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...
13
41%
No, 1 ≠ 0.999...
16
50%
Other
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9%
 
Total votes : 32

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

Postby Ecmandu » Tue Jan 14, 2020 2:12 am

Magnus Anderson wrote:
Ecmandu wrote:The reason Silhouette sees no difference between .000...1 and zero is because the 1 in 0.000...1 is never arrived at. It’s ALWAYS zero!


\(0.000\dotso1\) represents \(\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots\).

This infinite product never attains \(0\).


Actually, the only way it can NEVER equal a zero is if it adds 1/10th sequentially. Otherwise, it’s a zero.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Jan 14, 2020 2:15 am

Ecmandu wrote:Actually, the only way it can NEVER equal a zero is if it adds 1/10th sequentially. Otherwise, it’s a zero.


What does it mean to add 1/10th sequentially?
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Tue Jan 14, 2020 2:32 am

Magnus Anderson wrote:
Ecmandu wrote:Actually, the only way it can NEVER equal a zero is if it adds 1/10th sequentially. Otherwise, it’s a zero.


What does it mean to add 1/10th sequentially?


1/10th. STOP * 1/10th STOP. * 1/10th STOP etc...
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Jan 14, 2020 2:35 am

I don't understand.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Tue Jan 14, 2020 2:37 am

Magnus Anderson wrote:I don't understand.


It’s means that 1/10th MOVES for EVERY 9 instead of the 1 never occurring (the “the end”)
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Jan 14, 2020 3:28 am

gib wrote:First, \(0.\dot9\) is like \(0.\dot3\) in that they both have infinite decimal expansions. But \(0.\dot3\) is not an infinite quantity. It's just one third. (I chose \(0.\dot3\) because I can't say \(0.\dot9\) is just 1 without you fighting me on it.) Mistaking the quantity represented with the representation is such a juvenile mistake.


Wikipedia proofs treat these symbols as representing infinite sums.

\(0.\dot9 = 0.9 + 0.09 + 0.009 + \cdots\)

Remember that \(\infty\) has an infinite sum equal to it which is \(1 + 1 + 1 + \cdots\). This means that if you can't do arithmetic with infinities (because they are qualities, as you say, and not quantities) then you can't do arithmetic with infinite sums either. (Which would invalidate Wikipedia proofs.)

Unless, for some strange reason, you don't think that infinity can be represented as an infinite sum. In that case, I could simply stop talking about infinities and start talking about. . . infinite sums. There would be no difference with regard to my argument.

Do you agree that infinite sums come in different sizes?

Do you agree that \( (1 + 1 + 1 + \cdots) + 1 > 1 + 1 + 1 + \cdots\)?

If you do, thank you very much.

But if you don't, this means that:

\((1 + 1 + 1 + \cdots) + 1 = 1 + 1 + 1 + \cdots\)

Do you agree that \(1 + 1 + 1 + \cdots = 1 + 1 + 1 + \cdots\)?

Remember that one of the Wikipedia proofs claims that \(0.9 + 0.09 + 0.009 + \cdots = 0.9 + 0.09 + 0.009 + \cdots\).

If you don't agree with this, you also don't agree with Wikipedia proofs.

If you do agree, let's subtract \(1 + 1 + 1 + \cdots\) from the above equation.

What do we get?

We get \(1 = 0\).

Do you agree with the conclusion?
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Re: Is 1 = 0.999... ? Really?

Postby gib » Tue Jan 14, 2020 8:13 am

Are we still doing this?

Magnus Anderson wrote:
gib wrote:First, \(0.\dot9\) is like \(0.\dot3\) in that they both have infinite decimal expansions. But \(0.\dot3\) is not an infinite quantity. It's just one third. (I chose \(0.\dot3\) because I can't say \(0.\dot9\) is just 1 without you fighting me on it.) Mistaking the quantity represented with the representation is such a juvenile mistake.


Wikipedia proofs treat these symbols as representing infinite sums.

\(0.\dot9 = 0.9 + 0.09 + 0.009 + \cdots\)


That's right.

