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## #101 2006-11-20 14:25:02

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

#### Ricky wrote:

Well, now what does the 9 on Digit Infinite, or the Infiniteth digit stand for?

This question has no meaning.  I might as well attempt to name the continents on the moon.

#### Ricky wrote:

What do you mean interpret?  It is a 0. and then an infinite number of 9's after it..

X'(y-Xβ)=0

## #102 2006-11-20 14:27:03

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

What do you mean interpret?  It is a 0. and then an infinite number of 9's after it..

-It is you who wrote nonsense.

X'(y-Xβ)=0

## #103 2006-11-21 01:35:45

Anthony.R.Brown
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### Re: 0.9999....(recurring) = 1?

So here is the Challenge!

Can anyone prove Infinite 0.9 = 1

The proof someone puts forward must be shown as a Mathematical Proof.
I can Prove Infinite 0.9 (Stage one) Starts with 0.9 which is less than 1 to Start with.
So for me to be proved wrong someone has to show the Mathematics on how to Bridge the Gap from 0.9 to 1.

This is all someone has to do?   Infinite 0.9 ( Bridge this Gap with Math Please! )  1

A.R.B

## #104 2006-11-21 10:02:22

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

#### Ricky wrote:

Infinity isn't part of the real numbers, yet you always try to include it as so.  Why?

Because you said infinite number of 9's, haven't you?????

Stating that there are an infinite number of 9's does not include infinity as a real number.  It just means the sequence of 9's is not finite in number.

However, trying to multiply and divide by infinity does.

"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..."

## #105 2006-11-21 10:02:46

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

#### Anthony.R.Brown wrote:

So here is the Challenge!

Can anyone prove Infinite 0.9 = 1

The proof someone puts forward must be shown as a Mathematical Proof.
I can Prove Infinite 0.9 (Stage one) Starts with 0.9 which is less than 1 to Start with.
So for me to be proved wrong someone has to show the Mathematics on how to Bridge the Gap from 0.9 to 1.

This is all someone has to do?   Infinite 0.9 ( Bridge this Gap with Math Please! )  1

A.R.B

Here you go, from post #74:

Let a = 0.999... throughout.

a is in A_0 = [0.9, 1]
a is in A_1 = [0.99, 1]
a is in A_2 = [0.999, 1]
...
a is in A_n = [0.999...n...9, 1]
...

Note that because of the infinite number of 9's, this statement will be true for any number of A_n.

But, A_0 is a subset of A_1 is a subset of A_2 is a subset of ... A_n is a subset of A_n+1 ...

Also, note that with any finite amount of 0.999..., we may get arbitrarily close to 1.  Thus, for any e (epsilon) greater than 0, we may find a finite amount of 9's such that 1 - e < 0.999...

This means that any real number less than 1 is not in one of the A_n sets listed above.  Also, any real number greater than 1 is obviously not.

However, by the nested interval property, and infinite series of intersections of closed intervals is non-empty.  That is:

A_0 intersected with A_1 intersected with A_2 ....

Is not empty.  But it contains no numbers less than 1 or greater than 1.  Thus, it contains 1 and only 1.  But remember that a is in every set A_0, ..., A_n, ... and thus, a = 1.

"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..."

## #106 2006-11-22 17:22:06

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

#### Ricky wrote:

Infinity isn't part of the real numbers, yet you always try to include it as so.  Why?

Because you said infinite number of 9's, haven't you?????

Stating that there are an infinite number of 9's does not include infinity as a real number.  It just means the sequence of 9's is not finite in number.

However, trying to multiply and divide by infinity does.

Great! This is what I have been waiting for, Ricky. And without explicitly using the concept of infinity I still can figure out the dilema.

Not finite in number? This stating is a step from perfect defination- Let us define it as not only nonfinite in number but also LARGER than any number, and it has to be so otherwise 9's  less than any finite 9's obviously cannot defeat sufficient 9's after the digit point   begining the interval [x,1]. And it is nothing wrong to name the quantity of 9's as a superquantity q that has the property q>N, in which N is any integer, is it?

However, what does the 9 after q-1 9's represent? You won't say this is invalid, will you? For if it is, not finite 9's is even not possible or feasible. Back to the topic, this 9 represents a quantity 1/10q because it is situated as the qth digit from the digit point. You wouldn't say it does not represent so, would you? (If you would, please tell me what does it represent- a 0 is only fine.)

