Solving 0.999999999... in Fractional Form

[bs]

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!

 

[AKG]

because it is 1

 

[bs]

?
how can u say tat 0.999999999... equals to 1
it just approaches to 1
but yet it can't b considered as 1

 

[matt grime]

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.

 

[mustafa]

Find out 1 - 0.999999999...
Isn't it 0.000000000... ?

 

[HallsofIvy]

?
how can u say tat 0.999999999... equals to 1
it just approaches to 1
but yet it can't 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.

 

[Zurtex]

?
how can u say tat 0.999999999... equals to 1
it just approaches to 1
but yet it can't 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.

 

[Icebreaker]

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.

 

[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?

 

[lurflurf]

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.

 

[Curious3141]

Is it the same troll beating the same dead horse an infinite number of times ?

 

[matt grime]

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.

 

[Icebreaker]

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.

 

[Gokul43201]

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 $0.\overline{9}$.

 

[lurflurf]

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.

 

[Zurtex]

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.

Eh?

What has any of this got to do with rounding?

 

[James R]

The fact that 0.999... = 1 is an artifact of our base 10 representational system.

Suppose we work in base 3, for example. Then, counting from one to ten would look like this:

1, 2, 10, 11, 12, 20, 21, 22, 100, 101.

The number one in base 3 can also be represented as 0.222...

i.e. in base 3 we have 1 = 0.222..., just as in base 10 we have 1 = 0.999...

 

[Gokul43201]

There really isn't a problem with the long division algorithm at all...it can easily be tweaked to generate recurring 9s ( by replacing the odd "$\leq$" in the common version with a "<" and make other related changes).

 

[lurflurf]

Eh?

What has any of this got to do with rounding?

I have a feeling someones was sitting with their calculator and noticing 1/3=.333333... then .333333333...*3=1, but this person does not believe .9999999...=1 so when .99999 occurs it is rounded, but wen they do 1/1 the calculator does not show .999999999 hence confusion. Here is a stupid algorith to generat 1=.99999... we do silly long division 1=10/10=9/10+1/10=9/10+9/10^2+1/10^3+9/10+9/10^2+9/10^3+1/10^3. Just be use to throw away the 1/10^n terms are rounding error and don't round up.

 

[beanybag]

Another way to prove this that may help you understand is if take x=.9r, and multiply by 10, u get 10x=9.9r. if you then take 10x-x, u'd get 9x=9, therefore x=1 and 1=.9r.