Ordering on 1-Based Natural Numbers is Total Ordering
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Theorem
Let $\N_{> 0}$ be the $1$-based natural numbers.
Let $<$ be the strict ordering on $\N_{>0}$.
Then $<$ is a (strict) total ordering.
Proof
From Ordering on $1$-Based Natural Numbers is Trichotomy we have that $<$ is trichotomy.
From Ordering on $1$-Based Natural Numbers is Transitive we have that $<$ is transitive.
From Trichotomy is Antireflexive it follows that $<$ is antireflexive.
It follows by definition that $<$ is a strict ordering.
By the trichotomy law it follows that $<$ is a strict total ordering.
$\blacksquare$
Sources
- 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $2.2$: Theorem $2.10$