Ordering on 1-Based Natural Numbers is Compatible with Multiplication

Theorem
Let $\N_{> 0}$ be the $1$-based natural numbers.

Let $\times$ denote multiplication on $\N_{>0}$.

Let $<$ be the strict ordering on $\N_{>0}$.

Then:
 * $\forall a, b, n \in \N_{>0}: a < b \implies a \times n < b \times n$

That is, $<$ is compatible with $\times$ on $\N_{>0}$.