Okay. 1/9 = 0.1 rec. 2/9 = 0.2 rec. 3/9 = 1/3 =.6 rec. 6/9 = 2/3 = .6 rec.
1/3 + 2/3 = 1
1/3 + 2/3 =0.3 rec + 0.6 rec = 0.9 rec. Thus 1= 0.9 rec. This is not entirely perfect, but is rhetorically easier than:
0.9 rec = lim(n->infinity) (10^n - 1)/10^n = lim(n->infinity) (1-10^-n) =1.
That is a mathematically sound proof. Genius, you are familiar w/ limits, correct?
Originally posted by royalchickenthe third proof (i.e. from CK) was the one that i'd heard before, but i suppose yours would work too... and i have no idea about limits (although i think we did something on them in maths saying that a grath tended to infinity or summant 😛)
Okay. 1/9 = 0.1 rec. 2/9 = 0.2 rec. 3/9 = 1/3 =.6 rec. 6/9 = 2/3 = .6 rec.
1/3 + 2/3 = 1
1/3 + 2/3 =0.3 rec + 0.6 rec = 0.9 rec. Thus 1= 0.9 rec. This is not entirely perfect, but is rhetorically easier than:
0.9 rec = lim ...[text shortened]... tically sound proof. Genius, you are familiar w/ limits, correct?
Don't overlook this proof.
Observe that 0.999... may be viewed as a geometric series as such:
T = .9 + .09 + .009 + ...
Recall the derivation of the formula for geometric series:
S = a + ar + ar^2 + ... + ar^n + ... where |r| < 1.
Then:
Sr = ar + ar^2 + ar^3 + ... + ar^(n+1) + ...
Subtracting the second series from the first cancels all terms on the right-hand side except one:
S - Sr = a
S(1 - r) = a
S = a/(1 - r)
For the number in question, a = 0.9 and r = 0.1:
S = 0.9/(1 - 0.1)
S = 0.9/0.9
S = 1.0
-Ray.
Originally posted by rgoudie==============================================
Don't overlook this proof.
Observe that 0.999... may be viewed as a geometric series as such:
T = .9 + .09 + .009 + ...
Recall the derivation of the formula for geometric series:
S = a + ar + ar^2 + ... + ar^n + ... where |r| < 1.
Then:
Sr = ar + ar^2 + ar^3 + ... + ar^(n+1) + ...
Subtracting the second series from the first cancel ...[text shortened]... er in question, a = 0.9 and r = 0.1:
S = 0.9/(1 - 0.1)
S = 0.9/0.9
S = 1.0
-Ray.
The above is a proof because it defines what an infinitely recurring decimal IS.
.