Multiplication on 1-Based Natural Numbers is Cancellable

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Theorem

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

Let $\times$ be multiplication 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 $\N_{>0}$.


Proof

By Ordering on $1$-Based Natural Numbers is Trichotomy, one and only one of the following holds:

$a = b$
$a < b$
$b < a$


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 $\N_{>0}$.

From Natural Number Multiplication is Commutative and Right Cancellable Commutative Operation is Left Cancellable:

$\forall , b, c \in \N_{>0}: a \times b = a \times c \implies b = c$


So $\times$ is both right cancellable and left cancellable on $\N_{>0}$.

Hence the result.

$\blacksquare$


Sources

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