Fall2004 Final

Fall2004 Final - Faculty of Mathematics University of...

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Unformatted text preview: 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 1 767 1 472 1- 1 295 1- 1 2 177 1 2- 3 118 1- 3 5 59 1 8- 13 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 + n b d =- 399 + n 472 59 =- 399 + 8 n y = y- 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 . 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....
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This note was uploaded on 10/21/2010 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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Fall2004 Final - Faculty of Mathematics University of...

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