Exploring the Convergence of 0.999... and the Concept of Infinity

[hello3719]

with my current knowledge i have already spotted flaws in the system and come up with solutions for them that work.

Can you show us and explain us these "flaws" in our system?
Show me 1 contradiction in our system and not something you don't like because you think it is illogical without even backing it out.

 

[ram2048]

okay this is just stupid.

you: show me where we're wrong
me: <here> see?
you: you've proven nothing show me where it's wrong
me: wtf <here>
you: you know nothing, show me where we're wrong

i'm not showing you any more until you can come up with a disproof for my theory that consists of more than "that's not how we do things, you must be wrong" or "there is no end, you must be wrong"

my numbers are the EXACT same ones in subtraction as the ones gotten from multiplication. yours are NOT "blah blah 8.999~ IS 9..." if it was 9 then it would have calculated out to BE 9.

 

[Zurtex]

ram2048 that is total rubbish. We have shown you over and over again why it is what it is and shown you over and over again why your so called disproofs are wrong. But you just keep posting over the same thing over and over again.

 

[ram2048]

that consists of more than "that's not how we do things, you must be wrong" or "there is no end, you must be wrong"

no, not really

 

[Zurtex]

no, not really

Then at least reply to the last few posts once again showing how unbelievably wrong you.

Edit: Why have you quoted yourself AGAIN? And disagreed with yourself.

 

[ram2048]

Then at least reply to the last few posts once again showing how unbelievably wrong you.

translation: dance for us while we give you theoretical problems way over your head even though they have nothing to do with the issue at hand.

prove me wrong first.

i wasn't quoting myself i was re-iterating my current stance and the disagreement was with YOUR statement as a response.

if you would at least read and understand what I'm telling you, you wouldn't ask such dumb questions :D

 

[matt grime]

Ok, here's a proof that 0.99999...=1

0.9999... is clearly the limit of the sequence x_n=0.9..9 with n nines in it, do we agree on that, since that isby definition what the symbol 0.9999... means, there can be no argument there.

consider the sequnce y_n=1 for all n.

y_n must tend to 1, yes?


consider x_n-y_n, then in abs values this is less than 1/10^n, hence x_n-y_n converges to zero, agreed? thus by the definition of the real numbers, these sequences lie in the same equivalence class of cauchy sequences, hence they are equal as real numbers.

now, point by point go through that and state where it is wrong, not where you think it is wrong, but where you can prove it is mathematically wrong.

 

[ram2048]

0.9999... is clearly the limit of the sequence x_n=0.9..9 with n nines in it, do we agree on that, since that isby definition what the symbol 0.9999... means, there can be no argument there.

.999~ has a limit of 1 but a SUM of .999~

it TENDS to 1, but it will never have a value of exactly 1.

.333~ has a limit of 1/3 but a SUM of .333~

it tends to 1/3 but will never have a value of exactly 1/3

 

[matt grime]

then you don't understand analysis do you? And we can see where you've gone wrong. No finite partail sum will reach the limit, but so what? this is your issue with a DEFINITION, sorry, you don't like it, but the point of analysis is to allow one to take limits.

you've not disproved anything, just shown you don't accept a definition. we can't alter that, but that's your problem.

 

[Hurkyl]

too much at once. just pick ONE

Asking ram decimals to represent the square root of 2 is probably going to be the most challenging "simple" question we could ask.

One problem is that you need to be able to forbid certain decimals from being in your system (such as the standard meaning to 0.999~ or 0.333~). However, you need to retain enough flexibility to represent irrational numbers like sqrt(2).

However, the most disasterous part, I think, is that without the nicely repeating patterns of rational numbers to help you out, there may be no way to "connect" the end of a decimal to the beginning of a decimal.

Note that your system already has limited forms of this problem with rational numbers; for instance, for the decimal 0.121212~, is the last digit a 2 (with remainder 12/99) or is it a 1 (with remainder 21/99)? This problem is solvable; I think the irrational version is not (at least not without using some difficult mathematics).

by the way, what's the accepted calculation for obtaining digits of pi ?

