Is 1 = 0.9999......

[silverpig]

This thread will go on forever.... with a consensus at the end.

 

[BigNeko]

RossGr,
Caught that after I posted, thus the edit.

In each of your relationships you are neglecting an infinte tail on 9s.

No, I did not neglect that. I was pointing out that for each nine we add a zero.
See edited post above for continuance of this debate.


When this thread reaches an infinite number of posts, will someone still be able to bump it?

 

[RossGr]

At EXACTLY what point does .000...1 become a less viable number than .999...

Go back a few posts and make an effort to understand my attempts at an explaintion of what a real number is.

Your number fails to be a real number because you cannot assign an integer to it. Every digit of a Real number has a location represented by some integer. This what I mean by D(n)10^-n. Since the 1 is a digit in must have an associated integer, ie it is not at infinity, but at some finite location.

in .999.... D(n)=9 for ALL n.

 

[BigNeko]

You add a nine, I add a zero.

Back and forth.

On and on.

FOREVER.

My equation will ALWAYS be true.

.999... does not equal 1.

 

[BruinEd03]

Wow...looks like a holy war in here.

All started by:

Kyteland
Member

Posts: 40
Joined: Dec 2002

Quite an impression he's made


(And yes I believe one side is correct but I won't fan the flame any more )

-Ed

 

[Haircut]

Originally posted by: BigNeko
You add a nine, I add a zero.

Back and forth.

On and on.

FOREVER.

My equation will ALWAYS be true.

.999... does not equal 1.

The thing is, not matter how many nines you add you will never reach 0.999...

 

[BigNeko]

BruinEd03,
Flame, what flame? No one is getting upset except the kids.

RossGR,
I did read your explanation. Do not imply my attempt at understanding was not serious. Seriously.

Haircut,

The thing is, not matter how many nines you add you will never reach 0.999...

I think you just killed the thread!

 

[Haircut]

I think you just killed the thread!

I damn well hope so!

 

[RossGr]

You add a nine, I add a zero.

There's where you go wrong. I do not "add" a 9. The nines are ALL there. That is what .999... means. It is NOT a limiting process were you tack on more nines. This is the sum to infinity, all the nines are in place simply add them up.


So for any 10^-n you add on you can find the correct number of 9s to make 1, but still have an INFINITLY many more which sum to a number greater then 0. so .999.... + 10^-n > 1 for ALL n in the integers.

Go back and give my proof a careful read.

 

[MadRat]

Originally posted by: ElFenix

Originally posted by: MadRat
The value at the infinite position

there is no infinith position. thats why the ellipsis used in the middle of the number means that theres a finite number of decimal places, and that the numbers after that are whats at the end. why is that so hard to comprehend?

Excuse me, but if we have a .999... then there is an infinite position, or do you not understand vectoring?

Originally posted by: TuxDave

What if I throw in a monkey wrench and concede that 0.000...r1 does exist only if it's equal to zero.

No problem. But then again that would be conceding then that .999... does not exist.

 

[MadRat]

Originally posted by: RossGr

Go back and give my proof a careful read.

I have and your logic mistake is a glaring distraction from it being any sort of proof.

 

[MadRat]

Originally posted by: BigNeko
I think MadRat understands what I am trying to say. Credit to the guy (or gal) back on page one or two who did this.
.9+.1=1
.99+.01=1
.999+.001=1
JosphII said

you can not have an infinite string of 0's, and then have a number after that.

At EXACTLY what point does .000...1 become a less viable number than .999...
If you cannot define that point, then my equation is still correct;
and .999... does not equal 1.

Some plain don't understand that the points they argue about either both exist or neither exists.

Originally posted by: BigNeko
P.S. This is a rather good debate, so let's skip the name calling, even though you did add a nine to your infinite nines, then turn around and say to me that I had one too many zeroes before my one.

So simple yet some cannot grasp the truth.

 

[RossGr]

have and your logic mistake is a glaring distraction from it being any sort of proof.

You need to be a bit more specific then that.

Again, EVERY digit in a real number is indexable by an integer. What is the integer at the infinte position?

 

[BigNeko]

RossGR,
I will concede that my EXPRESSION of the concept .000...1 is technically incorrect,
however, I am asking you to overlook the expression (in our finite system) and look more to the concept itself.

If one number can go on forever, so can the other. It is so simple.

The nines are already there? Okay.
So are the zeroes, with that pesky one at the non-end.

Come to think of it, my expression is technically correct,
because my zeroes are merely place-holders for a little one that
is infinitely to the right of the decimal point. You may not be able to see it,
but its way-y-y-y down there.

