What about the classic f_n(x) = x^n on [0,a] with 0 <= a <= 1?

If you use fractional exponents you can write the sixth root of 256 as (2^{8/6}) because 256 is 2^8 thus splitting it up you can write it as 2^{4/3} or 2^{3/3}*2^{1/3} i.e. 2*the cube root of 2. Its in simplest form because you have split it into an integer exponent and a fractional exponent.

Yes, but because it only needs to be positive semi-definite it restricts the how negative they can be, and gets harder in higher dimensions, i think if you had for example [[4,-2,-1],[-2,3,-1],[-1,-1,2]] that'd work

You can also just relog and it should pop up

Simple one but a gold rush is quite good.