There can be particular methods for particular numbers, but there is a most general method to constructing a (mathematical) decimal, which I will demonstrate to construct the first few digits of sqrt(6).

2 < sqrt(6) < 3, so the first digit is 2
2.4 < sqrt(6) < 2.5, so the second digit is 4
2.44 < sqrt(6) < 2.45, so the third digit is 4
2.449 < sqrt(6), so the fourth digit is 9
2.4494 < sqrt(6) < 2.4495, so the fifth digit is 4
...

(see 1: below)

This recursively defines the n-th digit (for any positive integer n). You'll have to devise some clever trick, though, to extend this definition to go beyond integer positions, and such extensions are not generally easy, especially with recursions that strongly depend on knowing the previous term.

"blah blah 8.999~ IS 9..." if it was 9 then it would have calculated out to BE 9.

Do you also assert that the fraction 2/4 cannot be equal to the fraction 1/2?
(After all, if it was 1/2, then it would've been calculated out to BE 1/2...)



1: For simplicity, I was not considering the possibility that any of these partial decimals may be equal to the target number, but you wouldn't be happy with such a more complete method anyways, because it would yield yet another direct proof that 0.999~ = 1

 

[ram2048]

then you don't understand analysis do you? And we can see where you've gone wrong. No finite partail sum will reach the limit, but so what? this is your issue with a DEFINITION, sorry, you don't like it, but the point of analysis is to allow one to take limits.

you've not disproved anything, just shown you don't accept a definition. we can't alter that, but that's your problem

see the problem lies within that limit of infinity.

you: the limit is 1, therefore when infinity is hit, it equals 1.
me: the limit is 1, but even when infinity is hit it doesn't equal 1.

logically i am correct. but you win out with practicality because there's no good reason to go "beyond" infinity.

you define your lim as the sum of the set to infinity, but there's no point in you saying that because you've already limited yourself mentally at infinity.

 

[hello3719]

so still think that infinity +1 > infinity,
how can you go beyond infinity if infinity is not a number?

 

[ram2048]

Do you also assert that the fraction 2/4 cannot be equal to the fraction 1/2?
(After all, if it was 1/2, then it would've been calculated out to BE 1/2...)

no because 2/4 can be simplified to a rational base expression. if you could simplify ([8/1].[9/1][9/1][9/1]~[1/1]) any of those terms then they should be simplified... :D

i'm just messin with ya, I'm working on sqrt2 at the moment

 

[Hurkyl]

you: the limit is 1, therefore when infinity is hit, it equals 1.

When did we say "when infinity is 'hit' " or anything similar?

As long as you are in the mindset that our definitions are literally involving the behavior at infinity, you will never understand them.

 

[matt grime]

we haven't limited ourselves at all to stopping at infinity, which in the sense you mean is not what we're doing anyway as we never reach infinity, what with it not being an integer and everything. if you'd read anything about transfinite induction as i'd suggested rather than dismissing it contemptuously you'd realize

 

[ram2048]

well that makes no sense.

.999~ tends to 1. but since 1 can never be reached it IS 1?

instead of

.999~ tends to 1. but since 1 can never be reached it never becomes 1.

you seriously believe that?

and don't say you're not limiting yourself at infinity when you clearly are.

.999~ having infinite number of digits 9. if i add one more at the end i come EVEN CLOSER to 1. but you won't allow that so hence LIMIT.

 

[matt grime]

if you want to add another nine after the infinite number of them already there then you aren't using decimal notation so you aren't making any sense.

every mathematical object is limited by its definitions, but that you interpret that lmit in a negative way is your own problem: there is no square root of -1 in the real numbers, does that imply they are analytically deficient?

