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

[ram2048]

How do you figure? If I do it with 9/10s, then Zeno considers each of these intervals of position [0, 9/10], [9/10, 99/100], [99/100, 999/1000] ...
Putting all of these intervals together yields the interval [0, 1).

how is that different from [0, 1/2], [1/2, 3/4], [3/4, 7/8]...

 

[ram2048]

i'm still not understanding your wording on this

I accept that; the destination is reached after the "steps" contemplated by Zeno

if every step is half the distance (how the problem is defined) then after the "steps" still leaves a whole destination left to cover. if every step is a fraction of 1 whole then somewhere along you believe 1/2 = 1

the only thing i can think of to account for this is your belief that infinity is always equal to itself no matter how you transform it such that 1/2 x ∞ = 1 x ∞.

something like that would make a last step possible but it's not logical at all.

 

[Zurtex]

the only thing i can think of to account for this is your belief that infinity is always equal to itself no matter how you transform it such that 1/2 x ∞ = 1 x ∞.

something like that would make a last step possible but it's not logical at all.

Infinity isn't a real number and doesn't act like that one, so yes that would be correct.

Last possible step? What are you on about?

 

[Hurkyl]

how is that different from [0, 1/2], [1/2, 3/4], [3/4, 7/8]...

Put all of those together and you also get the interval [0, 1).

i'm still not understanding your wording on this

My intent was to state this: the destination is not reached during the sequence of steps considered by Zeno, but that does not imply that the destination cannot be reached at some time that occurs later than the steps considered by Zeno.


However, this reminded me of some of the things I used to point out in Zeno's paradox discussions; the following statement is also true:

At any particular point in time, if it can be said that all of the steps considered by Zeno have occurred, then it is also true that the destination has been reached.

But allow me to emphasize; the destination is not reached during the steps considered by Zeno. In particular, in order to make this statement, I do not make the assumption that one of these steps covers the entire remaining distance. Additionally, I need not use "infinity" anywhere to prove my claim.

the only thing i can think of to account for this is your belief that infinity is always equal to itself no matter how you transform it such that 1/2 x ∞ = 1 x ∞.

I "believe" it because that is how multiplication by ∞ is defined in the extended real numbers.

something like that would make a last step possible but it's not logical at all.

You mean that it conflicts with your common sense. It is entirely logical because it can be proven rigorously from the definitions and axioms.


But, it shouldn't conflict with common sense. Consider this; when things grow without bound, they approach infinity, right? E.G. the sequence 1, 2, 3, 4, ... approaches infinity, and the sequence 2, 4, 6, 8, ... does as well.

Furthermore, multiplication is a continuous operation. If a sequence approaches a limit, then if I double every number in the sequence, then the limit gets doubled.

Now, 2, 4, 6, 8, ... is the double of the sequence 1, 2, 3, 4, ..., so the limit of 2, 4, 6, 8, ... is double the limit of 1, 2, 3, 4, ..., thus suggesting that ∞ = 2 * ∞ should be correct.

 

[ram2048]

the destination is not reached during the sequence of steps considered by Zeno, but that does not imply that the destination cannot be reached at some time that occurs later than the steps considered by Zeno.

well since zeno considers to infinity, you believe that "beyond infinity" there lies a step such that 1/2 the remaining distance = the whole distance?

what I'm saying is infinity or not you believe you can take 1/2 a distance and it equals the whole distance. basically going back to what Integral said about the smallest thing being a point, what's the 'length" of a point there isn't a length so points cannot be used to fill a "gap" of any distance.

because going backwards in steps you'd also have to have 2x"the length of a point distance left to cover such that you could 'fill' it with your 1x point.

the convergence would look something like...