Magnus Anderson wrote:Remember that \(\infty\) has an infinite sum equal to it which is \(1 + 1 + 1 + \cdots\). This means that if you can't do arithmetic with infinities (because they are qualities, as you say, and not quantities) then you can't do arithmetic with infinite sums either. (Which would invalidate Wikipedia proofs.)


Except that \(0.9 + 0.09 + 0.009 + \cdots\) does not equal \(\infty\). It equals \(0.\dot9\), which you can easily do arithmetic with.

Your confusing the infinite sum with the result of the infinite sum. 1 + 1 + 1 + ... is not just an infinite sum, it also equals \(\infty\). \(0.9 + 0.09 + 0.009 + \cdots\), though it is an finite sum, does not.

What I'm saying is that you can't do arithmetic with \(\infty\), but you can do as much arithmetic as you want with an infinite number of terms.

I explained all this right after the snippet you quoted above.

Magnus Anderson wrote:Unless, for some strange reason, you don't think that infinity can be represented as an infinite sum. In that case, I could simply stop talking about infinities and start talking about. . . infinite sums. There would be no difference with regard to my argument.

Do you agree that infinite sums come in different sizes?


Sure, \(0.9 + 0.09 + 0.009 + \cdots\) < 1 + 1 + 1 + ...

Magnus Anderson wrote:Do you agree that \( (1 + 1 + 1 + \cdots) + 1 > 1 + 1 + 1 + \cdots\)?


I don't know whether to agree or disagree. I don't know how to make sense of that.

Magnus Anderson wrote:But if you don't, this means that:

\((1 + 1 + 1 + \cdots) + 1 = 1 + 1 + 1 + \cdots\)

Do you agree that \(1 + 1 + 1 + \cdots = 1 + 1 + 1 + \cdots\)?

Remember that one of the Wikipedia proofs claims that \(0.9 + 0.09 + 0.009 + \cdots = 0.9 + 0.09 + 0.009 + \cdots\). <-- Did they really have to do that.

If you don't agree with this, you also don't agree with Wikipedia proofs.

If you do agree, let's subtract \(1 + 1 + 1 + \cdots\) from the above equation.

What do we get?

We get \(1 = 0\).

Do you agree with the conclusion?


The problem, Magnus, is that (1 + 1 + 1 + ... ) + 1 doesn't mean anything different than 1 + 1 + 1 + ... So yeah, they're equal.

If you just had a finite sum of 1's, like this: (1 + 1 + 1) and you added another 1 to it, you'd get this: (1 + 1 + 1) + 1. <-- But in that case, why keep the brackets? You can just drop them: 1 + 1 + 1 + 1. <-- Now if that was an infinite set of 1's, you'd write it: 1 + 1 + 1 + ... But isn't that the same as the result above? Just because you have an infinite set of 1s in the brackets doesn't mean there's any more reason to keep the brackets. You can drop them for the same reason you can drop them in the finite case (because brackets aren't needed in sums).

Magnus Anderson wrote:We get \(1 = 0\).


And that, my friend, is why you don't do arithemtic with infinity.
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Re: Is 1 = 0.999... ? Really?

Postby Pedro I Rengel » Tue Jan 14, 2020 9:39 am

What?

No, 1 does not then = 0

If they were equal, which they are, because they are both infinite sums of 1, then you have to equally subtract any amount you subtract from one side from the other. You can't add an extra 1 to the infinite sum, because it is already infinite. You can write a 1 outside of the brackets, but this changes nothing. If you wanted that argument to make sense, you would have to actually write an infinite amount of 1s, and then add one. Which makes no sense, because they are infinite. Math is a notation of quantities, not a representation. Otherwise we would just write 1s until we get the desired amount for any given operation.

The arithmetic holds. 1+(1+1+1+...) = 1+1+1+...

If you take away (1+1+1+...) you get,

0 = 0

Or

1 = 1

Or however many 1s you arbitrarily decide to leave.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Jan 14, 2020 11:39 am

gib wrote:Except that \(0.9 + 0.09 + 0.009 + \cdots\) does not equal \(\infty\).


