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Maurog
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So Pi doesn't equal 3.14159... then? Because one is a symbol and another is a bunch of digits.

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01-31-2008 at 08:32 AM
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Beef Row
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Maurog wrote:
So Pi doesn't equal 3.14159... then? Because one is a symbol and another is a bunch of digits.

Actually, considering his concern about proving 1 = 1, perhaps the problem is establishing that equals equals equals?

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01-31-2008 at 08:45 AM
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Rabscuttle
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Syntax wrote:
If indeed it is the case that 1 = 0.999..., why then has the same entity 2 different names?

Because the positional system of representing numbers has redundancies.

If you want each number to have a unique representation, then you can either
* don't allow a number to end with a repeated 9
OR
* don't allow terminating decimals (replace them with repeated 9s)

You'd also have to not allow extra 0s on the start or end.


edditional: If you do this, it'd be better to not allow any terminating numbers, then you don't run into trouble with things like 0.(3) * 3.

[Last edited by Rabscuttle at 01-31-2008 10:29 AM]
01-31-2008 at 08:59 AM
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Mattcrampy
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Anyone else find it ironic exactly who is arguing about syntactical representation?

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01-31-2008 at 10:31 AM
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Syntax
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Mattcrampy wrote:
Anyone else find it ironic exactly who is arguing about syntactical representation?
:lol If only my nick was "=nottrue" ;)

As for the reply given by eytanz, I do not consider 0.999... or 1 as any form of transformation (algebraic or otherwise). They represent 2 separate entities.

If they were indeed equal, they would be represented by the same symbol.
01-31-2008 at 12:16 PM
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Syntax
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Beef Row wrote:
Maurog wrote:
So Pi doesn't equal 3.14159... then? Because one is a symbol and another is a bunch of digits.

Actually, considering his concern about proving 1 = 1, perhaps the problem is establishing that equals equals equals?
And yes... The way I see it is equals represents equality.
0.999... does not equal 1 in my opinion.
01-31-2008 at 12:18 PM
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Znirk
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Syntax wrote:
If they were indeed equal, they would be represented by the same symbol.
So, in that semantic universe, would you say that "equals" equals "is identical to", or that "equals" is identical to "is identical to"?
01-31-2008 at 01:04 PM
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Rabscuttle
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Syntax wrote:
If they were indeed equal, they would be represented by the same symbol.

This is where you are wrong.

Unless you're not talking about the real numbers. Or thinking in base 11 or something.

If you object to the tranformations, how about 1.0 vs 1.00? Or 2/4 vs 1/2 (fractions)?

Also, I wouldn't call 0.(9) a symbol, it's a collection of symbols that represent a number.

[Last edited by Rabscuttle at 01-31-2008 01:14 PM]
01-31-2008 at 01:09 PM
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Syntax
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Rabscuttle wrote:
Syntax wrote:
If they were indeed equal, they would be represented by the same symbol.

This is where you are wrong.

Unless you're not talking about the real numbers. Or thinking in base 11 or something.

If you object to the tranformations, how about 1.0 vs 1.00? Or 2/4 vs 1/2 (fractions)?

Also, I wouldn't call 0.(9) a symbol, it's a collection of symbols that represent a number.
2/4 and 1/2 is a transformation by division. They're equivalent.

I apologise for the use of the word symbol. I meant representation.
01-31-2008 at 02:44 PM
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Maurog
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How about 1/1 = 1? One is a fraction, the other is an integer.
How about 1+0i = 1? One is a complex number, the other is an integer.
0.999... = 1; one is an infinitecimal representation, the other is an integer.
9/10+9/100+9/1000+9/10000... = 1; One is a converging geometric series...

What makes it different for you Syntax? Which of the above do you accept and which you don't, and why?



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01-31-2008 at 03:14 PM
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DiMono
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Maurog wrote:
So Pi doesn't equal 3.14159... then? Because one is a symbol and another is a bunch of digits.
No, because one is a number and the other is meant to be cooked and eaten.

