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

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As we all know, there exist many arguments that 0.999...=1.

All of the proofs I have seen are based upon arguments drawn from mid-level mathematics.  As a philosopher, this bothers me because, when proofs of a basic-level proposition depend upon mid-level arguments, there is the obvious danger that these arguments, if analysed and deconstructed in sufficient detail, will be found to be question-begging;  that's to say, at some point, we may find that they have assumed the conclusion as a premise.

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?

The elephant in the sitting room, it appears to me, is the axiomatic principle that two numbers cannot have the same successor;  the numbers in question being, of course, 9 and 10.

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3 hours ago, amplitude said:

All of the proofs I have seen are based upon arguments drawn from mid-level mathematics.

Multiply 0.9999... by 10

Subtract 0.999999... to get 9

but 10X -X =9X

so 9 times 0.999999.... is 9

so

0.999........ =1

I hope you are now happy that you can prove it with elementary school maths.

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Nice one John, I was going to say something complicated but yours is better.  +1

Edited by studiot
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What I don't understand is why everyone thinks it's somehow complicated, or controversial.

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The fallacy in 10x-x = 9x is that truly infinite numbers do not behave properly in mathematics.  In order to use mathematics for this you have to truncate the number someplace.  When you do that, and multiply a finite version of 0.999... by 10, then try to subtract the original 0.999..., you have a result a tiny bit smaller than 9 because the first digit is 8 and the last is 1.

for example:    lets use 0.999999999999999.  10 x = 9.99999999999999  (14 digits to the right of the decimal).  The original number, 0.999999999999999, has 15 digits to the right of the decimal.  When you do the subtraction, the result is 8.999999999999991, which is not 9.

In order to make the result be 9, you have to suppose that the "infinite" number you multiply by 10 has one more digit than the "infinite" number that you subtract from it.  Two infinite numbers with a different number of digits is, of course, ridiculous, but mathematics will not work without that condition.

You have to be very careful drawing conclusions in math using numbers that allegedly have an infinite number of digits.  This is related to the prohibition against dividing by zero-- it's just another property of infinite numbers that math can't handle properly.

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19 minutes ago, OldChemE said:

The fallacy in 10x-x = 9x is that truly infinite numbers do not behave properly in mathematics.  In order to use mathematics for this you have to truncate the number someplace.

Huh?

There are no infinite numbers here.

And arithmetic works perfectly well with an numbers with an infinite number of digits in the decimal expansion. Otherwise it would not be possible to divide something by 3.

21 minutes ago, OldChemE said:

This is related to the prohibition against dividing by zero-- it's just another property of infinite numbers that math can't handle properly.

Citation needed.

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1 hour ago, OldChemE said:

The fallacy in 10x-x = 9x is that truly infinite numbers do not behave properly in mathematics.  In order to use mathematics for this you have to truncate the number someplace.  When you do that, and multiply a finite version of 0.999... by 10, then try to subtract the original 0.999..., you have a result a tiny bit smaller than 9 because the first digit is 8 and the last is 1.

for example:    lets use 0.999999999999999.  10 x = 9.99999999999999  (14 digits to the right of the decimal).  The original number, 0.999999999999999, has 15 digits to the right of the decimal.  When you do the subtraction, the result is 8.999999999999991, which is not 9.

In order to make the result be 9, you have to suppose that the "infinite" number you multiply by 10 has one more digit than the "infinite" number that you subtract from it.  Two infinite numbers with a different number of digits is, of course, ridiculous, but mathematics will not work without that condition.

You have to be very careful drawing conclusions in math using numbers that allegedly have an infinite number of digits.  This is related to the prohibition against dividing by zero-- it's just another property of infinite numbers that math can't handle properly.

But the point is that the string of nines is definitively infinite- so one digit fewer is still infinite and the subtraction works just fine.

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The point is that when you talk about an infinite string of numbers you get into the topic of limits.  The Limit of 0.999... as the number of digits goes to infinity is indeed 1, but 0.999... itself is not 1.  Limits and integers are not the same thing.  Try subtracting 0.99999..... from 1.0000.....  There will be an infinitely small non-zero result.  That result approaches zero as the number of digits increases, but is never actually zero.

