theorem Th56:
  a>0 & b>0 implies (a*b) #Q p = a #Q p * b #Q p
proof
  assume that
A1: a>0 and
A2: b>0;
A3: a #Z numerator(p) >= 0 by A1,Th39;
A4: b #Z numerator(p) >= 0 by A2,Th39;
  thus (a*b) #Q p = (denominator(p)) -Root (a #Z numerator(p) * b #Z numerator
  (p)) by Th40
    .= a #Q p * b #Q p by A3,A4,Th22,RAT_1:11;
end;
