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<P><FONT size=2><BR></FONT><FONT size=3 face=Arial>As promised, I've checked
754. The final sentence of 5.2 (Decimal exponent calculation)
has:<BR><BR> If x is infinite, Q(x) is +∞.<BR><BR>Therefore the
result of quantum(x) where x is infinite should be +∞ (not x).
This is consistent in that if x is finite the result is also not affected by the
sign of x.<BR><BR>For when x is qNaN or sNaN I can think of no reason why the
results would be different from any other single-argument function or
general operation.<BR><BR>Mike </FONT></P></BODY></HTML>