Square root and double rounding

[Samuel A. Figueroa]

It's good to see someone else confirm my results.  For me, one interesting
thing about this problem is that only 2p bits are required to handle cases
where the square root of a p-digit number is just barely greater than a
half-way case, but 2p+2 bits are needed to handle cases where the square
root is just barely less than a half-way case.  (I don't think this is
evident from Priest's proof, but if anyone is interested in seeing an
alternative proof, just let me know.)  By the way, can anyone confirm if
a proof for this problem (and for the other arithmetic operations) are
contained in Kahan's lecture notes?  (I don't have access to them.)
--
Sam Figueroa
figueroaacs.nyu.edu