Archimedean_Property

Theorem

  1. Given any number $x \in \mathbb{R}$, there exists an $n \in \mathbb{N}$ satisfying $n > x$.

  2. Given any real number $y > 0$, there exists an $n \in \mathbb{N}$ satisfying $1/n < y$

Proof

Proceed via proof_by_contradiction. Assume, for contradiction, that $N$ is bounded above. By the axiom_of_completeness (AoC), $\mathbb{N}$ should then have a least upper bound, and we can set $\alpha = \sup\mathbb{N}$. If we consider $\alpha - 1$, then we no longer have an upper_bound, and therefore there exists an $n \in \mathbb{N}$ satisfying $\alpha - 1 < n$ But this is equivalent to $\alpha < n + 1$. Because $n + 1 \in \mathbb{N}$, we have a contradiction to the fact that $\alpha$ is supposed to be an upper_bound for $\mathbb{N}$. (Notice that the contradiction here depends only on the axiom_of_completeness and the fact that $\mathbb{N}$ is closed under addition.)