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: $x > 0$ and $ y > 0$. $|xy|=xy=|x||y|$ $x > 0$ and $ y < 0$. $|xy|=(xy)(-1)=x[y(-1)]=|x||y|$ $x < 0$ and $ y > 0$. $|xy|=(xy)(-1)=[(-1)x]y=|x||y|$ $x < 0$ 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 […]
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