1873_solutions

Lying between x and sup a since x is less than the

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Unformatted text preview: B is the least upper bound of A  B, as promised. Exercises on the Archimedean Property of the System R 1. Prove that if A is the set of all rational numbers in the interval 0, 1 then sup A  1. Quite simply, the largest member of the set A is 1. 2. Suppose that A 1 , 1 , 1 , 123  1 n n Z . Prove that inf A  0. The number 0 is clearly a lower bound of A. Now given any number x  0 we know that there exists a positive integer n such that 1  x and, consequently, that x cannot be a lower bound of A. n Therefore 0 is the greatest lower bound of A. 3. A nonempty set G of real numbers is said to be a subgroup of R if whenever x and y belong to G then so do the numbers x  y and x y. a. Determine which of the following sets are subgroups of R. 1, 0, 1 Z  0 Q  R Q 2n Z n R Z Q 2n mn 2 n m Z Z and n Z The set is not a subgroup of R because the definition requires a subgroup to be nonempty. The set 1, 0, 1 is not a subgroup because 1  1 does not belong to this set. The set Q is a subgroup of R. The set Z is a subgroup of R. The set Z  is not a subgroup of R because 2 4 does not belong to Z  . The same argument applies to the sets Q  and 2n n Z  . The set 2n n Z is a subgroup of R. The set 0 is a subgroup of R. The set R is a subgroup of R. The set R Q is not a subgroup of R because the number 2 2 does not belong to R Q. 82 The set mn 2 m Z and n Z is a subgroup of R. b. Explain why every subgroup of R must contain the number 0. Show that if G is any subgroup of R other than 0 then G must contain infinitely many positive numbers. Suppose that G is a subgroup of R. Using the fact that G is not empty we choose a member x of G. Since 0  x x we conclude that 0 G. Now suppose that the subgroup G is not 0 . Choose a member x of G such that x 0. Since the infinite set nx n Z is an infinite subset of G we know that G is an infinite set. c. Suppose that G is a subgroup of R other than 0 , that p  inf x G x  0 and that the number p is positive. Prove that either p G or the set G must contain at least two different members between p and 3p/2. Solution: We assume that p does not belong to the group G. Now since p is the greatest lower bound of G and p  3p/2 we know that 3p/2 fails to be a lower bound of G. Using this fact we choose a member x of G such that 3p x . 2 p 0 x 3p 2 Since p does not lie in G we know that p  x and so x is not a lower bound of G. Choose a member y of G such that y  x. In this way we have found two members of G lying between p and 3p/2. py 0 x 3p 2 d. Suppose that G is a subgroup of R other than 0 , that p  inf x G x  0 and that the number p is positive. Prove that p is the smallest positive member of G and that G  np n Z . Solution: Given any members x and y of the set G we know that if x  y then y x p. Therefore no two different members of G can both lie between p and 3p/2 and we conclude from part (c) that p G. We therefore know that G np n Z . Now suppose that x is any member of G. We define n to be the largest integer...
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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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