Given a finite set of real numbers A, not containing 0 and 1 and possessing the property: if the number a belongs to A, then numbers 1/a and 1-a also belong to A. How many numbers are in the set A?
Performing either of those operations twice in a row obviously just returns you to the original number, so we just need to check how many new numbers can be produced by alternating the operations.
So given a number n that is in A, we have the following that belong to A:
Start: n
Subtr: (1-n)
Recip: 1 / (1-n)
Subtr: 1 - 1/(1-n) = n/(n-1)
Recip: (n-1)/n
Subtr: 1 - (n-1)/n = 1/n
Recip: n
So we've completed the loop. And note that if instead we started with the reciprocal operation, it would just produce the same values in reverse order (since we ended with n after performing the reciprocal).
So if a given number n is in A, then there are up to 5 additional related numbers also in A (for n=2 some of those values are not distinct).
What I'm not clear about is how we determine how many numbers are in the set altogether, e.g. A might just be the set (2, -1, 0.5) in which case it has 3 numbers. Or A could be the set (3, -2, -1/2, 3/2, 2/3, 1/3) in which case it has 6 numbers. Or it could be the union of those sets, for cardinality 9. And it seems like you could just add arbitrarily many "seed" numbers, each of which would add up to 6 more members to the set.
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Posted by tomarken
on 2021-04-19 09:23:32 |
Partial solution?
