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

[Julio1]

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

 

[Euge]

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 $p \le x \le p + 1$.

 

[Julio1]

Thanks :)

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

 

[Euge]

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 $p \le x \le p + 1$.