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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 Moivres theorem for complex numbers. Solution: For any R , and n Z , (cos + i sin ) n = cos( n ) + i sin( n ) . 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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 Spring '11
 Phystago
 Math, Algebra

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