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

[ram2048]

That is your problem, you automatically assume you are right, you are unwilling to learn mathematics, you have not actually disproved anything and that makes you ignorant

i have disproved them, the ball is in your court. what purpose would it serve for me to disprove myself? unless you can disprove my disproofs then my way is correct. that's how this works.

when person A comes and tells me something and i prove him wrong, it doesn't lie on MY shoulders to disprove myself.

and I'm always willing to learn mathematics as long as they have some logical basis in reality. i will not blindly accept "Something equals nothing" unless convinced thoroughly that it is logically true.

actually, until you can come up with rational arguments as to why one of my proofs does NOT work, you have no right to call me ignorant. Come up with something on your own instead of monitoring the thread waiting to say "yea Hurkyl is right, you're stupid ram..." whenever he posts something.

 

[chroot]

ram2048:

You didn't disprove anythnig. You might feel you did, but that's because you don't understand what a proof even is.

- Warren

 

[Hurkyl]

no SPACE can exist between 1/∞ and 0

Where is (0 + 1/∞) / 2? If 0 and 1/∞ are different, then this quantity must lie between them. If there is nothing between them, then they must be equal.

infinity is not a number

Then what can 1/∞ possibly mean? Is that also not a number? If that's not a number, then how can the remaining distance (which is a number) be equal to 1/∞?

let's make 500 be the highest possible number there can be. let's now say 1/500. is 1/500 = 0? it is most certainly NOT, yet this is the closest POSSIBLE number to 0 you can get because of the limit imposed.

Ok. And?

well the point of zeno's exercise was that you could NEVER complete all the moves, that's why it was proposed as a paradox...

The problem is that Zeno never suggests any reason why you cannot.

you want to explain to me then since you know how to do it? none of that fancy math stuff, just logically spell it out for me how it could EVER be possible to cover a whole distance if you can only move in halves of the remaining distance.

You couldn't. It's quite fortunate that we're not restricted to moving in halves of remaining distance.

this is possibly the best place to find proper definintions for these terms

Then consider these definitions.

The usual definition of the real numbers used these days is along the lines of this one, slightly paraphrased from Buck's Advanced Calculus:

"The real numbers (R) constitute a complete simply ordered field"

In terms of axioms, this means:
R is a set of elements.
P is a subset of R (whose elements are called positive)
+ and * are two operations on elements of R.
0 and 1 denote particular elements in R

For any a, b, and c in R:
a + b is in R
a * b is in R
a + b = b + a
a * b = b * a
a + (b + c) = (a + b) + c
a * (b * c) = (a * b) * c
a * (b + c) = (a * b) + (a * c)
a + 0 = a
a * 1 = a
a + x = 0 can be solved for x
a * x = 1 can be solved for x if a is not zero
if a and b are positive, then so are a + b and a * b
either a is in P, -a is in P, or a = 0.

If A and B are nonempty subsets of R, and a <= b for any a in A and b in B, then there exists a number c such that a <= c <= b for any a in A and b in B.


The extended real numbers, as from Royce's Real Analysis is (I don't have my text handy, so I'm doing this from memory, and am probably saying the same thing in many more words):

The extended real numbers consist of the real numbers plus two additional elements, &infin; and -&infin;.
+&infin; is positive, and -&infin; is not.
For any extended real numbers a and b that are not equal to +&infin; or -&infin;:
a + b and a * b in the extended real numbers is the same as in the real numbers.
a + +&infin; = +&infin;
a + -&infin; = -&infin;
a - +&infin; = -&infin;
a - -&infin; = +&infin;
If a is positive, then a * +&infin; = +&infin; and a * -&infin; = -&infin;
If a is negative, then a *+&infin; = -&infin; and a * -&infin; = +&infin;
a / +&infin; = 0
a / -&infin; = 0
+&infin; + +&infin; = +&infin;
-&infin; + -&infin; = -&infin;
+&infin; - -&infin; = +&infin;
-&infin; - +&infin; = -&infin;


In particular, 1 / +&infin; = 0 simply because that's what is defined to equal, thus it is certainly not logical that 1 / +&infin; be inequal to zero.

