What is Proof by Cases?                      •Situational Examples          •Numerical Examples

**Proof by Cases** is a mathematical technique used to prove a statement that depends on several distinct possibilities or conditions. The idea is to break down the problem into separate cases, prove that the statement holds true in each case, and then conclude that the statement is true overall.

### Steps in Proof by Cases:
1. **Identify All Possible Cases**: Determine all the different scenarios or cases that could occur. These cases should be mutually exclusive (no overlap) and exhaustive (cover all possibilities).
   
2. **Prove Each Case Individually**: For each identified case, show that the statement is true under that specific condition.
   
3. **Conclude the Proof**: Once all cases are proven, conclude that the statement is true in general, since it holds in every possible scenario.

### Situational Examples

**Example 1: Proving the Parity of an Integer**
- **Statement**: Prove that the square of any integer is either even or odd.
- **Case 1**: The integer is even. If \( n \) is even, then \( n = 2k \) for some integer \( k \). The square is \( n^2 = (2k)^2 = 4k^2 \), which is even.
- **Case 2**: The integer is odd. If \( n \) is odd, then \( n = 2k + 1 \) for some integer \( k \). The square is \( n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 \), which is odd.
- **Conclusion**: In both cases, the square of \( n \) is either even or odd.

**Example 2: Triangles in Geometry**
- **Statement**: Prove that in any triangle, the sum of the angles equals 180°.
- **Case 1**: The triangle is acute (all angles < 90°).
- **Case 2**: The triangle is right (one angle = 90°).
- **Case 3**: The triangle is obtuse (one angle > 90°).
- **Conclusion**: In each case, the sum of the angles equals 180°, so the statement holds for all triangles.

### Numerical Examples

**Example 1: Parity Check**
- **Statement**: Prove that for any integer \( n \), \( n^2 + n \) is always even.
- **Case 1**: \( n \) is even. If \( n = 2k \) for some integer \( k \), then \( n^2 + n = (2k)^2 + 2k = 4k^2 + 2k = 2(2k^2 + k) \), which is even.
- **Case 2**: \( n \) is odd. If \( n = 2k + 1 \) for some integer \( k \), then \( n^2 + n = (2k + 1)^2 + (2k + 1) = 4k^2 + 4k + 1 + 2k + 1 = 2(2k^2 + 3k + 1) \), which is even.
- **Conclusion**: In both cases, \( n^2 + n \) is even.

**Example 2: Absolute Value**
- **Statement**: Prove that \( |x| = x \) if \( x \geq 0 \) and \( |x| = -x \) if \( x < 0 \).
- **Case 1**: \( x \geq 0 \). Here, \( |x| = x \) by the definition of absolute value.
- **Case 2**: \( x < 0 \). Here, \( |x| = -x \) by the definition of absolute value.
- **Conclusion**: The statement is true in both cases.

Would you like more details or have any questions?

### Five Related Questions:
1. How does Proof by Cases differ from Proof by Contradiction?
2. Can Proof by Cases be used to prove inequalities?
3. What are the limitations of using Proof by Cases?
4. How does Proof by Cases relate to conditional statements in logic?
5. Can Proof by Cases be applied to non-integer domains, like real numbers?

### Tip:
When using Proof by Cases, ensure that the cases are exhaustive and cover all possible scenarios. If you miss a case, your proof may be incomplete.

Explore the concept of Proof by Cases in mathematics with detailed examples and step-by-step explanations. Learn how to prove statements based on different scenarios and cases, covering situational and numerical examples.

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