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01-31-2008 at 03:16 PM
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Penumbra
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I am not usually one to post on forums often, as I feel my writing makes me appear to be a pedantic jerk. My apologies to anyone who thinks I am "correcting" them. I make no claim to be a mathematician, I just enjoy the discussion no matter how it turns out. I believe this to be one of the few places on the internet where this conversation will not degrade into baseless name calling.

I think I have found what my problem with the "quick" form of the proof is. It begins with the assumption that 0.9999~ is equivalent to 1. Without that assumption, 10*0.9999~ would not equal 9.9999~. Mind you I am not saying this to contradict the 0.9999~=1 statement, only that we are using the conclusion to prove it.

When I said multiplying by 10 was putting a 0 at the "end" of the number, I was not using any mathematical basis. It is a short hand we use to multiply quickly, just like the "move the decimal to the right." Multiplication is defined as repeated addition. So multiplying 0.9999~ by 10 is the same as
    0.9999~
    0.9999~
    0.9999~
    0.9999~
    0.9999~
    0.9999~
    0.9999~
    0.9999~
    0.9999~
  + 0.9999~
    9.9999~
If it were not assumed before this step that 0.9999~ were equal to 1, and not some infinitesimal nonzero number x less than 1, then 9.9999~ would be 10x less than 10.

This short hand we use to quickly multiplying a number by 10 is the key to why this proof "works" on a gut level. This is also why 10 is used, and not something else, like 2 or 736. Both these are just as valid, assuming 0.9999~=1
  x = 0.9999~
 2x = 1.9999~
 -x  -x
  x = 1


   x =   0.9999~
736x = 735.9999~
  -x    -x
735x = 735
   x =   1
Remember, I am not saying that 0.9999~ isn't equivalent to 1, just that those steps are only valid if it is. This is where the "proof" breaks down. It is a clever example using the equality, however.
01-31-2008 at 04:21 PM
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Maurog
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Nono, you're skipping the 9.9999~ = 9 + 0.9999~ step. There is no assumption of the equality we want to prove, and what's subtracted from both sides is 0.9999~, not x nor 1. Only after we subtract and get 9 we go "omg, we multiplied 0.9999~ by 10 and subtracted 0.9999~ and got 9 this means 0.9999~ * 9 is 9, this means 0.9999~ is 1".

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01-31-2008 at 04:51 PM
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Jutt
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Penumbra wrote:
I think I have found what my problem with the "quick" form of the proof is. It begins with the assumption that 0.9999~ is equivalent to 1
This assumption isn't made anywhere in the proof. The only thing assumed at the begining is the fact that we have a real number x, which is represented in decimal as 0.999…. This is only to be followed by legitimate manipulations of the equality x=0.999…, which results in x=1.

To repeat the proof once more in yet another slightly different form:
  x = 0.999…                   (we have a real number, no further assumptions made)
    = 0.9 + 0.09 + 0.009 + …   (the sum our number actually represents)

10x = 10 · (0.9 + 0.09 + 0.009 + …)
    = 10·0.9 + 10·0.09 + 10·0.009 + …
    = 9 + 0.9 + 0.09 + …
    = 9.999…                   (hey, multiplication with 10 shifts the digits to the left!)

 9x = 10x - x
    = 9.999… - 0.999… 
    = 9

  x = 1


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01-31-2008 at 04:53 PM
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Penumbra wrote:
I think I have found what my problem with the "quick" form of the proof is. It begins with the assumption that 0.9999~ is equivalent to 1. Without that assumption, 10*0.9999~ would not equal 9.9999~.
Then Penumbra wrote:
This short hand we use to quickly multiplying a number by 10 is the key to why this proof "works" on a gut level. This is also why 10 is used, and not something else, like 2 or 736. Both these are just as valid, assuming 0.9999~=1
  x = 0.9999~
 2x = 1.9999~
 -x  -x
  x = 1


