• flyos@jlai.lu
      ·
      1 year ago

      Out of curiosity, I guess pi cannot be rational in any p-adic set, right?

      • silent_water [she/her]
        ·
        edit-2
        1 year ago

        yeah, the remainder of pi after p^2 for p>2 will always be the same as the remainder after division by p (p^2 if p=2), so it truncates to the left of the radix, and p-adic rationals can't go off infinitely to the right of the radix. also, more trivially, pi isn't algebraic so it's not the solution to any polynomial with rational coefficients, so it's also not the solution to any polynomial mod p^n.

        • flyos@jlai.lu
          ·
          1 year ago

          Not sure I understand all of it, but the last part about pi not being algebraic made sense to me, at least! Thanks!

          • silent_water [she/her]
            ·
            1 year ago

            what's the remainder of pi after division by 3 and 3^2? notice how the remainder is the same -- i.e. there's no 3s digit needed. and the same will be true for all higher powers of 3. this is because pi is <3^n for all n > 1. but we failed to express the fractional part of pi. if we extend to the p-adic rationals, we can express it as an infinite expansion but that's not allowed for p-adics -- they can only have a finite number of digits in the part to the right of the radix.