M135F08A2S

# M135F08A2S - MATH 135 Assignment#2 Solutions Hand-In...

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MATH 135 Fall 2008 Assignment #2 Solutions Hand-In Problems 1. (a) TRUE (b) FALSE (c) TRUE (d) FALSE (e) TRUE (f) FALSE (g) TRUE (h) “If n 2 is odd, then n is odd” (i) “If n 2 is even, then n is even” (j) “There is an integer whose square is less than or equal to the squares of all integers.” OR “There exists an integer x such that for all integers y , x 2 is less than or equal to y 2 .” (k) x y , x = y + 7 2. (a) This statement says “For all real numbers x and for all real numbers y , x 2 + y 2 = 1”. This statement is FALSE, since if x = 2 and y = 2, x 2 + y 2 = 1. (b) This statement says “There exists a real number x such that there exists a real number y such that x 2 + y 2 = 1”. This statement is TRUE, since choosing x = 0 and y = 1 gives x 2 + y 2 = 1. (c) This statement says “There exists a real number x such that for all real numbers y , x 2 + y 2 = 1”. This statement is FALSE, since for a fixed choice of x , y 2 cannot be always equal to 1 - x 2 for all choices of y (choose y = 2, for example; y 2 = 4 cannot be equal to 1 - x 2 ). (d) This statement says “For all real numbers x there exists a real number y such that x 2 + y 2 = 1”. This statement is FALSE, since if we choose x = 2, we need to find a real number y such that 4 + y 2 = 1 or y 2 = - 3, which is impossible. (e) This statement says “For all real numbers x there exists a real number y such that x 3 + y 3 = 1”. This statement is TRUE, since no matter what x we choose, we can always choose y = 3 1 - x 3 (since the cube root of any real number can be calculated). 3. We prove the result by induction on n . Base Case If n = 1, the left side equals 1 2 = 1 and the right side equals 1(1)(3) 3 = 1 which are equal. Therefore, the result holds when n = 1.

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Induction Hypothesis Suppose that the result holds for n = k , for some k P , k 1. That is, suppose that 1 2 + 3 2 + 5 2 + · · · + (2 k - 1) 2 = k (2 k - 1)(2 k + 1) 3 . Induction Conclusion Consider n = k + 1. Then 1 2 + 3 2 + 5 2 + · · · + (2 k - 1) 2 + (2 k + 1) 2 = k (2 k - 1)(2 k + 1) 3 + (2 k + 1) 2 (by Induction Hypothesis) = (2 k + 1) 1 3 k (2 k - 1) + (2 k + 1) = 1 3 (2 k + 1) [2 k 2 - k + 6 k + 3] = 1 3 (2 k + 1) [2 k 2 + 5 k + 3] = (2 k + 1)( k + 1)(2 k + 3) 3 = ( k + 1)(2( k + 1) - 1)(2( k + 1) + 1) 3 so the result holds for n = k + 1. Therefore, the result holds for all n P by POMI. 4. Base Case n = 1 a 1 = 5 and 2(3 n ) - 1 = 2(3) - 1 = 5, so the formula is correct. Induction Hypothesis Suppose the result is true for n = k , for some k P , k 1. That is, suppose that a k = 2(3 k ) - 1. Induction Conclusion Consider n = k + 1. Then a k +1 = 3 a k + 2 = 3(2(3 k ) - 1) + 2 (by Induction Hypothesis) = 6(3 k ) - 3 + 2 = 2(3 k +1 ) - 1 as required. Therefore, the result holds for all n P , by POMI. 5. Base Case n = 2 LS = k r =1 1 - 1 r = 1 - 1 2 = 1 2 = RS Induction Hypothesis Suppose the result is true for n = k , for some k P , k 2. That is, suppose that k r =2 1 - 1 r = 1 k .
Induction Conclusion Consider n = k + 1. Then k +1 r =2 1 - 1 r = k r =2 1 - 1 r 1 - 1 k + 1 = 1 k k k + 1 (by Induction Hypothesis) = 1 k + 1 as required.

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