8 point lengths remaining (move (1/2 x 8) points)
4 point lengths covered

4 point lengths remaining (move (1/2 x 4) points)
2 point lengths covered

2 point lengths remaining (move (1/2 x 2) points)
1 point length covered

1 point length remaining (move (1/2 x 1) point. point is not divisible. move 1 point)
1 point length covered

i don't know if that makes anymore more sense at all, but that last consideration is the one i say wouldn't happen because it would be beyond the confines of the problem.

bear in mind i know a point has no "length" so don't lecture me on that. I'm using it to illustrate a point since it's the smallest conceivable thing.

so using that model, stuff has a limit of divisibility. the end step would break the rules set forth in the problem such that you'd be moving not 1/2 step but a whole step. while we calculate out with infinite steps we still reach that imposed limit caused by the point not being divisible.

but if it WERE divisible it STILL wouldn't be reached because we'd be faced with 1/2 a point 1/4 a point etc etc.

 

[ram2048]

Now, 2, 4, 6, 8, ... is the double of the sequence 1, 2, 3, 4, ..., so the limit of 2, 4, 6, 8, ... is double the limit of 1, 2, 3, 4, ..., thus suggesting that ∞ = 2 * ∞ should be correct

i'm not seeing how that proves infinity=2xinfinity. seems quite the opposite because one sequence is twice the other. how are you saying they are equal?

consider:

when you say something grows you have to apply a rate. if the rate of growth is equal among 2 sequences such that sequence 1 growing double that of sequnce 2. then it's easy to prove that within any given frame of reference in time, sequence one is larger than sequence 2. even at infinity.

1,2,3,4... ...100,101,102... ...1001,1002,1003...
2,4,6,8... ...200,202,204... ...2002,2004,2006...

such that any line cutting perpendicularly into these 2 parallel lines will always yield a number and a number 2xgreater than it.

although if you were to grow things without a rate of time, these number sequences would "exist" as any number within their range at any given time, it would be impossible to accurately assess any relation in value when comparing them. 1 could = 3 or 300,000.

 

[Hurkyl]

you believe that "beyond infinity" there lies a step such that 1/2 the remaining distance = the whole distance?

I believe there is nothing between Zeno's steps and the destination.


Consider that the sequence 2, 4, 6, 8, ... is a subsequence of 1, 2, 3, 4, ...,

  2   4   6   8 ...
1 2 3 4 5 6 7 8 ...

Since the former can be formed by removing terms from the latter, how can the former have a bigger limit?

within any given frame of reference in time

although if you were to grow things without a rate of time,

What does time have to do with anything?

even at infinity.

The sequences aren't even defined at infinity. (unless your "infinity" is a positive integer, in which case it is finite, thus making your use of the term highly misleading)

these number sequences would "exist" as any number within their range at any given time

At any given time, a number sequence is a number sequence. They do not "exist" as anything else.

 

[hello3719]

Ram2048, infinity is a PROCESS and not a number, it only means that you are constantly growing in relation to the certain set of numbers considered.

so you can see that infinity*2 is the same as infinite since in both cases the growing process is present which is what infinity means. growing twice as fast or growing is the same as growing.

 

[ram2048]

I believe there is nothing between Zeno's steps and the destination.

so the sum of that sequence will never equal its limit...

if this is NOT the case you need to explain to me how we're processing along towards the destination with computation and all of a sudden there's no distance left and we're there. There has to be a definable process that gets us there.

Consider that the sequence 2, 4, 6, 8, ... is a subsequence of 1, 2, 3, 4, ...,

  2   4   6   8 ...
1 2 3 4 5 6 7 8 ...

Since the former can be formed by removing terms from the latter, how can the former have a bigger limit?

What does time have to do with anything?

that's why i said growth has to be a function of time or reference in some way "rate" such that the steps can be measured. otherwise if sequence 1 "grew" at a rate 50,000 times slower than sequence 2, it wouldn't matter that they had the same limit or that sequence 1 was "twice the value" of sequence 2, [edit] at the beginning seq 1 would be greater than seq 2 but it would then be overtaken due to superior rate.

