theorem
  for m,n being natural Number holds
  m^2 <= n < (m+1)^2 implies [\ sqrt n /] = m
  proof
    let m,n be natural Number;
    assume m^2 <= n;
    then
A1: sqrt (m^2) <= sqrt n by SQUARE_1:26;
    assume n < (m+1)^2;
    then
A2: sqrt n < sqrt ((m+1)^2) by SQUARE_1:27;
A3: sqrt (m^2) = m by SQUARE_1:def 2;
    sqrt ((m+1)^2) = m+1 by SQUARE_1:def 2;
    then sqrt n - 1 < m+1-1 by A2,XREAL_1:14;
    hence thesis by A1,A3,Def6;
  end;
