How to prove the Archimedean property?

Question

The archimedean property states that

$$\boxed{~\forall~ ~a,b\in \mathbb{Z}^+~ \exists ~n~|~na\geq b~}$$

I started with disproving ..

Suppose $\forall ~\{n,a,b\} \subset \mathbb{Z}^+ , \text{na < b}$

$\implies b-na >0$

$\text{As this holds true for all n} \in \mathbb{Z}^+, \text{ let n = x}$

$\implies b-xa>0$

$\implies b-(x+1)a > 0\quad or\quad b-xa-a<0$

$b-ma>b-ma-a$

I dont know how to proceed any further. If i cancel b-ma, i get nothing but that a is positive. so how do we prove it. and also is there any false statement in my proof? Thank you!

Answer 1

How about $n = b$? Then $na = ba \geq b$.

Answer 2

The classical proof start from the lemma:

The set $\mathbb{N}$ of positive integers $\mathbb{N} = \{0, 1, 2, > \cdots\}$ is not bounded from above.

That can be proved by Peano axioms.

Now: $na\ge b$ is equivalent to : $n\ge \dfrac{b}{a}$ and, if such an $n$ does not exists than $n<\dfrac{b}{a}$ for all integers $n$ and this contradicts the lemma.