Archimedean Property Proof

WTS: $na>b⟺n>\frac{b}{a}$

Proof

WTS: $n>\frac{b}{a}$

• Note that $\frac{b}{a}$ is a real number So, we want to show that for any real number, there exists a natural number greater than it

Proof by contradiction: Assume for contradiction that $\exists x\in R$ such that $\forall n\in N,x\ge n$ Thus $N$ is bounded above.

• Note that $N$ is non-empty
• By the Completeness Axiom, $N$ has a supremum Then there must be a supremum $sup(N)=s$ Consider $s-1$. This is in the set $N$ as it is less than the supremum. Then, there must be a natural number greater than $s-1$ $\exists m\in N⟹m>s-1$ This means $m+1>s$ through re-arrangement. as $m\in N⟹m+1\in N$ and since $m+1>s$, we have shown that an element in the set $N$ is bigger than the supremum which is a contradiction. Thus $\exists n\in N,x<n$