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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.
 Fall '08
 STAFF
 Math, Calculus

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