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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 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 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 xt 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 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 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 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 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...
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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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