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HWSet-Chapter3 - Math 326 Fall 2010 Homework Set Chapter 3...

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Math 326 Fall 2010 Homework Set Chapter 3 Solutions 1
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3.3-3 Does 5 | 0? Yes. 0 = 0 · 5. 3.3-4 Is (3 k + 1)(3 k + 2)(3 k + 3) divisible by 3? Yes, because (3 k + 1)(3 k + 2)(3 k + 3) = 3(3 k + 1)(3 k + 2)( k + 1) by factoring a 3 from the last term. 3.3-5 Is 6 m (2 m + 10) divisible by 4? Yes, because 6 m (2 m + 10) = 2 · 3 m · 2( m + 5) = 4 · 3 m ( m + 5) 3.3-13 If n = 4 k + 3, does 8 divide n 2 - 1? Yes, because n 2 - 1 = (4 k + 3) 2 - 1 = 16 k 2 + 24 k + 9 - 1 = 16 k 2 + 24 k + 8 = 8(2 k 2 + 3 k + 1) 3.3-16 Prove the following directly from the definition of divisibility: For all integers a , b , and c , if a | b and a | c , then a | ( b - c ). Since a | b then k Z such that b = ka (def of divisibility) Since a | c then m Z such that c = ma (def of divisibility) Hence b - c = ka - ma (substitution) = ( k - m ) a (distributive) Let q = k - m . Then by closure of Z under subtraction, q Z . Hence b - c = qa . Hence q Z such that b - c = qa . Hence a | ( b - c ). 3.3-17 True (prove) or false (counterexample): The sum of any three consecutive integers is divisible by 3. Proof: Let k Z and let S = k + ( k + 1) + ( k + 2) = 3 k + 3 = 3( k + 1) Let p = k + 1 Z because Z is closed under addition. Hence S = 3 p . Hence 3 | S . Hence the sum of any 3 consecutive integers is divisible by 3. 3.3-18 True (prove) or false (counterexample): The product of any two even integers is a multiple of 4. Proof: Let p, q be any even integers. Then by the definition of even, m, n Z such that p = 2 m and q = 2 n . Hence pq = 2 m · 2 n = 4 mn . Since pq Z by closure of Z under multiplication, 4 | pq . 3.3-20 True (prove) or false (counterexample): A sufficient condition for an integer to be divisible by 8 is that it be divisible by 16. Another way to write this is ( a Z )(16 | a = 8 | a ). Proof: Suppose a Z and 16 | a . Then k Z such that a = 16 k = 2 · 8 · k . Then let p = 2 k . Since k Z , by closure of Z , p Z . By substitution, a = 8 p . Hence 8 | a . 2
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3.3-26 True (prove) or false (counterexample): For all integers a and b , if a | b then a 2 | b 2 . Proof: Let a, b Z such that a | b . By definition of divisibility, k Z such that b = ak . Squaring both sides of this equation gives b 2 = ( ak ) 2 = a 2 k 2 . But k 2 is an integer (being a product of the integer k times itself). Hence by definition of divisibility, a 2 | b 2 . 3.3-28 True (prove) or false (counterexample): For all integers a and b , if a | 10 b then a | 10 or a | b . Counterexample : Let a = 25 and b = 5. Then 10 b = 50. Since 25 | 50, a | 10 b . But neither a | 10, which is 25 | 10 nor a | b , which is 25 | 5, are true. 3.3-34 Use the unique factorization theorem to write the following integers in standard factored form. (a) 1176 = 2 3 · 3 · 7 2 (b) 5733 = 3 2 · 7 2 · 13 (c) 3675 = 3 · 5 2 · 7 2 3.3-35 Suppose that in standard factored form a = p e 1 1 p e 2 2 · · · p e k k where k is a positive integer; p 1 , p 2 , . . . are prime numbers; and e 1 , e 2 , . . . are positive integers. (a) What is the standard factored form for a 2 ? a 2 = p 2 e 1 1 p 2 e 2 2 · · · p 2 e k k (b) Find the least positive integer n such that 2 5 · 3 · 5 2 · 7 3 · n is a perfect square.
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