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0.9 (repeated) does NOT equal 1, with proof

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posted on Dec, 8 2006 @ 11:41 AM
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By standard result:


This being true, I can evaluate it down
0.9 = 9/10
(9(10^(-a-1) - 1)) / (10(10^(-1) - 1)) // 10^1 * 10^(-1) = 10^0 = 1
= (-9(10^(-a-1) - 1)) / (9) // (9)s cancel out
= -(10^(-a-1) - 1)
= 1 - 10^(-a-1)


Lim a->infinity, substitute infinity to get the value
= 1 - 10^(-i-1)
= 1 - 10^(-i) // negative infinity subtract a number is negative infinity
= 1 - (10 / i)
= 1 - (1 / i)

Can we agree that
Lim a->infinity = 1 - (1 / i)
All I did was simplify the equation by elimination, so there's nothing changed.

Assume,
0.9r = 1
(I write 0.9 repeated as 0.9r)

The use of geometric progression is the only way to write 0.9r.

So looking at this:



Therefore
1 - (1 / i) = .9r


Agree? Ok.
.9r = 1
.9r = 1 - (1 / i)
1 - (1 / i) = 1
1 / i = 0


FALSE. Division by infinity DOES NOT EQUAL ZERO. It's an infintesimally small number, NOT ZERO.

THEREFORE 0.9r CANNOT EQUAL 1

Kepp in mind: In all practical use, division by infinity is simply assumed to be zero, but we're not talking about practical use, we're talking about exact precision of a number.

I don't see anywhere which I have done anything erroneous. I took the geometric progression equation images from Wikipedia, and have worked with them before. What's the argument against me?




posted on Dec, 8 2006 @ 03:32 PM
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Originally posted by Zurahn
FALSE. Division by infinity DOES NOT EQUAL ZERO. It's an infintesimally small number, NOT ZERO.

THEREFORE 0.9r CANNOT EQUAL 1

Kepp in mind: In all practical use, division by infinity is simply assumed to be zero, but we're not talking about practical use, we're talking about exact precision of a number.


This statement is where your problem comes in. The division by infinity DOES equal zero. If you can find a proof for that statement then you have prooved me wrong but I dont think you can.

An infinitely small number is zero. Dividing by a very large number will get you a very small number but infinity isn't a number at all, its a concept. If you describe a infinitely small number as 0. followed by an infinite number of zeros then 1, then you will never get to that 1. The zeros continue forever and the infinitely small number is just 0.0000000000...



posted on Dec, 8 2006 @ 05:03 PM
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Originally posted by gfad

Originally posted by Zurahn
FALSE. Division by infinity DOES NOT EQUAL ZERO. It's an infintesimally small number, NOT ZERO.

THEREFORE 0.9r CANNOT EQUAL 1

Kepp in mind: In all practical use, division by infinity is simply assumed to be zero, but we're not talking about practical use, we're talking about exact precision of a number.


This statement is where your problem comes in. The division by infinity DOES equal zero. If you can find a proof for that statement then you have prooved me wrong but I dont think you can.

An infinitely small number is zero. Dividing by a very large number will get you a very small number but infinity isn't a number at all, its a concept. If you describe a infinitely small number as 0. followed by an infinite number of zeros then 1, then you will never get to that 1. The zeros continue forever and the infinitely small number is just 0.0000000000...


You can't divide a number to zero. Nomatter how large, you can always divide it further, just as on .9 repeated, you can always have another 9 on there. In all practical use as I said, division by infinity is assumed to be zero.

Here's my attempt at a proof (well, not a proof per se, but showing the problems with it equalling zero)



----- IGNORE DUE TO STUPIDITY


Assume i = infinity

1 - (1 / i) = 0
1 = 1 / i
i = 1

1 equals infinity? Let's try doing the same idea under my concept of it being greater than zero.

1 - (1 / i) > 0
1 > 1 / i
i > 1

So is infinity equal to one, or is it greater than one? I just threw this one together, but I'm pretty sure this legitimately backs up my point.

