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.
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.
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.