## Repeating decimals represented as fractions

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Adam Preston
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### Repeating decimals represented as fractions

Hey, I have forgotten how to represent irrational numbers repeating decimals as fractions, I think you multiply it by 1000 then take a (a being the irrational integer) and get a/999. I think thats correct, if anyone could show me the algorithm/method for doing it; it would be much appreciated Last edited by gmalivuk on Thu Mar 24, 2011 4:50 am UTC, edited 1 time in total.
Reason: fixed the confusion between irrational numbers and repeating decimals
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joshz
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### Re: Irrational numbers represented as fractions

Uh...I may be missing something here, but the definition of an irrational number is that it cannot be expressed as a fraction with integer numerator and denominator. And saying "irrational integer" is similarly nonsensical--all integers are rational.
Could you clarify what you mean?
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Necklesspen
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### Re: Irrational numbers represented as fractions

He's either trying to be funny or talking about continued fractions.

MartianInvader
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### Re: Irrational numbers represented as fractions

Perhaps the OP is referring to the technique to represent a repeating decimal as a fraction? For example, if we have

$a = .496149614961...$

we can get the expression for [imath]a[/imath] by multiplying both sides of the equation by 10000:

$10000a = 4961.496149614961...$

$10000a - a = 4961$

$9999a = 4961$

$a = \frac{4961}{9999}$

But it's important to note that these numbers are not irrational. Indeed, using the technique above we can see that any repeating decimal can be expressed as a fraction, and so all repeating decimals are rational numbers.
Last edited by MartianInvader on Wed Mar 23, 2011 7:35 pm UTC, edited 2 times in total.
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Dopefish
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### Re: Irrational numbers represented as fractions

I suspect they meaning repeating decimals of the form [imath]0.\bar{a}[/imath], where a could be any number of digits long, say, n.

I believe in the case that a above is n digits long, it's fractional form would be [imath]\frac{a}{10^n-1}[/imath], with the recipe given in the OP being specific to the case that a is 3 digits long.

If the integer part of the number isn't 0, as in [imath]b.\bar{a}[/imath], then you could just rewrite it as [imath]b+ 0.\bar{a}[/imath] and then combine them with a common denominator.

edit: Beaten to it, but may as well post anyway.

Cleverbeans
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### Re: Irrational numbers represented as fractions

I believe he's looking at rationals with repeating decimal representations. I'll work under that assumption for now.

Lets say you have a rational number, and after a while it's decimal representation starts to repeat. For a simple example consider 1.5662323... where only the the 23s repeat. Since we're trying to solve for some unknown fractional representation, it's useful to set it up as x = 1.5662323..., and make some changes and monitor the value of x. The first step is to get the repeating portion just to the right of the decimal point, which gives 10^3 * x = 1566.2323... and then we multiply by another power of 10^2 to move the pattern over one cycle length which gives 1^5 * x = 156623.2323.... which gives two equations for x each having a decimal with the repeating decimal. Now we subtract them, giving (10^5 - 10^3) * x = (156623 - 1566) and now everything is an integer so we can solve for x as a fraction. To generalize this idea you'll have to consider two lengths, the first being "to the start of the pattern" and the second being "length of the pattern".

If you've seen some shenanigans with some integer over an integer with all 9s as digits, I would encourage you to use the above method on those special cases where the pattern starts right after the decimal to begin with, as it should become clear how you can solve the problem quickly in those cases.
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Adam Preston
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### Re: Irrational numbers represented as fractions

ye sorry I was kinda tired when i wrote it, what i meant is how would i convert a repeating decimal into a fractions. Sorry for everyone who misunderstood.
He who will not economize will have to agonize - Confucius

Micali
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### Re: Repeating decimals represented as fractions

You can convert them using a limiting sum as well.
For example, the decimal 0.3333....
You'd write:
3/10 + 3/100 + 3/1000 + ....
Then a=3/10, r=1/10
So you'd use the limiting sum formula: S=a/(1-r)
S=(3/10)/(1-(1/10))
S=1/3

Adam Preston
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### Re: Repeating decimals represented as fractions

Thanks alot guys, after reading this I immediately got the hang of doing this, just to check that I am correctly doing this I'll post a question on my homework and see if you guys can correct me on anything :

A simple one, I have to convert 0.43(recursive) into a fraction :

0.43, the 4 and 3 are recursive, so a = 0.43 (recursive).
so I multiplied them with 100 to get 100a = 43.43 (recursive)
then I did 100a - a = 43.43 - 0.43
so 99a = 43.00
then to get the value of a I did
43 / 99 = a
Is this correct?
He who will not economize will have to agonize - Confucius

Adam Preston
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### Re: Repeating decimals represented as fractions

I hope i got this technique correct as I needed to use it in my maths exam.
He who will not economize will have to agonize - Confucius

Yesila
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### Re: Repeating decimals represented as fractions

Adam Preston wrote:I hope i got this technique correct as I needed to use it in my maths exam.

The example you did looks good. I hope you did well (or will do well) on the exam! Good luck!

Adam Preston
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### Re: Repeating decimals represented as fractions

Thank you very much, I also wish I did well .
He who will not economize will have to agonize - Confucius

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