Direct proof of Archimedean Property (not by contradiction)

Question

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$}$

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The following is a bit of what I found. All of these sources prove the property by contradiction:

Basic Real Analysis - Howland

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

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So, my question is: How to proof this theorem directly, without using proof by contradiction?

Answer 1

I think I can patch up Surb's idea (now deleted).

For simplicity, assume $y \ge 0$. Let $A = \{k \in \mathbb{N} : k \le y/x\}$. This set is bounded above by construction, and it is nonempty since $0 \in A$, so it has a least upper bound; call it $t$. If we can show $t$ is a natural number, then $n=t+1$ is the desired natural number.

Since $t = \sup A$, there exists $k_0 \in A$ with $t-1 < k_0 \le t$. Note that $k_0 \in \mathbb{N}$. Also, by definition of supremum, for any $0 < \epsilon < 1$ there exists $k \in A$ with $k > t-\epsilon$. However, then $|k-k_0| < 1$, and $k, k_0$ are both natural numbers, so $k=k_0$. Thus $k_0 > t-\epsilon$. As this is true for every $\epsilon > 0$, we must have $k_0 \ge t$. Hence $t = k_0$ and so $t \in \mathbb{N}$.

Answer 2

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.