$\forall n \in \mathbb{N}, 2 \leq n, \exists k \in \mathbb{N}$ such that $ 2^{k-1} \leq n \leq 2^{k}$

Question

I am confused on how to proceed on the Inductive step for this problem. (sorry in advance if my Latex is trash, im still learning) I.H: Assume that $P(2), P(3), ...... P(m)$ holds where $m \in \mathbb{N}$ I broke it down into two cases. case $1$: $m +1$ is even, case $2$: $m + 1$ is odd. case 1: By definition of $m + 1$ being even $m + 1 = 2 k_{1}$ ($k_{1} \in \mathbb{N}$ so $\frac{m+1}{2} = k_{1}$, and thus covered by I.H. so $2^{k} \leq m + 1 \leq 2^{k+1}$. However, I get stuck on the odd case as there can be composite odd numbers but those dont necessarily result in $2^{k-1} \leq m+1 \leq 2{k}$ form

Answer 1

Assume $P(m)$ so for some $k_0$ we have $2^{k_0-1}\leq m\leq 2^{k_0}$ then if $m+1\leq 2^{k_0}$ we are done since $2^{k_0-1}\leq m\leq m+1$ otherwise we had $m=2^{k_0}$ so $2^{k_0}\leq m+1\leq 2^{k_0}+1\leq 2^{k_0}+2^{k_0}=2^{k_0+1}$

Answer 2

Hint: The set $\{ m \in \mathbb N : 2^m \le n \}$ is finite and so has a maximal element, since it is not empty.