# 0.9 rec=1

genius
Posers and Puzzles 21 Jun '03 16:28
1. genius
Wayward Soul
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?
2. royalchicken
CHAOS GHOST!!!
21 Jun '03 19:00
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-&gt;infinity) (10^n - 1)/10^n = lim(n-&gt;infinity) (1-10^-n) =1.

That is a mathematically sound proof. Genius, you are familiar w/ limits, correct?
3. 22 Jun '03 10:37
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
4. genius
Wayward Soul
22 Jun '03 11:252 edits
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 ðŸ˜›)
5. 22 Jun '03 16:01
1/9 = .111 rec

So:
1/9*9 = .999 rec

But also:
1/9*9 = 1
6. royalchicken
CHAOS GHOST!!!
22 Jun '03 16:55
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...
7. 06 Jun '04 15:542 edits
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.

ðŸ™„
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8. TheMaster37
Kupikupopo!
06 Jun '04 16:38
No, it isn't;

1 = 9 * 1/9 = 9*0.11111111... = 0.9999999....
9. 07 Jun '04 01:14
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| &lt; 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.

10. 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.

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The above is a proof because it defines what an infinitely recurring decimal IS.

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