In this post, I prove that

$|xy| = |x||y|$

in which $x$ is a variable for any real number and $y$ is a variable for any real number. This is done by approximately following the steps in reference [1]. Then I provide a lot of commentary because that is what I like to do.

Proof:

Note that $x$ can be zero or nonzero, and $y$ can be zero or nonzero–this leads to four cases. The cases involving zero are investigated first.

If $x=0$, then $|xy| = |0| = 0 = |0||y| = |x||y|$.

If $y=0$, then $|xy| = |0| = 0 = |x||0| = |x||y|$.

Thus, if $x=0$ and/or $y=0$, then $|xy| = |x||y|.$

For nonzero $x$ and nonzero $y$, there are four cases:

1. $x > 0$ and $y > 0$.
• $|xy|=xy=|x||y|$
2. $x > 0$ and $y < 0$.
• $|xy|=(xy)(-1)=x[y(-1)]=|x||y|$
3. $x < 0$ and $y > 0$.
• $|xy|=(xy)(-1)=[(-1)x]y=|x||y|$
4. $x < 0$ and $y < 0$.
• $|xy|=xy=xy(-1)^2=[(-1)x][y(-1)]=|x||y|$

$\square$

Commentary:

Ok, maybe I should explain a little more. Sometimes equations can condense information so much that the amount of information becomes overwhelming and for practical purposes inaccessible to the human eye.

First I separate out the cases involving zero because it is easier to deal with $<$ and $>$ signs than $\le$ and $\ge$ signs. I recognize that it is possible for one or both of the variables $x$ and $y$ to be zero, so I consider each case separately. The first case involves $x$ being zero and $y$ being any real number which, of course, includes zero. The second case involves $y$ being zero and $x$ being any real number which, again, includes zero.

If $x$ is zero, then $xy$ is zero because any real number multiplied by zero is zero. The absolute value signs remain because I am just evaluating the quantity inside the vertical bars. Then I realize that the absolute value of zero is equal to zero, and that any real number multiplied by zero is still zero so I can multiply the zero by the absolute value of $y$ while maintaining the equality. Finally, I recall that zero is $x$, so I replace $|0|$ with $|x|$, and I am done.

Next, if $y$ is zero, then $xy$ is zero. The absolute value signs remain. Then I realize that the absolute value of zero is equal to zero, and that any real number multiplied by zero is still zero so I can multiply the zero by the absolute value of $x$ while maintaining equality. Finally, I recall that zero is $y$, so I replace $|0|$ by $|y|$, and I am done.

This is why $x$ being zero, $y$ being zero, or $x$ and $y$ being zero leads to $|xy| = |x||y|$. But what about nonzero numbers? This other option is addressed next.

The main idea to keep in mind is that absolute values are, by definition, not negative numbers. And since the cases involving zero have been addressed, I can simply use the fact that the absolute value of a nonzero number is a positive number. Note that each variable can be positive or negative, and there are two variables, so there are four cases.

In the first case, $|xy|=xy$ because when I multiply $x$ and $y$ together, the result is already positive, so I do not need to modify the result. And the absolute value of a positive number is a positive number, so I can write $x=|x|$ and $y=|y|$. This takes care of the first case.

The second case is slightly trickier because one of the variables is allowed to be negative. When I multiply $x$ and $y$ together, I get a negative number because $y$ is negative. So if I want to find the absolute value of this number, I can multiply $xy$ by $-1$, and this is what I did. Finally, I notice that $x=|x|$ since $x$ is positive in this case, and realize that $y$, a negative number, multiplied by $-1$ is equal to the absolute value of $y$. That takes care of the second case.

The third case is similar to the second, except this time $x$ is negative and $y$ is positive. The same kind of logic applies here, but I’ll got through it to be complete. To find the absolute value of $xy$, first I multiply $x$ and $y$ together to obtain a negative number, and then I multiply by $-1$ to change the sign of the number. Then I notice that $|y|=y$ and $x$–a negative number–multiplied by $-1$ is the absolute value of $x$, since multiplying by $-1$ converts a negative number to a positive number. The third case has been addressed.

Last but not least is the fourth case. This time $x$ and $y$ are negative. Since multiplying a negative number by another negative number produces a positive number as taught in school, the result of multiplying $x$ with $y$ is a positive number and therefore equal to the absolute value of $x$ and $y$. But $x$ alone is not equal to the absolute value of $x$ without multiplying $x$ by $-1$, and the same can be said for $y$; this is why there are two instances of $-1$ in the above proof. Then I realize that $-1$ times $x$ is $|x|$ and that $-1$ times $y$ is $|y|$. The fourth case is complete.

And the whole proof is done.

References:

[1] https://www.math.tamu.edu/~mike.stecher/171/absoluteValueFunction.pdf