And this 1/10q is conceivably less than any positive rationals(Why I use ratioanals instead of quantities will be illustrated later. However quantities will  equally function here). For  any rational α, there will be at least one N to insure that 1/10N<α. Hence 1/10q would be less than α too. What? You want to say that it isn't- do you want to state that this 9 represents sth larger than some 9 before it? Thereby it still represents a quantitiless. (It is only fine if you name quantitiless, instead, 0)

The same dilema again, you can forbid the expression of q/1,000,000,000, but you will still be unable to figure out from where 9's start to represent quantitilesses.

By the way, just by stating infinity is not finity doesn't solve any problem. Logically infinity and finity are completementory and together exhaustive. Stating "not finite" is the same as stating some "quantity" "infinite", as I abbreviated as q(you can use any other symbol but that would change nothing essential).

Further, if you deny this, you have an attitude problem I am afraid to say. Not to me, though, but to the concept of "infinite 9's" you've employed. In the "proof" on post 76, you employed this guy to outweigh any finite 9's after the digit point that starts the intervals like  [0.9,1], [0.99,1], [0.999,1] etc. But now when you face the dilema created by it, you fire it, and depose of it . If you haven't got it, reread from Paragraph 2.

Last edited by George,Y (2006-11-22 17:34:36)

X'(y-Xβ)=0

## #107 2006-11-22 17:30:20

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

Here are two links, in which you may find a way "out" by stating you are using potentially infinite 9's, though by doing so you admit recurring 9's is a Growing variable.

X'(y-Xβ)=0

## #108 2006-11-22 17:39:24

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

BTW, would you stop saying judgement words like "absurd", "nonsense"?That  should be considered both impolite and out of rational reasoning.

X'(y-Xβ)=0

## #109 2006-11-23 02:22:45

Anthony.R.Brown
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### Re: 0.9999....(recurring) = 1?

A.R.B
SO LETS TRY AGAIN!!.........................................

Here is the Challenge!

Can anyone prove Infinite 0.9 = 1

The proof someone puts forward must be shown as a Mathematical Proof.
I can Prove Infinite 0.9 (Stage one) Starts with 0.9 which is less than 1 to Start with.
So for me to be proved wrong someone has to show the Mathematics on how to Bridge the Gap from 0.9 to 1.
This is all someone has to do?

Infinite 0.9 ( Bridge this Gap with Math Please! )1

-------------------------------------------------------------

Ricky (the Moderator) has put forward

Let a = 0.999... throughout.
a is in A_0 = [0.9, 1]
a is in A_1 = [0.99, 1]
a is in A_2 = [0.999, 1]
a is in A_n = [0.999...n...9, 1]
Note that because of the infinite number of 9's, this statement will be true for any number of A_n.
But, A_0 is a subset of A_1 is a subset of A_2 is a subset of ... A_n is a subset of A_n+1 ...
Also, note that with any finite amount of 0.999..., we may get arbitrarily close to 1. Thus, for any e (epsilon) greater than 0, we may find a finite amount of 9's such that 1 - e < 0.999...
This means that any real number less than 1 is not in one of the A_n sets listed above. Also, any real number greater than 1 is obviously not.
However, by the nested interval property, and infinite series of intersections of closed intervals is non-empty. That is:
A_0 intersected with A_1 intersected with A_2 ....
Is not empty. But it contains no numbers less than 1 or greater than 1. Thus, it contains 1 and only 1. But remember that a is in every set A_0, ..., A_n, ... and thus, a = 1.
A.R.B
The problem with the above Proof?
Is that it does not show clearly! How to show...............

Infinite 0.9 ( Bridge this Gap with Math Please! )1

using the numbers above as a Math Equation etc.............
The only answer would be! What I have put forward below.

Infinite 0.9 ( Bridge this Gap with Math Please! )1
Infinite 0.9 (            + 0.1                  )= 1

The only problem with the answer above is that,it changes the Definition of an Infinite Number! ie having no end!
But its the only way to reach the Conclusion that Infinite 0.9 = 1 ?

So without using the above Answer?,can anyone show in a simple clear way!

Infinite 0.9 ( Bridge this Gap with Math Please! )1

## #110 2006-11-23 06:16:27

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

The only problem with the answer above is that,it changes the Definition of an Infinite Number! ie having no end!

This is the only criticism of my proof that I could find in your post.  And an infinite number has no end.  What is your problem with that?

"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..."

## #111 2006-11-24 00:22:45

Anthony.R.Brown
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### Re: 0.9999....(recurring) = 1?

Ricky!

In the real world, this is a problem.  But in mathematics, we can not define your PROOF? where this problem does exist.  So we'll go ahead and do that now..." and try to understand what you are saying!!

So Ricky do you agree that an Infinite Number has No End!

