An ordered field is a field together with a total ordering of its elements that is compatible with the field operations.
Assume that an ordered field like $(F,+,*,\le)$ is given.
Prove that from $1$, we can conclude $2$ .
- $\mathbb N_F$ is not bounded above.
- $\forall x \in F \space \space \space\exists n\in\mathbb Z_F \space n\le x \lt n+1$
Note : I know how to prove that for $x \gt 0$.
I define a set like $A:=\{m\in\mathbb N:x\lt m\}\subseteq \mathbb N_F$
$A$ has a minimum element like $m_0$ and i define $n:=m_0-1$
So we have :
$m_0 \in A \rightarrow x \lt m_0=n+1$
$m_0 -1 \lt m_0 \implies m_0 -1 \notin A$
$m_0 -1 \le x \implies n \le x$ $\space\space \space n \in\{0\}\cup\mathbb N_F$
So we are done.
What i'm stuck on is for $x \le 0$.
And by $\mathbb N_F$ i mean that every field like $F$ has $0_F$ and $1_F$ so we can define $\mathbb N_F$ inductively.
Let $x<0$ and $x\notin\mathbb{Z}$. Then $-x$ is positive, and your reasoning allows you to find $m\in\mathbb{N}$ such that $m<-x<m+1$. Then you get $-m-1<x<-m$. If $x$ is an integer, just take $n=x$.