**Theorem** For any [[Real numbers]] $r$, there exists $n \in \mathbb{N}$ such that $n>r$. **Remark** **So** [[There are no infinitesimal real numbers]]. **Proof** BWOC suppose $\exists r\in\mathbb{R}$ such that $n<r$ for all $n \in \mathbb{N}$ (i.e. $\mathbb{N}$ is [[Bound of Set of Reals|bounded above]]). Clearly $\mathbb{N}$ is non-empty so, by [[Real numbers|LUBA]], the set has a supremum $R$. Note that $n \in \mathbb{N} \implies n+1 \in \mathbb{N}$ so for every $n \in \mathbb{N}$, $n+1\leq R \implies n\leq R-1$. So $R-1$ is an upper bound for $\mathbb{N}$ which contradicts the fact that $R$ is the least upper bound, so our original assumption that $\mathbb{N}$ is bounded cannot be true. Corollary () Given $x\in \mathbb{R}$ with $x >0$, $\exists n \in \mathbb{N}$ such that $\frac{1}{n} <x$. **Proof** By AP, $\exists n \in \mathbb{N}$ such that $n > \frac{1}{x}$. Note that $\frac{x}{n}$ is also positive so multiplying both sides gives $\frac{1}{n}<x$. # Application - [[There are no infinitesimal real numbers]].