How Does the Axiom of Archimedes Prove Integer Density in Real Numbers?

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Discussion Overview

The discussion centers on the application of the Axiom of Archimedes to demonstrate the existence of integers within any real number interval, specifically proving that for any real number \( x \), there exists an integer \( p \) such that \( p \leq x \leq p + 1 \). The scope includes mathematical reasoning and exploration of foundational principles in real analysis.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests using the Axiom of Archimedes and the Principle of Well Order to show that for any real number \( x \), there exists an integer \( p \) such that \( p \leq x \leq p + 1 \).
  • Another participant defines the set \( S \) as the collection of integers \( n \) such that \( n \geq x - 1 \) and argues that this set is nonempty, citing the Axiom of Archimedes to support their reasoning.
  • A participant questions the conclusion regarding \( n \geq x - 1 \), indicating a lack of understanding about this step in the argument.
  • A follow-up response clarifies that the definition of \( S \) is not a conclusion but a setup for showing that \( S \) has a minimal element, which leads to the desired inequality involving \( p \).

Areas of Agreement / Disagreement

Participants express some confusion regarding the reasoning behind the definition of the set \( S \) and its implications, indicating that the discussion remains unresolved on this point.

Contextual Notes

The discussion does not resolve the assumptions underlying the definition of the set \( S \) or the application of the Axiom of Archimedes, leaving these aspects open for further exploration.

Julio1
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Show that $(\forall x\in \mathbb{R})(\exists p\in \mathbb{Z}):\, p\le x\le p+1.$Hello :). The Hint is use the Axiom of Archimedes and the Principle of Well Order
 
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Julio said:
Show that $(\forall x\in \mathbb{R})(\exists p\in \mathbb{Z}):\, p\le x\le p+1.$Hello :). The Hint is use the Axiom of Archimedes and the Principle of Well Order

Hi Julio,

Let $S$ be the set of all $n\in \Bbb Z$ such that $n \ge x - 1$. Then $S$ is nonempty since if $x \le 1$, $1\in S$ and if $x > 1$, the Axiom of Archimedes gives a $k\in \Bbb N$ such that $k \cdot \tfrac{1}{x + 1} > 1$ (so $k\in S$). Hence, by the Principle of Well Order, $S$ has a minimal element, $p$. Thus $p - 1 < x - 1 \le p$, which implies $p \le x \le p + 1$.
 
Thanks :)

But how conclude that $n\ge x-1$? I don't understand :(
 
Julio said:
Thanks :)

But how conclude that $n\ge x-1$? I don't understand :(

It's not a conclusion. I let $S = \{n\in \Bbb Z\,:\, n \ge x - 1\}$. Then I showed $S$ has a minimal element $p$, which satisfies $p \le x \le p + 1$.
 

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