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## #901 2007-12-27 16:16:02

John E. Franklin
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### Re: 0.9999....(recurring) = 1?

I wonder if there are more numbers from 0 to 1, or from 1 to 2, or from 523 to 524?
Maybe it depends what type of density mode you are in.
Thanks for the compliment, George.
Happy Holidays to both of you!!

igloo myrtilles fourmis

## #902 2007-12-28 05:06:25

George,Y
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### Re: 0.9999....(recurring) = 1?

#### Ricky wrote:

Neither of those posts made any sense, in both the English and mathematical sense.  What is your objection to my relabeling method?

You can relabel, OK? But you cannot simply state that your relabeled set is the same as a part of the original set. And you cannot then move on to use the same and new concept of large to disprove this concept because you haven't even reached the same condition yet.

Why have you made such a mistake in thinking them the same? Because you just see 2 in both, 3 in both, 4 in both, and elipse in both, and then you think that they are the same? I have pointed out this fallacy in Post 900.

And I have given the proof why they are not the same in Post 899. The relabeled
{2, 3, 4, ...}#
has one more member than the part
{2, 3, 4, ...}
in the original set
{1, 2, 3, 4, ...}
And this proof is based on your assumption that you have relabeled all the memebers of the original to the new {2, 3, 4, ...}#.
Since they have the same amount of member, a part of either one cannot have the same amount because the larger defination. (I will explain this in BTW)
And thus you cannot equate the relabeled to the part of the orginal because even they start at the same begining, they go on in seemingly the same pattern, they have different amount of elements after all!

Do you call
{2,3,4,5,...1001}
and
{2,3,4,5,...1000}
the same?

The same concept applies to this situation also. The only difference is that this is an infinite case. And in order to compare, I applied the larger concept.

BTW
The larger concept is quite equivalent to
the concept of
Part is smaller than the whole.
in philosophy.

(
The almost equivalence is as follows:
The whole has what all the part has, but also has something that it doesn't has, hence the whole is larger than the part. (or the part is smaller than the whole)
The back way:
A has all B has and has further something it doesn't. Select just the "all B has" in A and it is clearly a part in A because A has not only it but also something else. And this part equals B and hence is smaller than A. (Note this time the proof applies the equal concept, and that's why I said almost equivalence)
)

This concept is so far the most conservative defination in all the concepts of larger. I cannot find a more fundamental defination than this. If any other defination contradicts this one in some sense, we may examine that one is true or not.

X'(y-Xβ)=0

## #903 2007-12-28 05:28:55

luca-deltodesco
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### Re: 0.9999....(recurring) = 1?

you assume that B is not whole. why does it have to be a part?

its like saying.

im going to prove this apple is not a fruit by assuming it isnt a fruit, and showing that if it isnt a fruit, it is therefore not a fruit.

The Beginning Of All Things To End.
The End Of All Things To Come.

## #904 2007-12-28 05:45:11

TheDude
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### Re: 0.9999....(recurring) = 1?

George, you don't understand what infinity is.  Your last post proves this:

Do you call
{2,3,4,5,...1001}
and
{2,3,4,5,...1000}
the same?

The same concept applies to this situation also. The only difference is that this is an infinite case. And in order to compare, I applied the larger concept.

You're right in that the only difference between this example and
{1, 2, 3, 4, 5, ...} vs. {2, 3, 4, 5, 6, ...} is infinity.  You're wrong, however, in that the same concept does not apply.  Infinity is not a normal number.  Indeed, it is not a number at all.  You can't treat it like a number.  Infinity + 1 = Infinity doesn't make sense, but neither does Infinity + 1 != Infinity, because you're treating infinity like a number when it isn't one.  Until you accept the nature of infinity, discussion of this topic is pointless.

Wrap it in bacon

## #905 2007-12-28 06:14:45

George,Y
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### Re: 0.9999....(recurring) = 1?

Well I made a mistake in the back way, sorry.

It should be:

From A is larger than B
A has all the things B has, as well as something that B doesn't have.
Just label "all the things" in A as C. Clearly C is in A but A contains not only C. This means C is a part of A. (So far my part defination is the whole contains part, but not only part)

And from part<whole
C<A
since
C=B
then
B<A

The back way is from Part<Whole to the larger concept. The smaller label is passed on from the equivalent part to B. Thus in this way any A who contains not only all B has is proven as larger than B

through the concept that the A's part the same as B is smaller than A the whole.

