0.999999... =1?

Elio (n)

You know the rule of converting the periodic numbers in fractions I haven't yet understand his logic.

$0.9999.....=\frac{9}{9}=1$ but $0.777....=\frac{7}{9}$

$\frac{7}{9}$ is not equal with $\frac{9}{9}$

**Posted:** Mon Feb 11, 2008 11:02 am

asintota

**Posted:** Mon Feb 11, 2008 11:06 am

azjps

$0.\overline{9} - 0.\overline{8} = 0.\overline{1}$ , which isn't nothing Wink

Edit: My question was, how does $1 = 0.\overline{9}$ imply that $0.\overline{9} = 0.\overline{8}$ ? Indeed, $0.\overline{9} = \frac{9}{9} \neq 0.\overline{8} = \frac 89$

**Posted:** Mon Feb 11, 2008 11:08 am

Elio (n)

$\frac{9}{9}=1>\frac{8}{9}>\frac{7}{9}$ this make untrue your hypothesis

**Posted:** Mon Feb 11, 2008 11:12 am

asintota

**Posted:** Mon Feb 11, 2008 11:16 am

Elio (n)

Look this:
$0.99999......... + 0.0000000000......................................1 = 1$ Razz

While $0.888...... + 0.11111....... = 0.9999999......$

$0.0000000000000000 ................1 < 0.11111.....$

$0.999....$ is to close to $1$

jarrro · New Member Offline Joined: 11 Feb 2008 Posts: 7

**Posted:** Mon Feb 11, 2008 11:19 am

asintota

**Posted:** Mon Feb 11, 2008 11:30 am

Nerd_of_the_Ages

In general,

$0.x_1x_2x_3 ... x_kx_1x_2... x_k ... = n$

where n is a fraction, each x is a digit, multiply the equation by the number of digits in the repeating block, in the case k. We now have $x_1x_2...x_r.x_1x_2...x_kx_1x_2...x_k... = kn$
We now subtract n from both sides, leaving us with
$(k - 1)n = x_1x_2 ... x_k$ .
Now divide by (k - 1).

So in general, $n = \boxed{\frac{x_1x_2...x_k}{k - 1}}$

**Posted:** Mon Feb 11, 2008 12:33 pm

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worthawholebean

**Posted:** Mon Feb 11, 2008 12:50 pm

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**Posted:** Mon Feb 11, 2008 1:09 pm

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Elio (n)

Yeas maybe you are right but it was very funny when I thought.

**Posted:** Tue Feb 12, 2008 11:46 am

Temperal

The most rigorous proof I can think of is this:

$0.999\ldots = \lim_{n\to\infty}0.\underbrace{ 99\ldots9 }_{n} = \lim_{n\to\infty}\sum_{k = 1}^n\frac{9}{10^k} = \lim_{n\to\i...$

Unfortunately, no proof will convince my classmates...

**Posted:** Tue Feb 12, 2008 3:09 pm

Hustla25 · P versus NP Offline Joined: 11 Feb 2008 Posts: 42

0.999999999999.................... is truely equal to 1

say 0.999999999999.................... = x

10x = 9.999999999999....................

-(x = 0.999999999999.................... )
___________________________________(Add the last two equations)

9x = 9

or x=1

its truely weird but true

**Posted:** Tue Feb 12, 2008 3:13 pm

Nerd_of_the_Ages

**Posted:** Tue Feb 12, 2008 3:43 pm

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t0rajir0u

$0.000...1$ isn't well-defined notation at all. You can't tell me where the digit $1$ is. Temperal's argument is correct. The notation

$0.999...$

is defined to mean

$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{9}{10^k}$

which is equal to $1$ . Understanding this subtlety, however, requires a thorough understanding of the definition of a limit.

People who think $0.999... \neq 1$ make the implicit assumption that the map from a sequence of digits to a real number is a bijection. This is not, in fact, the case (as the fact that $0.999... = 1$ shows).

**Posted:** Tue Feb 12, 2008 4:45 pm

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In order to evaluate that limit, we are invoking a nontrivial property of the real numbers: the Archimedean property. The ordinary rules of arithmetic tell us that the limit of $10^{-n}$ can't be any nonzero number (it's ten times itself), but they don't necessarily tell us that the limit exists at all.
There are two consistent alternatives here: $0.99999\dots=1$ , or $0.99999\dots$ is not a number. The standard definitions of mathematics choose the first; we prefer to have more ways to write something down rather than fewer.

See also here.

**Posted:** Tue Feb 12, 2008 5:01 pm