1.4444444444444444444444444444444444444444444444444444444444444444444444444444444

Convert the following to fractional form in lowest terms. The answer is 13/9.
Can someone show me the steps to this. Thanks.

1.4 with a dot above the 4 to indicate and endless number of 4's.

Examples:
0.4 = 04/10 = 2/5.
3.68 = 368/100 = 184/50 = 92/25.
kadinAsked:
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d-glitchConnect With a Mentor Commented:

The general solution technique is to do whatever it takes to cancel the 
repeating decimal.


      Y  =  3.12727272727

Multiplying by 10 doesn't work because there are two repeating digits.
But multiplying by 100 does.

       100*Y  = 312.727272727
           Y  =   3.127272727
      ------------------------
        99*Y  = 309.6

           Y  = 3096/990   (which can be reduced further)

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d-glitchCommented:
Subtract    X = 1.4444     from    10*X = 14.44444
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sdstuberCommented:
1.444~  =  1.333~ +  0.111~
           
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Infinity08Commented:
As a rule, something of the form 0.xxxxxx... with an infinite amount of x'es (and x being any digit), it can be written in fractional form as x/9
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Infinity08Commented:
Sorry, d-glitch. If I would have seen your comment, I wouldn't have posted mine :)
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kadinAuthor Commented:
So your saying: 1.4/9 = 14/9 = 13/9 I remove a .1... any time I encounter and endless number?
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Infinity08Connect With a Mentor Commented:
>> So your saying: 1.4/9 = 14/9 = 13/9

No, because none of that is true ;)

I'm saying that if you have a number that looks like 0.xxxxxx..., with x any digit (x = 4 eg., which gives 0.444444...), then it can be written as x/9 in fractional form (4/9 eg.).


Have a look at d-glitch's comment. It might be easier to understand from that ...
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sdstuberCommented:
Maybe rearranging d-glitches post will make it clearer

   10x   =   14.4444
 -   x   =  - 1.4444
------      ---------
  ???    =     ???

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do both subtractions then solve for x
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sdstuberCommented:
oops!

sorry, too slow
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kadinAuthor Commented:
Thanks to all of you for your help. I understand it a little better which is good enough for now. I will likely come across this in the future and learn more then.
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