The sequences aren't even defined at infinity. (unless your "infinity" is a positive integer, in which case it is finite, thus making your use of the term highly misleading)

that statement was a conclusion stemming from the process. most processes that tend towards infinity have a very predictable output and logical conclusion. with the same growth "rate" and the same start point, the double sequence will hit infinity while the single will only be at 1/2 infinity.

this is the logical conclusion, nevermind that neither will EVER reach infinity,

At any given time, a number sequence is a number sequence. They do not "exist" as anything else.

not so, if no rate is applied to growth it means that it simultaneously exists as any allowable value within its field (start to limit)

----------------

and as a side note i want to know if this proof works by your math.

x=.999~
10x = 9.999~
10x - x = 9.999~ - .999~
9x = 9

x = 1
.999~ = 1

curious to see if that's allowed or if some crackpot made that up without running it by you guys.

 

[Hurkyl]

Ram2048, infinity is a PROCESS and not a number

That's at least half incorrect. In number systems with a single nonfinite number (or some otherwise especially identifiable nonfinite number), that nonfinite number is often labelled "infinity". Its other uses (such as in \sum_{i=0}^{\infty}) are as a formal symbol to denote something in particular (IMHO it would be misleading to call things so denoted "processes").

so the sum of that sequence will never equal its limit...

The sum of an infinite sequence is, by definition, the limit of the partial sums, so...

if this is NOT the case you need to explain to me how we're processing along towards the destination with computation and all of a sudden there's no distance left and we're there. There has to be a definable process that gets us there.

Zeno's steps cover every point between the start and destination (but not including the destination). If Achilles has completed all of Zeno's steps, where could he be if not at (or past) the destination?

A more analytical answer might go as follows: if p(t) is the position of Achilles at time t, then, according to Zeno's steps, p(0) = 0, p(1/2) = 1/2, p(3/4) = 3/4, p(7/8) = 7/8, et cetera.

Now, lim_{n \rightarrow \infty} (1 - 1/2^n) = 1, and since motion is continuous, we conclude

\begin{align*}<br /> p(1) &amp;= p(lim_{n \rightarrow \infty} (1 - 1/2^n)) \\<br /> &amp;= lim_{n \rightarrow \infty} p(1 - 1/2^n) \\<br /> &amp;= lim_{n \rightarrow \infty} (1 - 1/2^n) \\<br /> &amp;= 1\end{align}

that's why i said growth has to be a function of time or reference in some way "rate" such that the steps can be measured.

That seems necessary to maintain your stance, but I can't see any reason that this is necessary in general.

not so, if no rate is applied to growth it means that it simultaneously exists as any allowable value within its field (start to limit)

Rephrase yourself. It is absurd to say, for instance, 1 = <1, 2, 3, 4, ...>, but I am interpreting you as suggesting that this equality can be correct.

and as a side note i want to know if this proof works by your math.

x=.999~
10x = 9.999~
10x - x = 9.999~ - .999~
9x = 9

x = 1
.999~ = 1

Yes, this is a valid proof.

It is not a valid "first" proof, though, in the sense that this proof depends on the fact that the decimal numbers are a model of the real numbers, so it cannot be used to prove, in the first place, that the decimal numbers are a model of the real numbers.

 

[ram2048]

It is not a valid "first" proof, though, in the sense that this proof depends on the fact that the decimal numbers are a model of the real numbers, so it cannot be used to prove, in the first place, that the decimal numbers are a model of the real numbers.

darn cause i had a good disproof and proof on infinity+1≠infinity from that one :(

oh well

Zeno's steps cover every point between the start and destination (but not including the destination). If Achilles has completed all of Zeno's steps, where could he be if not at (or past) the destination?

he could still be in transit.

\begin{align*}p(1) &amp;= p(lim_{n \rightarrow \infty} (1 - 1/2^n)) \\&amp;= lim_{n \rightarrow \infty} p(1 - 1/2^n) \\&amp;= lim_{n \rightarrow \infty} (1 - 1/2^n) \\&amp;= 1\end{align}

i've always concluded that 1/∞ has a positive value. I've been trying real hard to understand you guys' "something turns into nothing" but it's not getting through :(

Rephrase yourself. It is absurd to say, for instance, 1 = <1, 2, 3, 4, ...>, but I am interpreting you as suggesting that this equality can be correct.

no no, that's not what i meant by it. i meant since it is a defined series of growth, without anything to denote the transition in steps such a rate or time value, it would be anyone of those numbers at the exact instance of transformation.

it's part Schrodinger in a way i guess. but it's kinda veering on a tangent away from the issue at hand hehe.

in any case I'm pretty sure if i can convince myself that 1/∞=0 everything will be clear to me :O

 

[Hurkyl]

he could still be in transit.

