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Q. Let $2^k$ be the highest power of $2$ in the set $\{1, 2, ..., n\}$. Show $2^k$ does not divide any other element in the set.

The proof is:

If $2^k$ divides any other element of the set, this implies that there is a multiple $d2^k$, $(d>1)$ which is $\leq n$ .

Thus, $$2^{k+1} \leq d2^k \leq n$$ In which case, $2^{k+1}$ would be the highest power of 2 which is a contradiction.

My misunderstanding is where the $2^{k+1}$ comes from. Can someone explain? How does that value lower bound the $d2^k$. I understand that the $d$ is greater than $1$ and thus must be at least $2$ but I'm not seeing how that follows through.

Mathstatsstudent
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    Since $d>1$ we have $2\leq d$, hence $2^{k+1}=2\cdot 2^k\leq d\cdot 2^k$. – Dave Apr 28 '19 at 02:47
  • oh so basically $2^k$ either divides another element in the set or is not the highest power of $2$ in the set and we know by hypothesis that it is so therefore cannot divide another element? Is my reasoning correct? – Mathstatsstudent Apr 28 '19 at 02:49
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    Well the proof you give is the reasoning: suppose $2^k$ is the highest power of $2$ in the set, and we assume by contradiction that $2^k$ divides another element in the set. Then we can show that this implies that $2^{k+1}$ is also in the set, which contradicts our assumption. – Dave Apr 28 '19 at 02:52
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    This is applied in this popular answer. – Bill Dubuque Apr 28 '19 at 03:00
  • ... and less trivially in the purely arithmetical proof that binomial coefficients are integers. – Bill Dubuque Apr 28 '19 at 03:15
  • Alternative proof : the set contains of positive integers smaller than $2^k$ , apart from $2^k$ itself , but a positive integer smaller than $2^k$ can of course not be dividisible by $2^k$. – Peter Apr 29 '19 at 11:17

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