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.