Is 0.999... Truly Equal to 1 in the Realm of Infinity?

[ram2048]

suppose that .999~ this number has (infinity(d)-1) of the digit nine by your definitions, then you can agree that .4999~ has (infinity(d)-2) of the digit nine right?

k then by your reasonning .999~ - .4999~ has to have 1 digit 9 right?

but in fact it has 0 digit nine since .999~ - .4999~ = .500~

no you had it right. look here. .999~ and .0999~ exhibit a has 1 more 9 than exhibit b.

the 9 in that place is destroyed by the 4 when you do the subtraction of .999~ - .499~

that's why there's no 9 left over. there is no contradiction, visually or otherwise.

 

[hello3719]

exactly, you were proving the problem visually, look again at your proof.

 

[ram2048]

not understanding you.

please point it out for me

 

[Hurkyl]

9.999~ consider this to have "default infinity" number of digits 9

Then what you mean by 9.999~ is not what mathematicians mean by 9.999~, unless you can somehow show that having a "default infinity" number of digits means exactly the same thing as saying that there is exactly one digit corresponding to each positive integer.

999~ this number has (infinity(d)-1) which is demonstrably true when you subtract .999~ from 9.999~ you are left with exactly 1 digit 9.

Where is that 1 digit 9?


And I forget if I asked this:
How does your system of decimals tell the difference between 9 + 0.999~ and 10 * 0.999~, seeing how they're both 9.999~?

both. it is both not logical and provably untrue as shown above.

It can't be both; nonsensical statements are just that, nonsensical, and cannot be logically manipulated. In particular, you can't prove them false. This is more evidence that you are not talking about logic.

because in base 10 calculations, multiplying by 10 creates a process which shifts ALL the digits. after defining a default infinity you can see this creates a distinguishable "gap" in comparison to the standard transient infinity.

Then, again, you aren't talking about the standard decimals. (Assuming "transient infinity" means the standard mathematical way of doing things)

This is called a "strawman argument". Your "decimals" are something different than the decimals of mathematics. While your statements may then be true about what you mean by "decimal", you are merely attacking a strawman; there are no grounds to suggest any of your statements also apply to the mathematical meaning of "decimal".

the 0 is at the "end" of an infinite number of 9's.

How can there be an end if every digit has another digit to its right?

And here's another stickler. If there's an "end" then what happens to the last 9 in 0.999~ when you divide by 10?

x = .666~
10x = 6.666~
10x - x = 6.666~ - .666~
9x=6
x=2/3

according to calculus. 10x - x = 6 BUT 9x = 5.999~4

This has nothing to do with calculus; it's arithmetic. But 9x is 5.999~ = 6, and 5.999~4 is something that has no meaning in standard mathematics.

no you had it right. look here. .999~ and .0999~ exhibit a has 1 more 9 than exhibit b.

If you consider .999~ to have "one more" 9 than .0999~, then why don't you consider 9.999~ to have "one more" 9 than .999~?

What about this proof?
x = .999~
x/10 = .0999~
x - x/10 = .999~ - .0999~ = .9
(9/10) x = .9
9x = 9
x = 1

 

[ram2048]

What about this proof?
x = .999~
x/10 = .0999~
x - x/10 = .999~ - .0999~ = .9
(9/10) x = .9
9x = 9
x = 1

assume .999~ in line 1 to be default infinity number of 9's
in line 2 assume x/10 would shift all decimals right one digit and create a 0, yet maintain the same "default infinity" number of digits 9
.999~ - .0999~ =.8999~1

9/10 = .9
9/10 x .999~ = .8999~1

 

[ram2048]

How does your system of decimals tell the difference between 9 + 0.999~ and 10 * 0.999~, seeing how they're both 9.999~?

they wouldn't be the same.

in case 1, 9.000000000~ even is added to .999~ . in case 2, 10 instances of .999~ are added.

if we were to set .999~(d) as default infinity number of 9's (from now on i'll use this notation) adding 9 would not cause the shift in digits, 10 x .999(d) would. such that 9+.999(d) - (10 x .999~(d)) = .000~9

 

[hello3719]

remember this?

9.999~ consider this to have "default infinity" number of digits 9
.999~ this number has (infinity(d)-1) which is demonstrably true when you subtract .999~ from 9.999~ you are left with exactly 1 digit 9.

you SAW that there is 1 digit 9 left, this is visual and doesn't mean that "infinity(d)" - ("infinity(d)"-1) = 1

assume .999~ in line 1 to be default infinity number of 9's
in line 2 assume x/10 would shift all decimals right one digit and create a 0, yet maintain the same "default infinity" number of digits 9
.999~ - .0999~ =.8999~1

9/10 = .9
9/10 x .999~ = .8999~1

k so you just admitted that .999~ and .0999~ have the same "default infinity" number of digits 9. You are now giving your infinity the same property as our infinity.So "infinity(d)" - ("infinity(d)"-1) isn't anymore equal to 1.

