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I'm self-studying the Concrete Mathematics book and I'm trying to work through the following problem:

Consider the following statement:

$ P(n): x_1 ... x_n \leq \left( \frac{x_1 + ... + x_n}{n}\right)^n$ , if $ x_1, ..., x_n \geq 0 $

This is true when $n = 2$, since $(x_1 + x_2)^2 - 4x_1x_2 = (x_1 - x_2)^2 \geq 0$.

a) By setting $x_n = \frac{x_1 + ... + x_{n-1}}{n-1}$, prove that $P(n) \implies P(n-1)$ whenever $n > 1$.

b) Show that $P(n)$ and $P(2)$ imply $P(2n)$

c) Explain why this implies the truth of $P(n)$ for all $n$.

I've already done the first two parts, but I'm trying to reason about why (c) is true. I guess it's because the question asks us to explain why this implies truth for all $n$ instead of prove it, but I'm not satisfied with the book's answer. I've attempted to come up with my own rigorous proof to convince myself of this assertion inspired by this post. I'm wondering, would this be a valid proof by contradiction to show that $P(n)$ is true for all $n$?

Proof by Contradiction

Assume there is a non-empty set $S = \lbrace s \in \mathbb{N} | P(s) \text{ is false } \rbrace $.

Due to the well-ordering principle, we know that $S$ must have a minimum if it is non-empty. Let's take the minimum to be $s_0 = \text{min}(S)$. Because we know the base case P(2) is true, $s_0 > 2$.

Since $s_0$ is the minimum of set $S$, $\forall x \in [2, n), P(x)$ is true.

Without loss of generality, assume that $s_0$ is odd as it can be easily proven if $s_0$ is even. $P \left( \frac{s_0+1}{2} \right)$ is true since $\frac{s_0+1}{2} < s_0$ for $s_0 > 1$ and we've established that $s_0 > 2$.

Since $ P(n) \land P(2) \implies P(2n)$, $P(\frac{s_0+1}{2}) \land P(2) \implies P(s_0 + 1)$. Thus, $P(s_0+1)$ must be true.

Because $ P(n) \implies P(n-1)$ when $n > 1$, $P(s_0+1) \implies P((s_0 + 1) - 1) = P(s_0)$. Thus, $P(s_0)$ must be true.

However, this contradicts our assumption that there is a minimum of $S$, $s_0 = \text{min}(S)$. Thus, if there is no minimum of $S$, $S = \emptyset$ and there are no $x \in \mathbb{N}$ such that $P(x)$ is false.

vegeta1
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  • Your proof is fine, but a little overboard for the task. A simpler proof would be as follows: Suppose that there exists some $m$ for which $P$ doesn't hold. Then, let $n$ be the smallest such natural. There are two such cases: If $n$ is even, then $P(n/2)$ is true by minimality, and hence, $P(n)$ is true, a contradiction. If $n$ is odd, then $P((n+1)/2)$ is true since $n>1$ and hence, $n>(n+1)/2$. So, $P(n+1)$ is true, and so, $P(n+1-1)=P(n)$ is true, a contradiction. – Rushabh Mehta Jan 23 '21 at 21:39
  • Why do you claim that it can be easily proved if $s_0$ is even. – José Carlos Santos Jan 23 '21 at 21:40

1 Answers1

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Your argument is correct, if a bit over-complicated. And I would not say that you can assume without loss of generality that $s_0$ is odd: the argument in the even case is pretty trivial, but an argument is required.

Alternatively, one can observe that it follows from (b) by a trivial induction that $P(2^n)$ holds for all $n\in\Bbb Z^+$, and it follows from (a) by an equally easy (downward) induction that if $2\le m\le n$, and $P(n)$ holds, then $P(m)$ holds. Since for each $m\ge 2$ there is an $n\in\Bbb Z^+$ such that $m\le 2^n$, the result follows.

Brian M. Scott
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