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...
10
33%
No, 1 ≠ 0.999...
15
50%
Other
5
17%
 
Total votes : 30

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

Postby Ecmandu » Tue Jan 07, 2020 6:50 pm

Magnus Anderson wrote:
Ecmandu wrote:Nah, unfortunately, that doesn’t work.

10-infinity is not a quantity that’s defined.

It’s undefined in this equation.

Think of it this way... walk up to anyone, even mathematicians and say, “dude! 10-infinity” or “dude! 10 minus the power of infinity!”

They’ll be like, “dude! what the fuck are you talking about?”

You’ll have to explain this to everyone because it makes no sense.


Doesn't work because you don't understand it?

You can replace \(10^{-\infty}\) with \(\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots\). It's the same quantity expressed differently. And it should have been obvious. \(10^{-\infty}\) is a compact representation and it's akin to a hyperreal number \(10^{-\omega}\).


Alright, so let’s look at it a different way.

When does the 1 in 1/10 ever occur? If it always occurs in 1:1 correspondence then that means that .111... equals .0...1.

Problem is, you have the carry from THE LAST DIGIT!!

Slight problem with that, there is no last digit.

Carrying doesn’t work from the first digit
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Tue Jan 07, 2020 8:19 pm

Ecmandu wrote:
Magnus Anderson wrote:
Ecmandu wrote:Nah, unfortunately, that doesn’t work.

10-infinity is not a quantity that’s defined.

It’s undefined in this equation.

Think of it this way... walk up to anyone, even mathematicians and say, “dude! 10-infinity” or “dude! 10 minus the power of infinity!”

They’ll be like, “dude! what the fuck are you talking about?”

You’ll have to explain this to everyone because it makes no sense.


Doesn't work because you don't understand it?

You can replace \(10^{-\infty}\) with \(\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots\). It's the same quantity expressed differently. And it should have been obvious. \(10^{-\infty}\) is a compact representation and it's akin to a hyperreal number \(10^{-\omega}\).


Alright, so let’s look at it a different way.

When does the 1 in 1/10 ever occur? If it always occurs in 1:1 correspondence then that means that .111... equals .0...1.

Problem is, you have the carry from THE LAST DIGIT!!

Slight problem with that, there is no last digit.

Carrying doesn’t work from the first digit


Oh, by the way QED.

I already pm’d prom and told him he can collect his reward.
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Re: Is 1 = 0.999... ? Really?

Postby Silhouette » Tue Jan 07, 2020 11:38 pm

At this point the same errors are recurring indefinitely, like this new \(10^{-\infty}\) notation still treating infinity like a finite quantity.

As Ecmandu correctly pointed out, there is no last \(10^{-1}\) in what that's supposed to denote, just the same as there is no last \(1\) in \(0.\dot0{1}\)

Until something comes up that hasn't already been disproven \(\lim_{n\to\infty}10^n\) times over, I'm gonna go ahead and leave the nonsense peddlers to it.
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Re: Is 1 = 0.999... ? Really?

Postby MagsJ » Wed Jan 08, 2020 12:03 am

Silhouette wrote:At this point the same errors are recurring indefinitely, like this new \(10^{-\infty}\) notation still treating infinity like a finite quantity.

As Ecmandu correctly pointed out, there is no last \(10^{-1}\) in what that's supposed to denote, just the same as there is no last \(1\) in \(0.\dot0{1}\)

Until something comes up that hasn't already been disproven \(\lim_{n\to\infty}10^n\) times over, I'm gonna go ahead and leave the nonsense peddlers to it.

Does reading up on last minute math make you an expert as you go along, Sil/Twit? It seems not!

I await both your responses with baited breath! Show me Math! Teach me! I doubt you can, but I know you’ll try.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Wed Jan 08, 2020 12:09 am

MagsJ wrote:
Silhouette wrote:At this point the same errors are recurring indefinitely, like this new \(10^{-\infty}\) notation still treating infinity like a finite quantity.

As Ecmandu correctly pointed out, there is no last \(10^{-1}\) in what that's supposed to denote, just the same as there is no last \(1\) in \(0.\dot0{1}\)

Until something comes up that hasn't already been disproven \(\lim_{n\to\infty}10^n\) times over, I'm gonna go ahead and leave the nonsense peddlers to it.

Does reading up on last minute math make you an expert as you go along, Sil/Twit? It seems not!

I await both your responses with baited breath! Show me Math! Teach me! I doubt you can, but I know you’ll try.


