DiMono
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I have to disagree with you. We aren't dealing with different infinities in that proof, we're dealing with the same infinity; the infinite quantity of 9s after the decimal point. Consider the following:

∞ + 1 = ∞ (a)
10x = 9.99999 has  (∞)  9s to the right of the decimal (b)
  x = 0.99999 has (∞+1) 9s to the right of the decimal (c)

Because of (a), the ∞ 9s in (b) and (c) are the same, which means they line up exactly. 9-9=0 no matter how many times you do it, even if it's an infinite quantity. That means the proof holds.

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[Last edited by DiMono at 01-29-2008 07:23 PM]

Penumbra
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You are talking about the 9's at the end of the list "lining up." It is indeterminate and just as undefined as x/0.

As for ( ∞ + 1 = ∞ ) The two "equal" infinities aren't really equal. Infinity plus one is still infinity. But infinity isn't a number. So (not a number) + (a number) = (not a number). Once you include (not a number) you can't go back again.

Even something simple, such as ∞ = ∞ isn't really the algebraic use of the "=" sign. That means both quantities on either side of the equation are infinite, not necessarily equal.

Maurog
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Two infinities are considered equal in size if there exists a complete inversible function from one to the other.

Anyway, the axiom r - r = 0 holds true for any real number r, including 0.999... Therefore 9.999... - 0.999... = (9 + 0.999...) - 0.999... = 9 + (0.999... - 0.999...) = 9 + 0 = 9

I don't see how can anyone see a fallacy here.

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Penumbra
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To say that 0.9999~ = 1, we must first look at what the decimal representation of 0.9999~ actually means.

The number 1234 represents the sum of ( 1 ⋅ 1000) + (2 ⋅ 100 ) + ( 3 ⋅ 10 ) + ( 4 ⋅ 1 ).

0.999 represents ( 9 ⋅ 1/10 ) + ( 9 ⋅ 1/100 ) + ( 9 ⋅ 1/1000 ) or, with x equal to 3,

 x    9
 ∑   ---  
n=1  10ⁿ

0.9999~ represents the summation of an infinite series, which is calculated with limits.

      x    9
lim   ∑   ---  
x→∞  n=1  10ⁿ

This is read as the summation as x approaches infinity. This series is convergent meaning it will be bounded by a single real number(which happens to be one). Whether or not this ever actually equals one depends on semantics and your personal feelings on Zeno's Paradox ( i.e. the math never gets there, but that turtle gets caught )

Timo006
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Let's try it another way.

1/3=0.3333333..
<=>(1/3)*3=(0.333333..)*3
<=>1=0.999999..

This does work sometimes if you put it in a calculator. (depending on how it works)

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[Last edited by Timo006 at 01-29-2008 09:32 PM]

zex20913
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Penumbra wrote:

zex20913 wrote:

    x= .9999999...
-(10x=9.9999999...)
  -9x=-9
    x=1

That proof,*

Exactly why I said it wasn't a proof. It's an argument that seems true, but falls short of a true proof. A geometric series argument works a lot better.

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Penumbra
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Maurog wrote:
Two infinities are considered equal in size if there exists a complete invertible function from one to the other.

Anyway, the axiom r - r = 0 holds true for any real number r, including 0.999... Therefore 9.999... - 0.999... = (9 + 0.999...) - 0.999... = 9 + (0.999... - 0.999...) = 9 + 0 = 9

I don't see how can anyone see a fallacy here.

There is a fallacy, and it lies in the mathematical definitions.

      x= .9999999...
  -(10x=9.9999999...)
    -9x=-9
      x=1

There are two 0.9999~ in this equation at the top, x and 0.9999~. It could be rewritten as:

       0.9999~ = 0.9999~

  10 * 0.9999~ = 9.9999~
                -0.9999~

In line 1, we have the same "number" of infinite 9's after the decimal place. There exists, as you said, a complete invertible function from one to the other. This would be, for any n-th 9 of the left side, we have a corresponding 9 on the right. Once you multiply it by ten and "shift" all the 9's to the left one place, your mapping function breaks down. For each "extra" 9 you have to add to complete the mapping, you will still have one more 9 unaccounted.

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[Last edited by Penumbra at 01-29-2008 09:42 PM]

Jutt
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Once you multiply it by ten and "shift" all the 9's to the left one place, your mapping function breaks down. For each "extra" 9 you have to add to complete the mapping, you will still have one more 9 unaccounted.

That's nonsense, the mapping still matches perfectly. Really, both sets of nines have the same cardinality (= 'size') and the same order type, so the given mapping is perfectly bijective (= 'invertible').

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Rabscuttle
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zex20913 wrote:

Penumbra wrote:

zex20913 wrote:

    x= .9999999...
-(10x=9.9999999...)
  -9x=-9
    x=1

That proof,*

Exactly why I said it wasn't a proof. It's an argument that seems true, but falls short of a true proof. A geometric series argument works a lot better.

The decimal representation is just a way of representing the geometric series. They are the same thing. Although expressing it as a geometric series is more explicit.

Rabscuttle
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Penumbra wrote:

Maurog wrote:
Two infinities are considered equal in size if there exists a complete invertible function from one to the other.

