The Real And Complex Number Systems

In addition, since f1 f1 by f0 0, we have
f1 f 1 m .. m
1
mf
1 m
f 1 m
mtimes
f1
m *’
Thus we have
f m n f1/m ...1/m ntimes nf1/m m n f1 by (*) and (*’) **
So, given any x R, and thus choose a sequence x n Q with x n x. Then
fx f lim n
x n
lim n
fx n by continuity of f on R
lim n
x n f1 by (**)
xf1
ax.
Remark: There is a similar statement. Suppose that fx y fxfy for all real x
and y.
(1) If f is differentiable and non-zero, prove that fx e cx ,wherec is a constant.
Proof: Note that f0 1sincefx y fxfy and f is non-zero. Since f is
differentiable, we define f 0 c. Consider
fx h fx fh f0
fx fxf
h
h
0 cfx as h 0,
we have for every x R, f x cfx. Hence,
fx Ae cx .
Since f0 1, we have A 1. Hence, fx e cx ,wherec is a constant.
Note: (i) If for every x R, f x cfx, then fx Ae cx .
Proof: Since f x cfx for every x, we have for every x,
f x cfxe cx 0 e cx fx 0.
We note that by Elementary Calculus, e cx fx is a constant function A. So,fx Ae cx
for all real x.
(ii) Suppose that fx y fxfy for all real x and y. Iffx 0 0forsomex 0 , then
fx 0 for all x.
Proof: Suppose NOT, then fa 0forsomea. However,
0 fx 0 fx 0 a a fx 0 afa 0.
Hence, fx 0 for all x.
(iii) Suppose that fx y fxfy for all real x and y. Iff is differentiable at x 0 for
some x 0 , then f is differentiable for all x. And thus, fx C R.
Proof: Since