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
42%
No, 1 ≠ 0.999...
15
48%
Other
3
10%
 
Total votes : 31

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

Postby phyllo » Fri Jan 24, 2020 3:15 am

You moved out of the decimal system to make your argument, you switched back to whole numbers, which is the same as never using decimal expression in the first place.
No I didn't.

Do I need to rewrite it for you?
It is simple.

1.00/9.00= 0.111... Equation 1 (ordinary division by 9)

(0.111... )*9.00=0.999... Equation 2 (ordinary multiplication by 9)

(1.00/9.00)*9.00=1.00 Equation 3 (ordinary division and multiplication)

Substitute for (1.00/9.00) in Equation 3 with the equivalent (1.00/9.00) from Equation 1 :

(0.111...)*9.00=1.00 Equation 4

The left side of Equation 2 is exactly the same as the left side of Equation 4. Therefore whatever is on the right side of equation 2 is exactly equal to whatever is on the right side of Equation 4.

Therefore :

1.00=0.999...

QED

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

Postby Ecmandu » Fri Jan 24, 2020 3:17 am

phyllo wrote:
You moved out of the decimal system to make your argument, you switched back to whole numbers, which is the same as never using decimal expression in the first place.
No I didn't.

Do I need to rewrite it for you?
It is simple.

1.00/9.00= 0.111... Equation 1 (ordinary division by 9)

(0.111... )*9.00=0.999... Equation 2 (ordinary multiplication by 9)

(1.00/9.00)*9.00=1.00 Equation 3 (ordinary division and multiplication)

Substitute for (1.00/9.00) in Equation 3 with the equivalent (1.00/9.00) from Equation 1 :

(0.111...)*9.00=1.00 Equation 4

The left side of Equation 2 is exactly the same as the left side of Equation 4. Therefore whatever is on the right side of equation 2 is exactly equal to whatever is on the right side of Equation 4.

Therefore :

1.00=0.999...

QED

Happy now?


Nope. You did. Step three like I said before is where you switched from decimals to whole numbers exclusively.

It’s not about making me happy, it’s about what’s true.
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Re: Is 1 = 0.999... ? Really?

Postby phyllo » Fri Jan 24, 2020 3:19 am

It's only 5 steps with very basic math operations and you can't follow the reasoning. #-o
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Fri Jan 24, 2020 3:24 am

phyllo wrote:It's only 5 steps with very basic math operations and you can't follow the reasoning. #-o


Nice try. You may as well as typed the proof without using decimals at all (which is how standard Mathematicians argue it). At some point, you dropped the decimals and went to only whole numbers.

Yes. That proof always works with only whole numbers.
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Re: Is 1 = 0.999... ? Really?

Postby gib » Fri Jan 24, 2020 3:59 am

Magnus Anderson wrote:You'd have to explain why you're limiting yourself to integers.


Try it without integers. It doesn't work.

Let's take an example. Let's say n = 2.5.

First, we'd have to define how one even does sums with non-integers, how to increment the value of i. Obviously you start with the initial value (1 in the cases we've been dealing with), but what's the next number? Do you go:

1, 2, 2.5?

Do you go:

1.0, 1.5, 2.0, 2.5?

Do you go:

1.0, 1.1, 1.2 ... 2.4, 2.5?

Do you stop at 2 (because you can't jump by a whole integer amount from 2 to 2.5)? Do you stop at 3 (because maybe n = 2.5 means don't stop as long as you're under 2.5)?

But for argument's sake, let's say you go: 1, 2, 2.5. Then the sum is:

\(\frac{9}{10^{1}} + \frac{9}{10^2} + \frac{9}{10^{2.5}} = \frac{9}{10} + \frac{9}{100} + \frac{9}{316.227766} = 0.9 + 0.09 + 0.02846 = 1.01846\)

Magnus Anderson wrote:Basically, what you're saying is that \(\infty\) does not refer to a specific number in the same way that the statement "An integer greater than 10" does not refer to a specific integer since the quantity that it represents can be 11 or 12 or 100. However, you have yet to explain why it's invalid for n (the upper bound of summation) to be a non-specific number.


