theorem Th28:
  a>=0 & a<b & n>=1 implies n -Root a < n -Root b
proof
  assume that
A1: a>=0 and
A2: a<b and
A3: n>=1 and
A4: n -Root a >= n -Root b;
  (n -Root a) |^ n = a by A1,A3,Th19;
  then
A5: (n -Root a) |^ n < (n -Root b) |^ n by A1,A2,A3,Lm2;
  n -Root b > 0 by A1,A2,A3,Def2;
  hence contradiction by A4,A5,Th9;
end;