Edited by OldChemE
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I would assume we're talking real numbers here, in which case infinitesimals don't exist. There's no "infinitely small non-zero result".

Or in other words, do you think these are not actually equal? How would you write the difference?

1 - 0.999... = 10 - 9.999...

Relatedly, between any two real numbers, are an infinite number of reals, e.g. between 0.9 and 1.0 is 0.95; what's between 0.999... and 1.0?

Don't confuse the process with the number. You'd never write down all the digits of 0.999... in long form, but they are all in there, infinite never ending 9's in the "...". You showed that earlier when you had to artificially limit 0.999... to 15 decimal places to try to make a point. As soon as you truncate at some number of decimal places, you've lost sight of the infinite decimals therein.

Edited by pzkpfw
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Let us assume the hypothesis that 0.99.... = 1

Then, by the rules of mathematics, 1-0.99.... = 0, which is a definable number in mathematics.  Or, to say it differently, a defined result proper to mathematics.

if we subtract 0.9 from 1.0, the result is 0.1, which is 1/10, which is 1/(10)^1

if we subtract 0.99 from 1.00 the result is 0.01, which is 1/(10)^2

Generalizing, 1 - 0.99.....  = 1/(10)^Infinity

But the result of division by an infinite number is undefinable in mathematics.

Therefore, 1 - 0.9999......  produces an undefined result

This falsifies the assumption that there is a definable result of zero

This falsifies the  original hypothesis.

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No, it shows your generalisation is inapplicable.

You're trying to make an infinitesimal.

Like the people who think 0.999... is less than 1 by 0.000...1

Which is a contradiction, by trying to put something at the end of an infinite sequence.

(It occurs you think you have a special case where { some real number } minus { some real number } = { undefined }. Do you really think that's valid?)

Edited by pzkpfw
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3 hours ago, OldChemE said:

Let us assume the hypothesis that 0.99.... = 1

That is not a hypothesis, it is a fact.

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3 hours ago, Strange said:

That is not a hypothesis, it is a fact.

Still treating people with passive aggressive superiority I see.  I suppose you think "facts" are unchangeable?  Case in point....Greek ellipses...case in point flat earth.....facts can be challenged....OldchemE has the right to do so.  Sure if he hadn't have argued with you after your first reply things would have been great huh Strange?

That said .999...is = 1.....in my opinion.  But I appreciate and will continue to listen to the op and his/her ideas with out being a jerk.

Edited by conway
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3 minutes ago, conway said:

I suppose you think "facts" are unchangeable?

They are in mathematics. This isn't a matter of opinion.

Quote

facts can be challenged....OldchemE has the right to do so.  Sure if he hadn't have argued with you after your first reply things would have been great huh Strange?

I don't think he did argue with me, did he? But I don't really care either way. I will leave others explain why he is wrong, as it bothers you that I have commented.

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45 minutes ago, Strange said:

They are in mathematics. This isn't a matter of opinion.

I don't think he did argue with me, did he? But I don't really care either way. I will leave others explain why he is wrong, as it bothers you that I have commented.

What do you think Greek ellipses are?  They ARE mathematics.  It WAS considered FACT.   Yet it was still wrong wasn't it.

Edited by conway
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45 minutes ago, conway said:

What do you think Greek ellipses are?

I have no idea. What are they?

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1 hour ago, conway said:

I suppose you think "facts" are unchangeable?  Case in point....Greek ellipses...case in point flat earth.....facts can be challenged...

Those are not ''facts''. Those are just hypotheses or wild guesses. The majority consensus is not a fact. You need to understand what a fact is. Nothing (afaik) in science outside of math is considered factual. The highest degree is a theory, like relativity or evolution, hence the remodelling and re-thinking.

Also, I don't know what a greek elipse is. Google shows zero results.

11 hours ago, pzkpfw said:

Relatedly, between any two real numbers, are an infinite number of reals, e.g. between 0.9 and 1.0 is 0.95; what's between 0.999... and 1.0?

I think this is the best example for what the OP is looking for, described with words and not symbols. There must be an infinite amount of real numbers between any two numbers. Seeing how no defined number is in between 0.999... and 1, one must conclude that 0.999 = 1.

