> [!summary] This proof uses proof by cases and definitions of an absolute value to prove the inequalities rule > **Key Result:** $|x| \leq A \Rightarrow -A \leq x \leq A$ > $|x| > A \Rightarrow x<-A \space\lor x>A$ >[!info]+ Read Time **⏱ 2 mins** # Mathematical Proof > [!warning] Assumptions To prove the absolute inequality rule by [[Proof by Cases|proof by cases]] for both directions, assume the following: > - The definition of an absolute value is $ |x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases} > $ > - A is a non-zero [[Real Numbers|real number]] Prove that $|x| \leq A \Rightarrow -A \leq x \leq A$ $ \begin{array}{c} \text{Case 1: $x \geq 0$} \\ \\ \text{By definition if $x \geq 0 \Rightarrow x$} \\ \text{Then if x < A and A > 0 and $-A \leq 0\leq x$ then by definition:}\\ -A \leq x\leq A \\ \\ \text{Case 2: $x < 0$}\\ \\ \text{By definition if $x<0 \Rightarrow -x$} \\ \text{Then if $-x <A\Rightarrow x>-A$ and $x<0<A$ then by definition:}\\ -A \leq x\leq A \end{array} $ Prove that $|x| > A \Rightarrow x<-A \space\lor x>A$ $ \begin{array}{c} \text{Case 1: $x\geq 0$} \\ \\ \text{By definition if $x\geq 0 \Rightarrow x$} \\ \text{Then if $x>A$ by definition:} \\ x>A \\ \\ \text{Case 2: $x<0$} \\ \\ \text{By definiton if $x<0 \Rightarrow -x$} \\ \text{Then if $-x<A\Rightarrow x<-A$ so by definition:}\\ x<-A \end{array} $ # Resources <iframe width="560" height="315" src="https://www.youtube.com/embed/X7GhczgUy7c?si=7BTmC4MrsjB099Zp" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe> --- > 📚 Like this note? [Star the GitHub repo](https://github.com/rajeevphysics/Obsidian-MathMatter) to support the project and help others discover it! ---