## #112 2006-11-24 01:28:15

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

Anthony.R.Brown, if you didn't already know, that is Ricky's signature, which appears at the bottom of all his posts.

Just so you know.

## #113 2006-11-24 05:27:59

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

Then Anthony, what is your problem with my proof?

George, I have a fairly long post to write responding to what you have recently posted and the problem in general.  Hopefully I'll have it done by tomorrow or Saturday.

"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..."

## #114 2006-11-24 17:12:13

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

George, as promised, I will be answering your posts either tomorrow or Saturday, so don't worry, they won't go ignored.  I ask you (as a friend in a debate, not as a moderator) not to post until I have, otherwise I fear there will just be utter confusion.

Till now, you have disagreed on decimal expansion on the real numbers, and I always figured the problem was higher up, for example definition of the convergence of an infinite series, which would intuitively come after a consensus of real numbers and decimal expansions have been reached.  Let me address this now.

I believe we all agree on the following things:

1. A decimal expansion of any real number r takes the form of:

2. For any real number r, given an epsilon (e), we can find a finite number of decimal digits d such that:

But we have problems with this.  Specifically we can't write all real numbers or even all rational numbers with a finite number of digits.  So what do we do about this?

One thing is to do nothing.  Accept this failure of decimals.  Personally, I feel unsatisfied with this.  Is there no way we can save decimal expansion of infinite decimals?  Certainly, we can't "see" infinitely many decimals.

But wait!  We have this really cool method of dealing with infinite things invented/discovered by Cauchy et. al.  Limits.  They allow us to handle infinite sequences of numbers.  And in the end, isn't this all we're dealing with?

So let's just explore this possibility for the time being.  How can we use limits?  Certainly it may not be possible, but lets just give it a try.

For any real number r, let us define a sequence of real numbers (using the same notation as before):

It should be clear that:

For any integer n.  So this sequence is monotonely increasing, as well as bounded.  By the monotone-bounded convergence theorem, this sequence must converge.  As we (hopefully) agreed in #2 (see above), it does in fact converge to r.

So where are we at?  Given any real number, we can use the decimal expansion to approach it, and get arbitrarily close, a limit per say.

It is with this in mind that we make the following definition:

Note that the notation above simply means an infinite amount of decimal points.

The results of this definition are the following:

1. No contradiction with any math that I'm aware of
2. Every real number has at least one decimal expansion equal to it.
3. Some decimal expansions are not unique.

3 can be considered a problem.  I certainly do.  But I argue (with only opinion, not pure logic) that having multiple decimal expansions are a lesser problem than not being able to represent some real numbers with a decimal expansion.

An immediate consequence of this definition is that 0.999... = 1.

Most people accept this definition by intuition alone.  You're the first person I've ever seen which did not.

Edit: I don't think my signature has every applied to any one of my posts greater than this one.

"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..."

## #115 2006-11-24 23:46:47

Anthony.R.Brown
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### Re: 0.9999....(recurring) = 1?

Ricky!

The problem is! you have not answered the question below,using only the numbers I have put forward! its that simple!

Infinite 0.9 ( Bridge this Gap with Math Please! )1

lets see your Math to show how 0.9 (Jumps) to become 1 ? and please dont say it already is! because as we all know Infinite 0.9 (stage one) starts its life smaller than 1 why should! or how is it possible to change its value,when it's an Infinite Number!

## #116 2006-11-25 01:03:03

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

The problem is! you have not answered the question below,using only the numbers I have put forward! its that simple!

Anthony, my post #74 starts with a premise of 0.999..., and ends with the conclusion that 0.999... = 1.  It has logic in between with each step.  Please explain which step of my logic is invalid.

If you can not level a criticism of my proof, then how can you complain it's invalid?  Also, you keep saying "using on the the numbers I have put forward".  Why must I stick to numbers?  Real analysis is all about the properties of numbers, not the numbers themselves.

lets see your Math to show how 0.9 (Jumps) to become 1 ? and please dont say it already is! because as we all know Infinite 0.9 (stage one) starts its life smaller than 1 why should! or how is it possible to change its value,when it's an Infinite Number!

Numbers don't "start life" nor do they change value.  I believe you aren't understanding the concept behind a decimal expansion, but I have no clue where to begin.  How much math have you had?  I'm assuming you've never taken a course in real analysis, is that correct?

"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..."

## #117 2006-11-26 02:38:59

Anthony.R.Brown
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### Re: 0.9999....(recurring) = 1?

In the real world, this is a problem.  But in mathematics, only you can just define a place where this problem doesn't exist.  So why dont you answer MY QUESTION! in a way that I have asked you to ANSWER IT! and dont just say things like "Why must I stick to numbers?" Do your best! im sure if you know what you are talking about you can do it!!

