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I wrote a proof for this but am not completely sure if this is valid. Specifically, can I go from $m\leq k+0.5$ to $m\leq k$ because $m,k\in\mathbb Z^+$ with no further explanation? Any feedback would be greatly appreciated!

Proof. Let $n=\lfloor\log_2(2k+1)\rfloor$. Then, $$\begin{align*} n&\leq\log_2(2k+1)\lt n+1&&\text{(since $\forall x\in\mathbb R$, $\lfloor x\rfloor\leq x\lt x+1$),}\\ \implies2^n&\leq2k+1\lt2^{n+1}&&\text{(by algebra).}\\ \end{align*}$$ Observe that $2^n\in2\mathbb Z^+$ so $2^n=2m$ for some $m\in\mathbb Z^+$. Thus, $$\begin{align*} 2m&\leq2k+1&&\text{(since $2m=2^n\leq2k+1$),}\\ \implies m&\leq k+0.5&&\text{(by algebra),}\\ \implies m&\leq k&&\text{(since $m,k\in\mathbb Z^+$),}\\ \implies 2m&\leq2k&&\text{(by algebra),}\\ \implies 2^n&\leq2k&&\text{(since $2^n=2m$),}\\ \implies 2^n&\leq2k\lt2^{n+1}&&\text{(since $2k+1\lt2^{n+1}\rightarrow2k\lt2^{n+1}$),}\\ \implies n&\leq\log_2(2k)\lt n+1&&\text{(by algebra),}\\ \therefore n&=\lfloor\log_2(2k)\rfloor&&\text{(since $\forall x\in\mathbb R$, $\lfloor x\rfloor\leq x\lt x+1$).} \end{align*}$$ $$\tag*{$\blacksquare$}$$

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  • $\begingroup$ In your latter paragraph it seems you can just go from line $1$ to lines $4/5$ $\endgroup$ – David P Dec 3 '19 at 21:29
  • $\begingroup$ Your proof looks good to me! $\endgroup$ – Bram28 Dec 3 '19 at 21:37
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Note that for all $n \in \mathbb{N_{>0}} $ there exist $m \in \mathbb{N}$ such that \begin{eqnarray*} 2^m \leq n < 2^{m+1}. \end{eqnarray*} This value $m$ is $\lfloor\log_2(n)\rfloor$ And $n$ increases passed a power of $2$ the value $m$ will increase by $1$.

If $n$ increases (by $1$) from an even value to an odd value the $m$ will not change (provided $n>1$). So \begin{eqnarray*} \lfloor\log_2(2k+1)\rfloor = \lfloor\log_2(2k)\rfloor. \end{eqnarray*}

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