 

[Hurkyl]

.999~ = 1 because:

For every positive real number e,
there exists a positive integer N such that,
for every integer m bigger than N,
|1 - 0.9..9| < e (there are exactly m 9's in that number)


Notice, in particular, that this definition is entirely in terms of finite examples. Infinity doesn't inter into this at all.

 

[ram2048]

.999~ = 1 because:

For every positive real number e,
there exists a positive integer N such that,
for every integer m bigger than N,
|1 - 0.9..9| < e (there are exactly m 9's in that number)


Notice, in particular, that this definition is entirely in terms of finite examples. Infinity doesn't inter into this at all.

i have no idea where you're trying to go with that one, I'm putting positive reals in for e but it doesn't seem to be telling me anything useful :(

 

[hello3719]

1/9 = 1/10 + 1/100 + 1/1000 + ...

then 9 * 1/9 = 9/10 + 9/100 + 9/1000 +...
9/10 + 9/100 + 9/1000 +... = .999...
9/9 = .999...
1 = .999...

tell me where do you see a contradiction in this proof.( show me the illegal step )

 

[ram2048]

1/9 = 1/10 + 1/100 + 1/1000 + ...

right there. convergent sum to infinity.

.111~ has a limit of 1/9 but a sum of .111~ [edit] forgot the ~ thingy

in my notation 1/9 = .111~ r1/9
9 x .111~ r1/9 = .999 r9/9 = 1
and
9x 1/9 = 1

perfectly.

yours

9x 1/9 = 1

BUT

9x.111~ = .999~

 

[Hurkyl]

Were you able to find an appropriate N for every positive real you plugged in for e?


What if we asserted 0.999~ was something other than one... like, maybe, 1.01, which (by definition!) would mean:

For every positive real number e,
there exists a positive integer N such that,
for every integer m bigger than N,
|1.01 - 0.9..9| < e (there are exactly m 9's in that number)

I bet you can find an e that is a counterexample!

 

[ram2048]

where the heck is n in that equation.

damn don't play hide the variables with me :|

 

[hello3719]

1/9 = 1/10 + 1/100 + 1/1000 + ...

right there. convergent sum to infinity.

.111~ has a limit of 1/9 but a sum of .111

in my notation 1/9 = .111~ r1/9

Remember we are talking about OUR system not your system, you are using your definitions here to contradict the proof, don't introduce your notations in our system. Use our definitions of infinity and real numbers to find a contradiction. If you can't then this is the proof is completely valid and 1= .999... Well before asking this, do you know how do we define infinity and do we treat it ?

 

[ram2048]

i'm using your system and it doesn't work that's why i introduced my system :P

9x.111~ in your system = .999~

9x1/9~ in your system = 1

this leads to TWO possible conclusions.

.999 is equal to 1

OR

1/9 ≠.111~

you have chosen the 1st possibility and not proven it, i have chosen the second possibility and HAVE proven it.

dunno what more i can tell you...

 

[hello3719]

i'm using your system and it doesn't work that's why i introduced my system :P

9x.111~ in your system = .999~

9x1/9~ in your system = 1

this leads to TWO possible conclusions.

.999 is equal to 1

OR

1/9 ≠.111~

you have chosen the 1st possibility and not proven it, i have chosen the second possibility and HAVE proven it.

dunno what more i can tell you...

k this means that you didn't understand the proof and the properties of our system integrated into it. 1/9 = .111... isn't a conlusion you can take from the proof but a CONSEQUENCE OF OUR SYSTEM if you didn't understand that yet.
You really don't know how to analyse a proof. 1/9 is equal to .111... in our system by using the definition of our infinity in the concept.

 

[Hurkyl]

Let's go over the (usual) definition of the limit of a sequence again.

$$\lim_{n \rightarrow \infty} a_n = L$$

means:

For every positive real number e
there exists a positive integer N
such that for every integer m bigger than N,
$|L - a_m| < e$.


