Prove that the absolute value of a product is the product of the absolute values of factors.

Question

Theorem. $|a||b|=|ab|$

Proof. Applying the definition of absolute value, the left hand side of the equation could be either $a\times(-b)$ or $(-a)\times(b)$ or $a\times b$ or $(-a)\times(-b)$. For this reason we could have either $-ab$ or $ab$.

If we apply the definition of absolute value to the right hand side of our equation, we discover that we could have either $-ab$ or $ab$. So the theorem is proved.

Is this proof valid? Is it elegant from a mathematical viewpoint?

Thank you.

Edit: The definition of absolute value I am using above is the following: $$|x|=\begin{cases}x, & \text{if $x\ge0$}\\-x,&\text{if $x<0$}\end{cases}$$

Answer 1

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

1. $a \geq 0$, $b \geq 0$
2. $a \geq 0$, $b < 0$
3. $a < 0$, $b \geq 0$
4. $a < 0$, $b < 0$

separately, and verify that the equation $|ab| = |a||b|$ holds in each case.

Another way to prove this is to use the fact that $|x| = \sqrt{x^2}$.

Answer 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, $$|a||b|=a\cdot b=ab=|ab|.$$