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Posers and Puzzles
Your Blackened Sky
Joined 12 Mar '02 Moves 15128 0.9 recurring (i.e. 0.999999999999999999...9) is equal to one-can anyone work out a proof for this?
Elsewhere
Joined 29 Nov '02 Moves 17317 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?
Copenhagen, Denmark
Joined 21 Mar '03 Moves 53738 Another classic proof of this is:
x=0.9 rec.
10x=9.9 rec.
9x=10x-x=9
x=1
therefore x= 1 AND x= 0.9 rec.
so 0.9 rec. = 1
Your Blackened Sky
Joined 12 Mar '02 Moves 15128 Originally posted by royalchicken
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? the 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 😛)
Norway
Joined 19 Dec '02 Moves 483 1/9 = .111 rec
So:
1/9*9 = .999 rec
But also:
1/9*9 = 1
Elsewhere
Joined 29 Nov '02 Moves 17317 All of these work....and depend on different fundamental details, from the definition of a recurring decimal (the limits proof), to the multiplicative inverse property of the real numbers (zamba's). Good going all...
Originally posted by zamba
1/9 = .111 rec
So:
1/9*9 = .999 rec
This is not really a proof.
It's just moving the goalposts.
🙄
.
Out of my mind
Joined 25 Oct '02 Moves 20443 No, it isn't;
1 = 9 * 1/9 = 9*0.11111111... = 0.9999999....
Joined 08 Dec '03 Moves 3140 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.
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The above is a proof because it defines what an infinitely recurring decimal IS.
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