# 1=0.999... infinities and box of chocolates..Phliosophy of Math...

From above:

Yes, a pizza can be divided equally into 2 pieces, or 4 pieces.

Is it possible, then, to divide a day equally into 2 pieces of equal duration?

If so, how long would each piece of the day be?

So, according to MD, an infinite series never reaches the limit that it tends toward... (which seems to be the point of this "debate").

I therefore say that MD put his money where his mouth is and stand against a wall to be shot at.
(yes, use a paint gun if you must )
Afterall, the bullet has to go half of the way. (1/2)
Then a further half of the remaining distance. (1/4)
Then a further half of the remaining distance. (1/8)
And so on.

If MD is right then the bullet will not reach him.
If he is wrong then... well... ouch.

But it should be interesting.

And before you dispute it, MD, the logic you are using in your argument above is exactly the same as should lead you to conclude that the bullet will never reach you.
To agree that the bullet will reach you, (as I sincerely hope you concede would be the case) you need to accept that the sum of an infinite series does indeed equal the number it tends towards.

Correct, I will never accept that 99.999...%=100%. Never!
There you have it. But the point of a conversation is not to just state your opinion, but to argue for it.
0.999... = 1 explicitly because of the properties given to real numbers by those that defined them (i.e. mathematicians).

If Motor Daddy were to speak of some other number system then he would be making progress. But if this system is strong enough to calculate x = 1 - 0.999... and x is not zero, then is x a number such that x + x > x and 1/x < 1 + 1/x ? That would be an interesting thing to see. But that requires Motor Daddy to have a principled argument about a concrete concept of number and not simple denial about theorems proved from existing definitions of the real numbers.

So, according to MD, an infinite series never reaches the limit that it tends toward... (which seems to be the point of this "debate").

I therefore say that MD put his money where his mouth is and stand against a wall to be shot at.
(yes, use a paint gun if you must )
Afterall, the bullet has to go half of the way. (1/2)
Then a further half of the remaining distance. (1/4)
Then a further half of the remaining distance. (1/8)
And so on.

If MD is right then the bullet will not reach him.
If he is wrong then... well... ouch.

But it should be interesting.

And before you dispute it, MD, the logic you are using in your argument above is exactly the same as should lead you to conclude that the bullet will never reach you.
To agree that the bullet will reach you, (as I sincerely hope you concede would be the case) you need to accept that the sum of an infinite series does indeed equal the number it tends towards.

Not only is your math garbage, but your idea of motion and locations and time and distance is all out of whack! You don't have a clue about what you are talking about! When you care to discuss math and physics with me again, then it will be your analysis of this. When you can demonstrate to me an understanding of those numbers and locations, and tell me how it works, then we'll talk, but not until then! K?

The same goes for everyone. I'm not gonna waste my time playing musical chairs with you. Show me that you are smart enough to understand what is done in that diagram. Show me that you're an honest person, and not some religious fanatic sticking to your pseudo scientific beliefs. SHOW ME!

There you have it. But the point of a conversation is not to just state your opinion, but to argue for it.
0.999... = 1 explicitly because of the properties given to real numbers by those that defined them (i.e. mathematicians).

If Motor Daddy were to speak of some other number system then he would be making progress. But if this system is strong enough to calculate x = 1 - 0.999... and x is not zero, then is x a number such that x + x > x and 1/x < 1 + 1/x ? That would be an interesting thing to see. But that requires Motor Daddy to have a principled argument about a concrete concept of number and not simple denial about theorems proved from existing definitions of the real numbers.

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Hansda, consider $$0.999... = \sum_{k\geq 1} \frac{9}{10^k}$$
If you agree that the left side is a number, then the right side must also be just a number -- that's what the equals sign means.
So what is the number? I'll pretend I don't know so I'll call it $$S$$.
$$0.999... = S = \sum_{k\geq 1} \frac{9}{10^k}$$
That didn't do much but first it saves some typing and it lets me write the family related quantities $$S_n = \sum_{k=1}^{n} \frac{9}{10^k}$$.
Now every n specifies how many terms we use before we stop the sum. $$S_1 = \frac{9}{10}, \; S_2 = \frac{9}{10} + \frac{9}{100}, \; S_3 = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000}, \; \dots$$
And for every n greater than or equal to 1, we can prove $$S_{n+1} = S_n + \frac{9}{10^{n+1}}$$ and $$\frac{9}{10^{n+1}} \gt 0$$ so $$S_{n+1} \gt S_n$$.
So presumably $$S_1 \lt S_2 \lt S_3 \lt \dots \lt S_{n+5}\lt S_{n+6}\lt S_{n+7} \lt \dots \lt S$$ and indeed this is provable from the rules of inequalities.
So any member of the family of $$S_n$$ is less than $$S$$, in other words each member is a lower bound on what the value of $$S$$ could be. But none is the greatest possible lower bound on what $$S$$ could be.

