Archimedean Property

$\text{I can get arbitrary large natural numbers.}$ Let $x\in \mathbb{R},x>0$. Then $\forall y\in \mathbb{R},\exists n\in \mathbb{N}$ such that $nx>y$.

In particular (and weakly), $\forall y\in \mathbb{R},\exists n,m\in \mathbb{Z}$ such that $n>y>m$.

Proof

We note the second claim follows from the first. Indeed, let $x=1$. By AP, $\exists n\in \mathbb{N},n=n\cdot 1>y$. Repeat with $-y$ to get (with AP) some $m\in \mathbb{N}$ such that $m>-y⟺y>-m\in \mathbb{Z}$. So, $-m<y<n$ Now, we prove AP.

---|---|---|--- ... ----|--->
   x   2x  3x           y

Suppose for the sake of contradiction, that AP fails. This means $\forall n\in \mathbb{N},nx\le y$. So, $y$ is an upper bound for $nx$. Let

$$A:=\{nx\mid n\in \mathbb{N}\}$$

It is clear $y$ is a upper bound for $A$, and $A$ is bounded above. Then, $A\ne \varnothing$ as $x\in A,2x\in A$. This, by the Completeness Axiom, the $\sup (A)$ exists. Let $a:=\sup (A)$. Hence, $\forall n\in \mathbb{N},nx\le a$

In particular,

$$\begin{array}{rl}\forall n\in \mathbb{N},& (n+1)x\le a\\ \forall n\in \mathbb{N},& nx\le a-x\end{array}$$

Which is a contradiction of the definition of the Supremum because we have a smaller upper bound.

Corollary ($\mathbb{Q}$ is dense in $\mathbb{R}$)

Let $x,y\in \mathbb{R}$ with $x<y$. Then $\exists r\in \mathbb{Q}$ such that $x<r<y$.

• Think of squeezing in a rational number between any arbitrarily small interval of two real numbers.
• We want $r=\frac{m}{n}\in \mathbb{Q}$ such that $x<\frac{m}{n}<y$

WLOG, $n>0,m\in \mathbb{Z},n\in \mathbb{N}$, $x<\frac{m}{n}<y⟺nx<m<ny$. So our goal is to actually find $n\in \mathbb{N}$ such that $nx<m<ny$.

<----|----|----|---- ... ----|---|---|-->
    -1    0    1             nx  m   ny     

Proof:

We want $r=\frac{m}{n},n>0$ such that $x<\frac{m}{n}<y$. Note that $x<y⟹y-x>0$. So by AP, $\exists n\in \mathbb{N}$ such that $n(y-x)>1⟺ny-nx>1$. So BY RUDIN, $\exists m\in \mathbb{Z}$ such that $ny<m<nx$.