I'm not just saying \(\infty\) does not refer to a specific number, I'm saying it doesn't refer to any number. It can't. You've heard me say it before and I'll say it again. Infinity means endless. It refers to a property of sets. It's not that we're not being specific about which number it refers to, it's that we (or I) are not referring to something that could be a number.

Magnus Anderson wrote:My claim is that \(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) for any \(n\) that is greater than \(0\). It does not have to be an integer. The only condition is that it's greater than \(0\). Do you think there is \(n\) greater than \(0\) such that the sum is not less than \(1\)? No. You obviously do (otherwise, you'd not say that \(0.\dot9 = 1\).)


Like I said, \(\sum_{i=1}^{\infty} \frac{9}{10^i}\) is not a special case of \(\sum_{i=1}^{n} \frac{9}{10^i}\). It's a case that lies outside all specific cases of \(\sum_{i=1}^{n} \frac{9}{10^i}\) (because \(\infty\) is not a valid value for n). \(\sum_{i=1}^{\infty} \frac{9}{10^i}\) is saying: don't pick a value for n. Let the sum run forever.

So I agree that there is no n such that \(\sum_{i=1}^{n} \frac{9}{10^i} \geq 1\), but \(\infty\) is not a valid option for n. Plugging \(\infty\) into the upper bound of the sum isn't picking a value for n. It's refusing to pick a value for n. Only in that case does the sum equal 1.

Magnus Anderson wrote:\(\infty\) is not an integer. It's a number greater than every integer.

And \(\sum_{i=1}^{\infty} \frac{9}{10^i}\) is not a completely different statement. The only difference between \(\infty\) and integers is 1) \(\infty\) is a bigger number and 2) it's non-specific. The fact that it's non-specific shouldn't be a problem at all. And if you absolutely hate non-specific numbers, you can switch to some specific infinity.


You're so close to biting, Magnus. Wanna talk about hyperreals?
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Re: Is 1 = 0.999... ? Really?

Postby phyllo » Fri Jan 24, 2020 1:30 pm

Ecmandu wrote:
phyllo wrote:It's only 5 steps with very basic math operations and you can't follow the reasoning. #-o


Nice try. You may as well as typed the proof without using decimals at all (which is how standard Mathematicians argue it). At some point, you dropped the decimals and went to only whole numbers.

Yes. That proof always works with only whole numbers.
Show how it fails.

Show the cases where it doesn't work.
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Re: Is 1 = 0.999... ? Really?

Postby Carleas » Fri Jan 24, 2020 5:04 pm

Apologies in advance for the drive-by, I haven't followed this thread but Ecmandu asked for my response.

Ecmandu wrote:1.) series don’t converge! If a series converges, this means that every counting number is an infinite sequence (proof through contradiction).

This seems question begging of the subject of this thread: if .999... = 1, then every counting number is an infinite sequence (.9 + .09 + .009 + .0009 + ...)

Ecmandu wrote:I use this order symbolically (not using numbers) to disprove this!

A.) rational number
B.) uncounted infinity
C.) different rational number
D.) different uncounted infinity

Etc...

This places all the “lower” infinities in 1:1 correspondence with the “highest order of infinity”, thus there are no orders of infinity.

I don't follow this proof. Are you rejecting the idea that each rational number represents a different infinity? Can you cast the proof in terms of Hilbert's Hotel?
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Fri Jan 24, 2020 6:06 pm

Carleas wrote:Apologies in advance for the drive-by, I haven't followed this thread but Ecmandu asked for my response.

Ecmandu wrote:1.) series don’t converge! If a series converges, this means that every counting number is an infinite sequence (proof through contradiction).

This seems question begging of the subject of this thread: if .999... = 1, then every counting number is an infinite sequence (.9 + .09 + .009 + .0009 + ...)

