1873_solutions

Exercise into one that is a shade more interesting by

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Unformatted text preview: mathematical induction which is available in this book in an optional section then you should use mathematical induction to do this exercise. 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 11  k which contradicts the way in which the integer k was chosen. R, that f 1  1 and that the equation f xt  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 00  f 0 f 0 we have f 0  0. Now we know from Exercise 15 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. 16. Given that f : Z R, that f 1  1 and that the equation f xt  f x f t holds for all rational numbers x and t, prove that f x  x for every rational number x. 17. Given that f : Q 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 195 18. Given that f is a continuous function from R to R, that f 1  1 and that the equation f xt  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 Exercise 17), the desired result follows at once from Exercise 13. 19. Given that f is an increasing function from R to R, that f 1  1 and that the equation f xt  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...
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