Solving 0.999999999... in Fractional Form

  • Thread starter bs
  • Start date
In summary, the conversation discusses the concept of recurring decimals and their corresponding fractional forms. It is agreed that decimals such as 0.333... can be represented as fractions, but there is confusion about whether 0.999999... can be represented as 1. Some argue that it is simply a different representation of the same number, while others question the use of rounding in calculations involving 1. The conversation also brings up the idea of different numerical systems, such as base 3, and how they can affect the representation of numbers. Ultimately, it is concluded that 1 and 0.999999... are indeed
  • #1
bs
4
0
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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  • #2
because it is 1
 
  • #3
?
how can u say tat 0.999999999... equals to 1
it just approaches to 1
but yet it can't b considered as 1
 
  • #4
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.
 
  • #5
Find out 1 - 0.999999999...
Isn't it 0.000000000... ?
 
  • #6
bs said:
?
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.
 
  • #7
bs said:
?
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,

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

These approach 1 as n approaches infinity, 0.9999999... is 1.
 
  • #8
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.
 
  • #9
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?
 
  • #10
bs said:
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.
 
  • #11
Is it the same troll beating the same dead horse an infinite number of times ?
 
  • #12
bs said:
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.
 
  • #13
bs said:
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 [tex]0.142857142857[/tex], but rather [tex]0.\overline{142857}[/tex]. Similiarly, 1/1 is not [tex]0.99999[/tex], but [tex]0.\bar{9}[/tex].

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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  • #14
matt grime said:
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 [itex]0.\overline{9} [/itex].
 
  • #15
Gokul43201 said:
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 [itex]0.\overline{9} [/itex].
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.
 
  • #16
lurflurf said:
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?
 
  • #17
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...
 
  • #18
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 "[itex]\leq[/itex]" in the common version with a "<" and make other related changes).
 
  • #19
Zurtex said:
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.
 
  • #20
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.
 

1. What is the fractional form of 0.999999999...?

The fractional form of 0.999999999... is 9/9 or 1.

2. How do you prove that 0.999999999... is equivalent to 1?

One way to prove this is to use the concept of infinite geometric series. We can represent 0.999999999... as the sum of the infinite series 0.9 + 0.09 + 0.009 + ... which can be simplified to 9/10 + 9/100 + 9/1000 + ... Using the formula for the sum of an infinite geometric series, we get 9/10 / (1-1/10) = 9/10 x 10/9 = 1.

3. Is 0.999999999... a rational or irrational number?

0.999999999... is a rational number because it can be expressed as a ratio of two integers, 9/9 or 1.

4. Why is the concept of 0.999999999... often confusing?

The concept can be confusing because we are used to thinking of numbers as finite and the idea of an infinite number of 9's seems counterintuitive. Additionally, the decimal representation of 0.999999999... does not terminate or repeat, which can be difficult to grasp.

5. Can 0.999999999... be represented in any other forms besides decimal and fractional?

Yes, it can also be represented in other forms such as a repeating decimal (0.9̅) or a limit of a series (lim n→∞ 9/10 + 9/100 + 9/1000 + ... + 9/10n).

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