The first proof is A>B => Whole>Part.
since a part of A is an eligible candidate for B, it is a B. Hence part(B)<whole(A)
Why is it qualified as B?
Since the whole has all the things its part has, and something else.
You see, you can just switch whole here to A, and the part here to B, and A and B are in the defination of A>B.

X'(y-Xβ)=0

## #906 2007-12-28 06:30:39

George,Y
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### Re: 0.9999....(recurring) = 1?

"Infinity + 1 = Infinity doesn't make sense, but neither does Infinity + 1 != Infinity, because you're treating infinity like a number when it isn't one.  Until you accept the nature of infinity, discussion of this topic is pointless."

Yes, you are switching infinity concept to a docile and flexible one. Well initially I assume you perceive it as stable and cannot change.

Well it is just simpler to disprove infinity in this sense.
Suppose you think the amount of elements in this
{2,3,4,5,...}
can vary.
(Infinity+k=Infinity)
And I wonder which element(s) can vary so that the whole amount of all the elements can vary?

Since if it has got infinity elements or infinity+1 elements doesn't matter at all. I wonder what is the +1 element that is totally disposable yet addible and doesn't make a difference at all?

BTW
When you have infinity+1=infinity
you assume
infinity1+1=something
And this something qualifies as infinity too.
However, is it just the same as infinity1?
No, it can be strictly qualified as non-finite and infinite,
But this doesn't guarantee you can treat them as equal.
Because infinity is not necessarily single and unique.
And we have just got two infinities here
One infinity1 and the other infinity1+1
The latter strictly literally contains not only the former but also 1 more and is not equal to the former (in fact larger than).

And the fact that it is also non-finite doesn't lead to the conclusion that they are all the same.

It's just the same argument that we are both human, which doesn't prove we are all the same one.

X'(y-Xβ)=0

## #907 2007-12-28 10:16:53

Ricky
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### Re: 0.9999....(recurring) = 1?

Thanks Dude.  George seems to have read some very old math books (pre 1700's) or perhaps philosophy when infinity was not as rigorous as it is today.  Nowhere other than his posts have I seen "stable" infinity or "growing" infinity.  Perhaps they aren't from old books, but just his interpretation and in the end, creation of what he thinks when others say infinity.

This was in response to TheDude:

Yes, you are switching infinity concept to a docile and flexible one. Well initially I assume you perceive it as stable and cannot change.

I realize now that while you are reading our posts, you are taking what we say, flipping it around inside your head to what you want us to say.  TheDude only said that you can't treat infinity as a number, and nothing more.  How you cooked this up, I'm not certain, other that the fact that it wasn't in his post.

It's this kind of thing as well as other posts here which make me believe you, George, come here to confirm your preconceived notions rather than to discuss.  Till you actually wish to understand what we are saying, this conversation is the equivalent of talking to a brick wall.  And I'm not going to take part in it anymore.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #908 2007-12-28 13:54:15

John E. Franklin
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### Re: 0.9999....(recurring) = 1?

George, I really enjoy your thinking and I have been thinking along these lines to some extent.  For example, I have a hunch there might be three worlds of numbers:
The zero numbers, the infinite numbers, and the numbers in between.
Take all the non-zero-non-infinite numbers and multiply them by zero, and
you get the zero numbers.
Take all the non-zero-non-infinite numbers and multiply them by infinity, and you get the infinite numbers.
This is just a simplified version of something more beautiful that I don't fully understand, but I can see that the three ranges of numbers makes a beautiful arrangement!!
The multiplying by zero or infinity in the above statements is just
a suggestion more than an actual equation, as the details are not
well understood.
Neat idea, huh?

igloo myrtilles fourmis

## #909 2008-01-02 18:52:07

Kargoneth
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### Re: 0.9999....(recurring) = 1?

#### George,Y wrote:

{1, 2 , 3, 4, 5, ...}

You have a set of all natural numbers...

#### George,Y wrote:

matching by +1=
{2, 3,  4, 5, 6, ...}

You have a set of all natural numbers excluding 1...

#### George,Y wrote:

above plus the set {1}=
{1, 2, 3, 4, 5, 6,...}#

You have a set of all natural numbers...

I realize that although both sets are infinite, technically, {1, 2, 3, ...} has one more element than {2, 3, 4, ...}. This is a most confusing thing...

I don't understand what you are trying to prove.

