Irwin's TheoryDate: 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 denominators. 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) Then 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 and 30/15 - 2/5 = 32/20 which is a correct equation. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/