No, I did not neglect that. I was pointing out that for each nine we add a zero.In each of your relationships you are neglecting an infinte tail on 9s.
At EXACTLY what point does .000...1 become a less viable number than .999...
The thing is, not matter how many nines you add you will never reach 0.999...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.
I think you just killed the thread!The thing is, not matter how many nines you add you will never reach 0.999...
You add a nine, I add a zero.
Originally posted by: ElFenix
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?Originally posted by: MadRat
The value at the infinite position
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.
Originally posted by: RossGr
Go back and give my proof a careful read.
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
At EXACTLY what point does .000...1 become a less viable number than .999...you can not have an infinite string of 0's, and then have a number after that.
If you cannot define that point, then my equation is still correct;
and .999... does not equal 1.
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.
You need to be a bit more specific then that.have and your logic mistake is a glaring distraction from it being any sort of proof.
It's not way down anywhere.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.
But we're not doing.But I still don't see how you can apply rule of forever to one number and not the other.
But I still don't see how you can apply rule of forever to one number and not the other.
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
Originally posted by: RossGr
You need to be a bit more specific then that.have and your logic mistake is a glaring distraction from it being any sort of proof.
Again, EVERY digit in a real number is indexable by an integer. What is the integer at the infinte position?
Ah, but can't you use a regular expression to construct such a number?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?