1873_solutions

# B we mean a set of the form x y x a where y is any

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Unformatted text preview: holds for some n. Therefore, whenever u  x there are members of A and also members of B between u and x and it follows from the fact that A and B are closed that x A B. Finally we need to observe that A B is closed. Suppose that x A B . We know that 1 either x A or x B. We assume, without loss of generality that x A. 1 1 1 Choose u  x such that no member of A lies between u and x. Of course, no member of A B can lie between u and x. Ý b. Prove that if A n is a closed uncountable subset of 1 for every positive integer n, then the set  n1 A n is a closed uncountable subset of 1 . The proof of this stronger assertion is similar to the one given in part a. However, in order to Ý show that  n1 A n has no upper bound we start with an arbitrary member y of 1 and we define the members x n as follows: x 1 is the least member of A 1 greater than y. x 2 is the least member of A 2 greater than x 1 . x 3 is the least member of A 1 greater than x 2 . x 4 is the least member of A 2 greater than x 3 . x 5 is the least member of A 3 greater than x 4 . x 6 is the least member of A 1 greater than x 5 . x 7 is the least member of A 2 greater than x 8 . x 8 is the least member of A 3 greater than x 7 . x 9 is the least member of A 4 greater than x 8 . x 10 is the least member of A 1 greater than x 9 . The sets A n appear in this sequence as A1, A2, A1, A2, A3, A1, A2, A3, A4, A1, A2, A3, A4, A5, A1, A2, A3, A4, A5, A6, A1,  The purpose of this arrangement of the sets is that, for each positive integer k, the condition x n A k holds for infinitely many values of n. A less intuitive but more constructive way of arranging the sets A k is make x n A k whenever n can be expressed in the form 2 k 3 m for some positive integer m and making x n A 1 whenever n can’t be expressed in the form 2 k 3 m . 5 The Real Number System Some Exercises on Surds 1. Given that a, b and c are surds, that a 0 and that b 2 4ac 0, explain why the solutions of the equation ax 2  bx  c  0 are surds. This fact follows at once from the formula x b b2 2a 4ac for the solutions of the equation ax 2  bx  c  0. 2. A detailed discussion of the material of this exercise can be found in the document on cubic equations. Please refer to that discussion for a detailed solution to this exercise. a. Verify by direct multiplication that for any numbers u, v and x, u  v  x u 2  v 2  x 2 uv ux vx  u 3  v 3  x 3 73 3uvx. b. Given numbers x, a and b, show that the expression x 3  ax  b can be written in the form x 3  ax  b  x 3  u 3  v 3 3uvx by solving the simultaneous equations 3uv  a Now show that as long as 27b 2  u 3  v 3  b. 0 these equations can be solved giving 4a 3 u 3 3 b  27b 2  4a 3 63 3 and v 3 3b 3 c. Deduce that if a and b are surds and 27b 2  4a 3 27b 2  4a 3 . 63 0 then the solutions of the cubic equation x 3  ax  b  0 are also surds. Why doesn’t this fact contradict the claim made earlier that the solutions of the equation 8x 3 6x 1  0 are not surds? d...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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