If he's still in transit, he has to be someplace between the start and finish.
If he's someplace between the start and finish, then he has not done all of Zeno's steps.

Thus, if he has done all of Zeno's steps, then he's not in transit.



The key thing about the real numbers, that you are not using, is that they are complete. Intuitively, that means they have no "holes"; this is usually stated as follows:

If A and B are nonempty collections of real numbers, such that A lies entirely on the left of B, then there is a number c seperating them.
More precisely,
Let A and B are nonempty sets of real numbers
For all a in A and b in B: a < b
Then, there exists c such that
For all a in A and b in B: a <= c <= b


This can be used to prove the Archmedian property; every number is smaller than some integer (and is thus finite) as follows:

Let A be the set of everything smaller than some integer. (Note that all integers are in A, because n < n+1)
Let B be the set of everything bigger than all integers.
Assume B is nonempty.
Then, there exists some c seperating them.

Thus, a <= c <= b for all a in A and b in B.

Now, c is either bigger than all integers (and thus in B), or it's not (and thus in A).

If c is in A, then it is smaller than some integer. Call it n.
Because c < n, c + 1 < n + 1. Thus, c + 1 is in A.
This is a contradiction because c + 1 > c, but because c + 1 is in A, c+1 <= c. Thus c cannot be in A.

Thus, c is in B.
c - 1 < c, so c - 1 must be in A. c - 1 must be smaller than some integer. Call it n. Because c - 1 < n, we have c < n + 1. However, n + 1 is an integer and is in A, so n + 1 <= c, which is a contradiction.

Thus c cannot be in B.

So our assumption that B is nonempty led to a contradiction: there's a hole between the finite and infinite numbers, but by definition the real numbers have no holes.

Thus, we conclude that there are no infinite numbers; every number is smaller than some integer, and thus the Archmedian property is proven.


Now, we turn back to your question. Suppose 1 / &infin; = e > 0. (whatever 1 / &infin; may happen to mean).

By the Archmedian property, 1 / e is smaller than some integer. Call it n.
Because 1/e < n, e > 1/n.

So, if you insist on maintaining that 1 / &infin; > 0, then there exists an integer n such that 1/n is smaller than 1/&infin;, which is absurd!


Roughly the same argument is used to prove that \lim_{n \rightarrow \infty} 1/2^n = 0.


(More directly, if you maintain that &infin; is a real number, then I could simply apply the Archmedian property to produce an integer bigger than &infin;)

 

[matt grime]

ram, have you considered stepping back, taking a deep breath and admitting to yourself that you don't understand the mathematics involved? that, in the extended numbers, infinity*2=infinity follows from the definitions of that system, and 1/infinity is zero. in the surreals, 1/w is not zero, but w is still not a real number.

As for zeno and trying to understand how you can 'go beyond infinity' as is being talked about, consider trying to learn about transfinite systems, and limit ordinals, you also need to stop thinking in terms of moving from one spot to the next, but in terms of having done all the previous moves (a vague introduction to transfinite induction).

If you tried to understand the limit concept properly instead of presuming you know what it is then none if this nonsense would happen.

