Do you think 0.9999999~ =1 , that is infinite 9s. [Archive] - Page 9

Bogie

2008-May-03, 07:31 PM

No. Grapes' questions first. They've been on the table far, far longer.Moose, here’s the thing. After you learn how some people view you, you realize that they are committed to some agenda that you’re not going to like. You don’t know what it is but you can bet you’re not going to like it. You become suspicious. You feel that by playing the nice guy again for them you give them an advantage to work you over one more time, and so you are disadvantaging yourself by doing that. Do you see? So why would I ever believe that you have any good intentions? Why would I disadvantage myself by playing into some hidden agena of yours?

If there is no hidden agenda and you really don't hold me in low esteem as I think you do, you answer my question if you have any good intentions. I'm sure I can lead you through a series of steps to show my point.

And who is Grape?

worzel

2008-May-03, 07:35 PM

I was blackballed by a 20 plus poster and so I am no longer the Con spokesperson.
You weren't black balled. It would be as presumptuous of me (or anyone else here) to accept your spokesmanship for the doubting lurkers as it would be for you to presume to speak for them.

Who is really missing the point?
Who's failing to make one? Please get over your persecution complex and just get to the point. I repeat my earlier question:

In what way do you disagree with the equality if you accept it mathematically?

Oh, and please answer Grapes' question honestly and directly.

Disinfo Agent

2008-May-03, 07:41 PM

And who is Grape?Grapes is hhEb09'1, the poster formerly known as Grapes of Wrath.

worzel

2008-May-03, 07:43 PM

And who is Grape?
Grapes is hhEb09'1. Answer his question!

Bogie

2008-May-03, 07:53 PM

Grapes is hhEb09'1. Answer his question!I could go through the posts and come up with ten questions that I have asked that haven't been answered.

Also I have explained myself enough times for you to get it.

I have had to respond to some posts in a manner I don't like to have to use but that is truthful. They go unacknowledged. And yet the gang insists that I answer their questions. I don't think you really know what equality is ...~.

hhEb09'1

2008-May-03, 07:57 PM

I was trying to answer one of your questions here (http://www.bautforum.com/off-topic-babbling/14593-do-you-think-0-9999999-1-infinite-9s-67.html#post1233208), but I asked for clarification.

Red Dawn

2008-May-03, 08:07 PM

Are you saying that you thought I'd dodge the question and were surprised when I didn't. Or are you saying you couldn't find my answer in my posts?

The latter. I can't find the answer in your posts, because it's not there.

The P. T. Barnum act is getting old.

Bogie

2008-May-03, 08:20 PM

I was trying to answer one of your questions here (http://www.bautforum.com/off-topic-babbling/14593-do-you-think-0-9999999-1-infinite-9s-67.html#post1233208), but I asked for clarification.Earlier, a few days ago, I tried answering everyones questions and at the same time I asked questions of my own. Five people had their own version of the arithmetic proof that they wanted me to walk through with them. I naively answered a few of those questions even though my questions we being ignored. Moose turned that around on me with disdain. Now I should trust him?

Currently there are a few throwing questions, demands and comments all at the same time. Just like it asked Moose about why I should disadvantage myself that way again, I ask you ... acknowledge my questions, acknowledge my expressed perception of the attitude of several I have been forthright with and who totally ignore my forthrightness, and answer my question:

What do you get when you divide an infinite by any number?

I'm confident if you answer me I can step you through a serious of posts that will prove my point.

The fact that there a several people at once who are throwing the same question at me seems to confirm my doubts about a hidden agenda to again express disdain.

Bogie

2008-May-03, 08:24 PM

The latter. I can't find the answer in your posts, because it's not there.

The P. T. Barnum act is getting old.Good one.

I think Disinfo Agent saw my answer. If he did not he could say otherwise. If he did, and you make statements like "The P. T. Barnum act is geting old" because you can't see the answer, then I don't know what to tell you.

Disinfo Agent

2008-May-03, 08:47 PM

I think Disinfo Agent saw my answer. If he did not he could say otherwise.I'm afraid I don't know what is the question. Isn't it simpler just to reply again?

Red Dawn

2008-May-03, 08:52 PM

I'm afraid I don't know what is the question.

The question I asked, was whether the value of:

0.9+0.11-0.011+0.0011-0.00011+0.000011-0.0000011+...

was greater than one or less than one. Bogie said we would take it upon himself as representative of the cons to answer my question, but has quite persistently declined to do what he said he would do.

Isn't it simpler just to reply again?

Or for the first time :)

This isn't the best web site I have seen, but it's definitely the funniest :)

hhEb09'1

2008-May-03, 09:07 PM

Currently there are a few throwing questions, demands and comments all at the same time. Just like it asked Moose about why I should disadvantage myself that way again, I ask you ... acknowledge my questions, acknowledge my expressed perception of the attitude of several I have been forthright with and who totally ignore my forthrightness, and answer my question:

What do you get when you divide an infinite by any number?

I'm confident if you answer me I can step you through a serious of posts that will prove my point.That was the question that I asked for clarification about.

If by "infinite" you mean, a real number represented by an infinite decimal expansion, then dividing an infinite by any real number results in a real number.

Did I interpret your question correctly?

Bogie

2008-May-03, 09:15 PM

That was the question that I asked for clarification about.

If by "infinite" you mean, a real number represented by an infinite decimal expansion, then dividing an infinite by any real number results in a real number.

Did I interpret your question correctly?Are real numbers all infinite decimal expanions (just answer this, I know it sounds like a stupid question but it is one of the steps) and it is an easy answer, can you say yes or no?.

hhEb09'1

2008-May-03, 09:28 PM

Your turn to answer a question I think

What do you mean by 0.999... divided by 3? Does it have any meaning at all for you?

Bogie

2008-May-03, 09:49 PM

Your turn to answer a question I think

What do you mean by 0.999... divided by 3? Does it have any meaning at all for you?I don't think so.

Did you read this post:
http://www.bautforum.com/off-topic-babbling/14593-do-you-think-0-9999999-1-infinite-9s-68.html#post1233313

Moose, here’s the thing. After you learn how some people view you, you realize that they are committed to some agenda that you’re not going to like. You don’t know what it is but you can bet you’re not going to like it. You become suspicious. You feel that by playing the nice guy again for them you give them an advantage to work you over one more time, and so you are disadvantaging yourself by doing that. Do you see? So why would I ever believe that you have any good intentions? Why would I disadvantage myself by playing into some hidden agena of yours?

If there is no hidden agenda and you really don't hold me in low esteem as I think you do, you answer my question if you have any good intentions. I'm sure I can lead you through a series of steps to show my point.
And I asked you directly:
http://www.bautforum.com/off-topic-babbling/14593-do-you-think-0-9999999-1-infinite-9s-68.html#post1233271

Currently there are a few throwing questions, demands and comments all at the same time. Just like I asked Moose about why I should disadvantage myself that way again, I ask you ... acknowledge my questions, acknowledge my expressed perception of the attitude of several I have been forthright with and who totally ignore my forthrightness, ...What part of that do you not understand?

And you do deserve an atta boy :D for the "real" number answer.

Moose

2008-May-03, 10:10 PM

First, I have exactly one agenda in this thread: correct math. For exactly the same reason I don't want ATM posts outside of ATM, I don't want bad math floating around BAUT, confusing newbies. That's it. Anything else you think you sense from me is your own imagination.

And you do deserve an atta boy :D for the "real" number answer.

And second, I have to ask you to clarify because your choice of language in that sentence is just a hair ambiguous. You are aware Grapes is referring to the set of real numbers (http://en.wikipedia.org/wiki/Real_number), right, and not numbers that are real (in the "exists" sense)?

Third, if this is what you're asking me:

What do you get when you divide an infinite by any number?

You're going to need to clarify what you mean by "an infinite". That's non-standard terminology, and as a colloquialism, it's ambiguous. Do you mean infinity or an infinite repeating progression?

Infinity / 2 = infinity. It's weird, it's non-intuitive, but there it is.

An infinite repeating progression (x) is x/2. Until and unless you cite me a specific case, that's the only correct answer you can be given. If you're asking about 0.999~ / 2, it's 0.4999~ = 0.5.

Your turn:

When you say 0.999~ / 3 has no meaning for you, are you saying 0.999~ / 3 has no answer, or are you declining to give one?

Bogie

2008-May-03, 10:23 PM

First, I have exactly one agenda in this thread: correct math. For exactly the same reason I don't want ATM posts outside of ATM, I don't want bad math floating around BAUT, confusing newbies. That's it. Anything else you think you sense from me is your own imagination.

And second, I have to ask you to clarify because your choice of language in that sentence is just a hair ambiguous. You are aware Grapes is referring to the set of real numbers (http://en.wikipedia.org/wiki/Real_number), right, and not numbers that are real (in the "exists" sense)?

Third, if this is what you're asking me:

You're going to need to clarify what you mean by "an infinite". That's non-standard terminology, and as a colloquialism, it's ambiguous. Do you mean infinity or an infinite repeating progression?

Infinity / 2 = infinity. It's weird, it's non-intuitive, but there it is.

An infinite repeating progression (x) is x/2. Until and unless you cite me a specific case, that's the only correct answer you can be given. If you're asking about 0.999~ / 2, it's 0.4999~ = 5.

Your turn:

When you say 0.999~ / 3 has no meaning for you, are you saying 0.999~ / 3 has no answer, or are you declining to give one?OK Moose. You are clearly over you disdain for me. My suspicions have been alleviated.

Why don't we leave it at that. I don't' think the Twelve Angry Men scenario applies when what seems to be defined in limit theory is not considered a definition. I understand that the whole inner structure of math depends on denying that there are definitions and rules, otherwise admitting to rules in math wouldn't be so hard for all of you.

And I won't be able to get you or others to even understand me if I don't use the jargon, so I'll leave with the acknowledgment that you are all too smart for me.

Moose

2008-May-03, 10:31 PM

OK Moose. You are clearly over you disdain for me. My suspicions have been alleviated.

My personal opinion of you is unchanged, unexpressed, uninvolved, and utterly irrelevant to this discussion.

Answer Grapes' question: do you have an answer for 0.999~ / 3?

Bogie

2008-May-03, 10:44 PM

My personal opinion of you is unchanged, unexpressed, uninvolved, and utterly irrelevant to this discussion.

Answer Grapes' question: do you have an answer for 0.999~ / 3?You just said I couldn't speak your language and I admit it.

You said it is not a rule and went into a discussion of how math must be internally consistent and not based on rules. That means that math has changed since I was in school, I remember a bunch of rules in math. I'm not qualified to discuss math with you and so if by acknowledging the equality in the OP based on rules in math is not good enough I don't know what to say.

Just for the record, what answer are you driving for? Just Yes it is an equality without any caveats?

If I add the caveat about it being a rule then my yes is not good enough for you?

Moose

2008-May-03, 11:02 PM

If I add the caveat about it being a rule then my yes is not good enough for you?

I want for you what I want for everybody who comes through this thread: to understand why 0.999~ = 1. When I'm sure you've understood, it'll be 'good enough'.

Bogie

2008-May-03, 11:08 PM

I want for you what I want for everybody who comes through this thread: to understand why 0.999~ = 1. When I'm sure you've understood, it'll be 'good enough'.Are you just saying that it is not a rule that an infinite progression is equal to the limit that it approaches?

Moose

2008-May-04, 12:01 AM

Are you just saying that it is not a rule that an infinite progression is equal to the limit that it approaches?

Pretty much.

As I've explained before, it really isn't a rule, it's a consequence. "Rule" implies arbitrariness, and that's simply not the case. This isn't arbitrary at all. 0.999~ = 1 is a property, a consequence, of how the "real" number set was defined. (It's probably fair to call negative numbers, zero, integers, decimals and infinity 'arbitrary definitions' or 'rules' at some level.)

Nobody decided to make 0.999~ = 1. That property had to be discovered, proven, then generalized as what we now understand as limit theory, and the theory itself proven.

But that neat property of 1 existed from the moment the "real" number set and its basic operators were defined. Lying dormant, undiscovered, but always there. That's what I mean when I say it's a consequence.

What I find pretty about 0.999~ = 1 is that you can show it to kids, and they can amaze their friends with it. Done right, it's a confidence builder (considering this is technically university-level math.) Maybe build up some interest for math in the hopes of insulating them from the unperformance-pressures of middle-school.

Ultimately, if 0.999~ = 1 can be thought of as a rule, it's only because the entirety of mathematics, including our understanding of the value of '1', is part of the very same rule.

Moose

2008-May-04, 12:20 AM

I just wanted to quote something JayUtah wrote in the CT forum last night. It seems appropriate to this discussion:

Specialized (http://www.bautforum.com/1233197-post110.html) vocabularies develop among practitioners out of necessity. I cannot be effective as an engineer without being able to communicate concisely and accurately to my colleagues according to our conventions and standards. Doctors, lawyers, sailors, musicians, and other highly trained professionals constantly express the same need. Detailed and complex subjects require precise nomenclature in order to avoid confusion. Nothing prevents newcomers from learning to understand specialized vocabularies.

worzel

2008-May-04, 05:08 AM

I could go through the posts and come up with ten questions that I have asked that haven't been answered.
List the outstanding questions you wish answered and I will give my answer to each one.

The only question from you that I've seen is "What do you get when you divide an infinite by any number?" You've been asked for clarification on what you mean by "an infinite". Once you've clarified then I can answer. You can't expect an answer when I don't know what you mean, can you?

Also I have explained myself enough times for you to get it.
Your position has changed over the course of the thread, because you started out by disagreeing with the mathematical equality (you were trying to find fault with the mathematical proofs offered). Following your change in position my question was:

In what way do you disagree with the equality if you accept it mathematically?

Could you please either answer it or quote the post in which you think you have already answered it. Any post prior to your acceptance of the mathematical equality is obviously irrelevant to that question.

worzel

2008-May-04, 05:15 AM

Bogie, it is comments like this:

Yes, in mathematical terms. But in reality you cannot produce anything that is equal to .999~ because infinity can not be attained in the real world. Lucky that we have minds that can grasp math so we can talk about the difference between math and reality.

that I want clarification on. You have yet to explain what reality there is to 0.999~ outside of mathematics.

TheHalcyonYear

2008-May-04, 07:44 AM

I just wanted to quote something JayUtah wrote in the CT forum last night. It seems appropriate to this discussion:

Specialized vocabularies develop among practitioners out of necessity. I cannot be effective as an engineer without being able to communicate concisely and accurately to my colleagues according to our conventions and standards. Doctors, lawyers, sailors, musicians, and other highly trained professionals constantly express the same need. Detailed and complex subjects require precise nomenclature in order to avoid confusion. Nothing prevents newcomers from learning to understand specialized vocabularies.

Exactly... there are plenty of calculus books for those who are interested.

worzel

2008-May-04, 08:10 AM

hhEb09'1

2008-May-04, 09:39 AM

Are you just saying that it is not a rule that an infinite progression is equal to the limit that it approaches?
There is a more natural way of approaching the issue that doesn't rely on jargon or arbitrary rules or advanced mathematical reasoning, that I thought might appeal to you, based on some of the objections you raised in previous posts. I was hoping to present that, but you seem to think that the dialogue I started was somehow to your disadvantage. So, I'll just wrap it up in this post.

You answering that 0.444... divided by 2 was 0.222... (in this post (http://www.bautforum.com/showthread.php?p=1229233#post1229233)) was consistent with your later post (http://www.bautforum.com/showthread.php?p=1232003#post1232003), where you presented a dialogue with students. Those students agree that 3/9 is 0.333..., probably because when they sit down to compute 3/9 on paper, using long division, they quickly realize that the process will produce a string of 3s without terminating.

0.999... is more problematic, because there is no long division that will produce that, using the usual procedure. And you end your post with a justification for the 0.999... that seems to depend upon an arbitrary rule, and you ask:

So the proof in reality is not possible and requires a rule.

Where am I wrong?
However, as you said, dividing 0.444... by 2 is easy with the long division procedure. It's clearly 0.222....

The question that I have been asking ("What 0.999... is divided by 3?"), clearly leads to 0.333..., using similar reasoning, if one conditionally accepts the possibility that 0.999... even exists.

