# Concept of Quantity

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• 03-01-2011
User Name:
Quote:

Originally Posted by Mario F.
It follows on the same concept that oo does not equal oo, brewbuck. It's another of infinity paradoxes; also somewhat explored by Hippasus when he proved that the square root of 2 is both odd and even.

Infinity is not a number. .999... is. Therefore, this is irrelevant. I'm not sure how sqrt(2) relates, but I'll let it pass.

Quote:

Originally Posted by Mario F.
Conceptually you know very well that for every precision, no mater how large, that you can think for 0.(9), you need always to add another decimal place. You'll do this indefinitely. However, for 0.(9) to become 1 you need to add 0.(0)1. You are required an infinitesimal to reach 1. You are thus required another infinity.

You're considering finite expansions. Finite amounts are irrelevant, because .999... is an infinite expansion. The fact that the distance(amount you must add) between them is an infinitesimal means they are equal. Since, apparently, brewbuck's statement isn't enough, here's a link:
Quote:

Theorem 0.22: [The rationals are dense in the reals.]
Between any two distinct real numbers there is a rational number. (In fact,
there are infinitely many such rational numbers.)
Name ONE rational between .999... and 1. That's all I want. Anything else is intuition, and math is not based on intuition.

Quote:

Originally Posted by Mario F.
But I also believe your axiom to be wrong:

"Axioms" are true by definition. I would use "proof" or "statement" in this case, IMHO.

Quote:

Originally Posted by Mario F.
Let X = 0.999
Thus 10 * x = 9.99.
Thus 10 * x - x = 9.99 - 0.999
Oops!

Yeah, oops, we've reached the end of an infinite expansion and shifted it. Big oops.

Quote:

Originally Posted by Mario F.
But oo-oo != 0. It's undefined. It's both 0 and 1, or 12 and 13. Or any other number you can imagine. One of the properties of infinity is that it does not equal itself. Mathematically your expression becomes undefined at this point. The thought is that we know there is a number for your expression, but we can never reach it. As such it's undefined.

"indeterminate" is the word you're looking for, and 1-.999... isn't indeterminate. Again, .999... is a number, not infinity.

Quote:

Originally Posted by Mario F.
I suppose I can be corrected somewhere since my math skills are tremendously limited. But more than my gross attempt at correcting you, I think it stands that 0.(9) is the representation of a number that never reaches 1. On the right side of an interval, this number would be represented as "1[" and this is a crucial hint to the nature of this number and the fact it does not equal 1.

It never reaches anything, it's a constant. 1 doesn't reach 1, it is 1. Refer to the quote of the theorem above for a rigorous proof.
• 03-02-2011
whiteflags
Quote:

Infinity is not a number. .999... is. Therefore, this is irrelevant. I'm not sure how sqrt(2) relates, but I'll let it pass.
I tried to prove that 1 = 0.999... a different way after that post you replied to, and Mario responded to me with something about how 1/3 is irrational. That's how sqrt 2 came into the discussion, because in saying that 1/3 is rational, I wanted to give an example of an actually irrational number. Just something for Mario to think about because he couldn't think of any integer fractions a/b to express it. As quickly as it entered the discussion, sqrt 2 leaves it.

I knew what Mario wanted to say, so I responded to that as well.

Whether my proof is rigorous or not is another thing, but I work as well as I can.

BTW one of the benefits of being friends with me User Name is that PMs open up, so you can do that now.
• 03-02-2011
brewbuck
Quote:

Originally Posted by Mario F.
They don't really. Not for the "last" decimal place, represented by an infinitesimal. If we replace the above for a finite quantity it's easy to see why:

Let X = 0.999
Thus 10 * x = 9.99.
Thus 10 * x - x = 9.99 - 0.999
Oops!

I'm not sure I understand how there can be a last digit in an infinite string of digits. I also have heard that this is a common sort of misunderstanding when thinking about this problem (imagining that there can somehow be a "last" digit even if arbitrarily chosen).

How about a proof that 1 + 2 + 4 + 8 + 16 + ... is equal to -1?

X = 1 + 2 + 4 + 8 + 16 + ...
X = 1 + 2 * (1 + 2 + 4 + 8 + 16 + ...)
hey, the thing in parentheses is X again
X = 1 + 2 * X
X - 2 * X = 1
-X = 1
X = -1

There are such proofs for absurdities like "0 = 1" but they mostly involve on steps which are obviously problematic such as dividing by zero. There's nothing obviously wrong with the above... Or is there? ;)
• 03-02-2011
EVOEx
Quote:

Originally Posted by brewbuck
I'm not sure I understand how there can be a last digit in an infinite string of digits. I also have heard that this is a common sort of misunderstanding when thinking about this problem (imagining that there can somehow be a "last" digit even if arbitrarily chosen).

