Square root and double rounding

[Samuel A. Figueroa]

Not having a copy of Kahan's lecture notes from his indoctrination course
a few years ago (which I understand contains material relevant to the subject
at hand), I have been trying to figure out how many significant digits double
precision needs to have in order for the square root of a number computed to
double precision (using the IEEE Standard's strict accuracy requirements) and
then rounded to single precision to yield the same value as if the square root
had been computed to single precision in the first place.  I have been able to
prove that if the significands of single precision numbers consist of p digits,
double rounding does not change the answer if the significands of double
precision numbers consist of at least 2p+2 digits.  Is it possible to prove
that double rounding cannot change the answer if the significands of double
precision numbers consist of less than 2p+2 digits?  (For some reason, I
thought the magical number 2p+1 was the lower bound for the number of digits
in double precision in order for double rounding to not be a problem with any
of the arithmetic operations.  Certainly for addition, 2p+1 is the lower
bound.)  I think, but haven't tried to prove, that I can come up with numbers
with p significant (binary) digits, for any p greater than 1, such that if
their square roots are computed to 2p+1 digits of accuracy and then rounded to
p digits, the final value will not be the same as if their square roots had
been computed to p digits in the first place.