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  n1 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  n1 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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