But, if the student accepts

0.999... /3 equals 0.333...

and has already accepted that

0.333... equals 3/9

it's an elementary step from those two equations, to the following equation:

0.999.../3 equals 3/9

If 0.999... exists, of course. Multiplying both sides by 3 results in

0.999... equals 9/9

And we've accomplished that without jargon or arbitrary rules or advanced mathematical reasoning. Do you agree?

Moose

2008-May-04, 10:01 AM

I think it's quite reasonable to ask Bogie what he means by "an infinity" in his question without requiring him to conform to any jargonistic norms.

I also think it is quite reasonable for people to enter into these sorts of discussions with out them first having to read up on the jargon.

Was in no way suggesting he had to, although an effort to try to use terms correctly is always appreciated.

I quoted Jay mostly to call attention to words like "set", "real", "definition" which have very specialized meanings in math, so they should be used carefully. It also goes very much to why I've been objecting to the word "rule" in the colloquial context.

HenrikOlsen

2008-May-04, 10:21 AM

Bogie has stated clearly that 0.999~ = 1, and that he's only fluffing around on whether it's because it just is or whether it's because someone defined it to be.

This has resulted in several pages of troll feeding which is doing absolutely nothing to clarify the question for anyone, least of all Bogie who is by his own words taking sides depending on who he likes rather than on any merit of the arguments.

This should stop now.

Bogie, if you post anything deliberately trollish in this thread again, you'll get a suspension.

mickal555

2008-May-04, 01:02 PM

Ha ha

oh wow.

Another 20 pages added. This will never end :D

At least no one is claiming pi isn't a number though.

Kaptain K

2008-May-04, 04:44 PM

To paraphrase my earlier post (which, itself was a paraphrase of ELP) "Welcome back my friends, to the thread that's about to end."

korjik

2008-May-04, 06:05 PM

Moose

2008-May-04, 06:27 PM

Mmm. I'd agree with that. Is there really anything more to say that hasn't already been said a few dozen times?

Bogie

2008-May-04, 07:23 PM

I apologize to the thread for giving the appearance of trolling.
I apologize to the posters on the thread for any posts that were out of order or antagonistic.
I wish I could change my vote.
I agree that the thread should be closed.

Chuck

2008-May-04, 08:37 PM

This has been dicuessed to death. There must be something else to argue about. We've done The Doomsday Argument and The Monty Hall Problem. Have we done Hempel's Ravens (http://en.wikipedia.org/wiki/Raven_paradox) yet?

hhEb09'1

2008-May-04, 10:00 PM

Make the hurting stop, please!
No problem. Just stop doing that. :)

Seriously, this thread serves a useful purpose.

Neverfly

2008-May-04, 10:46 PM

No problem. Just stop doing that. :)

Seriously, this thread serves a useful purpose.

I agree.
And I disagree with those who pop up occasionally commenting on how even a child can understand the difference. It takes a bit of discussion before it clicks into place.

Moose and myself both admitted to being on the other side of the fence before. I don't know how many others were before it was explained and they slapped their forehead and said "Doh!"

agingjb

2008-May-05, 08:41 AM

I think the 108 votes for "no", even if a few of them might be changed with a more flexible polling feature, do serve a useful purpose. I suspect that this utility is not entirely to the benefit of the reputation of Baut.

Neverfly

2008-May-05, 08:45 AM

Ok so close this thread and open a new poll. Everyone then go there and vote "yes" and it will all be good.
Then slather it in BAUT Bug Juice so that the Google Slurp Spider goes and slurps that thread first.

Ouso

2008-May-05, 09:58 AM

fight the good fight

Disinfo Agent

2008-May-05, 01:21 PM

Ok so close this thread and open a new poll. Everyone then go there and vote "yes" and it will all be good.Been there, done that... (http://www.bautforum.com/off-topic-babbling/16806-where-do-you-stand-now.html)

TheHalcyonYear

2008-May-05, 04:31 PM

I agree.
And I disagree with those who pop up occasionally commenting on how even a child can understand the difference. It takes a bit of discussion before it clicks into place.

Moose and myself both admitted to being on the other side of the fence before. I don't know how many others were before it was explained and they slapped their forehead and said "Doh!"
If you have difficulty with a basic concept like this one, you are gunna be in for quiet a shock when you hit differential equations!!

Disinfo Agent

2008-May-05, 04:44 PM

Actually, I think that it's quite possible to grasp complicated concepts well, and get stuck on a seemingly simple one like this. My guess is that somehow fully understanding what is meant by decimal expansions requires a higher level of abstraction than solving a (simple) differential equation. It's similar to how some people can go through an entire calculus course, and come out thinking that the limit of a convergent sequence can never be attained, or that a series doesn't quite equal the limit of its partial sums.

This is why I agree that these discussions are useful, though they can get repetitive and tiresome. I argued early in the thread (I think it was in this thread, but I didn't go back to check) that this expression, "0.999~", isn't really important in mathematics. We could do without it -- we could make the convention of always using the symbol "1" for the number one, and never the symbol "0.999~", in other words we could refuse to define "0.999~" like we refuse to define "0/0". But I think it's good that we don't outlaw the symbol "0.999~". The confusion it generates is a good opportunity to clear some common misunderstandings about the concept of number.

agingjb

2008-May-05, 05:55 PM

You could "outlaw" point-nine-recurring or however you want to write it, but the limit it represents would still exist, and be equal to 1 (unless, of course, you simply throw away classical analysis - good luck with astronomy if you do).

Coral

2008-May-05, 07:07 PM

This is why I agree that these discussions are useful, though they can get repetitive and tiresome. I argued early in the thread (I think it was in this thread, but I didn't go back to check) that this expression, "0.999~", isn't really important in mathematics. We could do without it -- we could make the convention of always using the symbol "1" for the number one, and never the symbol "0.999~", in other words we could refuse to define "0.999~" like we refuse to define "0/0". But I think it's good that we don't outlaw the symbol "0.999~". The confusion it generates is a good opportunity to clear some common misunderstandings about the concept of number.

I agree you could just declare that "1" is the "official" way to write this number (and we pretty much do that in practice), but convergent series would still play an important role in mathematics and sciences, and would still have be dealt with in other contexts. And even in this contect, sometimes it can't be avoided; for example, in Cantor's diagonal proof that the real numbers are not a countable set, you have to deal explicitly with the fact that some real numbers have multiple decimal representations.

Mister Earl

2008-May-05, 07:20 PM

Neverfly

2008-May-05, 07:27 PM

Been there, done that... (http://www.bautforum.com/off-topic-babbling/16806-where-do-you-stand-now.html)

Ah ok- I went over there and redeemed myself. a bit.

korjik

2008-May-05, 07:56 PM

Let's keep this thread open, even if for no other reason than to watch it occasionally resurface like a brain-hungry zombie. Those new here that stumble upon it get a bit of a mental workout, and that's never a bad thing.

You are a sick, sick man :)

Here I am trying to think up a good way to get this thread closed without getting myself banned, just so this thread will finally die, and you are wanting a Lazarus act from it :(

worzel

2008-May-05, 10:10 PM

Try unsubscribing from it.

agingjb

2008-May-05, 10:27 PM

While there is such a significant, and I would say, unbridgeable gap between the groups saying "yes" and "no" to this question, I don't see how the debate can end.

And in any case the fact that such a large proportion of the BAUT contributors take either view (which ever way we may think is the proper view) is something that should be kept perpetually present to those that seek this site as a place to get reliable information about astronomy.

ToSeek

2010-Oct-14, 10:01 PM

There is now a paper (http://www.math.umt.edu/TMME/vol7no1/TMME_vol7no1_2010_article1_pp.3_30.pdf) on the subject, and it's being discussed on Slashdot. (http://news.slashdot.org/story/10/10/14/135219/Proving-0999-Is-Equal-To-1)

Swift

2010-Oct-15, 01:23 AM

I know you have to practice for your Goddard zombie training, but did you have to bring this zombie thread back to life.
IT EATS BRAINS!!!!

http://www.booktalks.org/forums/images/smilies/panic.gif

;)

grapes

2010-Oct-15, 03:25 AM

There is now a paper (http://www.math.umt.edu/TMME/vol7no1/TMME_vol7no1_2010_article1_pp.3_30.pdf) on the subject.So, so sloppy. Notice, how the discussion ranges over arguments presented in the 18th century, on into the 21st century--but some of the references are dated with ('75) and ('77) and referred as such in the text even. Shameful. It'd be as if I referred to a post to this thread from November.

Jens

2010-Oct-15, 03:37 AM

While there is such a significant, and I would say, unbridgeable gap between the groups saying "yes" and "no" to this question, I don't see how the debate can end.

Well, just give me a bit more time. I haven't finished counting yet.

AndreasJ

2010-Oct-15, 07:19 AM

This? Here?

http://andjo.free.fr/shakehead.gif

HenrikOlsen

2010-Oct-15, 07:49 AM

This? Here?

http://andjo.free.fr/shakehead.gif
Sure, it offers insights into mathematical reasoning that is useful for students of mathematics as well as for teachers of those students, plus allows for the release of incoherent babbling for those who need that.:D

Heid the Ba'

2010-Oct-15, 12:56 PM

This? Here?

http://andjo.free.fr/shakehead.gif

If you think this thread has daft comments you should read some of the economics threads.

peteshimmon

2010-Oct-15, 02:31 PM

I just thought I would add my name to
such a historic thread! I'll go with
the vote.

:)

Chuck

2010-Oct-15, 04:00 PM

Anyone going to his or her computer to post in this thread first has to move 90% of the way to the computer, then 99% of the way, then 99.9% of the way, then 99.99% of the way, etc. If 0.999... is less than 1 then the computer can never be reached. So people who claim than 0.999... is less than 1 do so immediately after proving that they are equal, thus contradicting themselves.

jfribrg

2010-Oct-15, 04:15 PM

May the bannings resume :)

Solfe

2010-Oct-15, 04:17 PM

Arise my dark army of Infinite Nines! Back to the living, breakers the peace, sowers of distention err... dissension, Infinite Nines! :)

grapes

2010-Oct-15, 04:30 PM

Grey

2010-Oct-15, 07:05 PM

Ah, this thread brings back memories. As for the paper, I'm not convinced by his arguments. He seems to be arguing that because students are introduced to the idea that 0.9999... is equal to 1 long before they've tackled the subjects of limits and the real numbers, so they don't have the groundwork to really grasp the idea. But then he suggests talking about transfinite mathematics, an even more complicated subject. It just seems likely to leave them even more confused.

grapes

2010-Oct-15, 07:14 PM

He's also arguing for non-standard analysis, which allows such things as 0.999... < 1.0, and fears stifling such mathematical geniuses. That entails avoiding 0.333... = 1/3 as well, though, which is a whole nother can of wormholes.

pzkpfw

2010-Oct-15, 08:38 PM

If 0.9999999... = 1

Is there a number that 3.141592653589793238462643383... equals?

a1call

2010-Oct-15, 08:47 PM

If 0.9999999... = 1

Is there a number that 3.141592653589793238462643383... equals?
Of course, 10 at base pi. The amazing part is that it does so even though 0.99999....<1:lol:

HenrikOlsen

2010-Oct-15, 08:48 PM

If 0.9999999... = 1

Is there a number that 3.141592653589793238462643383... equals?
No.

It is implied in the first case that all subsequent digits are 9's, there's no such implication in the second, which means your question is really about the range of numbers from and including 3.1415926535897932384626433825 to but not including 3.1415926535897932384626433835.
That one of those numbers is pi is just a coincidence:D

a1call

2010-Oct-15, 08:52 PM

I stand corrected, but 0.9999..... Is still less than 1.

Strange

2010-Oct-15, 08:54 PM

If 0.9999999... = 1

Is there a number that 3.141592653589793238462643383... equals?

Itself?

HenrikOlsen

2010-Oct-15, 08:59 PM

Of course, 10 at base pi.
On the other hand, 30.121201110022... base pi ~= 10 base 10.
I suspect that can also be shown to be non-repeating.

HenrikOlsen

2010-Oct-15, 09:00 PM

I stand corrected, but 0.9999..... Is still less than 1.
Do you also think that 0.111111... is less that 1/9?

a1call

2010-Oct-15, 09:12 PM

Do you also think that 0.111111... is less that 1/9?
No, not in base pi.

Chuck

2010-Oct-15, 10:14 PM

I'm not immediately convinced. What if you changed all the "9"s to "8"s? :)It would have worked better in binary. Clarifying it would destroy the simplicity so I'll just abandon it instead.

Tobin Dax

2010-Oct-15, 10:18 PM

This must be the BAUT equivalent of pi-in-the-face slapstick.

pzkpfw

2010-Oct-15, 10:45 PM

Do you also think that 0.111111... is less that 1/9?

That seems different.

If you divide 1 by nine, you get the remainder that you divide by nine etc. To me, 0.111... does equal 1/9. Pretty much by defintion.

But if you ask is "0.999... = 1"; using your question that becomes "is 0.999... = 1/1".

And 1/1 very quickly simply becomes just 1. I don't know how to put it into maths, but that seems a very different siutation.

grapes

2010-Oct-15, 11:01 PM

And 1/1 very quickly simply becomes just 1. I don't know how to put it into maths, but that seems a very different siutation.I was just thinking about this situation.

In long division, if you divide 53 by 13 and make an error, the error is selfcorrecting: If you decide the first digit of the answer is 3, the remainder will be 14, bringing down a zero makes it 140, and 13 goes into that ten times. At that point you have two options: back up and repeat the first step with a higher guess, or place the 10 in the second digit place of the answer--which involves putting down a zero and carrying the one. Either way works fine.

In fact you could bullheadedly keep insisting that the result is less than 10 and you'd get 3.99999.... but the remainder would build and build. Taken together though, 3.99999.... plus the remainder divided by 13 placed in the appropriate decimal place is equal to 53/13. Obviously, having the remainder get larger and larger is a problem--but it doesn't invalidate the division. Making sure that the remainder never gets bigger than the divisor avoids that problem, right?

Well, that's what we can do with 1/1. The remainder would always be 1, and the answer is 0.99999...., continued out forever, until there is no remainder. Same thing we do with 1/3 = 0.333.....

whimsyfree

2010-Oct-15, 11:59 PM

It's stupefying how many respondents, including moderators, got this question wrong.

pzkpfw

2010-Oct-16, 12:06 AM

... continued out forever, until there is no remainder.

If you carry it on forever, when/why does the remainder become zero?

jfribrg

2010-Oct-16, 03:47 AM

It's stupefying how many respondents, including moderators, got this question wrong.

I find it interesting that so many people don't understand the explanation that 0.999.... = 1. Admittedly, this does seem counter-intuitive when you first encounter it. At the same time, there are so many other counter-intuitive concepts in physics. Heisenberg Uncertainty, wave-particle duality, space-time warping. I don't see as many people arguing that these well-established concepts are wrong, but with the math question, everyone seems to consider themselves an expert. I think the reason for this is that it seems to not require anything more than middle school math to understand. In actuality though, a good understanding of the real number system is required, and that usually doesn't occur until the Calculus sequence in college.

grapes

2010-Oct-16, 04:31 AM

If you carry it on forever, when/why does the remainder become zero?Not become zero, just no remainder at all. :)

The remainder would always be 1, and the answer is 0.99999...., continued out forever,
And by "forever" I don't mean until the end of time, I mean an infinite number of times--there just is no remainder.

I find it interesting that so many people don't understand the explanation that 0.999.... = 1. Admittedly, this does seem counter-intuitive when you first encounter it. At the same time, there are so many other counter-intuitive concepts in physics. Heisenberg Uncertainty, wave-particle duality, space-time warping. I don't see as many people arguing that these well-established concepts are wrong, but with the math question, everyone seems to consider themselves an expert. I think the reason for this is that it seems to not require anything more than middle school math to understand. In actuality though, a good understanding of the real number system is required, and that usually doesn't occur until the Calculus sequence in college.Yes and no.

Is 9.999... just ten times 0.999...? That's grade school math, right? What's 9.999... minus 0.999...? Also grade school math. It's the same trick they use in grade school to determine what fraction is equal to 0.703703703...

a1call

2010-Oct-16, 04:44 AM

0.999...=1 => 0.000...1=0 => 1/∞=0

let x=1 and y=∞, then we can write 1/∞=0 as:

x/y=0 => x=0*y => x=0
for any x and y where y<>0 (Read y not equal to 0).

