@Kolyo,
First Proof:
Quote:Let x = .999...
Then 10x = 9.999...
9x = 10x - x = 9.999... - 0.999... = 9
So x = 1.
It's hard to argue with that proof, it was creative that you thought that up, but I can't help thinking there is a flaw in that proof. Something is wrong with the internal logic of that proof. Tomorrow I will graph that equation and manipulate it.
x= 3.33 repeated
10x = 33.33 repeated
10x-x= 33.333 - 3.333
x(10-1)= 30
9x = 30
x = 30/9
x= 3.33 repeated
well it worked for 3.33 repeated
x= .99998
10x= 9.9998
10x- x = 9.9998-.99998
9x = 8.99982
x=8.99982/9
x=.99998
Well it works, but I can't help thinking your breaking some mathematical rule in that equation, by just manipulating one number in an equation like that. Maybe I'm wrong.
x = 0
10x = 0
10x-x = 0
9x = 0
x = 0/9 = 0 huh
x= 1/3= .33 repeated
1/9 (x)= 1/27
1/9 x - x =1/27- 1/3
-8/9(x) = -8/27
x = 1/3
huh
x = .99 repeated
1/10 (x) =.099 repeated
1/10 (x) - x = -.899 repeated with a 1 at the end
-.9 x = -.89999999999999999999999999...99991
I don't believe this equation is solvable, that may provide a hint. The fact that the logic this equation is based upon doesn't work in this instance, may prove that the logic is flawed. you can't have the 9s go on for infinity and than add a 1. This equation may only be valid with numbers that are capable of being represented by a fraction.
It just seems like .9999 goes towards infinity difference between this number and 1 become vanishingly (I just made that word up, I'm like Sarah Palin) infinitesimally small but it doesn't quite close in on 1.
with .9 you are .1 away from 1.
with .99 you are .01 away from 1.
with .9999999999999999999999999999999999999999
you are .0000000000000000000000000000000001 away from 1.
There will always be that 1 in the difference no matter how far you take the .9s
Anyway I commend you Kolyo. We'd probably both be better off debating this, rather than debating social and political topics.