1=.999999...?

[hoplitejoe]

But as you note yourself, 0.999.. can't reach an end. Which is just another way of saying that it can never be one.

It can never reach one if we restrict ourselves only to your non-mathematical world.

With maths we are able to use techniques (most which have already explained in far more detail than I am capable of) to find what 0.999... is equal to in this theoretical limit.


I'm still kinda confused where your objection is, so perhaps you can clarify for me with the question I asked brennan,

Do you accept 0.999... (an infinitely repeating decimal) to be a meaningful concept,

A)within mathematics
B)within whatever mathematics free world you existing in?

 

[Akka]

I am not disputing that. That is rather my point. If 0.999... never ends - it will also never reach one, even if it goes on forever, which is all 0.999.. means and I just illustrated how that works.

No you didn't. See :

Hence it cannot be one. Otherwise the ancestors would reach it, since what they are carrying out for eternity is exactly what 0.999... represent: an infinite process.

You continue to not grasp what "infinity" when you say "the ancertors would reach it". They wouldn't. That's the whole concept of "infinity", it has NO END. "reaching" means it's ending. Infinite means it's neverending. Any time you try to argue that there is a final result, you're failing at treating the number of 9 as infinite.

It can not be a simple number exactly because it is infinite, but allegedly, the opposite was true. Which makes no sense what-so-ever.

It makes sense, you're just smacking your head against the concept of infinity.
You're right that the 0,999... can be seen as the INFINITE process of adding 9 at the end of the number. Where you're wrong is that you try equate it with a necessarily finite physical representation of it.

The human mind has an impossible time with grasping infinity. When we say "forever" or "eternity", we only think as humongous amount, but I don't think we ever grasp what "neverending" truly, really means. An infinite process, having no end, is effectively equal to its limit (if it were in any way not strictly equal, it would mean the process stops at one finite point, which is not the case as it's infinite).

 

[Lohrenswald]

This is the third (fourth?) time I'm asking.

I would like to field a second question.

Does "0.999...8" also equal 1? Why or why not?

There is no final 9 in 0,999...

 

[onejayhawk]

Not the limit of the infinite sum, but the limit of the finite sums. There's no difference between the limit of the finite partial sums and "the infinite sum". The latter is defined to be the former.

Why it's so? Because originally you have just a sum of two objects. You can extend it to more recursively, for example you can define a+b+c = (a+b) + c. That is, first sum a and b and then sum the result with c. This way you can define sum for all finite number of summands, but not for infinite. That's why the infinite sum requires a different definition. Notice also that it's different from a finite sum in many other ways too: finite sums are always defined, infinite not necessarily. Finite sums can be summed in any order, infinite sums not.

You can even go further and define sum for an uncountable amount of numbers, but need a different definition for that too.

EDIT: Forgot the second question. I don't understand what you're exactly getting at. It's easy to show that there's no biggest number in ]0,1[ for example: By definition ]0,1[ = { real x: 0<x<1}. If there would be the biggest number in ]0,1[, call it a, these would be true:
a < 1
For every x in ]0,1[ x < a.
Now, choose b = (a+1)/2, i.e. the average of a and 1. It's easy to see that a<b<1, so b is in ]0,1[ and on the other hand b>a. That is a contradiction with a being the biggest element of ]0,1[.

Besides, if an open interval included it's endpoints, how would it differ from the closed interval? In no way. It would be exactly the same thing. The difference between them is however huge, as you can see already in the elementary maths, for example a continuous function is bounded and reaches it's maximum and minimum on a closed interval, but not necessarily on an open one.

The proof is the same method as has been used many times. Once again, I think it applies to a discrete set inbedded in the continuous set.

Are you familiar with the so called popcorn function? If x is rational, written a/b, then f(x) = 1/b. Otherwise f(x) = 0. It is discontinuous at all rational numbers, but continuous everywhere else. Those proofs will not work in the continuous parts.

J

 

[Borachio]

Can it be discontinuous at a rational number though? You've defined it to actually exist at that point (assuming b is an integer, though you've not said so). So it's not discontinuous there, imo.

 

[onejayhawk]

Yes, b is an integer. It must be discontinuous. Since f(x) = 0 for all non rational numbers, any &#949; < b suffices. For any other rational number, x', written c/d, with c,d being integers. x - x' = a/b - c/d = (ad-cb)/bd. Choose &#949; < abs[(ad-cb)/bd].

This method should be very familiar by now. it has been used constantly in this thread. For f(x) = 0, proving continuity is trivial. Any test works.

J

 

[Borachio]

Yeah. I know the method well enough.

It's just that I don't see the function being discontinuous at a point where it's defined to exist. Which is what you seemed to be saying.

Still, I expect you're right. It's a long time since I looked at this sort of thing.

 

[onejayhawk]

Yeah. I know the method well enough.

It's just that I don't see the function being discontinuous at a point where it's defined to exist. Which is what you seemed to be saying.

Why? It is discontinuous, not undefined.

This is a piece constructed function. Another example: f(x) = 1 for x =/= 0, f(0) = 0. It is defined everywhere, but discontinuous at 0.

J

 

[Borachio]

Oh. Then I've misunderstood what undefined and discontinuous mean.

 

[Atticus]

The proof is the same method as has been used many times. Once again, I think it applies to a discrete set inbedded in the continuous set.

Are you familiar with the so called popcorn function? If x is rational, written a/b, then f(x) = 1/b. Otherwise f(x) = 0. It is discontinuous at all rational numbers, but continuous everywhere else. Those proofs will not work in the continuous parts.

