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In article <AGLEW.89Oct3123439@chant.urbana.mcd.mot.com> aglew@urbana.mcd.mot.com (Andy-Krazy-Glew) writes:
 > Eg.
 >     If I compute (x-y)*z
 >     and x-y produces a denorm,
 >     then, instead of relative precision related to 1/2^M, 
 > where there are M bits in the mantissa,
 >     do you not have relative precision related to 1/2^D, 
 > where there are D valid bits in the denormalized difference.
This is only true if the multiplication produces a denorm too.
Note that if underflow to zero occurs the relative precision would be 0.
Numerical analysis has always to take care of underflow and near underflow.
But the use of denorms simplifies matters in a number of cases (and does not
make it more difficult in other cases).  The major advantage of denorms is,
in my opinion, that statements like:
	if(x-y != 0.0) z = x/(x-y);
will not trap with a divide by zero trap but gives an (approximately) correct
result.
-- 
dik t. winter, cwi, amsterdam, nederland
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