I would think that 0.09999... + 0.9 has the same number of nines as 0.9...

I like my rationalization of 0.9... lol

Last edited by Kargoneth (2008-01-02 18:59:14)

## #910 2008-02-17 03:26:58

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### Re: 0.9999....(recurring) = 1?

0.9999..... can never be equal to 1 unless you convrert the values to some decimal places

## #911 2008-02-17 10:40:56

MathsIsFun

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### Re: 0.9999....(recurring) = 1?

Sure?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #912 2008-04-15 15:02:51

Stumpe
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### Re: 0.9999....(recurring) = 1?

As long as we're working with non-existent numbers here:

1
-
0.999...
_______
0.000...1

Ooooo, sorry. I went ahead and posted a topic about this and didn't realize it was already here (although, why wouldn't it be...). I'll just paste my squabble in here and you can all ignore my separately posted topic.

Ctrl+V!!
____________________________________________________________

So my roommate and I were talking the other night and ended up arguing for 3 hours about why or why not "0.999..." = 1. We came up with some interesting points. I'd like to point out why I believe it doesn't.

1:

1 X 2 = 2
0.9 X 2 = 1.8
0.99 X 2 = 1.88
0.999 X 2 = 1.888 and so on and so forth, so
0.999... X 2 = 1.888...
because no matter how many 9's you add to it, you will always end up adding that many 8's .
Okay, I was tired... and didn't think about what I was typing just there...

2:

The structure of numbers.
For any number, let's say "5", you can add anything to the places after it and no matter how many of anything you add you will never end up changing the numbers place previous to whatever you are adding at that time.
5
53
534
5341
53417
No matter how far I go, nothing I do can change the initial 5. Ever.
Saying that you're adding an infinite number of 9's to the end of "0.9" can never change the initial 9 laid down there. You will always end up with a "0.9" with a long string of numbers after it. Even infinity can't change that.

What I've heard is that there is no definable difference between "0.999..." and 1, therefore they are equal. I must agree that there can be no difference because no matter how many times you subtract "0.999..." from 1 you will always get "undefined". This does not mean their difference is "0", it only means we cannot calculate with "0.999..." because it is not a number and can never be a number. Infinity can't exist no matter how many ways we want to try and comprehend it. So introducing it into a number is about as useful as saying, "Chair = 8".
However, if you'd like to think of it as a real number, the arguments above apply and therefore there is a difference!

In conclusion, we cannot make assumptions about a number just because we cannot comprehend it *yet... The world is still not flat.

Last edited by Stumpe (2008-04-16 03:12:54)

## #913 2008-04-15 20:42:14

JohnnyReinB
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### Re: 0.9999....(recurring) = 1?

#### Stumpe wrote:

As long as we're working with non-existent numbers here:

1
-
0.999...
_______
0.000...1

Ooooo, sorry. I went ahead and posted a topic about this and didn't realize it was already here (although, why wouldn't it be...). I'll just paste my squabble in here and you can all ignore my separately posted topic.

I would like to comment on this one. I read in a similar topic before that you can't have a 1 in the end of the difference, because there is no end.

#### Stumpe wrote:

No matter how far I go, nothing I do can change the initial 5. Ever.
Saying that you're adding an infinite number of 9's to the end of "0.9" can never change the initial 9 laid down there. You will always end up with a "0.9" with a long string of numbers after it. Even infinity can't change that.

What I've heard is that there is no definable difference between "0.999..." and 1, therefore they are equal. I must agree that there can be no difference because no matter how many times you subtract "0.999..." from 1 you will always get "undefined". This does not mean their difference is "0", it only means we cannot calculate with "0.999..." because it is not a number and can never be a number. Infinity can't exist no matter how many ways we want to try and comprehend it. So introducing it into a number is about as useful as saying, "Chair = 8".
However, if you'd like to think of it as a real number, the arguments above apply and therefore there is a difference!

In conclusion, we cannot make assumptions about a number just because we cannot comprehend it *yet... The world is still not flat.

Like you said, we cannot make assumptions. There are a lot of undefined in Mathematics, and we just have to accept that:)

"There is not a difference between an in-law and an outlaw, except maybe that an outlaw is wanted"

Nisi Quam Primum, Nequequam

## #914 2008-04-15 20:55:59

Maelwys
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### Re: 0.9999....(recurring) = 1?