Ecmandu wrote:I use this order symbolically (not using numbers) to disprove this!

A.) rational number
B.) uncounted infinity
C.) different rational number
D.) different uncounted infinity

Etc...

This places all the “lower” infinities in 1:1 correspondence with the “highest order of infinity”, thus there are no orders of infinity.

I don't follow this proof. Are you rejecting the idea that each rational number represents a different infinity? Can you cast the proof in terms of Hilbert's Hotel?


It’s totally on topic for the thread (thanks for stopping by btw!)

Think about what this means. It means that 1 (a definitely finite number)

IS!!! ( bolding important )

An infinite sequence (which is a disproof through contradiction of definitions)

As for the second part. I’m sorry it didn’t make sense.

All I’m trying to say is that ALL infinities can be paired in 1:1 correspondence. The argument that they cannot, is put forth that you cannot enumerate all infinities in one single list.

So I CHEATED =)

I just list them (symbolically (not numerically)) as “uncounted numbers”

Math uses abstract symbols constantly that people complain about!! But, from my perspective, my CHEAT is fair game!

As far as Hilbert hotel is concerned, this is already a 1: 1 correspondence, you’re not doing “dimensional flooding” in that thought experiment. “ Dimensional flooding” is when every number in a single list has an infinite amount of numbers that cannot be put in each slot of the list.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Fri Jan 24, 2020 8:21 pm

Carleas,

I also wanted to add to my last post.

Lots of people think number theory is dry, boring and useless.

Little do they know, I’m actually making a theological argument.

If there are no orders of infinity, then that means that no being is greater than another being (including hypothetical “god”)

I think everyone in this thread can all agree that they are “math perverts”. Math always has philosophic implications, in fact, it is the purest language we have for discussions.

That’s what I wanted to add. Again, thanks for joining the thread.
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Re: Is 1 = 0.999... ? Really?

Postby Carleas » Fri Jan 24, 2020 10:19 pm

Ecmandu wrote:It means that 1 (a definitely finite number)

IS!!! ( bolding important )

An infinite sequence (which is a disproof through contradiction of definitions)

Two things:

1) I think "is" may be ambiguous in this context. I would say that natural numbers can be expressed as the sum of an infinite series. I think that is synonymous with the mathematical 'is'. (Aside: my understanding is that category theory is an attempt to rebuild math without using the '=' operator (or without using it the way it's traditionally used. Unfortunately, I don't understand it any further than that, and I don't know its implications for questions like these.)

2) You need a contradiction to create a proof by contradiction. You've assumed that a finite number can't be expressed as the sum of an infinite series, and I agree that if we take that as a given, we could use it to create a proof by contradiction. But we don't have to take that as a given: there is no contradiction between something being both a finite number and the sum of an infinite series.

Ecmandu wrote:I just list them (symbolically (not numerically)) as “uncounted numbers”

This assumes something that isn't true for uncountably infinite sets. You can't list them, even in theory. There is no next largest real number.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Fri Jan 24, 2020 11:05 pm

Carleas wrote:
Ecmandu wrote:It means that 1 (a definitely finite number)

IS!!! ( bolding important )

An infinite sequence (which is a disproof through contradiction of definitions)

Two things:

1) I think "is" may be ambiguous in this context. I would say that natural numbers can be expressed as the sum of an infinite series. I think that is synonymous with the mathematical 'is'. (Aside: my understanding is that category theory is an attempt to rebuild math without using the '=' operator (or without using it the way it's traditionally used. Unfortunately, I don't understand it any further than that, and I don't know its implications for questions like these.)

2) You need a contradiction to create a proof by contradiction. You've assumed that a finite number can't be expressed as the sum of an infinite series, and I agree that if we take that as a given, we could use it to create a proof by contradiction. But we don't have to take that as a given: there is no contradiction between something being both a finite number and the sum of an infinite series.

