M135F08A4S

# M135F08A4S - MATH 135 Assignment #4 Solutions Hand-In...

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MATH 135 Fall 2008 Assignment #4 Solutions Hand-In Problems 1. In each part, we must ensure that the remainder r satisﬁes 0 r < | b | = 12. (a) Since 261 = 21(12) + 9, then the quotient is 21 and the remainder is 9. (b) Since - 261 = - 22(12) + 3, then the quotient is - 22 and the remainder is 3. (c) Since 261 = ( - 21)( - 12) + 9, then the quotient is - 21 and the remainder is 9. (d) Since - 261 = 22( - 12) + 3, then the quotient is 22 and the remainder is 3. 2. We prove the result by contradiction. Assume that a, b, c are consecutive terms in a geometric sequence. Thus, b = ar and c = ar 2 for some real number r . (Since none of the three terms equals 0, then a 6 = 0 and r 6 = 0.) So our quadratic equation becomes ax 2 + arx + ar 2 = 0. Since a 6 = 0, we can divide through by a to obtain x 2 + rx + r 2 = 0. The discriminant is Δ = r 2 - 4(1)( r 2 ) = - 3 r 2 which is strictly less than 0 since r 6 = 0. Since the discriminant is negative, the quadratic equation cannot have real roots. This is a contradiction, as we are told that the quadratic equation does have real roots. Therefore, a, b, c cannot be consecutive terms in a geometric sequence. 3. (a) Using the Euclidean Algorithm, 1989 = 7(251) + 232 251 = 1(232) + 19 232 = 12(19) + 4 19 = 4(4) + 3 4 = 1(3) + 1 3 = 3(1) + 0 so gcd(1989 , 251) = 1, the second last remainder. (b) Reversing the Euclidean Algorithm above, gcd(1989 , 251) = 1 = 4 - 3 = 4 - (19 - 4(4)) = 5(4) - 1(19) = 5(232 - 12(19)) - 1(19) = 5(232) - 61(19) = 5(232) - 61(1(251) - 1(232)) = 66(232) - 61(251) = 66(1989 - 7(251)) - 61(251) = 66(1989) - 523(251) so one integer solution to 1989 x + 251 y = gcd(1989 , 251) is ( x, y ) = (66 , - 523).

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4. We answer both parts together using the Extended Euclidean Algorithm. 1 0 686 0 1 511 1 - 1 175 1 - 2 3 161 2 3 - 4 14 1 - 35 47 7 11 73 - 98 0 2 Therefore, gcd(686 , 511) = 7. Also, one solution to 686 x + 511 y = gcd(686 , 511) is ( x, y ) = ( - 35 , 47). 5. Since ac | bc , then bc = qac for some integer q . Regrouping, bc - qac = 0 or c ( b - qa ) = 0. Since c 6 = 0, then b - qa = 0 or b = qa . Therefore by deﬁnition, a | b . 6. We prove the contrapositive: “If m | n , then m 2 | n 2 .” Since m | n , then n = qm for some q Z . Thus, n 2 = ( qm ) 2 = q 2 m 2 . Since q 2 Z , then m 2 | n 2 by deﬁnition. Since the contrapositive is true, then the original statement is true. 7. Let d = gcd( a, b ). Since d | a and d | b , then nd | an and nd | nb .
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## This note was uploaded on 11/01/2008 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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M135F08A4S - MATH 135 Assignment #4 Solutions Hand-In...

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