puzzle with bfloat and radcan
- Subject: puzzle with bfloat and radcan
- From: Richard Fateman
- Date: Mon, 14 May 2012 16:31:31 -0700
I hadn't thought about it so much, but it is not so hard to fool bfloat.
fpprec:30
bfloat(g(10^41+1)-g(10^41));
returns 0.0b0