Noone said that.

The point is that both \(0.\dot9\) and \(\infty\) can be represented as infinite sums.

Your confusing the infinite sum with the result of the infinite sum.


I am not.

1 + 1 + 1 + ... is not just an infinite sum, it also equals \(\infty\).


That's correct. (And yes, you are right, there are infinite sums that evaluate to a finite number. But that's irrelevant.)

I take this to mean that you agree that infinity can be represented as an infinite sum. In other words, you agree that \(\infty = 1 + 1 + 1 + \cdots\).

\(0.9 + 0.09 + 0.009 + \cdots\), though it is an [in]finite sum, does not.


That's something that has to be proven. One way to prove it is by doing arithmetic with infinite sums. If you can't do arithmetic with \(1 + 1 + 1 + \cdots\), because it equals to \(\infty\) and because \(\infty\) is not a quantity but a quality, what makes you think you can do arithmetic with \(0.9 + 0.09 + 0.009 + \cdots\)?

What I'm saying is that you can't do arithmetic with \(\infty\), but you can do as much arithmetic as you want with an infinite number of terms.


But \(\infty\) can be represented as an infinite number of terms.

I don't know whether to agree or disagree. I don't know how to make sense of that.


Are you telling me you don't really know how to do arithmetic with infinite sums?

Remember that one of the Wikipedia proofs claims that \(0.9 + 0.09 + 0.009 + \cdots = 0.9 + 0.09 + 0.009 + \cdots\). <-- Did they really have to do that.


They had to. One of their proofs is based on the assumption that \(0.\dot9 - 0.\dot9 = 0\). That's the same kind of assumption that leads to \(1 = 0\).

\(\infty + 1 = \infty\) // subtract \(\infty\) from both sides
\(\infty + 1 - \infty\ = \infty - \infty\) // substitute \(\infty - \infty\) with \(0\)
\(1 = 0\)

What's wrong here is that due to the meaning of the symbol \(\infty\), it does not necessarily follow that \(\infty - \infty\) equals \(0\). It's indeterminate.

This proof does exactly that. It assumes that the difference between the two underlined infinite sums is equal to zero. Hence the wrong conclusion.

The problem, Magnus, is that (1 + 1 + 1 + ... ) + 1 doesn't mean anything different than 1 + 1 + 1 + ... So yeah, they're equal.


Fine. In that case, you have to accept the conclusion that \(1 = 0\). But you don't because you don't like it -- I don't like it either. We already know that \(1 \neq 0\), so this is an indication that the proof is fallacious. But where's the mistake?

And that, my friend, is why you don't do arithemtic with infinity.


Well, in that case, you disagree with this Wikipedia proof because it does arithemtic with infinite sums.

The only way out, it appears to me, is to claim that you can do arithmetic with SOME infinite sums but not ALL of them e.g. you can say you can only do arithmetic with convergent sums (such as \(0.\dot9\)). But in that case, you'll have to explain why. And you can't say it's because you don't like the consequences of doing arithmetic with divergent series. That's not an acceptable answer.
Last edited by Magnus Anderson on Tue Jan 14, 2020 12:23 pm, edited 2 times in total.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Jan 14, 2020 11:59 am

Pedro I Rengel wrote:What?

No, 1 does not then = 0

If they were equal, which they are, because they are both infinite sums of 1, then you have to equally subtract any amount you subtract from one side from the other. You can't add an extra 1 to the infinite sum, because it is already infinite. You can write a 1 outside of the brackets, but this changes nothing. If you wanted that argument to make sense, you would have to actually write an infinite amount of 1s, and then add one. Which makes no sense, because they are infinite. Math is a notation of quantities, not a representation. Otherwise we would just write 1s until we get the desired amount for any given operation.

The arithmetic holds. 1+(1+1+1+...) = 1+1+1+...

If you take away (1+1+1+...) you get,

0 = 0

Or

1 = 1

Or however many 1s you arbitrarily decide to leave.