(When I say "means", I really mean it; I don't mean "is assumed to be", "is proven to be", or "is approximately")


So, for example, if I claim:

$$\lim_{n \rightarrow \infty} 1/n = 0$$

That means that I claim

For every positive real number e
there exists a positive integer N
such that for every integer m bigger than N,
|0 - 1/m| < e


A proof of this would go as follows:
Let e be any positive real number.
Select N to be the first integer bigger than 1/e.
Let m be any integer bigger than N.

Then, we have:
m > N
N > 1/e
m > 1/e
1/m < e
|-1/m| < e
|0 - 1/m| < e

So we have proven:
for any positive real number e,
there exists a positive integer N such that
for any integer m bigger than N,
|0 - 1/m| < e

Thus, it is true that

$$\lim_{n \rightarrow \infty} 1/n = 0$$

Because this means precisely what we have proven.

 

[ram2048]

can you please run some numbers through that for me? still can't find the n. is it in another equation or is n the same thing as m?

 

[Hurkyl]

Maybe the "game" intepretation will help you.


Suppose we play this game:
You pick a positive real number (call it e).
I pick a positive integer (call it N).
You pick a positive integer bigger than N (call it m).
If $|L - a_m| < e$ I win, otherwise you win.

If I can win all the time, then it is true that:
$$\lim_{n \rightarrow \infty} a_n = L$$
If you can win all the time, then it is false.


So, for example, let's consider the case where L = 0 and $a_n = 1/n$. Then, the "game" is:

You pick a positive real number (call it e).
I pick a positive integer (call it N).
You pick a positive integer bigger than N (call it m).
If $|0 - 1/m| < e$ I win, otherwise you win.

If I win every time we play then $\lim_{n \rightarrow \infty} 1/n = 0$ is true.

(And I will win every time. My strategy is to pick N to be the first integer bigger than 1/e)

 

[Zurtex]

translation: dance for us while we give you theoretical problems way over your head even though they have nothing to do with the issue at hand.

prove me wrong first.

Done it, many times. Go back to the first page and you shall see.

 

[matt grime]

The bit where you don't understand what the definition of the real numbers is is the bit that I meant you should perhaps take in from all this.

In the proof i offered that 0.99...=1 in the real numbers used the definition of what the rel numbers are - they are not decimal expansions, they are equivalence classes of cauchy sequences of real numbers. (or dedekind cuts, but that's less usefulo here, but they are equivalent.) When you've learned all the meaning of all those terms perhaps then you ought to reconsider your position. and seeing as you don't even follow the idea of a limit we culd be here some time.

x_n=0.9...9 n nines
y_n=1 for all n

are different cauchy sequences, but they have the same limit, and thus by the defintion of real numbers represent the sam real number. if not then you have the problem in your system that 1/n does not tend to zero (since you insist that these infinitesimals are real numbers), indeed you can't do analysis at all because practically nothing converges or has a decimal expansion, moreover not even multiplication is continuous., and subsequences of convergent sequence tend to different limits then the sequence's limit.

 

[ram2048]

and what you don't seem to understand is that the number .999~ is on the same slate as infinity and thus is subject to your same limitations.

1. there is no "Archemedian" number closer to 1 than .999~ because 9 is the digit closest to 10 and you are not allowing the addition of more 9's on the end.

2. Archemedian properties wouldn't apply to numbers subject to a limit in the first place. for example let's limit 500 to be the highest possible number. what's the closest number to 0 that is not 0. 1/500 is definitely NOT zero, but you cannot make anything closer. By archemedian then 1/500 is EQUAL to 0. which we KNOW is false.

3. .999~ is not the closest number to 1 in base 11. i don't have a calculator handy. The number closest would be .aaa~ where a is the 11th digit in that base.

4. subsequences of my set do not converge to different limits, they converge to the same one, but at a different rate/staggered interval if started at the same "time"

 

[Hurkyl]

1. there is no "Archemedian" number closer to 1 than .999~ because 9 is the digit closest to 10 and you are not allowing the addition of more 9's on the end.

