A strict non-standard inequality .999 < 1

[Count Iblis]

None of the discussions on PF on this topic have discussed this result:

http://arxiv.org/abs/0811.0164

A strict non-standard inequality .999... < 1


Authors: Karin Usadi Katz, Mikhail G. Katz

(Submitted on 2 Nov 2008 (v1), last revised 24 Feb 2009 (this version, v8))


Abstract: Is .999... equal to 1? Lightstone's decimal expansions yield an infinity of numbers in [0,1] whose expansion starts with an unbounded number of digits "9". We present some non-standard thoughts on the ambiguity of an ellipsis, modeling the cognitive concept of generic limit of B. Cornu and D. Tall. A choice of a non-standard hyperinteger H specifies an H-infinite extended decimal string of 9s, corresponding to an infinitesimally diminished hyperreal value. In our model, the student resistance to the unital evaluation of .999... is directed against an unspoken and unacknowledged application of the standard part function, namely the stripping away of a ghost of an infinitesimal, to echo George Berkeley. So long as the number system has not been specified, the students' hunch that .999... can fall infinitesimally short of 1, can be justified in a mathematically rigorous fashion.

 

[CompuChip]

Yes, that's a good idea.
Instead of trying to justify that 0.999... = 1 by applying a rigorous limit argument which nicely shows how a mathematical proof works, let's talk to students about non-standard analysis!

If I'm not mistaken, the argument is:
We can define a quantity $\epsilon$ which is infinitesimally small but non-zero, and then $1 - \epsilon < 1$.
Right?

 

[Count Iblis]

Yes, but the calculus of infinitesimals has been rigorously developed only in the 20th century. So, I don't think the statement is trivial (I don't know much about the rigorous formulation of nonstandard analysis).

 

[matt grime]

None of the discussions has touched on that because none needs to. The discussions here are (almost) always by people refusing to accept that in the standard decimal representation of real numbers 0.999.. and 1 are the same (equivalence class).

No one would argue that there are situations in which 0.99.. and 1 represent different objects (not least if it were base b for b>10 for example).

Introducing non-standard analysis (or something like it) is immaterial to the typical discussion.

 

[maze]

I think the main problem is that the result is presented to students before they have sufficient background to understand what it means (ie: basics of real analysis).

 

[Hurkyl]

Yes, but the calculus of infinitesimals has been rigorously developed only in the 20th century.

The most important property of non-standard analysis is the transfer principle -- every theorem of real analysis is also a theorem of non-standard analysis.

In particular, if we use 0.999... to denote the repeating decimal that has a 9 in every place, then it's a theorem of nonstandard analysis that 0.999... = 1.



Although not in these words, the author identifies that some students fail to understand "a decimal number with a 9 in every place to the right of the decimal point" who are instead conceptualizing "a decimal number with a large¹, fixed, but unspecified amount of 9's to the right of the decimal point".

His response? Tell them they're right!


1: Of course, I mean this number to remain finite (or hyperfinite, as appropriate)

 

[Yoo]

Is 0.999... even anywhere close to what is supposed to be meant by 1 minus an infinitesimal in non-standard analysis?

 

[Hurkyl]

Is 0.999... even anywhere close to what is supposed to be meant by 1 minus an infinitesimal in non-standard analysis?

0.999... is infinitessimally greater than 1 minus an infinitessimal.

 

[CompuChip]

His response? Tell them they're right!

Precisely what I thought when I read it, although you worded it so nicely

That aside, he is not providing a new viewpoint in the discussion, he is avoiding the discussion.
Suppose a student tells you that he is convinced that 1 = 0. Instead of demonstrating how that must be false and lead to contradictions, you say: "I'm glad you ask, because people always say that that's false, but in modulo 1 calculations it is actually true."

 

[Count Iblis]

Well, a lot of mathematics has its origins in vague intuitions that clashes with an existing formalism. In this case the idea that 0.9999 is somehow less than 1, even though in the standard formalism they are the same, can be used to motivate infinitessimals.

Similarly, if you read Dirac's book "The Principles of Quantum Mechanics", he argues that the standard result that the derivative of Log|x| is 1/x should be modified by adding a term proportional to the delta function (which he just invented a few pages back).

The theory of distributions was developed later, motivated by the need in physics.

 

[Tac-Tics]

Well, a lot of mathematics has its origins in vague intuitions that clashes with an existing formalism. In this case the idea that 0.9999 is somehow less than 1, even though in the standard formalism they are the same, can be used to motivate the of infinitessimals.

0.9999 is less than 1 ;-)

The problem comes with students being taught to think like engineers and not mathematicians. To most people, pi is something like 3.14159265 and nothing more. It's just understood that a number is just a list of digits that fit on a calculator. And since on a calculator, 0.9999999, the closest number you can get on an 8-digit display, is less than 1, then 0.9999... is also less than 1.

But it's not. And you don't even need nonstandard analysis to prove it. And if you were going to use an advanced technique anyway, nonstandard analysis would be a lame way to prove it, because nonstandard analysis doesn't deal as well as regular calculus with infinite sums.

 

[HallsofIvy]

Tac-tics, you may have missed the point of this thread. The assertion is, I think, that in "non-standard analysis", 0.999... is less than 1.