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In the regular type of math "0.999..." is the same thing as the value 1.

In some other different kind of math they say that "0.999..." is not the same thing as the value 1. Where 1 is > "0.999..."

So 1 - "0.999..." = "0.000...1"

My question is what is the benefit of having "0.000...1" Which I am calling infinitesimal. Is there some new theorems or ideas you can have by allowing that kind of number? How is a mathematics that allows such a number to exist to be different from the standard "elementary + high school" math?

Neil
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    $0.999\ldots$ is always equal to $1$. – Brian M. Scott Apr 10 '15 at 21:54
  • The numbers are just a representation of the rules you are using. It's very easy to make new rules. For example I heard there is a kind of math that deals with knots in strings where 1 +1 =3. Because in order to connect two strings you have to tie a knot. – Neil Apr 10 '15 at 22:05
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    So give us precisely the axioms for your modified real numbers where $0.999...$ is not equal to $1$. And ideally prove that it is not equal to $1$ carefully with your axioms. I share @BrianM.Scott's skepticism. A version of this question has been asked a few times this week. It mostly hinges on what $0.999...$ means. In any reasonable interpretation that set of symbols has to be equal to $1$. – Simon S Apr 10 '15 at 22:06
  • Simon If I could decipher the Wikipedia page on infinitesimals then I wouldn't be asking here. But the point I was making is that 1+1 isn't always 2. Another example is that triangles don't always have 180 degrees, like if you draw one on a sphere. They were talking about surreal numbers and hyperreal numbers so I'm sure it's allowed. http://en.wikipedia.org/wiki/Infinitesimal – Neil Apr 10 '15 at 22:11
  • Yes, there are structures where those relations are correct according to the axioms being used. The sum of angles of a triangle not being equal to $180$ degrees is famously in geometries where the fifth postulate of Euclid is removed but the first four remain. The question then is: what is the modified number system you want? – Simon S Apr 10 '15 at 22:13
  • Simon is there more than one system where 1 is > "0.999..." ? – Neil Apr 10 '15 at 22:14
  • I know of no such system. Even in (stadard) 'non-standard' analysis $0.999.... = 1$. – Simon S Apr 10 '15 at 22:14
  • If you write $0.000\ldots1$ then you imply that your number has a $1$ in some digit. The same digit and the digit to the right of that in $1 - 0.000\ldots1$ cannot both be $9$, because if they were, then you would have $(1 - 0.000\ldots1) + 0.000\ldots1 \geq 1.000\ldots09 > 1$. Therefore, at some point in your number $1 - 0.000\ldots1$ you must stop writing $9$s. So you really have something like $1 - 0.000\ldots1 = 0.999\ldots90.$ – David K Apr 10 '15 at 22:16
  • Students easily relate to the intuitive notion of an infinitesimal difference 1-"0.999...", where "0.999..." differs from its standard meaning as the real number 1, and is reinterpreted as an infinite terminating extended decimal that is strictly less than 1.[14][15] from the Wikipedia article – Neil Apr 10 '15 at 22:17
  • Ah, so you really do mean $0.999\ldots9000\ldots$ after all. (The final $000\ldots$ represents the digits after your decimal terminates.) – David K Apr 10 '15 at 22:18
  • After about nine or ten pages of http://www.math.umt.edu/tmme/vol7no1/TMME_vol7no1_2010_article1_pp.3_30.pdf, it seems they do not really support the argument for which Wikipedia cites them. What they do say is that pre-calculus students do not easily relate to the notion that $1=0.999...$, because the concept of a non-terminating decimal has not properly been defined for them. – David K Apr 10 '15 at 22:33

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