## Is 1 = 0.999... ? Really?

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## Is it true that 1 = 0.999...? And Exactly Why or Why Not?

Yes, 1 = 0.999...
12
32%
No, 1 ≠ 0.999...
22
59%
Other
3
8%

### Re: Is 1 = 0.999... ? Really?

I disagree and I understand.

The following is an argument against your claim:

The Initial Argument
(Version 3)

1) Let $$X$$ be a class of decimal / base-10 numerals that have no fractional part and that stand for a positive number.

2) Any two decimal expressions that are instances of $$X$$ represent one and the same number if and only if every explicit and implicit digit in one expression is the same as the explicit or implicit digit located at the same place in the other expression.

3) $$999\dotso$$ is an instance of $$X$$.

4) $$999\dotso$$ has no explicit digit associated with $$10^0$$ but it has an implicit one and that digit is $$0$$.

5) The decimal expression of $$\sum_{i=0}^{i->\infty} 9\times10^i$$ is an instance of $$X$$.

6) The decimal expression of $$\sum_{i=0}^{i->\infty} 9\times10^i$$ has an explicit digit associated with $$10^0$$ and that digit is $$9$$.

7) Therefore, $$999\dotso$$ and $$\sum_{i=0}^{i->\infty} 9\times10^i$$ represent two different numbers.
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:4) $$999\dotso$$ has no explicit digit associated with $$10^0$$ but it has an implicit one and that digit is $$0$$.

Disagree.

The representation 999... refers only to the digit "9" being repeated endlessly.
There is no $$0$$ digit involved in 999... implicit or otherwise.
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### Re: Is 1 = 0.999... ? Really?

Alright, you stated that you disagree with (4).

Are you going to present an argument against (4)?

Or perhaps you think that this . . .

The representation 999... refers only to the digit "9" being repeated endlessly.
There is no $$0$$ digit involved in 999... implicit or otherwise.

. . . is an argument against (4)?
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:Alright, you stated that you disagree with (4).

Are you going to present an argument against (4)?

Or perhaps you think that this . . .

The representation 999... refers only to the digit "9" being repeated endlessly.
There is no $$0$$ digit involved in 999... implicit or otherwise.

. . . is an argument against (4)?

How else can we argue that something isn't present other than to simply state that it isn't - implying that if you still contend that it is there - show it.
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### Re: Is 1 = 0.999... ? Really?

Are you saying that proving a negative is an impossible task?

I thought you don't subscribe to such popular but nonetheless erroneous ideas.
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:Are you saying that proving a negative is an impossible task?

I thought you don't subscribe to such popular but nonetheless erroneous ideas.

I didn't say that it is impossible. I said that when someone simply says that something is there and we do not see it there - there is nothing to say but - "It is not there - so show us why you think it is there".

If you insist that it is there without being able to provide proving evidence - we jump down on the chart to a categorical disagreement assignment ("Responder Expires" on the next to last flowchart line) - meaning a new class of people who can choose whichever branch of the debate they prefer to believe. Future debates can take it up from there.

Alternatively the responder can offer a new argument, expiring only that one argument first offered but can offer a different one in a similar manner as before (looping back up toward the top of the chart).
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### Re: Is 1 = 0.999... ? Really?

obsrvr524 wrote:I didn't say that it is impossible. I said that when someone simply says that something is there and we do not see it there - there is nothing to say but - "It is not there - so show us why you think it is there".

How about presenting an argument that concludes with "Therefore, it is not there"?

If you insist that it is there without being able to provide proving evidence - we jump down on the chart to a categorical disagreement assignment ("Responder Expires" on the next to last flowchart line) - meaning a new class of people who can choose whichever branch of the debate they prefer to believe. Future debates can take it up from there.

So you think the burden of proof is entirely on responders? Responders have to prove the proposer wrong whereas the proposer merely has to propose ideas and accept/reject counter-arguments he is offered?

One way to argue against (4) is by arguing in favor of your belief that the digit associated with $$10^0$$ in $$999\dotso$$ is $$9$$.
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:
obsrvr524 wrote:I didn't say that it is impossible. I said that when someone simply says that something is there and we do not see it there - there is nothing to say but - "It is not there - so show us why you think it is there".

