Dividing Fractions

Date: 10/16/97 at 10:41:16
From: Jimmy De Armas
Subject: Pre-Algebra

I was reading over the explanation on dividing fractions and why 
we have to flip the second fraction, and I am still confused. 
I understand that we have to do it in order to do the problem, 
but I want to know the reason why we have to flip it. I want a 
simpler explanation.

Date: 10/16/97 at 12:12:36
From: Doctor Rob
Subject: Re: Pre-Algebra

In a fraction its value represents the number of things of size 
measured by the denominator you add up to get one thing of size
measured by the numerator. For example, 31/11 is the number of
11-pound objects you put together to to get one 31-pound object.
When the denominator is itself a fraction, as in your situation,
this does not change.

For example, (14/3)/(2/5) is the number of objects, each weighing
2/5 pound, which are put together to get a weight of 14/3 pounds.
How many does it take to make one pound? Answer: 5/2, each
weighing 2/5 pound, will make one pound. How did we get 5/2?  
By inverting 2/5, or, in other words, finding its "reciprocal."  
Why is this the right answer?  Because (5/2)*(2/5) = (5*2)/(2*5) 
= 10/10 = 1.  Then to get 14/3 pounds it will take (14/3)*(5/2)
objects, each weighing 2/5 pound. (Of course this equals 35/3,
so you'll need 11 and 2/3 objects each weighing 2/5 pound to make
14/3 pounds.)

There is nothing special about 14, 3, 2, and 5. They could be replaced 
by any four numbers - except zero: remember, you can't divide by zero!

Another way to look at this is to start with your original compound 
fraction, and multiply the numerator and denominator of the fraction 
by 5/2.  You get:

   14     14   5     14   5
   --     -- * -     -- * -
    3      3   2      3   2    14   5
  ---- = -------- = -------- = -- * -
    2      2   5        1       3   2
    -      - * -
    5      5   2

Why did we pick 5/2?  Because 5/2 is the reciprocal of 2/5, the
denominator, and when you multiply any number by its reciprocal,
you get 1, which is what we want to create in the denominator.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/