Actually, I know Magnus well enough that he’s going to take the last two posts very seriously ... you’re embarrassing yourself here.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Wed Jan 08, 2020 12:26 am

Silhouette wrote:At this point the same errors are recurring indefinitely, like this new \(10^{-\infty}\) notation still treating infinity like a finite quantity.


If you don't like the notation (merely because you don't understand it, I must say), you may consider its equivalent which is \(\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots\). That's a "valid" numerical representation, isn't it?

I've already stated it but you conveniently ignored it opting instead to focus on what you don't understand.

As Ecmandu correctly pointed out, there is no last \(10^{-1}\) in what that's supposed to denote, just the same as there is no last \(1\) in \(0.\dot0{1}\)


It seems like you're one of those rare people who can understand Ecmandu :)

There is no last \(1\) in \(0.\dot01\). The product \(\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots\) is infinite. The end is merely in the symbol.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Wed Jan 08, 2020 12:35 am

Magnus Anderson wrote:
Silhouette wrote:At this point the same errors are recurring indefinitely, like this new \(10^{-\infty}\) notation still treating infinity like a finite quantity.


If you don't like the notation (merely because you don't understand it, I must say), you may consider its equivalent which is \(\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots\). That's a "valid" numerical representation, isn't it?

I've already stated it but you conveniently ignored it opting instead to focus on what you don't understand.

As Ecmandu correctly pointed out, there is no last \(10^{-1}\) in what that's supposed to denote, just the same as there is no last \(1\) in \(0.\dot0{1}\)


It seems like you're one of those rare people who can understand Ecmandu :)

There is no last \(1\) in \(0.\dot01\). The product \(\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots\) is infinite. The end is merely in the symbol.


“The end is merely in the symbol”

There is no end!

I’m going to explain this as simply as possible!

You need a last digit to carry.

There’s is no last digit to 0.0...1.

The 1 never gets expressed.

In order for the 1 to get expressed. The added decimal HAS to be 0.111...

The problem with this, is that you have to carry from the first digit!

You can’t carry from the first digit!

Last digit = contradiction
First digit = contradiction

You’re now left with no wiggle room
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Wed Jan 08, 2020 1:34 am

A couple of questions for Ecmandu.

Ecmandu wrote:When does the 1 in 1/10 ever occur?


What does it mean to say that \(1\) occurs in \(\frac{1}{10}\)?

If it always occurs in 1:1 correspondence then that means that .111... equals .0...1.


I am not sure what it means for \(1\) to always occur in \(\frac{1}{10}\) in one-to-one correspondence.

What I know is that \(0.\dot1\) does not equal \(0.\dot01\) and that it's very easy to demonstrate this.

Problem is, you have the carry from THE LAST DIGIT!!


Why do you mention carrying?
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Wed Jan 08, 2020 1:44 am

Magnus Anderson wrote:A couple of questions for Ecmandu.

Ecmandu wrote:When does the 1 in 1/10 ever occur?


What does it mean to say that \(1\) occurs in \(\frac{1}{10}\)?

If it always occurs in 1:1 correspondence then that means that .111... equals .0...1.


I am not sure what it means for \(1\) to always occur in \(\frac{1}{10}\) in one-to-one correspondence.

What I know is that \(0.\dot1\) does not equal \(0.\dot01\) and that it's very easy to demonstrate this.

Problem is, you have the carry from THE LAST DIGIT!!


Why do you mention carrying?


My post about 1/10 ever being expressed implied your argument of 1/10*1/10*1/10.... etc....

I thought I could have used short hand, but it doesn’t bother me that we’re trying to be precise here, so, I apologize for the shorthand.

So here’s the deal: “carrying” is when you carry numbers from right to left. This is first grade math.

If I have a number like 99+2... the 2 plus the last 9 is 11. The one gets deposited in the solution column and the 10 gets carried over to the left in the tens column as a 1. 1+9 equals 10, which you put in front of the one that you already have the the solution column, to equal 101.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Wed Jan 08, 2020 1:56 am

Ecmandu wrote:So here’s the deal: “carrying” is when you carry numbers from right to left. This is first grade math.


That's correct.

If I have a number like 99+2... the 2 plus the last 9 is 11. The one gets deposited in the solution column and the 10 gets carried over to the left in the tens column as a 1. 1+9 equals 10, which you put in front of the one that you already have the the solution column, to equal 101.


Sure. But how is that relevant? That was my question.

Why do you mention carrying?

It's obvious that you're talking about the standard algorithm of adding numbers together. The question is: why?

