_ 0.9 must equal 1, however, at least in the real numbers. It may not in a system with infinitesimal/differential numbers but in _ the reals 0.9 = 1, OK?
I'd bet $1,000,000 that your proof contains some sort of logical fallacy. Based on what bit of reading of it I did it seems like you are assuming a system where infinitesimals exist, but such a system is NOT the real numbers no matter how much you'd like it to be.
"Again, the problem is introduced because decimal representation is not able to express or perceive a distance between a number that is infinite in position beyond the decimal point relative to other non-infinite decimals."
Such a number is an infinitesimal, and thuse do not exist in the _ reals. So 0.9, as a representation of a _real number_, *IS* equal to one.
The other problem is that you seem to thin the definition of decimal system must be different. The sum
sum_{n=-infinity...infinity} d_n B^n, 0 <= d_n < B
is the _definition_ of the base-B repesentation of a real number! _ For 0.9, d_n = 0 for n >= 0 and 9 for n < 0. The DEFINITION of infinite summation is
This LIMIT equals ONE when a_n = 9 * 10^n when n < 0 and 0 when n >= 0. Since this Limit ***DEFINES*** the infinite sum and the infinite sum ***DEFINES*** the decimal then _ 0.9 = 1, by DEFINITION.
