Direct Proof and Counterexample IV Lecture 17 Section 3.4 Wed, Feb 14, 2007

The Quotient-Remainder Theorem Theorem: Let n and d be integers, d  0. Then there exist unique integers q and r such that n = qd + r and 0  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, then q = a/b and 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)*b for 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? Is a = (a/b)*b + (a % b) still true when a or b is negative?

Example: QuotientRemainder.cpp

Example: 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 Therefore, 6 | (r3 – r) in general. In every case, 6 | (r3 – r). Therefore, 6 | (r3 – r) in general. Therefore, 6 | (n3 – n) for all integers n.

max(x, y) = ((x + y) + |x – y|)/2. Example: Max(x, y) Theorem: For all real numbers x and y, max(x, y) = ((x + y) + |x – y|)/2. Proof:

max(x, y) = ((x + y) + |x – y|)/2. 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, z)?