Not really.

If we say that \(a = b\) this means that \(a - b = 0\) and that we can substitute every occurrence of \(a - b\) with \(0\). If you can't do that, then \(a \neq b\). But I guess you want to have it both ways. You want \(a\) and \(b\) to be both equal and not equal. Sort of like how Hilbert's hotel is both full and not full.

The problem is that the way mathematicians define \(\infty\) implies that \(\infty - \infty = 0\) is not necessarily true. This means you can't substitute \(\infty - \infty\) with \(0\). It also means you can't substitute \(0.\dot9 - 0.\dot9\) with \(0\).
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Re: Is 1 = 0.999... ? Really?

Postby obsrvr524 » Tue Jan 14, 2020 1:23 pm

Pedro I Rengel wrote:If they were equal, which they are, because they are both infinite sums of 1, then you have to equally subtract any amount you subtract from one side from the other. You can't add an extra 1 to the infinite sum, because it is already infinite. You can write a 1 outside of the brackets, but this changes nothing. If you wanted that argument to make sense, you would have to actually write an infinite amount of 1s, and then add one. Which makes no sense, because they are infinite.

So if you have an infinite line, you have an infinity of points on that line {1+1+1+...}, right? If you then add a dot or point (.) next to that line, you have added something that wasn't there before, a another point. But you are saying that you cannot add anything because you already have an infinity of points.

An infinite line and a point;
<_______________________________>
        .

How many points are there? {1+1+1+...} + 1
(1+1+1+...) = an infinite line
1 = another point




Take away the infinite line;
        .

How many points are there? 1
Take away the line, -(1+1+1+...)
The 1 remains.




({1+1+1+...} + 1) - {1+1+1+...} = 1

But you are saying that can't be done.
              You have been observed.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Jan 14, 2020 3:08 pm

He probably thinks that in order to add \(1\) to an infinite sum of \(1\)'s you have to actually reach the end of that infinite sum and place it precisely at that point.

The fact is, you can add that \(1\) at any position in the sum. Why are these people obsessing over the non-existent end of infinite sums?

Of course, there is no last position, so you can't add it at the end of the sum. But you can add it anywhere else. It makes no difference.

I suppose the problem is that, since all of the terms are equal, the standard way of representing infinite sums makes the result of \(\infty + 1\) look the same as \(\infty\).

What do you get when you take \(1 + 1 + 1 + \cdots\) and add \(1\) to it? Well, you get \(1 + 1 + 1 + \cdots\). The symbols are exactly the same but they are representing different things.

One way to solve this problem is to maintain the identity of terms. So instead of writing \(1 + 1 + 1 + \cdots\), we could say \(One_1 + One_2 + One_3 + \cdots\) so that when we add \(One_{new}\) to this sum, we could represent the result as \(One_{new} + One_1 + One_2 + One_3 + \cdots\).

But since we're comparing sums that are infinite, even if we do this, it would still be very easy to pretend that they are not different.
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Re: Is 1 = 0.999... ? Really?

Postby phyllo » Tue Jan 14, 2020 3:32 pm

One way to solve this problem is to maintain the identity of terms. So instead of writing 1+1+1+⋯, we could say One1+One2+One3+⋯ so that when we add Onenew to this sum, we could represent the result as Onenew+One1+One2+One3+⋯.
That's not possible because you don't have a number to attach to the last One. If you did then the series would be finite and the last term would be number n.

Similarly you don't have number to attach to the New One.

That's why 1+1+1+... is the same as (1+1+1+...)+1. There is no way to tell them apart. Parentheses give the illusion of containing a fixed number of 1s.

Infinity isn't a quantity. It's not a particular quantity of 1s.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Jan 14, 2020 4:11 pm

phyllo wrote:That's not possible because you don't have a number to attach to the last One. If you did then the series would be finite and the last term would be number n.