How about presenting an argument that concludes with "Therefore, it is not there"?

Ok -

• If there was an implicit 0 digit there, I would see it
• I don't see an implicit 0 digit indicated.
• Therefore there is no implicit 0 digit present.

That is my evidentiary argument.

It is not an issue of finding logic flaws when dealing with a premise. Premises must be agreed to by both parties. If the responder cannot change the proposer's mind or vise versa, the argument ends there. A new argument must be presented by the responder.

Magnus Anderson wrote:One way to argue against (4) is by arguing in favor of your belief that the digit associated with $$10^0$$ in $$999\dotso$$ is $$9$$.

You haven't proposed that is what you are talking about. If it is, then that should be your response. Otherwise your statement that a 0 digit is implied is simply not agreeable.
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### Re: Is 1 = 0.999... ? Really?

I've realized something. (4) is wrong though not entirely wrong. It is possible for the digit associated with $$10^0$$ in $$999\dotso$$ to be $$9$$, it is merely not necessarily so. The key insight is that the position of the first digit in $$999\dotso$$ is not specified. It can be literally any. (I think it was phyllo who hinted at something similar in the past.)

For example, if the index of the first digit is $$0$$, which means that its weight is $$10^0$$, then the digit associated with $$10^0$$ is $$9$$. The only problem is that it is not necessarily so. For example, the index of the first digit can also be $$infA^2$$, in which case, if the number of $$9$$s is $$infA$$, the digit associated with $$10^0$$ would be $$0$$ since the endless chain of $$9$$s would never reach that position.

(There's another problem too. If the weight of the first digit is $$10^0$$ then what is the weight of the next $$9$$? There must be "the next $$9$$" since we're dealing with an endless series of $$9$$s in the direction of the least significant digit. It should be $$10^{-1}$$ but that would mean that $$999\dotso$$ has a fractional part which we have previously agreed that it does not have. Thus, in such a case, the expression would represent a conceptually impossible number. And even if it were not a conceptually impossible number, it still wouldn't be $$\dotso999$$ simply because the digit associated with $$10^{-1}$$ in $$\dots999$$ is $$0$$.)

With all that said, I have to revise my argument. Like so:

The Initial Argument
(Version 4)

1) Let $$X$$ be a class of decimal / base-10 numerals that have no fractional part and that stand for a positive number.

2) Any two decimal expressions that are instances of $$X$$ represent one and the same number if and only if every explicit and implicit digit in one expression is the same as the explicit or implicit digit located at the same place in the other expression.

3) $$999\dotso$$ is an instance of $$X$$.

4) $$999\dotso$$ has no explicit digit associated with $$10^0$$ but it has an implicit one and that digit can be $$0$$.

5) The decimal expression of $$\sum_{i=0}^{i->\infty} 9\times10^i$$ is an instance of $$X$$.

6) The decimal expression of $$\sum_{i=0}^{i->\infty} 9\times10^i$$ has an explicit digit associated with $$10^0$$ and that digit is $$9$$.

7) Therefore, $$999\dotso$$ and $$\sum_{i=0}^{i->\infty} 9\times10^i$$ represent two different numbers.
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### Re: Is 1 = 0.999... ? Really?

Certainly real wrote:Long-division algorithm is a term I am hearing for the first time. I'm also pretty sure we were never taught exactly why a fraction like 1/3 amounts to .333... . I'm from the UK. Maybe this is something they don't teach us here, or maybe it's just something they didn't teach me at my high school. Alternatively, they taught me and I forgot (sorry if this is true).

Probably no longer taught. When I'm made emperor, first thing I'm going to do is send all the math education bureaucrats to a reeducation camp. Math education is a disaster. Not the students' fault.

Certainly real wrote:Given your use of finite in the above quote, the only way I can see this hold, is if we said that pi is also a finite number.

Of course pi is finite. It's a real number between 3 and 4 on the number line.

Code: Select all
0   1   2   3   4   5  ...
^
|
pi

You are right that the decimal representation of pi has infinitely many digits. But that doesn't make pi infinite. In fact pi only contains a finite amount of information. For example, the famous Leibniz series for pi is

$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{3} + \dots$$

This formula was discovered by Leibniz in 1676 and before that by the Indian mathematician Madhava of Sangamagrama in the 14th century.

