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?

  • $\begingroup$ See here $\endgroup$ – user1981 Jun 2 '14 at 18:09
  • $\begingroup$ 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 $S$. $\endgroup$ – André Nicolas Jun 2 '14 at 18:15
  • $\begingroup$ @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.\ \ $ $\endgroup$ – Gone Jun 2 '14 at 19:24

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.