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 $\N_{>0}$: $c \times n \in \N_{> 0}$ |
$\blacksquare$
Sources
- 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $2.2$: Theorem $2.11 \ \text{(ii)}$