It can be written in finite form as

$$\displaystyle \pi = 4 \sum_{n \in \mathbb N} \frac{(-1)^n}{2k + 1}$$.

Note that I completely expressed pi using only 15 characters. In other words if I wanted to completely transmit all the digits of pi to a friend on Mars, I would not have to send infinitely many digits; I'd only have to send the 15 character formula and they could use that to calculate as many digits as they want.

Certainly real wrote:But according to my quick session of google, pi is not a finite number.

Google says nothing of the kind. Pi is a finite number, it's a real number between 3 and 4.

Certainly real wrote: But I think I get where you're coming from and I still think it semantically inconsistent to say the nature of .333... is finite and yet equal to 1/3 at the same time.

Well 1/3 is exactly expressed with only three characters. Again you are confusing a number with its representation.

Certainly real wrote:I understand the move behind saying when you have an infinity of 3s after 0. , you have 1/3.

If you prefer, and I explained this several times in my previous post and urge you to slow down and comprehend this point: I do not need to talk about infinitely many 3's at all, and from now on I won't do that anymore.

What I am saying is that I can get AS CLOSE AS I WANT to 1/3 by taking a large enough FINITE number of 3's, such as 0.333333333 or 0.3333333333333333. I never have to talk about infinitely many 3's and from now on I won't do so.

Certainly real wrote:I am questioning whether this is a semantically justified move.

It's not one that I'm making. I am not talking about infinitely many 3's at all. I already explained that several times. I am asking you to slow down and focus on what I am saying to you. I am NOT talking about infinitely many 3's.

Certainly real wrote:Indeed, if we could have an infinity of 3s,

I am no longer talking about that because it's more precise to talk about arbitrary closeness of finitely many 3's.

Certainly real wrote: it would follow from this that we have absolutely hit 1/3. It’s complete, it’s finished, so 1/3 has been hit.

1/3 is never 'hit." Rather, we can get as close as we want to 1/3 by taking a sufficiently large but still FINITE number of 3's.

I am asking you to slow down and engage with what I'm saying.

Certainly real wrote:Yes I do. However, changing to base 3 to say 1/3 = 0.1, to me, is just a different way of representing the semantic of 1/3 of 0.3. It doesn't really change anything semantically as far as I can see, and my issue here is one in terms of semantics, not representation.

Well, 0.1$$_3$$ is as finite a representation as can be. Why don't you take that to heart?
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:1) Let $$X$$ be a class of decimal / base-10 numerals that have no fractional part and that stand for a positive number.

2) ... (not relevant here)

3) $$999\dotso$$ is an instance of $$X$$.

Not so. Any instance of X contains only finitely many digits. That's because the natural numbers are generated by starting with 0 and repeatedly adding 1. Every number so generated has only finitely many digits.

Once you stipulate that your expressions "stand for a positive number," you are limited to finite-length strings of decimal digits. Those are the rules for positive numbers.

Of course you can make up your own notation if you can make sense of it (you haven't yet done so) but you can't call them positive numbers as if they're natural numbers. They're not.
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:I've realized something. (4) is wrong though not entirely wrong. It is possible for the digit associated with $$10^0$$ in $$999\dotso$$ to be $$9$$, it is merely not necessarily so. The key insight is that the position of the first digit in $$999\dotso$$ is not specified. It can be literally any. (I think it was phyllo who hinted at something similar in the past.)

I think the left most digit is necessarily indexed at $$infA$$ because of your "...". But that is another issue.

Magnus Anderson wrote:For example, if the index of the first digit is $$0$$, which means that its weight is $$10^0$$, then the digit associated with $$10^0$$ is $$9$$. The only problem is that it is not necessarily so. For example, the index of the first digit can also be $$infA^2$$, in which case, if the number of $$9$$s is $$infA$$, the digit associated with $$10^0$$ would be $$0$$ since the endless chain of $$9$$s would never reach that position.

Your index $$i$$ cannot exceed $$infA$$ because it has to remain countable - the natural numbers. The "..." indicates a countable natural number series only.

