final-soln-s05 for exam

# final-soln-s05 for exam - Math 135 Final Examination 1 Math...

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Unformatted text preview: Math 135 - Final Examination 1 Math 135 Algebra Final Examination Solutions August 4, 2005 9:00 – 11:30 a.m. Instructor: Ashwin Nayak Question 1. (a) [4 marks] Write truth tables for the two statements: P = ⇒ ( Q = ⇒ R ) , and ( P = ⇒ Q ) = ⇒ R . Solution: P Q R Q = ⇒ R P = ⇒ ( Q = ⇒ R ) T T T T T T T F F F T F T T T T F F T T F T T T T F T F F T F F T T T F F F T T P Q R P = ⇒ Q ( P = ⇒ Q ) = ⇒ R T T T T T T T F T F T F T F T T F F F T F T T T T F T F T F F F T T T F F F T F (b) [2 marks] Are the two statements in part (a) equivalent? Give reasons. Solution: The statements are not equivalent, since they do not have the same truth value for every assignment to the variables P, Q, R . When P , Q and R are all false, the first statement is true whereas the second is false. Question 2. (a) [2 marks] Define a prime number. Math 135 - Final Examination 2 Solution: An integer p > 1 is prime if its only positive integer divisors are 1 and p . (b) [2 marks] Describe a rule to check whether an integer expressed in decimal digits (base 10) is divisible by 9. Solution: The integer is divisible by 9 iff the sum of all its digits in the decimal representation is divisible by 9. (c) [4 marks] State the Chinese remainder theorem (for two moduli). Solution: Let m , m 1 ∈ Z , such that m , m 1 > 1, and gcd( m , m 1 ) = 1. The simultaneous linear congruences x ≡ a (mod m ) x ≡ a 1 (mod m 1 ) always have a solution for any a , a 1 ∈ Z . If x ∈ Z is one solution, then all the solutions are given by x ≡ x (mod m m 1 ). (d) [2 marks] Give an expression for the multiplicative inverse of a complex number z in terms of ¯ z and | z | . Solution: z- 1 = ¯ z | z | 2 . (e) [2 marks] State De Moivre’s theorem for complex numbers. Solution: For any θ ∈ R , and n ∈ Z , (cos θ + i sin θ ) n = cos( nθ ) + i sin( nθ ) . Question 3(a). [5 marks] Let a, b, q, r ∈ Z such that b = aq + r . Prove that gcd( b, a ) = gcd( a, r ) . Solution: This is the same as proposition 2.21 in the text book. Note that b = a = 0 iff a = r = 0. In this case, both the GCDs are 0. Otherwise, let d = gcd( b, a ) > 0 and c = gcd( a, r ) > 0. Since d divides both b, a , it follows that d | ( b- qa ), i.e., d | r . Thus, d is a common divisor of a and r . So by definition, d ≤ c = gcd( a, r ). Similarly, c | ( qa + r ), i.e., c | b . So it is a common divisor of b and a . So c ≤ d = gcd( b, a ). Math 135 - Final Examination 3 Putting the two together, d = gcd( b, a ) = gcd( a, r ) = c ....
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## This note was uploaded on 06/21/2011 for the course MATHEMATIC 135 taught by Professor Phystago during the Spring '11 term at Waterloo.

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final-soln-s05 for exam - Math 135 Final Examination 1 Math...

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