Clarifying Fraction Notation

Date: 07/02/97 at 00:04:39
From: Paul
Subject: HELP


The problem:

                      83 1/3 
    .83 1/3 =    ---------------------    =  ?
                        100

The answer is supposed to be

                5
               ---
                6

Please help me. My nephew is trying to stump me. Is this a trick 
problem?

Date: 08/29/97 at 14:57:06
From: Doctor Sonya
Subject: Re: HELP

Hi Paul:

It is hard to tell what this notation means. We don't usually include 
a fraction such as 1/3 after a decimal point. It confuses the meaning
of a decimal, a fraction whose denominator is a power of 10, not 10
and 3 as here.


The number of digits after the decimal place is the same as the power 
of 10 in the denominator.  For example, 3/10 = 0.3, 40/100 = 0.40, 
333/1000 = 0.333, etc. You could rewrite 0.83 1/3 as a series of 
decimals by separating the digits like this: 

     0.83 1/3 = 0.8 + 0.03 + 0.00 1/3

which can be written 

     8/10 + 3/100 + (1/3)/100. 

The sum of these terms will simplify correctly into the answer you 
want.

But there is a more direct way to solve your nephew's problem, and 
it involves a different interpretation of the notation. The "decimal" 
part of the number has two decimal places. Therefore, you can think 
of 0.83 1/3 as the compound fraction (83 1/3)/100, where the mixed 
fraction 83 1/3 is the numerator, and 100 is the denominator.

Rewrite the mixed fraction in the numerator as the compound fraction 
250/3. The expression now looks like (250/3)/100.

Dividing by 100 is the same as multiplying by 1/100. Therefore, 
(250/3)/100 is equivalent to (250/3) x 1/100. Multiply the two 
fractions: The product is 250/300 which reduces to 25/30 = 5/6. 

There's also an algebraic way to solve the problem, if you're 
interested. It is the most sophisticated, but also the most 
elegant. The number 1/3 as a decimal is 0.3333..., where the ellipsis 
after the 3 means that the 3s never stop coming. Then, write 
0.83 1/3 as 0.83333.... with lots of trailing 3s implied.

Let   x = 0.833333... Multiply both sides by 10:
    10x = 8.3333...

Subtract the first equation from the second and all the repeating 3s 
disappear.

   10 x = 8.3333...
    1 x = 0.8333...
  _______________
    9 x = 7.5 (Divide both sides by 9.)
      x = 7.5/9 = 75/90 = 5/6.

Whew! This was a fun problem!

-Doctors Chita and Sonya,  The Math Forum

 http://mathforum.org/dr.math/