That's correct. [-11 7>
Cool! Is that because Q(√5) is the smallest field that contains the noble numbers, and all elements of Q(√5) can be obtained as products and quotients of 0+√5, 1+√5, 2+√5, 3+√5, 4+√5, ... which are the primes of Q(√5)?Well, you'd now have a new series of primes involving the square root of five. In particular, 5 itself would now be a composite number, and √5 would map to 19827 half-tinas. That's one way to get half-tinas, I guess, but then you'd have to figure out what 1 + √5, 2 + √5, etc. correspond to.Dave Keenan wrote: ↑Wed Apr 01, 2020 4:10 pm Here's a wild thought. What if the odd half-tinas were not used to notate ratios, but to notate noble numbers? How could that be made to work?
Could it be, that to get the nobles, we only need the odd primes: 1+√5, 3+√5, 5+√5, ... ? That would be because nobles are of the form (i+mϕ)/(j+nϕ) where ϕ = (1+√5)/2 and i, j, m, n are whole numbers such that | in-jm | = 1.
[Edit: I have now answered these questions for myself (in the negative), and given a correct list of the primes of ℚ(√5) as needed for the nobles, here: Noble frequency ratios as prime-count vectors in ℚ(√5).]
I assume Q(√2) doesn't relate to the nobles, but to the "silver aristocratic" numbers (where the nobles are the gold aristocratics). Is that correct?Similarly, you can use the field Q(√2) and map √2 to 8539 half-tinas, I think.
I'd be happy to just have the nobles, as I understand they maximally avoid nearby simple ratios.