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20 hours ago, OldChemE said:

The fallacy in 10x-x = 9x is that truly infinite numbers do not behave properly in mathematics.  In order to use mathematics for this you have to truncate the number someplace.

Truncation is not necessary, you just have to correctly apply the rules taught in high school for sequences and series.

13 hours ago, OldChemE said:

The point is that when you talk about an infinite string of numbers you get into the topic of limits.  The Limit of 0.999... as the number of digits goes to infinity is indeed 1, but 0.999... itself is not 1.  Limits and integers are not the same thing.  Try subtracting 0.99999..... from 1.0000.....  There will be an infinitely small non-zero result.  That result approaches zero as the number of digits increases, but is never actually zero.

Again it is a question of following the rules, limits are for sequences, when we talk about series we use the rules for convergence.

So the limit of the following sequence is zero

$\mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{9}{{{{10}^n}}}} \right\} = 0$

That is

$\frac{9}{{10}},\frac{9}{{100}},\frac{9}{{1000}},\frac{9}{{10000}} \to 0$

However the  infinite series is convergent so has a sum to infinity

$\sum\limits_{n = 1}^\infty {\frac{9}{{{{10}^n}}}} = \frac{9}{{10}} + \frac{9}{{100}} + \frac{9}{{1000}} + \frac{9}{{10000}}............. = 1$

As John showed.

Edited by studiot
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On 17-10-2017 at 3:02 PM, amplitude said:

All of the proofs I have seen are based upon arguments drawn from mid-level mathematics.

1 = a number

2 = a number

1 = 2

Now where's my nobel-prize ??

In highschool i seemd to be have  mind for maths and not languages,

since internet i 've learned more different languages then i can count xDD.

If a language (of logic)| does not yield logical results, for example by seeing two different meanings covered by the same term,

you should assume that this particular language does not use the terms in a way for you to get meaningfull results ##### Share on other sites

7 hours ago, pzkpfw said:

No, it shows your generalisation is inapplicable.

You're trying to make an infinitesimal.

Like the people who think 0.999... is less than 1 by 0.000...1

Which is a contradiction, by trying to put something at the end of an infinite sequence.

(It occurs you think you have a special case where { some real number } minus { some real number } = { undefined }. Do you really think that's valid?)

Wow.

Thank you, everybody, for the courtesy and thoughtfulness of your replies, first of all.  I had no idea that my post would provoke this kind of response.  This is clearly an issue upon which there are strongly-held opinions!

I don't know how this will go down, but, in terms of the axioms and definitions of arithmetic, what number is the successor of 0.999...?  This seems to me to be the crux of the issue.  Because, of course, the axioms dictate that two numbers cannot have the same successor.

So do 0.999... and 1 have the same successor?  Nobody has denied that they are different numbers;  but arguments purport that they are of equal absolute value;  does this mean that they have the same successor?

By the way, I believe that 0.000...1 is a number.  It's just an infinite set ordered by diminishing magnitude, with the "1" defined as occupying a decimal place which is smaller than any other.

Argument:

(1) Since the irrational numbers imply an infinite set of decimal places, all numbers imply such a set;  including the whole numbers: in which case, every decimal place is presumed to be occupied by 0.

(1) So, for example, although 1 is a whole number, after the number 1, there is implied an infinity of decimal places, each of which is presumed to be occupied by a 0. The fact that every decimal place is occupied by a 0 may be considered as a proof that the number we are thinking about is really "1".

(2) In set theory, identical members of the set would normally collapse, but, in this case, the 0's do not collapse because they are not the defining property of the set;  the defining property is the relative value of each decimal place;  the fact that every decimal place is occupied by a 0 is incidental.

(3) conclusion:  an infinite set of 0's, following a decimal point, with each 0 denoting a decimal place which is 1/10 of its predecessor, is a mathematically respectable example of an infinite set.

But for any infinity, infinity+1=infinity. (sorry I don't know how to represent the proper infinity symbol).

Given that this particular infinity is ordered by decimal-place magnitude,  to an infinite collection of 0's, we can add an extra member  There's no problem about that, in principle.  And we stipulate that the extra member will be a 1.

