theorem Th58:
  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;
  thus (a/b) #Q p = (a*(1/b)) #Q p .= a #Q p * (1/b) #Q p by A1,A2,Th56
    .= a #Q p * (1/b #Q p) by A2,Th57
    .= a #Q p / b #Q p;
end;
