theorem Th50:
  for n be Nat holds n>=1 & p = n" implies a #Q p = n -Root a
proof
  let n be Nat;
  assume that
A1: n>=1 and
A2: p = n";
  reconsider q=n as Rational;
A3: p=1/n by A2;
  then denominator(p)=numerator(q) by A1,RAT_1:44;
  then
A4: denominator(p) = n by RAT_1:17;
  numerator(p)=denominator(q) by A1,A3,RAT_1:44;
  then numerator(p) = 1 by RAT_1:17;
  hence thesis by A4,Th35;
end;
