1873_solutions

1873_solutions

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: e. Otherwise we can use the method of proof by contradiction as follows: To obtain a contradiction, suppose that their are positive integers n for which the equation f n  n fails to hold and define k to be the least of these positive integers. Since we are given that f 1  1, we know that k  1. Therefore k 1 is a positive integer that is less than k and we conclude that f k 1  k 1. Therefore f k  f k 1 1  f k 1 f 1  k 11  k which contradicts the way in which the integer k was chosen. R, that f 1  1 and that the equation f xt  f x f t holds for all integers x and t, prove that f x  x for every integer x. First we observe that, since f 0  f 00  f 0 f 0 we have f 0  0. Now we know from part b that f x  x for every positive integer x. If x is a negative integer then, since f x  x we have 0  f 0  f x  x  f x  f x  f x x, and so f x  x. c. Given that f : Z d. Given that f : Q R, that f 1  1 and that the equation 224 f xt  f x f t holds for all rational numbers x and t, prove that f x  x for every rational number x. Solution: By the preceding exercise we know that the equation f x  x holds whenever x is an integer. We shall now show that if x is any real number and n is a positive integer then f nx  nf x . Once again, you can use the method of proof by mathematical induction if you are familiar with it but we shall use the method of proof by contradiction. Suppose that x is any real number and, to obtain a contradiction, suppose that there are positive integers n for which the equation f nx  nf x fails to hold. We define k to the least of these integers. Since we know that the equation f nx  nf x holds when n  1, we know that k  1. Therefore k 1 is a positive integer less than k and so f kx  f k 1 x  x  f k 1 x  f x  k 1 x  x  kx which contradicts the way in which k was chosen. To complete the exercise, suppose that x is any rational number and choose integers m and n such that n  0 and x  m/n. We see that f x  1 nf x  1 f nx  1 f m  m . n n n n e. Given that f is a continuous function from R to R, that f 1  1 and that the equation f xt  f x f t holds for all rational numbers x and t, prove that f x  x for every real number x. In view of the fact that f x  x for every rational number x (by part d), the desired result follows at once from Exercise 17b. f. Given that f is an increasing function from R to R, that f 1  1 and that the equation f xt  f x f t holds for all rational numbers x and t, prove that f x  x for every real number x. Solution: Suppose that x is any real number and choose two sequences a n and b n of rational numbers such that an x bn for each n and lim a nÝ n  n Ý b n  x. lim Since the function f is increasing we know that for each n we have fx f bn bn an  f an and so it follows from the sandwich theorem for limits that f x  x. 19. Given that f : R R and that for all numbers x and t we have |f x f t | |x t | 2 , prove that the function f must be constant. Note that althoug...
View Full Document

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.

Ask a homework question - tutors are online