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0.000...1 = 0

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smack Down

Diamond Member
Sep 10, 2005
4,507
0
0
Originally posted by: mugs
There is no such thing as 0.000....1. You can't have something that comes after something that repeats infinitely.
Sure it is. It is what you get when subtract .99999... from 1.
 

Ika

Lifer
Mar 22, 2006
14,267
3
81
Originally posted by: smack Down
Originally posted by: mugs
There is no such thing as 0.000....1. You can't have something that comes after something that repeats infinitely.
Sure it is. It is what you get when subtract .99999... from 1.
No, that would be 0.
 

PowerEngineer

Diamond Member
Oct 22, 2001
3,318
350
126
Originally posted by: Aflac
Originally posted by: smack Down
Originally posted by: mugs
There is no such thing as 0.000....1. You can't have something that comes after something that repeats infinitely.
Sure it is. It is what you get when subtract .99999... from 1.
No, that would be 0.
Yes, that would be zero...and that is exactly his point.

We've had long (and painful to read) threads arguing whether or not .99999... = 1. It should be obvious that 1 - 0.99999... = 0.000...1, and that if .99999... = 1, then 0.000...1 = 0.

Here we go again.

Edit: yes, 0.000...2 is also zero. 0.000...X where X is anything will be zero.
 

Ika

Lifer
Mar 22, 2006
14,267
3
81
Originally posted by: PowerEngineer
Originally posted by: Aflac
Originally posted by: smack Down
Originally posted by: mugs
There is no such thing as 0.000....1. You can't have something that comes after something that repeats infinitely.
Sure it is. It is what you get when subtract .99999... from 1.
No, that would be 0.
Yes, that would be zero...and that is exactly his point.

We've had long (and painful to read) threads arguing whether or not .99999... = 1. It should be obvious that 1 - 0.99999... = 0.000...1, and that if .99999... = 1, then 0.000...1 = 0.

Here we go again.
I know, I was just trying to stir the pot.
 

Hyperion042

Member
Mar 23, 2003
53
0
0
I'm nipping this in the bud:

Consider the number 0.1.
This is equivalent to 10^(-1).
Then 0.01 = 10^(-2), 0.001=10^(-3), etc.
Then if there are n 0s between the decimal and the one, the number is 10^(-(n-1)), clearly.
Now we increase n arbitrarily, to 0.000...1:
lim (n->infinity) 10^(-(n-1)) = 10*lim (n->infinity) 10^(-n).
Now, consider ANY arbitrarily small number, call it a.
Then 10^(-n) < a if and only if n > -log a.
Since for any arbitrarily small number a, there is a FINITE n for which 10^(-n) < a, we then have that the limit as n->infinity of 10^(-n) <= 0.
But 10^(-n) > 0 for all real n. Then the limit as n-> infinity of 10^(-n) <=0 and >= 0.
Then limit as n-> infinity of 10^(-n) = 0.

Problem done. This solution is mathematically and logically complete, sound, and excepting any basic typos I've made, irrefutable. If this thread grows to 5 pages after this, I'm disowning people.
 

smack Down

Diamond Member
Sep 10, 2005
4,507
0
0
Originally posted by: Aflac
Originally posted by: PowerEngineer
Originally posted by: Aflac
Originally posted by: smack Down
Originally posted by: mugs
There is no such thing as 0.000....1. You can't have something that comes after something that repeats infinitely.
Sure it is. It is what you get when subtract .99999... from 1.
No, that would be 0.
Yes, that would be zero...and that is exactly his point.

We've had long (and painful to read) threads arguing whether or not .99999... = 1. It should be obvious that 1 - 0.99999... = 0.000...1, and that if .99999... = 1, then 0.000...1 = 0.

Here we go again.
I know, I was just trying to stir the pot.
Isn't everyone?
 

spidey07

No Lifer
Aug 4, 2000
65,469
5
76
1 - 0.0000...1 does not equal zero.

end thread.

Join the club, subscribe to the newsletter! 9/11 was a deliberate act by the US government, man never set foot on the moon, and indeed there is a number in our minds called 0.000...747. Infinity is a concept used by the man to hide THE TRUTH!!!

OMG! Look at the coincidence!
7+4=11.
7+(1/2 *4) = 9.

9....
1.....
1.....

 
Oct 20, 2005
10,978
44
91
Originally posted by: PowerEngineer
Originally posted by: Aflac
Originally posted by: smack Down
Originally posted by: mugs
There is no such thing as 0.000....1. You can't have something that comes after something that repeats infinitely.
Sure it is. It is what you get when subtract .99999... from 1.
No, that would be 0.
Yes, that would be zero...and that is exactly his point.

We've had long (and painful to read) threads arguing whether or not .99999... = 1. It should be obvious that 1 - 0.99999... = 0.000...1, and that if .99999... = 1, then 0.000...1 = 0.

Here we go again.

Edit: yes, 0.000...2 is also zero. 0.000...X where X is anything will be zero.
you can't have 0.000....X ever...you can never tack on an "X" to the end b/c the end keeps on going forever.
 

Ruptga

Lifer
Aug 3, 2006
10,247
206
106
Originally posted by: Sentrosi2121
If something has the smallest amount of a 'thing' in it, it no longer weighs 0 (insert preferred weight measurement here)

But that's just an observation.
It's close enough. Your notation is wrong, but if you're asking something like "after a decimel point there's an infinite number of zeros followed by a 1, does this equal zero" then my answer is yeah, it's close enough to be insignificant to anything and everything, and thus may as well be zero.
 

BrownTown

Diamond Member
Dec 1, 2005
5,314
0
0
asking this question and posting the term "0.000...1" shows a complete lack of understanding of the notion of infinity. There is no such number as "0.000...1".
 

IronWing

No Lifer
Jul 20, 2001
62,079
15,292
136
This is why numbers are bad. We should just stick to booleans. Q = 1 or Q = 0 and there ain't nothing else to worry about.


Hmmm, I'm feeling like a .... Republican.
 

Special K

Diamond Member
Jun 18, 2000
7,098
0
76
Originally posted by: Hyperion042
I'm nipping this in the bud:

Consider the number 0.1.
This is equivalent to 10^(-1).
Then 0.01 = 10^(-2), 0.001=10^(-3), etc.
Then if there are n 0s between the decimal and the one, the number is 10^(-(n-1)), clearly.
Now we increase n arbitrarily, to 0.000...1:
lim (n->infinity) 10^(-(n-1)) = 10*lim (n->infinity) 10^(-n).
Now, consider ANY arbitrarily small number, call it a.
Then 10^(-n) < a if and only if n > -log a.
Since for any arbitrarily small number a, there is a FINITE n for which 10^(-n) < a, we then have that the limit as n->infinity of 10^(-n) <= 0.
But 10^(-n) > 0 for all real n. Then the limit as n-> infinity of 10^(-n) <=0 and >= 0.
Then limit as n-> infinity of 10^(-n) = 0.

Problem done. This solution is mathematically and logically complete, sound, and excepting any basic typos I've made, irrefutable. If this thread grows to 5 pages after this, I'm disowning people.
It doesn't matter. Do a search for the old thread - some people won't be convinced no matter how many sound proofs you post. Not coincedentally, these people admit to not having a strong background in math, although they still reason that they are correct, and the rest of the math world is wrong.
 

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