 

[ram2048]

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1

Assume .999~ has infinity number of digits 9.
when we multiply this by 10, 1 of 2 things happens:

1. a 0 is added to the end and all digits of 9 are shifted up one digit

OR

2. all digits of 9 are shifted up one digit and a digit of 9 is "created" to maintain that "infinite" number of 9's

out of these two possibilities, your proof favors the second notion, that all digits are shifted and an extra 9 is "created" such that the .999~ in the 9.999~ can be perfectly canceled out to = 9 exactly.

now note that the expression .999~ having "infinity" number of digits 9. therefore 9.999~ MUST have "infinity+1" number of digits since this is proven when the subtraction takes place and a whole number digit 9 is left over.

9.999~ has "infinity+1" digits of 9
.999~ has "infinity" digits of 9

the creation of the extra digit of 9 causes inequalities in the infinities used in both numbers.

in short 9.999~ would have the same number of digits of 9 as .999~ and subtracting it would NOT yield the integer 9. using rational and logical step 1 instead it would output something like this: 8.999~1

going backwards we see this is absolutely true. take 9x which is 9 x .999~ and multiply the digits out 8.1 +.81 + .081 ... such that the sum EQUALS 8.999~1 [edit] woops added one too many steps there

QED

 

[matt grime]

0.999... certainly doesn't have a finite number of nines so it must have an infinite number of them, and?

multiply by ten and we shift one place to the right and add one 'at the end' to maintain this infinite number? what utter utter nonsense.

we have shown you where you are wrong yet you refuse to recognize this fact, but then this is all opinion isn't it and our opinion is no more valid than yours, or so you seem to think... again, wrong, these are based on definitions of what these symbols mean mathematically, something we understand and that you do not.

and round it goes in ever decreasing circles.

 

[Hurkyl]

1. a 0 is added to the end and all digits of 9 are shifted up one digit

There is no end.

2. all digits of 9 are shifted up one digit and a digit of 9 is "created" to maintain that "infinite" number of 9's

Where is the "created" digit placed?


How can you maintain your position is logical when it requires there to be a rightmost digit... but every digit has another digit to its right?

 

[ram2048]

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

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.

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

There is no end.

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?

multiply by ten and we shift one place to the right and add one 'at the end' to maintain this infinite number?

that's what I'm talking about. this is visual proof that transformations using default infinity can yield perfect calculations. it's completely based off what you guys believe. 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~ meaning it is NOT the same number as (10x.999~)

the proof is in the pudding when you go back and multiply the numbers out and acquire the exact same thing as you would subtracting them longhand.

9 x .999~ = (8.1 + .81 + .081...) = 8.999~1
10x - x = (9.999~0 - .999~) = ([9-1].[18-9][18-9][18-9]~[10-9]) = 8.999~1

 

[Zurtex]

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?

Shift the digits? Create a 0? Once again you show your astounding ability not to understand even the most basic mathematics.

 

[hello3719]

Ram2048, can you give us an exact definition of your "default infinity" and the definition of "infinity' ?

 

[ram2048]

Shift the digits? Create a 0? Once again you show your astounding ability not to understand even the most basic mathematics.

good work, Zurtex.

you caught me. apparently i have NO IDEA how this is resolved

(10 x 8.16) - 3.57 =
81.6 (shift digits) - 3.57 =

  81.60 (create 0)
-  [u]3.57[/u]
  78.03

yes... NO IDEA WHATSOEVER...

 

[ram2048]

Ram2048, can you give us an exact definition of your "default infinity" and the definition of "infinity' ?

well i would probably define infinity as a transient variable such that the variable will always be a value greater than any number it is compared against.

default infinity would be a mathematical concept or technique in which the user establishes a base infinity for computations and any transforms based upon it are weighed mathematically to discern logical outcomes as to the nature of numbers defined using infinity.

instead of just saying infinity + 1 = infinity, which is slightly true because adding 1 to infinity means it's still infinite.

we could say ∞(d) + 1= ∞(e) and then extract meaningful values like ∞(e) > ∞(d) and ∞(e) - 1 = ∞(d)

 

[ram2048]

also when using default infinity you could say things like

∞(d) - ∞(d) = 0

∞(d) / ∞(d) = 1

 

[hello3719]

So how is it different from real numbers ?

