Dipper Operation is Commutative

Theorem

Let $m, n \in \Z$ be integers such that $m \ge 0, n > 0$.

Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:

$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$

The dipper operation on $\N_{< \paren {m \mathop + n} }$ is commutative.

Proof

Recall the definition of the dipper operation on $\N_{< \paren {m \mathop + n} }$ defined as:

$\forall a, b \in \Z_{>0}: a +_{m, n} b = \begin{cases}

a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$

where $k$ is the largest integer satisfying:

$m + k n \le a + b$

Let $a + b < m$.

Then:

\(\ds a +_{m, n} b\)

\(=\)

\(\ds a + b\)

Definition of $a +_{m, n} b$

\(\ds \)

\(=\)

\(\ds b + a\)

Natural Number Addition is Commutative

\(\ds \)

\(=\)

\(\ds b +_{m, n} a\)

Definition of $b +_{m, n} a$

Otherwise $a + b \ge m$.

Then:

\(\ds a +_{m, n} b\)

\(=\)

\(\ds a + b - k n\)

Definition of $a +_{m, n} b$, for some $k \in \N$

\(\ds \)

\(=\)

\(\ds b + a - k n\)

Natural Number Addition is Commutative, for that same $k \in \N$

\(\ds \)

\(=\)

\(\ds b +_{m, n} a\)

Definition of $b +_{m, n} a$

So in both cases:

$a +_{m, n} b = b +_{m, n} a$

and the result follows by definition of commutative operation.

$\blacksquare$

Also see

• Dipper Operation is Associative

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.8 \ \text{(b)}$