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If: $0.\overline{9999999} \equiv 1$

Then how would you represent a value that is infinitesimally close to one, but not quite one?

i would have thought: $1-\frac 1 \infty $

But i would take that to be: $0.\overline{9999999} = 1$

Or do i have to subtract an infinitesimal amount from one?

$ 1 - 0.\overline{000000}1$

$ 1 - 1 \times 10 ^{-\infty}$

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    @IanBoyd: $1-x = 0$ if & only if $x = 1$. If you meant to ask whether $1 - (1-x) = 0$ for "small enough" values of $x$, note that $1 - (1-x) = x$, and $x$ equals zero (oddly enough) only when it *actually equals zero*.2012-09-22

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The real numbers do not have any infinitesimals, so there is no need to represent such a number. There are other fields that do allow them, such as the hyperreal numbers and the surreal numbers and they have ways of representing them. They give up some of the properties of the reals that many find convenient, such as completeness.

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    @Gerry, Well, they have Gonshor's sign expansions, so that a good portion of them have something very much like binary. All dots are binary points, not decimal points: "+"=1, "+-"=.1, "+-+"=.11, "+-++"=.111, "+-+++"=.1111, ... *However* "+-++++++..." refers to a well defined Surreal less than 1 but greater than all of 1/2, 3/4, 7/8,...2013-02-07