>[!summary] Proof by cases applies conditional statements where you create cases in the hypothesis that return to the same conclusion. >[!info]+ Read Time ⏱ **2 mins** # Definition Proof by cases is very similar to [[Types of Proofs/Direct Proof.md|direct proof]] in that it applies [[Logic/Conditional Statements.md|conditional statements]], but the step structure differs slightly: 1. Assume the hypothesis from proposition is true (**Hypothesis has cases, e.g. Hypothesis is an integer and irrational number**) 2. Then using definitions, proven results or facts to justify why the conclusion must be true The highlighted part of the steps are the difference between proof by cases and [[Types of Proofs/Direct Proof.md|direct proof]] ## Examples >[!example] Absolute value function for all real numbers x For all real numbers x, > $ |x| = \begin{cases} x, & \text{if } x \ge 0 \\ -x, & \text{if } x < 0 \end{cases} >$ **Proof by cases:** > Case 1: $x \geq 0$ Then the $|x| = x$ by definition > Case 2: $x \leq 0$ Then the $|x| = -x$ by definition since x is negative and taking the negative of a negative returns positive --- >[!example] For any [[Integers|integer]] $n, \space n^2$ is even if and only if $n$ is [[Even & Odd Numbers|even]] > **Proof by cases:** > Case 1: $n$ is even If n is even assume $n = 2k$ where k is an integer > Then $n^2 = 4k^2$ which will always be even > Case 2: $n$ is [[Even & Odd Numbers|odd]] If n is odd assume $n = 2k + 1$ where k is an integer > Then >$ \begin{array}{c} n^2 = (2k+1)^2 = 4k^2 + 4k+1 = 2(2k^2 + 2k ) +1 \\ \text{Assume $2k^2 + 2k = m$ where $m$ is also an integer by defintion} \\\\ 2(2k^2 + 2k ) +1 = 2m +1 \end{array} >$ Is in the form of an odd number so $n^2$ returns an odd number, proving the statement. --- 📂 Want to see more structured notes like these? Help grow the project by [starring Math & Matter on GitHub](https://github.com/rajeevphysics/Obsidian-MathMatter). ---