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

[Hurkyl]

The extended real numbers, as from Royce's Real Analysis is

I was wrong; division by infinity is not allowed in the definition presented by Royce.

no. by comparing the 2 numbers in such a fashion a "default infinity" is defined such that transformations are measurable.

Would you care to present a definition?

it it absolutely logical that if you shift the decimal point down, that the number now has (infinity-1) number of 9's after the decimal point.

Then present the logic.

it doesn't make any sense to say it still has exactly "infinite" number of 9's when compared to our established "default infinity"

Are you saying this statement doesn't make "sense", or that it is false?

it doesn't MATTER if there's an end or not, with all real numbers, performing such a calculation WOULD shift digits and create a 0. what makes you think that this case would be different?

Because you can't tell me where the 0 is.


And, heuristically speaking, why would you think it makes sense to "create" a 0 but not to "create" a 9?

if both numbers STILL had "infinity" number of 9's AFTER the decimal allowing you to cancel them out EXACTLY, then 9.999~ had 1 MORE digit 9 TOTAL than .999~

Where is that extra 9?

logical flaws and contradictions.

Would you care to present such a flaw and contradiction? As in a proof, and not your constant labelling of your intuition as logic?

 

[Hurkyl]

Let's see if you can do something useful with your system of infinities. Let's start with some basic geometric "facts"; mathematicians once struggled to figure out how talk about them, let alone prove anything about them:

What does it mean to say that the collection of all numbers between 0 and 1 is a "continuum"? How do you prove it is a continuum?

How does one show that the collection of numbers that are either between 0 and 1 or 2 and 3 is not a continuum?

How does one say that the function f(x) = x is continuous?

And how would you prove it?

 

[matt grime]

Ram, why do you want to treat infinity as a real number? I mean, in considering infininty minus infinity and so on? Why are you so fixed on it having a place in the real numbers, why do you think it is a number for a start? what is it that prevents you from thinking about and learning about all these other different things such as the extended numbers, the surreal,s the hyper reals, ordinals, cardinals, and so on. try it you might, well, learn something, rather than spouting the same ignorant garbage that every other idiotic crank can't deal with either

 

[hello3719]

.

by the current definition of infinity:

infinity + 1 = infinity
infinity + 2 = infinity
infitity + 3 = infinity
...

1 = 2 = 0 = 519,249 = any other number added to it.

So you think that infinity - inifinity = 0 by our definition of infinity ?
( unless you meant something else by "current definition")

and if you think that there is a problem in having infinity-infinity being indeterminate this means that you don't know how to use the infinity concept. Understand the use of limits and you'll see how it is completely LOGICAL that "inifinity - infinity" is indeterminate. It is not at all a problem of the current mathematical system.

If you still think that our system has some major flaws,and your concept of "infinity" would solve problems, then try finding, for example, the area under the curve of some function using your definitions.

 

[pallidin]

OK, ram, I will give SOME credence to your notions. But only some.
This might be of interest to you: I developed a theory some time ago that I called the "h-factor"
What it encompasses is beyond the scope of this discussion, but parts of it may be relevant here. I will attempt to illustrate those points:

Take an orange and weigh it.
Cut the orange in half.
Weigh the 2 separate pieces and sum the result.
The result will be less than the whole.
Why?
Because juice from the orange is left on the table.
The "juice on the table" is the expression of the energy lost from division transformation.

Not convinced of where this is leading to or it's importance?

Take a diamond and weigh it.
Cut it in half and weigh the separate diamonds.
The summed weight is less than the whole.
The diamond "dust" left over during division is the "missing" weight.

Still not convinced of serious mathematical implications concerning this?

Let's take this to the ultimate.
Take a theorectical rod of certain construction and nature such that when, broken in half, does not produce ANY fracture "dusting"
Surely, then, the two halfs will sum in weight to equal the weight of the original whole.
Correct!
But, what is forgotten here?
Energy is used AND DISSIPATED to enable that ideal event.