   x =   0.9999~
736x = 735.9999~
  -x    -x
735x = 735
   x =   1

I believe you are mistaken. 10 isn't used because it's convenient, 10 is used because the numbers in the series 10^n are the only ones with which you can perform the proof without assuming that 0.999~ = 1, and 10 is the smallest of these > 1. Multiplying by 10^n has the functional result of moving the decimal point to the right a number of digits equal to n, regardless of how many digits are in the number or what those digits are. For this reason, mutliplying by 10^n is the only way to carry out this proof without manipulating the digits in 0.999~.

However, if it will ease your mind:
      x = 0.999~

     2x = x + x
     2x = 0.999~ + 0.999~
     2x = 1.999~

     3x = x + x + x
     3x = 0.999~ + 0.999~ + 0.999~
     3x = 2.999~

3x - 2x = 2.999~ - 1.999~
      x = 1


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01-31-2008 at 06:37 PM
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Syntax, Equal means Equivalent. In geometry, this means that they have the same size, shape, and location. They do not have to have the same name. In arithmetic, this means two numbers have the same value. Again, they do not have to have the same name. Perhaps it would help if I set up a hierarchy.

Hold valid in Arithmetic and Geometry:
Identical- Same name, value, size, shape, location, etc.
Equal- Identical to identical, but with a different name (In other words, equal to it) and without the stipulation that the name has to be the same.
Hold vaild only in Geometry, have no meaning in Arithmetic:
Congruent- Equal to Equal, but without the stipulation that it has to be in the same place.
Similar- Equal to Congruent, but without the stipulation that it must be the same size.

The fact that two items do not have to be called the same thing to be equal is NECESSARY for all of mathematics to work. Equal is not equal to identical, let alone identical. Get it? Got it? Good.

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[Last edited by Sillyman at 02-01-2008 12:10 AM]
02-01-2008 at 12:08 AM
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Penumbra
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DiMono wrote:
Multiplying by 10^n has the functional result of moving the decimal point to the right a number of digits equal to n, regardless of how many digits are in the number or what those digits are.
The key word there is functional and not the definition of multiplication by 10.

Let me try and state my point another way. Whether or not 0.9999~ is equal to 1 or not, I think it is fair to assume, up front that 0.9999~ ≤ 1 regardless of your thoughts on the issue. With this assumption, we can assert that 0.9999~ = 1 - ε for some very small value of ε. In order to prove that 0.9999~ is equal to 1, we must prove that ε is equal to 0.

To repeat the steps....
   Assumption: 0.9999~ = 1 - ε

     x =  1 - ε
   10x = 10( 1 - ε )
   10x = 10 - 10ε
 -(1-ε ) -(1 -  ε )
    9x =   9 - 9ε
    9x = 9( 1 - ε )
     x =  1 - ε
You will see, that the unknown quantity of ε has not been canceled. In fact, we haven't really shown anything at all. The only way these steps can be true is if ε is equal to 0 from the very beginning.

02-01-2008 at 03:26 AM
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Sillyman
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Okay, I have a proof that can be gotten from that.
Now, let us assume that motion is possible. After all, I moved to type this message. Also, the laws of physics dictate that it is. People have recorded the fact that motion has happened many times in the past as well. So you can't argue this point. It's an axiom.
For that to be true, infinitessimal must equal 0. (See Zeno's Paradox)
ε is infinitessimal.
Therefore, ε must equal 0.
Therefore, .9 repeating must equal 0, which was to be demonstrated.

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[Last edited by Sillyman at 02-01-2008 03:36 AM]
02-01-2008 at 03:30 AM
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Penumbra wrote:
DiMono wrote:
Multiplying by 10^n has the functional result of moving the decimal point to the right a number of digits equal to n, regardless of how many digits are in the number or what those digits are.
The key word there is functional and not the definition of multiplication by 10.

