (*) \iff 0 > b \wedge a = 0 \iff b < 0 \wedge a = 0
$$
so it depends on $b$ if there are infinite many solutions $x$ with statement $(*)$ true or not.
Your book is mostly correct, it only misses the case $a=b=0$, where there is also no solution $x$ which makes $(*)$ true.
A: Your book is correct.
When $a=0$, our inequality is of the form $$0\cdot x-b>0.$$ 
If $b<0$, then $-b>0$. Now how many values of $x$ satisfy $0\cdot x -b>0$? No matter what we plug into $x$, we get that the inequality holds. Say $x=34$, then $0\cdot34-b=-b>0$. Since this is true no matter what we choose for $x$, there are infinitely many solutions. 
If $b>0$, however, then $-b<0$. Now can you plug anything into $0\cdot x-b>0$ to make the statement true? No matter what you plug in for $x$, you get that $-b>0$, but this is false since we assumed that $-b<0$. So there is no solution $x$ that satisfies the inequality.
