1

Let $n$ be a non zero natural number.

We say that $n$ has the property $P$ if there exists a sequence of numbers $a_k$,$a_1$,$a_2$...$a_m$ strictly positive rational numbers (not necessarily distinct), we get : $\sum_{k=1}^m a_k$ $=$ $n$ and $\sum_{k=1}^m \frac{1}{a_k} = 1$

Question : Suppose $n$ has the property $P$.

Show that $N(n+N-1)$ too has the property $P$ for every non zero natural numbers $N$.

This probably is a very easy proof but i don't really know the correct answer to this because i'm completely new to Sums and Proofs.

Any hints/tips are appreciated.

I know that for the $P$ to work for $n$, $a_k$ has to be equal to $\sqrt n$ and $m$ has to be equal to $\sqrt n$. or maybe i'm wrong here too.

Mario SOUPER
  • 149
  • 1
  • 7
  • 1
    Surely, you've misstated the conditions. Perhaps you meant this: $$$$ We say a positive integer $n$ has property $P$ if there exists a finite sequence of positive rational numbers whose sum is $n$, and the sum of whose reciprocals is $1$. $$$$ – quasi Sep 19 '17 at 20:06
  • Yes yes, it's 'if there exists', sorry for the mistake, i forgot to mention that it's fine if $a_1$=$a_2$ and so on.. – Mario SOUPER Sep 19 '17 at 20:14
  • I edited it with proper conditions (i hope). – Mario SOUPER Sep 19 '17 at 20:16
  • The basic idea is to assume there is a set of $a$s for $n$ and produce a set of $b$s for $N(n+N-1)$. For example, assume there is a set of $a$s for $n$ and we want to show there is a set of $b$s for $N^2n$ I would point out that if we let $b_i=Na_i$ we would have $\sum_{k=1}^mb_k=nN$ and $\sum_{k=1}^m \frac 1{b_k}=\frac 1N$. Then if we take $N$ copies of each of the $b_i$ we get the sum to be $N^2n$ and the sum of the reciprocals to be $1$, so we have demonstrated a solution for $N^2n$ – Ross Millikan Sep 19 '17 at 21:06

1 Answers1

3

Let $S$ be the set of positive integers $n$ such that there exists a finite sequence of positive rational numbers whose sum is $n$, and the sum of whose reciprocals is $1$.

Let $n \in S$, and let $N$ be a positive integer.

We want to show $N(n+N-1) \in S$.

Since $n \in S$, there exist positive rational numbers $a_1,...,a_m$ such that \begin{align*} \sum_{i=1}^m a_i &= n\\[4pt] \sum_{i=1}^m \frac{1}{a_i} &= 1\\[4pt] \end{align*} Let$\;q = m + N -1,\;$and let$\;b_1,...,b_q\;$be defined by $$ b_i = \begin{cases} Na_i&\text{if$\;\,1\le i \le m$}\\[4pt] N&\text{if$\;\,m < i \le q$}\\[4pt] \end{cases} \\[4pt] $$ \begin{align*} \text{Then}\;\;\sum_{i=1}^q b_i &=\sum_{i=1}^m b_i +\sum_{i=m+1}^q b_i\\[4pt] &=\sum_{i=1}^m Na_i +\sum_{i=m+1}^q N\\[4pt] &=N\left(\sum_{i=1}^m a_i\right) +N\left(\sum_{i=m+1}^{m+N-1} 1\right)\\[4pt] &=N(n) + N(N-1)\\[4pt] &=N(n+N-1)\\[14pt] \text{and}\;\;\sum_{i=1}^q \frac{1}{b_i} &=\sum_{i=1}^m \frac{1}{b_i}+\sum_{i=m+1}^q \frac{1}{b_i}\\[4pt] &=\sum_{i=1}^m \frac{1}{Na_i}+\sum_{i=m+1}^q \frac{1}{N}\\[4pt] &=\frac{1}{N}\left(\sum_{i=1}^m \frac{1}{a_i}\right)+\frac{1}{N}\left(\sum_{i=m+1}^{m+N-1} 1\right)\\[4pt] &=\frac{1}{N}(1) + \frac{1}{N}(N-1)\\[4pt] &= 1\\[4pt] \end{align*} It follows that $N(n+N-1) \in S$, as was to be shown.

quasi
  • 58,772