Here $$S_n = \sum_{k=1}^{n} \frac{9}{10^k}$$. Consider $$S_n = T_1 + T_2 + T_3 + ... + T_n$$. So the n-th term in this series will be $$T_n = \frac{9}{10^n}$$; where $$T_1 = \frac{9}{10^1}$$, $$T_2 = \frac{9}{10^2}$$, $$T_3 = \frac{9}{10^3}$$ ...and so on.

$$S_n$$ gets ugly to compute the hard way -- imagine the pain if n = 1000. And yet most of all counting numbers are actually larger than 1000. So I want to prove that:
$$S_n = \frac{9}{10} \times \frac{1 - 10^{-n}}{1 - 10^{-1}}$$ for all counting numbers, n.​
First let me prove that $$S_1 = \frac{9}{10} \times \frac{1 - 10^{-1}}{1 - 10^{-1}}$$. The proof writes itself: $$S_1 = \frac{9}{10} = \frac{9}{10} \times 1 = \frac{9}{10} \times \frac{1 - 10^{-1}}{1 - 10^{-1}}$$.
Next, I want to prove that if for a particular counting number k, that $$S_k = \frac{9}{10} \times \frac{1 - 10^{-k}}{1 - 10^{-1}}$$ is true, that $$S_{k+1} = \frac{9}{10} \times \frac{1 - 10^{-(k+1)}}{1 - 10^{-1}}$$ follows as a result. So from the definition of the family, I know that $$S_{k+1} = S_k + \frac{9}{10^{k+1}}$$ and from assumption: $$S_{k+1} = \frac{9}{10} \times \frac{1 - 10^{-k}}{1 - 10^{-1}} \, + \frac{9}{10^{k+1}}$$, So we have $$S_{k+1} = \frac{9}{10} \times \frac{1 - 10^{-k}}{1 - 10^{-1}} \, + \; \frac{9}{10} \times \frac{1}{10^k} = \frac{9}{10} \times \left( \frac{1 - 10^{-k}}{1 - 10^{-1}} + 10^{-k} \right) = \frac{9}{10} \times \left( \frac{1 - 10^{-k}}{1 - 10^{-1}} + \frac{10^{-k} - 10^{-k-1}}{1 - 10^{-1}} \right) = \frac{9}{10} \times \frac{1 - 10^{-k} + 10^{-k} - 10^{-(k+1)}}{1 - 10^{-1}} = \frac{9}{10} \times \frac{1 - 10^{-(k+1)}}{1 - 10^{-1}}$$ which is what I wanted to prove.
Using the principle of finite induction, I now know that $$S_n = \frac{9}{10} \times \frac{1 - 10^{-n}}{1 - 10^{-1}}$$ is true for all counting numbers, n.
But because $$\frac{9}{10} = 1 - 10^{-1}$$ this also proves that $$S_n = 1 - 10^{-n}$$ for all counting numbers, $$n$$.

Thus our lower bounds on what $$S$$ can be looks like this:
$$1 - 10^{-1} \, \lt \, 1 - 10^{-2} \, \lt \, 1 - 10^{-3} \, \lt \dots \lt \,1 - 10^{-(n+5)} \, \lt \, 1 - 10^{-(n+6)} \, \lt \, 1 - 10^{-(n+7)} \, \lt \dots \lt S$$