Magnus Anderson wrote:(There's another problem too. If the weight of the first digit is $$10^0$$ then what is the weight of the next $$9$$? There must be "the next $$9$$" since we're dealing with an endless series of $$9$$s in the direction of the least significant digit. It should be $$10^{-1}$$ but that would mean that $$999\dotso$$ has a fractional part which we have previously agreed that it does not have. Thus, in such a case, the expression would represent a conceptually impossible number. And even if it were not a conceptually impossible number, it still wouldn't be $$\dotso999$$ simply because

Again the "..." is the natural numbers - not negative. So your index could not become negative even if it began at $$infA$$ counting down.

Magnus Anderson wrote:With all that said, I have to revise my argument. Like so:

The Initial Argument
(Version 4)

1) Let $$X$$ be a class of decimal / base-10 numerals that have no fractional part and that stand for a positive number.

2) Any two decimal expressions that are instances of $$X$$ represent one and the same number if and only if every explicit and implicit digit in one expression is the same as the explicit or implicit digit located at the same place in the other expression.

3) $$999\dotso$$ is an instance of $$X$$.

4) $$999\dotso$$ has no explicit digit associated with $$10^0$$ but it has an implicit one and that digit can be $$0$$.

So I have to still disagree with (4).

I still cannot identify an implicit 0 digit possible anywhere (other than your index $$i$$).

Want to go for version 5?
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### Re: Is 1 = 0.999... ? Really?

Your index i cannot exceed infA because it has to remain countable - the natural numbers. The "..." indicates a countable natural number series only.
Isn't this a fundamental problem?

If infA is infinite, then it's not countable.

If infA is countable, then it's a natural number - not infinite.
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### Re: Is 1 = 0.999... ? Really?

phyllo wrote:
Your index i cannot exceed infA because it has to remain countable - the natural numbers. The "..." indicates a countable natural number series only.
Isn't this a fundamental problem?

If infA is infinite, then it's not countable.

If infA is countable, then it's a natural number - not infinite.

Since $$infA$$ is an infinity, the index certainly can't count beyond it.
$$infA$$ as an index is a different issue that we haven't got to yet.
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### Re: Is 1 = 0.999... ? Really?

Since infA is an infinity, the index certainly can't count beyond it.
If it's an infinity, then you also can't count to it.

And you also can't count almost to it ... ie infA-1 would seem to be logically nonsensical.

That's why standard math uses the concept of limits.
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### Re: Is 1 = 0.999... ? Really?

phyllo wrote:
Since infA is an infinity, the index certainly can't count beyond it.
If it's an infinity, then you also can't count to it.

And you also can't count almost to it ... ie infA-1 would seem to be logically nonsensical.

That's why standard math uses the concept of limits.

We haven't said anything about counting to it.
Every index less than $$infA$$ is "countable".
So "$$infA-1$$" is technically countable - just not identifiable.
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### Re: Is 1 = 0.999... ? Really?

obsrvr524 wrote:So I have to still disagree with (4).

I still cannot identify an implicit 0 digit possible anywhere

There are implicit $$0$$s everywhere. For example, there's an implicit digit right before the first digit in $$99\dot9$$. That digit is $$0$$. There is also an implicit digit at $$-1$$ (which is the index of the first digit after the decimal point.) That digit is also $$0$$.

I assume what you're saying here is that you do not think that the digit associated with $$10^0$$ is $$0$$ but rather $$9$$.

Can I assume that the following argument of yours is applicable to Version 4 of my argument?

https://www.ilovephilosophy.com/viewtop ... 0#p2804739

obsrvr524 wrote:
• If there was an implicit 0 digit there, I would see it
• I don't see an implicit 0 digit indicated.
• Therefore there is no implicit 0 digit present.

You said that you still disagree with (4), so it should be applicable.

If it's applicable, I am ready to present my agreements/disagreements as well as my counter-argument.
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:
obsrvr524 wrote:So I have to still disagree with (4).

I still cannot identify an implicit 0 digit possible anywhere

There are implicit $$0$$s everywhere. For example, there's an implicit digit right before the first digit in $$99\dot9$$. That digit is $$0$$. There is also an implicit digit at $$-1$$ (which is the index of the first digit after the decimal point.) That digit is also $$0$$.