But you don't need a number. All you need is a symbol that is different from every other symbol used in the sum.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Tue Jan 14, 2020 4:47 pm

Magnus Anderson wrote:
phyllo wrote:That's not possible because you don't have a number to attach to the last One. If you did then the series would be finite and the last term would be number n.


But you don't need a number. All you need is a symbol that is different from every other symbol used in the sum.


That’s called “adding dimensionality”

So first we agree that you can’t add an infinite amount of “new ones”, only a finite amount.

Problem is, that means the ones weren’t infinite (endless) in the first place... you push the “last” one up one more place, which is the same as adding a 1 at the end.
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Re: Is 1 = 0.999... ? Really?

Postby phyllo » Tue Jan 14, 2020 5:17 pm

Magnus Anderson wrote:
phyllo wrote:That's not possible because you don't have a number to attach to the last One. If you did then the series would be finite and the last term would be number n.


But you don't need a number. All you need is a symbol that is different from every other symbol used in the sum.
Okay, let's say for the sake of argument that you can have more symbols than your number system can count. As if that makes any sense in itself.
You can manipulate those symbols using the rules of math but at the end you are forced to convert the result into a number.

If the result has more symbols than your number system can count, then you end up with an infinity.

You still don't have a quantified number of 1s.

Try this example :

Let's say that you have a counting system (1,2,3,4,5). Anything over 5 is infinity.

1+1+1+... is infinity : over 5.

(1+1+1+...)+1 is over 5. Therefore it is also equal to infinity in your number system.

(1+1+1+...)+2 is over 5. Therefore equals infinity.

infinity+infinity is over 5. Therefore equals infinity.
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Re: Is 1 = 0.999... ? Really?

Postby gib » Tue Jan 14, 2020 6:33 pm

Magnus Anderson wrote:
gib wrote:Except that \(0.9 + 0.09 + 0.009 + \cdots\) does not equal \(\infty\).


Noone said that.


Magnus Anderson wrote:
\(0.9 + 0.09 + 0.009 + \cdots\), though it is an [in]finite sum, does not.


That's something that has to be proven. One way to prove it is by doing arithmetic with infinite sums. If you can't do arithmetic with \(1 + 1 + 1 + \cdots\), because it equals to \(\infty\) and because \(\infty\) is not a quantity but a quality, what makes you think you can do arithmetic with \(0.9 + 0.09 + 0.009 + \cdots\)?


I need to be absolutely clear about what you're saying. Are you actually saying that:

\(0.9 + 0.09 + 0.009 + \cdots\) = \(0.\dot9\) = \(\infty\)

Magnus Anderson wrote:
What I'm saying is that you can't do arithmetic with \(\infty\), but you can do as much arithmetic as you want with an infinite number of terms.


But \(\infty\) can be represented as an infinite number of terms.


You can do arithmetic with finite terms in order to create an infinite sum. But once it's created, you can't do arithmetic with that.

I'll explain (like that will have any effect):

You can do arithmetic with a series of 1s:

1 + 1 + 1 = 3

You can do arithmetic with a larger series of 1s:

1 + 1 + 1 + 1 + 1 + 1 = 6

There is no limit to the number of 1s you can do arithmetic with. You can do arithmetic with an infinite number of 1s:

1 + 1 + 1 + ... = \(\infty\)

But now you can't take that and do further arthemtic with it:

(1 + 1 + 1 + ...) + n <-- Can't do.

\(\frac{(1 + 1 + 1 + ...)}{n}\) <-- Can't do.

\((1 + 1 + 1 + ...)^{n}\) <-- Can't do.

^ Note that not all infinite sums give you infinity. So we can do the same thing with the 9s:

0.9 + 0.09 + 0.009 = 0.999

0.9 + 0.09 + 0.009 + 0.0009 + 0.00009 + 0.000009 = 0.999999

0.9 + 0.09 + 0.009 + ... = \(\dot9\)

^ Unless you think \(\dot9\) = \(\infty\), you have not derived \(\infty\) here. Therefore, you can do further arithmetic with it.

