theorem
  a>=0 & b>0 & n>=1 or b<>0 & n is odd implies
  n-root (a/b) = n-root a / n-root b
proof
  assume
A1: a>=0 & b>0 & n>=1 or b<>0 & n is odd;
 now per cases by A1;
    suppose
A2:   a>=0 & b>0 & n>=1;
      hence n-root (a/b) = n -Root (a/b) by Def1
        .= n -Root a / n -Root b by A2,PREPOWER:24
        .= n-root a / n -Root b by A2,Def1
        .= n-root a / n-root b by A2,Def1;
    end;
    suppose
A3:   b<>0 & n is odd;
      thus n-root (a/b) = n-root (a*(1/b))
        .= n-root a * n-root (1/b) by A3,Th11
        .= n-root a * (1 / n-root b) by A3,Th12
        .= n-root a / n-root b;
    end;
  end;
  hence thesis;
end;