Are you trying to use the standard algorithm to add \(0.999\dotso\) and \(0.000\dotso1\) together? If so, why?

Do you think that you can prove that there is no difference between \(0.999\dotso\) and \(1\) by proving that the standard algorithm of adding numbers together can't be used to add \(0.999\dotso\) and \(0.000\dotso1\) together?
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Wed Jan 08, 2020 1:58 am

Magnus Anderson wrote:
Ecmandu wrote:So here’s the deal: “carrying” is when you carry numbers from right to left. This is first grade math.


That's correct.

If I have a number like 99+2... the 2 plus the last 9 is 11. The one gets deposited in the solution column and the 10 gets carried over to the left in the tens column as a 1. 1+9 equals 10, which you put in front of the one that you already have the the solution column, to equal 101.


Sure. But how is that relevant? That was my question.

Why do you mention carrying?

It's obvious that you're talking about the standard algorithm of adding numbers together. The question is: why?

Are you trying to use the standard algorithm to add \(0.999\dotso\) and \(0.000\dotso1\) together? If so, why?

Do you think that you can prove that there is no difference between \(0.999dotso\) and \(1\) by proving that the standard algorithm can't be used to add \(0.999dots\) and \(0.000\dotso1\) together?


Ok, fine. Notice I don’t back up to former arguments, I always use new ones.

Let me ask you this from a utilitarian perspective.

Is there anything that 0.9... could be used for that 1 cannot ?
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Re: Is 1 = 0.999... ? Really?

Postby Silhouette » Wed Jan 08, 2020 2:00 am

MagsJ wrote:Does reading up on last minute math make you an expert as you go along, Sil/Twit? It seems not!

I await both your responses with baited breath! Show me Math! Teach me! I doubt you can, but I know you’ll try.

Actually I've known all this math since over half my life ago, and I have been regularly keeping up with advanced level maths to this day for fun - I actually really enjoy it.

The problem here though is if others aren't close enough to the same level, they aren't going to recognise this and will just assume it's just another bogus claim from some randomer on the internet because they don't know any better.

I'm dealing with a self-professed non-mathematician here and it shows. All he has left anymore is "you just don't understand", and when I say/prove that I do, he can just respond with incredulity and keep repeating the accusation. That's all this thread is now - non-mathematicians claiming long-time competent mathematicians don't understand, whilst repeating the same schoolboy mistakes over and over. Other competent mathematicians here recognise these mistakes too, and keep advising the non-mathematicians that they're wrong and I'm right. But our old-friend "the backfire effect" is just making them double down harder because admitting you're wrong is hard for people since it requires emotional maturity and intellectual honesty to overcome the cognitive dissonance. That's why there's no longer any point in me trying to help them anymore - they don't want it.

I only recently learned all the MathJax code to express the math nicely on this forum though - I already thanked Magnus for his only contribution so far of bumping the post from a couple of years ago that mentioned it and linked to its documentation. I also unequivocally accept the utility of exploring what happens when you take something that doesn't work mathematically, and treat it as something that does work - as I've mentioned several times by now. All I'm doing is explaining exactly why it doesn't work mathematically - I'm not even against doing it anyway because that's where new math can be born, but there's still a couple of people left who find even that too much to accept.

See, I'm humble enough to give credit where credit is due, and I'm simply honest about people being wrong when they are.
There's plenty of topics on this forum that I leave alone because I don't know enough about them - you won't find me reading up last minute on those topics and claiming expertise there, and you'll only find me claiming expertise when I actually have it, such as here. There's far better mathematicians out there than me, but the dissenters here most definitely aren't one of them.
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Re: Is 1 = 0.999... ? Really?

Postby MagsJ » Wed Jan 08, 2020 2:05 am

Ecmandu wrote:
MagsJ wrote:
Silhouette wrote:At this point the same errors are recurring indefinitely, like this new \(10^{-\infty}\) notation still treating infinity like a finite quantity.

As Ecmandu correctly pointed out, there is no last \(10^{-1}\) in what that's supposed to denote, just the same as there is no last \(1\) in \(0.\dot0{1}\)

Until something comes up that hasn't already been disproven \(\lim_{n\to\infty}10^n\) times over, I'm gonna go ahead and leave the nonsense peddlers to it.

Does reading up on last minute math make you an expert as you go along, Sil/Twit? It seems not!

I await both your responses with baited breath! Show me Math! Teach me! I doubt you can, but I know you’ll try.


