theorem Th30:
  0<=a & a<1 & n>=1 implies a <= n -Root a & n -Root a < 1
proof
  assume that
A1: 0<=a and
A2: a<1 and
A3: n>=1;
A4: now
    per cases by A1;
    suppose
      a>0;
      hence a |^ n >= 0 by Th6;
    end;
    suppose
      a=0;
      hence a |^ n >=0;
    end;
  end;
  now
    per cases by A1;
    suppose
      a>0;
      hence a |^ n <= a by A2,A3,Th14;
    end;
    suppose
      a=0;
      hence a |^ n <= a by A3,NEWTON:11;
    end;
  end;
  then n -Root (a |^ n) <= n -Root a by A3,A4,Th27;
  hence a <= n -Root a by A1,A3,Th19;
  n -Root a < n -Root 1 by A1,A2,A3,Th28;
  hence thesis by A3,Th20;
end;