This implies that if 0.999...=1 then 1=0 which is false, so 0.999... can not equal to 1.:naughty:

grapes

2010-Oct-16, 05:05 AM

1/∞Just as division by zero is not allowed, neither is division by infinity :)

a1call

2010-Oct-16, 05:31 AM

Just as division by zero is not allowed, neither is division by infinity :)

While you are 100% correct since infinity is not a real number, there are many who will disagree with passion. These are generally the ones who claim 0.999...=1:lol:
So it's easy to prove them wrong.:dance:

ETA: remember this thread:
http://www.bautforum.com/showthread.php/64419-question-about-quot-infinity-quot-in-math/page2

grapes

2010-Oct-16, 05:54 AM

While you are 100% correct since infinity is not a real number, there are many who will disagree with passion. These are generally the ones who claim 0.999...=1
So it's easy to prove them wrong.So prove away!

ETA: remember this thread:Yes, one of those forgettable ones.

Strange

2010-Oct-16, 08:32 AM

If you divide 1 by nine, you get the remainder that you divide by nine etc. To me, 0.111... does equal 1/9. Pretty much by defintion.

OK. And what do you get if you multiply 0.111... by 9 ?

Frog march

2010-Oct-16, 09:05 AM

like I said in a thread OP a few years ago, if 0.999..... and 1 are different then one take the other must equal something other than naught producing a range of numbers between 0.999.... and 1, but nobody seemed able to supply a number between 0.999.... and 1.

HenrikOlsen

2010-Oct-16, 09:18 AM

That seems different.

If you divide 1 by nine, you get the remainder that you divide by nine etc. To me, 0.111... does equal 1/9. Pretty much by defintion.

But if you ask is "0.999... = 1"; using your question that becomes "is 0.999... = 1/1".

And 1/1 very quickly simply becomes just 1. I don't know how to put it into maths, but that seems a very different siutation.
But it's quite obvious that if 0.111... = 1/9, then since 9*0.111...=0.999... and 9*1/9=9/9=1 this gives you 0.999...=1.
Which is why I asked my question, to get your head around the idea that 0.999... IS =1 :)

SeanF

2010-Oct-16, 03:13 PM

0.999...=1 => 0.000...1=0 => 1/∞=0
0.000...1 is nonsensical. The ... indicates infinity. You can't put a 1 after it, because there is no such thing as "after it."

like I said in a thread OP a few years ago, if 0.999..... and 1 are different then one take the other must equal something other than naught producing a range of numbers between 0.999.... and 1, but nobody seemed able to supply a number between 0.999.... and 1.
Another way to say the same thing is just to ask what 1.999.../2 is. If 0.999... is less than 1, then (1+0.999...)/2 must be between them, just like (4+5)/2 is 4.5.

If you do the division of 1.999.../2, you'll see that it's 0.999... And, if (A+B)/2=B, then A=B.

a1call

2010-Oct-16, 04:25 PM

0.000...1 is nonsensical. The ... indicates infinity. You can't put a 1 after it, because there is no such thing as "after it."

Perhaps you would prefer the scientific-notation of the same notion:
1 x 10-∞

HenrikOlsen

2010-Oct-16, 04:42 PM

Perhaps you would prefer the scientific-notation of the same notion:
1 x 10-∞

We'll prefer if you stop talking nonsense.

grapes

2010-Oct-16, 04:42 PM

like I said in a thread OP a few years ago,In an OP!? :)

0.000...1 is nonsensical. The ... indicates infinity. You can't put a 1 after it, because there is no such thing as "after it."
ToSeek's paper addresses that (and, by ToSeek's paper, I mean the paper that ToSeek linked). To get an idea what might be possible, imagine writing a decimal point above 0 on the number line, then a zero above each point (1-10-n) (that is, above the points .9, .99, .999, etc.), then a one above 1.

Perhaps you would prefer the scientific-notation of the same notion:
1 x 10-∞Are you trying to use ∞ as a number again?

a1call

2010-Oct-16, 04:52 PM

Are you trying to use ∞ as a number again?
Yes, I figure no one in the "=" camp will dare to object.:dance:

grapes

2010-Oct-16, 04:56 PM

Yes, I figure no one in the "=" camp will dare to object.I already did. So, where's your proof?

HenrikOlsen

2010-Oct-16, 04:57 PM

Yes, I figure no one in the "=" camp will dare to object.:dance:
And you'd be wrong, because people in the "=" camp, as you call it, are the people who knows mathematics and they know that ∞ is not a number, it is a limit.

agingjb

2010-Oct-16, 05:36 PM

I suppose if I were to suggest that ATM mathematics should have the same rules on BAUT as ATM astronomy or ATM physics, I'd get an infraction for not saying it a report or PM.

a1call

2010-Oct-16, 05:38 PM

How about:
limn=>-∞ 1 x 10n

Would that be more proper (given the imitation in typing mathematical notations on this board)?

HenrikOlsen

2010-Oct-16, 05:47 PM

That would be proper.

BTW limn=>-∞10n=0

And since limn=>-∞10n=0
limn=>-∞1-10n=1

Which is yet another way of showing that 0.9999...=1.

a1call

2010-Oct-16, 05:52 PM

That would be proper.

BTW limn => -∞ 10n=0
No argument there.

a1call

2010-Oct-16, 06:10 PM

That would be proper.

And since limn => -∞ 10n=0
limn => -∞ 1-10n=1

No argument there either, Since the statement:
Limit of f(x) converges to y as x approaches infinity
is not quite the same as the statement:
f(x) = y

Which is yet another way of showing that 0.9999...=1.

That on the other hand is quite a different story. Refer to post #2092 (http://www.bautforum.com/showthread.php/14593-Do-you-think-0.9999999-1-that-is-infinite-9s.?p=1804526#post1804526) and substitute with proper notations.
0.999... is not an (proper) expression of a limit but a real number less than 1.

grapes

2010-Oct-16, 06:54 PM

Refer to post #2092 (http://www.bautforum.com/showthread.php/14593-Do-you-think-0.9999999-1-that-is-infinite-9s.?p=1804526#post1804526) and substitute with proper notations.The substitutions are not clear to me. Can you explain?
0.999...=1 => 0.000...1=0 => 1/∞=0

let x=1 and y=∞, then we can write 1/∞=0 as:

x/y=0 => x=0*y => x=0
for any x and y where y<>0 (Read y not equal to 0).

HenrikOlsen

2010-Oct-16, 07:33 PM

0.999... is not an (proper) expression of a limit but a real number less than 1.
And now we're back to square one. This is getting to be reminiscent of ATM.

Ok, please try to define what 0.9999... is if it isn't another way of writing limn=>-∞1-10n?

sumn=1 to ∞(9*10-n)? (that's also 1 in case you're wondering)

agingjb

2010-Oct-16, 07:48 PM

Well a reply to my report would have been encouraging. Sorry folks, I do happen to know what mainstream mathematics is, but BAUT doesn't. So much the worse for BAUT.

a1call

2010-Oct-16, 08:17 PM

Every rational number is either a terminating or repeating decimal
(http://www.google.com/m/search?hl=en&aq=f&oq=&aqi=-k0d0t0&fkt=9329&fsdt=20894&q=%22Every+rational+number+is+either+a+terminating +or+repeating+decimal%22)

grapes

2010-Oct-16, 08:17 PM

Well a reply to my report would have been encouraging. Sorry folks, I do happen to know what mainstream mathematics is, but BAUT doesn't. So much the worse for BAUT.We muddle through. What are you using for your reference, is it online? I'd like to see it.

agingjb

2010-Oct-16, 08:52 PM

pzkpfw

2010-Oct-16, 09:08 PM

But it's quite obvious that if

0.111... = 1/9, then since

9 * 0.111... = 0.999... and

9 * 1/9 = 9/9 = 1 this gives you

0.999... = 1.

Which is why I asked my question, to get your head around the idea that 0.999... IS =1 :)

(My line breaks and spacing)

Thank you (and Strange et al too). That gave me one of those "aha" moments.

------

I think the bit that's hard to "get" is whether 0.9999... actually means "1 minus an infinitesimally small number" or not. As implied here:

(my bold)

0.999... = 1 => 0.000...1 = 0 => 1/∞=0 (snip)

That's also how it seemed to me. The notation 0.9999... implied (to me) 1 minus "a little bit". Even if that "little bit" was infinitely small that seemed to make it simply not equal 1. And if it did, then that "little bit" must be zero.

I was thinking, "what if you removed 2 times that infinitely small bit"? Or more?

e.g. using this notation:

limn=>-∞1-10n=1

limn=>-∞1 - (2 x 10n) = 1 ? (Does removing 2 infinitely small bits make a diff?)

limn=>-∞1 - (20 x 10n) = 1 ? (What about 20 of them?)

[Edit: limn=>-∞1 - ((n/2) x 10n) = 1 ? (I would expect 0.5, not 1; but why?) not sure what I was thinking.]

Further "aha" moments gratefully received.

(Edit: Further, I thought "lim" meant "nearly, but never quite, gets there". Am I wrong here too?)

Frog march

2010-Oct-16, 09:30 PM

In an OP!? :)

Yes, it was the thread, I started, that led to JPax2003 and SciFi Chick's temporary ban, and the thread was locked and I can't find it so it probably wasn't archived.

Old link:http://www.badastronomy.com/phpBB/viewtopic.php?t=20344&

HenrikOlsen

2010-Oct-16, 10:06 PM

(Edit: Further, I thought "lim" meant "nearly, but never quite, gets there". Am I wrong here too?)
Yep, at least for the definition.
In math speak, the limit of f(n) as n goes to infinity is x, doesn't mean that f(n) "nearly, but never quite, gets to" x.
Instead it means that no matter how small a number e you pick, as n goes along there's a point after which f(n) is always less than e from x, or in other words: if you can show that there exists an x, such that for any e>0, there exists a number p, such that for all n>p, x-e<=f(n)<=x+e (the "<=" here is less-or-equal), then that x is the limit.

There are similar definitions for n going to -infinity and for n going to a specific number, from either or both sides.

If no such x exists but it can be shown that for any e, there is a p, such that f(x)>e for all n>p, then the limit is ∞, with the obvious variations for the other signs.

So in English, it's really "will get arbitrarily close".

Hlafordlaes

2010-Oct-17, 01:44 AM

I must say, this is the nicest zombie encounter I've ever had.

Would have answered the poll incorrectly, but now have had my little "aha."
Thx.

grapes

2010-Oct-17, 03:43 AM

I suppose that Spivak's "Analysis" is as good as any. Page 70 (in the edition I have) says (speaking of infinite decimal expansions): "Every decimal ending in a string of 9's is equal to another ending in a string of 0's". I'm sure there are rather a lot of other references.Is that Michael Spivak? Real Analysis?

But really, if BAUT moderators do not regard the fundamental mathematics of the reals as given, then what credence can we give to anything here?That's the point of ToSeek's paper, that there are valid systems other than the reals. Bizarre, to some minds, though they may be, they're just as accepted to the mathematical "mainstream" (I'm not even certain that exists) as hyperbolic geometry or complex numbers--which aren't reals either.

grapes

2010-Oct-17, 03:59 AM

Yes, it was the thread, I started, that led to JPax2003 and SciFi Chick's temporary ban, and the thread was locked and I can't find it so it probably wasn't archived.

Old link:http://www.badastronomy.com/phpBB/viewtopic.php?t=20344&O yeah. Here is the BA post (http://www.bautforum.com/showthread.php/17009-JPax2003-SciFi-Chick-me-and-the-BABB)about it (probably where you got that invalid link). Since his post is in OTB, and this thread as well, so probably was yours. I think there were a lot of OTB/FnG aka BABBling threads that were purged.

On further reflection (googling), probably not. Here is another 0.999... poll in OTB started by SciFi Chick in 2005:
http://www.bautforum.com/showthread.php/16806-Where-do-you-stand-now
And one by Sticks in 2005:
http://www.bautforum.com/showthread.php/18175-Bad-Math
And one by Normandy6644 in 2005:
http://www.bautforum.com/showthread.php/16961-Two-reasons-why-.999...-1

Ours of course was started by VTBoy in 2004:
http://www.bautforum.com/showthread.php/14593-Do-you-think-0.9999999-1-that-is-infinite-9s.

Frog march

2010-Oct-17, 05:43 AM

I suppose my thread didn't make it as it was locked.

It had a poll where people who thought that 0.999... and 1 were different should then provide a number between 0.999... and 1...people voted for that option but no body came up with a number.

Jens

2010-Oct-17, 06:11 AM

Well a reply to my report would have been encouraging. Sorry folks, I do happen to know what mainstream mathematics is, but BAUT doesn't. So much the worse for BAUT.

Without arguing one way or the other, I think you are part of BAUT, so how can you be saying what "BAUT" knows and doesn't know? Who exactly are you talking aBAUT?

pzkpfw

2010-Oct-17, 06:21 AM

Let me be clear; that this thread is here, is not any kind of endorsement by BAUT forum of any claim (ATM or not (i.e. that isn't the point)) contained herein. And please, let's all stick to the topic - not meta discussion about the thread.

Reports and PMs have been received and it's being discussed by the mods.

agingjb

2010-Oct-17, 06:40 AM

Spivak's "Calculus", not "Analysis", sorry.

And I believe that the notation .999 recurring clearly denotes that the number system being discussed is the real field (not that the question would have another answer in any non-standard extension of the reals that remains a field).

grapes

2010-Oct-17, 07:18 AM

Spivak's Calculus is a fine book, but I don't think it addresses issues of what is mainstream or not. I've met, and had personal discussions, with Michael Spivak, and I think he'd be fine with the discussion in this thread (yellow pig! :) ). But I could be wrong.

It's not clear to me that the discussion is limited to the reals--that paper that ToSeek linked certainly isn't, and quite a few people have introduced things that are not "real". That's where I've focussed my efforts--if the disponent has thought about the problem enough that they can make the arguments, they can also listen to the arguments and some have changed their minds, over the years decades.

agingjb

2010-Oct-17, 07:30 AM

So the OP's premise in the poll did not necessarily address numbers in the classical real field? Hmm.

grapes

2010-Oct-17, 07:54 AM

You mean, restrict it to reals? I would say, by the nature of the poll itself, it does not, no?

HenrikOlsen

2010-Oct-17, 08:05 AM

Most expansions of reals don't change the properties of the reals themselves but rather expand the set of numbers, either by adding an orthogonal set, like the complex numbers, or by including the infinities and/or infinitesimals as actual numbers as e.g. hyperreals, superreals and surreals do.

I suspect many of the people answering no have a strong intuition about numbers, but it's closer to the hyperreals than to the reals, thus giving wrong results when applied to questions about infinitesimals and infinities, though it's perfectly good for answering questions about numbers in general.

a1call

2010-Oct-17, 04:46 PM

I suspect many of the people answering no have a strong intuition about numbers, but it's closer to the hyperreals than to the reals, thus giving wrong results when applied to questions about infinitesimals and infinities, though it's perfectly good for answering questions about numbers in general.

IMHO the issue is in essence very simple and in-abstract. It is a matter of the concept of infinity being undefined and "people who knows mathematics" engaging in an endless argument based on presumptions which are matter-of-convention to some and unproven-assumptions to the others.

At it's simplest form the issue boils down to some arguing that:

0.999... = 0.666... + 0.333... = 2/3 + 1/3 = 1

And others arguing that 1/3 = 0.333... is a matter of convention based on the concept of infinity (undefined). As such the equality is undefined and unresolved. No one can put infinite 3s in front of 0. and an overline or 3 small dots is not going to change that fact.
IMHO any attempts at complicating the matter by mappings to hyper-real fields is an uncalled for complication.

Moose

2010-Oct-17, 05:32 PM

No one can put infinite 3s in front of 0. and an overline or 3 small dots is not going to change that fact.

Only if you're stuck thinking about it as the iterative process it isn't. The whole point of limit theory is that it gets us past the very problem you (and everybody else so far) has been stuck upon, precisely so we can consider the complete expansion, with all infinite digits already in place.

a1call

2010-Oct-17, 05:43 PM

How would you define a limit in terms which does not include infinity (which is undefined)?
What can be concluded about a definition based on an undefined concept?

