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

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: $$|x|=\begin{cases}x, & \text{if x\ge0}\\-x,&\text{if x<0}\end{cases}$$

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 $|x| = \sqrt{x^2}$.

• What if the question had the theorem |a||b||c|=|abc| how many cases would it take to prove? Sep 17, 2015 at 15:07
• You can use the theorem above to prove it: $|abc| = |(ab)c| = |ab||c| = |a||b||c|$ Sep 17, 2015 at 17:11

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

• Thanks for contributing. I added formatting to your post. All it takes, really, is putting dollar signs around each equation (double dollar signs if you want a displayed equation). I did a couple other fancy things (you can see if you click "edit" on your post), but just putting dollar signs around will get you most of the way there. Oct 13, 2016 at 0:30

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