 

[ram2048]

i don't think you're understanding that i can take any "infinity" and define that as my "default infinity" and work calculations from there.

default infinity isn't the same for every calculation, it MUST be designated to be used.

all it does is create a base to work from, it doesn't matter what you use as a base, the logical transformations from there stem outward and resolve to the same conclusion.

the notation system does need work i admit that, but there are no contradictions in the system, visually or otherwise.

 

[hello3719]

So in fact it is useless since you can't prove anything usefull about your system.

 

[Hurkyl]

i know nothing of continuums.

not going to even go there

Ok, I rememebred some easier to describe problems that have similar trappings:

Suppose I have a circle. Point A is inside the circle and point B is outside the circle. Can you prove the line segment joining A and B intersects the circle?

Prove that x^2 = 2 has a solution. (In other words, prove that √2 exists)

 

[Hurkyl]

Back to your "default infinity":

You suggest that ∞(d+1) = ∞(d) + 1.
Is this true in general? That ∞(a + b) = ∞(a) + b?
If so, note that we could define ω := ∞(0), then ∞(d) = ω + d.

Do any of the following make sense? If so, what are they?
∞(1/2)
2 * ∞(0)
(1/2) * ∞(0)
π * ∞(0)
∞(0)^2
∞(0)^π
√∞(0)
2^∞(0)
sin ∞(0)
∞(∞(0))

If this question makes sense, is &infin;(0)^2 < 2^&infin;(0)? How would you prove your answer?

To what default infinity does this sequence approach:
0, 2, 2, 4, 4, 6, 6, 8, 8, 10, 10, 12, ...
and what about this sequence:
1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, ...
Which of the two infinities is bigger? Or are they the same?

What do you have to say about this sequence:
0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...
and how does it compare to
1, 2, 3, 4, 5, 6, ...
or to
1/2, 1, 3/2, 2, 5/2, 3, ...?


Can I define a number x to have a 9 in every finite position, but a 0 in every infinite position? If so, how does x compare to 10x - 9? If not, why not?

Can I define x to have 9 in every position and do the same thing? That means not only does it have a 9 in every finite position, but also in every infinite position. (So, for instance, there is no d so that the &infin;(d+1)-th position doesn't have a 9 in it)
[/size]

 

[ram2048]

Ok, I rememebred some easier to describe problems that have similar trappings:

Suppose I have a circle. Point A is inside the circle and point B is outside the circle. Can you prove the line segment joining A and B intersects the circle?

i don't remember how to plot a circle, but I'm assuming this "proof" would involve me computing an exact value of π

Prove that x^2 = 2 has a solution. (In other words, prove that √2 exists)

a square with side length 1 has a corner to opposing corner diagonal or exactly √2 length using pythagorean. I'm not sure what you want me to do about it though, visually and logically they both exist, but can they be rationally measured?

i'll work on that but don't expect results anytime soon ;D

 

[Hurkyl]

a square with side length 1 has a corner to opposing corner diagonal or exactly &radic;2 length using pythagorean. I'm not sure what you want me to do about it though, visually and logically they both exist, but can they be rationally measured?

Blarg, right, square roots aren't a problem; the problem I meant is showing x^3=2 has a solution!


Actually, there is some issue related to square roots, I'll see if I can remember what it is.

 

[Hurkyl]

Given a point O, and another point R:

Any point X is said to be:
Inside the circle if X = O.
Inside the circle if the segment OX is shorter than the segment OR.
On the circle if the segment OX is the same size as the segment OR.
Outside the circle if the segment OX is longer than the segment OR.


And since you bring up pi, here's another problem which is much trickier than the above one:

Given a point P, does there exist a point Q such that the line segment PQ has length equal to the circumference of a circle of radius of length 1?

 

[Hurkyl]

Oh, I almost forgot to ask this again: how does your decimal system represent irrational numbers, like &radic;2 or &pi;?

 

[Hurkyl]

Blarg, you got me with the red herring.

You've presented no reason whatsoever for one to suppose that your number system has anything to do with ordinary Euclidean geometry. (which is part of the reason why I was asking these particular questions!)

Thus, I am still interested in seeing you prove that your decimal system contains a number whose square is 2. (Since the proof from Euclidean geometry does not suffice)


Also, I really should rephrase my question about circle continuity:

In your system, I'm defining the "ram plane" to be ordered pairs of "ram decimals"; so each point has an x and y coordinate that is one of your decimal numbers.

