still struggling!!!!

normally when a number v repeated digit
such as 0.333.... can b expressed in fractional form,ie 1/3 for this case

for more examples
0.142857142857... = 1/7
0.090909090909....= 1/11
0.285714285714.... = 2/7

n now this is my quest..
what is 0.999999999... in the fractional form
i try to solve it by using the method of sum to infinity,S=a/(1-r)
but it gives me the ans of 1..
y does it so!!
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 ???? how can u say tat 0.999999999... equals to 1 it just approaches to 1 but yet it cant b considered as 1

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still struggling!!!!

search these very forums for this, but save yourself the time. they are equal, just like 1/2 adn 2/4 are equal. this is a mathematical property of the real numbers and the decimal expansions as *representations* of them. they are different representations of the same real number nothing more nor less. but please don't spend time argunig against this fact.
 Find out 1 - 0.999999999... Isn't it 0.000000000... ???

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 Quote by bs ???? how can u say tat 0.999999999... equals to 1 it just approaches to 1 but yet it cant b considered as 1
It isn't "approaching" anything- it isn't changing. Any infinite decimal is, by definition of the base 10 number system, the limit of the "partial sums". You might well say that the sequence {.9, .99, .999, .9999, ...} is "approaching" 1 but the number 0.999999... is, by definition, the limit of that sequence, 1.

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 Quote by bs ???? how can u say tat 0.999999999... equals to 1 it just approaches to 1 but yet it cant b considered as 1
As stated above,

$$1 - 10^{-n} \quad \text{or} \quad \sum_{r=1}^{n} \left( 9 \cdot 10^{-r}\right)$$

These approach 1 as n approaches infinity, 0.9999999... is 1.
 I find it odd that you accept the other repeating decimals as their respective fractions yet do not accept 1 as 0.9r. Especially considering you know that 0.333... = 1/3 ---> 3*0.333... = 3*(1/3) = 0.999... = 3/3 = 1.
 how if i reverse the calculation: 1/7 = 1 divided by 7 = 0.142857142857... 1/11 = 1 divided by 11 = 0.090909090909.... 2/7 = 2 divided by 7 = 0.285714285714.... 1 = 1 divided by 1 = 1 ( y not 0.9999999......) can u explain?

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 Quote by bs how if i reverse the calculation: 1/7 = 1 divided by 7 = 0.142857142857... 1/11 = 1 divided by 11 = 0.090909090909.... 2/7 = 2 divided by 7 = 0.285714285714.... 1 = 1 divided by 1 = 1 ( y not 0.9999999......) can u explain?
.99999999999999999999...=1
It is just two ways of writing the same thing.
1-10^-n<.9999999...<1 for all n=1,2,3,...
The only real number that can do that if 1.
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 Quote by bs how if i reverse the calculation: 1/7 = 1 divided by 7 = 0.142857142857... 1/11 = 1 divided by 11 = 0.090909090909.... 2/7 = 2 divided by 7 = 0.285714285714.... 1 = 1 divided by 1 = 1 ( y not 0.9999999......) can u explain?

explani what? there's nothin wrong here. you just think there is.decimals are representations of numbers just like the symbols 1/2 2/4 are representations of rational numbers. it is perfectly alright for their to be two representations of certain decimal numbers. not all have this property. oddly, there are infinitely many representations for EVERY rational yet you probably have no problem accepting that 1/2 and 2/4 are the same. your attitude is very common. but you are attempting to read things into these representations that simply need not be true and indeed cannot be true.

 Quote by bs how if i reverse the calculation: 1/7 = 1 divided by 7 = 0.142857142857... 1 = 1 divided by 1 = 1 ( y not 0.9999999......) can u explain?
Notice, for example, that 1/7 does not equal to $$0.142857142857$$, but rather $$0.\overline{142857}$$. Similiarly, 1/1 is not $$0.99999$$, but $$0.\bar{9}$$.

I know "..." was your way of representing recurring decimals, but you feel somehow it's "OK" for 1/7 to be a recurring decimal, yet not 1/1.

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 Quote by matt grime explani what?
I think what troubles bs here is that while the other decimal representations can be generated as outputs of an algorithm (ie: long division) whose input parameters are the numerator and denominator of the rational fraction, this doesn't "seem to" work for $0.\overline{9}$.

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 Quote by Gokul43201 I think what troubles bs here is that while the other decimal representations can be generated as outputs of an algorithm (ie: long division) whose input parameters are the numerator and denominator of the rational fraction, this doesn't "seem to" work for $0.\overline{9}$.
And also the fact that even if one developed a (necessisarily stupid) algorithm to generate .999999... one would have to avoid rounding to prevent getting 1.

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