Ecmandu wrote:I just list them (symbolically (not numerically)) as “uncounted numbers”

This assumes something that isn't true for uncountably infinite sets. You can't list them, even in theory. There is no next largest real number.


For the first part, I think the word “represents” is the most accurate English word you have available to you.

Like: you’re a lawyer. Representing a client does not mean that you are the client!

In saying this, I must reassert that equality has an extremely narrow and well documented definition in mathematics (per the op)

As for number two “there is no next largest real number”.

This is the myth I’m dispelling with my symbolic cheat (non-enumerated!)

Take the series...

1.) real number
2.) next largest real number
3.) next, next largest real number
4.) next, next, next largest real number

Etc...

I’m using a substitution symbol (extremely common in math!)

What’s wrong with it?
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Fri Jan 24, 2020 11:33 pm

Carleas,

Before you answer my last post. Consider that I pointed out that the best English word for you in argument 1 was “represents”

It’s the same in argument 2. I misspoke and said “substitutes”. However, if I use the word “represents” as well, it changes the meaning and ability to refute it substantially
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Sat Jan 25, 2020 1:53 am

gib wrote:Try it without integers. It doesn't work.

Let's take an example. Let's say n = 2.5.

First, we'd have to define how one even does sums with non-integers, how to increment the value of i. Obviously you start with the initial value (1 in the cases we've been dealing with), but what's the next number? Do you go:

1, 2, 2.5?


If \(i\) must start with \(1\), end with \(n\) and increase by \(1\), then \(2.5\) is not a valid value for \(n\).

On the other hand, a sum of natural numbers \(1 + 2 + 3 + \cdots + n\) is equal to \(\frac{n(n+1)}{2}\). If we plug \(2.5\) into that, we get \(4.375\) as a result. We might also be able to find a sum with an index that starts with certain number, ends with \(2.5\) and increases by certain value.

Have you considered limiting \(n\) to numbers that have no fractional component but wihtout limiting it to integers?

gib wrote:But for argument's sake, let's say you go: 1, 2, 2.5. Then the sum is:

\(\frac{9}{10^{1}} + \frac{9}{10^2} + \frac{9}{10^{2.5}} = \frac{9}{10} + \frac{9}{100} + \frac{9}{316.227766} = 0.9 + 0.09 + 0.02846 = 1.01846\)


Most likely, you are breaking the pattern.
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Re: Is 1 = 0.999... ? Really?

Postby gib » Sat Jan 25, 2020 10:18 am

Magnus Anderson wrote:If \(i\) must start with \(1\), end with \(n\) and increase by \(1\), then \(2.5\) is not a valid value for \(n\).

Exactly!

On the other hand, a sum of natural numbers \(1 + 2 + 3 + \cdots + n\) is equal to \(\frac{n(n+1)}{2}\). If we plug \(2.5\) into that, we get \(4.375\) as a result. We might also be able to find a sum with an index that starts with certain number, ends with \(2.5\) and increases by certain value.

All good.

Have you considered limiting \(n\) to numbers that have no fractional component but wihtout limiting it to integers?


Like \(\infty\)?

Magnus Anderson wrote:
gib wrote:But for argument's sake, let's say you go: 1, 2, 2.5. Then the sum is:

\(\frac{9}{10^{1}} + \frac{9}{10^2} + \frac{9}{10^{2.5}} = \frac{9}{10} + \frac{9}{100} + \frac{9}{316.227766} = 0.9 + 0.09 + 0.02846 = 1.01846\)


Most likely, you are breaking the pattern.


I am breaking the pattern. That was my point.

But you want to put \(\infty\) in there. So let's talk about \(\infty\) as a fuzzy blob of impossibly big numbers, not as limitlessness, as the spectrum of numbers you would find if you transcended the limits of the limitless.
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Re: Is 1 = 0.999... ? Really?

