# Axioms, definitions, and 0.999...=1

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28 minutes ago, Outrider said:

For the crowd that believes 0.999 = 1.

Does this mean 0.999 is useless?

And.

Does .0991 also equal 1?

Nobody thinks 0.999 = 1, that's 0.001 less than 1.

0.999... = 1. That's an infinite never ending sequence of 9 digits.

No idea why you'd think 0.999 (nor 0.999...) is "useless".

No, 0.0991 does not equal 1. It's 0.9009 less than 1.

However, note that the infinite sequence of 9 digits does have the same value in all places.

e.g. 0.24999... = 0.25

e.g. 5.123999... = 5.124

e.g. 6.999... = 7

Edit: while I don't want to invoke argument from authority, I thought I'd better reply to "For the crowd that believe". Be very clear, 0.999... = 1 is the standard understanding in math. There certainly are people who don't "agree" (discussion of this topic is common on the internet), but they are the (ahem) outriders.

Edited by pzkpfw

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20 minutes ago, Outrider said:

For the crowd that believes 0.999 = 1.

Does this mean 0.999 is useless?

No. 0.999 does not equal 1.

0.999...however, equals 1. There is an infinite difference between the two. The three dots imply that the nines continue into infinity. Since you cannot identify a real number in between 0.999... and 1, the conclusion must follow that 0.99... = 1.

20 minutes ago, Outrider said:

Does .0991 also equal 1?

In the same vain, no. Not even 0.0991... equals 1. For example, 0.991... < 0.992 < 1, therefore, it doesn't equal 1, since there is a number which is higher than it but lower than 1.

EDIT: Beaten to it by pzpkfw.

Edited by Lord Antares

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Oh dear I have neen confused about the very nature of the problem for a numberof years. Thank you both for the education.

About the useless part. If 1 and 0.999... are the same thing what would you need 0.999... for?

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7 minutes ago, Outrider said:

If 1 and 0.999... are the same thing what would you need 0.999... for?

I like to think of it simply as another way to write the same number. Like 6/3 and 2 both represent the same number, just written in different ways.

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18 minutes ago, Outrider said:

Oh dear I have neen confused about the very nature of the problem for a numberof years. Thank you both for the education.

About the useless part. If 1 and 0.999... are the same thing what would you need 0.999... for?

It's not about needing it, it's simply about the way the system of real numbers works.

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7 hours ago, pzkpfw said:

Edit: while I don't want to invoke argument from authority, I thought I'd better reply to "For the crowd that believe". Be very clear, 0.999... = 1 is the standard understanding in math. There certainly are people who don't "agree" (discussion of this topic is common on the internet), but they are the (ahem) outriders.

As the Speaker of the UK Parliament (Bercow?) once said: "They are, of course, entitled to their own opinion. However, they do suffer from the not inconsiderable disadvantage of being wrong."

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Bignose and pzkpfw, +1 for more than one good posts apiece.

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8 hours ago, Outrider said:

About the useless part. If 1 and 0.999... are the same thing what would you need 0.999... for?

It's mainly to avoid ambiguity or counting (fractional) numbers twice, since 1.000... = 0.999... etc

I suppose 0.999... is generally used since unlike 1.000... it's clearer that writing 1 instead would not be useful in some contexts.

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On 10/20/2017 at 9:23 AM, Bignose said:

If we know that some a does not equal b (let a > b), then there always exists some number c such that a > c > b.

If 0.99999... is not equal to 1, what is that number that comes between them?

I have never seen someone who 'doesn't believe' 0.99999... = 1 give any kind of meaningful answer to this.

BN, I don't think you've understood the original question and you appear to be unfamiliar with its context in the axioms and definitions of arithmetic.  In the number line of the natural integers (0, 1, 2, 3, ...n), 9 < 10, but there is no number between them.  This is a part of the paradox which provoked my original question. There can be no number between 0.999... and 1, for the same reason that there is no natural integer between 9 and 10, because 10 is the immediate successor of 9.

But nobody denies that 0.999... and 1 are different numbers (hence the requirement for proofs of their equivalence).