Now I want to prove that if $$X$$ is less than 1 then there are members of the family $$S_n$$ which are larger than it.
Say $$X \lt 1$$. Then $$0 \lt 1 - X$$. Then $$1 - X$$ is a positive number. Then there is a real number $$\epsilon$$ such that $$e^{-\epsilon} = 1 - X$$ because $$f(x) = e^x$$ is a continuous function takes on all positive real values. Also $$f(x \ln 10) = e^{x \ln 10} = 10^x$$. So $$X = 1 - e^{-\epsilon} = 1 - 10^{- \frac{\epsilon}{\ln 10}}$$. So the question of if there are members of the family $$S_n$$ which are greater than $$X$$ reduces to the question of asking if there are counting numbers which are larger than the real number $$- \frac{\epsilon}{\ln 10} = - \frac{\ln ( 1 - X) }{\ln 10}$$ and the answer is yes, most of them.
So every number less than 1 is less than some (or all!) members of the family $$S_n$$ and thus less than $$S$$.

Thus 1 is greatest lower bound on what S could be, because all numbers less than 1 fail to qualify.

One of the biggest principles of the real numbers is that if a non-empty set of numbers has a upper (or lower!) bound then it has a least upper bound (or a greatest lower bound). By this, we see that the set of positive rational numbers whose square is less than 2 has greatest lower bound of 0 and a least upper bound of √2. 0 is not positive and √2 is not rational, but they are the minimal bounds of that particular set of positive rationals. That's why $$S$$ which is the least upper bound of the family of $$S_n$$ need not be a member of the family and can be 1.

Likewise $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} (1 - 10^{-n}) = 1 - \lim_{n \to \infty} 10^{-n} = 1$$ provides a concise line of reasoning for what S is.

Here you are considering $$\lim_{n\to \infty} 10^{-n} = 0$$. In that case the term $$T_n = 9 \times 10^{-n} = 0$$.

This violates the infinity nature of the series.

Not only is your math garbage, but your idea of motion and locations and time and distance is all out of whack! You don't have a clue about what you are talking about! When you care to discuss math and physics with me again, then it will be your analysis of this. When you can demonstrate to me an understanding of those numbers and locations, and tell me how it works, then we'll talk, but not until then! K?
How does that link have any bearing on your inability to understand that 0.999... = 1?
:shrug:
How is my idea of "motion and locations and time and distance" all out of whack?
If you disagree with my assessment of your position, do show where you disagree, as what I have posted is a logical conclusion of your argument.
Disagree with that conclusion and you disagree with your own argument.

How does that link have any bearing on your inability to understand that 0.999... = 1?
:shrug:

It shows how you avoid the issue at all cost because you know you're wrong. It shows, 100% that you are a dishonest person! Prove me wrong!

Oh, if you couldn't understand my french like I can't understand rpenner's wall of cool symbols then I'll spell it out a little more clearly for you:

There is not 1 point in the entire infinite volume of space that can escape this diagram. Not 1! Not 1 point! Can you imagine that??

But there is no last digit, if there was the sequence would be finite instead of infinite.
Consider, 0.999... = 0.9 + 0.09 + 0.009 + 0.0009 + 0.00009 + ...... upto infinite terms.
Can you tell me, which term in the RHS above is not having the last digit as 9?
Hansda, consider $$0.999... = \sum_{k\geq 1} \frac{9}{10^k}$$
If you agree that the left side is a number, then the right side must also be just a number -- that's what the equals sign means.
So what is the number? I'll pretend I don't know so I'll call it $$S$$.
$$0.999... = S = \sum_{k\geq 1} \frac{9}{10^k}$$
That didn't do much but first it saves some typing and it lets me write the family related quantities $$S_n = \sum_{k=1}^{n} \frac{9}{10^k}$$.
Here $$S_n = \sum_{k=1}^{n} \frac{9}{10^k}$$. Consider $$S_n = T_1 + T_2 + T_3 + ... + T_n$$. So the n-th term in this series will be $$T_n = \frac{9}{10^n}$$; where $$T_1 = \frac{9}{10^1}$$, $$T_2 = \frac{9}{10^2}$$, $$T_3 = \frac{9}{10^3}$$ ...and so on.

For what purpose do you define the family $$T_n$$? $$S_n$$ exists so I can more easily say "for every counting number, n, the statement $$S_n \lt S$$ is true" without the clutter of a bunch of complicated symbols that are subject to typos.