Anyway, the axiom r - r = 0 holds true for any real number r, including 0.999... Therefore 9.999... - 0.999... = (9 + 0.999...) - 0.999... = 9 + (0.999... - 0.999...) = 9 + 0 = 9

I don't see how can anyone see a fallacy here.

There is a fallacy, and it lies in the mathematical definitions.

      x= .9999999...
  -(10x=9.9999999...)
    -9x=-9
      x=1

There are two 0.9999~ in this equation at the top, x and 0.9999~. It could be rewritten as:

       0.9999~ = 0.9999~

  10 * 0.9999~ = 9.9999~
                -0.9999~

In line 1, we have the same "number" of infinite 9's after the decimal place. There exists, as you said, a complete invertible function from one to the other. This would be, for any n-th 9 of the left side, we have a corresponding 9 on the right. Once you multiply it by ten and "shift" all the 9's to the left one place, your mapping function breaks down. For each "extra" 9 you have to add to complete the mapping, you will still have one more 9 unaccounted.

So you'd also argue that there are more positive even numbers than positive integers?

Mattcrampy
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Penumbra wrote:
For each "extra" 9 you have to add to complete the mapping, you will still have one more 9 unaccounted.

This is where you've slipped up. It's an infinite series of 9s here, so you can't 'add' any extra nines. Infinity + 1 = infinity, as does infinity - 1.

In fact, I'm not even sure you get zero when you subtract infinity from infinity.

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Penumbra
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I was a little hasty in my mapping example. I was merely trying to point out that it is indeterminate. When multiplying the series by 10, to "shift" the numbers, you put a 0 on the "end." This can't happen. You can't shift them all up. "Something" happens, but you can't specifically say what it is. There are many things what are indeterminate with infinity. I mentioned ∞ - ∞ = 0 as one of those undefined cases.

Sillyman
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1=.9999999... holds true because Zeno's Paradox is resolved. If Zeno's Paradox were not resolved, it would not be true. So yeah, it does depend on Zeno's Paradox.

Edit: Actually, what I said isn't quite true. What I meant was that the two problems rely on the same fact: Infinitessimal=0.

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[Last edited by Sillyman at 01-30-2008 02:55 AM]

Rabscuttle
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Penumbra wrote:
I was a little hasty in my mapping example. I was merely trying to point out that it is indeterminate. When multiplying the series by 10, to "shift" the numbers, you put a 0 on the "end." This can't happen. You can't shift them all up.

You're actually shifting the decimal point one space right.

But even if you are shifting the numbers, why can't you shift them all up? It's not as though you have to do them individually (in a finite amount of time.)

And how could you multiply any repeating decimal by 10?
0.3333... * 10 = 3.3333... with a 0 on the end?

Maurog
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By the way, why don't anyone have a problem with 1.000... * 10 = 10.000... with an extra 0 in the end?

After all, you shift the 0s left. And of course you don't have to get tangled in infinity to prove that 1 = 0.999..., see my original proof based on properties of real numbers.

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Jutt
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Penumbra wrote:
…you put a 0 on the "end."

No you don't, because you can't. Technically you never do when working with real numbers, because any real number has an infinite number of digits behind the decimal point. But in the case they're all zero after a certain point they're usually left out in notation.

Penumbra wrote:
You can't shift them all up.

You can shift them—you just follow the rules for multiplication which have been proven correct, indeed also for infinite representations. The fact that this proof of correctness is based on the infinite series a decimal real number represents, is usually forgotten.


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DiMono
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0.12 * 10 = 1.2

I see no extra 0.

Also, as a random addition, the terminating real numbers are countably infinite, not uncountably infinite. They can be mapped to the set of integers. It's all in how you count them. Rather than going by the value of the number, go by how many digits you have to write to represent it on paper. Counting them in this way looks something like this (presented in shorthand to save space):

1, 2, 3, 4, 5, 6, 7, 8, 9, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 10-99, 1.1-1.9, 2.1-2.9, ... 9.1-9.9, .01-.09, .11-.19, ... .91-.99, and so on.

In my first year at University I was able to express the number of terminating reals as a Sigma function, but the paper with that function is at home. Note that this only works for terminating real numbers; non-terminating numbers are uncountably infinite, and non-terminating non-repeating numbers are even more uncountably infinite.

Edit: I've re-generated the Sigma series:

x represents number of digits

x=1 - (1-9), (.1-.9)
x=2 - (10-99), (1.1-1.9 ... 9.1-9.9), (0.01-0.09 ... 0.91-0.99)
x=3 - (100-999), (10.1-10.9 ... 99.1-99.9), (1.01-1.09 ... 1.91-1.99), (0.101-0.109 ... 0.991-0.999)

x=1 - 9 + 9 = 9*10^0 + 1(9*1)
x=2 - 90 + 81 + 81 = 9*10^1 + 2(9*9)
x=3 - 900 + 810 + 810 + 810 = 9*10^2 + 3(9*90)
x=n n>1 - 9*10^(n-1) + 9n(9*10^(n-2))

Therefore the number of terminating reals > 0 is:

18 + Sum as x goes from 2-∞ of (9*10^(x-1) + 9x(9*10^(x-2)))

If you want all terminating reals, double it and add 1. I could swear I was able to figure out a single function that describes x=1 as well, but I can't figure it out right now. It relies on being able to create a single series that begins 1, 9, 90.

