All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 The Sum of the Parts Is Less Than And More Than the Parts (Posted on 2004-10-15)
Given a number of different fractions, create a new fraction whose numerator is the sum of all those fractions' numerators and whose denominator is the sum of the denominators. Call it y. Call the smallest of the original fractions x and the largest z.

Prove that for all cases, x < y < z.

 No Solution Yet Submitted by Victor Zapana Rating: 2.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Another Practical Use | Comment 10 of 11 |
(In reply to Practical use by Tristan)

If one wants to make a gear train to rotate a shaft at a different angular rate from another shaft, such adding of numerators and denominators is of use in getting a workable set of gears.  By the way, there's no real need for all the fractions to be different. Even one that's different assures the inequality.  In gear trains, also, we do not worry if the result falls slightly outside the bounds so even negative values are ok.   Here's an example.

Suppose we have a clock shaft that rotates once per day and want a shaft driven off it that rotates once every lunar month: 29.530588 days.  Most moon dials use a ratio of just 59/2, but we want to be more accurate.

The first thought is to use continued fractions, which yield the following sequence: 29/1, 30/1, 59/2, 443/15, 502/17, 1447/49, 23654/801, 25101/850, ... . The last four of these result in 29.52941..., 29.53061..., 29.5305867... and 29.5305882..., so the last could be considered good enough for our purposes.  However 25101 is 3*3*2789, so its use would necessarily entail using a gear with at least 2789 teeth--an impossibly large number.

But if we take 23654/801 and 25101/850, each as many times as necessary to get "good" results (in small integer values) we can get a number between the two fractions (which is good because one is too small and the other just slightly too big).  In fact the next value in the sequence of continued fractions would be just one such combination, but there are others that would not be quite as good as the next one in the continued fraction sequence, but good enough for the level of accuracy we were going for.

If we take 9 occurrences of 23654/801 and 4 occurrences of 25101/850, by adding the numerators and denominators we get 313290/10609, which comes out to about  29.5305872..., which we might consider good enough. Broken down into primes, this is 2*3*3*5*59*59/103*103.  Translated into a gear train, we might consider ratios of 2*59/103, 3*59/103 and 3*5*k/k, where the numerators are the number of teeth one of a pair of meshing gears and the denominators the other member of the pair. We'd have to make k sufficiently large, say 10, to have reasonable numbers of teeth on each side, or separate that pair into, say 60/20 and 90/18.

If nothing is suitable, or greater accuracy is sought, consideration would then shift to the next member of the continued fraction sequence, and work similarly from there.  This is an instance then where we might even accept a value outside the bounds of an already overly accurate pair.  And if the parity of the number of gear interfaces is wrong (a consideration for the direction of rotation), just add a pair with ratio 1/1.

An example of continuing on, and taking a negative number of times, would be if the continued fractions were taken to 124,058/4,201 and 149,159/5,051 (29.530587955... and 29.530588002... respectively).  The 149,159 itself is prime.  But if we take 3 occurrences of 124,058 and negative 1 of the 149,159/5,051, we get  223,015/7,552, which evaluates as 29.53058792..., which is good enough, though outside the bounds of the two fractions given. It factors as 5*13*47*73/(2*2*2*2*2*2*2*59).

Edited on October 16, 2004, 5:23 pm
 Posted by Charlie on 2004-10-16 16:54:51

 Search: Search body:
Forums (0)