Moose

2010-Oct-17, 06:01 PM

How would you define a limit in terms which does not include infinity (which is undefined)?

I wouldn't.

What can be concluded about a definition based on an undefined concept?

That your premise, and thus your conclusions based on that premise, is wrong.

In mathematics, "infinity" is often treated as if it were a number (i.e., it counts or measures things: "an infinite number of terms") but it is not the same sort of number as the real numbers. [...] [Infinity is] a number greater than any real number.

Your insistence upon trying to treat it like a real number is precisely why you're getting stuck on the concept. You're trying to pin it down to a precise value, and you can't. And so you're trying to make this a semantics argument rather than realize there's absolutely no reason to try pinning infinity down to a value you can wrap your mind around. Infinity is simply a number that's "greater than any real number". That's enough of a definition to make limit theory both correct and useful.

0.999~ = 1 is a property of the real number system quite simply because infinity is not (strictly) a real number. Merely the upper bound.

a1call

2010-Oct-17, 07:12 PM

You do not seem to understand my point based on the fact that any conclusion is based on underlying presumptions and definitions. I maintain that this argument can be proven either way.
The fact that this and simillar problems have been discussed on this board for years must be a hint to this.

0.999~ = 1 is a property of the real number system quite simply because infinity is not (strictly) a real number. Merely the upper bound.

No, by definition it can't be. Please refer to my post #2114 (http://www.bautforum.com/showthread.php/14593-Do-you-think-0.9999999-1-that-is-infinite-9s.?p=1804733#post1804733)

1.0 can not be both a terminating as well as a repeating rational number in the decimal base.

cjl

2010-Oct-17, 07:45 PM

You do not seem to understand my point based on the fact that any conclusion is based on underlying presumptions and definitions. I maintain that this argument can be proven either way.
The fact that this and simillar problems have been discussed on this board for years must be a hint to this.

No, by definition it can't be. Please refer to my post #2114 (http://www.bautforum.com/showthread.php/14593-Do-you-think-0.9999999-1-that-is-infinite-9s.?p=1804733#post1804733)

1.0 can not be both a terminating as well as a repeating rational number in the decimal base.

Just because you found a few google links for your assertion doesn't make it true. Every terminating decimal can also be represented as a repeating decimal (though the reverse is not true).

Moose

2010-Oct-17, 07:47 PM

If by trees, you mean leprechauns, then yes there are no trees.

Yes, by redefining inconvenient words, you can seem to invalidate anything, at least superficially. And yet limit theory works, remains useful, and exists despite your strange disapproval. Deal with it.

a1call

2010-Oct-17, 07:49 PM

Every terminating decimal can also be represented as a repeating decimal (though the reverse is not true).
I would be very interested in any reference you or anyone else would be able to provide to that effect.

ETA: Am I the only one who sees an oxymoron in that sentence.

astromark

2010-Oct-17, 08:06 PM

So... the two worlds collide. In mathematics 0.99999~ is never equal to 1.

For the purest thinker. It nearly does. Almost and not quite. understanding that ~ is of some value.

Why is this a argument ? Where can there be a question of different view.

astromark

2010-Oct-17, 08:14 PM

a1call

2010-Oct-17, 08:19 PM

As the use of the ~ (infinite) Immediately disqualifies the number as defined. It is not defined if its infinite is it ?

Bingo.

caveman1917

2010-Oct-17, 08:28 PM

Numbers are numbers. Proper whole or rational. Have you a irrational number ? Maybe we do.

As the use of the ~ (infinite) Immediately disqualifies the number as defined. It is not defined if its infinite is it ?

It's actually the other way around Mark. Every number is defined as the limit (to infinity) of a series.

astromark

2010-Oct-17, 08:41 PM

SeanF

2010-Oct-17, 09:01 PM

I would be very interested in any reference you or anyone else would be able to provide to that effect.
A simple Google search will turn up a lot of them. I expect none of them will satisfy you, though, because they will all be examples of infinite repeating 9s.

1 = 0.999...
2.5 = 2.4999...
3.7382 = 3.7381999...

ETA: Am I the only one who sees an oxymoron in that sentence.
Do you also see an oxymoron in the sentence, "Every square is a rectangle (though the reverse is not true)"?

a1call

2010-Oct-17, 09:32 PM

A simple Google search will turn up a lot of them. I expect none of them will satisfy you, though, because they will all be examples of infinite repeating 9s.

1 = 0.999...
2.5 = 2.4999...
3.7382 = 3.7381999...

Do you also see an oxymoron in the sentence, "Every square is a rectangle (though the reverse is not true)"?

Yes, I can see such references and some very close to cjl's claim:

A terminating decimal can be written as a decimal fraction: 317/200 = 1585/1000. However, a terminating decimal also has a representation as a repeating decimal, obtained by decreasing the final (nonzero) digit by one and appending an infinitely repeating sequence of nines. 1 = 0.999999... and 1.585 = 1.584999999... are two examples of this.

Source (http://wapedia.mobi/en/Repeating_decimal)

So where does the conflicting definitions lead us?

Oxymoron was a reference to a repeating representation of terminating(at least when I thought about it:lol:).

HenrikOlsen

2010-Oct-17, 09:53 PM

How would you define a limit in terms which does not include infinity (which is undefined)?
What can be concluded about a definition based on an undefined concept?
I did, in post #2119 (http://www.bautforum.com/showthread.php/14593-Do-you-think-0.9999999-1-that-is-infinite-9s.?p=1804772#post1804772), where the limit as n goes to infinity is defined without using infinity as a number.

I can repeat it too:

The limit of f(n) as n goes to infinity is x if (and only if), for any e>0, there exists a number s such that x-e<=f(n)<=x+e for all n>s.
"<=" meaning less or equal.

In plain words what it means is than no matter how close you want it to be, as n increases there will be a point after which the function will always be closer.

This BTW is the mainstream definition of limits and it works quite well without using infinity as a number and also without having to do a process infinitely many times, "all" you have to do is show a way to reliably find an s given e such that the proposition is true.

Or to say it another way, there exists no other numbers between 0.9999~ and 1 and since one of the fundamental attributes of the real numbers is that there are infinitely many numbers between any two non-equal numbers (this is even true of the rationals), that means they can't be non-equal.

HenrikOlsen

2010-Oct-17, 09:57 PM

So... the two worlds collide. In mathematics 0.99999~ is never equal to 1.
If by ~ an infinite repeat of the last digit is meant, and you're talking reals, then in mathematics it is always=1

Sorry but if you claim otherwise you're not talking about real numbers.

Numbers are numbers. Proper whole or rational. Have you a irrational number ? Maybe we do.
We do have irrational numbers, they are those real numbers that aren't rational numbers.
Pi is one of them.
Your comments sounds almost like you haven't been properly introduced to real numbers yet.

HenrikOlsen

2010-Oct-17, 10:09 PM

1.0 can not be both a terminating as well as a repeating rational number in the decimal base.
Why not?
9/9=1

caveman1917

2010-Oct-17, 10:12 PM

Hmmm... I am surprised to be saying this for and at 'caveman1917'...

No. When I use a number. That is a definite point. Some place between 0 and ~.

I do not like being told what I mean... I know.

Conveyance of it as a number should not be anything other than the value stated.

I meant that's how the real numbers are defined in the field of mathematics (and by extension science in general). You are of course free to use any definition you see fit, i did not mean to argue that, just to add that it is defined differently in mathematics.

For my argument try this... Half of ten is five. That the use of ~ would disqualify halving of it.

Why would it? 0.50~ x 10.0~ = 5.0~

HenrikOlsen

2010-Oct-17, 10:13 PM

IMHO any attempts at complicating the matter by mappings to hyper-real fields is an uncalled for complication.
The thing is that you are arguing based on concepts from the hyper-reals that just aren't applicable to reals, which was what I was trying to draw attention to.
Your use of infinity as if it's a number essentially means that you've dropped talking about reals and is talking hyperreals because that's where infinity is a number.

a1call

2010-Oct-17, 11:18 PM

The limit of f(n) as n goes to infinity is x if (and only if), for any e>0, there exists a number s such that x-e<=f(n)<=x+e for all n>s.

Thanks for the added clarification. Wouldn't have been able to connect the dots otherwise.:lol:

Doesn't that hold through only when n is infinity though?

HenrikOlsen

2010-Oct-18, 12:03 AM

Doesn't that hold through only when n is infinity though?
Which part of "for all n greater than s" don't you get?
n is never infinity, infinity is not a number.

Now let's try to see what happens when we look at limn→∞(1-10-n).
If I suggest an s=-log10(e), then it's clear that since 1-10-s=1-10log10(e)=1-e, and the 10-n term is smaller than 10-s for all n larger than s, then I have shown that for any e>0, there exists a number s (given e I can always make s=-log10(e)) such that 1-e<1-10-n<1+e for all n larger than s, hence the definition of the limit is satisfied and I have proven that limn→∞(1-10-n)=1

What I'm not sure about is if the problem is that you can't see that 0.9999... is just another way of writing limn→∞(1-10-n).

PS Normally "ε" is used where I wrote "e", it was just a bit much to put in everywhere:)

a1call

2010-Oct-18, 12:44 AM

Which part of "for all n greater than s" don't you get?
n is never infinity, infinity is not a number.

Now let's try to see what happens when we look at limn→∞(1-10-n).
If I suggest an s=-log10(e), then it's clear that since 1-10-s=1-10log10(e)=1-e, and the 10-n term is smaller than 10-s for all n larger than s, then I have shown that for any e>0, there exists a number s (given e I can always make s=-log10(e)) such that 1-e<1-10-n<1+e for all n larger than s, hence the definition of the limit is satisfied and I have proven that limn→∞(1-10-n)=1

What I'm not sure about is if the problem is that you can't see that 0.9999... is just another way of writing limn→∞(1-10-n).

PS Normally "ε" is used where I wrote "e", it was just a bit much to put in everywhere:)

No the problem is not that at all. I think I mentioned it before(remember the no argument there bit?).

The use of e did make it a bit more difficult than it had to be. I can follow the satisfaction of the condition.
I'd be interested in a short plane English and informal proof/explanation of why the condition is sufficient to prove that f(n) will converge. I did major in Mathematics but that was a lifetime ago.:lol:

HenrikOlsen

2010-Oct-18, 12:55 AM

No the problem is not that at all. I think I mentioned it before(remember the no argument there bit?).

The use of e did make it a bit more difficult than it had to be. I can follow the satisfaction of the condition.
I'd be interested in a short plane English and informal proof/explanation of why the condition is sufficient to prove that f(n) will converge. I did major in Mathematics but that was a lifetime ago.:lol:
The condition is sufficient to show convergence because that condition is how convergence is actually defined:lol:

Informal version: if it can be shown that, no matter how tight a range (x-ε to x+ε) you define around x, f(n) will eventually stay within that range as n grows and never stray outside again, then we say that f(n) converges to x as n→∞ aka limn→∞f(n)=x.

Again, note that n never equals ∞, we only talk about something that is true for all n larger than some number and that this number exists for all ε.

a1call

2010-Oct-18, 01:03 AM

What I'm not sure about is if the problem is that you can't see that 0.9999... is just another way of writing limn→∞(1-10-n).

That was sneaky.:lol:

ETA:
Perhaps a formal reference on the web would be better then. Bypassing the in-definition of infinity in defining a limit is profoundly important to me.

HenrikOlsen

2010-Oct-18, 01:31 AM

That was sneaky.:lol:
Thanks.:)

ETA:
Perhaps a formal reference on the web would be better then. Bypassing the in-definition of infinity in defining a limit is profoundly important to me.
It was profoundly important for all mathematicians when it was first formalized too, because it meant limits became something that could be handled rigorously, without requiring word arguments involving infinities, thus making analysis possible.

http://en.wikipedia.org/wiki/Limit_of_a_function#Limits_involving_infinity is one reference.

You'll note that I only mentioned the case of the function converging to a number as n→∞ and ignored the other (but similarly defined) cases of lim f(n)=∞ (where instead of showing that f gets closer than any ε no matter how small, you instead show that f gets larger than any S no matter how large) and of n→c (where you show that you can always pick a number δ>0 such that f is closer/larger for all n where c-δ<n<c+δ (in words, "close enough" always exists)), since those cases aren't relevant for the current question and I thought they would confuse matters.

Perhaps a matter of confusion for some people regarding limits and infinities is if they've been taught the sloppy way of reading limn→∞f(n)=x as "f(n) goes to x as n goes to infinity" instead of the rather better "f(n) goes towards x, as n goes towards infinity", which doesn't seem to imply that infinity will eventually be reached or that infinity can just be plugged in for n to give x.

dgavin

2010-Oct-18, 02:36 AM

I ended up voting no. Even though i admit that many mathmatical forms show it being equal, 0.9999~ is actualy infinately close to being equal to 1, but it's just not quite there.

for examples 3 * 1/3 while expressed as 3 * .3333~ does appear to equal .9999~, if you solve it another way it does not. 3 * 1/3 resolves down to 3/3 which reluts in a 1. So the proper answer is when dealing with infinates is to renormalize them out, if possible, first.

a1call

2010-Oct-18, 04:23 AM

Thanks.:)

It was profoundly important for all mathematicians when it was first formalized too, because it meant limits became something that could be handled rigorously, without requiring word arguments involving infinities, thus making analysis possible.

http://en.wikipedia.org/wiki/Limit_of_a_function#Limits_involving_infinity is one reference.

You'll note that I only mentioned the case of the function converging to a number as n→∞ and ignored the other (but similarly defined) cases of lim f(n)=∞ (where instead of showing that f gets closer than any ε no matter how small, you instead show that f gets larger than any S no matter how large) and of n→c (where you show that you can always pick a number δ>0 such that f is closer/larger for all n where c-δ<n<c+δ (in words, "close enough" always exists)), since those cases aren't relevant for the current question and I thought they would confuse matters.

Perhaps a matter of confusion for some people regarding limits and infinities is if they've been taught the sloppy way of reading limn→∞f(n)=x as "f(n) goes to x as n goes to infinity" instead of the rather better "f(n) goes towards x, as n goes towards infinity", which doesn't seem to imply that infinity will eventually be reached or that infinity can just be plugged in for n to give x.
:wall::wall::wall:
Thank you for all that. I will need some time to reflect.

pzkpfw

2010-Oct-18, 04:41 AM

for examples 3 * 1/3 while expressed as 3 * .3333~ does appear to equal .9999~, if you solve it another way it does not. 3 * 1/3 resolves down to 3/3 which reluts in a 1. So the proper answer is when dealing with infinates is to renormalize them out, if possible, first.

Which way is that?

Perikles

2010-Oct-18, 09:03 AM

Apologies if this has been asked before in this thread, but is this issue equivalent to asking the question whether there is a fundamental difference between positive integers and a decimal (or any other base) representation of a measurement?

grapes

2010-Oct-18, 09:09 AM

Strange

2010-Oct-18, 12:38 PM

I'll echo pzkpfw's question, but add the question, do you think 0.333... is equal to 1/3?

Good point. Presumably people who vote 'no' must think that 0.333... is only an approximation to 1/3.

adapa

2010-Oct-18, 01:31 PM

.9999999... is exactly equal to 1.

If x=.9999999...
then 10x=9.9999999...

subtract the first equation from the second and you get:
9x=9
This means that x=1. This is elementary algebra and it doesn't get much simpler than this.

SeanF

2010-Oct-18, 01:34 PM

...0.9999~ is actualy infinately close to being equal to 1...
What's the difference between "infinitely close to being equal" and "is equal"? If there's any difference, than it's not really infinitely close, is it? :)

mike alexander

2010-Oct-18, 02:53 PM

I wonder if this thread should be merged with the "What Causes War" thread? Insights might be found in both directions.

