Chapter 3: Elementary Number Theory and Methods of Proof. January 31, 2010


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1 Chapter 3: Elementary Number Theory and Methods of Proof January 31, 2010
2 3.4  Direct Proof and Counterexample IV: Division into Cases and the QuotientRemainder Theorem
3 QuotientRemainder Theorem Given any integer n and positive integer d, there exist unique integers q and r such that n = dq + r and 0 r d. d is the divisor, n is the dividend, q is the quotient, and r is the remainder Ex. Let n = 37 and d = 5. Then 37 = 5 (7) + 2 (where q = 7 and r = 2). Ex. Let n = 52 and d = 23. Then 52 = 23 ( 3) + 17 (where q = 3 and r = 17).
4 DEFINITION: Given a nonnegative integer n and a positive integer d, n div d = the integer quotient q when n is divided by d. n mod d = the integer remainder r when n is divided by d. In other words: Suppose n, d N, q, r Z, 0 r d n = dq + r n div d = q and n mod d = r [ If n = dq + r, then d (n r) and we say n is congruent to r ] modulo d and often use the notation n r mod d.
5 Most standard programming languages (C++, Java, Pascal, etc) have built in functions that compute mod and div, given a nonnegative integer n and a positive integer d. Ex. Compute 62 div 7 and 62 mod 7: Since 62 = n = dq + r we have 62 div 7 = 8 n div d = q and 62 mod 7 = 6 n mod d = r
6 Ex. Suppose today is Wednesday and you know that neither this year nor the next year are leap yeats. What day of the week will it be in 342 days time? Solution: Labels the days as: 0 =Sunday, 1 =Monday,..., 6 =Saturday. Today is Wednesday (day 3), so we want to work out what day of the week = 345 will be. We find 345 mod 7 = 2 [i.e. 345 = 7 (49) + 2] So in 342 days time, it will be Tuesday.
7 Representations of Integers Given any integer n and positive integer d, we can represent n as n = dq + r 0 r d for some integers q, r. For d = 2: n = 2q + r, 0 r 2 which is the same as either n = 2q + 0 or n = 2q + 1 The parity property: every integer is either even or odd.
8 Theorem: Any two consecutive integers have opposite parity. Proof: Let n, n + 1 be two arbitrary, but consecutive integers. By the parity property, n is either odd or even. Case 1: Assume n is even. Then n = 2k, for some k Z. Then n + 1 = 2k + 1 and by definition, n + 1 is odd. Case 2: Assume n is odd. Then n = 2k + 1, for some k Z. Then n + 1 = 2k + 2 = 2(k + 1) and as k + 1 is an integer, by definition, n is even. Therefore, regardless of whether n is odd or even, the theorem is true.
9 Ex. Prove that every integer can be written in one of the four forms for some integer q: Proof: n = 4q, n = 4q + 1, n = 4q + 2, n = 4q + 3 Let n be any integer. We apply the QuotientRemainder Theorem to n with d = 4: n = 4q + r, 0 r 4 We list all the possibilities of r to prove the result: n = 4q, n = 4q + 1, n = 4q + 2, n = 4q + 3
10 Theorem: The square of any odd integer has the form 8m + 1 for some integer m. Proof: We want to show: odd integers n, integer m such that n 2 = 8m + 1. Let n be an arbitrary odd integer. Using the previous result (with the QuotientRemainder Theorem and d = 4, n can be written in one of four forms: n = 4q, n = 4q + 1, n = 4q + 2, n = 4q + 3 for some integer q. Since n is odd, it cannot have either of the forms 4q or 4q + 2 (since they are both even). Therefore, n = 4q + 1, or, n = 4q + 3
11 Case 1: Assume n = 4q + 1. Then n 2 = (4q + 1) 2 (substitution) = 16q 2 + 8q + 1 = 8(2q 2 + q) + 1 Let m = 2q 2 + q. Then we have proven that n 2 = 8m + 1. Case 2: Assume n = 4q + 3. Then n 2 = (4q + 3) 2 (substitution) = 16q q + 9 = 16q q = 8(2q 2 + 3q + 1) + 1 Let m = 2q 2 + 3q + 1. Then we have proven that n 2 = 8m + 1. We have proven that for any odd integer n, n 2 = 8m + 1 for some integer m.