 

[hello3719]

since you've defined default infinity as being bigger than every other value compared to it then "default infinity" is bigger than "default infinity" +1 ?

then the contradiction arises here, (d) infinity + 1 must then be equal to inifinity since both are being compared and each of them must be bigger than the other.

 

[ram2048]

it's not much different from real numbers at all which is what makes it useful.

standard "calculus" definition of infinity creates a lot of "undefined" processes that lead to computational "dead-ends". like infinity minus infinity is unresolvable because it's not defined which infinity is the same, the natural inclination is to believe that they ARE the same and the result would be 0. default infinity provides a way to establish the relations of infinities we're using within a given mathematical process.

Infinity(d) - Infinity(d) = 0
Infinity(d) - Infinity(e) = unresolvable

define infinity(e) = Infinity(d) - 5

Infinity(d) - infinity(e) = 5

math
infinity(d) - (infinity(d) - 5) =
infinity(d) -1(infinity(d) + -1(-5) =
infinity(d) - infinity(d) + 5 =
5

something like that...

 

[ram2048]

since you've defined default infinity as being bigger than every other value compared to it then "default infinity" is bigger than "default infinity" +1 ?

no i defined it as a variable "value" not a variable "number". it is bigger than all NUMBERS compared against it, not values.

but it allows trasformations using numbers to act on it such that numbers will cause its value to increase or decrease.

such that:

infinity(d) + 1 = infinity(e) (defining infinty(e)
infinity(e) > infinity(d) (logical comparison)

BUT

infinity(e) ≯ infinity(d) + 2 (adding the value of the number 2 to our default infinity)

 

[hello3719]

Then there is no advantages of you're definition of inifinity over the system of real numbers by seeing the properties you are giving it.

well i would probably define infinity as a transient variable such that the variable will always be a value greater than any number it is compared against.

Example, If I take the number 1 , then by your definition of "inifinity" it must be 1 value greater than 1 so,
"inifinity" = 1+1 = 2

If I take the number 2, then inifnity = 2 +1 = 3

etc...

you can see we are constructing real numbers, but then you can see how our definition of inifinity helps in saying how much of your "inifinities" there is.

 

[ram2048]

not so.

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.

the logical conclusion is that you CANNOT change the value of infinity in the current system so infinty - infinity MUST ALWAYS BE 0

and

infinity / infinity = 1

but according to current definitions infinity - infinity = unresolvable, infinity/infinity = unresolvable.

that's by the current system. that's the system with the logical flaws and contradictions.

i'm just creating a system that defines infinity to clear things up

 

[Zurtex]

not so.

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.

the logical conclusion is that you CANNOT change the value of infinity in the current system so infinty - infinity MUST ALWAYS BE 0

and

infinity / infinity = 1

but according to current definitions infinity - infinity = unresolvable, infinity/infinity = unresolvable.

that's by the current system. that's the system with the logical flaws and contradictions.

i'm just creating a system that defines infinity to clear things up

Wrong, if infinity can be greater than itself in either finite terms or infinite terms then infinity - infinity does not have to equal 0. Also infinity / infinity does not have to equal 1.

It is NOT A REAL NUMBER AND DOES NOT ACT LIKE ONE. Get used to it or don't bother doing maths.


Edit: If you want to create a system for classifying all infinites and works with real numbers go and do one and come back when you have it.

 

[Zurtex]

good work, Zurtex.

you caught me. apparently i have NO IDEA how this is resolved

(10 x 8.16) - 3.57 =
81.6 (shift digits) - 3.57 =

  81.60 (create 0)
-  [u]3.57[/u]
  78.03

yes... NO IDEA WHATSOEVER...

ROFL, it doesn't create a 0 there were an infinite number of 0s already there. Whoever taught you maths was either really bad or you weren't listening.

Consider this:

10 \cdot (3) = 30

Another way of writing it is:

10 \cdot (...0000003.0000000 \overline{0}...) = ...00000030.000000 \overline{0}...

Do you see anymore than an infinite number of 0s? I thought it was that your terminology was bad with using words like "shift" and "create" but now I see you actually think them.

 

[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.