A.R.B

## #118 2006-11-26 04:48:25

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

Why can't you all just call it undetermined and leave it be like how they dealt with

?

## #119 2006-11-26 06:23:01

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

#### Anthony.R.Brown wrote:

In the real world, this is a problem.  But in mathematics, only you can just define a place where this problem doesn't exist.  So why dont you answer MY QUESTION! in a way that I have asked you to ANSWER IT! and dont just say things like "Why must I stick to numbers?" Do your best! im sure if you know what you are talking about you can do it!!

A.R.B

Now you just aren't making sense.  Again, what is wrong with my proof in post #74?  You have still not answered this question.

"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..."

## #120 2006-11-26 08:02:28

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

#### Toast wrote:

Why can't you all just call it undetermined and leave it be like how they dealt with

?

Surely the zero thing depends which axioms you decide to apply to it

## #121 2006-11-26 14:00:30

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

#### Ricky wrote:

1. A decimal expansion of any real number r takes the form of:

2. For any real number r, given an epsilon (e), we can find a finite number of decimal digits d such that:

But we have problems with this.  Specifically we can't write all real numbers or even all rational numbers with a finite number of digits.  So what do we do about this?

One thing is to do nothing.  Accept this failure of decimals.  Personally, I feel unsatisfied with this.  Is there no way we can save decimal expansion of infinite decimals?  Certainly, we can't "see" infinitely many decimals.

But wait!  We have this really cool method of dealing with infinite things invented/discovered by Cauchy et. al.  Limits.  They allow us to handle infinite sequences of numbers.  And in the end, isn't this all we're dealing with?

So let's just explore this possibility for the time being.  How can we use limits?  Certainly it may not be possible, but lets just give it a try.

For any real number r, let us define a sequence of real numbers (using the same notation as before):

It should be clear that:

For any integer n.  So this sequence is monotonely increasing, as well as bounded.  By the monotone-bounded convergence theorem, this sequence must converge.  As we (hopefully) agreed in #2 (see above), it does in fact converge to r.

So where are we at?  Given any real number, we can use the decimal expansion to approach it, and get arbitrarily close, a limit per say.

It is with this in mind that we make the following definition:

Note that the notation above simply means an infinite amount of decimal points.

The results of this definition are the following:

1. No contradiction with any math that I'm aware of
2. Every real number has at least one decimal expansion equal to it.
3. Some decimal expansions are not unique.

3 can be considered a problem.  I certainly do.  But I argue (with only opinion, not pure logic) that having multiple decimal expansions are a lesser problem than not being able to represent some real numbers with a decimal expansion.

An immediate consequence of this definition is that 0.999... = 1.

Most people accept this definition by intuition alone.  You're the first person I've ever seen which did not.

Edit: I don't think my signature has every applied to any one of my posts greater than this one.

Again, you haven't got my idea, Ricky, which is a big waste both for your posts and mine. First, you didn't know what you were talking about an infinite number of 9's. And now you don't understand the meaning of my objection as a whole, and even what infinite digits stands for.

In several posts, my objection is to reject the validity of an expression with infinite digits. So changing  the application of it won't change the wrong starting. Again you use abitary defination again, and you state that "we" define it as something. This is not a proof. Further you proposed " most people accept it" as a "proof". I have to say that you just make a common logic mistake " Appeal to the people". You haven't proven anything but have tried to suppress the minority's opinion by the majority's. If anything can be defined as true or false, I think there is no need for us to receive education- we can define anything we want. Or math isn't as correct as we thought.

Moreover, you express decimal expansion simply as a number form. Frankly I would like to note you that each decimal represents a rational number with some exponent of 10 as the denominator, and all of them together make a sum. My disproof is to explore the inconsistency of what each decimal represents,and it is useful in defying any infinite digits.

The most tricky part, however, is that you even don't figure out what "infinite" means. Some times you say " not finite", and the other times you say " no end". Apparently these mean two kinds- one constant while the other growing; one "real infinity" while the other "potential infinity". If you haven't got it, please google " Aristotle potential infinity" or check the two links I offered.

X'(y-Xβ)=0

## #122 2006-11-26 14:08:30

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

#### Toast wrote:

Why can't you all just call it undetermined and leave it be like how they dealt with

?

Perhaps because Ricky believes what he has leant from books or what he has been taught are correct.

