1873_solutions

Any number x 0 we know that there exists a positive

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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 so-called 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.

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