MATH135 - Assignment2 Solutions

MATH135 - Assignment2 Solutions - MATH 135 Assignment #2...

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MATH 135 Fall 2007 Assignment #2 Due: Wednesday 26 September 2007, 8:20 a.m. N.B. Assignments 3 to 9 will not be distributed in class. You must download them from the course Web site. Hand-In Problems 1. A sequence of integers is defined by x 1 = 2, x 2 = 86, x m +2 = - 2 x m +1 + 15 x m for m 1. Prove that x n = 4 · 3 n + 2( - 5) n for all n P . 2. A sequence of integers is defined by y 1 = 2, y 2 = 11, y 3 = 2, y m +3 = 3 y m +1 - 2 y m for m 1. Prove that y n = ( - 2) n + 3 n + 1 for all n P . 3. Suppose that x, y R . Prove that x 2 - 4 xy + 5 y 2 - 4 y + 4 = 0 if and only if ( x, y ) = (4 , 2). 4. Prove by contradiction that log 10 3 cannot be written in the form m n where m and n are positive integers. 5. Prove that if x 6 = y and a 6 = 0, then x y 6 = x + a y + a . 6. To prove a statement of the form “If H then C 1 or C 2 ” (where H, C 1 , C 2 are mathematical statements), we can assume that H is TRUE and that C 1 is FALSE, and prove that C 2 is TRUE. (If C 1 happened to be TRUE, we would be done, so we assume that C 1 is FALSE and prove that C 2 must be TRUE.) Suppose that m and n are integers. (a) Prove that if 3 m 2 + 6 m + 5 n 2 + 3 is odd, then m is even or n is odd. (b) Is the converse TRUE or FALSE? Either prove or give a counterexample. 7. An expression of the form a n x n + a n - 1 x n - 1 + ··· + a 1 x + a 0 with n 0 and a n , a n - 1 , . . . , a 1 , a 0 Q (or R ) is called a polynomial in x with coefficients from Q (or R ). If a n 6 = 0, the polynomial is said to have degree n . (The degree of a polynomial is the highest power of x that has a nonzero coefficient.) If all of the coefficients are 0, the polynomial is called the zero polynomial and its degree is not defined. (a) In each part, state whether the expression is a polynomial. If it is a polynomial, state its degree. i. 2 x 3 - πx + 3 ii. 2( x ) 3 - πx + 3 iii. 4 (b) In each part, perform the calculation, simplifying as much as possible. i. ( x 2 - 2 x + 1)( x 2 + 2 x + 1) ii. ( x 2 + 3 x + 1)( x 3 + x 2 - 2) + 3 x 4 + 4 x 2 + 5 ...continued
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8. A pile starts with n sticks, with n 2. The pile is split into two piles, and the product of the sizes of the piles is calculated. One of these new piles is then split into two piles, with the product of the sizes of these two new piles calculated and added to (the sum of) the previous product(s). This process continues until we have n piles with 1 stick each. For example, starting with 5 sticks, we could split 2, 3 (giving 6), then 2, 2, 1 (giving 6 + 2), then 1, 1, 2, 1 (giving 6 + 2 + 1), then 1, 1, 1, 1, 1 (giving 6 + 2 + 1 + 1 = 10). Prove using strong induction that, no matter how the pile is split, the total at the end is always 1 2 ( n 2 - n ). 9. In the country of Alwynnia, there are 10 cent, 20 cent and 50 cent coins. In her purse, Maddie
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MATH135 - Assignment2 Solutions - MATH 135 Assignment #2...

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