Direct Proof and Counterexample IVThe Quotient-Remainder TheoremThe C Divide and Mod OperatorsSlide 4Slide 5Example: QuotientRemainder.cppExample: Proof by CasesProof continuedSlide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Example: Max(x, y)Slide 17Other FormulasDirect Proof and Counterexample IVLecture 17Section 3.4Wed, Feb 14, 2007The Quotient-Remainder TheoremTheorem: Let n and d be integers, d  0. Then there exist unique integers q and r such that n = qd + rand0  r < |d|.q is the quotient and r is the remainder.The C Divide and Mod OperatorsThe operators / and % in C are based on this theorem.If a and b are positive integers, thenq = a/band r = a % b, where 0  r < b.The C Divide and Mod OperatorsThus, a = (a/b)*b + (a % b) is true for all positive integers a and b.Therefore,a % b = a – (a/b)*bfor all positive integers a and b.The C Divide and Mod OperatorsWhat are a/b and a % b when a or b is negative?Isa = (a/b)*b + (a % b) still true when a or b is negative?Example: QuotientRemainder.cppExample: Proof by CasesTheorem: For any integer n, n3 – n is a multiple of 6.Proof:Divide n by 6 to get q and r: n = 6q + r, where 0  r < 6.That is, r = 0, 1, 2, 3, 4, or 5.Substitute: n3 – n = (6q + r)3 – (6q + r).Proof continuedExpand and rearrange: n3 – n = 6(36q3 + 18q2r + 3qr2 – q) + (r3 – r).Therefore, 6 | (n3 – n) if and only if 6 | (r3 – r).Proof continuedConsider the 6 possible cases:•Case 1: r = 0. r3 – r = 03 – 0 = 0 = 6 0.Proof continuedConsider the 6 possible cases:•Case 1: r = 0. r3 – r = 03 – 0 = 0 = 6 0.•Case 2: r = 1. r3 – r = 13 – 1 = 0 = 6 0.Proof continuedConsider the 6 possible cases:•Case 1: r = 0. r3 – r = 03 – 0 = 0 = 6 0.•Case 2: r = 1. r3 – r = 13 – 1 = 0 = 6 0.•Case 3: r = 2. r3 – r = 23 – 2 = 6 = 6 1.Proof continuedConsider the 6 possible cases:•Case 1: r = 0. r3 – r = 03 – 0 = 0 = 6 0.•Case 2: r = 1. r3 – r = 13 – 1 = 0 = 6 0.•Case 3: r = 2. r3 – r = 23 – 2 = 6 = 6 1.•Case 4: r = 3. r3 – r = 33 – 3 = 24 = 64.Proof continuedConsider the 6 possible cases:•Case 1: r = 0. r3 – r = 03 – 0 = 0 = 6 0.•Case 2: r = 1. r3 – r = 13 – 1 = 0 = 6 0.•Case 3: r = 2. r3 – r = 23 – 2 = 6 = 6 1.•Case 4: r = 3. r3 – r = 33 – 3 = 24 = 64.•Case 5: r = 4. r3 – r = 43 – 4 = 60 = 610.Proof continuedConsider the 6 possible cases:•Case 1: r = 0. r3 – r = 03 – 0 = 0 = 6 0.•Case 2: r = 1. r3 – r = 13 – 1 = 0 = 6 0.•Case 3: r = 2. r3 – r = 23 – 2 = 6 = 6 1.•Case 4: r = 3. r3 – r = 33 – 3 = 24 = 64.•Case 5: r = 4. r3 – r = 43 – 4 = 60 = 610.•Case 6: r = 5. r3 – r = 53 – 5 = 120 = 620.Proof continuedIn every case, 6 | (r3 – r).Therefore, 6 | (r3 – r) in general.Therefore, 6 | (n3 – n) for all integers n.Example: Max(x, y)Theorem: For all real numbers x and y,max(x, y) = ((x + y) + |x – y|)/2.Proof:Example: Max(x, y)Theorem: For all real numbers x and y,max(x, y) = ((x + y) + |x – y|)/2.Proof:Consider two cases…Other FormulasWhat is a similar formula for min(x, y)?What is a formula for max(x, y,