This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View
Full
Document
This note was uploaded on 06/21/2011 for the course MATHEMATIC 135 taught by Professor Phystago during the Spring '11 term at Waterloo.
 Spring '11
 Phystago
 Math, Algebra

Click to edit the document details