Composition of Addition Mappings on Natural Numbers
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Theorem
Let $a \in \N$ be a natural number.
Let $\alpha_a: \N \to \N$ be the mapping defined as:
- $\forall x \in \N: \map {\alpha_a} x = x + a$
Then:
- $\alpha_{a + b} = \alpha_b \circ \alpha_a$
Proof
\(\ds \alpha_{a + b}\) | \(=\) | \(\ds x + \paren {a + b}\) | Definition of $\alpha$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x + a} + b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map {\alpha_a} x} + b\) | Definition of $\alpha$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\alpha_b} {\map {\alpha_a} x}\) | Definition of $\alpha$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {\alpha_b \circ \alpha_a} } x\) | Definition of Composition of Mappings |
$\blacksquare$
Sources
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $7$