I looked at the proof of Archimedean Property in several places and, in all of them, it is proven using the following structure (proof by contradiction), without much variation:
If $\space x \in \mathbb{R} \space,\space y \in \mathbb{R},$ and $x > 0$, then there is at least one natural number $n$ such that $nx > y$. $\bf{(*)}$
Proof:
Let $A$ be the set of all $nx$, where $n$ runs through the positive integers.
$$A=\left\{nx \space| \space n \in \mathbb{N}\right\}$$
If $\bf{(*)}$
were false, which is equivalent to suppose that: $$nx \leq y, \space \forall n \in \mathbb{N}$$ then $y$ would be an upper bound of $A$. But then $A$ has a least upper bound in $\mathbb{R}$. Put $\alpha = \sup A$. Since $x > 0$, $\alpha – x < \alpha$, and $\alpha – x$ is not an upper bound of $A$. Hence $\alpha – x < mx$ for some positive integer $m$. But then $\alpha < (m+1)x \in A$, which is impossible, since $\alpha$ is an upper bound of $A$.
Then, by contradiction, there exist an natural number n such that $nx > y$.
$\tag*{$\blacksquare$}$
The following is a bit of what I found. All of these sources prove the property by contradiction:
Mathematical Analysis – Browder
A First Course in Analysis – Donald Yau
Principles of Mathematical Analysis – Rudin
A Course in Calculus and Real Analysis – Ghorpade & Limaye
Elementary Real Analysis – Thomson, J. Bruckner & A. Bruckner
So, my question is: How to prove this theorem directly, without using proof by contradiction?
Best Answer
We can modify the proof to be a direct proof by defining $A$ as
$A = \{nx|n \in \mathbb N, nx \le y\}$.
Then we don't need say the contradiction is equivalent to this being unbounded; we can simply note that $y$ is an upper bound of $A$. However we have to consider if $A$ is empty. (which could happen if $y < x$.)
If $A$ is empty we are done. For any natural $n$ then $nx \not \in A$ so $nx > y$.
If $A$ is not empty .... then the proof continues as it was given.
($A$ has a $a = \sup A$ and $a-x$ is not an upper bound so there exists an $mx$ so that $a-x < mx \le a$ so $(m+1)x > a =\sup A $. No need to say "But that's impossible" as we made no contrary assumption to begin with. Instead we point out so $(m+1)x \not \in A$ so $(m+1) x > y$.)
The contradiction was unnecessary. But I think it may have been to ease the student into just how broad and powerful the least upper bound property actually is.