theorem
  a>=0 & b>0 & n>=1 implies n -Root (a/b) = n -Root a / n -Root b
proof
  assume that
A1: a>=0 and
A2: b>0 and
A3: n>=1;
  thus n -Root (a/b) = n -Root (a*b")
    .= n -Root a * n -Root (b") by A1,A2,A3,Th22
    .= n -Root a * n -Root (1/b)
    .= n -Root a * (1/n -Root b) by A2,A3,Th23
    .= n -Root a * 1 / n -Root b
    .= n -Root a / n -Root b;
end;
