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Fall2004 Final

Fall2004 Final - Faculty of Mathematics University of...

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Faculty of Mathematics University of Waterloo MATH 135 FINAL EXAM Fall 2004 Friday 17 December 2004 14:00 - 17:00 Solutions 1. (a) Determine the complete solution to the Linear Diophantine Equation [6] 767 x + 472 y = 7847 Solution: First, we use the Extended Euclidean Algorithm to calculate the gcd of 767 and 472: 767 x +472 y = r q i 1 0 767 0 1 472 1 - 1 295 1 - 1 2 177 1 2 - 3 118 1 - 3 5 59 1 8 - 13 0 2 Therefore, gcd(767 , 472) = 59 and 767( - 3) + 472(5) = 59. Also, 59 divides into 7847 giving a quotient of 133, so the original Linear Diophantine Equation has solutions. A particular solution is found by multiplying both sides of 767( - 3) + 472(5) = 59 by 133 to obtain 767( - 399) + 472(665) = 7847. Therefore, the complete solution to 767 x + 472 y = 7847 is x = x 0 + n b d = - 399 + n 472 59 = - 399 + 8 n y = y 0 - n a d = 665 - n 767 59 = 665 - 13 n for all n Z (b) Determine all non-negative integer solutions to the Linear Diophantine Equation [4] 767 x + 472 y = 7847 Solution: Using the complete solution from (a), we want x 0 and y 0. From x 0, we obtain - 399 + 8 n 0 or 8 n 399 or n 399 8 = 49 7 8 , so n 50 since n Z . From y 0, we obtain 665 - 13 n 0 or 13 n 665 or n 665 13 = 51 2 13 , so n 51 since n Z .

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MATH 135, Final Exam Solutions Page 2 of 10 Combining these inequalities, we get 50 n 51. If n = 50, ( x, y ) = (1 , 15). If n = 51, ( x, y ) = (9 , 2). Therefore, the non-negative solutions are (1 , 15) and (9 , 2). 2. (a) Determine the truth table for the statement ( P = Q ) OR ( Q = R ). [4] Solution: P Q R ( P = Q ) ( Q = R ) ( P = Q ) OR ( Q = R ) T T T T T T T T F T F T T F T F T T T F F F T T F T T T T T F T F T F T F F T T T T F F F T T T (b) Find a statement which does not contain the connective NOT and is equivalent to the [3] statement NOT( x y, x y ) No justification is required. Solution: Using the negation rules, we obtain x y, x > y (c) Give examples of statements P ( x ) and Q ( x ), each with universe of discourse equal to the [2] set of real numbers R , which demonstrate that the following statements are not equivalent: x, ( P ( x ) AND Q ( x )) and ( x, P ( x )) AND ( x, Q ( x )) Give a brief explanation. Solution: For x, ( P ( x ) AND Q ( x )) to be true, there needs to be an x for which both P ( x ) and Q ( x ) are true. For ( x, P ( x )) AND ( x, Q ( x )) to be true, there needs to be an x for which P ( x ) and there needs to be a (possibly different) x for which Q ( x ) is true. If we choose P ( x ) = “ x > 0” and Q ( x ) = “ x < 0”, then the first statement is false, since there is no x for which both P ( x ) and Q ( x ) are true. However, the second statement is true, since there is an x (for example, x = 1) for which P ( x ) is true, and there is an x (for example, x = - 1) for which Q ( x ) is true. Therefore, the statements are not equivalent.
MATH 135, Final Exam Solutions Page 3 of 10 3. (a) Determine all z C such that [4] 2 iz 2 + (2 + 2 i ) z + 1 = 0 Give your answers in standard form. Solution: Using the quadratic formula, z = - (2 + 2 i ) ± (2 + 2 i ) 2 - 4(2 i )(1) 4 i = - (2 + 2 i ) ± 4 + 8 i + 4 i 2 - 8 i 4 i = - 2 - 2 i ± 0 4 i = - 1 2 i - 1 2 = - 1 2 + 1 2 i (b) If z = x + iy , with x, y R , give the definition of | z | , the modulus of z . [2] Solution: The modulus of z , or | z | , equals x 2 + y 2 .

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