hw05sol

hw05sol - M220(2009) SolutionsProblem Set 5 (c) 2009, UBC...

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Unformatted text preview: M220(2009) SolutionsProblem Set 5 (c) 2009, UBC Mathematics Department 1. (a) To show that f is one-to-one, pick any x 1 , x 2 R and suppose f ( x 1 ) = f ( x 2 ). Notice that p 1 + x 2 1 > 0 for each real x , so the sign of f ( x ) always agrees with the sign of x . It follows that x 1 and x 2 have the same sign, and that each line below implies the one after it: x 1 p 1 + x 2 1 = x 2 p 1 + x 2 2 x 2 1 1 + x 2 1 = x 2 2 1 + x 2 2 x 2 1 + x 2 1 x 2 2 = x 2 2 + x 2 1 x 2 2 x 2 1 = x 2 2 | x 1 | = | x 2 | . Since x 1 and x 2 have the same sign, the last line implies x 1 = x 2 . Therefore f is one-to-one. To show that f is onto, let any y (- 1 , 1) be given. We must show x R : f ( x ) = y. To do this, consider x = y/ p 1- y 2 . This makes sense, because the denominator is not 0, and calculation reveals the desired equation: f ( x ) = y/ p 1- y 2 p 1 + ( y 2 / (1- y 2 )) = y p 1- y 2 1 p 1 / (1- y 2 ) ! = y. Since f is both 1-1 and onto, it is a bijection. (b) The proof that f is onto implicitly provides the inverse function: f- 1 ( y ) = y p 1- y 2 y (- 1 , 1) . (c) The hard part is finding a bijection h : [- 1 , 1] (- 1 , 1). For this, enumerate the rationals in [- 1 , 1] by some scheme that gives Q [- 1 , 1] = { q 1 , q 2 , q 3 , . . . } , where q 1 =- 1 , q 2 = 1 . Then define h ( x ) = q n +2 , if x Q [- 1 , 1] has label x = q n , x, if x [- 1 , 1]- Q . Clearly h is a bijection, and g = f- 1 h is therefore a bijection from [- 1 , 1] to R , as required. 2. Chartrand et. al. , Chapter 10 (Cardinalities of Sets): #10.12, 10.30, 10.36. 10.12 Given A = { a 1 , a 2 , a 3 , . . . } and B = A-{ a n 2 : n N } , define g : B A by g ( x ) = x . Clearly g is one-to-one and A is denumerable, so B must be countable by a File hw05, version of 28 May 2009, page 1. Typeset at 17:21 4, 2009. UBC M220 Solutions for Problem Set #5 2 theorem given in class. (That theorem essentially stated the results in Exercise 10.34, below.) Its obvious that B is not a finite set, so B must be denumerable. Hence | B | = | A | . For a direct proof, imagine filling an infinite matrix with the positive integers in the pattern illustrated below: 1 2 5 10 17 4 3 6 11 18 9 8 7 12 19 16 15 14 13 20 25 24 23 22 21 . . . . . . . . . . . . . . . . . . (The next stage in this construction is to wrap the numbers 26 to 36 around the corner from top right to bottom left.) All the squares turn up in the first column. So delete the first column and use the diagonal process illustrated in the textbook to put the remaining entries into a sequence. Call this list p , so p = (2 , 5 , 3 , 10 , 6 , 8 , 17 , 11 , 7 , 15 , 26 , 18 , 12 , 14 , 24 , . . . ) ....
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hw05sol - M220(2009) SolutionsProblem Set 5 (c) 2009, UBC...

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