Is this getting clearer with regards to relevance in mathematics?
Look, in mathematics the division process is "pure", addressing adjunct concerns AFTER the "pure" division process.
Reality does not work this way.
In REALITY, division of a system requires the expenditure of energy upon INITIAL execution of the process.
Therefore(for simple example), 4/2=2 is INVALID because the process is held as being "pure", which is realistically impossible because energy MUST BE EXPENDED for the division process to occur even in this simple equation.

Got it?
Can ANYTHING be divided in the REAL WORLD(GR, SR, etc...) that does not require energy expenditure on INITIAL invocation of the event? No.
So, where is the energy expenditure in the equation 4/2=2 ?

Oh, some will say, that is all well and fine but does not apply to mathematics or when it does it is expressed in following equations.
Think again. It MUST be expressed in initial operand involvement because that is EXACTLY how REALITY works.

BTW, this is by no means subject only to division. All operations are subject to this.

Just some thoughts...

 

[ram2048]

it it absolutely logical that if you shift the decimal point down, that the number now has (infinity-1) number of 9's after the decimal point.

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.

it doesn't make any sense to say it still has exactly "infinite" number of 9's when compared to our established "default infinity"

Are you saying this statement doesn't make "sense", or that it is false?

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

it doesn't MATTER if there's an end or not, with all real numbers, performing such a calculation WOULD shift digits and create a 0. what makes you think that this case would be different?

Because you can't tell me where the 0 is.

And, heuristically speaking, why would you think it makes sense to "create" a 0 but not to "create" a 9?

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. in order to maintain the equality in infinities the gap is filled by 0 during the subtraction.

the 0 is at the "end" of an infinite number of 9's. it doesn't matter that "you cannot get there" because in the calculation you're pitting it against a previously "measured" infinity such that you can logically maintain that when the digits are shifted, such a gap is created.

adding a 9 is illogical because 9 is a digit that has value, such that the number you create by adding it is a different number than what you started with.

let's try this with another repeating decimal.

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

but using my system:
10x - x = 6.666~0 - .666~ = ([6-1].[15-6][15-6][15-6]~[10-6]) = 5.999~4
9x = 5.999~4

real numbers coming out perfectly. no approximation.

 

[hello3719]

real numbers coming out perfectly. no approximation.

where do you see any approximation in these calculations?
x = .666~
10x = 6.666~
10x - x = 6.666~ - .666~
9x=6
x=2/3

and if you do the long division of 2 over 3 you will get .66666~
so clearly 2/3 = .666~ and we had x = .666~ and x=2/3
I don't see any problem in our system, it seems consistent and logically complete.

 

[ram2048]

please read the post before replying

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

 

[ram2048]

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?

 

[hello3719]

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.

ok then, consider this:

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~

See the contradiction, interpreting the mathematical concepts purely by graphical representations of numbers is not a universally valid proof as you can easily fall into intuitive traps. Sorry, but in fact you didn't prove anything worthy of your system here.

 

[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 &radic;2 exists)

 

[Hurkyl]

Back to your "default infinity":

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

Do any of the following make sense? If so, what are they?
&infin;(1/2)
2 * &infin;(0)
(1/2) * &infin;(0)
&pi; * &infin;(0)
&infin;(0)^2
&infin;(0)^&pi;
&radic;&infin;(0)
2^&infin;(0)
sin &infin;(0)
&infin;(&infin;(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)

 

[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:

[tex]\pi = \sum_{k=0}^{\infty} \frac{{-1}^k}{2k + 1}[/tex]

 

[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 [itex]\pi[/itex] or [itex]\sqrt{2}[/itex]. Also please define this so called default infinity. Also please could you show us if multiplying 10 by [itex]0.\overline{9}[/itex] "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:

[tex]\frac{1}{\infty} = 0.\overline{0}1[/tex]

As all numbers of the form [itex]0.000...0001[/itex] can be written of the form:

[tex]\frac{1}{10^n}[/tex]

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:

[tex]\log_{10} \infty[/tex]

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 [itex]10\pi[/itex]. Also multiplying your default infinity by [itex]\pi[/itex] 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:

[tex]\infty \pi[/itex]

What is the number in the unit digit [itex]\pi[/itex]? Or another way of putting it is "what is the last digit" in [itex]\pi[/itex]?