HenrikOlsen

2010-Oct-18, 02:57 PM

I think this one has the greatest chance of being resolved through education of the participants.

dgavin

2010-Oct-18, 03:02 PM

I'll echo pzkpfw's question, but add the question, do you think 0.333... is equal to 1/3?

pzkpfw - I thought i had show that in my example, you don't divide the fraction 1/3 first, but multiply the fraction by 3 first, and then do the division, which yeilds 1. It eliminates the need of delaing with an infinately repeating decimal. Granted Renormalization like this is more complex in say QM, but it;s the same basic idea. Eliminate the infinities first.

grapes - from the point of view of decimal math yes it is equivlent, and mutiplying it by 3 yeilds .999~ However if you normalize the fraction out first, you get a 1. It's a case where math done different ways, yields different results, becuase of the infinities involved. QM itself deals with the same issue in that they constantly have to renormalize the equations to get rid of the infinities. Same priciple involved.

So my answer is 0.333~ is the decimal representation of 1/3. In decimal math it would be equivlent, in algebra you'd normalize it out first.

HenrikOlsen

2010-Oct-18, 03:23 PM

grapes - from the point of view of decimal math yes it is equivlent, and mutiplying it by 3 yeilds .999~ However if you normalize the fraction out first, you get a 1. It's a case where math done different ways, yields different results, becuase of the infinities involved.
You are trying to claim that if a=b, you can't multiply by 3 on both sides unless you first replace the b with a before multiplying because otherwise you'd get 3a=3b which you don't want to believe. Sorry, that's not how things work in mathematics.

Your only real argument for claiming that it is illegal not to normalize first is your belief that 0.999~ is not 1, which is circular.

BTW, what do you mean when you say normalization? It isn't a required part of any mathematical proofs I've ever seen and sounds like a convenience operation done when working with the rational numbers, which we aren't, we're working with reals.

Oh, and also BTW, what do you mean with "decimal math".

grapes

2010-Oct-18, 04:31 PM

grapes - from the point of view of decimal math yes it is equivlent, and mutiplying it by 3 yeilds .999~ Multiplying by three is not part of my argument. I was just going to remind you why 0.333... is equal to 1/3. It's because the sum of 3/10 + 3/100 + 3/1000 + ... is considered equal to 1/3, right?

However if you normalize the fraction out first, you get a 1. It's a case where math done different ways, yields different results, becuase of the infinities involved. QM itself deals with the same issue in that they constantly have to renormalize the equations to get rid of the infinities. Same priciple involved.

So my answer is 0.333~ is the decimal representation of 1/3. In decimal math it would be equivlent, in algebra you'd normalize it out first.Others may make the point that 3 times 0.333... is 0.999... but I am not. Still, renormalization is done to avoid otherwise intractable problems--I don't think we're dealing with that here. :)

gzhpcu

2010-Oct-18, 06:38 PM

Apologies as well, as this is a long thread.

0.9999.... is a sequence. As the number of 9's approaches infinity, the limit of the sequence is the real number 1. Limits are essential to calculus. If calculus is accepted as correct, then the limit defines an equality.

pzkpfw

2010-Oct-18, 07:13 PM

pzkpfw - I thought i had show that in my example, you don't divide the fraction 1/3 first, but multiply the fraction by 3 first, and then do the division, which yeilds 1. It eliminates the need of delaing with an infinately repeating decimal. Granted Renormalization like this is more complex in say QM, but it;s the same basic idea. Eliminate the infinities first.

Sorry, I didn't pick that up from your post. So what you are saying is that:

3 x 1/3 = 3 x 0.3333~ = 0.9999~

is different to

3 x 1/3 = 3/3 = 1

Yeah, I don't buy it, for the same reasons mentioned in posts above mine. 1/3 = 0.333~. The end.

---

To no-one in particular, but just to get some of my own thoughts written down to help me Grok them (it's that word from that book by that guy, I think):

It seems the problem with intuitively understanding this is that it's very hard to get past the idea that 0.9999~ is not less than 1. Those 9's shout "less than". It's not, and this is caused by infinities being freaky. Posts such as those by dgavin (just an example, there were others) say things like "Renormalization" to "Eliminate the infinities first", but that's putting the cart before the horse. The maths is correct, there's no need to bend the maths to fit intuition.

I started thinking:
Is 0.9 less than 1? Yes.
Is 0.99 less than 1? Yes.
Is 0.999 less than 1? Yes.

By induction (more or less) that would seem to imply that it does not matter how many times another 9 is added, the result will always be less than 1. But that's wrong.

Where that goes wrong is that at any time you make the comparison, you've stopped adding the 9's. 0.999~ means the 9's never stop getting added. At any point you stop and compare, sure your number is less than 1, but you can always go another 90% of the distance closer to 1 than you are, by adding another 9.

The induction actually works the other way (as shown by HenrikOlsen):

limn => -∞ 1 - 10n = 1

---

Another way to see this, as also noted by someone above, is that there is no number between 0.999~ and 1. There's simply no room for it.

If at any time, a number has a decimal digit that is not 9, it's always possible to add a 1 to that and truncate the number at that point, to get closer to the next integer.

e.g.

A = 0.999999999899999~
B = 0.9999999999

A < B < 1

But if all the decimal digits are 9 (and there's infinite of them), there's just no place to do that. You can't get closer to 1.

C = 0.99999999~

C < D < 1 : there is no D that satisfies this, so you must instead have C = D = 1; C = 1

---

A post above tried to show this...:

0.999~ = 1 - 0.0~01 (actually it used "..." not "~").

But that can't work. 0.0~1 was (I think) meant to imply an infinite number of zeros and then a 1 (the equivalent of 1 - 10n where n = -∞). It was used to try to illustrate the "0.999~ is 1 minus an infinitely small amount" concept. But that can't work. As soon as you put that "1" on the end, you've tried to define the number at some specific point. There is no end to an infinite series of digits.

The n never actually "equals" -∞.

---

0.999~ = 9 x 0.111~

0.999~ = 9 x 1/9

0.999~ = 9/9

0.999~ = 1

The only way around this, that I can see, is try to say 0.111~ does not equal 1/9; and I can't see how that would be done.

dgavin

2010-Oct-18, 07:29 PM

You are trying to claim that if a=b, you can't multiply by 3 on both sides unless you first replace the b with a before multiplying because otherwise you'd get 3a=3b which you don't want to believe. Sorry, that's not how things work in mathematics.

Your only real argument for claiming that it is illegal not to normalize first is your belief that 0.999~ is not 1, which is circular.

BTW, what do you mean when you say normalization? It isn't a required part of any mathematical proofs I've ever seen and sounds like a convenience operation done when working with the rational numbers, which we aren't, we're working with reals.

Oh, and also BTW, what do you mean with "decimal math".

No what I'm saying is this is one of those trick questions math teachers like to pull on people.

The proper solution lies in doing the math in the sequence that is most correct. Part of the trick on this one is the order of operations, if you do the division before the multiplication, you get a result of .9999~ = 1. If you do the multiplication before the division you get 1= 1.

The proper order of operations in basic algebra, is Multiplication first, Division next, Addition third, Subtration last. Which if you do it this way you get the correct answer of 1=1

the other parts of the trick question is that you should reduce (normalize) the problem algebraically first, by replacing the numbers with variables.

in algebraic terms:

given a = 3

b= a * 1/a, reduces to b = 1a/a, reduces to b = a/a, reduces to b = 1

ergo, no matter what value you use for A, the correct answer to the problem is it's always equal to 1. In fact in this case it even works even if a=~.

By "decimal math" I simply mean the process of resolving fractions into their decimal numbers first, before applying operators.

astromark

2010-Oct-18, 07:34 PM

pzkpfw

2010-Oct-18, 07:38 PM

...

I then was told that I need educating... That, I find insulting and unfair.

...

To the purest scientific mind 0.9999~ can not ever be 1.

...

It is also insulting to imply that those who agree with the mainstream position that 0.999~ = 1 are not somehow of "pure scientific mind".

If you wish to argue that 0.999~ is not 1, then do so using maths, not veiled insults or simply claiming "I don't believe".

pzkpfw

2010-Oct-18, 07:43 PM

... Multiplication first, Division next, ...

Multiplication and division are essentially the same thing. Order doesn't matter here.

e.g.

2 x (1/2) = (2 x 1) / 2 = 1

i.e. dividing by 2 is the same as multiplying by one half.

2 x (1/2) = 2 x 0.5 = 1

(2 x 1) / 2 = 2 / 2 = 1

edit: or even 2 x (1/2) = (2/1) x (1/2) = (2x1) / (1x2) = 2/2 = 1

(Just like subtraction is the same as addition - of a negative number

2 - 4 = 2 + (-4) = -2

)

Overall order is of course still important, but does not fly as a way to get around 3 x 0.333~ = 0.999~ = 1.

dgavin

2010-Oct-18, 08:18 PM

Multiplication and division are essentially the same thing. Order doesn't matter here.

e.g.

2 x (1/2) = (2 x 1) / 2 = 1

i.e. dividing by 2 is the same as multiplying by one half.

2 x (1/2) = 2 x 0.5 = 1

(2 x 1) / 2 = 2 / 2 = 1

edit: or even 2 x (1/2) = (2/1) x (1/2) = (2x1) / (1x2) = 2/2 = 1

(Just like subtraction is the same as addition - of a negative number

2 - 4 = 2 + (-4) = -2

)

Overall order is of course still important, but does not fly as a way to get around 3 x 0.333~ = 0.999~ = 1.

2 x (1/2) = (2 x 1) / 2 = 1

Where a = 2

a * (1/a) = a * 1/a Reduces to: 1/a = (a* 1/a) / a Reduces to 1 = a/a * (a* 1/a) reduces to 1 = 1 * (a/a) reduces to 1 = 1 * 1

Still works 1 = 1

pzkpfw

2010-Oct-18, 08:35 PM

2 x (1/2) = (2 x 1) / 2 = 1

Where a = 2

a * (1/a) = a * 1/a Reduces to: 1/a = (a* 1/a) / a Reduces to 1 = a/a * (a* 1/a) reduces to 1 = 1 * (a/a) reduces to 1 = 1 * 1

Still works 1 = 1

And?

Please add some line breaks to that, I've no idea what point you are making, nor how it's relevant to 0.999~ = 1.

HenrikOlsen

2010-Oct-18, 09:20 PM

No what I'm saying is this is one of those trick questions math teachers like to pull on people.

The proper solution lies in doing the math in the sequence that is most correct. Part of the trick on this one is the order of operations, if you do the division before the multiplication, you get a result of .9999~ = 1. If you do the multiplication before the division you get 1= 1.

The proper order of operations in basic algebra, is Multiplication first, Division next, Addition third, Subtration last.
Are you claiming that c/a*b is not = c*b/a?

Because if you are, you have got to go back to your math teacher and demand your money back because that's just plain wrong.

SeanF

2010-Oct-18, 09:24 PM

Are you claiming that c/a*b is not = c*b/a?
I will. Without parenthesis, you do multiplication first and division second.

c/a*b <> (c/a)*b

EDIT: Your point is still valid of course. (c/a)*b = (c*b)/a. :)

Grey

2010-Oct-18, 09:35 PM

I will. Without parenthesis, you do multiplication first and division second.

c/a*b <> (c/a)*bI think that the official rule is that multiplication and division happen at the same time, left to right (followed by addition and subtraction, again left to right), so Henrik is correct here. Of course, in any case where you're concerned that there might be ambiguity, adding in parentheses is a good plan.

HenrikOlsen

2010-Oct-18, 09:38 PM

I've always learned that division and multiplication are equal in rank and are evaluated left to right in the absence of parentheses.
That's also true of the vast majority of programming languages, so if you expect something different there, you're going to be surprised. :)

But anyway, dgavin's claim is that (a/b)*c<>(a*c)/b which may be a badly misremembered version of the same multiplication before division rule SeanF mentioned.

Incidentally in numerical analysis, i.e. computer programming, it is often a good idea to reorder operations so multiplication happens before division because that can give significantly smaller rounding errors, but that is not a good lesson to bring back to pure mathematics because it doesn't have rounding errors. :)

SeanF

2010-Oct-18, 09:41 PM

I stand corrected. :)

Anyhoo, like I said, it doesn't detract from your point.

adapa

2010-Oct-18, 11:39 PM

Pages ago I said that 0.9999~ is not and, is never equal to 1.

I then was told that I need educating... That, I find insulting and unfair.

As if I intended to write 1. then that is what I would do.

To the purest scientific mind 0.9999~ can not ever be 1.

By some ever so small fraction it just never makes it... and ~ indicating a repeating sequence infinitum ...

I extract some entertainment from the fact that there is any discussion... It for me is that clear.

Do you believe or not believe that .00000~ is equal to 0 ?

You may say that this question is irrelevant but it is essentially the same question as whether or not you believe that .99999~ = 1. Here is why:

If x=.99999~, then 10x=9.99999~ (Simple multiplication)

therefore 10x-x=9x which is equal to 9.00000~ (Basic subtraction)

Now if you believe that .00000~=0 then 9.00000~ = 9 which equals 9x and therefore x=1 (case closed).

Now, you do believe .00000~ is equal to 0 or do you not?

dgavin

2010-Oct-19, 12:37 AM

Algebra of Operations PEMDAS (parentheses, exponents, multiplication, division, addition, subtraction)

Some teach it under new math (yuk) as parentheses, exponents, multiplication and division (left to right) , addition and subtraction (left to right)

I was trained before the new math days so I always follow the first. Which one you want to call correct is based on if you are a new math proponent or not.

What I've been trying to point out is what you are seeing here is a math trick. It was used, along with a slew of others back in the 80's, as a way to help programmers to learn how to look for tiny commonly made mathematical errors. I actually learned the 1/3 times 3 one before programming though.

However you express it, the trick is again, to boil the problem down to an algebraic solution.

ergo:

2 * ( 1/2) = 1
3 * ( 1/3) = 1
4 * ( 1/4) = 1

Once you replace the variable number with a variable, a, the solution always equals 1 no matter what direct you perform the operators. a * (1/a) = 1.

However if you want to get technical about ~.9999 in one formula seeming equal to 1 in another there is a proper mathematic symbol to use other then equals.

in one of the examples provided 3 x (1/3) = (3 x 1) / 3 = 1 should be expressed as 3 x (1/3) ≈ (3 x 1) / 3 ≈ 1. (approximately equal)

Either way I adequately proved my point that by taking the formula into pure algebra twice now, you can always get the answer of one with a * (1/a) or a * 1/a when a is any number including 3. How? You reduce the formula to it's simplest form first a/a, and then substitute variable a back with 3. In-fact you technically don't even need to do the substitution as any number divided by itself, is always 1.

If you chose not to simplify, (reduce, normalize, etc...) then you get stuck with .9999~ = 1. Myself, I'd use the proper symbol and write it as .9999~ ≈ 1 however.

dgavin

2010-Oct-19, 12:46 AM

dgavin's claim is that (a/b)*c<>(a*c)/b which may be a badly misremembered....

Excuse me? where did I ever claim that? What I did was take it into algebra and reduce the equation to it's simplest form, which completely eliminates said error of ~.9999 = 1.

I don't need to really demonstrate again that the reduction of the formula fixes it?

pzkpfw

2010-Oct-19, 12:49 AM

...

However if you want to get technical about ~.9999 in one formula seeming equal to 1 in another there is a proper mathematic symbol to use other then equals.

in one of the examples provided 3 x (1/3) = (3 x 1) / 3 = 1 should be expressed as 3 x (1/3) ≈ (3 x 1) / 3 ≈ 1. (approximately equal)

Eh? "3 x (1/3) ≈ (3 x 1) / 3 ≈ 1"

You really say (3 x 1) / 3 is "only" equivalent to, but not equal to 1?

Why ≈ and not = ?

i.e. 3 x (1/3) = (3 x 1) / 3 = 1

I really don't see where you get this idea.

Either way I adequately proved my point that by taking the formula into pure algebra twice now, you can always get the answer of one with a * (1/a) or a * 1/a when a is any number including 3.

Sure. a x 1/a = 1. No question about that. Frankly I think it's pretty much irrelevant.

How? You reduce the formula to it's simplest form first a/a, and then substitute variable a back with 3. In-fact you technically don't even need to do the substitution as any number divided by itself, is always 1.

If you chose not to simplify, (reduce, normalize, etc...) then you get stuck with .9999~ = 1. Myself, I'd use the proper symbol and write it as .9999~ ≈ 1 however.

Sorry, but I still don't see why you say "stuck with" or feel that "≈" is the proper symbol.

