Prove that the absolute value of a product is the product of the absolute values of factors.
No, as far as I can see you have said that the left side is either $ab$ or $-ab$ and that the right side is either $ab$ or $-ab$, but you haven't argued that both the left and the right side are equal to $ab$ or $-ab$ simultaneously, depending on what the sign of $a$ and $b$ might be. For instance, you haven't proved the impossibility that the left side of the equation is $ab$ while the right side of the equation is $-ab$.
To make the proof correct, consider the four exhaustive cases
- $a \geq 0$, $b \geq 0$
- $a \geq 0$, $b < 0$
- $a < 0$, $b \geq 0$
- $a < 0$, $b < 0$
separately, and verify that the equation $|ab| = |a||b|$ holds in each case.
Another way to prove this is to use the fact that $|x| = \sqrt{x^2}$.
You need to break it up into four cases (as Svinepels pointed out) to have a rigorous proof :
ex) If $a\ge0,b\ge0$, then $ab\ge0.$
Hence, by the definition of absolute value, we have $$|ab|=ab, |a|=a,|b|=b.$$ Hence, $$|a||b|=a\cdot b=ab=|ab|.$$
Using definitions and results from the previous answers, strict inequalities and some extra words.
Three cases:
1) $a,b$ both positive, i.e. $a>0$ and $b>0$
2) $a,b$ both negative, i.e. $a<0$ and $b<0$
3) wlog $a>0$ and $b<0$
In either case 1 or 2 above $ab>0$ so $$|ab|=ab \tag{eq1}$$
case 1
Since $a>0$ and $b>0$ we have $|a|=a$ and $|b|=b$, so $ab=|a||b|$ and by (eq1) $|ab|=|a||b|$
case 2
Since $a<0$ and $b<0$, $|a|=-a$ and $|b|=-b$ (here is where the confusion may start). $|a||b|=-a(-b)=ab$ keeping in mind that -a>0 and -b>0 and also that the product of two negatives is a positive. However you look at it $|a||b|=ab$. Although $a<0$ and $b<0$, both sides of the previous equation are positive and equal in magnitude and by (eq1) $|ab|=|a||b|$
case 3
$a>0, b<0,$ so $ab<0$ and $$|ab|=-ab=a(-b) \tag{eq2}$$
Now $|a|=a$, and since $b<0$, $|b|=-b$
That is, $a(-b)=|a||b|$ and by (eq2) $a(-b)=|ab|$ so that $|ab|=|a||b|$.