This may be completely dumb, but is it always possible to solve x/y = a/b

Thanks for the comments. I have realized that my point is that solving for x/y - a/b = c/d is a complete doddle, yet if we extract from x/y = a/b + c/d we arrive at x/y = (ad+bc)/bd, therefore bd=Yy and ad+bd=Yx, which does not yield as easily as one might hope for. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com]On Behalf Of Christian G. Bower Sent: 29 October 2003 18:26 To: math-fun Subject: Re: [[math-fun] Simple EF] "Jon Perry" <perry@globalnet.co.uk> wrote: +

c/d, where all algebraic quantities are positive integers, for all x and y?

Maybe I'm missing something here, but how about: x/y = x/(2*y) + x/(2*y) ? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun