solutionshw3-554-2010

solutionshw3-554-2010 - a. True. If x + y rational, then...

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Solutions homework 3. Page 16 Problem 1 : Any b B is an upper bound for A , so A is bounded above and sup A b for all b B . Now sup A is a lower bound for B and thus sup A inf B . Page 16 Problem 5. For n even the numbers look like 1 + 1 n , which are bounded below by 1. For n odd the numbers look like - 1 + 1 n , which are bounded below by - 1. Hence inf { ( - 1) n + 1 n : n 1 } exists and is ≥ - 1. If m ≤ - 1 + 1 2 m - 1 for all m 1, then m ≤ - 1. Hence the inf is equal to -1. Page 18 Problem 2. If 0 < x y n , then x y 1 n for all n 1. This implies x y 0 and thus also x 0, which contradicts 0 < x . Hence there is no x in the intersection. Page 19 Problem 1. [2 . 3] = 2, [ - 2 . 3] = - 3, and [ - 2] = - 2. Page 19 Problem 2. a. Figure 1. f ( x ) = [ x ] 1
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b. Figure 2. f ( x ) = [2 x ] c. Figure 3. f ( x ) = [ - x + 3] Page 20 Problem 2.
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Unformatted text preview: a. True. If x + y rational, then also y = ( x + y )-x is rational. b. False. Take e.g. x =-√ 2 and y = √ 2. c. True. If xy rational, so is x = xy y . Page 20 Problem 3. First solution: By Theorem 2.4.1 we can rind a rational number r such that x < r < y . Now by the Archimedean property we can find n ∈ N such that √ 2 < n ( y-r ). Let t = r + √ 2 n . Then by the previous problem t is irrational and x < r < t < y . Second Solution: Let x < y . Then x-√ 2 < y-√ 2, so there exists a rational number r with x-√ 2 < r < y-√ 2. Take now t = r + √ 2....
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This note was uploaded on 02/05/2012 for the course MATH 554 taught by Professor Girardi during the Fall '10 term at South Carolina.

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solutionshw3-554-2010 - a. True. If x + y rational, then...

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