How about a proof that 1 + 2 + 4 + 8 + 16 + ... is equal to -1?

X = 1 + 2 + 4 + 8 + 16 + ...
X = 1 + 2 * (1 + 2 + 4 + 8 + 16 + ...)
hey, the thing in parentheses is X again
X = 1 + 2 * X
X - 2 * X = 1
-X = 1
X = -1

There are such proofs for absurdities like "0 = 1" but they mostly involve on steps which are obviously problematic such as dividing by zero. There's nothing obviously wrong with the above... Or is there? ;)

I can't actually find anything wrong in your proof that ISN'T done in the 0.9999... proof. However, I've worked on your proof, I find this one slightly more interesting, as it's easier and the result is even stranger:
Code:

```X = 1 + 1 + 1 + 1 + ... X = 1 + (1 + 1 + 1 + 1 + ...) X = 1 + X 0 = 1```
• 03-02-2011
ಠ_ಠ
Quote:

Originally Posted by EVOEx
I can't actually find anything wrong in your proof that ISN'T done in the 0.9999... proof. However, I've worked on your proof, I find this one slightly more interesting, as it's easier and the result is even stranger:
Code:

```X = 1 + 1 + 1 + 1 + ... X = 1 + (1 + 1 + 1 + 1 + ...) X = 1 + X 0 = 1```

why is that strange?
• 03-02-2011
EVOEx
Quote:

Originally Posted by ಠ_ಠ
why is that strange?

Because it means the world is square ;-).

Okay brewbuck, I've been working on my OWN proof that 0.999... = 1 (though there are many on wikipedia, but I'm sure about this one). Not using the same method that you used. Try to find something wrong with this:

