Ordering on 1-Based Natural Numbers is Compatible with Addition

Theorem

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

Let $+$ denote addition on $\N_{>0}$.

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

Then:

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

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

Proof

\(\ds a\)

\(<\)

\(\ds b\)

\(\ds \leadsto \ \ \)

\(\ds \exists c \in \N_{>0}: \, \)

\(\ds a\)

\(=\)

\(\ds b + c\)

Definition of Ordering on $1$-Based Natural Numbers

\(\ds \leadsto \ \ \)

\(\ds a + n\)

\(=\)

\(\ds \paren {b + c} + n\)

Definition of Binary Operation

\(\ds \)

\(=\)

\(\ds b + \paren {c + n}\)

Natural Number Addition is Associative

\(\ds \)

\(=\)

\(\ds b + \paren {n + c}\)

Natural Number Addition is Commutative

\(\ds \)

\(=\)

\(\ds \paren {b + n} + c\)

Natural Number Addition is Associative

\(\ds \leadsto \ \ \)

\(\ds a + n\)

\(<\)

\(\ds b + n\)

Definition of Ordering on $1$-Based Natural Numbers

$\blacksquare$

Sources

• 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $2.2$: Theorem $2.11 \ \text{(i)}$