Chapter 1 - Section 1.8 - Proof Methods and Strategy - Review Questions - Page 111: 16

Case 1: $x$ is negative and $y$ is negative Case 2: $x$ is negative and $y$is non-negative Case 3: $x$ is nonnegative and $y$ is negative Case 4: $x$ is nonnegative and $y$ is nonnegative

We want to proof $|x y|=|x| \cdot|y|$ with a proof by Cases. The absolute value of $x$ has two options: If $x$ is negative, then $|x|=-x$ If $x$ is non-negative, then $|x|=x$ The absolute value of $y$ has two options If $y$ is negative, then $|y|=-y$ If $y$ is nonnegative, then $|y|=y$ The cases of the proof by cases should then be any combination of the two possibilities of $x$ and $y$ Case 1: $x$ is negative and $y$ is negative Case 2: $x$ is negative and $y$ is non-negative Case 3: $x$ is nonnegative and $y$ is negative Case 4: $x$ is nonnegative and $y$ is nonnegative Hence with the help of these cases, we can say that $|x y|=|x| \cdot|y|$