Middle-level or 'synthetic' mathematics - that's to say, mathematics above the level of the axioms - offers several proofs that 0.999...=1.

I have personally had no luck trying to devise an additional proof based solely upon the axioms and definitions, and I was curious to know whether anybody had any ideas as to how such a proof might be constructed.  There are philosophical reasons why a proof based upon primitive statements would be more satisfactory than proofs based upon synthetic statements.

As I suggested in another post, I believe the central difficulty is the axiomatic requirement that two numbers cannot have the same successor.  If we expand the natural number line to include the fractions, this means that the successor to 0.999... cannot be equal to 0.999...

BTW, I "believe" nothing in relation to this question.  I am trying to follow an enquiry wherever it may lead.

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45 minutes ago, amplitude said:

But nobody denies that 0.999... and 1 are different numbers (hence the requirement for proofs of their equivalence).

All mathematicians and a large number of other people (who are familiar with mathematics) do deny exactly this. They are the same number.

45 minutes ago, amplitude said:

I have personally had no luck trying to devise an additional proof based solely upon the axioms and definitions, and I was curious to know whether anybody had any ideas as to how such a proof might be constructed.  There are philosophical reasons why a proof based upon primitive statements would be more satisfactory than proofs based upon synthetic statements.

I'm sure this is possible although it may not be practical. And seems a little pointless. Mathematics works as a hierarchy of proofs and derivations. You don't need to start from number theory to prove the Taniyama-Shimura conjecture. You build on what is already known about elliptical functions, etc.

45 minutes ago, amplitude said:

As I suggested in another post, I believe the central difficulty is ...

45 minutes ago, amplitude said:

BTW, I "believe" nothing in relation to this question.

45 minutes ago, amplitude said:

I believe the central difficulty is the axiomatic requirement that two numbers cannot have the same successor.

No, because this doesn't apply to reals. Although, the fact it doesn't apply to reals is a good proof that 0.999... = 1 (as I think someone already pointed out).

Edited by Strange

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I thought I had said goodbye to this thread days ago, but mathematics and philosophy are like malaria, once you've been bitten, it's in your blood for the rest of your life.

A couple of general comments:  glancing back over some of the posts, I think some confusion has arisen because it's not always easy to  maintain the proper distinction between  the natural numbers, the real numbers, and the imaginary numbers.  Maybe the way I've flitted from one to the other hasn't always been as clear as it could have been, either.

The question I raised is principally a problem in the natural number line.  There are moments when we need to think in terms of natural integers, as when we say, there is no number (ie natural integer) between 9 and 10, and there are moments when we need to expand the scale so as to include the natural fractions (as when we return to 0.999...=1).  Perhaps confusingly, there are also moments when it is useful to think in terms of the real numbers (eg which is correct according to the axioms of arithmetic, 1 - 0.999... = 0 or 1 - 0.999... = .000...1?  And the latter answer, theoretically the smallest possible number, would be an imaginary number, more usually represented as h;  but whether we use h or 0.000...1 is mainly a matter of convenience and does not reflect any significant philosophical point, except that many mathematicians think that h is just another name for 0, which frankly doesn't help).

55 minutes ago, Strange said:

All mathematicians and a large number of other people (who are familiar with mathematics) do deny exactly this. They are the same number.

I'm sure this is possible although it may not be practical. And seems a little pointless. Mathematics works as a hierarchy of proofs and derivations. You don't need to start from number theory to prove the Taniyama-Shimura conjecture. You build on what is already known about elliptical functions, etc.

No, because this doesn't apply to reals. Although, the fact it doesn't apply to reals is a good proof that 0.999... = 1 (as I think someone already pointed out).

There is a difference between a belief as to what is one's central difficulty, and a belief (or prejudice) as to the truth of the matter.

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57 minutes ago, amplitude said:

BN, I don't think you've understood the original question and you appear to be unfamiliar with its context in the axioms and definitions of arithmetic.  In the number line of the natural integers (0, 1, 2, 3, ...n), 9 < 10, but there is no number between them.  This is a part of the paradox which provoked my original question. There can be no number between 0.999... and 1, for the same reason that there is no natural integer between 9 and 10, because 10 is the immediate successor of 9.