12 3.5  Direct Proof and Counterexample V: Floor and Ceiling Floors and Ceilings just tell us whether to round a fraction up or down to the nearest integer.
13 DEFINITION: Given any real number x, the floor of x, denoted x is defined as: x = the unique integer n such that n x < n + 1. x = n n x < n + 1. We round down to the nearest integer. Ex. 3.2 = 3.2 Ex. 3.1 = 4 Ex. 7 = 7 Ex. Let n be an integer. n = n
14 DEFINITION: Given any real number x, the ceiling of x, denoted x is defined as: x = the unique integer n such that n 1 < x n. x = n n 1 < x n. We round up to the nearest integer. Ex. 3.2 = 4 Ex. 3.1 = 3 Ex. 7 = 7 Ex. Let n be an integer. n = n + 1
15 Ex. Boxes, each capable of holding 36 units, are used to ship a product from the manufacturer to a wholesaler. Express the number of boxes that would be required to ship n units of the product using either the floor or the ceiling notation. Which one is more appropriate? Solution: We need either n or n 36 boxes. Obviously, the ceiling notation is more convenient here.
16 Ex. Is the following statement true or false? For all real numbers x and y, x + y = x + y Solution: We will find a counterexample to prove this statement is false. Let x = 0.5 and y = 0.5 Then x + y = = 1 = 1 And x + y = = = 0 Therefore, if x = y = 0.5, then x + y = 1 0 = x + y
17 Theorem For all real numbers x and all integers m, x + m = x + m. Proof: Let x be an arbitrary real number and m be an arbitrary integer. Let x = n. This means n x < n + 1 (from the definition of the floor function). Add m to both sides of the inequality: n + m x + m < n + m + 1 Since n + m is an integer, by the floor function definition: x + m = n + m which is the same as x + m = x + m.
18 Theorem For any integer n, { n n = 2 if n is even 2 n 1 2 if n is odd [This value is required for binary search and mergesort algorithms.] Proof: Let n be an arbitrary integer. The QuotientRemainder Theorems says that n must be either odd or even, so we break the proof into two cases.
19 Case 1: Suppose n is odd. We wish to show that n 2 = n 1 2. As n is odd, by definition, n = 2k + 1 for some integer k. Then n 2 = 2k+1 2 = 2k = k = k But as n = 2k + 1, we can rearrange the equation to k = n 1 Thus, we have shown that n 2 = k = n 2, proving the result. 2.
20 Case 2: Suppose n is even. We wish to show that n 2 = n 2. As n is even, by definition, n = 2k for some integer k. Then n 2 = 2k 2 = k = k But as n = 2k, we can rearrange the equation to k = n 2. Thus, we have shown that n 2 = k = n 2, proving the result.
21 Recall the QuotientRemainder Theorem: for an integer n and positive integer d, there exist unique integers q, r such that n = dq + r and 0 r < d. We now prove a related theorem: Theorem: If n is any integer, d is a positive integer, q = n/d, and r = n d n/d, then n = dq + r and 0 r < d. Proof: Let n be any integer, d be a positive integer, q = n/d, and r = n d n/d.
22 ( ) Then dq + r = d n d + n d n d = n So n = dq + r and we simply need to show that 0 r < d. As q = n/d, by the definition of the floor function, q n d < q + 1. Multiply the inequality by d: dq n < dq + d Subtract dq from the inequality: 0 n dq < d Substitute r = n dq (we showed this above) into the inequality: 0 r < d and the proof is complete.
23 3.6  Indirect Argument: Contradiction and Contraposition
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