You appear to be saying:

3 x (1/3) = (3 x 1) / 3 = 3 / 3 = 1

is "better" than

3 x (1/3) = 3 x 0.333~ = 1

But all you are doing is taking the 0.999~ question and dividing it by 3, to make a similar complaint about 0.333~

Essentially, you are simply claiming that 0.333~ does not equal 1/3.

dgavin

2010-Oct-19, 01:01 AM

Eh? "3 x (1/3) ≈ (3 x 1) / 3 ≈ 1"

You really say (3 x 1) / 3 is "only" equivalent to, but not equal to 1?

Why ≈ and not = ?

i.e. 3 x (1/3) = (3 x 1) / 3 = 1

I really don't see where you get this idea.

Sure. a x 1/a = 1. No question about that. Frankly I think it's pretty much irrelevant.

Sorry, but I still don't see why you say "stuck with" or feel that "≈" is the proper symbol.

You appear to be saying:

3 x (1/3) = (3 x 1) / 3 = 3 / 3 = 1

is "better" than

3 x (1/3) = 3 x 0.333~ = 1

But all you are doing is taking the 0.999~ question and dividing it by 3, to make a similar complaint about 0.333~

Essentially, you are simply claiming that 0.333~ does not equal 1/3.

No, not claiming any such thing. I /showed/ with algebra, twice, that by reducing the fomula (or fomulas) to it's simplest form first, fixes the issue completely and always yields a 1.

pzkpfw

2010-Oct-19, 01:08 AM

It always yields a 1, because 0.999~ = 1.

Take C = A x B

It shouldn't matter how A and B are represented, multiplication is multiplication, and the answer will always be the same.

Using fractions:

A = 3 = (3/1)
B = 1/3

C = A x B = (3/1) x (1/3) = (3 x 1) / (1 x 3) = 3 / 3 therefore C = 1

Using decimals:

A = 3.0
B = 0.333~

C = A x B = 3.0 x 0.333~ = 0.999~ therefore C = 1 = 0.999~

That is, 3 x (1/3), however you prefer to calculate it, = 3 x 0.333~

(
This is of course different to truncation.

3 x 0.33334 d.p. = 0.9999 which does not equal 1.
)

0.333~ (repeated decimals) is either an exact (not equivalent) representation of 1/3, or not. And if not, I'd like to know why not.

dgavin

2010-Oct-19, 01:19 AM

It always yields a 1, because 0.999~ = 1.

C = A x B

It shouldn't matter how A and B are represented, multiplication is multiplication, and the answer will always be the same.

Using fractions:

A = 3 = (3/1)
B = 1/3

C = A x B = (3/1) x (1/3) = (3 x 1) / (1 x 3) = 3 / 3 therefore C = 1

Using decimals:

A = 3.0
B = 0.333~

C = A x B = 3.0 x 0.333~ = 0.999~ therefore C = 1 = 0.999~

That is, 3 x (1/3), however you prefer to calculate it, = 3 x 0.333~

(
This is of course different to truncation.

3 x 0.33334 d.p. = 0.9999 which does not equal 1.
)

0.333~ (repeated decimals) is either an exact (not equivalent) representation of 1/3, or not. And if not, I'd like to know why not.

0.333~ (repeated decimals) is either an exact (not equivalent) representation of 1/3, or not. And if not, I'd like to know why not.

Let me ask you instead, do you agree or disagree that a * (1/a) can be properly reduced to a/a in algebra?

pzkpfw

2010-Oct-19, 01:31 AM

Let me ask you instead, do you agree or disagree that a * (1/a) can be properly reduced to a/a in algebra?

Of course I agree. I already did (post #2186).

a x (1/a) = (a x 1) / (1 x a) = a / a = 1

SeanF

2010-Oct-19, 01:46 AM

What I've been trying to point out is what you are seeing here is a math trick. It was used, along with a slew of others back in the 80's, as a way to help programmers to learn how to look for tiny commonly made mathematical errors. I actually learned the 1/3 times 3 one before programming though.
Careful with extending computer programming issues into the real world, Dgavin. If you tell a computer to divide 1 by 3 and then multiply the answer by 3, you quite likely will not get 1. But that's because the computer is rounding, not because 1 divided by 3 and then multiplied by 3 doesn't actually equal 1.

The computer's wrong, not the math.

dgavin

2010-Oct-19, 01:49 AM

Of course I agree. I already did.

a x (1/a) = (a x 1) / (1 x a) = a / a = 1

There is your solution. Reduce the equasion first. No more .9999~ = 1. Thats one of the nice things of algebra, in that it lets you clean up, short comings, with some mathmatical issues.

If I remember correctly if you drop down to binary system, or octal, you get similar issues but with different fractions, again until you reduce the fomulas, with the other number systems. I can't remember any of those specific fractions though as it was ages ago. I could probably figure them out again, but thats a bit beyond the OP.

grapes

2010-Oct-19, 01:53 AM

0.333~ (repeated decimals) is either an exact (not equivalent) representation of 1/3, or not. And if not, I'd like to know why not.

Let me ask you instead, do you agree or disagree that a * (1/a) can be properly reduced to a/a in algebra?
Of course I agree. I already did (post #2186).OK, now that we have that answered to everyone's satisfaction, I'd like an answer to a question I asked before.

But, first I want to clear something up. In this context, "equal" and "equivalent" mean the same thing. It may be a cultural thing, so I want to clear that up now, before we go on.

The question, though, is:

Multiplying by three is not part of my argument. I was just going to remind you why 0.333... is equal to 1/3. It's because the sum of 3/10 + 3/100 + 3/1000 + ... is considered equal to 1/3, right?

dgavin

2010-Oct-19, 01:54 AM

Careful with extending computer programming issues into the real world, Dgavin. If you tell a computer to divide 1 by 3 and then multiply the answer by 3, you quite likely will not get 1. But that's because the computer is rounding, not because 1 divided by 3 and then multiplied by 3 doesn't actually equal 1.

The computer's wrong, not the math.

Very true, I was tryign to indicate that I had some practical training with how to deal with this very specific issue however.

dgavin

2010-Oct-19, 02:02 AM

OK, now that we have that answered to everyone's satisfaction, I'd like an answer to a question I asked before.

But, first I want to clear something up. In this context, "equal" and "equivalent" mean the same thing. It may be a cultural thing, so I want to clear that up now, before we go on.

The question, though, is:

But, first I want to clear something up. In this context, "equal" and "equivalent" mean the same thing. It may be a cultural thing, so I want to clear that up now, before we go on.

I'm afraid I'm going to hold to the algebraic line on this one and say that = (equal) and ≈ (approximately equal or equivalent to) are not the same.

*edit to add*

Ok so I was over trained in Algebra perhaps.

pzkpfw

2010-Oct-19, 02:11 AM

There is your solution. Reduce the equasion first. No more .9999~ = 1. Thats one of the nice things of algebra, in that it lets you clean up, short comings, with some mathmatical issues.

No, that's not a "solution" because there is no "short coming"; and it certainly does not show that 0.999~ does not equal 1.

There is no reason to want 0.999~ to not equal 1, because it does equal 1.

Doing the calculation one way instead of another, makes no difference as long as the mathematical operations are equivalent.

Have you answered...

Is 0.333~ = 1/3 ?

If I remember correctly if you drop down to binary system, or octal, you get similar issues but with different fractions, again until you reduce the fomulas, with the other number systems. I can't remember any of those specific fractions though as it was ages ago. I could probably figure them out again, but thats a bit beyond the OP.

This is a different issue; that of representation of numbers in different bases, in limited precision. Some fractons that are easily represented in decimal are not so easily represented in binary. That's a different issue.

dgavin

2010-Oct-19, 02:18 AM

No, that's not a "solution" because there is no "short coming"; and it certainly does not show that 0.999~ does not equal 1.

There is no reason to want 0.999~ to not equal 1, because it does equal 1.

Doing the calculation one way instead of another, makes no difference as long as the mathematical operations are equivalent.

Have you answered...

Is 0.333~ = 1/3 ?

This is a different issue; that of representation of numbers in different bases, in limited precision. Some fractons that are easily represented in decimal are not so easily represented in binary. That's a different issue.

I would form it as .3333~ ≈ 1/3 for the main reason that not matter how long I calcualted 1/3, I'd be pushing up daisies before I could actualy prove or disprove matmatically that .3333~ is actualy equal to 1/3.

grapes

2010-Oct-19, 02:25 AM

I'm afraid I'm going to hold to the algebraic line on this one and say that = (equal) and ≈ (approximately equal or equivalent to) are not the same.
I don't know of any context where "equivalent" has any connotation of "approximately". Instead, it means, "essentially equal".

3 and 9/3 are not equal, by dint of the "9/", but they are equivalent. As numbers, they also are equal.

So, about those 0.333... questions...

dgavin

2010-Oct-19, 02:30 AM

No, that's not a "solution" because there is no "short coming"; and it certainly does not show that 0.999~ does not equal 1.

Yes it is a solution. I showed exactly how, matmatically, to avoid the issue where it seems that .9999~ = 1. If you want to ignore the math, thats your choice, it doesn't invalidate my point however.

3 * (1/3) = .9999~. Reduced to 3/3 it always equals one. I'd use the reduced form always because that is entirely permissable and considered advisable in math. Reduction first, then solve.

SeanF

2010-Oct-19, 02:37 AM

3 * (1/3) = .9999~. Reduced to 3/3 it always equals one. I'd use the reduced form always because that is entirely permissable and considered advisable in math. Reduction first, then solve.
Are you suggesting that a failure to reduce first will sometimes result in the wrong answer, even if you do all the (non-reduced) math correctly?

dgavin

2010-Oct-19, 02:48 AM

Are you suggesting that a failure to reduce first will sometimes result in the wrong answer, even if you do all the (non-reduced) math correctly?

Not suggesting any such thing. I was trained to always reduce first (if possible) and solve after. This was later reinforced in life by speeding up math in code by also reducing formulas to thier simplest form.

If someone asked me to automate a calcualtion of say input b + (input a * (1/input a) it would wind up being coded as either input b + ( input a / input a) or better (and faster) yet as input b + 1.

dgavin

2010-Oct-19, 02:53 AM

I don't know of any context where "equivalent" has any connotation of "approximately". Instead, it means, "essentially equal".

3 and 9/3 are not equal, by dint of the "9/", but they are equivalent. As numbers, they also are equal.

So, about those 0.333... questions...

Reposting my answer:

I would form it as .3333~ ≈ 1/3 for the main reason that not matter how long I calcualted 1/3, I'd be pushing up daisies before I could actualy prove or disprove matmatically that .3333~ is actualy equal to 1/3.

Frog march

2010-Oct-19, 03:03 AM

wouldn't 10 divided by 3 always produce the answer 3 with one remainder?

pzkpfw

2010-Oct-19, 04:01 AM

Yes it is a solution. I showed exactly how, matmatically, to avoid the issue where it seems that .9999~ = 1. If you want to ignore the math, thats your choice, it doesn't invalidate my point however.

3 * (1/3) = .9999~. Reduced to 3/3 it always equals one. I'd use the reduced form always because that is entirely permissable and considered advisable in math. Reduction first, then solve.

I'm sorry, but there's lots of math etc. that you are ignoring.

In this case, I'm not ignoring your maths, I'm rejecting that it makes a difference.

I totally agree that: 3 x (1/3) can be done as (3 x 1) / 3 giving 3 / 3 = 1.

What I reject is the distinction between doing it that way and doing the division first.

After all, you start by saying: 3 x (1/3) can be done as (3 x 1) / 3; you are saying they are equal.

Doing the (1/3) first can't produce a different answer. (It can result in a different notation.)

Just like: 4 x (6 / 2) = 4 x 3 = 12 = (4 x 6) / 2 = 24 / 2 = 12. The order does not change the math here.

Do you not agree that: (a / b) x (c / d) = (a x c) / (b x d) ?

If the thing on the left equals the thing on the right, how can it matter which you work out first?

If the thing on the left gives an answer and the thing on the right gives an answer, how can those answers be different?

What you are showing is that 0.999~ = 1.

By rejecting that result you are in effect rejecting that 0.333~ is the decimal representaton of the fraction 1/3. (As you admitted/posted in answer to my question).

That, I believe, is way off mainstream. *

(Edit: * just to clear, that's a "member comment" not a "moderator comment").

a1call

2010-Oct-19, 06:26 AM

In this thread I have presented a proof that 0.999... = 1 (http://www.bautforum.com/showthread.php/14593-Do-you-think-0.9999999-1-that-is-infinite-9s.?p=1804945#post1804945).

I also suggested proofs/definitions to the contrary which were rejected.

Here is another attempt at the latter to drive home the point that all this is a matter of which convention or definition you happen to choose as valid:

In mathematics, a rational number is any number that can be expressed as the quotient a/b of two integers, with the denominator b not equal to zero. (http://en.wikipedia.org/wiki/Rational_number)

Since 0.999... is a rational number we will "express" it as:

999.../1000...

The rational number a/b is not an algebraic integer unless b divides a (http://lcni.uoregon.edu/~mark/Geek_math/Discrete_math/Math_backgound-discrete-math.html)

So if 999.../1000... is an algebraic integer then 1000... divides 999...

if a divides b and b divides c then a must divide c

A number (i.e., integer) expressed in the decimal numeral system is even or odd according to whether its last digit is even or odd. That is, if the last digit is 1, 3, 5, 7, or 9, then it's odd; otherwise it's even. (http://en.wikipedia.org/wiki/Parity_%28mathematics%29)

Since 1000... is even, it is dividable by 2 and since it divides 999..., then 999... is divisible by 2. However 999... is by definition odd and is not divisible by 2.

Thus 999.../1000... can not be an integer which means if it is equal to 1 then 1 is not an integer either since a number can not be both an integer and a non-integer.

However 1 is an integer(I will let you prove that on your own).:lol:

I will also let you conclude what this discrepancy entails just to keep the neighbors from being disturbed by the yelling which will inevitably peruse.:lol:

dgavin

2010-Oct-19, 06:39 AM

I'm sorry, but there's lots of math etc. that you are ignoring.

In this case, I'm not ignoring your maths, I'm rejecting that it makes a difference.

I totally agree that: 3 x (1/3) can be done as (3 x 1) / 3 giving 3 / 3 = 1.

What I reject is the distinction between doing it that way and doing the division first.

After all, you start by saying: 3 x (1/3) can be done as (3 x 1) / 3; you are saying they are equal.

Doing the (1/3) first can't produce a different answer. (It can result in a different notation.)

Just like: 4 x (6 / 2) = 4 x 3 = 12 = (4 x 6) / 2 = 24 / 2 = 12. The order does not change the math here.

Do you not agree that: (a / b) x (c / d) = (a x c) / (b x d) ?

If the thing on the left equals the thing on the right, how can it matter which you work out first?

If the thing on the left gives an answer and the thing on the right gives an answer, how can those answers be different?

What you are showing is that 0.999~ = 1.

By rejecting that result you are in effect rejecting that 0.333~ is the decimal representaton of the fraction 1/3. (As you admitted/posted in answer to my question).

That, I believe, is way off mainstream. *

By rejecting that result you are in effect rejecting that 0.333~ is the decimal representaton of the fraction 1/3. (As you admitted/posted in answer to my question).

(Edit: * just to clear, that's a "member comment" not a "moderator comment").

And you seem to be pulling in straw-man arguments by putting words into my mouth that I clearly, and pointly, did not say.

My reply to your pointed question, about 1/3 = .3333~, which was not even part of the OP I might add, which I answered properly as .3333~ ≈ 1/3.

And explained why, but i'll explain it again.

It would take an infinate amount of time and an infinate amount of calculations, to actually prove that 1/3 = .3333~. It's not provable (or disporvable), because the calcualtion never stops. There for the expresion .3333~ ≈ 1/3. is the correct one. There is a reason for the ≈ symbol in math, and this is one good example. Another one would be Pi ≈ 3.1415926.... (no matter how many decimial places you go out to).

Even by your logic of division first with 3 * (1/3) its yeilds the approximal of 1/3 stated as .3333~. Multiplying an approximal by any number yeilds another approximal. in this case .9999~. So once again you have an approximal .9999~ ≈ 1. Hence the reason for my no vote that .9999~ = 1.

