Is 1 = 0.9999......

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mchammer187

Diamond Member
Nov 26, 2000
9,114
0
76
Originally posted by: MadRat
Originally posted by: mchammer187
are you telling me Pi is not a real number so the ratio between the area of a circle to the radius does not exist because it has an infinite number of digits

Pi does not have a fixed value, correct, so it does not exist on any exact place on a number line. We can approximate the location of Pi on a numberline, but we cannot find its exact value. As supercomputers become more powerful then the accepted value of Pi is extended further.

just because it cannot be approximated with a finite number of digits does not mean it does not exist

Pi by definition is an IRRATIONAL number do you mean to say that all irrational numbers do not exist
 

mchammer187

Diamond Member
Nov 26, 2000
9,114
0
76
Originally posted by: MadRat
I said Pi wasn't finite, not that it does not exist.

i take it when you say it doesnt exist on the number line to saying it does not exist

i assert that there is a spot on the number line because there are an INFINITE number of points between any two points on the number line
 

MadRat

Lifer
Oct 14, 1999
11,965
279
126
Go ahead and define it for us, otherwise agree it is not fixed... unless you have some new way to define Pi we should know about!
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Pi not finite? Pi is less than 4 and greater than 3. It exists in a finite interval on the number line. It has to be finite.

Does 1/3 not exist on the number line because you can't write it as a decimal?

Look:

|(0)------------(Pi)---------->

There's Pi on a number line.

Futher evidence you have no idea what you're talking about:

3 a : less than an arbitrary positive integer and greater than the negative of that integer b : having a finite number of elements <a finite set>

That's a definition of finite. Let b = 4. Pi is less than 4 and greater than negative 4. It is therefore finite.

At least you can agree that 0.999... is less than 2. It is also greater than -2, therefore it too is finite.
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: MadRat
Go ahead and define it for us, otherwise agree it is not fixed... unless you have some new way to define Pi we should know about!

Ummm arctan(1/3) + arctan(1/2) I believe...
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: bleeb
At least you can agree that 0.999... is less than 2. It is also greater than -2, therefore it too is finite.

Have you been smokin CRACK?? 0.9999... is NOT finite.

I know you're just playing a devil's advocate on crack just to keep this thread going so I know I don't have to really respond to this post.

I guess there's no real point to this so put this one down as a nef.
 

bleeb

Lifer
Feb 3, 2000
10,868
0
0
Yeah I have been playing devils advocate, but you are saying 0.9999... is finite. For which it is NOT.
 

mchammer187

Diamond Member
Nov 26, 2000
9,114
0
76
Originally posted by: LackCash
.999999........ never converges w/ 1. It becomes infinitly closer but still can never reach 1.

again you are taking .99999 to be a sum of something im guess .9 + .09 + .009 but this is not the case because it is not a sum it is just a number

it does not converge to anything it is .99999999999999.....

.999...... is not some limit

how can something that does not change its value over x converge to anything

y= .999999........... does not converge to anything it is a straight horizontal line that has the same value at x -> -infinity as it does @ x -------> +infinity
 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
Originally posted by: MadRat
I said Pi wasn't finite, not that it does not exist.

Sweet, now PI > 10^100

Something that is not finite must be infinite. So you claim that all irrational numbers are infinite.

LOL

Insane rodent, why not just find something else to do with your time, you do not have any clue about the number system.


By my definition, if we agree that .999... exists then .999... < 1 --- that is, remember, if we agree it exists at all.

just a bit ago you were telling me that .999... was infinite now it is less then 1! So 1 is greater then infinity! Wow, this is an insane (dare I say the "mad") number system!



Adding .000...1 to .999... simply sums to 1

Don't you mean sums to 1 + .000....999...

you droped the infinite tail of nines, your single one must exist somewhere, like the energizer bunny, the nines keep right going.

Once again,

you CANNOT reach infinity, there is NO point at infinity. You keep on insisting on doing things "at the infinte" digit, this is impossible because there is NO such digit. Each digit in the decimal representation of a real number has a integer assigned to it. You keep talking about something beyond the largest integer, since the largest integer does not exist there can be nothing "beyond" it.
 

Chu

Banned
Jan 2, 2001
2,911
0
0
Originally posted by: MadRat
If .999... has no end then it has no segments and therefore has no finite value therefore is not a real number. So it must not exist on a number line which means you're wrong saying .999...=1.

I haven't read the replies to this yet, but I assume you are being flamed for your very loose definitions of "real number," "finite value," and "segment." I will just make one comment. How do you define a real number? I am curious because the construction of the Real Numbers from the Rational Numbers is excruciatingly complex, and it involves the concept of "cuts," which I have a fealing might be close to your definition of segment. Anyways . . . I will take a different path.

No matter what definition you assign to the above words, I assume we both agree that the number "0.1" is a very "real" number. Now convert it to binary. Is it no longer real since it no longer has a finite representation?

How about pi? By your definition, it is not real. Now let's go to base pi. In base pi, pi=1. Is it now real?

Definitions are incredibly important. If your using them in a nonstandard way, please define them.