Magnus Anderson wrote:
The problem, Magnus, is that (1 + 1 + 1 + ... ) + 1 doesn't mean anything different than 1 + 1 + 1 + ... So yeah, they're equal.


Fine. In that case, you have to accept the conclusion that 1 = 0.


Ooooor... I can just avoiding doing arithmetic with it.

Magnus Anderson wrote:Well, in that case, you disagree with this Wikipedia proof because it does arithemtic with infinite sums.


I didn't say you can't do arithmetic with infinite sums, I said if the infinite sum gives you \(\infty\), THEN you can't do arithmetic with it. \(0.\dot9 \neq \infty\). (<-- Is this seriously lost on you?)

Magnus Anderson wrote:\(\infty + 1 = \infty\) // subtract \(\infty\) from both sides
\(\infty + 1 - \infty\ = \infty - \infty\) // substitute \(\infty - \infty\) with \(0\)
\(1 = 0\)

...

But where's the mistake?


I agree with step 1. The mistake is not there. It's not in any of the steps. It's the fact that you took steps to begin with. You should have stopped at \(\infty + 1 = \infty\). Why? Take a guess, Magnus. Because you can't do what with infinity?

I will say this: Another rendition of what I'm trying to say is that you can do arithmetic with infinity, but then different rules apply. So the rule that adding 1 to something gives you a greater number no longer applies. Instead, adding 1 to \(\infty\) still gives you \(\infty\).

So don't do arithmetic with \(\infty\) or accept that different rules apply. <-- Take your pick.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Tue Jan 14, 2020 7:30 pm

Magnus,

I’m going to explain a very sophisticated concept to you and I hope you can take it in.

Your fallback position is that planes are infinity larger than lines. This is a mistake James made.

Have you ever seen a “2D” optical illusion that moves when you look at it, or gives you a three dimensional image if you look at it correctly.

The same thing can be done in unary with 2D images.

What I’m trying to explain to you here is that planes aren’t a greater infinity than lines.

I hope you understood all that.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Jan 14, 2020 10:37 pm

gib wrote:I need to be absolutely clear about what you're saying. Are you actually saying that:

\(0.9 + 0.09 + 0.009 + \cdots\) = \(0.\dot9\) = \(\infty\)


Not really.

I'm merely asking why do you think that you can do arithmetic with some infinite sums (such as \(0.9 + 0.09 + 0.009 + \cdots\)) but not all (such as with \(1 + 1 + 1 + \cdots\)). Is it because the latter evaluates to \(\infty\) or because you think the former equals to a finite number or because of something else? And how did you come to your conclusion?

Why is it okay to say that \(10 \times (0.9 + 0.09 + 0.009 + \cdots) = 9 + 0.9 + 0.09 + \cdots\) but not okay to say that \(10 \times (1 + 1 + 1 + \cdots) = 10 + 10 + 10 + \cdots\)?

^ Unless you think \(\dot9\) = \(\infty\), you have not derived \(\infty\) here. Therefore, you can do further arithmetic with it.


If I understand you correctly, we can do arithmetic with \(0.9 + 0.09 + 0.009 + \cdots\) because it's not equal to \(\infty\).

Why is that so?

I agree with step 1. The mistake is not there. It's not in any of the steps. It's the fact that you took steps to begin with. You should have stopped at \(\infty + 1 = \infty\). Why? Take a guess, Magnus. Because you can't do what with infinity?

I will say this: Another rendition of what I'm trying to say is that you can do arithmetic with infinity, but then different rules apply. So the rule that adding 1 to something gives you a greater number no longer applies. Instead, adding 1 to \(\infty\) still gives you \(\infty\).

So don't do arithmetic with \(\infty\) or accept that different rules apply. <-- Take your pick.


But you didn't explain why.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Jan 14, 2020 10:48 pm

gib wrote:I agree with step 1. The mistake is not there. It's not in any of the steps. It's the fact that you took steps to begin with. You should have stopped at \(\infty + 1 = \infty\). Why? Take a guess, Magnus. Because you can't do what with infinity?


The mistake must be in one of the steps.