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[Last edited by DiMono at 01-30-2008 03:53 PM]

Jutt
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DiMono wrote:
… non-terminating numbers are uncountably infinite, and non-terminating non-repeating numbers are even more uncountably infinite.

Aren't the non-terminating non-repeating (= irrational) numbers a subset of the non-terminating numbers and therefore not 'more' uncountably infinite? I'd say both sets have the same size.

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DiMono
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Jutt wrote:

DiMono wrote:
… non-terminating numbers are uncountably infinite, and non-terminating non-repeating numbers are even more uncountably infinite.

Aren't the non-terminating non-repeating (= irrational) numbers a subset of the non-terminating numbers and therefore not 'more' uncountably infinite? I'd say both sets have the same size.

Ah, you are of course correct. My mistake. Also I tried to rediscover my counting the reals Sigma function in my post above yours. I'm about half a step away, but I can't see the landing.

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zex20913
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Actually, the set of non-terminating, repeating decimals is countable.

If the decimal repeats infinitely, it can be represented as a fraction.

The set of rational numbers is countable.

Therefore, such numbers as described above are countable.

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stigant
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Exactly why I said it wasn't a proof. It's an argument that seems true, but falls short of a true proof. A geometric series argument works a lot better.

I object. The geometric series argument (related above by Penumbra) relies on the fact that if 0 < r < 1 then sum(i = 0 to inf)(r^i) = 1/(1-r). The proof of THAT fact (at least the one I've seen) goes like this:

Let S = sum(i = 0 to inf)(r^i) = 1 + r + r^2 + r^3 + r^4 + ...
Then rS = r + r^2 + r^3 + r^4 + ....
S - rS = 1
S = 1 / (1-r)

This relies on the same sort of lining up and cancelling that the proof of .9999... = 1 does. Its just hidden behind an accepted theorem so that you don't notice it.


To be sure, I accept both facts as proven since the sequences in question (ie the ones that are "cancelling out termwise") are both countable (ie there does in fact exist a 1-to-1 correspondence between them which can be used to set up the cancelling in a more rigorous manner) and therefore we don't run into the problem of different infinities.


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[Last edited by stigant at 01-30-2008 10:03 PM]

Beef Row
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Timo006 wrote:
Let's try it another way.

1/3=0.3333333..
<=>(1/3)*3=(0.333333..)*3
<=>1=0.999999..

This does work sometimes if you put it in a calculator. (depending on how it works)

This.

penumbra, do you have any problems with this version of the proof?

Also, this adding up 1/3s approach makes this more intuitive than multiplying by ten and subtracting does.

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zonhin
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Seriously, I have no idea what the point of this thread is. Yes, 0.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999... Does equal 1. What's the point of discussing new, more obtuse proofs if an intuitive one is already (relatively) well known.

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zex20913
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Because an alternate method of proof may lend itself to an easy proof of another proposition.

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Sillyman
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And because some people just don't get it.

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Mr. Slice
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I agree with Zonhin. The main point of how 0.99999999999... equaled 1, so why are we continuing this?

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coppro
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Here's another way:

Find x:

1 - x = 0.99999...

The answer is 0.00000... or in other words, 0, since there are an infinite many zeroes.

Sillyman
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Exactly! There is nowhere for any infinitessimal piece to go, it just doesn't work. Infinitessimal=0, always.*

*Well, not ALWAYS. Look up the mathimatical definition of "Almost Never".

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[Last edited by Sillyman at 01-31-2008 03:24 AM]

Syntax
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I'm with Penumbra on this one though my analysis is rather different.

If indeed it is the case that 1 = 0.999..., why then has the same entity 2 different names?

I wait for the day TI has 2 keypads for input, just in case you meant the *other* same thing.

Look at subject line itself, and you will see that they are not the same. One has a 1 in it, the other has loads of 9s. Proving that 1 = 1 is hard enough, as that is still relative to semantic context. Two things are only equal if they point at the exact same result. Not an equivalent one... the same one.

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[Last edited by Syntax at 01-31-2008 07:49 AM]

eytanz
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Syntax wrote:
I'm with Penumbra on this one though my analysis is rather different.

If indeed it is the case that 1 = 0.999..., why then has the same entity 2 different names?

I wait for the day TI has 2 keypads for input, just in case you meant the *other* same thing.

Look at subject line itself, and you will see that they are not the same. One has a 1 in it, the other has loads of 9s. Proving that 1 = 1 is hard enough, as that is still relative to semantic context. Two things are only equal if they point at the exact same result. Not an equivalent one... the same one.

That's a fallacy.

Note that the same number has an infinite amount of names; it can also be called "3-2", or "the positive square root of 1" or "the positive square root of the positive square root of 1", or "-56+57" or many other names.

By your logic, "3-2" does not equal "1" because the first has a 3 and a 2 in it, and "1" does not.

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