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Irwin's Theory

Date: 07/06/98 at 08:25:39
From: David Sim
Subject: Add fractions without changing denominators

Hello Dr. Math,

I have a question about adding fractions. 

When you subtract 1/4 from 9/12, the answer is 1/2. Right?  
But according to Irwin's theory, he adds the numerators and the 
denominators together and also gets 1/2 without changing the 

So I just want to know under what conditions such fractions can be 
added using Irwin's theory, and how must I apply this theory?

     Thank you!

Date: 07/06/98 at 10:55:12
From: Doctor Rob
Subject: Re: Add fractions without changing denominators       

Let the fractions to be subtracted be a/b and c/d, with all variables
representing positive integers. Then, according to Irwin's Theory,

   a/b - c/d = (a+c)/(b+d)

Clear fractions, and you get

   a*d*(b+d) - b*c*(b+d) = b*d*(a+c)
   a*b*d + a*d^2 - b^2*c - b*c*d = a*b*d + b*c*d
   a*d^2 - 2*b*c*d -b^2*c = 0
   d/b = (c +- sqrt[c*(a+c)])/a

Thus it must be true that c*(a+c) is a perfect square,

   c*(a+c) = z^2

for some integer z > 0.  Then we can write

   c = e*y^2,  a + c = e*x^2

for some integers e > 0 and x > y > 0, and z = e*x*y, so

   a = e*(x-y)*(x+y)


   d/b = (e*y^2 +- e*x*y)/e*(x-y)*(x+y) = y*(y+-x)/(x-y)*(x+y)

Now the - sign in the numerator is impossible, since x > y > 0 and d/b 
is positive, so

   d/b = y/(x-y)

This implies that there is a positive integer f such that

   d = f*y,
   b = f*(x-y)

For simplicity, let w = x - y > 0, so x = y + w.  Rewrite everything 
in terms of e, f, w, and y:

   a = e*w*(w+2*y)
   b = f*w
   c = e*y^2
   d = f*y

You can check that for any positive integers e, f, w, and y, you get a
solution to the original equation:

   a/b - c/d = e*(w+2*y)/f - e*y/f = e*(w+y)/f
   (a+c)/(b+d) = e*(w+y)^2/f*(w+y) = e*(w+y)/f

Example: e = 2, f = 5, w = 3, y = 1. Then a = 30, b = 15, c = 2, d = 5

   30/15 - 2/5 = 32/20

which is a correct equation.

- Doctor Rob, The Math Forum
Associated Topics:
Middle School Fractions

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