AJAY PATHAK (MATH)

Solution: Triangle Inequality Statement: For all real numbers x and y, |x + y| ≤ |x| + |y| (a) Verification Case (i): x = 2, y = 3 |x + y| = |2 + 3| = |5| = 5 |x| + |y| = |2| + |3| = 2 + 3 = 5 ∴ |x + y| = |x| + |y| Case (ii): x = −2, y = −3 |x + y| = |−2 − 3| = |−5| = 5 |x| + |y| = |−2| + |−3| = 2 + 3 = 5 ∴ |x + y| = |x| + |y| Case (iii): x = −2, y = 3 |x + y| = |−2 + 3| = |1| = 1 |x| + |y| = |−2| + |3| = 2 + 3 = 5 ∴ |x + y| ≤ |x| + |y| (b) Proof for all real numbers We prove the inequality by considering different cases. Case 1: x ≥ 0, y ≥ 0 |x + y| = x + y = |x| + |y| Hence, |x + y| ≤ |x| + |y|. Case 2: x ≤ 0, y ≤ 0 |x + y| = −(x + y) = −x − y = |x| + |y| Hence, |x + y| ≤ |x| + |y|. Case 3: x and y have opposite signs In this case, the sum x + y is reduced in magnitude. Therefore, |x + y| < |x| + |y| Hence, the inequality holds. Therefore, the Triangle Inequality holds for all real numbers x and ...

Max and Min of Two Numbers Using Absolute Value Maximum and Minimum of Two Numbers Using Absolute Value The maximum and minimum of two real numbers a and b can be expressed using the absolute value function. These formulas are very useful in mathematical analysis and proofs. 1. Formula for Maximum max(a, b) = (a + b + |a − b|) / 2 Proof Case 1: a ≥ b a − b ≥ 0 ⇒ |a − b| = a − b (a + b + |a − b|) / 2 = (a + b + (a − b)) / 2 = a Hence, max(a, b) = a . Case 2: b > a a − b < 0 ⇒ |a − b| = b − a (a + b + |a − b|) / 2 = (a + b + (b − a)) / 2 = b Hence, max(a, b) = b . 2. Formula for Minimum min(a, b) = (a + b − |a − b|) / 2 Proof Case 1: a ≤ b a − b ≤ 0 ⇒ |a − b| = b − a (a + b − |a − b|) / 2 = (a + b − (b − a)) / 2 = a Hence, min(a, b) = a . Case 2: b < a a − b > 0 ⇒ |a − b| = a − b (a + b − |a − b|) / 2 = (a + b − (a − b)) / 2 = b Hence, min(a, b) = b . Importa...