Triangle Inequality in Number Theory
Triangle Inequality in Number Theory Triangle Inequality in Number Theory For any real numbers a and b , the Triangle Inequality states: |a + b| ≤ |a| + |b| Proof We will consider different cases based on the signs of a and b : Case 1: a ≥ 0 and b ≥ 0 |a + b| = a + b = |a| + |b| Case 2: a ≥ 0 and b < 0 (or vice versa) If a + b ≥ 0, then: |a + b| = a + b ≤ a - b = |a| + |b| If a + b < 0, then: |a + b| = -(a + b) = -a - b = |a| + |b| Case 3: a < 0 and b < 0 |a + b| = -(a + b) = -a - b = |a| + |b| Algebraic Proof Using the definition of absolute values: |a + b| 2 ≤ (|a| + |b|) 2 |a + b| 2 = (a + b) 2 = a 2 + 2ab + b 2 (|a| + |b|) 2 = |a| 2 + 2|a||b| + |b| 2 Since |a| 2 = a 2 and |b| 2 = b 2 : a 2 + 2ab + b 2 ≤ a 2 + 2|a||b| + b 2 ...