The only remaining problem is,  to ensure that the 1 will always appear to be at the far boundary of the infinity;  ie, the number we are talking about will always appear as 0.000...1.

We can achieve this by means of a definition.  Definitions are perfectly respectable entities in mathematics and, in this case, we will define the 1 as occupying a decimal position which is smaller than any other decimal position.  (there will always be a decimal position which is smaller than any other position;  if we didn't specify that it must be occupied by a 1, it would be occupied by a 0 anyway).

This is a great forum.  All of the replies I deeply appreciate;  all are thoughtful and intelligent.    So different to my experiences elsewhere.  Thank you.

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5 minutes ago, amplitude said:

I don't know how this will go down, but, in terms of the axioms and definitions of arithmetic, what number is the successor of 0.999...?  This seems to me to be the crux of the issue.  Because, of course, the axioms dictate that two numbers cannot have the same successor.

This is a logical red herring.

0.9 recurring is an unending decimal number with no 'successor'.

Successors refer to the construction of the positive integers, by one particular axiom scheme.

Note that axiomatic schemas for arithmetic are not taught at the elvel you requested a reply.

A good (university) book which uses the successor scheme and comapres it with others is

The Number System by Professor H. A. Thurston.

Edited by studiot
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16 minutes ago, studiot said:

However the  infinite series is convergent so has a sum to infinity

1

While it is a theoretical nit (perhaps), it should be noted that the infinite sum of 9/10 + 9/100 .... does not actually reach 1, it converges toward 1.

9/10 is 9 tenths of the distance on the number line from 0 to 1

9/100 is nine tents of the distance on the number line from 9/10 to 1

Every term in the infinite sum adds to the sum 9/10 of the remaining distance on the number line between its previous term and the total of 1.  Because bo term ever adds more than 9/10 of the remaining distance on the number line, we never actually reach 1.

This is the situation of the old puzzle about a person who in each unit of time walks exactly half the remaining distance to his/her destination.  With each succeeding term we add to the infinite sum we travel 9 tenths of the remaining distance to 1.  Again-- it is a Limit, not an equality.

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36 minutes ago, Roamer said:

Now where's my nobel-prize ??

Sadly, there is no Nobel Prize for mathematics. 34 minutes ago, amplitude said:

Nobody has denied that they are different numbers

Then let me fix that: they are the same number, just written in different ways.

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27 minutes ago, studiot said:

This is a logical red herring.

0.9 recurring is an unending decimal number with no 'successor'.

Successors refer to the construction of the positive integers, by one particular axiom scheme.

Note that axiomatic schemas for arithmetic are not taught at the elvel you requested a reply.

A good (university) book which uses the successor scheme and comapres it with others is

The Number System by Professor H. A. Thurston.

Thanks Studiot, I will try to source that reference.

With regard to the assertion that 0.999... has no successor, can you provide further arguments?  Cantor's Diagonal Argument, for instance, implies that the successor of 0.999... would be 1.

The Diagonal Argument asks us (as a thought-experiment) to imagine a table which lists every fraction between 0 and 1.  The clear implication, though as far as I know Cantor doesn't appeal to it directly, is that the first number would be 0.000...1 and the last number would be 0.999... (0.000...1 being what you are left with, if you subtract 0.999... from 1, according to the axioms of arithmetic).

it's worth pointing out, perhaps, that the decimal point is itself a red herring... it could occur at any position within the number, or there could be no decimal point at all;  the essential problem in mathematical philosophy remains unaltered.  So it doesn't matter if the equation is 0.999...=1, or 999...=1{000}... (pls allow for my probably-erroneous symbolism here).

In relation to someone else's earlier post - I assume that there can be NO number between 0.999... and 1 because, in the natural number line, there is no number between 9 and 10.  10 is defined as the successor of 9, which rules out the possibility of any intermediate number.   It might be possible to stipulate an imaginary number, but I have no idea how such an argument might be set out.

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59 minutes ago, amplitude said:

So do 0.999... and 1 have the same successor?  Nobody has denied that they are different numbers;  but arguments purport that they are of equal absolute value;  does this mean that they have the same successor?

There are no successors to any real numbers. 0.99999... is no different than, say, 7 in that regard. Which real number comes after 7?

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