I've now show three different ways that you can use algebra to solve this without getting a result of .9999~ = 1 . All proper ways. if you want to keep arguing against math go right ahead.

But don't accuse people of beign ATM when thier ansering you questions with valid math, thats just out there and it's also a straw-man argument.

I've been very civil and patient making my case, and have even answered questions beyond the OP topic (which I might add I didn't have to according the forum rules). I don't apprecite being called ATM over it. I'm sorry you don't seem to like the answers, but math is like that at times.

With now three different ways shown that you can resolve the problem without winding up with .9999~ = 1, I think I've put a nail in the coffin on my case. .9999~ ≈ 1 is correct. .9999~ = 1 is not.

Nicolas

2010-Oct-19, 09:16 AM

You use this assumption that the infinite decimals are an approximation to prove that...infinite decimals are an approximation. That's not a proof. Not even after three times.

There is a reason for the ≈ symbol in math, and this is one good example.

No, this is a bad example.

0.333 ≈ 1/3
0.33333333333333333333 ≈ 1/3
0.333... = 1/3

That's the whole mathematical concept of infinite decimals. It's a limit result. The value where the limit goes to.

Strange

2010-Oct-19, 09:39 AM

in one of the examples provided 3 x (1/3) = (3 x 1) / 3 = 1 should be expressed as 3 x (1/3) ≈ (3 x 1) / 3 ≈ 1. (approximately equal)

So what is the value of 3 x (1/3) - (3 x 1) / 3 ?
What si the value of 1 - 3 x (1/3) ?
What is the value of 1 - (3 x 1) / 3 ?

Are these three differences equal or only approximately so?

grapes

2010-Oct-19, 09:53 AM

Reposting my answer:

I would form it as .3333~ ≈ 1/3 for the main reason that not matter how long I calcualted 1/3, I'd be pushing up daisies before I could actualy prove or disprove matmatically that .3333~ is actualy equal to 1/3.
There are mathematical techniques for handling such situations, that don't take a lifetime. It's true that some people never learn them in their lifetime, but that is not what we are talking about here.

My reply to your pointed question, about 1/3 = .3333~, which was not even part of the OP I might add, which I answered properly as .3333~ ≈ 1/3.

And explained why, but i'll explain it again.

It would take an infinate amount of time and an infinate amount of calculations, to actually prove that 1/3 = .3333~. It's not provable (or disporvable), because the calcualtion never stops. There for the expresion .3333~ ≈ 1/3. is the correct one. There is a reason for the ≈ symbol in math, and this is one good example. Another one would be Pi ≈ 3.1415926.... (no matter how many decimial places you go out to).
The only formulas we have to compute pi involve sums of an infinite number of terms, that can be mathematically proven to converge to a particular number--in this case, pi. Similarly, there are formulas for calculating 1/3 that have an infinite number of terms. We don't have to calculate them all to get a good idea of the size of the number itself, but we can prove, in a finite amount of time, that they are equal.

With now three different ways shown that you can resolve the problem without winding up with .9999~ = 1, I think I've put a nail in the coffin on my case. .9999~ ≈ 1 is correct. .9999~ = 1 is not.That's OK. You have a right to that opinion.

However, avoiding such constructs is not our goal. By definition, the real numbers contain numbers that have an infinite decimal expansion that cannot be understood except as an infinite number of terms. Mathematically, I can't afford to avoid them. :)

pzkpfw

2010-Oct-19, 09:54 AM

... My reply to your pointed question, about 1/3 = .3333~, which was not even part of the OP I might add, which I answered properly as .3333~ ≈ 1/3. ...

That came up, as 1/3 = 0.333~ was part of a proof that 1 = 0.999~. Since you rejected that, that made it something to discuss.

(See your own post, #2157.)

(Also, I reject your characterisation of that as the "proper" answer.)

... It would take an infinate amount of time and an infinate amount of calculations, to actually prove that 1/3 = .3333~. It's not provable (or disporvable), because the calcualtion never stops.

No, that's irrelevant. For example, you are ignoring the entire field of limits.

(
e.g. the kind of thing HenrikOlsen showed with: limn=>-∞ 1 - 10n = 1

Mathematicians happily use limits, without actually calculating out the things to an infinite number of steps!
)

There for the expresion .3333~ ≈ 1/3. is the correct one. There is a reason for the ≈ symbol in math, and this is one good example. Another one would be Pi ≈ 3.1415926.... (no matter how many decimial places you go out to).

No. .3333~ is not an approximation of 1/3, it is 1/3. There's no need to claim it's equivalent ( ≈ instead of = ).

Even by your logic of division first with 3 * (1/3) its yeilds the approximal of 1/3 stated as .3333~.

No, it's not the "approximal", it's equal.

Multiplying an approximal by any number yeilds another approximal. in this case .9999~. So once again you have an approximal .9999~ ≈ 1. Hence the reason for my no vote that .9999~ = 1.

That's circular, as your assumption of "approximal" results in a further "approximal".

I've now show three different ways that you can use algebra to solve this without getting a result of .9999~ = 1 . All proper ways. if you want to keep arguing against math go right ahead.

No, you haven't.

You've shown that one approach results in 1 written as 1. You've not disproved the other approach, which results in 1 written as 0.999~.

But don't accuse people of beign ATM when thier ansering you questions with valid math, thats just out there and it's also a straw-man argument.

Which part of my argument is the strawman?

(Describing the effect or result of your statement, is not the same as saying that you are intentionally saying that thing, by the way).

I've been very civil and patient making my case, and have even answered questions beyond the OP topic (which I might add I didn't have to according the forum rules). I don't apprecite being called ATM over it. I'm sorry you don't seem to like the answers, but math is like that at times.

We've both been civil and patient, so that's no "point". However, it does seem to me that you ignore the bulk of my posts as you don't actually reply in detail to any of it, you simply repeat your assertion. "Math is like that at times" is something that I could just as easily say to you.

With now three different ways shown that you can resolve the problem without winding up with .9999~ = 1, I think I've put a nail in the coffin on my case. .9999~ ≈ 1 is correct. .9999~ = 1 is not.

What three ways?

Besides, I've agreed that yes, you can result in a nice clean "1" if you do the math a certain way (in the specific example given).

3 x 1/3 = (3 x 1) / 3 = 3 / 3 = 1 : Yes! I Agree! You are correct!

But you still have not shown why that way is somehow the "best" or "better" than the other way.

3 x 1/3 = 1 x 0.333~ = 0.999~ = 1 : Why is this "wrong"? (Perhaps try to prove that 1/3 = 0.333~ is "wrong")

The simple existence of the first way to approach the problem does not show the other way to be wrong (unless you start with the assumption that 0.999~ is somehow "wrong").

4 x 8 / 2 = (4 x 8) / 2 = 32 / 2 = 16
4 x 8 / 2 = 4 x (8 / 2) = 4 x 4 = 16

You get the same result both ways. "math is like that at times". There may be operational reasons (rounding, capacity of variable type chosen in a computer language etc.) to choose one way over the other, but mathematically they are the same thing, both ways.

grapes

2010-Oct-19, 09:56 AM

No, this is a bad example.

0.333 ≈ 1
0.33333333333333333333 ≈ 1
0.333... = 1
That's a very bad example. :)

grapes

2010-Oct-19, 09:59 AM

No. .3333~ is not an approximation of 1/3, it is 1/3. There's no need to claim it's equivalent.
JOOC, where is this idea of "equivalent" meaning "approximate" coming from?

pzkpfw

2010-Oct-19, 10:28 AM

JOOC, where is this idea of "equivalent" meaning "approximate" coming from?

I'm using it as dgavin did, in post #2167, as the post was addressed to him.

I'm not denying your later correction of that words' correct usage.

(I've clarified in the post).

Strange

2010-Oct-19, 10:37 AM

Since 0.999... is a rational number we will "express" it as:

999.../1000...

That will be infinity / infinity so, as the young people round here say: FAIL

Ivan Viehoff

2010-Oct-19, 02:34 PM

Here is another attempt at the latter to drive home the point that all this is a matter of which convention or definition you happen to choose as valid:

In mathematics, a rational number is any number that can be expressed as the quotient a/b of two integers, with the denominator b not equal to zero.
Since 0.999... is a rational number we will "express" it as:
999.../1000...

You are correct that this is all a matter of convention or definition. But the question only makes sense within the category of Real Numbers, and in the category of real numbers it is quite clear that these are two notations for the same thing. There is no issue of whether they are equal or equivalent, they are exactly the same thing.

You give a definition of a rational number, and then present something calling it a rational number, when it does not have the form required by your definition.

You have to be very careful as to what what set you are acting.

If you are acting solely within the set of rational numbers, then the thing you mention is not one.

If you embed the rational numbers within the real numbers, then in the real numbers 0.999... is notation which is defined to refer to precisely the same object as 1.000... . According to the embedding, that is the rational number 1/1. That is the sense in which 0.999.... is a rational number.

Let's explore that in more detail

Consider the set Z x Z+, ie the set of ordered pairs (a,b) where a is an integer and b is a strictly positive integer.
Define an equivalence relation E such that (a,b) E (c,d) if ad=bc.
Define the symbol a/b to mean the equivalence class containing (a,b).
The Rational Numbers Q is the set of symbols a/b as defined above.
Note that a/b is a non-unique notation, because a/b and c/d refer to the same equivalence class if (a,b) E (c,d). So within the context of the rational numbers, 2/3 and 4/6 are merely two notations for the same thing. No need to get into arguments about equals and equivalent to. They are defined as the same thing.

Now consider what the Real Numbers are. They can be defined as follows.
Consider the set comprising the following members
r=(q1, q2, q3, ....)
where q1, q2, q3, etc are rational numbers. This is the set of all sequences of rational numbers.
Now consider the subset of those which are all those which are bounded above. Call this set S.
For each r define Ur as {rational q: q>=qi for all i}
Now consider the equivalence relation F such that r F s if Ur=Us.
The real numbers R is the set of equivalence classes of S under relation F.
The real number notation a.bcd... can be taken to mean the sequence (a, a.b, a.bc, a.bcd, ...)
So if r = 0.999..., then its sequence is (0.9, 0.99, 0.999,...) and if s=1 then its sequence is (1, 1, 1, ...)
The upper bound sets in each case are (q:q>=1) So they are members of the same equivalence class. That means that they are just two notations for the same member of R.

Clearly the symbols 0.999.... and 1.000... are different symbols. We can no doubt define a set in which these symbols mean something different from each other. However this set would not be Real numbers, and I can't think of any use for such a set. Within the Real Numbers, these two things are the same because they are defined to be the same, they are equivalent notations for the same thing in the same way that 2/3 and 4/6 are equivalent notations for the same thing in the Rational numbers.

dgavin

2010-Oct-19, 03:15 PM

The point i've been making, is that there are many ways to get .9999~ to = 1 that work (sequence sets, no real number between .9999~ and 1, inifinte limits, algebra) there are also ways to avoid it completely. In otherwards if .9999~ was = 1 then the various methods should be consistant, which they are demonstratably not.

The best argument for .9999~ to = 1 , no real number between .9999~ and 1, involves an imaginary number. However that can be repesented too with a formula as .9999~ - 1^(10*~ + 1), granted it is still an imaginary number, however it it can be expressed, if not solved.

So you can't really say conclusively that .9999~ = 1, when there are other ways available to avoid that out come. In which case expressing it as .9999~ ≈ 1 is correct. 1/.9999~ also yeilds an irrational number which is goes against .9999~ = 1.

And there is the quandry yet again. What can be used to say .9999~ = 1 can also be used to say it's not.

If two formulas, that should return the same exact result in all cases, do not, then the best that can be hoped for in math is an approximal result.

a1call

2010-Oct-19, 03:26 PM

That will be infinity / infinity so, as the young people round here say: FAIL

Keeping in mind that we have already established that infinity is not a number, be it real, rational or integer:

*- Are you saying that the set of integers contains only numbers with finite number of digits? And any number with infinite number of digits is therefor not an integer?:naughty::naughty:

Strange

2010-Oct-19, 03:27 PM

And there is the quandry yet again. What can be used to say .9999~ = 1 can also be used to say it's not.

But you haven't shown they are not equal. All you have shown is that you can avoid 0.999~, which proves nothing (except that the two numbers are equal).

Just because you can do a calculation in a way that produces the result 1, does not invalidate the equivalent calculation that produces the same result that demonstrates 0.999~ = 1.

The best argument for .9999~ to = 1 , no real number between .9999~ and 1, involves an imaginary number.

I assume you mean a non-existent number rather than imaginary (no sqrt(-1) here). It is the fact that the number is non-existent that proves the two representations are equal.

However that can be repesented too with a formula as .9999~ - 1^(10*~ + 1), granted it is still an imaginary number, however it it can be expressed, if not solved.

I'm not sure what that notation is supposed to mean. But if you mean the equivalent of 0.000~1 then that is meaningless. Or at least, it simply means 0 because there is no "1" at the end, because there are an infinite number of zeros. It is just 0.000~

As you have claimed that 3 x (1/3) ≈ (3 x 1) / 3, what is the value of (3 x (1/3)) - ((3 x 1) / 3) ?

Strange

2010-Oct-19, 03:30 PM

Keeping in mind that we have already established that infinity is not a number, be it real, rational or integer:

*- Are you saying that the set of integers contains only numbers with finite number of digits? And any number with infinite number of digits is therefor not an integer?:naughty::naughty:

Er, no. A number can have an infinite number of digits and still be finite. Some obvious examples are Pi and 1.

ETA: The difference is, you had an infinite number of digits before the decimal point (or at least, that is how I read your notation).

grapes

2010-Oct-19, 03:38 PM

Strange

2010-Oct-19, 03:41 PM

Oh yes, I missed that. But then I don't understand the point of the questions (and the naughty icons)...

grapes

2010-Oct-19, 03:41 PM

In otherwards if .9999~ was = 1 then the various methods should be consistant, which they are demonstratably not.
You haven't demonstrated any inconsistancy, though, only a way to maybe avoid the second representation. As you say, that still leaves the other proofs valid, that 0.999... = 1.0

grapes

2010-Oct-19, 03:42 PM

Oh yes, I missed that. But then I don't understand the point of the questions (and the naughty icons)...Perhaps they think integers can have an infinite number of digits.

Ivan Viehoff

2010-Oct-19, 04:00 PM

The point i've been making, is that there are many ways to get .9999~ to = 1 that work (sequence sets, no real number between .9999~ and 1, inifinte limits, algebra) there are also ways to avoid it completely. In otherwards if .9999~ was = 1 then the various methods should be consistant, which they are demonstratably not.
The problem with all these approaches is that they just heuristics, not rigorous arguments. Rigour requires that you define the symbols you are using, the sets that they are contained in, the arithmetical operations. This requires going back to the axiomatic set theory. As I demonstrated above, in the axiomatic set theory, you already have that WITHIN THE REAL NUMBERS RIGOROUSLY DEFINED 0.999... is just a notation for precisely the same thing as 1.000... before you even get as far as defining the arithmetical operations. The "demonstrations" that 0.999... = 1.000... by various means, eg multiply it by 9 on one side and 10-1 on the other, are just heuristic arguments to make people comfortable about it, without introducing them first to the necessary axiomatic set theory which would be necessary to make their arguments rigorous. Outside the real numbers rigorously defined, you can invent your own symbols and set such that 0.999... ¬= 1.000..., But this is not what anyone had in mind when they asked if these things are the same.

a1call

2010-Oct-19, 04:00 PM

The set of integers only contains numbers with a finite number of digits.

Are we gonna make up rules as we go along (http://www.google.com/search?hl=&q=The+set+of+integers+only+contains+numbers+with+a +finite+number+of+digits&sourceid=navclient-ff&rlz=1B3GGGL_enCA230CA231&ie=UTF-8#sclient=psy&hl=en&rlz=1B3GGGL_enCA230CA231&q=%22The+set+of+integers+only+contains+numbers+wit h+a+finite+number+of+digits%22&aq=f&aqi=&aql=&oq=%22The+set+of+integers+only+contains+numbers+wi th+a+finite+number+of+digits%22&gs_rfai=&pbx=1&fp=9937e1fec510269a)?