Maybe
infinity >= 1




----------END STUPIDITY
[edit on 8-12-2006 by Zurahn]

[edit on 8-12-2006 by Zurahn]



posted on Dec, 8 2006 @ 07:56 PM
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The first thing I want to point out to the previous poster, Zurahn:

"1 - (1 / i) = 0
1 = 1 / i
i = 1"

1 - 0 = 0 is not a true statement... 1 doesn't equal 0. 1/infinity is zero.

Second, I'm having alot of trouble understanding the purpose of this post. It's obvious 0.999999999999999999999999 doesn't equal to one - you round up. In mathematics fractions, such as 9/10, are only assumed to be 0.9 because it's division by two rational numbers, so in return you should get a rational number.

Maybe restate the question and hopefully I can understand what you're trying to convey...



posted on Dec, 8 2006 @ 09:01 PM
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Oh, haha, what a dumb mistake on that.

The purpose of the post is that I've seen it all over that "0.9r is equal to 1"

It even has its own Wikipedia page. This is an attempt at a clear refutal of that.

Back to 1 / i.

gfad, you claim it equals zero, if so, the same is also true

1 = 0i

I'm under the impression zero times infinity is, well, zero, as there's none of them.

I don't think I can provide a clear proof of 1 / i does not equal zero any more that you can provide a proof that it does.

[edit on 8-12-2006 by Zurahn]

[edit on 8-12-2006 by Zurahn]

[edit on 8-12-2006 by Zurahn]



posted on Dec, 9 2006 @ 02:05 PM
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1 = 0*infinity is a false statement as well. 0 times infinity is an indeterminate mathematical form - others include:

0/0
infinity/infinity
0*infinity
infinity-infinity
0^0
infinity^0
1^infinity

Dealing with infinities is tricky business because you must realize infinity is not a number - it's more of a benchmark.

Do you have any links to where people have "proved" that 0.9r equals to one? I'd like to see their technique and what they did. (I have taken a course in Real Analysis, so I could probably tell you if what they're doing is considered legal)



posted on Dec, 9 2006 @ 04:40 PM
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I linked the words "Wikipedia page" to the actual page in my previous post.

I know of the indeterminate values, but 0*infinity, while contested, has to be zero. Even if there are conflicting rules of infinity * x = infinity, if there are zero of anything, then there's nothing left to discuss. Infinity zeroes, regardless, is nothing. I don't think infinity needs to be a number for 0 multiplied by it to mean zero. 0*apples is still no apples.

Anyway, that's not what I'm using as a counter-proof, it's my main post I'm most concerned with.



posted on Dec, 9 2006 @ 10:15 PM
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Originally posted by gfad

Originally posted by Zurahn
FALSE. Division by infinity DOES NOT EQUAL ZERO. It's an infintesimally small number, NOT ZERO.

THEREFORE 0.9r CANNOT EQUAL 1

Kepp in mind: In all practical use, division by infinity is simply assumed to be zero, but we're not talking about practical use, we're talking about exact precision of a number.


This statement is where your problem comes in. The division by infinity DOES equal zero. If you can find a proof for that statement then you have prooved me wrong but I dont think you can.

An infinitely small number is zero. ...


A positive infinitesimal is defined as:
Any number greater than zero AND
Any number less than all real numbers greater than zero.

en.wikipedia.org...

These numbers were used to define Calculus (before Limits existed), but were not proven until Abraham Robinson wrote the Non-Standard Analysis in either the 1960s or 70s. Thus, Limits were proven first and that is what Calculus is taught with to this day.

Basically, you are wrong. An infinitely small number is an infinitesimal... NOT ZERO.



posted on Dec, 9 2006 @ 10:22 PM
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Here is the proof that .999 = 1

x = .999
x * 10 = .999 * 10
10x = 9.999
10x-x = 9.999 - .999
9x = 9
9x/9 = 9/9
x = 1



posted on Dec, 9 2006 @ 10:57 PM
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That's not the one to which I'm referring. The problem with the "old proof" you posted is that 10 * .999 = 9.99 not 9.999. This is for .9 repeated, which is more complicated.



posted on Dec, 10 2006 @ 01:11 AM
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Originally posted by Zurahn
That's not the one to which I'm referring. The problem with the "old proof" you posted is that 10 * .999 = 9.99 not 9.999. This is for .9 repeated, which is more complicated.


