That is what I'm saying. Try both of those solutions with \\( m = 1, n = 0 \\).

Using Proposition 1.81, it suffices to prove \\( f(g(n)) \le n \le g(f(n)) \\) for all \\( n \in \mathbb{N} \\) with \\( f(n) = \lfloor \frac{n+1}{2}\rfloor \\) and \\( g(n) = 2n \\).

This gives \\( \lfloor \frac{2n+1}{2} \rfloor \le n \le 2 \lfloor \frac{n+1}{2} \rfloor \\) which checks out.