Can you prove that it's safe to ignore irrational solutions? Here's a (non-optimal) irrational solution for f(3), using 5 component values: 1/3 = sqrt(2)/8 + (8 - 3*sqrt(2))/24 1/3 = (4 - sqrt(2))/8 + (3*sqrt(2) - 4)/24 1/3 = 1/3 1/2 = sqrt(2)/8 + (4 - sqrt(2))/8 1/2 = 1/3 + (8 - 3*sqrt(2))/24 + (3*sqrt(2) - 4)/24 Can we always do as well (or better) using only rational component values? I suspect the answer is yes. Note that the example above has the general form: 1/3 = a + (1/3 - a) 1/3 = (1/2 - a) + (a - 1/6) 1/3 = 1/3 1/2 = a + (1/2 - a) 1/2 = 1/3 + (1/3 - a) + (a - 1/6) This works for any value of "a" that doesn't produce zero or negative values. In particular, "a" can be given a rational value. Is it the case that any solution involving irrational values can be decomposed into something like this, which in turn can be converted into a rational solution? Tom Allan Wechsler writes:
I was challenging myself to try to think of any finite algorithm, be it ever so expensive and cumbersome, that would settle the question of whether p pieces suffices for a given n. A useful lemma would be that if it can be done in p pieces, then it can be done with pieces whose denominaters divide n!. (LCM(1,...,n) would be even stronger, but for my purposes it doesn't matter how weak this lemma is.)
Then, we could say, "Iterate over all partitions of n! into p parts, and for each such partition, check to see if it works, using the partition as the numerators of fractions whose denominators are all n!." This would be heinously expensive but it would at least provide an upper bound on how much computational work we have to do.
But I can't even prove the lemma.