But nobody denies that 0.999... and 1 are different numbers (hence the requirement for proofs of their equivalence).

Middle-level or 'synthetic' mathematics - that's to say, mathematics above the level of the axioms - offers several proofs that 0.999...=1.

I have personally had no luck trying to devise an additional proof based solely upon the axioms and definitions, and I was curious to know whether anybody had any ideas as to how such a proof might be constructed.  There are philosophical reasons why a proof based upon primitive statements would be more satisfactory than proofs based upon synthetic statements.

As I suggested in another post, I believe the central difficulty is the axiomatic requirement that two numbers cannot have the same successor.  If we expand the natural number line to include the fractions, this means that the successor to 0.999... cannot be equal to 0.999...

BTW, I "believe" nothing in relation to this question.  I am trying to follow an enquiry wherever it may lead.

Amplitude, with the greates respect, I don't think you have properly looked at the replies offered here.

I agree that at school level few proofs are offered, which is why John Cuthber's idea was so neat.

But you haven't replied to it.

Yes there are many things we have to take on trust at a lower level, which then become explained as we progress in education.
I have always thought it to be a sort of 'spiral process' whereby each year we look back at what we simply accepted last year in the light of the new stuff we have now learned.

You are bringing in concepts from higher levels and frankly, you do not understand all of these yet so are making basic but fundamental errors.

The first and worst is not listening when someone points these out.

Successors, do not apply to real numbers.

Integers do not have a 'number line', they are a discontinuous or discrete set of disconnected or isolated points.
Real number are not and we can properly talk of a number line for them.

But these are all material from more advanced (university ) mathematics.

Why is it Not OK for the answer to your question to be couched in the terms of higher maths, but OK for you to introduce it in your attempts to refute a proposition?

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28 minutes ago, amplitude said:

The question I raised is principally a problem in the natural number line.  There are moments when we need to think in terms of natural integers, as when we say, there is no number (ie natural integer) between 9 and 10, and there are moments when we need to expand the scale so as to include the natural fractions (as when we return to 0.999...=1).  Perhaps confusingly, there are also moments when it is useful to think in terms of the real numbers (eg which is correct according to the axioms of arithmetic, 1 - 0.999... = 0 or 1 - 0.999... = .000...1?  And the latter answer, theoretically the smallest possible number, would be an imaginary number, more usually represented as h;  but whether we use h or 0.000...1 is mainly a matter of convenience and does not reflect any significant philosophical point, except that many mathematicians think that h is just another name for 0, which frankly doesn't help).

You are using "imaginary" in a completely different sense than it is usually used in mathematics, which really doesn't help.

23 minutes ago, studiot said:

I agree that at school level few proofs are offered, which is why John Cuthber's idea was so neat.

The OP is not looking for a "school-level" or simple proof. He is looking for a proof from the axioms of number theory.

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12 hours ago, Outrider said:

For the crowd that believes 0.999 = 1.

Does this mean 0.999 is useless?

And.

Does .0991 also equal 1?

This is not an argument I'm just curious. I have no idea who is right and I'm ok with that.

This post raises an interesting point.  Because there is an infinite number of fractions between 0 and 1.  What is the interval between each?  It is difficult to imagine any other answer than h (the smallest imaginable number).  But h = 0.  Any questions?

In one way this thread has typified an important aspect of the history of mathematics since the 19th C.  On the one hand, there are those who argue, "hey, it works, what other proof do you need?"  And on the other (historically speaking) you had Cantor, Frege, Russell et al, who said, "yes, but we need to know why it works..."

I really am saying goodnight this time, chaps.  I thank those of you who have helped me to clarify my thoughts since my initial post;  but we haven't made any real progress in answering the original question, so I think it's probably time to close this thread.

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34 minutes ago, amplitude said:

(eg which is correct according to the axioms of arithmetic, 1 - 0.999... = 0 or 1 - 0.999... = .000...1?  And the latter answer, theoretically the smallest possible number, would be an imaginary number,

How many times do you need to be told that 0.000...1 is invalid? It is illogical. You cannot define a ''theoretically smallest possible number) because it does not exist. In the same vain, you could write something like 36.232323...6 and it would make even less sense.

