21 Jun '03 16:28

0.9 recurring (i.e. 0.999999999999999999...9) is equal to one-can anyone work out a proof for this?

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Your Blackened Sky- Joined
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Elsewhere21 Jun '03 19:00Okay. 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?- Joined
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Your Blackened Sky22 Jun '03 11:252 edits

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 ðŸ˜›)*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?- Joined
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Out of my mind- Joined
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07 Jun '04 01:14Don'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.

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07 Jun '04 03:19

==============================================*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.

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