X'(y-Xβ)=0

## #123 2006-11-26 14:46:14

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

I am sorry to say that I may break my promise to disprove Reals because the defination of Reals developed by Georg Cantor and the properties of it developed by Dedkind, Hilbert and other mathematicians require hard work and a detailed survey to disprove or to reexamine. However I list so far I can proceed as follows:

Any quantity such as 1, 2, 2/3 cannot reduce to quantitiless(an amount smaller than any quantity but at the same time not 0) smoothly, nor can the latter grow to the former. Because such changes require a change of category or a change of nature.

The best example for the above would be my several posts against infinite 9's after digit point.

Further, it is ok to say a quantitiless cannot reproduce or add itself to some quatity.

Consider this delima, an old German one:
One wheat is weighless. Two wheats are weighless. But when do some wheats start to bend a strong cow?  The premise is around quantitiless and the whole story is to question how quantity is built upon collective quantitilesses.

So the same question applies to the axiom " A line is made up of infinite points together".  Consider building a line segment by multiplying points: One points, two points, four points... Or trying to break a line segment down: infinite points, one half of infinite points, one quater of infinite points...
This premise itself is inconsistent and misleading.

Therefore there is no need to consider the square root of 2 because it does not exist in practice. And there is no need to define Reals.

Further, I happened to hear that each point on an axis is one-to-one to a real number. And this is a good reason to say that the Real system is equavantly wrong.

Anyway, I haven't post a solid disproof for Reals yet. But I don't like to remain in this swamp now, so I won't post for a long time...Sorry

X'(y-Xβ)=0

## #124 2006-11-26 16:46:42

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

George, we need to define what we mean by "decimal expansion" for the question, "Does 0.999... = 1?" to ever make sense.  You find my definition arbitrary, so can you please provide one that isn't?  In fact, all definitions are arbitrary.  That's why they are definitions and not theorems.

My definition is for defining the question, not the solution.  By intuition alone, I believe that:

1/3 = 0.333...
1/9 = 0.111...
square root of 2 = 1.41421...

I made a definition supporting this view.  My definition treats the digits of a decimal expansion as a sequence of numbers, and it is the limit of this sequence which we treat as the number itself.  With this definition, I claim that every real number has at least one decimal expansion.

You do not believe that such infinite decimal expansions exist.  As a result, not all real numbers, or even all rational numbers, have a decimal expansion.  You may not see this as a problem.  I do.

Now let me state the following very carefully because you have not been understanding my words.

Mathematical spaces are formed by definitions.  Every set of axioms can form a mathematical space, and each are just as valid as others.  Your rejection of my definition is just as valid of my acceptance of it.  Neither of us can say that one is right and one is wrong.  However, there is a trend in mathematics for definitions that get us somewhere.  When we find a set of definitions which can solve interesting and useful problems, people tend to flock to those definitions.  It does not mean they are any more valid, only popular.

The definition I use are the popular ones.  But I have also provided reasons for my definition.  I want a decimal expansion for every real number.  Frankly, I can't understand why you don't.  But that doesn't really matter in the end.

Now, as to your answer to the question, "Does 0.999... = 1", it should not be, "No."  Rather, is should just be, "The question does not make sense with my definitions of the real numbers and decimal expansion."  Those are two entirely different answers, and if you don't accept infinite decimal expansions, then how could you ever answer the question with anything other than "It doesn't make sense"?  And that is a perfectly valid answer.  For example, I have proved elsewhere that if you accept the real numbers as a field, then division by 0 just doesn't make sense.  And I find that as a valid conclusion.

But I'm curious.  Do you see any problems with my definition?  You have stated that you see problems with my use of infinity.  I use the same concept of infinity as used in the limit of an infinite sequence, and in fact, I set up the digits of a decimal expansion as an infinite sequence.  Does this make sense?

To be honest, I am rather insulted that you claimed that I was trying to, "suppress the minority's opinion by the majority's."  I come to this forum to help others, have fun, and because there are some rather smart people here worth talking to about math.  And I don't take claims that I'm trying to suppress someones opinion lightly.  I am not having this conversation/debate for my health, but because it interests me.  Do you really think I'm trying to suppress your opinion by having an open debate on a public forum?

"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..."

## #125 2006-11-28 02:13:01

Anthony.R.Brown
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### Re: 0.9999....(recurring) = 1?

LET,S MAKE THIS AS SIMPLE AS POSSIBLE!
CAN ANYONE SHOW HOW! USING MATH IT IS POSSIBLE TO MAKE A NUMBER THAT IS SMALLER THAN ANOTHER NUMBER BIGGER! WITHOUT USING ADDITION!
AS IN THE PROBLEM BELOW!

Infinite 0.9 (stage one) ( some how we have to make this part larger! ) = 1