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Unformatted text preview: Math 326 Fall 2010 Homework Set Chapter 3 Solutions 1 3.33 Does 5  0? Yes. 0 = 0 5. 3.34 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. 3.35 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) 3.313 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) 3.316 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 ). 3.317 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. 3.318 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 . 3.320 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 3.326 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 . 3.328 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. 3.334 Use the unique factorization theorem to write the following integers in standard factored form....
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This note was uploaded on 06/12/2011 for the course MATH 103 taught by Professor Wouters during the Spring '08 term at Wisc Oshkosh.
 Spring '08
 WOUTERS
 Algebra, Factoring

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