Trust me.

 

[RossGr]

I can specify an, n, for each 9, indeed for any 9, all I need do is name an integer, there is a nine in that slot. What integer can you assign to your 1? It has to have one or we are not talking about a Real number, since it has a coresponding integer, it is not at infinity.

 

[Haircut]

Originally posted by: BigNeko
RossGR,
I will concede that my EXPRESSION of the concept .000...1 is technically incorrect,
however, I am asking you to overlook the expression (in our finite system) and look more to the concept itself.

If one number can go on forever, so can the other. It is so simple.

The nines are already there? Okay.
So are the zeroes, with that pesky one at the non-end.

Come to think of it, my expression is technically correct,
because my zeroes are merely place-holders for a little one that
is infinitely to the right of the decimal point. You may not be able to see it,
but its way-y-y-y down there.

Trust me.

It's not way down anywhere.
Let's suppose we have a circle. Each time we go an arbitrary amount around the circumference of the circle we add another 0 to the expression 0.0 to 0.00 to 0.000 etc.

Now the 1 has to be placed when we reach the end of the circumference.
The 1 in this number cannot exist as we will never reach the end of the circumference.

We can't say that the 1 in 0.000...1 is just a long way down, or anything like that. It simply cannot exist.

 

[BigNeko]

Pondering further,

we are dancing around the main question: Can we add one to infinity?
By the nature of my previous arguments, I say that we can.

In order for you to prove otherwise, you would need an infinite amount of time.
Using your concept of infinite is close enough, that means you would need forever to prove otherwise.
Which is the same as saying you could NEVER prove otherwise.

Which means that not only am I right, I am right FOREVER.

 

[BigNeko]

Previous post was all in good fun.

It's not way down anywhere.

That is a pretty good way of looking at it (1.999... post above this one)
But I still don't see how you can apply rule of forever to one number and not the other.

Well, I will just agree to disagree with you on this and call it a night (which it is where I am).

 

[RossGr]

No, we don't need a lot of time, all we need is the definiton of infinity.

By the definition of infinity,

inifinty + x = infinity for all x>0

Infinity is an extension of the Real number line, it's properties are carefully defined when it is tacked on.

Basicly the definition is, that infinity is greater then x for all x in the Real numbers. Then how it behaves under each operation is defined. I do not have the formal statement in front of me and will not attempt to drege it up out of long term memory. If I have the time I will dig it out and post it later today.

 

[Haircut]

But I still don't see how you can apply rule of forever to one number and not the other.

But we're not doing.
With 0.999... we have a string of nines that never ends, I'm not saying that there is a last digit that is nine.
With 0.000...1 we have to assign a last digit. As I showed above we cannot have a last digit, we could have infinite 0s but then we can't have anything at the end.

 

[RossGr]

But I still don't see how you can apply rule of forever to one number and not the other.

What puzzels me is how us saying 9s forever is the same as saying 0s forever then a 1. When I say forever I am done, you say forever then a 1. How can the 1 be there if it is indeed 0s forever. Don't you see that that statement is self condraticding?

 

[Kyteland]

Originally posted by: BruinEd03
Wow...looks like a holy war in here.

All started by:

Kyteland
Member

Posts: 40
Joined: Dec 2002

Quite an impression he's made


(And yes I believe one side is correct but I won't fan the flame any more )

-Ed

I've been lurking on and off since ~2000 so I feel justified in in starting a flame war early on im my posting career on ATOT.

 

[Skyclad1uhm1]

If 1/infinity is 0, can 1/(1/infinity) be infinity? After all you'd divide by 0

What is 1-1/infinity?

(Just trying to prolong the discussion )

 

[HomeBrewerDude]

Originally posted by: RossGr

have and your logic mistake is a glaring distraction from it being any sort of proof.

You need to be a bit more specific then that.

Again, EVERY digit in a real number is indexable by an integer. What is the integer at the infinte position?

there is NO infinite position. that is the point.

 

[Kyteland]

Originally posted by: RossGr

But I still don't see how you can apply rule of forever to one number and not the other.

What puzzels me is how us saying 9s forever is the same as saying 0s forever then a 1. When I say forever I am done, you say forever then a 1. How can the 1 be there if it is indeed 0s forever. Don't you see that that statement is self condraticding?

Ah, but can't you use a regular expression to construct such a number?

0.0*1

S -> 0A
A -> .B
A -> .C
B -> 0B
B -> 0C
C -> 1

Can't this number have an orbitrary number of 0's? If that's the case then their concept of 0.00....r1 would be described by this.

Too bad a Turing machine would never recognize it....