Convert repeating decimal to improper fraction

[oddjobmj]

Homework Statement

Convert 23.588 (the 88 is repeating) to an improper fraction.

Homework Equations

I don't have any.

The Attempt at a Solution

I'm not sure of the best way to go about this so I've taken a method that I've seen on some problems that were more simple as they didn't have the 23.5 in front of the repeating decimal.

I see that this can be broken down into:

23 + .5 + 8/10 + 8/10(1/10) + 8/10(1/10)^2...etc.

Although, I'm really not sure how to collapse the infinite progression of 8/10(1/10)^n into an actual number.

I've tried plugging 8/10 in for 'a' and 1/10 in for 'r' in the infinite progression equation:

a/(1-r)

This results in 8/9 then adding 1/2 to that we get 25/18 and then adding 23 to that I get 439/18

However, this result is not correct.

Any help/suggestions are welcome. Thank you for your time!

 

[jtyler05si]

you have:
23 + .5 + 8/10 + 8/10(1/10) + 8/10(1/10)^2...etc.

needs to be:
23 + .5 + 8/100 + 8/100(1/10) + 8/100(1/10)^2...etc.

This corresponds to a geometric series with a=8/100 and r=1/10

so the fraction would be the sum of the series (using a/(1-r)) plus 23.5

23.5+ (8/100)/(1-1/10)=23.5+8/90=47/2+8/90=2123/90

 

[oddjobmj]

Ohh, of course I'd make a silly mistake like that...

Thanks!

 

[brainy kevin]

There's an easier way, as a matter of fact.

Let a=23.5888... (I'm using ellipses to represent repeating decimals.)
Then 10a=235.888...
and 100a=2358.888...
Subtraction will get rid of our repeating decimal.
Therefore, 90a=2123
a=2123/90.

23.5888...=2123/90

I got this from Numbers, Rational and Irration by Ivan Niven, which I recommend. It doesn't use any advanced mathematics.

 

[oddjobmj]

How did you jump from 100a=2358.888 to 90a=2123?

 

[jtyler05si]

How did you jump from 100a=2358.888 to 90a=2123?

100a-10a


good call on the easier way.

 

[oddjobmj]

a=8/10

10*a=8

You'd obviously get 90a on the left but how do I know to use 10 and multiply by 100 and how does 2358.888888888888 - 8 = 2123?

 

[Mentallic]

a=23.5888...
not 8/10.

I have no idea where you got that from.

Anyway, if

a = 23.588...
10a = 235.88...
100a = 2358.88...

Notice now that if we do 100a-10a, the 8's repeating will cancel each other out.

2358.888...
-235.888...
=2123.00...
=2123

But 100a-10a=90a, so if 90a=2123 then a=2123/90.

Try it in the calculator

 

[Integral]

If you have a single digit repeating infinitely it can be reduced to finding the representation of .111...

You should be able to find that with a bit of playing.

 

[oddjobmj]

Awesome :D Thank you jtyler, brainy, mentallic, and integral!

These techniques will all help me significantly on my test/exam coming up.

 

[brainy kevin]

No problem. Glad to help.