But your index can't reach either of those positions - so they are not implied.
1) You restricted the number to be non-fractional and also
2) your "..." restricts you from exceeding your upper "9" to any supposed preceding "0" (which wouldn't be relevant anyway).

Magnus Anderson wrote:I assume what you're saying here is that you do not think that the digit associated with $$10^0$$ is $$0$$ but rather $$9$$.

3) Certainly -- $$(9x10^0) = 9$$.

Magnus Anderson wrote:Can I assume that the following argument of yours is applicable to Version 4 of my argument?

https://www.ilovephilosophy.com/viewtop ... 0#p2804739

obsrvr524 wrote:
• If there was an implicit 0 digit there, I would see it
• I don't see an implicit 0 digit indicated.
• Therefore there is no implicit 0 digit present.

You said that you still disagree with (4), so it should be applicable.

Yes. It is still applicable - along with what I just stated (1-3).

Magnus Anderson wrote:If it's applicable, I am ready to present my agreements/disagreements as well as counter-argument.

Do it.
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### Re: Is 1 = 0.999... ? Really?

So "infA−1" is technically countable - just not identifiable.
Can you explain this distinction?
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### Re: Is 1 = 0.999... ? Really?

phyllo wrote:
So "infA−1" is technically countable - just not identifiable.
Can you explain this distinction?

I thought I just did that.

We cannot identify $$infA-1$$ because we cannot know what number or position it would be other than not quite being at infinity (the numerically non existent end of the endless). It is something less - but what we can't tell. And it is technically "countable" merely because it is less than the total number of natural numbers.
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### Re: Is 1 = 0.999... ? Really?

That doesn't explain why you think it's "technically countable".

If you can count it, then you can identify it. Countable and identifiable seem to go together.
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### Re: Is 1 = 0.999... ? Really?

phyllo wrote:That doesn't explain why you think it's "technically countable".

If you can count it, then you can identify it. Countable and identifiable seem to go together.

Wikipedia wrote:Countable
In mathematics, a countable set is a set with the same cardinality as some subset of the set of natural numbers. A countable set is either a finite set or a countably infinite set.
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### Re: Is 1 = 0.999... ? Really?

The text in red is inserted by me.

obsrvr524 wrote:
• If there was an implicit 0 digit there, I would see it
(Disagree.)
• I don't see an implicit 0 digit indicated.
(Agree.)
• Therefore there is no implicit 0 digit present.
(Disagree, of course.)

I agree that the conclusion follows from the premises and I agree with the second premise. What I disagree with is the first premise. That said, it is up to me now to present an argument against it. Wait for it

1) What is there is not necessarily seen.

2) Therefore, if there was a $$0$$ associated with $$10^0$$ in $$99\dot9$$, obsrvr524 would not necessarily see it.

Take THAT!
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### Re: Is 1 = 0.999... ? Really?

Wikipedia wrote:
Countable
In mathematics, a countable set is a set with the same cardinality as some subset of the set of natural numbers. A countable set is either a finite set or a countably infinite set.
I don't think that you are using it in that sense. Basically by writing infA-1, you're counting backwards from infA. That seems like a no no. You know, counting back from where exactly??
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### Re: Is 1 = 0.999... ? Really?

Magnus Anderson wrote:The text in red is inserted by me.

obsrvr524 wrote:
• If there was an implicit 0 digit there, I would see it
(Disagree.)
• I don't see an implicit 0 digit indicated.
(Agree.)
• Therefore there is no implicit 0 digit present.
(Disagree, of course.)

I agree that the conclusion follows from the premises and I agree with the second premise. What I disagree with is the first premise. That said, it is up to me now to present an argument against it. Wait for it

1) What is there is not necessarily seen.

2) Therefore, if there was a $$0$$ associated with $$10^0$$ in $$99\dot9$$, obsrvr524 would not necessarily see it.

Take THAT!

But I was hoping that you would skip that obvious argument.

If that is where you want to go and have nothing else to offer then you have expired your defense on that issue by not convincing me of your premise (4) - so the rest of that argument (version 4) is moot.

Do you have a different argument to make addressing the proposal? If so - it's time to post it.
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