The annoying old 0.999 vs 1, hopefully with a twist

  • Thread starter miscellanea
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In summary, the question of whether 0.999... is equal to 1 has been a source of debate and confusion. While most standard proofs show that they are indeed equal, the introduction of non-standard analysis and hyperreal numbers presents a different perspective. In non-standard analysis, 0.999... is not equal to 1 by definition, as it can be expressed as the sum of a geometric series with a different limit. This raises questions about the validity of non-standard analysis and the acceptance of infinitesimals and infinitely large numbers in mathematics. Ultimately, the answer to this question may depend on whether one allows for non-standard analysis or not.
  • #1
miscellanea
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This question is not new by any definition of the word. However, it's been bugging me for quite a while now and I've done some reading up on the issue.

The question: Is 0.999... (infinite number of 9s) equal to 1?

The answer is, of course, yes. There are numerous proofs showing this. Here are a couple of simpler proofs I remember seeing:

1/3 = 0.333... and 2/3 = 0.666... and 0.333... + 0.666... = 0.999... and 1/3 + 2/3 = 3/3 = 1

Then there's:

10x = 9.999...
x = 0.999...

10x - x = 9x
9.999... - 0.999... = 9

9x = 9, so 0.999... is the same as 1

And ultimately there's the proof which says that 0.999... can be expressed as the sum of a geometric series and the sum of a geometric series is, by definition, equal to its limit.

Which is where I have a bit of a problem. All other proofs seem to be referring to this one bit: by definition 0.999... as the sum of a geometric series is equal to its limit, 1.

I did some reading up and, remembering some advanced maths courses I took at the uni, looked to hyperreal numbers and non-standard analysis.

The difference seems to be that in non-standard analysis, the sum of a geometric series that would give us 0.999... is not equal to its limit. The sum would actually be different from the limit by an infinitesimal number.

In non-standard analysis, by definition, 0.999... is not equal to 1. Which gives us funky things like infinitesimals and their reciprocals, infinitely large numbers.

Some mathematicians seem to not like non-standard analysis all that much. The main criticism seems to be that it allows for things that are not Lebesgue measurable. Then again, Zermelo-Fraenkel set theory with the Axiom of Choice added allows for non-measurable sets, giving rise to things like the Banach-Tarski paradox, yet these results seem to be accepted for, well, practical purposes.

Basically, my theory is that whether 0.999... is equal to 1 or not depends on if you allow for non-standard analysis or not. If you allow for infinitesimals, then 0.999... is not equal to 1, by definition. If you use real analysis, then 0.999... is equal to 1.

And, basically, both answers seem to be equally valid.

What do you think? Discuss!
 
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  • #2
But 0.999... is not a hyperreal number, it is a real number.
 
  • #3
Hyperreal numbers are a proper extensions of the reals. You get reals for free and if you add non-standard analysis, you also get infinitesimals and infinitely large numbers.
 
  • #4
Let me rephrase: 0.999... is a real number (and yes, technically also a hyperreal number - I was thinking of "hyperreal" like "irrational", ie, those that are hyperreal but not real, but you're correct), so it cannot differ from 1 by an infinitessimal, since the difference of any two real numbers is another real number, and there are no real infinitessimals besides 0.
 
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  • #5
Hyperreals have been mentioned before on these forums in relation to this matter of often heated contention.

Though I don't remember it's name, there is a theorem which states any result true in normal analysis is true for non-standard analysis. If 0.9r = 1 in standard analysis then 0.9r* = 1* in non-standard analysis where 0.9r* and 1* are the elements in hyper reals. While 1 = 1*, 0.9r* is different in construction to 0.9r. Do a search to get a better explanation from someone who actually knows this stuff.
 
  • #6
AlphaNumeric said:
Though I don't remember it's name, there is a theorem which states any result true in normal analysis is true for non-standard analysis. If 0.9r = 1 in standard analysis then 0.9r* = 1* in non-standard analysis where 0.9r* and 1* are the elements in hyper reals. While 1 = 1*, 0.9r* is different in construction to 0.9r. Do a search to get a better explanation from someone who actually knows this stuff.

Yes, I've heard of this theorem as well, though I don't know its name. That theorem is one of the reasons I posted here in the first place -- this difference shouldn't actually exist, but everything I've seen suggests that it does.

Which is really strange.
 
  • #7
No, there is no difference between 0.9999... and 1. 0.9999... is a regular real number, 1. It has nothing to do with "hyper-reals". The "definition" y0u are talking about is simply the definition of the base 10 number system.
 
  • #8
In the reals, we have:

[tex]\sum_{i = 1}^{+\infty} 9 \cdot 10^{-i} = 1[/tex]

and in the hyperreals, we also have

[tex]
{}^{\star}\sum_{i = {}^{\star}1}^{+\infty} {}^\star 9 \cdot (^\star 10)^{-i} = {}^\star 1
[/tex]

(Bleh, I'm going to stop putting stars on things)

One fact that may be being overlooked as that these two sums are summing over a different set of terms. The first one is the (real) sum of [itex]9 \cdot 10^{-i}[/itex] for every positive natural number i. The latter is the (hyperreal) sum of [itex]9 \cdot 10^{-i}[/itex] for every hypernatural number i.


The interesting fact that demonstrates a reason why might want to study nonstandard analysis is that we can consider the hyperfinite hyperreal sum:

[tex]
\sum_{i = 1}^H 9 \cdot 10^{-i}
[/tex]

where H is some transfinite hypernatural number. (But still, [itex]H < +\infty[/itex], by definition of [itex]+\infty[/itex]). This sum is the one that differs from 1 by a positive infinitessimal.


Another angle is to look at the hyperdecimal numbers. If we have a transfinite, but hyperfinite number of 9's (that is, the number of 9's is a transfinite hypernatural number) in 0.999...9, then this does, in fact, denote a number infinitessimally close, but unequal, to 1. However, this is a terminating hyperdecimal. The nonterminating hyperdecimal 0.999... is, in fact, equal to 1.
 
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  • #9
I've moved the posts not dealing with nonstandard analysis into a new thread.
 
  • #10
Why? I would have thought it should be the responses that DO deal with non-standard analysis that should be moved. The original post had nothing to do with non-standard analysis!
 

What is the difference between 0.999 and 1?

The difference between 0.999 and 1 is very small, but it is a difference nonetheless. 0.999 is a repeating decimal, meaning it goes on forever, while 1 is a whole number.

Why is the debate between 0.999 and 1 important?

The debate between 0.999 and 1 is important because it challenges our understanding of mathematics and the concept of infinity. It also has practical applications in fields such as calculus and number theory.

Can 0.999 ever equal 1?

Yes, mathematically, 0.999 can equal 1. This is because in the decimal system, there is no number between 0.999 and 1, so they are considered to be equal.

How do you prove that 0.999 is equal to 1?

There are several ways to prove that 0.999 is equal to 1. One way is to use algebra and show that 0.999 is equal to the fraction 9/10, which is then equal to 1. Another way is to use the concept of limits in calculus to show that as the number of 9s in 0.999 increases, it approaches 1.

Can we ever truly resolve the debate between 0.999 and 1?

This debate can never truly be resolved because it ultimately depends on how we define and understand numbers. Some people may argue that 0.999 and 1 are different because of their different representations, while others may argue that they are equal because they have the same value. It is ultimately a philosophical question rather than a purely mathematical one.

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