Postby phyllo » Sat Jan 25, 2020 3:30 pm

On the other hand, a sum of natural numbers 1+2+3+⋯+n is equal to n(n+1)2. If we plug 2.5 into that, we get 4.375 as a result. We might also be able to find a sum with an index that starts with certain number, ends with 2.5 and increases by certain value.

All good.
2.5 is not a natural number so this makes no sense.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Sun Jan 26, 2020 9:58 am

gib wrote:Like \(\infty\)?


Yes.

I am breaking the pattern. That was my point.


Why would you break the pattern? (Assuming you know what I mean.) Obviously, if you break the pattern, you might not get a number that is less than \(1\).

But you want to put \(\infty\) in there. So let's talk about \(\infty\) as a fuzzy blob of impossibly big numbers, not as limitlessness, as the spectrum of numbers you would find if you transcended the limits of the limitless.


In the case of \(\sum_{i=1}^{n}\), \(i\) starts with \(1\), increases by \(1\) and ends with \(n\). The number of terms is \(n\), so the sum stops (is complete) after \(n\) number of terms.

In the case of \(\sum_{i=1}^{\infty}\), \(i\) starts with \(1\), increases by \(1\) and does not end. The fact that \(i\) does not end tells us that the number of terms is \(\infty\). This means the sum stops (is complete) after an infinite number of terms. (I assume you're one of the people in this thread who have no problem with the concept of "actual or completed infinity".)

\(\sum_{i=1}^{\infty}\) does not mean that \(i\) ends with \(\infty\). In other words, \(\sum_{i=1}^{\infty} i\) is not equal to something like \(1 + 2 + 3 + \cdots + \infty\). (Note that such a sum would have more than \(\infty\) terms.)

The value of \(i\) is always a natural number. So if we are asking a question such as "What's the value of the sum of terms whose index is \(i, 1 \leq i \leq x\)?" then \(x\) cannot be anything other than a natural number since the range of \(x\), in such a case, must be the range of \(i\) -- and this means that \(x\) can't be \(\infty\). But if we're asking a question such as "What's the value of the sum after \(x\) number of terms?" then the range of \(x\) goes from \(1\) to the number of terms of the sum. If the number of terms is infinite, then \(x\) can be \(\infty\). And it is precisely this question that we're asking.

What is the value of the sum \(\sum_{i=1}^{\infty} \frac{9}{10^i}\) after an infinite number of terms?

You're saying it's \(1\), I am saying it's less than \(1\).

My argument (which is basically James's argument) is that the pattern of this sum prohibits its value after \(x\) number of terms to be equal to \(1\) for any \(x > 0\). (You can limit the value of \(x\) to numbers that have no fractional component, if you want.) Since \(\infty\) is greater than \(0\), it applies to \(\infty\) as well.

By what logic does the value of this sum become \(1\) after an infinite number of terms?
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Sun Jan 26, 2020 3:45 pm

\(\sum_{i=1}^{n} \frac{9}{10^i} = 1 - \frac{1}{10^n}\)

In order for \(\sum_{i=1}^{n} \frac{9}{10^i}\) to be equal to \(1\), there must be \(n\) such that \(1 - \frac{1}{10^n} = 1\).

This means that there must be \(n\) such that \(\frac{1}{10^n} = 0\).

Suppose that \(x\) is such a number: \(1 \div 10^x = 0\).

This means that \(0 \times 10^x = 1\).

But zero times any number is zero, right?
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Sun Jan 26, 2020 5:44 pm

Magnus Anderson wrote:\(\sum_{i=1}^{n} \frac{9}{10^i} = 1 - \frac{1}{10^n}\)

In order for \(\sum_{i=1}^{n} \frac{9}{10^i}\) to be equal to \(1\), there must be \(n\) such that \(1 - \frac{1}{10^n} = 1\).

This means that there must be \(n\) such that \(\frac{1}{10^n} = 0\).

Suppose that \(x\) is such a number: \(1 \div 10^x = 0\).

This means that \(0 \times 10^x = 1\).

But zero times any number is zero, right?


