M325K_HW06_Soln - 2 p , p is not divisible by 2. So p...

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M325K HW 06 - Solutions to Graded Questions (Out of 25 pts; 5pts free) Section 3.3 16. (4pts) Proof. Suppose a, b, and c are integers and b a and c a . By definition of divisibility, there exist integers r and s such that ar b = and as c = .Then ) ( s r a as ar c b = = . Since r and s are integers, s r is an integer as well. So c b is a times an integers, and hence ) ( c b a by definition. 36. (4pts) a) k e k e e p p p a 3 3 2 3 1 3 ..... 2 1 = b) 11 7 3 2 2 2 = k and the resulting product as a cube= 3 3 2 2 3 3 6 6 2 5 4 2772 ) 11 7 3 2 ( 11 7 3 2 11 7 3 2 = = = k 37b. (4pts) Question: If x and y are integers and y x 9 10 = , does y 10 ? Does x 9 ? Explain. The first part only. Yes y 10 . Both 2 and 5 are prime factors of 10 x and so by unique factorization theorem, both 2 and 5 must occur in the prime factorization of 9 y . Since neither 2 nor 5 are factors of 9, they must be among the factors of y . Section 3.4 42. (4pts) Proof. Let p be any prime number except 2 and 3. By quotient-remainder theorem with 6 = d , we get k p 6 = , 1 6 + k , 2 6 + k , 3 6 + k , 4 6 + k , or 5 6 + k . Since p is a prime and
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Unformatted text preview: 2 p , p is not divisible by 2. So p cannot be equal to 6 k , 6 k +2, 6 k +4 (because these are divisible by 2). Similarly, p is a prime and 3 p , p is not divisible by 3, and hence 6k+3 is not possible. Thus, we have only two cases; namely, 1 6 + = k p or 5 6 + = k p . 45. (4pts) They are equal. Proof. Suppose m, n , and d are integers and ) ( n m d . By definition of divisibility, dk n m = for some integer k . Hence dk n m + = . Also denote d n mod by r . Then by definition of mod, we have r dq n + = where q and r are integers and d r < . By substitution, r k q d r dk dq dk r dq dk n m + + = + + = + + = + = ) ( ) ( ) ( . Since k q + is an integer and d r < , the above equation means the quotient of the division of n by d is ( q+k ) and the remainder is r . That is r d m = mod . This shows that d m mod and d n mod are equal....
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