0.00...1 is paradoxical and nonsensical. For the last time, when you put that 1 at the end (or any number would do the job, really), you immediately make the number of zeroes before it finite.

1 minute ago, amplitude said:

I really am saying goodnight this time, chaps.  I thank those of you who have helped me to clarify my thoughts since my initial post;  but we haven't made any real progress in answering the original question, so I think it's probably time to close this thread.

You are pretentious.

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15 minutes ago, amplitude said:

but we haven't made any real progress in answering the original question,

On 10/17/2017 at 2:02 PM, amplitude said:

So my question is:  is there any way by which we can mount an argument that 0.999...=1 by arguing "upwards" from the axioms and definitions of arithmetic, rather than "downwards" from mid-level mathematics?

and as far as I can see we have several answers.

I particularly like the  one about "if 0.9999... isn't 1 then what's the number in between them?" idea. (Thanks Bignose)

It has the interesting merit of needing even less mathematical "work" than the proof  I put forward.

The "rebuttal" of that proof seemed to rest on the problems of "infinite numbers", but whether 0.999... and 1 are the same or not, they are clearly between 0 and 2 and thus obviously finite.

If you introduce the (slightly odd) concept of "the smallest number that isn't zero" and call it h, that doesn't really help.

The number half way between 0.999... and 1 is 1-(h/2).

But that's a contradiction because h/2 is smaller than "the smallest number" because dividing a number by two makes it smaller- unless the number is zero- and h isn't zero by definition.

Edited by John Cuthber
Giving credit to Bignose.

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14 minutes ago, Lord Antares said:

How many times do you need to be told that 0.000...1 is invalid? It is illogical. You cannot define a ''theoretically smallest possible number) because it does not exist. In the same vain, you could write something like 36.232323...6 and it would make even less sense.

0.00...1 is paradoxical and nonsensical. For the last time, when you put that 1 at the end (or any number would do the job, really), you immediately make the number of zeroes before it finite.

You are pretentious.

Yes, there is no 'end' to put it at.

But we can still handle the totality of the unending sequence mathematically (consistently), if we follow the rules.

+1

John, I liked your 'proof' because it follows the same path that is used in standard derivations of sum to infinity of several standard series, although doubling is more usual than multiplying by 10.

Edited by studiot

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33 minutes ago, amplitude said:

Because there is an infinite number of fractions between 0 and 1.  What is the interval between each?

There is no interval between them. That is why it called the continuum.

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2 hours ago, studiot said:

Yes, there is no 'end' to put it at.

But we can still handle the totality of the unending sequence mathematically (consistently), if we follow the rules.

+1

I think, by using his logic, I could write down the number 0.000...01 which is lesser than his 0.000...1 number! I am not breaking any rules he isn't. If he can put a 1 at the ''end'' of the infinite sequence, why can't I put 01? I know neither of these numbers are valid, but it just further demonstrates why this concept is invalid.

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5 hours ago, amplitude said:

BN, I don't think you've understood the original question and you appear to be unfamiliar with its context in the axioms and definitions of arithmetic.  In the number line of the natural integers (0, 1, 2, 3, ...n), 9 < 10, but there is no number between them.  This is a part of the paradox which provoked my original question. There can be no number between 0.999... and 1, for the same reason that there is no natural integer between 9 and 10, because 10 is the immediate successor of 9.

The first 'paradox' you have to get over then is trying to understand 0.99999... and the natural numbers. The naturals are just just the positive integers. Once you write that decimal point, you've already gone outside the bounds.

Secondly, the concept of an inequality is axiomatic. And what I wrote above comes directly as a consequence of using that axiom on the set of reals, not natural numbers. On the reals, and the limitless amount of numbers that can be found between any given two numbers that aren't equal (which comes directly from the definition of the reals) you can always find a number between them. The c in the a > c > b I wrote above.