Code:

```0.9999... is, per definition: 0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) (that's the definition of ..., agreed?) If so: 1 - lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) = lim(x -> inf) ( 1 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) = lim(x -> inf) ( 1*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) = lim(x -> inf) ( 1*10^-2 + 9*10^-3 + ... + 9*10^-x ) ... = lim(x -> inf) ( 1*10^-x ) = 0 So: 1 - 0.9999... = 0 1 = 0.9999...```
Is that a better proof for you?
• 03-02-2011
laserlight
Quote:

Originally Posted by whiteflags
But I see you're point: 0.333... is an approximation of 1/3, just as 0.999... would be of 1. As an approximation, it couldn't be exactly equal, but the difference is infinitesimally small, so by all means of proof it would be equal.

It is not an approximation; it is another representation. 0.333333333333, on the other hand, is an approximation of 1/3.
• 03-02-2011
Mario F.
Quote:

Originally Posted by User Name:
Name ONE rational between .999... and 1. That's all I want. Anything else is intuition, and math is not based on intuition.

How can I possibly name a rational, when the interval for which you require me to do so is bounded by irrationals? Explain to me how can I name a rational in [0.(9), 1[ and, more importantly even, how my physical inability to do so does in any way mean that 0.(9) is the same as 1. Do you deny irrationals? Why don't you ask me for an irrational when you know perfectly well that's the only number representation I can give you?

As for the axiom you introduced, I'm afraid at this point you are possibly taking advantage of my basic knowledge of mathematics. I'm pretty sure that it may be valid for some sort of calculations where the infinite string of decimals required by 0.(9) does not need to be factored in. But it cannot be valid here where the proposition 0.(9) == 1 is being discussed to an absolute certainty. And because we are discussing the infinite nature of rational numbers, you cannot honestly say 0.(9) equals one without introducing an error margin.

So, it's my turn to play cat and mouse with you: Name me a rational representation of 0.(9) where it becomes equal to 1 and the error margin is 0. Dishonest question, I know. But you made me one too.

EDIT: BTW, I took a long and closer look at the axioms you linked me to. And nowhere I can see where anything even resembling your proposed proof that 0.(9) == 1 is being discussed, particularly where it relates to this statement of yours: "The fact that the distance(amount you must add) between them is an infinitesimal means they are equal". This is a very strong thing to say. Nowhere it is being discussed there and never before I heard such a statement. It's particularly confusing, because those axioms reach a point where it is established that "Between any two distinct real numbers there is a rational number. (In fact, there are infinitely many such rational numbers.)". This pertains also to Cantor's proof that shown there are more irrationals in the Real set of numbers, than there are Rationals (another issue of differently sized infinities).

The question I pose you is then, given there's an infinite amount of rational numbers between any two real numbers, how does one represent one such rational number between irrational limits? Our inability to name such a number really means there isn't any? If there isn't (and then 0.(9) equals 1), how come there's in fact an infinite number of rationals between these two limits?
• 03-02-2011
Mario F.
Quote:

Originally Posted by brewbuck
I'm not sure I understand how there can be a last digit in an infinite string of digits. I also have heard that this is a common sort of misunderstanding when thinking about this problem (imagining that there can somehow be a "last" digit even if arbitrarily chosen).

Hence why I was careful by placing last between quotes. I was hoping you understood it wasn't meant to be talking literally. Instead, if you wish, "for an infinite number of digits there will always be one next digit that doesn't allow you to zero out the 9s like you did in your example".

This is all to do with the proportions of infinity. Let me give you an example with natural numbers:

Let's look at an infinite series of natural numbers; 1, 2, 3, 4, ....
Let's remove from that series all even numbers; 2, 4, 6, ...

We now have two infinite series. One with all odd numbers and one with all even numbers. It could be said that these two infinities are smaller than the infinity of natural numbers. But all are infinite. We will also have an hard time explaining which of the odd or even infinities is larger. One should be larger than the other because we took each number alternatively from an infinite series. For a given natural number placed on the upper limit, we can immediately gauge at the proportions. But without that possibility, we cannot say which 3 infinities are larger or smaller. Infinity does not have a proportion. And as such you cannot do what you proposed to do. For every 9 you zero out in 10*x-x, there will always be another 9 next in line needing to be zeroed out. 10*x and X are both infinities in this context. But they aren't of the same size.

You can pretend to treat them the same, but for the purpose of proving that 0.(9) equals 1, you failed for this reason: For every single rational x, up to infinity, your calculations will fail. So how do you propose your proof to be valid when it fails every single time, all the way to infinity? Is this a case of many wrongs do a right? ;)
• 03-02-2011
Yarin
Quote:

Originally Posted by whiteflags
0.333... is 1/3.

Is it? Or is 1/3 actually 0.(3) + 0.(0)1. (That is, it simply can't be accurately represented as a single decimal).
Just as 1/3 * 3 = 1 != 0.(9)... 0.(3) + 0.(0)1 = 1 != 0.(9).

How can 0.(9) be a number any more than infinity or 0.(0)1? I understand that infinity really can't be treated as a number without big problems, but the same seems to be so for 0.(9) as well. 1 - 0.(9) = 0.(0)1 and 1 / 0.(0)1 = oo.
If one were to accept the proofs as valid (at least, EVOEx's proof, even after reading up on limits I still don't understand them enough to get what I've seen of them here), then one should have to also accept the proof: oo + 197 = oo = oo + 2.(9), therefor 1 = 0.(9) and 197 = 2.(9). Which we know doesn't work.
• 03-02-2011
ಠ_ಠ
Quote:

Originally Posted by Yarin
Is it? Or is 1/3 actually 0.(3) + 0.(0)1. (That is, it simply can't be accurately represented as a single decimal).

please reread that statement and explain to me why you think it's not completely retarded.

• 03-02-2011
ಠ_ಠ
Quote:

Originally Posted by EVOEx
Because it means the world is square ;-).

no, it means that ∞ - ∞ = ∞, something your equation doesn't account for
• 03-02-2011
GReaper
Quote:

Originally Posted by ಠ_ಠ
no, it means that ∞ - ∞ = ∞, something your equation doesn't account for

But ∞ - ∞ = ?
• 03-02-2011
whiteflags
Quote:

Originally Posted by Yarin
Is it? Or is 1/3 actually 0.(3) + 0.(0)1. (That is, it simply can't be accurately represented as a single decimal).
Just as 1/3 * 3 = 1 != 0.(9)... 0.(3) + 0.(0)1 = 1 != 0.(9).

But it can be represented in decimal. It just takes infinite places.

0.0... is zero. 0.0... * 1 = 0, so I dunno what you mean to say but 0.333... + 0 is still 0.333...

It doesn't matter how many places you carry zero out to, it's going to be zero still. Unless I totally misunderstand what you mean by 0.(0)1 -- but I think if you think it means something else, then you are probably wrong.
• 03-02-2011
GReaper
Quote:

Originally Posted by whiteflags
Unless I totally misunderstand what you mean by 0.(0)1

He surely meant 0.(0)1 as 1/∞
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