On 2016-11-08 08:45, mclaren wrote:
Technically speaking, we're also dealing with transcendental tuplets rather than irrationals here, since the square root of pi is a transcendental number (I think. Pi is known to be transcendental, but is the square of pi proven to be transcendental?).
Algebraic numbers form a field. So yes - if sqrt(pi) were algebraic, so would be pi. Obviously, they are also closed under root extraction (more precisely, powers to rational exponents); hence pi^2 is transcendental, too.
Cheers, Alexander _______________________________________________ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
