'This may be completely dumb, but is it always possible to solve x/y = a/b + c/d, where all algebraic quantities are positive integers, for all x and y?' 'Isn't the general solution x = k(ad + bc), y = kbd ??' No. We get to x/y = (ad+bc)/bd. Let bd=ky, then ad+bc=kx But as x and y are pre-determined, we must select a k and therefore some partition of bd such that bd=ky, and with these values of k,b and d, then find an a and c (assuming that they exist for the chosen b,d and k values, which does not have a deterministic solution - unlike the simple rearrangement into x/y - a/b = c/d, where picking any valid a,b combo yields the values of c and d immediately. 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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

'x/y = (ad+bc)/bd. Let bd=ky, then ad+bc=kx' Additionally, both b and d must be less than kx. 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

Boiling the problem down a bit, if we view the rearrangement via minus as making use of the the group properties of the rationals, and we also gain the knowledge that for any given x,y, the range of valid a,b is mapped onto c,d, but we gain no insight as to the nature of this mapping. Approaching the problem via the addition method, we do gain an insight into the nature of the mapping, but the solutions seem more difficult to acquire. So the 'missing link' is in how we connect these two features of the same object together. 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

Extending previous ideas, I propose zop = zero or positive ("nonnegative") number, x>=0 zon = zero or negative number zopi=zero or positive integer zoni=zero or negative integer Zop could instead be zom (zero or more) and Zon could instead be zol . Usage: Let x be zopi (let x be zero or a positive integer) (zopi is an adjective), or let x be a zopi (zopi is a noun). Either would be ok, but no "a" makes more sense if you expand zopi. So far as I know, none of these putatively new words are previously used. They are all easily pronounced and unambiguous, and none sounds too silly. Steve Gray

On Friday, October 31, 2003, at 11:02 AM, Jon Perry wrote:

Does there always exist, for any x and y, a partition bd of ky for some k such that kx = ad+bc for some positive integers a and c?

Sylvester's coin problem is relevant here: given coins of values b and d which are relatively prime, you can make any amount of money greater than bd-b-d. (I call this the Houston Eulers problem: a football team with a perfect kicker always gets 3 or 7 points at a time, so can end the game with any score larger than 11.) You'll have to make kx larger, since you want a and c to both be positive, not just, er, zop. But if you make kx large enough, some a,c always exist, and then you can ensure positive ones by trading b d's for d b's.

Jon Perry

Jon: in 24 hours, you've just sent ten messages to this list each with at most one sentence of content. First, this borders on list abuse -- math-fun aims for high content-per-message. Second, even aside from this, your current practice will probably lead to no one reading your messages. To join the community, please pay more attention to the culture of this mailing list: think first, then post when you have something to say. --Michael Kleber kleber@brandeis.edu