General Operation from Binary Operation

== Theorem ==
Let $\struct {S, \oplus}$ be a magma.
Then there a unique sequence $\sequence {\oplus_k}_{k \mathop \ge 1}$ such that:
:$(1): \quad \forall n \in \N_{>0}: \oplus_n$ is an $n$-ary operation on $S$ such that:
:$(2): \quad \forall \tuple {a_1, \ldots, a_k} \in S^k: \map {\oplus_k} {a_1, \ldots, a_k} = \begin{cases}
a : & k = 1 \\
\map {\oplus_n} {a_1, \ldots, a_n} \oplus a_{n + 1} : & k = n + 1
\end{cases}$
In particular, $\oplus_2$ is the same as the given binary operation $\oplus$.
The $n$th term $\oplus_n$ of the sequence $\sequence \oplus$ is called the $n$-ary operation defined by $\oplus$.
== Proof ==
Let $\Bbb S = \leftset {\odot:}$ for some $n \in \N_{>0}$, $\odot$ is an $n$-ary operation on $\rightset S$.
Let $s: \Bbb S \to \Bbb S$ be the mapping defined as follows.
Let $\odot$ be any $n$-ary operation defined on $\Bbb S$.
Then $\map s \odot$ is the $\paren {n + 1}$-ary operation defined by:
:$\forall \tuple {a_1, \ldots, a_n, a_{n + 1} } \in S^{n + 1}: \map {\map s \odot} {a_1, \ldots, a_n, a_{n + 1} } = \map \odot {a_1, \ldots, a_n} \oplus a_{n + 1}$
By the Principle of Recursive Definition, there is a unique sequence $\sequence {\oplus_k}_{k \mathop \ge 1}$ such that:
:$\oplus_1$ is the unary operation defined as $\map {\oplus_1} a = a$
and:
:$\oplus_{n + 1} = \map s {\oplus_n}$ for each $n \in \N_{>0}$.
Let $A$ be the set defined as the set of all $n$ such that:
:$\oplus_n$ is an $n$-ary operation on $S$
and:
:by the definition of $s$, $(2)$ holds for every ordered $n + 1$-tuple in $S^{n + 1}$.
It will be demonstrated by the Principle of Mathematical Induction that $A = \N$.
=== Basis for the Induction ===
We have that:
:$\oplus_1$ is the unary operation defined as $\map {\oplus_1} a = a$
By the Principle of Recursive Definition:
:$\oplus_2 = \map s {\oplus_1}$
is the unique operation on $S$ such that:
By the definition of a magma, $\oplus$ is a operation defined such that:
:$\forall \tuple {a_1, a_2} \in S \times S: map \oplus {a_1, a_2} \in S$
where $\oplus: S \times S \to S$ is a mapping.
$a_1 \oplus a_2$ is defined as:
:$\forall \paren {a_1, a_2} \in S \times S: a_1 \oplus a_2 := \map \oplus {a_1, a_2}$
So $1 \in A$.
This is our basis for the induction.
=== Induction Hypothesis ===
It is to be shown that, if $m \in A$ where $m \ge 1$, then it follows that $m + 1 \in A$.
This is the induction hypothesis:
:$\oplus_m$ is an $m$-ary operation on $S$
and:
:by the definition of $s$, $(2)$ holds for every ordered $m + 1$-tuple in $S^{m + 1}$.
It is to be demonstrated that it follows that:
:$\oplus_{m+1}$ is an $m+1$-ary operation on $S$
and:
:by the definition of $s$, $(2)$ holds for every ordered $m + 2$-tuple in $S^{m + 2}$.
=== Induction Step ===
This is our induction step:
By the Principle of Recursive Definition:
:$\oplus_{m + 1} = \map s {\oplus_m}$
is the unique operation on $S$ such that:
By the definition of a magma, $\oplus$ is a operation defined such that:
:$\forall \tuple {a_1, a_2} \in S \times S: \map \oplus {a_1, a_2} \in S$
where $\oplus: S \times S \to S$ is a mapping.
By the induction hypothesis:
:$\map {\oplus_m} {a_1, a_2, \ldots, a_m} \oplus a_{m + 1} \in S$
So by definition:
$\map {\oplus_{m + 1} } {a_1, a_2, \ldots, a_m, a_{m + 1} } \in S$
Let $b := \map {\oplus_{m + 1} } {a_1, a_2, \ldots, a_m, a_{m + 1} }$.
But then by the basis of the induction:
:$b \oplus a_{m + 2} \in S$
That is:
:$\map {\oplus_{m + 1} } {a_1, a_2, \ldots, a_m, a_{m + 1} } \oplus a_{m + 2} = \map {\oplus_{m + 2} } {a_1, a_2, \ldots, a_m, a_{m + 1}, a_{m + 2} } \in S$
So $m \in A \implies m + 1 \in A$ and the result follows by the Principle of Mathematical Induction:
== Also see ==
* Definition:N-Ary Operation Defined by Binary Operation
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.1$
Category:Abstract Algebra