
theorem Th22:
for a,b be Real, n be Nat st a < b holds
  a <= b-(b-a)/(n+1) < b & a < a+(b-a)/(n+1) <= b
proof
    let a,b be Real, n be Nat;
    assume a < b; then
A1: 0 < b-a & 0 < n+1 by XREAL_1:50; then
A2: (b-a)/(n+1) <= (b-a)/1 by NAT_1:11,XREAL_1:118; then
    a+(b-a)/(n+1) <= b by XREAL_1:19;
    hence a <= b-(b-a)/(n+1) < b by A1,XREAL_1:19,44;
    thus a < a+(b-a)/(n+1) <= b by A1,A2,XREAL_1:19,29;
end;
