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Unformatted text preview: Solutions to Homework 8  Math 2000 All solutions except 4.22,4.24, and 4.36 may be found in the book. (# 4.22) (a) Prove that if r is a real number such that 0 < r < 1, then 1 r (1 r ) 4 . (b) If the real number r in part (a) is an integer, is the implication true in this case? Explain. Solution . Note that 1 r (1 r ) 4 is true (multiply both sides of the equation by r (1 r )) only if 1 4 r (1 r ) is true (add 4 r 2 4 r to both sides) only if 4 r 2 4 r + 1 0 is true (factor) only if (2 r 1) 2 0 is true . However, this last equation is clearly true for all r R since the square of any real number is always nonnegative. Thus we could reverse all the steps to prove what we want. That is, if r R , then it follows that (2 r 1) 2 0. Now we have (2 r 1) 2 0 (expand out) 4 r 2 4 r + 1 0 (subtract 4 r 2 4 r from both sides) 1 4 r 4 r 2 = 4 r (1 r ) (divide by r (1 r ) as long as r 6 = 0 , 1) 1 r (1 r ) 4 (as long as r 6 = 0 , 1) ....
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This note was uploaded on 11/08/2009 for the course MATH 2000 taught by Professor Dd during the Spring '09 term at District of Columbia.
 Spring '09
 dd
 Math

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