Are you kidding me? Really? I can't believe you just said that.

What, you think there's a proof for .999 (literal)? Go get some sleep and read it tomorrow... and never post that again without thinking.



posted on Dec, 10 2006 @ 01:24 AM
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.9 repeating equals 1

There are about a million proofs

I can't believe this is even being questioned or discussed (again
).


edit:

You don't even need a mathematical proof per se, you can just do it by logic and mathematical definition.

A rational number is defined as any number than can be expressed as a fraction a/b where b is not 0 or alternatively by a terminating or repeating decimal. Therefore .9 repeating must be a rational number since it repeats.

There is no way to express .9 repeating as a fraction in any other way than something that reduces to 1 such as 9/9 etc. such as you can do with .3 repeating (1/3) .6 repeating (2/3) etc.

By definition .9 repeating is exactly 1.


[edit on 12/10/2006 by djohnsto77]



posted on Dec, 10 2006 @ 02:18 AM
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Why do people try to disprove fact on this board randomly. Are you just looking for attention?

lim_(m --> oo) sum_(n = 1)^m (9)/(10^n) = 1

0.9999... = 1

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.

Is the proof that I always used. When you multiply .99999... by 10, it will move the decimal point over one, so you get 9.99999....



posted on Dec, 10 2006 @ 06:52 AM
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This is the proof one..I can prove it easily


x = 0.9999...
x+1 = 1.9999...
x+1- 0.9999... = 1.0
x+0.0000... = 1.0
x = 1.0000... - 0.0000...


Infinity has no end...
So 1.0000... relates with 0.9999... likewise.



posted on Dec, 10 2006 @ 07:50 AM
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Algebraic proofs are not solid mathematical proofs due to the nature of infinities used in a repeating decimal. It would be more appropriate to use limits, sequences/series, and best of all uniform convergence.

However, I do not fully grasp the concept of 0.9r equally to one - I understand it property wise but the techniques used to prove it on Wikipedia aren't exactly the strongest proofs I can think about. I'm also thinking about mapping - I am no expert in the field as I've only taken one Analysis course so I'm going to ask an Analysis professor about this.

By the way, I think it's unfair to flame someone for attempting to understand a concept. You may have a million ways to prove something, but nothing is greater than only one way to disprove something.



posted on Dec, 10 2006 @ 04:24 PM
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Originally posted by T_Jesus
Algebraic proofs are not solid mathematical proofs due to the nature of infinities used in a repeating decimal. It would be more appropriate to use limits, sequences/series, and best of all uniform convergence.

However, I do not fully grasp the concept of 0.9r equally to one - I understand it property wise but the techniques used to prove it on Wikipedia aren't exactly the strongest proofs I can think about. I'm also thinking about mapping - I am no expert in the field as I've only taken one Analysis course so I'm going to ask an Analysis professor about this.

By the way, I think it's unfair to flame someone for attempting to understand a concept. You may have a million ways to prove something, but nothing is greater than only one way to disprove something.


Which is why I used limits to begin with. The only way to write .9 repeated without decimals is as a limit. I took the equation and cut it down until I had 1 - (1 / infinity) through arithmetic (as shown in my original post).

All I'm saying here is that for .9 repeated to equal 1, (1 / infinity) must equal zero, which I have stated and another has supported, is not true. I want comment on THAT, which no one is doing. I am glad that at least you, T_Jesus, are adding something to the discussion.



posted on Dec, 10 2006 @ 04:39 PM
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To those who merely post the following:

0.9999... = 1

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.

There IS a reason for my original post. Writing a number as a decimal is an imprecise way to prove something.

