theorem
  -f/g = (-f)/g & f/(-g) = -f/g
proof
  thus -f/g = (-f)/g by Th39;
  thus f/(-g) = f (#) ((-g)^) by Th38
    .= f (#) ((-1r) (#) (g^)) by Lm2,Th35,COMPLEX1:def 4
    .= -(f (#) (g^)) by Th18,COMPLEX1:def 4
    .= -(f/g) by Th38;
end;