Actually, I know Magnus well enough that he’s going to take the last two posts very seriously ... you’re embarrassing yourself here.

Fuck off, Tramp!
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Wed Jan 08, 2020 2:07 am

Silhouette wrote:The problem here though is if others aren't close enough to the same level, they aren't going to recognise this and will just assume it's just another bogus claim from some randomer on the internet because they don't know any better.



If you ask me, you spend a little bit too much time telling us what you think about yourself.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Wed Jan 08, 2020 2:10 am

Ecmandu wrote:Let me ask you this from a utilitarian perspective.

Is there anything that 0.9... could be used for that 1 cannot ?


That's exactly the point of this thread. \(0.\dot9\ \neq 1\) means the answer to your question is "Yes".
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Wed Jan 08, 2020 2:12 am

Magnus Anderson wrote:
Ecmandu wrote:Let me ask you this from a utilitarian perspective.

Is there anything that 0.9... could be used for that 1 cannot ?


That's exactly the point of this thread. \(0.\dot9\ \neq 1\) means the answer to your question is "Yes".


That’s my side! What the fuck are we arguing about?

Oh! That’s right! You think if you can add something to .9... that it doesn’t need to be added to 1!

So here we are!
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Wed Jan 08, 2020 2:17 am

Ecmandu wrote:That’s my side! What the fuck are we arguing about?


Yes, that's where we agree. Where we disagree is whether or not infinities come in different sizes. You think that they don't (that's where you agree with SIlhouette.)

Oh! That’s right! You think if you can add something to .9... that it doesn’t need to be added to 1!


Not sure what that means.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Wed Jan 08, 2020 2:24 am

Magnus Anderson wrote:
Ecmandu wrote:That’s my side! What the fuck are we arguing about?


Yes, that's where we agree. Where we disagree is whether or not infinities come in different sizes. You think that they don't (that's where you agree with SIlhouette.)

Oh! That’s right! You think if you can add something to .9... that it doesn’t need to be added to 1!


Not sure what that means.


So let’s get this straight!

You, sil and I all agree that 0.9... /= to 1

*phew*

Glad we got that out of the way!

I’m only speaking for myself, but I think both Sil and I thought you were defending the equality.

Cool! So let’s talk about orders now!

In response to your last question, if 1=0.9....

Then whatever you add (infA) to one, you have to add to the other (which makes the inequalities)

Does that clarify?
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Wed Jan 08, 2020 2:31 am

Ecmandu wrote:So let’s get this straight!

You, sil and I all agree that 0.9... /= to 1


I think that Silhouette thinks that \(0.\dot9 = 1\).

Here's Silhouette's "proof" that \(9.\dot9 = 10\).

Silhouette wrote:But enough about the visual test you're proposing - the only "exact" methods are going to have to transcend the practical into the theoretical, for which you have to accept the math with its concepts such as "one-to-one correspondence" in infinite series:

$$s=\sum_{x=0}^\infty \frac9{10^x}\to \frac{s}{10}=\sum_{x=1}^\infty \frac9{10^x}$$
$$s-\frac{s}{10}=\sum_{x=0}^0 \frac9{10^x}=9=s(1-\frac1{10})$$
$$s=\frac9{0.9}=10=\sum_{x=0}^\infty \frac9{10^x}=9.\dot 9$$


Ecmandu wrote:*phew*

Glad we got that out of the way!


I'm afraid we didn't (:

Cool! So let’s talk about orders now!

In response to your last question, if 1=0.9....

Then whatever you add (infA) to one, you have to add to the other (which makes the inequalities)

Does that clarify?


Not quite.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Wed Jan 08, 2020 2:37 am

Am I misunderstanding everyone?!?!

That wouldn’t be the first time.

Let’s let sil answer.

As for the last part:

If 0.999... plus 0.0...1 equals 1, then!

If 0.9... equals 1, then!

1 plus 0.0...1 must equal as well (which it doesn’t)
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Re: Is 1 = 0.999... ? Really?

Postby MagsJ » Wed Jan 08, 2020 2:43 am

Ecmandu wrote:Am I misunderstanding everyone?!?!

That wouldn’t be the first time.

Let’s let sil answer.

As for the last part:

If 0.999... plus 0.0...1 equals 1, then!

If 0.9... equals 1, then!

1 plus 0.0...1 must equal as well (which it doesn’t)

Piece of shit scum!

Don’t bother reply to me!
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Wed Jan 08, 2020 3:02 am

MagsJ wrote:
Ecmandu wrote:Am I misunderstanding everyone?!?!

