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Assorted Theorems Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem Assorted Theorems Alice E. Fischer CSCI 1166 Discrete Mathematics for Computing March 19-29, 2018 Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem 1 Integer properties 2 Parity: Even and Odd 3 The Quotient-Remainder Theorem Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem N: The Natural Numbers N = Z nonneg = 0; 1; 2; 3;::: Thm 1. There is no largest integer: 8x 2 Z 9y 2 Z; x < y There is no largest natural number: 8x 2 N 9y 2 N; x < y There is no smallest integer: 8x 2 Z 9y 2 Z; x > y There IS a smallest natural number: 0. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem Theorem 1: There is no greatest integer. Proof: Just suppose there is a greatest integer, N. Therefore 8n 2 Z; N ≥ n. Let M = N + 1. M is an integer since it is a sum of two integers, and M > N. Thus M is an integer greater than N, our supposed greatest integer. So we have a contradiction, N is the greatest and it is not. Therefore the original supposition is false, meaning that the theorem is true. Theorems 2 and 3 are proved similarly. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem Even and Odd Definition: Even integer. An integer n is even iff 9x 2 Z; n = 2x Definition: Odd integer. An integer n is odd iff 9x 2 Z; n = 2x + 1. Theorem 2: An integer n is odd iff 9x 2 Z; n = 2x − 1. 2x − 1 = 2x − 2 + 1 = 2(x − 1) + 1 Let y = x − 1. Then n = 2y + 1, where y is an integer. ) n is odd. You may use this theorem in your proofs. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem Parity Definition: The parity of an integer refers to whether the integer is even or odd. Even numbers have even parity, odd numbers have odd parity. Clearly, \parity" is related to the ability to organize things into pairs. For example, suppose you have 13 gray socks. The number 13 has odd parity, so you have 6 pairs and one odd sock. If you have 10 white socks, you have 5 pairs. The number 10 has even parity, so it comes out even, with no socks left over. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem Parity of Consecutive Integers Theorem 3: Consecutive integers have opposite parity. Proof by division into cases: Let m be any integer, and let n = m + 1 Case 1: Suppose m has even parity. Then m = 2x for some integer x. So n = m + 1 = 2x + 1 and n is therefore odd. Case 2: Suppose m has odd parity. Then m = 2x + 1 for some x 2 Z. So n = m + 1 = 2x + 1 + 1 = 2x + 2 = 2(x + 1) and n is therefore even. m and n have opposite parity in both cases, and there are no other cases. ) consecutive integers have opposite parity. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem Even and Odd Theorem Summary List of theorems about even and odd: 4 Zero is even. 5 −1 is not even. 6 If a and b are even, then a + b is even. 7 If a and b are odd, then a + b is even. 8 If a and b are even, then ab is even. 9 If a and b are odd, then ab is odd. 10 For all integers, n, if n2 is even, then n is even. 11 For all integers, n, if n2 is odd, then n is odd. 12 8m8n 2 Z, if m × n is even, then either m or n is even. 13 8m8n 2 Z, if m + n is even, then m and n are both even or both odd. 14 Every integer is either even or odd. Proofs for all are given below. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem Proofs about Even and Odd 4 Zero is even. By definition, 0 is even iff 9x 2 Z; 0 = 2x. So take x = 0; then 2x = 0. 5 −1 is not even. Because 8x 2 Z; −1 6= 2x. 6 If a and b are even, then a + b is even. Being even, a = 2x and b = 2y for some x and y. a + b = (2x + 2y) = 2 × (x + y) and (x + y) is an integer due to closure of addition. 7 If a and b are odd, then a + b is even. Being odd, a = 2x + 1 and b = 2 × y + 1 for some x and y. a + b = (2x + 1 + 2y + 1) = (2x + 2y + 2) = 2 × (x + y + 1) so (a + b) is an even integer. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem Proofs about Even and Odd, continued 8 If a and b are even, then ab is even By definition, a = 2x and b = 2y for some x and y. ab = (2x × 2y) = 2 × (2xy) and (2xy) is an integer due to closure of multiplication. 9 If a and b are odd, then ab is odd. By definition, a = 2x + 1 and b = 2 × y + 1 for some x and y. ab = (2x + 1) × (2y + 1) = (4xy + 2x + 2y + 1) = 2 × (2xy + x + y) + 1 and (2xy + x + y) is an integer due to closure. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem 10. To Be Even: Proof by Contraposition 8n 2 Z, if n2 is even, then n is even. Since this is a universal statement involving a condition, it is also possible to prove this by contraposition. The contrapositive is: 8n 2 Z, if n is NOT even, then n2 is NOT even. For all integers, n, if n is odd, then n2 is odd, since all integers are even or odd. Since n is odd, it can be represented as n = 2k + 1. So, n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. And since 2k2 + 2k is an integer via closure, this indicates that n2 must be odd by definition. So the contrapositive statement is true, therefore the original theorem is true. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem 10. To Be Even: Proof by Contradiction For all integers, n, if n2 is even, then n is even. A proof by contradiction. Suppose there is an integer, n, where n2 is even, but n is not. Since all integers are either even or odd, n must be odd, and can be represented as n = 2k + 1. So, n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. And since 2k2 + 2k is an integer via closure, this indicates that n2 must be odd, by definition. But n2 was assumed to be even. This contradiction indicates the original theorem is true. Proof by contradiction. Suppose there is an integer, n, where n2 is odd, but n is not. Since all integers are either even or odd, n must be even, and can be represented as n = 2k. So, n2 = (2k)2 = 4k2 = 2(2k2). And since 2k2 is an integer via closure, this indicates that n2 must be even by definition. But n2 was assumed to be odd. This contradiction indicates the original theorem is true. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem 11. To Be Odd For all integers, n, if n2 is odd, then n is odd. Try proving this yourself. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem 11. To Be Odd For all integers, n, if n2 is odd, then n is odd. Try proving this yourself. Proof by contradiction. Suppose there is an integer, n, where n2 is odd, but n is not. Since all integers are either even or odd, n must be even, and can be represented as n = 2k. So, n2 = (2k)2 = 4k2 = 2(2k2). And since 2k2 is an integer via closure, this indicates that n2 must be even by definition. But n2 was assumed to be odd. This contradiction indicates the original theorem is true. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem 12. Even Products For all integers, m and n, if m × n is even, then m is even or n is even. Proof by contraposition. The contrapositive: For all integers, m and n, if m is odd and n is odd, then m × n is odd. By definition, m = 2k + 1 and n = 2j + 1, for integers k and j. So, m × n = (2k + 1)(2j + 1) = 4kj + 2k + 2j + 1 = 2(2kj + k + j) + 1. And since 2kj + k + j is an integer via closure, this indicates that m × n must be odd by definition. Since the contrapositive is true, the original theorem is true. Outline Integer properties Parity: Even and Odd The Quotient-Remainder Theorem 13. Even Sums 8m; n 2 Z, if m + n is even, then m and n are both even or both odd. Proof by contradiction. Suppose m and n, are an odd integer an even integer. Case 1: m is even and n is odd. 1 By definition, m = 2k and n = 2j + 1, for integers k and j. 2 So, m + n = 2k + 2j + 1 = 2(k + j) + 1. 3 And since k + j is an integer via closure, this indicates that m + n must be odd by definition. 4 But m + n was assumed to be even, so this is a contradiction.
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