 

[ram2048]

So, if you insist on maintaining that 1 / ∞ > 0, then there exists an integer n such that 1/n is smaller than 1/∞, which is absurd!

that's because of the "self defined" upwards limit on infinity itself. BECAUSE you believe it to be all inclusive such that any value added to infinity has no meaning, you're presented with the upwards or downwards limits as such. but this still does NOT prove that the number would equal zero. consider 2 apples next to each other touching. there is no space in between them but they are not the same apple. BECAUSE infinity has a limit such that nothing can be greater than it, no SPACE can exist between 1/∞ and 0, but that doesn't prove they're the same number. infinity is not a number so it doesn't function as one in relation to this "archemidien rule". let's apply infinities "limit" to a real number and see how it works. 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.

ram, have you considered stepping back, taking a deep breath and admitting to yourself that you don't understand the mathematics involved? that, in the extended numbers, infinity*2=infinity follows from the definitions of that system, and 1/infinity is zero. in the surreals, 1/w is not zero, but w is still not a real number.

i doubt myself all the time, but until i can find the answers that convey meaning to me in such a way that I'm completely convinced and can properly relay this information to another person that might be having the same problem, i cannot quit.

As for zeno and trying to understand how you can 'go beyond infinity' as is being talked about, consider trying to learn about transfinite systems, and limit ordinals, you also need to stop thinking in terms of moving from one spot to the next, but in terms of having done all the previous moves (a vague introduction to transfinite induction).

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...

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.

 

[russ_watters]

i can goto 5 sites and pull a different definition of real numbers, irrational numbers, and infinity.

and you talk to ME about inconsistency. I can't use your words because your definitions are "Mumbo jumbo"|

Quite frankly, the discussion should have ended here. If you can't even accept that there are specific definitions for these concepts, you can't ever hope to understand the definitions or how to apply them.

If you really do want to understand what we're trying to tell you, this is where you must start.

 

[ram2048]

Quite frankly, the discussion should have ended here. If you can't even accept that there are specific definitions for these concepts, you can't ever hope to understand the definitions or how to apply them.

If you really do want to understand what we're trying to tell you, this is where you must start.

that's BS, russ. i wouldn't have known there were 5 or more definitions for commonly used math terms if i hadn't taken the time to research it and try to find which one was the correct one. i even go through the trouble to annotate myself with "quotes" when I'm unsure as to the correct term. this is possibly the best place to find proper definintions for these terms as if someone brings up the wrong one he is immediately and punitively apprehended... like within the hour. you guys are harsh ;D

and russ, if you have nothing positive to contribute, just don't. really.

 

[matt grime]

these 'different' definitions of real numbers are all equivalent, that is they do the same thing, but that seems beyond your capability to understand.

teach your self transfinite induction, it isn't very hard, any idiot can understand it.

it has been explained to you patiently and accurately what the use of infinity is in mathematics, now it is time for you to go away and think about all that has been written and say what you don't understand, not what you think is worong because your mathematical spohistication is not such that you are at a point to say something is wrong, merely that you don't understand it, and that is a problem with you and not with mathematics.

we are not harsh, you are ignorant. that is not an insult, just a statement of fact.

 

[ram2048]

it has been explained to you patiently and accurately what the use of infinity is in mathematics, now it is time for you to go away and think about all that has been written and say what you don't understand, not what you think is worong because your mathematical spohistication is not such that you are at a point to say something is wrong, merely that you don't understand it, and that is a problem with you and not with mathematics.

so you can't explain it without your fancy math. which leads me to the conclusion that you only believe it because it is popular and fashionable. If you don't understand the math well to be able to rationalize if logically, then all you are doing is parroting back text.

we are not harsh, you are ignorant. that is not an insult, just a statement of fact.

i am not knowledgeable in the ways of "transfusion inducers" or whatever the heck you're talking about, that hasn't stopped me from disproving 4 "accepted" proofs. quite silly how it can all be taken down by an "ignorant" person.

 

[Zurtex]

i am not knowledgeable in the ways of "transfusion inducers" or whatever the heck you're talking about, that hasn't stopped me from disproving 4 "accepted" proofs. quite silly how it can all be taken down by an "ignorant" person.

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.

 

[chroot]

When an intelligent man argues at length with a fool, it becomes difficult to tell them apart.

- Warren

 

[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 arguements 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/&infin; and 0

Where is (0 + 1/&infin;) / 2? If 0 and 1/&infin; 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/&infin; 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/&infin;?

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.