Proof for Dividing Fractions

Date: 01/18/2002 at 11:35:12
From: Cathy
Subject: Proof - dividing fractions

I am sure there is a proof for dividing fractions. Could you please 
send it to me?

Date: 01/18/2002 at 12:06:45
From: Doctor Peterson
Subject: Re: Proof - dividing fractions

Hi, Cathy.

Do you mean an explanation as to why we divide fractions the way we 
do? Try the Dr. Math FAQ:

   Dividing Fractions
   http://mathforum.org/dr.math/faq/faq.divide.fractions.html   

An actual proof will depend somewhat on what definitions and axioms 
you want to base the proof on. Several of the links at the bottom of 
the FAQ give informal demonstrations that could be turned into proofs. 

But let's try doing it algebraically. Just to make sure I don't 
accidentally assume what I'm proving, I'll write "DIV" for the 
division sign rather than "/" as usual, which sort of implies the 
connection between fractions and division. I'll suppose that we have 
defined a fraction as a pair "a/b" for which equality, addition, and 
multiplication have been defined in the standard ways.

We want to determine the value of a/b DIV c/d. First, we have to know 
how we are defining division. I would define it as the inverse of 
multiplication, so that

    x DIV y = z  if and only if  x = y * z

So our answer, a fraction x/y, must satisfy

    a/b DIV c/d = x/y  if and only if  a/b = c/d * x/y

By the definition of fraction multiplication,

    c/d * x/y = (cx)/(dy)

Therefore,

    a/b DIV c/d = x/y  if and only if  a/b = (cx)/(dy)

Now, how do we define equality of fractions?

    x/y = u/v  if and only if  xv = uy

So we can say

    a/b DIV c/d = x/y  if and only if  ady = bcx

and again, reinterpreting it as a different pair of fractions,

    a/b DIV c/d = x/y  if and only if  x/y = (ad)/(bc)

But the right side, by the definition of multiplication, is 
(a/b)(d/c); so we've found that

    a/b DIV c/d = (ad)/(bc) = (a/b)(d/c)

But this means that we divide by c/d by multiplying by d/c.

Is that the sort of proof you want?

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/