But what is the special utility of $$T_n$$ (as $$\frac{9}{10^n}$$ is pretty concise and typo-resistant)?

$$S_n = \frac{9}{10} \times \frac{1 - 10^{-n}}{1 - 10^{-1}}$$ for all counting numbers, n.​
...
But because $$\frac{9}{10} = 1 - 10^{-1}$$ this also proves that $$S_n = 1 - 10^{-n}$$ for all counting numbers, $$n$$.

Another thing that is true is for every counting number, n, is "$$S_n + \frac{1}{9} T_n = 1$$."

So every number less than 1 is less than some (or all!) members of the family $$S_n$$ and thus less than $$S$$.
Thus 1 is greatest lower bound on what S could be, because all numbers less than 1 fail to qualify.
In other words, any positive number is more than some (or all!) members of the family $$T_n$$.
Likewise $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} (1 - 10^{-n}) = 1 - \lim_{n \to \infty} 10^{-n} = 1$$ provides a concise line of reasoning for what S is.

Here you are considering $$\lim_{n\to \infty} 10^{-n} = 0$$.
Yes, that is true in analysis and non-standard analysis (a subject which I recently read up on).

In that case the term $$T_n = 9 \times 10^{-n} = 0$$.
No, because $$T_n$$ isn't a number, it's a family of numbers and doesn't have a specific value until you specify what the counting number n is. By itself, $$T_n$$ is just a convenient way to talk about the whole family or some abstractly specified member of the family.

An there is no value of n which makes $$T_n = 9 \times 10^{-n} = 0$$ true, because $$T_n$$ is only defined for the counting numbers.
There are an infinite number of members of that family, one for each of the counting numbers, but all are positive, which is to say, greater than zero.

What is correct to write is $$\lim_{n \to \infty} T_n = 0$$ which builds on the previous observation that "any positive number is more than some (or all!) members of the family $$T_n$$. " -- Specifically it is saying for any positive number, there is a finite number, m, so that every member of the family $$T_n$$ where $$n \gt m$$ is closer to 0 than that chosen positive number. And since no matter how large m is, most counting number are larger than m, this means no matter how much you "zoom in" on 0 with a microscope, there are still an infinite number of the family of $$T_n$$ close to it.

Limits and suprema are two topics of analysis that follow from the properties of real numbers which in turn follow from the basic geometrical ideas that Euclid wrote about.

This violates the infinity nature of the series.
I don't know what this means, but I think you haven't yet understood the notation and reasoning of my earlier post. No member of the $$S_n$$ family is equal to a sum over more than a finite number of terms. In short, $$S_n = \sum_{k=1}^n T_k$$. However $$S$$ is equal to a sum over all members of the $$T_n$$ family and thus equal to a sum over more than any finite number of terms. $$S = \sum_{k \geq 1} T_k = \sum_{n \in \mathbb{N}} T_n = \lim_{n\to \infty} S_n = 1 - \lim_{n\to \infty} 10^{-n} = 1$$. But $$S$$, just like 0.999... and 1, is just a name for a number.

And it is the properties of numbers (specifically the completeness of the real numbers and the property of the counting numbers to go on forever) that are the main points of disagreement in this thread, even though these are mathematical properties that follow directly from the relevant definitions.

It shows how you avoid the issue at all cost because you know you're wrong. It shows, 100% that you are a dishonest person! Prove me wrong!

Oh, if you couldn't understand my french like I can't understand rpenner's wall of cool symbols then I'll spell it out a little more clearly for you:

There is not 1 point in the entire infinite volume of space that can escape this diagram. Not 1! Not 1 point! Can you imagine that??
Maybe it's 'cos I only joined in this fun a few pages okay, but wtf are you on about?
As far as I could tell this was about whether 0.999... = 1 or not.
So I have no idea, no matter how much you rant, about what you are referring to with the link to your diagram.
And I see no link to the matter of whether 0.999... = 1 or not.

arfa brane said:
But there is no last digit, if there was the sequence would be finite instead of infinite.
hansda said:
Consider, 0.999... = 0.9 + 0.09 + 0.009 + 0.0009 + 0.00009 + ...... upto infinite terms.
Can you tell me, which term in the RHS above is not having the last digit as 9?
Every term is a finite number and has a last digit, but there are infinite terms. So there is no last term on the RHS, and no last digit on the LHS.
You don't have to count them all or add them all together; the infinity of terms follows by induction.

