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I am having trouble with the following problem in number theory.

Let $S$ be the set of first $n$ natural numbers and $m$ be the largest power of $2$ in the set $S$. We are required to show that m does not divide any element of $S$ other than $m$ and, using this, to further show that the sum of harmonic series to first $n$ terms is not an integer for any natural number $n$.

Would someone help me, please?

Shaun
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inshu
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  • See here –  Jun 02 '14 at 18:09
  • For the fact that $m$ does not divide any member of $S$ other than $m$, suppose to the contrary that $km\in S$ where $k\ge 2$. Since $2m\le km$, we conclude that $2m\in S$, contradicting the fact that $m$ is the largest power of $2$ in $S$. – André Nicolas Jun 02 '14 at 18:15
  • @Andre i.e. the next $\rm\color{blue}{multiple}$ of $,2^k$ is the next $\rm\color{blue}{power}\ 2^{k+1}!,,$ a property which characterizes $,2.\ \ $ – Bill Dubuque Jun 02 '14 at 19:24

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Solution for the first part:

The number $m$ is of the form $2^k$.
Now if any number is to be divisible by $m$ it should be of the form $2^k \cdot x$ where $x$ is an integer.
Now if $x$ is $1$ it is the number $m$ itself.
Whereas if $x \geq 2$ then there exists a higher power of $2$ than $k$ which is a contradiction.

Bart Michels
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Normie
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