.999 = 1

[s7yl33]

hey saw this on the HTS messageboard.... thought it was really interesting.

Here are the proofs that 0.999~ (representing 0.9 recurring) = 1

Here are a few proofs. The last couple are easiest to understand if you don't understand more advance maths:



999~ = sigma(.9*[.1]^[n-1]).

.9999~ = .9 + .09 + .009 + .0009 ... ad infinum
sigma[i:0->inf.](.9*.1^i)
omega->sigma = .9/(1 - .1) = .9/.9 = 1
.999~ = 1

.999~ = .9 + .09 + .009 .... (ad infinum)
Infinite geometric progression: a + ar + ar^2... (ad infinum)
Let a, 1st term = .9
Let r, common ratio = 10^-1
Sum to infinity = a / (1 - r) = .9 / (1 - .1) = .9/.9 = 1
.999~ = 1

sigma(n=1 ; n->inf.) 9/(10^n) = (def. of geometric series)
9 * sigma(n=1 ; n->inf.) (1/10)^n = (property of a series)
9 * 1/9 = (r/[1-r], r = .1)
3/9 = 1/3
.999~ = 1

The sequences (.9, .99, .999... ad infinum), and (1, 1, 1... ad infinum) are equivalent, so they have the same limit, .999~.
S = .999~
S = 0.9 + 0.09 + 0.009 + 0.0009... ad infinum
S = 0.9 + (1/10)(0.9 + 0.09 + 0.009 + .0009... ad infinum)
S = 0.9 + (1/10)S
(9/10)S = .9
S = 1
.999~ = 1

.333~ = sigma(n=1 ; n->inf.) 3/(10^n) = (def. of geometric series)
3 * sigma(n=1 ; n->inf.) (1/10)^n = (property of a series)
3 * 1/9 = (common ratio, r/[1-r], r = .1)
3/9 = 1/3
1/3 = .333~
1/3 * 2 = 2/3
.333~ * 2 = .666~
2/3 = .666~
.333~ + .666~ = .999~
1/3 + 2/3 = 3/3
3/3 = .999~
3/3 = 1
.999~ = 1

.000~ = 0/9
.111~ = 1/9
.222~ = 2/9
.333~ = 3/9
.444~ = 4/9
.555~ = 5/9
.666~ = 6/9
.777~ = 7/9
.888~ = 8/9
.999~ = 9/9
9/9 = 1
.999~ = 1

x = 0.999~
10x = 9.999~
10x - x = 9.999~ - 0.999~
9x = 9
x = 1
0.999~ = 1

1/3=.33333~
3(.33333~)=.99999~
3(1/3)=1
therefor, 1=.99999~

Makes you question reality dosent it?