I define the square distance between the two points (x, y) and (u, v) to be (x - u)^2 + (y - v)^2

So then, a "ram circle" would be the defined as I did above: Given a point O and another point R in the ram plane, for any point X I say:

Let x be the square distance between O and X.
Let r be the square distance between O and R.
Then, if x < r, X is said to be inside the circle OR
Then, if x = r, X is said to be on the circle OR
Then, if x > r, X is said to be outside the circle OR

 

[ram2048]

i didn't MEAN to red herring you, as all it did was make more problems for me ;D

too much at once. just pick ONE

and my notational system for decimals only represents irrational decimals created from fractional values to create rational ones.

i could possibly devise a way to represent pi as an accurate decimal as well, but i don't intend to bite more than i can chew.

if by some miracle people accept THIS then we can work on new stuff

 

[ram2048]

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

 

[Zurtex]

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

Well ram2048 you have totally failed to answer most of the questions and any of the questions with answers that make sense because most of them do show you clearly wrong. There are many ways of calculating Pi, here is one:

\pi = \sum_{k=0}^{\infty} \frac{{-1}^k}{2k + 1}

 

[matt grime]

i know nothing of continuums.

not going to even go there.

me: <looks up definition of continuum> if a continuum is blah blah then you could blah blah and so blah blah
you: no you can't
zurtex: WRONG hahaha you are WRONG SEE? YOU SUCK HAHAHA
me: <sad>

how about no?

so, you're going to spout on at length about what 'our' mathematical system is, and isn't, yet you aren't going to learn anything about it that might help in your understanding?

 

[ram2048]

don't start, Grime

i'm taking this a step at a time, i don't need to be stretched in directions i don't have to unnecessarily.

and like I've said before, with my current knowledge i have already spotted flaws in the system and come up with solutions for them that work. i'll get to continuums when time and resources permit.

Well ram2048 you have totally failed to answer most of the questions and any of the questions with answers that make sense because most of them do show you clearly wrong.

hmm i am clearly wrong when i can do calculations that come up with the EXACT correct answer and you are RIGHT when you come up with approximations.

yes that makes ALOT of sense Zurtex.

 

[Zurtex]

hmm i am clearly wrong when i can do calculations that come up with the EXACT correct answer and you are RIGHT when you come up with approximations.

yes that makes ALOT of sense Zurtex.

You do realize "alot" isn't a real word?

As you say sums to infinity are wrong please give us how you would define \pi or \sqrt{2}. Also please define this so called default infinity. Also please could you show us if multiplying 10 by 0.\overline{9} "creates" a 0, which if you'd listened you'd know was absurd, where this 0 actually happens considering all the 9s go on forever.

These are just 3 things you can't do, your system isn't more accurate it is just plain wrong. I have tried to show you this as have many people, but you apparently haven't taken in a single point and just carry on spewing out rubbish.

 

[Zurtex]

So you see all infinities are different and there is some infinity which follows this:

\frac{1}{\infty} = 0.\overline{0}1

As all numbers of the form 0.000...0001 can be written of the form:

\frac{1}{10^n}

Where n is a natural number. Such that n is how many digits the significant number 1 is below the unit digit. So taking your previous unique infinity can you define for us:

\log_{10} \infty

So we know how far down these infinitesimal numbers are. Once again if you think about this you will realize how absurd this whole thing is.

 

[matt grime]

yes, ram, you've managed to sort out all the problems of the real number system without bothering to find out what the real numbers are, that's exactly what you've done. have you looked up any of these other things you've been told about where people have actually done these things properly? robinsonian analysis? infinitessimals, conways arithmetic of infinite cardinals?

for instance, associate a number n with a set of cardinality n, then addtion of n+m is the cardinality of the union of the sets, thus if we take infinity as aleph-0, for the sake of argument, then aleph-0+aleph-0= aleph-0=aleph-0 + 1.

of course w and w+1 are distinct in the ordinals (w is the first infinite ordinal).

but we wouldn't expect you to have checked that.

so go on, infinity is a 'real number' is it? which one?

 

[Zurtex]

If multiplying by 10 "creates" a 0 then this will surely apply so that it "creates" a 0 in 10\pi. Also multiplying your default infinity by \pi will give us a number such that it will only have 0s as digits below the unit digit "after the decimal point". So please tell us in the multiplication:

\infty \pi[/itex]<br /> <br /> What is the number in the unit digit \pi? Or another way of putting it is &quot;what is the last digit&quot; in \pi?

 

[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...