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

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  • $\begingroup$ What is the definition of absolute value that you are using? $\endgroup$ – Michael Albanese Jan 10 '14 at 12:03
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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 $|x| = \sqrt{x^2}$.

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  • $\begingroup$ What if the question had the theorem |a||b||c|=|abc| how many cases would it take to prove? $\endgroup$ – wizlog Sep 17 '15 at 15:07
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    $\begingroup$ You can use the theorem above to prove it: $|abc| = |(ab)c| = |ab||c| = |a||b||c|$ $\endgroup$ – Ulrik Sep 17 '15 at 17:11
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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, $$|a||b|=a\cdot b=ab=|ab|.$$

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

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  • $\begingroup$ 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. $\endgroup$ – 6005 Oct 13 '16 at 0:30

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