You say that \(\infty + 1 = \infty\) is true. You have no problem with that expression. Even though what you're doing here is what you say you can't do with \(\infty\), which is arithmetic, you have no problem evaluating its truth-value. But for some strange reason, you have a problem with \(\infty + 1 - \infty = \infty - \infty\)? You can't evaluate the truth-value of that expression?

Very curious.

The mistake is in the assumption that \(\infty - \infty\) equals to \(0\). It does not. Even though \(\infty = \infty\), \(\infty - \infty\) is not necessarily \(0\).
Last edited by Magnus Anderson on Tue Jan 14, 2020 11:08 pm, edited 2 times in total.
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Re: Is 1 = 0.999... ? Really?

Postby obsrvr524 » Tue Jan 14, 2020 10:58 pm

phyllo wrote:Okay, let's say for the sake of argument that you can have more symbols than your number system can count. As if that makes any sense in itself.
You can manipulate those symbols using the rules of math but at the end you are forced to convert the result into a number.

An infinite line plus a single dot yields more locations than the number system provides. Two infinite lines provides an infinity more than the number system provides. In neither case is the total quantity a "number".

phyllo wrote:If the result has more symbols than your number system can count, then you end up with an infinity.

You still don't have a quantified number of 1s.

Try this example :

Let's say that you have a counting system (1,2,3,4,5). Anything over 5 is infinity.

1+1+1+... is infinity : over 5.

(1+1+1+...)+1 is over 5. Therefore it is also equal to infinity in your number system.

(1+1+1+...)+2 is over 5. Therefore equals infinity.


infinity+infinity is over 5. Therefore equals infinity.

Relevance? The decimal system doesn't have a top or final number.


Ecmandu wrote:Have you ever seen a “2D” optical illusion that moves when you look at it, or gives you a three dimensional image if you look at it correctly.

The same thing can be done in unary with 2D images.

Incoherent.
              You have been observed.
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Re: Is 1 = 0.999... ? Really?

Postby promethean75 » Tue Jan 14, 2020 11:03 pm

i got your PM, ecmandu, but i don't yet feel comfortable with publishing. there still seems to be much disagreement among our mathematicians and i'd like to wait until i see unanimous agreement.

maybe you guys'll figure out if 1 = 0.999 THIS FUCKING YEAR, and i can proceed with the project.
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Re: Is 1 = 0.999... ? Really?

Postby iambiguous » Tue Jan 14, 2020 11:18 pm

promethean75 wrote:i got your PM, ecmandu, but i don't yet feel comfortable with publishing. there still seems to be much disagreement among our mathematicians and i'd like to wait until i see unanimous agreement.

maybe you guys'll figure out if 1 = 0.999 THIS FUCKING YEAR, and i can proceed with the project.


More to the point, any number of them here already have.

Go ahead, ask them. :wink:
He was like a man who wanted to change all; and could not; so burned with his impotence; and had only me, an infinitely small microcosm to convert or detest. John Fowles

Start here: viewtopic.php?f=1&t=176529
Then here: viewtopic.php?f=15&t=185296
And here: viewtopic.php?f=1&t=194382
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Re: Is 1 = 0.999... ? Really?

Postby Pedro I Rengel » Tue Jan 14, 2020 11:21 pm

Ok I see you all have moved on, but check it:

The a and b thing is true, but in the infinite sum operation you misidentified a and b. Because the sum is infinite, any added number to a is still a. In that case, there is no a +1. Because a is already all the 1s.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Tue Jan 14, 2020 11:26 pm

phyllo wrote:Okay, let's say for the sake of argument that you can have more symbols than your number system can count. As if that makes any sense in itself.


You think there aren't more symbols than there are natural numbers? The set of natural numbers is the set of all symbols?

How about \(\{a_1, b_1, c_1, a_2, b_2, c_2, a_3, b_3, c_3, \dotso\}\)? That's \(3\) times more than \(\{1, 2, 3, \dotso\}\).

But I understand. You don't think that \(3 \times \infty > \infty\).
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