Multiplication on 1-Based Natural Numbers is Cancellable for Ordering
Theorem
Let $\N_{> 0}$ be the $1$-based natural numbers.
Let $\times$ be multiplication on $\N_{>0}$.
Let $<$ be the strict ordering on $\N_{>0}$.
Then:
- $\forall a, b, c \in \N_{>0}: a \times c < b \times c \implies a < b$
- $\forall a, b, c \in \N_{>0}: a \times b < a \times c \implies b < c$
That is, $\times$ is cancellable on $\N_{>0}$ for $<$.
Proof
By Ordering on $1$-Based Natural Numbers is Trichotomy, one and only one of the following holds:
- $a = b$
- $a < b$
- $b < a$
Let $a \times c < b \times c$.
Suppose $a = b$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication:
- $a \times c = b \times c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a \times c < b \times c$.
Similarly, suppose $b < a$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication:
- $b \times c < a \times c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a \times c < b \times c$.
The only other possibility is that $a < b$.
So
- $\forall a, b, c \in \N_{>0}: a \times c = b \times c \implies a < b$
and so $\times$ is right cancellable on $\N_{>0}$ for $<$.
Let $a \times b < a \times c$.
Suppose $b = c$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication:
- $a \times b = a \times c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a \times b < a \times c$.
Similarly, suppose $c < b$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication:
- $a \times c < a \times b$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a \times b < a \times c$.
The only other possibility is that $b < c$.
So
- $\forall a, b, c \in \N_{>0}: a \times b < a \times c \implies b < c$
and so $\times$ is left cancellable on $\N_{>0}$ for $<$.
From Natural Number Multiplication is Commutative and Right Cancellable Commutative Operation is Left Cancellable:
- $\forall a, b, c \in \N_{>0}: a \times b = a \times c \implies b = c$
So $+$ is both right cancellable and left cancellable on $\N_{>0}$.
Hence the result.
$\blacksquare$
Sources
- 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $2.2$: Theorem $2.14 \ \text{(ii)}$