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Unformatted text preview: that does not exceed x/p
and we observe that, since
pn x p n 1
we have
0 x pn p.
Since the set G has no positive members that are less than p we conclude that x pn 0 and we have
shown that every member of G is an integer multiple of p. Therefore
G np n Z . e. Suppose that G is a subgroup of R other than 0 , and that
0 inf x G x 0 .
Prove that if a and b are any real numbers satisfying a b then it is possible to find a member x of the
set G such that a x b. Solution: Suppose that a and b are real numbers and that a b. Using the fact that b
83 a0 and that inf G 0, choose a member q of G such that
0 q b a.
We now define
S n Z nq a .
Since the inequality nq a is the same as saying that n a/q and since the set Z of integers is
unbounded above we know that S is nonempty. Furthermore, since the number a/q is a lower bound
of S we know that S is bounded below. We deduce from an order property of the system of integers
that the set S has a least member that we shall call m. Thus mq a but m 1 q a.
We observe that
a mq m 1 q q a b
and so the member mq of G must lie between a and b. ab 4. This exercise invites you to explore the socalled division algorithm which describes the process by which
an integer b can be divided into an integer a to yield a quotient q and a remainder r.
Suppose that a and b are positive integers and that
S n Z nb a .
a. Prove that S
and that S is bounded above.
The fact that S is bounded above is clear since the number a
b must be an upper bound of S. b. Prove that if the largest member of S is called q and we define r a qb then a qb r and 0 r b.
We need to recall that every nonempty set of integers wil,if it is bounded above, have a largest
member. If we define q to be the largest member of S then, since qb a we have a qb 0. On
the other hand, since q is the largest member of S we have
q1 b a
which gives us a qb b. Some Exercises on Suprema and Infima of Functions
1. Given that f x 1/x whenever x 0, prove that the function f is unbounded above. Prove that for every
number 0, the restriction of f to the interval , Ý is bounded.
To show that f is unbounded above, suppose that w is any positive number. We observe that
1
f
1w w
1w
and therefore w is not an upper bound of the function f.
1
Now suppose that 0. Given any number x we have f x
and so 1 is an upper bound of
the restriction of f to the interval , Ý .
2. Give an example of a bounded function on the interval 0, 1 that has a minimum value but does not have a
maximum value.
We could define
x if 0 fx x1 0 if x 1 3. A function f is said to be increasing on a set S if the inequality f t
f x holds whenever t and x belong to S
and t x. Prove that every increasing function on the interval 0, 1 must have both a maximum and a
minimum value.
The definition of an increasing function makes it clear that if f is an increasing function on
theinterval 0, 1 then f 0 and f 1 are the minimum and maxim...
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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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