That wouldn’t be the first time.

Let’s let sil answer.

As for the last part:

If 0.999... plus 0.0...1 equals 1, then!

If 0.9... equals 1, then!

1 plus 0.0...1 must equal as well (which it doesn’t)

Piece of shit scum!

Don’t bother reply to me!


You didn’t even get that I was defending Magnus the whole time either, neither did Magnus, and apparently neither did sil...

Apparently we had cross discussions occurring when I only thought it was one discussion, a triangulation of sorts:

Magnus : 0.9... /= 1, orders of infinity exist
Silhouette: 0.9... does equal 1, no orders of infinity exist

Ecmandu/me: 0.9.../=1, no orders of infinity exist
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Re: Is 1 = 0.999... ? Really?

Postby Silhouette » Wed Jan 08, 2020 3:19 am

Magnus Anderson wrote:If you ask me, you spend a little bit too much time telling us what you think about yourself.

I'm not asking you, because you're not showing rationality in dealing with me.

It's okay for you to tell us all about yourself and how you're not a mathematician, but when I inform someone who isn't you that I am - that's a little too much for you...

And I'm only just bringing up myself now because I've finished trying to help you with the topic and I'm explaining something to someone else. But any excuse to take a swipe at me since you failed whilst we were on topic, right? So low.

FYI, this "standard algorithm" of addition that you're trying to bring into doubt is known as "addition". You'll even blame the literal math itself as lacking because it can't find this non-existent gap that your non-mathematician intuitions can't let go of. If you want to detract from one of the fundamental bases of mathematics, I suggest doing what even the best mathematicians that have ever lived still haven't done to this day and "improve" addition itself just to bring your abstract and metaphorical notions (for which you say you don't have a rigorous proof because you don't think it's necessary) into literal existence.

Ecmandu, I am saying \(1 = 0.\dot9\).
You were literally just explaining that the difference between the values never arrives because the infinite zeroes never ever get to that mythical 1 "at the end of infinity" (a contradiction in terms)...
In the same way that this difference turns out to not actually exist, \(1 + "0.\dot01"\) will just be \(1\) too.

Has nobody actually bothered to check what all the professional mathematicians have concluded on this subject?
I'll assume not, because if you do a quick search you'll find they all side with me that \(1 = 0.\dot9\). It's almost as if I wasn't kidding when I said I knew what I was talking about, and I haven't just been posturing all along like the dissenters here.
But I guess all the amateurs here can reject that because they know better, huh? This is just an appeal to authority after all, so it can be ignored if it doesn't match your misunderstandings right? :icon-rolleyes:
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Wed Jan 08, 2020 3:27 am

Alright Silhouette,

Now that we all know our respective sides and have been talking past each other for pages now...

I submit the same question that I submitted to Magnus ...

From a utilitarian perspective, is 0.9... useful in a way that 1 is not?

I know you’ll see the difference there.

So I have this to ask of you, “is your dogged defense of this rational given that “every mathematician” agrees with you.

What’s ultimately funny about this thread to me (now that I’ve got my bearings) is that we’re all debating each other about something!

May the best argument(s) win.
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Re: Is 1 = 0.999... ? Really?

Postby gib » Wed Jan 08, 2020 3:27 am

Magnus Anderson wrote:An infinite number of inches is equal to a smaller infinite number of 90 inch long segments and an even smaller infinite number of 93.75 inch long segments. There is no contradiction. You're just doing to the math the wrong way.


So you agree that the length of the lines is the same even though there are fewer 90-inch segments than 1-inch segments. But that would imply that if you counted the segments in each, you would finish counting the 90-inch segments before the 1-inch segments (that's what "fewer" means). That in turn implies that at least the number of 90-inch segments is finite.

Magnus Anderson wrote:The only reason you think that length implies a beginning and an end is because you're used to working with line segments.


That's sorta kinda how we derive the definition. We work with line segments and say: "Hey, why don't we refer to the distance between one end and the other as the line's 'length'". If you want to expand the definition to mean something different with infinite lines, you're on your own.

Magnus Anderson wrote:The other problem is that you're confusing conceptual matters with empirical ones. The same mistake that Silhouette is making.


Yeah, 'cause heaven forbid we confuse conceptual matters with empirical ones. You'd get science!!! :o

Magnus Anderson wrote:It does not.


Does so! :teasing-neener:

Magnus Anderson wrote:
gib wrote:I think it went over your head

Very convenient.


I'm sure it was convenient for you.

Magnus Anderson wrote:Why is infinity not a number?


