Here is a cool way to come to terms with it though it is not a proof (I'll let the experts handle that).

1/9 = .111111111111...

2/9 = .222222222222...

.

.

.

8/9 = .888888888888...

so this would imply that .9999999999... = 9/9 = 1

Math Helper

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- April 4th 2005, 04:34 PM #1

- April 4th 2005, 05:09 PM #2

- April 4th 2005, 05:33 PM #3

- April 5th 2005, 11:27 AM #4

- April 8th 2005, 06:00 AM #5Originally Posted by
**Yankee77**

Performing a devision you'll never get a decimal periodic number with an indefinite sequence of 9 or 0 digit. So the problem is only formal.

2.

Remember how decimal periodic number are transformed in fractions?

For instance 2.(3) - where () means indefinite sequence of 3 as decimal digit

2.3 = (23-2)/9=21/9=7/3

This applies to all cases

0.(9) = 9/9 = 1

3.

1/3 is 0.333333333333...........

3 times 1/3 is 1

but 3 times 0.333333333333........... = 0.99999999999...........

4.

0.(9)=0.99999.....

develops as the infinite sum

9/10+9/100+9/1000

that is 9(1/10+1/10^2+1/10^3+...)

the infinite sum in the parenthesis is the geometric serie with ratio 1/10 and gives

9[1/(1-1/10) -1] = 9[1/(9/10)-1] = 9[10/9-1] = 9(1/9)=1

bye

- April 14th 2005, 09:02 PM #6

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Originally Posted by**Yankee77**

- April 15th 2005, 08:44 AM #7

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- April 16th 2005, 10:14 PM #8

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- July 24th 2005, 02:32 PM #9
## ?

Here 's a little proof, that doesn't move far into sequence territory:

Consider the difference a=|0.(9)-1|. (the parentheses mean period).

Let this number be positive. Observe that a is non-negative, and less than any number of the form 0.0(etc)01, with a random number of zeros. This contradicts a>0 (*), and so a=0.

(*) 'Cause, if a>0, there ought to be a number of the form 0.0...01 < a.

- August 5th 2005, 02:39 PM #10
Perhaps a helpful way to think about this is to separate the concepts of 'real number' and 'decimal expansion'. We can set up the system of real numbers in various ways, some of the more high-brow being Dedekind cuts in the real numbers or the completion of the rationals via Cauchy sequences modulo null sequences. Once you have a notion of real number, you can then describe a real number by a decimal expansion. It turns out that decimal expansions have this awkward property that sometimes two different decimals express the same real number: and this happens when one of them ends in 9 recurring. Of course one way of setting up the real number system is to start with decimals expansions.

- August 6th 2005, 02:46 PM #11
For a further discussion of this point, see the page by Tim Gowers (author of the excellent "Mathematics: a Very Short Introduction") on

*What is so wrong with thinking of real numbers as infinite decimals?*on his website http://www.dpmms.cam.ac.uk/~wtg10/decimals.html

- August 28th 2005, 05:29 AM #12

- August 28th 2005, 06:50 PM #13

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- November 14th 2005, 08:25 PM #14

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The answer is basic, DEFINITON: THE REPETION OF A DECIMAL EQUALS THE VALUE OF ITS CONVERGENT. Thus, what most people do not relized is this is not paradoxical (nothing in math is paradoxical) but rather it is a DEFINITION and not a theorem. It is proven by looking at it geometric sum.

- November 16th 2005, 10:51 AM #15

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