#### Stumpe wrote:

1:

1 X 2 = 2
0.9 X 2 = 1.8
0.99 X 2 = 1.88
0.999 X 2 = 1.888 and so on and so forth, so
0.999... X 2 = 1.888...
because no matter how many 9's you add to it, you will always end up adding that many 8's .

I'd recommend checking your basic math again, before trying to attack a complicated concept like 0.999...

0.99 x 2 = 1.98, not 1.88.
0.999 x 2 = 1.998, not 1.888.
No matter how many 9's you add to it, you're not adding that many 8's, you're adding that many 9's (minus 1). So all it works out to is 1.999....8, and a difference of 0.000...2, which is the same argument as 0.999...9 and a difference of 0.000...1. Which still faces the same problem of that "infinitieth" or "last" digit of an infinitely long string of digits. And because of the definition of infinity, there is NO last digit, so the 'number' 0.000...2 = 0, because you can't have a 2 in the 'last spot', since there's always more 0s still.

## #915 2008-04-16 03:10:06

Stumpe
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### Re: 0.9999....(recurring) = 1?

Wow, my bad. I can't believe I actually wrote that down... It was late, and I was tired.

But as for the difference being "0.000...1", I was merely saying that you can get any absurd answer you want when you're working with numbers that can't exist in the first place.
However, it is possible to 'reach' infinity if you do something for an infinite amount of time... like writing 9's on an infinitely long sheet of paper. No matter how many days, years, millenia, infinities you do that for, it will never ever change to one.
Never > Infinity.

Last edited by Stumpe (2008-04-16 03:13:33)

## #916 2008-04-16 03:22:55

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### Re: 0.9999....(recurring) = 1?

I believe they are equal, cuz you cant find any number  larger than 0.99999999.... and less than 1

## #917 2008-04-16 04:39:55

TheDude
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### Re: 0.9999....(recurring) = 1?

I have a question for you Stumpe.  What do you believe 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... is equal to?

Wrap it in bacon

## #918 2008-04-16 06:53:41

Ricky
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### Re: 0.9999....(recurring) = 1?

I think TheDude hit the nail on the head.  Analysts consider that when two things are arbitrarily close, that they are equal.  In fact, this is how we rigorously define the concept of a real number.  People don't seem to like that, and it isn't particularly intuitive.

The other thing is that when you look at a series of terms, you think of that series growing, which is a rather common thing to do.  But it isn't.  It's just a series of terms, and all terms are always there.  The partial sums is growing as you go up in terms, but they are called partial sums for a reason.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #919 2008-04-24 10:53:03

George,Y
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### Re: 0.9999....(recurring) = 1?

" But it isn't.  It's just a series of terms, and all terms are always there."

Hey Ricky, I have already reached the static infinity concept and showed its internal contraversity, have you forgotten that?

As simple as that
0.9+A/10 (How many of 9's are here? Also infinite? But this infinite amount has 1 more than the previous infinite amount and how do you locate the 1 more? I have the whole proof to locate the one more by whole>part.)

And tell me please what do you feel about the two static infinity yet different at the same time.

X'(y-Xβ)=0

## #920 2008-04-24 11:31:24

Ricky
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### Re: 0.9999....(recurring) = 1?

George, you still don't seem to grasp the concept of infinite as is defined by set theory.  Nor the term countable, apparently.  You are trying to go by what seems intuitive: that infinity and infinity+1 (in very loose terms) are different.  But the way mathematics was constructed, was defined, they are not.  Again, study set theory, including finite/infinite mappings, cardinality, pigeon hole principle, 1-1, onto, and bijections.  You will see what I'm talking about.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #921 2008-04-24 22:12:10

TheDude
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### Re: 0.9999....(recurring) = 1?

http://en.wikipedia.org/wiki/Infinity#Cardinality_of_the_continuum

This paragraph in particular is of interest:

Cardinal arithmetic can be used to show not only that the number of points in a real number line is equal to the number of points in any segment of that line, but that this is equal to the number of points on a plane and, indeed, in any finite-dimensional space. These results are highly counterintuitive, because they imply that there exist proper subsets of an infinite set S that have the same size as S.

A proper subset is a subset that has fewer elements than it's parent.  For example, let A be a set containing {1, 2, 3, 4}, B containing {1, 2, 3}, and C containing {1, 2, 3, 4}.  B is a subset of A because all of it's elements are contained in A.  It is also a proper subset because it does not contain every element of A, namely 4.  C is a subset of A because all of its members are contained in A.  However, C is not a proper subset of A because there are no elements in A that are not in C.

