Ordering on 1-Based Natural Numbers is Compatible with Multiplication

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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 $\N_{>0}$.


Proof

\(\ds a\) \(<\) \(\ds b\)
\(\ds \leadsto \ \ \) \(\ds \exists c \in \N_{>0}: \, \) \(\ds a\) \(=\) \(\ds b + c\) Definition of $<$ on $\N_{>0}$
\(\ds \leadsto \ \ \) \(\ds a \times n\) \(=\) \(\ds \paren {b + c} \times n\) Definition of Binary Operation
\(\ds \) \(=\) \(\ds \paren {b \times n} + \paren {c \times n}\) Natural Number Multiplication Distributes over Addition
\(\ds \leadsto \ \ \) \(\ds a \times n\) \(<\) \(\ds b \times n\) Definition of $<$ on $\N_{>0}$: $c \times n \in \N_{> 0}$

$\blacksquare$


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