This seems too strong to me (though maybe I am misinterpreting Tom's claim). For instance, the colors of 1/1, 2/1, and 3/2 would have to be distinct. So there would be some restrictions even if the coloring isn't uniquely determined. My guess is that, in the absence of 1/0, each interval [n,n+1] would introduce a new bit's worth of freedom. Jim Propp On Sunday, April 17, 2016, Tom Karzes <karzes@sonic.net> wrote:
Actually, I can make a stronger statement: If you don't consider (1/0) to be a rational (for the purposes of the coloring restrictions), then you can assign *any* color to the integers > 1 to obtain an alternate coloring of the rationals > 1, since none of them would be subject to any color restrictions.
Tom
Tom Karzes writes:
This statement is true for rationals between 0 and 1. You cannot obtain rationals greater than 1 if you only consider the half of the s-b tree between 0/1 and 1/1.
For rationals greater than 1, it still works if you throw (1/0) into the mix (i.e., treat it as a rational) and color it green. But if you don't consider (1/0) to be a rational, then you could break the pattern by coloring both (0/1) and (1/0) the same color, e.g. red. And you could then choose any color for (1/1), but if you force it to blue, then you get the same color for any a/b and b/a, as opposed to red/green reversal if you allow (1/0) as a rational.
Tom
James Propp writes:
For uniqueness, one may cite the theory of Farey fractions and the Stern-Brocot tree. Every rational can be obtained by an iterative process of inserting mediants, starting from 0/1 and 1/1. This means that at each stage one has no choice (that is to say, that one has 1 choice).
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