Now, re-read the Wikipedia quote.  There exists proper subsets of S (S being any infinite set, countable or otherwise) that have the same size as S.  One such example would be A = {1, 2, 3, 4, 5, ...} and B = {2, 3, 4, 5, 6, ...}.  B is clearly a proper subset of A, but it can also be shown that they have the same size.  It's counterintuitive, but it's been rigorously proven.  This is another Wikipedia page that may be helpful to study:

Wrap it in bacon

## #922 2008-04-25 02:55:09

George,Y
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### Re: 0.9999....(recurring) = 1?

"But the way mathematics was constructed, was defined, they are not."

Yeah, mathematicians rather define something that has no reason to exist.
(I have a definition- I call a pair of numbers that each is larger than another and I call them George Pairs. I and my followers wrote textbooks, and don't care whatsoever that it can exist or not. How about this, Ricky?)

The set theory is just an excuse to hide a growing infinity into the definition of a number.

Mapping
On last page you have showed "mapping" from {1,2,3...} to {2,3,4,...}
But your only rational is to map 1 to 2, 2 to 3, 3 to 4,..., you are using growing mapping and cannot control the whole set. Whereas I control the whole set and showed a different result.

Ricky you see, this is your problem. You always switch between a growing infinity and a static one:

Whenever someone challenges you the precision of something backed on infinity, you turn from growing infinity to static infinity.
And whenever someone (especially me) uses static infinity to challenge the internal logic bug of the static infinity itself, you turn to a growing method to "prove" your claim.

X'(y-Xβ)=0

## #923 2008-04-25 03:05:25

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### Re: 0.9999....(recurring) = 1?

I think 0.9999999... is best represented by

I dont like to consider 0.9999... as a number , rather an expression. Since the concept of infinity is not shown by putting it 0.9999...

George , I dont think there's static infinity. Of course, if you were referring that to upper/ lower bound, then there is

Last edited by Dragonshade (2008-04-25 03:15:04)

## #924 2008-04-25 03:43:17

Ricky
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### Re: 0.9999....(recurring) = 1?

Yeah, mathematicians rather define something that has no reason to exist.
(I have a definition- I call a pair of numbers that each is larger than another and I call them George Pairs. I and my followers wrote textbooks, and don't care whatsoever that it can exist or not. How about this, Ricky?)

Mathematicans define it that way because it's the only way that provides useful results.  If you can come up with a better one, I'm all ears.

(I have a definition- I call a pair of numbers that each is larger than another and I call them George Pairs. I and my followers wrote textbooks, and don't care whatsoever that it can exist or not. How about this, Ricky?)

Certainly, sounds like an interesting version of an ordering.  Can you do anything useful with them?

But your only rational is to map 1 to 2, 2 to 3, 3 to 4,..., you are using growing mapping and cannot control the whole set.

"..." does not mean growing.  Just "this pattern continues".

On another note, in ZFC set theory (aka, math) it can be proven that a set (we call them natural numbers) exists such that an "inductive property" covers all of them.  This was Peano's 5th postulate.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #925 2008-04-25 08:31:01

George,Y
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### Re: 0.9999....(recurring) = 1?

" it's the only way that provides useful results"

Really? It seems the wrong method is the only method to describe the truth? Interesting. Still, I doubt the obsessive part to eliminate the residue error necessary or not.

"it can be proven that a set (we call them natural numbers) exists such that an "inductive property" covers all of them.  This was Peano's 5th postulate."

Excuse me for a moment, on what basis does the ZCF and Peano's 5th postulate stand? Sheer logic? Or some unintuitive assumptions plus logic? Ironically, the assumptions themselves might be illogical. Hence, the whole rational may just start on an arbitary basis.

Why it needs to go so far as to distort a number as set of numbers in ZCF to get your "only" way to gain some results? Haven't you seen the absursity in it to introduce unnecessary clumsy assumptions similar to a white elephant?

And what is so wrong to go intuitive? After all, intuition observes the truth as best as possible. While arbitary assumption and definition coincide with the truth at best.

The link in 922 by thedude also tells few physicians believe in infinity and infinitesimal idealism. And they came up with results while disbelieving your only method. I guess at least they disagree with you, Ricky.

X'(y-Xβ)=0