[pedantic]The definition of multiplication = repeated addition only works for integers.[/pedantic]
Even so, adding 0.(9) together 10 times still gives you 9.(9)

How about I do away with multiplication of repeated decimals all together:
e.g.
x   = 0.9(9) = 0.(9)
x/2 = 0.4(9)
(subtract)
x/2 = 0.5
x   = 1


Let me try and state my point another way. Whether or not 0.9999~ is equal to 1 or not, I think it is fair to assume, up front that 0.9999~ ≤ 1 regardless of your thoughts on the issue. With this assumption, we can assert that 0.9999~ = 1 - ε for some very small value of ε. In order to prove that 0.9999~ is equal to 1, we must prove that ε is equal to 0.

To repeat the steps....
   Assumption: 0.9999~ = 1 - ε

     x =  1 - ε
   10x = 10( 1 - ε )
   10x = 10 - 10ε
 -(1-ε ) -(1 -  ε )
    9x =   9 - 9ε
    9x = 9( 1 - ε )
     x =  1 - ε
You will see, that the unknown quantity of ε has not been canceled. In fact, we haven't really shown anything at all. The only way these steps can be true is if ε is equal to 0 from the very beginning.

You're not doing anything there because you're not using the properties of the repeating decimal. ie. if you multiply 0.(infinite amount of 9s) by 10 (or add 10 of them together or whatever) you get 9.(infinite amount of 9s)

If you want to use 1-e
Let 0.(9) = 1 - e
    9.(9) = 10 - 10e
        9 = 9 - 9e
        0 = -9e
        e = 0


You seem to be arguing that the infinite 9s after the "0." is different from the infinite 9s after the "9."
That's kind of the opposite problem to Syntax's. The two are represented in the same way and are equal.
02-01-2008 at 04:45 AM
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Penumbra
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Rabscuttle wrote:
You seem to be arguing that the infinite 9s after the "0." is different from the infinite 9s after the "9."
That's kind of the opposite problem to Syntax's. The two are represented in the same way and are equal.
My argument is that if there were some difference in value between 0.9999~ and 1, than the multiplication of the values would likewise multiply the difference. Or dividing by two would make that half a difference. I am not arguing that 0.9999~ ≠ 1, I am arguing that those statements prove nothing of the equality. They are merely an interesting example.
02-01-2008 at 05:01 AM
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Rabscuttle
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But then that's completely irrelevant to the discussion. You may as well say if there's a difference between 1+1 and 2, then there will be a bigger difference between 2+2 and 4.

The fact that the repeated decimals line up show that there isn't a difference.
02-01-2008 at 05:05 AM
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Maurog wrote:
Let's define a function f(n) as follows:
f(1) = 0.9
f(2) = 0.99
f(3) = 0.999

Now, technically, f(infinity) is 0.999... Also, we all agree that the bigger the n the closer f(n) is to 1, right?

Lemma 1: When we have two real numbers A and B, there are only two options - either there are infinite other real numbers between them, starting with (A+B)/2 or there are zero real numbers between them, in the case of A=B.

Let there be a real number A between 0 and 1 different from 0.999... depicted by its decimals A=0.a1a2a3a4... Let am be the first decimal of A that isn't 9. It's easy to see that f(m) is closer to 1 than A, therefore 0.999... is closer to 1 than A.

We just proved that there is no real number between 0 and 1 that's closer to 1 than 0.999... according to lemma 1, this means they are one and the same. 1 = .999...

So, I'm confused as to why there's all of this arguing. We have a perfectly good proof that avoids the difficulty of people having to wrap their heads around repeating decimals or infinite geometric series. So there be no need for people to prove that the algebraic one is true or not, because we already have a proof.

Capiche?

P.S. This also happens to prove the algebraic one true, because as .999... = 1, then every argument as to how they're different is wrong. Just, wrong.