Strange

2010-Oct-19, 04:08 PM

Are we gonna make up rules as we go along (http://www.google.com/search?hl=&q=The+set+of+integers+only+contains+numbers+with+a +finite+number+of+digits&sourceid=navclient-ff&rlz=1B3GGGL_enCA230CA231&ie=UTF-8#sclient=psy&hl=en&rlz=1B3GGGL_enCA230CA231&q=%22The+set+of+integers+only+contains+numbers+wit h+a+finite+number+of+digits%22&aq=f&aqi=&aql=&oq=%22The+set+of+integers+only+contains+numbers+wi th+a+finite+number+of+digits%22&gs_rfai=&pbx=1&fp=9937e1fec510269a)?

What does that mean? Clearly, an integer with an infinite number of digits (before the decimal point) is infinitely large and (by your own earlier argument, if I recall) not allowed in integer arithmetic.

grapes

2010-Oct-19, 04:10 PM

Are we gonna make up rules as we go along (http://www.google.com/search?hl=&q=The+set+of+integers+only+contains+numbers+with+a +finite+number+of+digits&sourceid=navclient-ff&rlz=1B3GGGL_enCA230CA231&ie=UTF-8#sclient=psy&hl=en&rlz=1B3GGGL_enCA230CA231&q=%22The+set+of+integers+only+contains+numbers+wit h+a+finite+number+of+digits%22&aq=f&aqi=&aql=&oq=%22The+set+of+integers+only+contains+numbers+wi th+a+finite+number+of+digits%22&gs_rfai=&pbx=1&fp=9937e1fec510269a)?A google search? :)

There is no integer with an infinite number of digits. If there were an infinite number of digits to the left of the decimal point, the number would be infinite, and as we've all apparently agreed before, infinity is not an integer.

Nicolas

2010-Oct-19, 04:15 PM

That's a very bad example. :)

And it took me waaaay too long to figure out why. :D

a1call

2010-Oct-19, 04:56 PM

What does that mean? Clearly, an integer with an infinite number of digits (before the decimal point) is infinitely large and (by your own earlier argument, if I recall) not allowed in integer arithmetic.

On my end I get:

No results found for "The set of integers only contains numbers with a finite number of digits".

Grapes, Please provide a reference for that claim other than your personal implication.
Thanks in advance.

grapes

2010-Oct-19, 05:03 PM

Please provide a reference for that claim other than your personal implication.
Personal implications? It was logic. :)

Another google (http://www.google.com/search?source=ig&hl=en&rlz=&q=%22every+integer+has+a+finite+number+of+digits%2 2&aq=f&aqi=&aql=&oq=&gs_rfai=Ce8Vkys69TJqqF4eazATJu8H6CQAAAKoEBU_Qfd1k)

a1call

2010-Oct-19, 05:24 PM

Well 5 results with a top result as a postulate isn't bad. But I hope that you would agree there are more numerous references to integers as an infinite set such as:

Like the natural numbers, the integers form a countably infinite set. (http://en.wikipedia.org/wiki/Integer)

Where countable here obviously does not mean finite.

agingjb

2010-Oct-19, 05:28 PM

Integers form an infinite set. No integer has an infinite number of digits. Is there a problem with that?

a1call

2010-Oct-19, 05:32 PM

Integers form an infinite set. No integer has an infinite number of digits. Is there a problem with that?

Yes, it sounds like an oxymoron to me.

grapes

2010-Oct-19, 05:46 PM

Oxymorons are allowed in mathematics, it's just language: binary decimal point

a1call

2010-Oct-19, 05:47 PM

HenrikOlsen

2010-Oct-19, 07:02 PM

But anyway, dgavin's claim is that (a/b)*c<>(a*c)/b
Excuse me? where did I ever claim that? What I did was take it into algebra and reduce the equation to it's simplest form, which completely eliminates said error of ~.9999 = 1.
a=1, b=3, c=3 You're claiming that (1/3)*3<>(1*3)/3 .

There is no error, that's the problem you apparently don't get. It doesn't need fixing.
There is no trick, that's the problem you apparently don't get. That's how things really works.

It looks almost as if you've been blinded by using a computer to learn mathematics and you apparently missed that the thing you call math teacher's tricks is the way they try to teach that a computer doesn't do mathematics, only approximations.
A computer is rubbish for developing a good feel for mathematics, exactly because it gives wrong results and only knows approximations.

With a computer it's entirely possible to get different results from (1/3)*3 and (1*3)/3, that's because the computer is wrong and only works with an approximation. In mathematics they are not approximated, so in mathematics we get the same results. Always.

What you are claiming is that (1/3)*3 and (1*3)/3 yields different results in mathematics, and that's just plain wrong.

If two formulas, that should return the same exact result in all cases, do not, then the best that can be hoped for in math is an approximal result.
In mathematics, if two mathematical formulas, that should return the same exact result in all cases, returns something that on the face of it looks to be different, then you'd better reconsider whether your idea about them being different is actually correct.
We're not talking about computers here.

Not suggesting any such thing. I was trained to always reduce first (if possible) and solve after. This was later reinforced in life by speeding up math in code by also reducing formulas to thier simplest form.
And you're talking numerical analysis again, approximations of mathematical values in a computer.
We have already covered that this it a situation where reduction yields better results because it invariably works with approximations.

Which part of "This is mathematics, not numbers in a computer" don't you get?

HenrikOlsen

2010-Oct-19, 07:09 PM

The best argument for .9999~ to = 1 , no real number between .9999~ and 1, involves an imaginary number. However that can be repesented too with a formula as .9999~ - 1^(10*~ + 1), granted it is still an imaginary number, however it it can be expressed, if not solved.
You're using "imaginary number" in a definitely nonstandard way here, please define what you mean by "imaginary".

pzkpfw

2010-Oct-19, 07:23 PM

This wiki article is a nice summary, not only of 0.999~=1, but why people sometimes don't accept it.

http://en.wikipedia.org/wiki/0.999...

Students are often "mentally committed to the notion that a number can be represented in one and only one way by a decimal." Seeing two manifestly different decimals representing the same number appears to be a paradox, which is amplified by the appearance of the seemingly well-understood number 1.

There's even a maths joke in there. (Don't quote it here, make it a reward for reading the article!)

Finally, it covers some alternate numbers systems where 0.999~ does not = 1.

(Wiki of course lets "anyone post" (yes, that's not quite true) so to be fair, here's some contesting stuff: http://en.wikipedia.org/wiki/User:ConMan/Proof_that_0.999..._does_not_equal_1 )

Dr Math is pretty good: http://mathforum.org/dr/math/faq/faq.0.9999.html (See also links at bottom of page).

dgavin

2010-Oct-19, 07:52 PM

You haven't demonstrated any inconsistancy, though, only a way to maybe avoid the second representation. As you say, that still leaves the other proofs valid, that 0.999... = 1.0

Then lets take it a step beyond just saying .9999~ = 1.0 and see what happenes (and where the another troublesome quandry really comes into play)

Given early paramaters that .9999~ = 1.0 then the following statement should also be true. 2 * .9999~ = 2 * 1. However when to try to solve the left half of the equasion on it's own 2 * .9999~ (no reduction, no rounding, no substitution of 1 for .9999~ etc...) you get into immediate trouble as it becomes an unsolvable situation without reduction, rounding, or substitution. Even if you were to do somehting radical and define a new symbol context for this by allowing additional digit after ~ sign, (and i'm not saying this is valid at all, just good for a visualization of the issue) 2 * .9999~ = 1.999~8...

You quickly realize that 2 * .9999~ no matter what you do, yeilds a non existant (or unsolvable, take your pick) number.

So basically if you take .9999~ = 1 to it's logical conclusion by working with .9999~ directly (2 * .9999~ = 2 * 1) , you wind up with a non existant (or unsolvable, take your pick) = 2 in the example.

It basically doesn't work, because math breaks down at the point of infinity, including infinate decimal places, without reduction, rounding, or substitution.

I think this makes my case, yet again, of why reduction, in the first palce, -is- an acceptable solution (and always yields a 1).

Anyway I will admit that (2 * .9999~ = 2 * 1) in my example is completely reducable as well, but I was using it as an an example of how to drive home the problem again of .9999~ = 1 by showing what happens if you actualy try to use .9999~ instead of 1. It doesn't work so well:D

pzkpfw

2010-Oct-19, 08:13 PM

pzkpfw

2010-Oct-19, 08:21 PM

...
You quickly realize that 2 * .9999~ no matter what you do, yeilds a non existant (or unsolvable, take your pick) number.
...

Just for fun:

2 x 0.999~
= 2 x 0 + 2 x 0.9 + 2 x 0.09 + 2 x 0.009 + ~
= 0 + 1.8 + 0.18 + 0.018 + ~
= 1.99~

Again, you are being circular. You call 2 x 0.999~ "a non existant (or unsolvable, take your pick) number", but that is only true if you already start with the assumption that 0.999~ is "a non existant (or unsolvable, take your pick) number".

[
Edit: Perhaps more formally (I'm rusty on this):

0.999~ = ∑n = -1 → -∞ 9 x 10n

2 x 0.999~ = ∑n = -1 → -∞ 18 x 10n
]

pzkpfw

2010-Oct-19, 08:54 PM

For example, your insistence that 3 x (1 / 3) be done as (3 x 1) x 3 is "better" has some merit operationally (in some circumstances), but has nothing at all do with mathematical certainty.

I wanted to explain what I meant by "(in some circumstances)", because I think it is relevant.

Someone (sorry, I forget who) mentioned a few pages back that yes it is true that is sometimes better to do the multiplication first, then the division, because it retains accuracy longer. No arguement there.

Our Government recently increased GST (Goods and Services Tax, like VAT or sales tax) to 15%.

They have issued the suggestion that the tax content of a price be worked out as "multiply by 23, then divide by 3" "multiply by 3, then divide by 23". The reason is that keeps the result more accurate than simply dividing by 7 2/3, because use of 7.6666~ will generally mean using the rounded 7.66667 or so.

However there are times when doing it the other way might be better. Let's say you are writing software that relies on 1 byte unsigned integers.

The calculation A x B / C could be done (A x B) / C or A x (B / C).

You might look at the problem domain and find that inputs such as A = 64, B = 6 and C = 2 are common.

This would mean that the intermediate result, of (A x B) would equal 384. This intermediate result is larger than your single byte unsigned integers can handle. In this case, you'd be wanting to do it as 64 x (6 / 2) = 64 x 3 = 192 : and it all fits in the 0 to 255 range.

Yes, this is a awfully contrived example, but the point is that the assumption that (A x B) / C is "better" than A x (B / C) is wrong; intermediate results can be important when looking at a computing problem. (I've programmed industrial control and monitoring equipment in the past; this kind of limitation is a real factor).

And in any case, both ways would give the same result (if we had a better type to store the numbers in).

64 x (6 / 2) = 64 x 3 = 192
(64 x 6) / 2 = 384 / 2 = 192

SeanF

2010-Oct-19, 09:12 PM

They have issued the suggestion that the tax content of a price be worked out as "multiply by 23, then divide by 3". The reason is that keeps the result more accurate than simply dividing by 7 2/3...
Significantly more accurate, I would think. ;)

Grey

2010-Oct-19, 09:31 PM

Our Government recently increased GST (Goods and Services Tax, like VAT or sales tax) to 15%.

They have issued the suggestion that the tax content of a price be worked out as "multiply by 23, then divide by 3". The reason is that keeps the result more accurate than simply dividing by 7 2/3, because use of 7.6666~ will generally mean using the rounded 7.66667 or so.Suggesting that people calculate the 15% that they owe you by multiplying by 23 and then dividing by 3 seems like it's adding a really big extra tax on people that are bad at math. :)

pzkpfw

2010-Oct-19, 09:48 PM

Suggesting that people calculate the 15% that they owe you by multiplying by 23 and then dividing by 3 seems like it's adding a really big extra tax on people that are bad at math. :)

Eh? That's to work out the tax content of a price, i.e how much of a tax inclusive price is the tax, not how much more is needed over and above the tax exclusive price. How would you do it?

SeanF

2010-Oct-19, 09:53 PM

Eh? That's to work out the tax content of a price, i.e how much of a tax inclusive price is the tax, not how much more is needed over and above the tax exclusive price. How would you do it?
So if the tax inclusive price is $115, I multiply by 23 ($2645) and then divide by 3 ($881.67) to determine how much is taxes?

Actually, now I need to edit my original response to your post, since clearly dividing by 7 2/3 would be more accurate than this. :)

pzkpfw

2010-Oct-19, 10:02 PM

Sigh. This is where a typo in a minor side point derails a thread. Call it dyslexia or whatever, I meant multiply by 3 then divide by 23, not multiply by 23 and divide by 3. The "2 shifted". (I actually did a little example in Excel before I typed that post, and it all worked, so convinced myself not to proof-read my own post. Doh!)

( http://www.ird.govt.nz/changes/news-updates/budget-news-calculate-gst.html )

100 x (115%) = 115 = 100 + 15

115 x 3 = 345

345 / 23 = 15

Does any of this change the point I was trying to make?

Strange

2010-Oct-19, 10:51 PM

let f(n) be the number of digits in the integer n. If there are infinite number of integers then f(n) can not be finite.

ETA: BTW IIRC, if you look in another thread that I linked to earlier, you will see a proof that I offer to the contrary. But that's a measure of the nature of infinity and not my weak resolution. I maintain that logic fails at infinity.

But this has nothing to do with 0.999... because infinities do not come into it. There is a big difference between an integer with an infinite number of digits (which would be infinite - and you foolishly tried to do arithmetic with two such values) and a real number which has an infinite number of digits in its decimal expansion. This is not an inifinte value. But we seem to be going round in circles here.

What is the value of 1-0.999~ ?

I assume you will say it is 0.000~1 ? But then you are bringing infinities (in this case infinitely small) into the argument and claiming that this has a value different from 0. Sigh.

Nicolas

2010-Oct-19, 10:59 PM

Anything behind the infinity symbol (in the decimals) has no meaning, because it is infinitely small. And in maths, infinitely small IS nothing. Because in maths, going to infinity means REACHING the limit value. And that's the point of the thread since the first post.

Maybe we should star discussing "maths" vs "math" for another 10 pages. Again. ;)

dgavin

2010-Oct-19, 11:09 PM

That a number has infinite digits, so that you couldn't write them all down if you did some maths with it, makes no difference.

Pi, for example, has infinite digits. That doesn't make it any less useful. If we use it to do an actual calculation we simply use a rounded number. If we use it in a mathematical proof, we simply use it's representation, "pi", or it's symbol, or "3.14~" or whatever we need.

Yes and the proper expression for that is Pi ≈ 3.14159~ Just like the proper expression for 1 = .99999~ is 1 ≈ .99999~, because an aproximal number is one that either is selected (or needs) to be rounded to use, or it is an irrational number and needs to be rounded to use. There is no rule that says approximates have to be both rounded and irrational. You just made my case, again, that 1 ≈ .99999~ is the proper expression as you can't use .9999~ in place of 1 without some sort of rounding, or truncating of an axiomatic set.

A x (B / C) = (A x B) / C

I've demonstrated (three times now) that this is handled by reduction of the formula (and that was agreed as being valid), which is a valid method of performing this equation and without getting a 1 = .9999~ result.

Now if you are asking me to not reduce the formula and then compute it I've already admited you wind out with a quirky result. People call that proff. I don't because: once you reduce the formula as is proper with algebra, that quirk goes away. Reduction eleminates the quirk, which is a more proper solution then building axiomatic sets to try to confirm a quirky result.

Finally, it covers some alternate numbers systems where 0.999~ does not = 1.

I mentioned this before that in other numbering systems you can prove a not =, and that point was tossed out. Appreciate you pointing out that I was right. (Granted that those other numberisng systems also have quirks very similar to this one but with differen fractions if I remember correctly.)

The point here being that in decimal this quirk occurs on decimal 'boundries', in other number systems they occur on differenent 'boundries'. Sorry can't think of a betetr term of it other then 'boundries', I'm sure there probably is one though.

So this is probably only place where new math (ewwwwww) may have been a valid point. What you are calling proof here, can and already has been disproven (by others) in different numbering systems.