Direct Proof and Counterexample IV

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1 Direct Proof and Counterexample IV
Lecture 14 Section 3.4 Fri, Feb 4, 2005

2 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.

3 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.

4 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.

5 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?

6 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).

7 Proof continued Expand and rearrange: n3 – n =
6(36q3 + 18q2r + 3qr2 – q) + (r3 – r). Therefore, 6 | (n3 – n) if and only if | (r3 – r).

8 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.

9 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.

10 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: Either x  y or x < y. Case I: Suppose x  y. Then |x – y| = x – y. So ((x + y) + |x – y|)/2 = ((x + y) + (x – y))/2 = x. Furthermore, max(x, y) = x.

11 max(x, y) = ((x + y) + |x – y|)/2.
Proof Continued Case II: Suppose x < y. Then |x – y| = -x + y. So ((x + y) + |x – y|)/2 = ((x + y) + (-x + y))/2 = y. Furthermore, max(x, y) = y. Therefore, max(x, y) = ((x + y) + |x – y|)/2.

12 Other Formulas What is a similar formula for min(x, y)?
What is a formula for max(x, y, z)?