How to prove, with both $a$ and $b \neq 0$ : if $-1\lt b \lt 1$ then $|\frac ab | \gt |a|$. I can prove 
(1) if $0 \lt b \lt 1$ then $\frac 1b\gt b $
(2) if $-1 \lt b \lt 0$ then $\frac 1b\lt b $
Proofs : 
(1) $0 \lt b \lt 1$ 
$\rightarrow \frac bb \lt \frac 1b$
$\rightarrow 1 \lt \frac 1b$
$\rightarrow \frac 1b \gt 1 \gt b$
$\rightarrow \frac 1b \gt b$
(2) $-1 \lt b \lt 0$ 
$\rightarrow \frac {-1}{b}\gt \frac bb$ ( dividing by a negative number reverses inequality) 
$\rightarrow \frac {-1}{b}\gt 1$ ( since $n/n$ =1 for all $n$) 
$\rightarrow \frac {1}{b}\lt -1\lt b$ ( multiplying both sides by $-1$). 
$\rightarrow \frac {1}{b}\lt b$ ( by transitivity of order)
But I can't find a way to prove that: if $-1 \lt b\lt 1$, then  $|\frac ab|\gt|a|$ 
Would you please give a hint? 
 A: You have the right idea in what you're doing, except you should use the limits of $1$ and $-1$ instead of $b$ as you did, so you have $\frac{1}{b} \gt 1$ and $\frac{1}{b} \lt -1$, respectively. Then, note that for the $b \lt 0$ case, you can multiply both sides by $-1$, so you need to switch the inequality, to get
$$\frac{1}{-b} \gt 1 \tag{2}\label{eq2A}$$
Now, since $-b \gt 0$, when you take absolute value signs, this inequality will not change, and likewise for the first case where $b \gt 0$, but note the inequality is the same, i.e., you get
$$\left|\frac{1}{b}\right| \gt 1 \tag{3}\label{eq3A}$$
Now, you just need to multiply both sides by $|a|$ to get the end result.
However, alternatively, it's somewhat shorter & simpler to do the following instead. First,
$$-1 \lt b \lt 1 \implies |b| \lt 1 \tag{4}\label{eq4A}$$
Since $b \neq 0$, you then have by dividing both sides by $|b|$ results in
$$1 \lt \left|\frac{1}{b}\right| \implies \left|\frac{1}{b}\right| \gt 1 \tag{5}\label{eq5A}$$
Next, as indicated earlier, since $a \neq 0$, multiplying both sides by $|a|$ gives
$$\left|\frac{a}{b}\right| \gt |a| \tag{6}\label{eq6A}$$
A: You can write:
$$\frac{|a|-|ab|}{|b|}>0\leftrightarrow\frac{|a|(1-|b|)}{|b|}>0$$
Then, because $a\neq 0 \land |b|  >0 \forall b \in R-\{0\}$, you can multiply by $|b|$ and so:
$$|a|(1-|b|)>0$$
Dividing by $1-|b|\neq 0 \land 1-|b|>0$, you obtain:
$$|a|>0$$
This is always true$\forall a \in R-\{0\}$, so your thesis is true.