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02-01-2008 at 09:07 AM
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It doesn't prove the algebraic argument true, though it makes it pointless. Why? Because a proof of a true statement can be flawed. Anyway, I agree that continuing to discuss the matter is pointless, but there are still some people who won't accept a good solid proof.

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02-01-2008 at 09:21 AM
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Sillyman wrote:
It doesn't prove the algebraic argument true, though it makes it pointless. Why? Because a proof of a true statement can be flawed. Anyway, I agree that continuing to discuss the matter is pointless, but there are still some people who won't accept a good solid proof.

Sadly, this was my entire point. :blush I only wish I could have been as succinct.
I feel the algebraic statement is a great way to get people to believe the actual proofs, of which there are many. Treating circular logic as a proof just bothers me I guess. I'll stop now, and go back to DROD. ;)
02-01-2008 at 03:45 PM
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IMO the problem this thread is seeing is that people are speaking different languages.

On one side you have the mathematicians using rigorous proofs, and on the other side you have people using what amounts to 'proof by common sense'. I don't think that the 10*.999...=9.999 proof would pass muster under formal review (although, to be fair, I abandoned formal mathematics 18 years ago). It happens to be true -- sometimes common sense will get you to the exact same place that rigorous proof gets you.

However, and this is why you have people blanching at 'common sense proofs', common sense can also lead you horribly astray.

(Is the set of all rational numbers bigger than the set of all natural numbers?)

To really delve into this, I prefer formal analysis, which requires, first and foremost, a definition of what exactly a number is. From there, proofs that 0.999...=1 become much more manageable.

(Anyway, my terminology may be off -- like I said, I'm out of practice. Don't hurt me.)

Josh

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02-01-2008 at 06:24 PM
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I'm no mathematician but, am I right in saying that 1 is a number, while 0.999999... is not? 0.9recurring is shorthand for lim(x to infinity)(1-(1/x)) which in turn equals 1.
Therefore, to say 0.9recurring=1 does not make sense since one is just a shorthand for another expression, and the other is actually a number.

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02-01-2008 at 06:32 PM
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jbluestein
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Jacob wrote:
I'm no mathematician but, am I right in saying that 1 is a number, while 0.999999... is not? 0.9recurring is shorthand for lim(x to infinity)(1-(1/x)) which in turn equals 1.
Therefore, to say 0.9recurring=1 does not make sense since one is just a shorthand for another expression, and the other is actually a number.

I have probably lost all of my rigor, but I would say that 0.999... is still a number. It's not self-evidently a member of the set of natural numbers, but it's quite clearly a member of the set of real numbers. Not every real number can be written. (In fact, it's provably true that there are more that can't be written than can.) So inability to express a number in decimal notation doesn't invalidate its...numbness.

Josh

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02-01-2008 at 06:38 PM
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Nor does it invalidate its numberness. It might invalidate the numbness, though, I'm not sure. Did you leave your numbers out in the cold too long?

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02-01-2008 at 07:14 PM
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Sillyman wrote:
Nor does it invalidate its numberness. It might invalidate the numbness, though, I'm not sure. Did you leave your numbers out in the cold too long?

Right now, it's cold enough where I live that they could have become numb inside.



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02-01-2008 at 07:59 PM
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icon Re: 1=.9999999... (0)  
Syntax wrote:
The way I see it is equals represents equality.
0.999... does not equal 1 in my opinion.
This really, seriously makes my head hurt. In part, because opinion has nothing to do with it; it's like saying "mrimer is not Mike Rimer, in my opinion". And in part because the objection is grounded in the view that "equals represents equality", which I can't help reading as "X represents X". It's like the argument here is "1 can't equal .999..., because equal means equal, and they're not equal; therefore, they can't be equal."

I'd love for Syntax to clarify the sense in which he doesn't find these equal. Perhaps more concretely, for him to explain why "3-2 = 1" is true, but ".9999... = 1" is false.
02-02-2008 at 07:36 PM
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