Cogman
Lifer
- Sep 19, 2000
- 10,283
- 134
- 106
Originally posted by: pray4mojo
new question.
does 0.000.... = 0?
No, it equals 1, duh!
Originally posted by: pray4mojo
new question.
does 0.000.... = 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.
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.
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.
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.
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.
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.
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.
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.