theorem Th57:
  a>0 implies (1/a) #Q p = 1/a #Q p
proof
  assume
A1: a>0;
  thus (1/a) #Q p = (denominator(p)) -Root (1/a #Z numerator(p)) by Th42
    .= (denominator(p)) -Root (a #Z (-numerator(p))) by Th41
    .= (denominator(p)) -Root (a #Z numerator(-p)) by RAT_1:43
    .= a #Q (-p) by RAT_1:43
    .= 1/a #Q p by A1,Th54;
end;