(I don't know what an Archmedian number is; I'm guessing that's your new name for the real numbers because they have the Archmedian property)

No... we don't allow the addition of more 9's because EVERY position (to the right of the decimal place) already has a 9; there's no digit that we could change into a 9.

Archemedian properties wouldn't apply to numbers subject to a limit in the first place. for example let's limit 500 to be the highest possible number.

I'm not even sure it makes sense to talk about the Archmedian property applying to this system, because it fails to have many basic properties of number systems. (e.g. addition of two numbers is frequently undefined)

But, indeed, I can say that this system has no integer which is larger than 500, so one might be inclined to say that it fails to have the Archmedian property.

what's the closest number to 0 that is not 0. 1/500 is definitely NOT zero, but you cannot make anything closer. By archemedian then 1/500 is EQUAL to 0. which we KNOW is false.

No... remember that the proof that (in the real numbers, or any ordered field) there is no smallest number larger than zero doesn't have anything to do with the Archmedian property; it goes like this:

Assume there is a smallest number larger than zero. Call it x. But then 0 < (1/2) x < x, which contradicts our assumption. Thus, there cannot exist a smallest number larger than zero.

This proof fails in the system you describe because we cannot multiply 1/500 by (1/2).

3. .999~ is not the closest number to 1 in base 11. i don't have a calculator handy. The number closest would be .aaa~ where a is the 11th digit in that base.

Correct. In base 11, .999~ [11] = 9/10 [11] = 9/11 [10] = .8181~ [10]. (numbers in brackets indicate the base in which the number is written)

And, incidentally, .aaa~ [11] = 1 [11]

subsequences of my set do not converge to different limits, they converge to the same one

This would suggest that there is only one positive infinite value...

If we take two sequences, a(n) and b(n), each of which converge to some positive infinite value, then we can make a new sequence c(n) by:
c(2n) = a(n)
c(2n+1) = b(n)
If c also converges to some positive infinite value, then, because a and b are subsequences of c, a and b must have the same limit.


I guess you can avoid this proof that there is only one positive infinite value (as far as ram limits are concerned) by forbidding some (most, actually) sequences that grow without bound from converging to infinite numbers.

 

[matt grime]

seeing as subsequences converge to the same ram number, we can demonstrate another way that there is no smallest non-zero positive ram number:

1/n converges to something, call i x, so does 1/n^2, and is must also converge to x^2, thus x=x^2, and x=0 or 1, so it must be 1. thus, given any non-zero ram number there is a number of the form 1/n less than it. oops.

of course this is assuming you are claiming your numbers are an improvement on the reals. and if not then what's you're point? have you read about robinsonian analysis yet? you might learn something if you weren't closed minded about maths like you are (we aren't by the way because we qualify all of our statements properly, and fully admit their inherent limitations, which is what generalizations are for).

why do you still keep pretending we think infinity is the largest real number? we've explained it isn't, if yo'ure are going to deliberately misrepresent us you'll always get it wrong

 

[ram2048]

No... remember that the proof that (in the real numbers, or any ordered field) there is no smallest number larger than zero doesn't have anything to do with the Archmedian property; it goes like this:

Assume there is a smallest number larger than zero. Call it x. But then 0 < (1/2) x < x, which contradicts our assumption. Thus, there cannot exist a smallest number larger than zero.

This proof fails in the system you describe because we cannot multiply 1/500 by (1/2).

0 < (1/2 x 1/∞) < 1/∞

same problem. same limit. which is why i was using that as an example.

you guys use limits all the time yet you don't understand one when you see it?

BECAUSE infinity is your limit, you cannot get a number closer to 1 than .999~ but that does not mean they're equal because by imposing that limit you also cancel out the validity of using Achemedian Property to assert that equality. With NO limit on infinity it's very easy to see that Infinity+1 digits of 9 is closer to 1 and infinity+2 digits is even closer... ad infinitum 1 can never be reached and you can archemedes all you want...