Since x<x', we have E(10^n*x)<E(10^n*x'), divide by 10^n and we are done. Or have I misunderstood?Now suppose x < x' and show that u(n) < u'(n).
Consider the serie:
E(x) means the integer part of x (E(5.2) = 5 for example)
The limit of the serie is x.
Now suppose x < x' and show that u(n) < u'(n).
u(n) is the serie of x and u'(n) the serie of x'.
I actually found a way, but it's not really good, because it contains a small contradiction. I wondered whether somebody knows who to get around that contradiction.
This is what I got when I started with the definitions of a serie:
x-x' < u(n) - u'(n) < x-x'
If I ignore the left part it works, but the way it stands there now, it's impossible since a number cannot be strictly smaller than itself.