9.9... - .9... only equals 9 if infinity is a constant, which it is not. The number of decimal places isn't necessarily equal, therefore .9... - .9... isn't necessarily zero, which is the problem with writing .9... as a decimal in the first place.

For example, here's what you are doing under a fractional form, using my 1 - (1/i) form

x = 1 - (1 / i)
10x = 10(1 - 1/i)
10x - x = 10 - 10/i - (1 - 1/i)
9x = 10 - 1/i - 1 + 1/i
9x = 10 - 1 + 1/i - 1/i

What you are doing is cancelling out +1/i - 1/i. Infinity subtract infinity isn't necessarily zero, nor is 1/i - 1/i, which is what the problem is with that proof. Or are you just going to express your disbelief that someone will dispute your almighty "proof".


EDIT: As a response to "and I can prove it easily"

x = 1 - 1/i
x+1 = 1 - 1/i + 1
x+1 = 2 - 1/i
x+1-(1 - 1/i) = 1
x+1-1+1/i = 1
x+1/i = 1
x = 1 - 1/i

[edit on 10-12-2006 by Zurahn]



posted on Dec, 10 2006 @ 06:13 PM
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Originally posted by Zurahn
9.9... - .9... only equals 9 if infinity is a constant, which it is not. The number of decimal places isn't necessarily equal, therefore .9... - .9... isn't necessarily zero, which is the problem with writing .9... as a decimal in the first place.


Not really. Infinity is not a value, but our value (.999...) is. In this case it would be defined strict enough that .999... could easily equal .999...
How?
.999... is not really infinity, it is an infinite sequence equivalent to...
."9 forever"
So if we do 9.999..., we are saying 9."9 forever"
Then, 9."9 forever" minus ."9 forever" equals 9
Using Cantor's idea of infinite sets with matching sizes (cardinalities), we are saying that both of our infinite sequences are of a single dimension and repeat the value of nine into infinity. So as long as we have the same cardinality for these infinite sequences, we should be OK in saying that they cancel one another perfectly AS LONG AS they are not of the "value" infinity or negative infinity.

Infinite sequences are not infinite sets are not infinite series are not infinity.
Write that down.

en.wikipedia.org...


Originally posted by Zurahn
For example, here's what you are doing under a fractional form, using my 1 - (1/i) form

x = 1 - (1 / i)
10x = 10(1 - 1/i)
10x - x = 10 - 10/i - (1 - 1/i)
9x = 10 - 1/i - 1 + 1/i
9x = 10 - 1 + 1/i - 1/i


You're making mistakes:

x = 1 - (1 / i)
10x = 10(1 - 1/i)
10x - x = (10 - 10/i) - (1 - 1/i)
9x = (10 - 10/i) + (-1 + 1/i) ... now combine like terms
9x = 9 - 9/i
9x / 9 = (9 - 9/i) / 9
x = 1 - (1/i)



posted on Dec, 10 2006 @ 06:36 PM
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I always used this approach to prove for myself that .9r = 1

1 / 3 = .3r

3 x .3r = .9r

=> .9r = 1

But that has been said earlier already in other words :-)



posted on Dec, 10 2006 @ 06:38 PM
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But your saying 9 an infinite number of times, when infinity isn't constant. That's my concern there. For it to be a cardinality, the sets have to be the same size, which is infinity, which isn't constant, which means they aren't necessarily the same. Basically it's saying A = [9, 9, 9, ...] B = [9, 9, 9, ...]
|A| = infinity
|B| = infinity
|A| - |B| = 0
which isn't true. If I have interpreted you incorrectly, sorry, please clarify.

And I know that you can combine like terms, but that's not illustrating what that decimal proof is doing. When it moves the decimal point over by multiplying by 10, it's the equivalent of reducing 10/i down to 1/i.

Doing it the other way by just eliminating out the multple is doing the same as

x = .9...
10x = 10(.9..)
10x = 10x
x = x
x = .9...

Unless, that is, you can tell me what my mistake specifically was. If so, please point it out.



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