Direct Proof Division into Cases
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1 Direct Proof Division into Cases Lecture 16 Section 4.4 Robb T. Koether Hampden-Sydney College Mon, Feb 11, 2013 Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
2 1 The Quotient-Remainder Theorem 2 The Operators / and % in C 3 Proof by Cases 4 Leap Years 5 Assignment Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
3 Outline 1 The Quotient-Remainder Theorem 2 The Operators / and % in C 3 Proof by Cases 4 Leap Years 5 Assignment Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
4 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. Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
5 Outline 1 The Quotient-Remainder Theorem 2 The Operators / and % in C 3 Proof by Cases 4 Leap Years 5 Assignment Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
6 The Operators / and % in C 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. Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
7 The Operators / and % in C Thus, a == (a/b)*b + (a % b) is true for all positive integers a and b. Therefore, a % b == a - (a/b)*b is true for all positive integers a and b. Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
8 The Operators / and % in C What are a/b and a % b when a or b is negative? C requires that the expression a % b == a - (a/b)*b be true for all integers (where b!= 0)). Certainly, a/b < 0 if a < 0 and b > 0 or if a > 0 and b < 0. a/b > 0 if a < 0 and b < 0. Thus, the value of a % b is determined by the expression a % b == a - (a/b)*b Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
9 The Operators / and % in C Determine the values of the following expressions. 52 % % % % -11 Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
10 Outline 1 The Quotient-Remainder Theorem 2 The Operators / and % in C 3 Proof by Cases 4 Leap Years 5 Assignment Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
11 Proof by Cases Theorem If n is odd, then 8 (n 2 1). Proof. Let n be an odd integer. Then n = 8q + r for some integers q and r with 0 r < 8. First, note that n 2 1 = (8q + r) 2 1 = 64q qr + r 2 1 = 8(8q 2 + 2qr) + (r 2 1). Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
12 Proof by Cases Proof. Thus we need only show that 8 (r 2 1). Because n is odd, we know that r = 1, 3, 5 or 7. We will consider these four cases. Case 1: If r = 1, then r 2 1 = 1 1 = 0 = 8 0. Case 2: If r = 3, then r 2 1 = 9 1 = 8 = 8 1. Case 3: If r = 5, then r 2 1 = 25 1 = 24 = 8 3. Case 4: If r = 7, then r 2 1 = 49 1 = 48 = 8 6. Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
13 Proof by Cases Proof. In every case, 8 (r 2 1). Therefore, 8 (n 2 1) for all odd integers n. Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
14 Proof by Cases Theorem For any odd integer n, n 3 n is a multiple of 12. Proof. Let n be an odd integer. Then n = 12q + r for some integers q and r with 0 r < 12. First, note that n 3 n = (12q + r) 3 (12q + 4) = (12 3 q q 2 r qr 2 + r 3 ) (12q + r) = 12(12 2 q q 2 r + 3qr 2 q) + (r 3 r). Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
15 Proof by Cases Proof. Thus we need only show that 12 (r 3 r). Because n is odd, we know that r = 1, 3, 5, 7, 9 or 11. We will consider these six cases. Case 1: If r = 1, then r 3 r = 1 1 = 0 = Case 2: If r = 3, then r 3 r = 27 3 = 24 = Case 3: If r = 5, then r 3 r = = 120 = Case 4: If r = 7, then r 3 r = = 336 = Case 5: If r = 9, then r 3 r = = 720 = Case 6: If r = 11, then r 3 r = = 1320 = Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
16 Proof by Cases Proof. In every case, 12 (r 3 r). Therefore, 12 (n 3 n) for all odd integers n. Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
17 Outline 1 The Quotient-Remainder Theorem 2 The Operators / and % in C 3 Proof by Cases 4 Leap Years 5 Assignment Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
18 Leap Years A year y is a leap year if y is a multiple of 4, but not a multiple of 100 (a century year), or y is a multiple of 400, but not a multiple of 100. Write an expression that will be true if and only if y is a leap year. Today (Feb 11, 2013) is a Monday. What day of the week will be Feb 11, 2113? Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
19 Outline 1 The Quotient-Remainder Theorem 2 The Operators / and % in C 3 Proof by Cases 4 Leap Years 5 Assignment Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
20 Assignment Assignment Read Section 4.4, pages Exercises 1, 2, 5, 6, 14, 15, 19, 21, 23, 29, 36, 41, 47, 48, page 189. Robb T. Koether (Hampden-Sydney College) Direct Proof Division into Cases Mon, Feb 11, / 20
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