Let $\mathbb{N}=\{1,2,3,\dots\}$ be equipped with its usual strict order $<$, and write $n+1$ for the successor of $n \in \mathbb{N}$. Then, for every $n \in \mathbb{N}$, there is no $k \in \mathbb{N}$ such that
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\begin{align*}
n<k<n+1.
\end{align*}
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Moreover, if $n \in \mathbb{N}$ and $n \ne 1$, then there exists $m \in \mathbb{N}$ such that $n=m+1$; for this $m$, there is no $k \in \mathbb{N}$ such that