First paragraph: I don't see why you would think so. I made no assumptions on any numbers being rational (except the end point 1, but that's obviously irrelevant).

Second paragraph: I don't see how the popcorn function is related to the question at hand. I know there are many different kinds of exotic functions with weird behaviour.

Also, irrationals don't form "a continuous part" of the reals. R\Q is as "non-continuous" as Q is. There's always a rational number between two separate irrational numbers.

 

[onejayhawk]

Also, irrationals don't form "a continuous part" of the reals. R\Q is as "non-continuous" as Q is. There's always a rational number between two separate irrational numbers.

Q is fully discrete. Q', R\Q if you prefer, is not. To say it is as "'non-continuous' as Q" is simply wrong. The popcorn function above is a disproof.

Oh. Then I've misunderstood what undefined and discontinuous mean.

I understand. f(x) = 1/x is both discontinuous and undefined at x = 0. I can see how someone might conflate the two.

J

 

[TheMeInTeam]

If 0.999... = 1

Is 0.998... = 0.999...?

The problem here seems to be that 0.999... is being confused for a number when it is a procedure.

If x=0.999...

And 10x=9.999...

Then it pretty clear that 0.999... is not a number, if it was then 10x would be 9.990

The quoted misconception originates from what I believe is a difficulty in picturing "infinite". You can hold one apple, and any fraction of that apple is going to add up to the whole, no matter how small you slice it (even if you slice it into subatomic particles). But, even those particles are finite...the 9's keep going right past that. They go forever.

Try to picture "infinitely many apples", or anything else in infinite quantities. That's not something you can visualize. People get lost in the scope of large finite numbers! Infinity isn't something the mind handles well.

.998.... =/= .999 by the way. .998999... would equal .999 (with the 9s stopping, assuming that I'm indicating 9's forever properly). We don't have a neat single-digit representation of .998... (just as we don't for say .555...), but we do have one for .999....

More to the general concept of the thread:

When you mix scope insensitivity and arguments over definitions you get some silly stuff, like people defining something as wrong and using that definition to conclude that it is therefore wrong by definition. The evidence we do have points to .999...=1, and I haven't seen evidence that should lead us to believe against that assertion.

Ignoring evidence "because logic" is the opposite of logic.

 

[onejayhawk]

Try to picture "infinitely many apples", or anything else in infinite quantities. That's not something you can visualize. People get lost in the scope of large finite numbers! Infinity isn't something the mind handles well.

Uncountable infinities are even worse. Even serious mathematicians have problems.

J

 

[Atticus]

Q is fully discrete. Q', R\Q if you prefer, is not. To say it is as "'non-continuous' as Q" is simply wrong. The popcorn function above is a disproof.

In what sense Q is fully discrete? Every neighbourhood of a rational contains infinitely many rationals.

The idea of Q as discrete and R\Q as "the continuous part" comes from R\Q making "the bulk" of the reals. You'd think that Q is just separate points every here and there, and in between is "the continuous part". That's a misunderstanding, at least in the sense that there are no connected segments in the R\Q.

Also, you seem to think that the popcorn function is constant in some neighbourhood of an irrational point. Otherwise you wouldn't hold this as an evidence. That's also a misconception that rises easily, it's the exact reason that makes the function amazing.

The popcorn function is similar to the function that is x on irrational x:s and 0 elsewhere, and which is continuous only in 0: The continuity is a result of the "dampening" of the non-constant parts of the function near the point examined.

 

[uppi]

An infinite process is just that. A process, no goal line. That does not at all mean that it does not exist. That would be like saying that space can not be infinite because we could never reach the end... The opposite is true, of course. Never reaching the end means space's infinity is real (supposing one continuously try's to reach the end).
So my illustration is fully fine with 0.999... existing.
.

As I said, the problem is with the definitions. In math 0.99... is defined as the goal line, not the process.
The process would be lim(n->inf) sum(1->n) 9x10^-n

If you were using math definitions, the question you should be asking is
lim(n->inf) sum(1->n) 9x10^-n ?= 0.99...

If you are using other definitions, then you have to tell us, before we can start a discussion that is not pointless.

 

[Hygro]

If there is a "final 8", then the number of 9 is finite, then it's not 0,999... anymore. I honestly don't know how you can imagine it's a legitimate comparison, when the whole concept of 0,999... is that it's an INFINITE amount of 9.

Again, people have no idea what infinity IS.

If you cannot conceive a mental construct that appends an 8 following an endless series of 9s the problem of comprehending infinity rests with you, not me.

I can understand when I am told that it ending in an 8 means that that value doesn't have a defined place (since it is at an endpoint that does not exist) so it can't be considered a real number according to the rules of numbers. This is seperate.

 

[Akka]

If you cannot conceive a mental construct that appends an 8 following an endless series of 9s the problem of comprehending infinity rests with you, not me.

You can only "follow" after an "end", and if there is an "end" then it's not "endless".
Either you're making some kind of tongue-in-cheek parody, or you don't even understand the basic meaning of the words you're using.

 

[Hygro]

You can only "follow" after an "end", and if there is an "end" then it's not "endless".
Either you're making some kind of tongue-in-cheek parody, or you don't even understand the basic meaning of the words you're using.

You will never know the sound of one hand clapping.

 

[Lohrenswald]

If you cannot conceive a mental construct that appends an 8 following an endless series of 9s the problem of comprehending infinity rests with you, not me.

Bold claim

0,999...998 has both a begining and an end. How could there be infinitely many digits in between?

 

[brennan]

How could there not be?