-Chu
 

Chu

Banned
Jan 2, 2001
2,911
0
0
Out of curiosity, MadRat, have you ever had a class in Calculus? From reading your last 2 page of replies I have the a gut fealing you were never introduced to what the concept of a "limit" is, while most people pulling their hair out trying to convince you that .9999... = 1 seem to assume you are fine with its definition.

-Chu
 

Looney

Lifer
Jun 13, 2000
21,938
5
0
LOL i just started reading this thread Madrat, but are you being serious or are you just acting it to rile up all the geeks here??
 

Chu

Banned
Jan 2, 2001
2,911
0
0
Originally posted by: Moralpanic
LOL i just started reading this thread Madrat, but are you being serious or are you just acting it to rile up all the geeks here??

I wouldn't be surprised if Madrat was just a troll, but considering how many people have trouble getting this through their brain, I wouldn't be surprised if he was not. Hell, there was a huge thread on sci.math recently over the "box and balls" problem. Some Math Majors here have trouble getting their heads around that problem.

-Chu
 

JoeKing

Lifer
Oct 9, 1999
10,641
1
81
blah blah, too long of a thread. Just remeber you can never reach a destination in true space without aproximation. Half of a half of a half etc. Even if you take a digital number to almost infinity it can never truly represnt the analog.
 

AIWGuru

Banned
Nov 19, 2003
1,497
0
0
Some people seem to be assuming that 0.99 somehow approaches 1 by getting closer to it with each decimal place. Obviously this is not the case. Something cannot grow in size infinitely without surpassing a static larger number.
What you are seeing with each decimal place is an increased granularity to the detail of the number. Sort of like resolution. It is not getting bigger.

Obviously, 0.99... is larger than 0.98 and smaller than 1.0. Therefore, technically, they are not equal.

However, in the world of calculus there is the principle of limits which says that if the granularity of detail on the number is large enough that it will produce no difference to the outcome of your calculation, then it can be regarded as the same as 1.

If we were to implement the technically accurate statement that 0.99...!=1 then we would never be able to use Pi since we cannot fully integrate the number into a calculation. Of course, for each calculation, only a certain number of decimal places of Pi are going to have any affect on the calc therefore that is all that is needed. Likewise, with 0.99... Infinity provides such a fine granularity that you might just as well consider it a 1. It's the difference between correctness and practicality to get something done.

So all the people who are frantically pointing and yelling that 0.99... = 1 are completely wrong.

They would do much better to say that for the practical purposes of calculus, 0.99? can be represented as 1 because of it's infinite granularity which renders the difference between it and 1 irrelevant for any calculation.
 

Chu

Banned
Jan 2, 2001
2,911
0
0
So all the people who are frantically pointing and yelling that 0.99... = 1 are completely wrong.

I think i'm going to cry There is a difference between really, really, stupidly, hugely big and infinite. People keep confusing the former with the latter. They are FUNDEMENTALY DIFFERENT. .9999... is shorthand for:

SUM(9*10^-k, k=1..infinity)

This is 1. Google will turn up plenty of proofs.

.999... is NOT shorthand for

SUM(9*10^-k, k=1..j)

which is really, really, stupidly, hugely close to 1, BUT IS NOT 1 because for any value of j, I can give you a value that takes you closer to 1, i.e. the result by plugging in j+1.

These are REALLY FUNDEMENTALLY IMPORATANT CONCEPTS. I'm not saying they are obvious concepts -- before Fourier forced us to take a hard look at what limits really are there was certainly a lot of debate over it, but I don't know how anyone who has admitted to taking calculus using the modern (i.e. cira 1900) definition of the limit can say that they are essentially the same.

-Chu
 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
Originally posted by: AIWGuru
Some people seem to be assuming that 0.99 somehow approaches 1 by getting closer to it with each decimal place. Obviously this is not the case. Something cannot grow in size infinitely without surpassing a static larger number.
What you are seeing with each decimal place is an increased granularity to the detail of the number. Sort of like resolution. It is not getting bigger.

Obviously, 0.99... is larger than 0.98 and smaller than 1.0. Therefore, technically, they are not equal.

However, in the world of calculus there is the principle of limits which says that if the granularity of detail on the number is large enough that it will produce no difference to the outcome of your calculation, then it can be regarded as the same as 1.

If we were to implement the technically accurate statement that 0.99...!=1 then we would never be able to use Pi since we cannot fully integrate the number into a calculation. Of course, for each calculation, only a certain number of decimal places of Pi are going to have any affect on the calc therefore that is all that is needed. Likewise, with 0.99... Infinity provides such a fine granularity that you might just as well consider it a 1. It's the difference between correctness and practicality to get something done.

So all the people who are frantically pointing and yelling that 0.99... = 1 are completely wrong.

They would do much better to say that for the practical purposes of calculus, 0.99? can be represented as 1 because of it's infinite granularity which renders the difference between it and 1 irrelevant for any calculation.


This is pretty sad, clearly you thought you understood calculus, unfortunatly you don't

Beyond that, believe it or not there is much more to math then calculus. If you get to the details you will find that .999... is not approximately 1 it IS 1. They are merely different representations of the same point on the real number line.

You need to go back to your old calc book and actually atempt to UNDERSTAND the meaning of a convergent sequence.
 
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