So, if 0.9999... (just to be clear, the ... means infinite 9s) does not equal 1, what is the c that lies between them? 0.999....1 is meaningless because you've already stated INFINITE 9s. It is as meaningless of a number as 'dog leg' or 'flooblie' or 'sasquach'. Sure, you can write something down there, but it has no meaning. Or, use your own phrasing here, 0.9999....1 isn't axiomaic because none of the accepted axioms give any meaning to something written in such a way.  In much the same way that starting with the natural numbers, 2.3 has no meaning. The natural numbers do not know what a decimal point is; it is against the rules you started with. There is  no rule for the reals giving any meaning to 0.999...1

In short, you need some clarity on defining your problem and your terminology. Because we're trying to answer it as best we can, but you're not playing along with the rules you've tried to set out.

Edited by Bignose

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7 hours ago, amplitude said:

I have personally had no luck trying to devise an additional proof based solely upon the axioms and definitions, and I was curious to know whether anybody had any ideas as to how such a proof might be constructed.

Yes, here is the bottom-up proof you are looking for.

Modern math rests on first-order predicate logic. [All references below]

Above that, we assume the axioms of Zermelo-Fraenkel set theory (ZF).

In particular, we assume the Axiom of Infinity which says that there is an infinite, inductive set.

Given this foundation, we can let 0 be the empty set, and for any set X, the successor of X is defined as X unioned with {X}. The resulting collection is an inductive set as given by the Axiom of Infinity, and it's a model of the Peano axioms of the natural numbers.

Part of your confusion is that the Peano axioms are NOT the axioms for the real numbers. They don't apply to the real numbers. There are no successors in the real numbers. You can't apply the Peano axioms to the real numbers. Also, the Peano axioms are not strong enough to model the real numbers as there are no infinite sets in the Peano axioms. There are infinitely many natural numbers 0, 1, 2, 3, ... but there is no completed set of them. For that, we need the Axiom of Infinity.

In fact it's precisely the Axiom of Infinity that makes the conceptual leap from Aristotle's potential to actual infinity. The Peano axioms give us potential infinity. The sequence 0, 1, 2, 3, 4, ... is never completed, but can always be extended. It's the Axiom of Infinity that says we may collect ALL the natural numbers into a completed infinity. Thus the real numbers require more powerful axiomatic principles than do the natural numbers.

Once we have the natural numbers modeled within ZF, we can construct the integers (positive and negative naturals and 0). Then we can construct the rationals via a construction known as the fraction field of an integral domain. Then we construct the real numbers as the set of limits of sequences of rationals. This is traditionally done via a construction like Dedekind cuts or equivalence classes of Cauchy sequences. There are alternate constructions besides these.

The point of all this is that we can build up a model of the real numbers within ZF. That is, the reals are an example of a complete, ordered field. Complete in this context means that there are no "holes" in the real numbers. Every nonempty set that's bounded above has a least upper bound. It's the completeness property that makes .999... = 1 work. In fact 1 is provably the least upper bound of the set {.9, .99, .999, .9999, ...}.

Once we do that, the traditional proof via infinite series given by Uncool a few posts back is valid and is now proven from first principles. That is, if you'll grant me the empty set and the axioms of ZF, along with the inference rules of first-order predicate logic; along with the definition of the limit of a sequence, and the definition of the sum of an infinite series; I can then prove rigorously from first principles that .999... = 1

One more thing I wanted to mention. There is no smallest positive real number. Indeed if you claim that x is such, I'll just give you back x/2, which is a smaller positive real.

Likewise the expression .000...1 makes no sense. The reason is because of the definition of decimal expressions. Each decimal position is indexed by a positive integer. We have decimal places 1, 2, 3, 4, ... There is no "last" decimal place for the exact same reason that there is no last positive integer. You can always take the successor of a natural number. There's no largest natural number and there's no "last" decimal position, since the natural numbers index the decimal positions.

Edited by wtf

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13 minutes ago, wtf said:

Yes, here is the bottom-up proof you are looking for.

Excellent summary.

Although, I suspect the response will be, "yeah but ..."

The Wikipedia article describing the axiomatization of the reals, linked to Tarski's approach: https://en.wikipedia.org/wiki/Tarski's_axiomatization_of_the_reals

I think the first three axioms by themselves make it pretty clear that 0.999... = 1.

We can leave the complete proof as an exercise for the OP....