Really???

Magnus Anderson wrote:It's not an unknown. And the word "infinity" does not mean "the highest number possible".


Exactly! You're arguing my point.

Magnus Anderson wrote:Meters, centimeters, inches, feet, etc are not undetermined things. \(1m = 100cm\) is not the same as \(1 \times x = 100 \times y\) where \(x\) and \(y\) are unknowns. Units \(\neq\) unknowns.


You're right about that, but variables (unknowns) can be treated like units, and visa-versa. 3X means 3 times the quantity X, but X here can be thought of as a unit, as in 3 times some object we call X (where this object is a set of X things). It works the other way around too: units can be treated like variables: 3cm = 3 x (100mm). You're right that we know how much a centimeter is and how much a millimeter is, but if I tell you 1cm = n gwackometers, it suddenly becomes unknown (you don't know how much a gwackometer is). So 1cm = an unknown number of gwackometers. The fact that you can divide a centimeter into any number of arbitrarily long segments means that it is an unknown in the sense that we don't know how many of these arbitrary segment there are.

The central insight in algebra is that it doesn't matter what the variables (or units) stand for--they might as well be unknowns--its the rules for manipulating them that matter. And that was my point about plugging \(\infty\) into algebraic equations--it doesn't matter what it stands for, it's just a symbol--manipulate to your hearts content. You don't end up prooving anything.

Magnus Anderson wrote:That's what it means with respect to line segments. And if we assume that the concept of length applies only to line segments, it does not follow that rays don't have something similar. Either way, you've got nothing.


Whatever it is you think this "similar" length is with respect to rays, you are once again on your own (and strictly speaking, we're talking about terms like "shorter" and "longer").

Magnus Anderson wrote:The symbol \(\infty\) represents an infinite quantity; believe it or not, even when I'm doing arithmetic with it. It certainly does not play the role of an unknown because the quantity is known -- it's not unknown. And yes, you can use any symbol you want (e.g. you can use \(\omega\) if you're into hyperreal numbers or \(\text{infA}\) if you prefer James's notation) but that does not mean we've somehow changed what we're representing. We're representing the same thing: infinite quantity.


"Infinite quantity" isn't really a quantity. That's just a phrase we use to talk about infinity. If you want to say you know what \(\infty\) stands for in the equation (namely infinity), then you're simply doing something you can't do. It doesn't stop it from being a symbol which you can plug into the equation and manipulate according to the rules of algebra. I can do the same with the symbol for male:

\(\frac{{6}\times{♂}}{3}\) = \({2}\times{♂}\)

Does this prove anything about males? Do you think it proves males are quantities?

Magnus Anderson wrote:Saying "1 of something symbolized by ∞" is exactly the same as saying "an infinite number" (such as an infinite number of train carts) because ∞ means "infinite number".


Then you're saying \(\infty\) is both the unit and the quantity at the same time. I take back what I said. I don't agree that you can do this. You can still manipulate the symbols according to the rule of algebra, but you're doing nothing but shuffling symbols around. You're not actually deriving any insight into the nature of infinity.

The only time I've seen \(\infty\) used in an equation is:

\(\frac{1}{0}\) = \(\infty\)

...and \(\infty\) in this case explicitly means "undefined"--as in "you can't do that!" You're grade school teacher taught you not that dividing by zero gives you a really, really, really big number, but that you can't divide by zero.

Magnus Anderson wrote:That's exactly what ∞ means. It means "an infinite number of things". The fact that I'm treating it like a unit DOES NOT change that fact.


Ok, Magnus, you're insisting that we treat \(\infty\) as an infinite quantity. Fine. I actually agree, \(\infty\) does mean "an infinite number of things". But then I deny your right to use it in math. You're trying to sneak it into math by, simultaneously, treat it as a unit--which, as I said, you can do but not while also treating it as a quantity. What number does 3cm equal? That is, how do you get rid of the units so that you're left with only a number? 3cm = X. What is X? <-- That's what you're trying to do with \(\infty\). You're trying to say 3\(\infty\) = X, and since \(\infty\) is an infinitely large number, you can plug that number into \(\infty\), multiply it by 3, and solve for X.

Magnus Anderson wrote:When I use the symbol m to represent one hundred centimeters, does it suddently stop representing one hundred centimeters? Of course not. THAT'S EXACTLY THE PURPOSE OF SYMBOLS.


What about the formula E = mc^2? We know what c is (speed of light), but what is m? It's a symbol for mass, but do you know what the mass is?
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