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

 

[blue_raver22]

.

Hello! Thank you for your question. The property of density is a fundamental concept in mathematics that relates to the distribution of elements within a set. In this case, we are looking at the set of real numbers, denoted by $\mathbb{R}$. The statement provided states that for any real number $x$, there exists an integer $p$ such that $x$ lies between $p$ and $p+1$.

To prove this statement, we can use the Axiom of Archimedes and the Principle of Well Order. The Axiom of Archimedes states that for any two positive real numbers $x$ and $y$, there exists a positive integer $n$ such that $nx>y$. We can use this axiom to show that for any real number $x$, there exists an integer $p$ such that $p\le x$.

Let us assume that there is no such integer $p$ for a given real number $x$. This means that all integers are either greater than $x$ or less than $x$. This contradicts the Axiom of Archimedes, as there would be no positive integer $n$ that satisfies $nx>x$, since all integers are either greater or less than $x$. Therefore, there must exist an integer $p$ such that $p\le x$.

Similarly, we can use the Principle of Well Order to show that there exists an integer $p'$ such that $x\le p'+1$. The Principle of Well Order states that every non-empty set of positive integers has a least element. In this case, the set of positive integers greater than $x$ is non-empty and therefore has a least element, denoted by $p'+1$. This means that $x$ is less than or equal to $p'+1$.

Combining these two results, we can conclude that there exists an integer $p$ such that $p\le x\le p+1$. This proves the statement provided, using the Axiom of Archimedes and the Principle of Well Order.