And what else follows is that some people "have difficulty" abstracting. Some people just don't get it.
It's interesting how it doesn't seem to matter how many times you show their ideas lead to contradictions, they cling to them anyway.
What is that, philosophically speaking?

So I have no idea, no matter how much you rant, about what you are referring to with the link to your diagram.

Right. That's what I said, you were clueless. I'm glad you acknowledge that. Now, the real question is, are you intelligent/honest enough to give your opinion on the diagram? I'm simply asking you to comment on the diagram in a constructive way. Tell me where I go wrong. Please!

Right. That's what I said, you were clueless.
Clueless as to why you suddenly started referring to a diagram that had not been raised in my discussion thus far, yes.
Clueless with regard 10 = 9.999...? No, we leave that for you.
Now, the real question is, are you intelligent/honest enough to give your opinion on the diagram? I'm simply asking you to comment on the diagram in a constructive way. Tell me where I go wrong. Please!
Honesty has nothing to do with it.
And this is the first time you have actually asked anyone, or at least me (I may have missed where you asked someone else), to look at the diagram.
Previously you just said that the next time you talk maths with me is when I have analysed that diagram.
That's not really a request to comment on it.

Why don't you raise another thread for it if you want people to comment on it?
This thread seems to be about whether or not people can accept that 0.999... = 1.

Clueless as to why you suddenly started referring to a diagram that had not been raised in my discussion thus far, yes.
Clueless with regard 10 = 9.999...? No, we leave that for you.
Honesty has nothing to do with it.
And this is the first time you have actually asked anyone, or at least me (I may have missed where you asked someone else), to look at the diagram.
Previously you just said that the next time you talk maths with me is when I have analysed that diagram.
That's not really a request to comment on it.

Why don't you raise another thread for it if you want people to comment on it?
This thread seems to be about whether or not people can accept that 0.999... = 1.

Blah, Blah, Blah! Where is your analysis? Either admit you don't know what you're talking about, or tell me where I go wrong. Don't come here talking pseudo trash about how you think your incomplete pizza is the same as my complete pizza. Non-sense! Don't you know the difference between a whole and a fraction of a whole? Or are you saying a fraction of a whole is equal to a whole? Just what the heck are you saying, other than you think your incomplete pizza is the same as my complete pizza?

So I have no idea, no matter how much you rant, about what you are referring to with the link to your diagram.
As a service to this thread, I have created a place to discuss Motor Daddy's diagram.

That seems to be a common idea: you can't just have a number, it has to be something real like a pizza, or a piece of paper.

Ergo, since you can't divide a pizza infinitely, you can't do it with numbers either.
But once again, if physics is mathematical, the converse is not true (since that's a case of affirming the consequent, which is a logical error). Numbers certainly are not physical, and when we want to represent them, their "information" is necessarily something physical and observable. To compute with numbers, they must have some encoding in space and time.

Divide a pizza made of 100 atoms by 3 equal pieces?

Just like if I try divide $100 in 3 equal pieces.$100 / 3 =
A1 = $33 A2 =$33
A3 = $33 and we have a remainder of$1 so we divide again...

$1 / 3 = A1 =$33.33
A2 = $33.33 A3 =$33.33
and we have a remainder of 1c so we divide again...

1c / 3 =
A1 = $33.333 A2 =$33.333
A3 = $33.333 and we have a remainder of 0.10c so we divide again... etc... There is infinitely a remainder and thats why$100 / 3 = $33.333... even though the ... signifies an infinity there is always still a remainder that can be divided.$33.333 x 3 = $99.999$33.333... x 3 = $99.999... The operation is not complete because it doesn't =$100

If 0.999... = 1 then
0.888... = ?
0.777... = ?
0.666....= ?
0.555... = ?
etc???

fill in the ?'s

If you say that 0.999... = 1 then you are saying
0.888... = 0.888...9
0777... = 0.777...8
0.666... = 0.666...7
0.555... = 0.555...6
etc...

If you say that 0.999... = 1 then you are saying
0.888... = 0.888...9
0777... = 0.777...8
0.666... = 0.666...7
0.555... = 0.555...6
etc...
that's an interesting thought!