Addition on 1-Based Natural Numbers is Cancellable
Theorem
Let $\N_{> 0}$ be the $1$-based natural numbers.
Let $+$ be addition on $\N_{>0}$.
Then:
- $\forall a, b, c \in \N_{>0}: a + c = b + c \implies a = b$
- $\forall a, b, c \in \N_{>0}: a + b = a + c \implies b = c$
That is, $+$ 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 Addition:
- $a + c < b + c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a + c = b + c$.
Similarly, suppose $b > a$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition:
- $b + c < a + c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a + c = b + c$.
The only other possibility is that $a = b$.
So
- $\forall a, b, c \in \N_{>0}: a + c = b + c \implies a = b$
and so $+$ is right cancellable on $\N_{>0}$.
From Natural Number Addition is Commutative and Right Cancellable Commutative Operation is Left Cancellable:
- $\forall a, b, c \in \N_{>0}: a + b = a + c \implies b = c$
So $+$ is both right cancellable and left cancellable on $\N_{>0}$.
Hence the result.
$\blacksquare$
Sources
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 4$: The natural numbers: $\text{A} 3$
- 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $2.2$: Theorem $2.12$