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Prove that, for any positive integer $n,$ we must be able to find a permutation of the numbers $1, 2, . . . , n$ such that the mean of any two numbers in the permutation will not appear between these two numbers.

I hope that someone can help me. Thanks!

Bill Dubuque
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train
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Mar 31 '24 at 11:22
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    Aside from the lack of effort, it's not clear (to me) what you are asking. The mean of two distinct integers will always lie between the two, so what are you asking? Please edit for clarity. Including clear, explicit examples for small $n$ would help. – lulu Mar 31 '24 at 11:24
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    This is the original question:证明：对于任意正整数 n，我们一定能找到数 1, 2, . . . , n 的一个排列，使得在排列中这些数任何两个数的均值都不会 出现在这两个数之间。 – train Mar 31 '24 at 11:43
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    Thanks to Google Translate the problem actually reads "For any $n\in \mathbb N$ show that we can find a permutation of ${1, 2, \cdots, n}$ such that the mean of any two of those numbers does not fall between them." Thus, for $n=3$ we might take $132$. It is easily verified that this permutation satisfies the rule. $123$ would not work because the mean of $1, 3$ is $2$ and in this permutation, $2$ falls between $1,3$. – lulu Mar 31 '24 at 11:49
  • Yes! Your understanding is correct! Thanks! – train Mar 31 '24 at 11:59
  • Please update the question. As it stands, the body of your question is not the question you intended to ask. – Adam Rubinson Mar 31 '24 at 12:02
  • Thank you for your reminder! I have updated my question. – train Mar 31 '24 at 12:05
  • Please change it into English. – Adam Rubinson Mar 31 '24 at 12:08
  • The edited question now makes sense, but your post still contains no effort. What have you tried? Writing out "good" permutations for the first $6$ or $7$ cases seems like a good place to start. Note that deleting $n$ from a good permutation must result in a good permutation for $n-1$, so maybe there is a simple rule for adding $n$ in. Can every good permutation of $n-1$ be extended to a good permutation of $n$? – lulu Mar 31 '24 at 12:35
  • here is a duplicate, and here is a related problem (which refers to real numbers, not integers) – lulu Mar 31 '24 at 12:46

1 Answers1

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I am assuming you're asking

Let $n$ be a possitive integer , prove that the mean let's call it $m$ of $0<a,b <n$ , $m \notin \{1,2,...,n\}$ ,$\forall a,b \in\{1,2,...,n\}$

If that is what you mean it is clearly wrong

Let $n=10$ and we chose $a=1 , b=9 $ the mean of $a,b$ is $\frac{1+9}{2}=5$ and $5 \in \{1,2,3,...,10\} $

If I got it wrong please edit your post or make a comment with more details and/or explanations and I'll be happy to help you further!

Antony Theo
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  • The OP has put the original (Chinese) question in a comment, and it turns out that the question is actually asking about permutations. Specifically, we seek a permutation of the first $n$ integers such that the mean of any two does not fall between the two in the sequence. This was, of course, impossible to deduce from the actual post. – lulu Mar 31 '24 at 11:54
  • I'm sorry that I didn't express my meaning accurately and this is the original question:证明：对于任意正整数 n，我们一定能找到数 1, 2, . . . , n 的一个排列，使得在排列中这些数任何两个数的均值都不会 出现在这两个数之间。 – train Mar 31 '24 at 12:01