Magnus! Honestly! *sigh*

This is what I was trying to explain to you before.

If you halve 1, it’s 1/2 + 1/2

If you halve THAT!

It’s 1/4+1/4+1/4+1/4. Etc....

If you keeps doing this:

At infinity (convergence)

EVERY!!! Whole number solves as zero.

I still think you are confusing all your arguments. You switch back and forth when it suits you without understanding (or caring) about the implications.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Sun Jan 26, 2020 5:53 pm

Well, you didn't say much. What you said is basically:

\(1 = \frac{1}{2} + \frac{1}{2} = 2 \times \frac{1}{2}\)
\(1 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 4 \times \frac{1}{4}\)
\(\dotso\)
\(1 = 0 + 0 + 0 + \cdots = \infty \times 0\)

How can that be the case when zero times anything is zero?

Properly speaking, \(1 = \frac{1}{\infty} + \frac{1}{\infty} + \frac{1}{\infty} + \cdots = \infty \times \frac{1}{\infty}\).
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Sun Jan 26, 2020 6:15 pm

Magnus Anderson wrote:Well, you didn't say much. What you said is basically:

\(1 = \frac{1}{2} + \frac{1}{2} = 2 \times \frac{1}{2}\)
\(1 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 4 \times \frac{1}{4}\)
\(\dotso\)
\(1 = 0 + 0 + 0 + \cdots = \infty \times 0\)

How can that be the case when zero times anything is zero?

Properly speaking, \(1 = \frac{1}{\infty} + \frac{1}{\infty} + \frac{1}{\infty} + \cdots = \infty \times \frac{1}{\infty}\).


I said a lot! I said that it’s impossible for convergence theory to be true (standard calculus and other mathematics)

Ultimately it equals (according to convergence theory)

0+0+0+0 (etc...) at infinity! (Convergence)

That means that every whole number equals zero. (Contradiction)
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Sun Jan 26, 2020 6:57 pm

How's that relevant to anything of what I'm saying?
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Sun Jan 26, 2020 7:01 pm

Magnus Anderson wrote:How's that relevant to anything of what I'm saying?


How’s what you said relevant to what I said?!?!

1/infinity is a meaningless mathematical phrase.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Sun Jan 26, 2020 7:44 pm

You quoted me. I guess it's somehow relevant to what I said in that quote.

\(\frac{1}{\infty}\) is a number greater than \(0\) but smaller than every number of the form \(\frac{1}{n}, n \in N\). That's what it means (so it's not really meaningless.)

\(0 < \frac{1}{\infty} < \cdots < \frac{1}{4} < \frac{1}{3} < \frac{1}{2} < \frac{1}{1}\)
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Sun Jan 26, 2020 7:57 pm

Magnus Anderson wrote:You quoted me. I guess it's somehow relevant to what I said in that quote.

\(\frac{1}{\infty}\) is a number greater than \(0\) but smaller than every number of the form \(\frac{1}{n}, n \in N\). That's what it means (so it's not really meaningless.)

\(0 < \frac{1}{\infty} < \cdots < \frac{1}{4} < \frac{1}{3} < \frac{1}{2} < \frac{1}{1}\)


1/infinity is not a number. It’s hypothetically, a procedure. It is most definitely not a number.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Sun Jan 26, 2020 8:04 pm

It's not a procedure.

\(\infty\) represents a number greater than every number of the form \(n, n \in N\).

Similarly, \(\frac{1}{\infty}\) represents a number greater than \(0\) but less than every number of the form \(\frac{1}{n}, n \in N\).

\(0 < \frac{1}{\infty} < \cdots < \frac{1}{3} < \frac{1}{2} < \frac{1}{1} < 2 < 3 < \cdots < \infty\)

Also:
\(0.\dot9 + 0.\dot01 = 1\)

Note that \(0.\dot01\) or \(\frac{1}{10^\infty}\) is actually smaller than \(\frac{1}{\infty}\).
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