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Definition. We call length of a positive integer $a\in \mathbb{Z}^+$, and we write $\ell(a)$, the number of $2$-adic digits of $a$. That is $\ell(a):=\lfloor \log _2 a\rfloor +1$.

I found on my textbook the following properties but without proof. So, I tried to prove the myself.

  1. $\ell (a)=k\iff 2^{k-1}\leq a<2^k$.
  2. We suppose that $a,b\in \mathbb{Z}^+$. Then, $a\leq b \implies \ell(a) \leq \ell(b)$. Is the inverse $``\Longleftarrow"$ direction true?
  3. Also, can we claim that $a=b\iff \ell(a)=\ell(b) ?$

My attempt. 1. We have that \begin{equation} \begin{split} 2^{k-1}\leq a<2^k & \iff k-1 \leq \log_2 a <k \\ &\iff \lfloor \log _2 a\rfloor =k-1 \\ & \iff k=\lfloor \log _2 a\rfloor +1=\ell(a). \end{split} \end{equation}

  1. We suppose that $a \leq b$. If $\ell(a)>\ell(b)$ then:

\begin{equation} \begin{split} \ell(a)>\ell(b) & \iff \lfloor \log_2 a \rfloor +1 > \log_2 b \rfloor +1\\ & \iff \lfloor \log_2 a \rfloor > \log_2 b \rfloor \\ & \implies \log_2 a > \log_2 b \\ & \iff a>b, \text{ contradiction. } \end{split} \end{equation}

Are these proofs correct? Are there other ways to prove these properties (especially the second) ?

Thank you in advance.

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    $\begingroup$ Your proof of 1 is correct, and the rest follows from there. Try to see that opposite implication is not true for 2. 1 should also tell you that 3 does not hold. $\endgroup$ – ms08030 Aug 24 '18 at 18:31
  • $\begingroup$ @ms08030 Thank you for your comment. Could you please write down the full answer? $\endgroup$ – Chris Aug 24 '18 at 20:55
  • $\begingroup$ @Chris You can't claim the $3^{rd}$ claim. $l(a)=l(b)$ does not implies $a=b$. $\endgroup$ – tarit goswami Aug 24 '18 at 21:12
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    $\begingroup$ @taritgoswami Thank you for your comment. A very simply thought is the following. If we take $a:=(1010)_2$ and $b:=(1111)_2$, then $a\neq b$, but $\ell (a) = \ell (b)$. Also, if $a=(a_1 \cdots a_k)=b=(b_1\cdots b_k)$ then $\ell(a)=\ell(b)=k$. Right? $\endgroup$ – Chris Aug 24 '18 at 21:17
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    $\begingroup$ Yeah, correct .. $\endgroup$ – tarit goswami Aug 24 '18 at 21:20
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The first one is OK.

That $a \le b$ implies $\ell(a) \le \ell(b)$ follows from the facts that $\log_2$ is a non-decreasing function and $\lfloor \cdot \rfloor$ is too:

$a \le b \implies \log_2(a) \le \log_2(b) \implies \lfloor \log_2(a) \rfloor \le \lfloor \log_2(b) \rfloor \implies \ell(a) = \lfloor \log_2(a) \rfloor + 1 \le \lfloor \log_2(b) \rfloor + 1 = \ell(b)$.

But $8 = (1000)_2$ and $15=(1111)_2$ so $\ell(15) = \ell(8) = 4$ while $15 \not \le 8$. This is a counterexample both to the reverse implication of the second as to the reverse implication of the third question.

What is true though is that $\ell(a) < \ell(b)$ implies $a < b$. (i.e the equal part is the problem in this reverse implication): Suppose that $\ell(b) = k$. This implies that $2^{k-1} \le b$ while $\ell(a) < \ell(b)$ implies $\ell(a) \le k-1$ so that $a < 2^{k-1}$, hence $a < b$ then holds.

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