nyuuzyou/wikis huggingface dataset:
Independent Subset Contains No Dependent Subset/Corollary 1

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $B \subseteq S$ be any base of $M$.
Then:
:No dependent subset $D$ of $M$ is a subset of $B$.
== Proof ==
By definition of matroid base:
:$B$ is an independent subset of $M$
From Independent Subset Contains No Dependent Subset:
:No dependent subset $D$ of $M$ is a subset of $B$.
Category:Independent Subset Contains No Dependent Subset
Independent Subset Contains No Dependent Subset/Corollary 2

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $X \subseteq S$ be any independent subset of $M$.
Then:
:No circuit $C$ of $M$ is a subset of $X$.
== Proof ==
Let $C$ be a circuit of $M$.
By definition of matroid circuit:
:$C$ is a dependent subset of $M$
From Independent Subset Contains No Dependent Subset:
:$C$ is not a subset of $X$.
The result follows.
Category:Independent Subset Contains No Dependent Subset
Independent Subset Contains No Dependent Subset/Corollary 3

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $B \subseteq S$ be any base of $M$.
Then:
:No circuit $C$ of $M$ is a subset of $B$.
== Proof ==
By definition of matroid base:
:$B$ is an independent subset of $M$
From Independent Subset Contains No Circuit:
:No circuit $C$ of $M$ is a subset of $B$.
Category:Independent Subset Contains No Dependent Subset
Independent Subset is Base if Cardinality Equals Rank of Matroid

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\rho: \powerset S \to \Z$ be the rank function of $M$.
Let $B \in \mathscr I$ such that:
:$\size B = \map \rho S$
Then:
:$B$ is a base of $M$.
=== Corollary ===
== Proof ==
Let $Z \in \mathscr I$ such that:
:$B \subseteq Z$
From Cardinality of Subset of Finite Set:
:$\size B \le \size Z$
By definition of the rank function:
:$\size Z \le \map \rho S$
Then:
:$\size Z = \size B$
From the contrapositive statement of Cardinality of Proper Subset of Finite Set:
:$B = Z$
It follows that $B$ is a maximal independent subset by definition.
That is, $B$ is a base by definition.
Category:Matroid Independent Subsets
Category:Matroid Bases
Category:Matroid Rank Functions
Independent Subset is Base if Cardinality Equals Rank of Matroid/Corollary

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $B \subseteq S$ be a base of $M$.
Let $X \subseteq S$ be any independent subset of $M$.
Let $\card X = \card B$.
Then:
:$X$ is a base of $M$.
== Proof ==
From All Bases of Matroid have same Cardinality:
:$\card B = \map \rho S$
where $\rho$ denotes the rank function on $M$.
Hence:
:$\card X = \map \rho S$
From Independent Subset is Base if Cardinality Equals Rank of Matroid:
:$X$ is a base of $M$.
Category:Matroid Independent Subsets
Category:Matroid Bases
Independent Subset is Contained in Base

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\mathscr B$ denote the set of all bases of $M$.
Let $A \in \mathscr I$.
Then:
:$\exists B \in \mathscr B : A \subseteq B$
== Proof ==
Consider the ordered set $\struct {\mathscr I, \subseteq}$.
From Element of Finite Ordered Set is Between Maximal and Minimal Elements:
:$\exists B \in \mathscr I : A \subseteq B$ and $B$ is maximal in $\struct {\mathscr I, \subseteq}$.
By definition of a base:
:$B \in \mathscr B$
Category:Matroid Independent Subsets
Category:Matroid Bases
Independent Subset is Contained in Maximal Independent Subset

== Theorem ==
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $A \subseteq S$.
Let $X \in \mathscr I$ such that $X \subseteq A$.
Then:
:$\exists Y \in \mathscr I : X \subseteq Y \subseteq A : \size Y = \map \rho A$
where $\rho$ is the rank function on $M$.
== Proof ==
By definition of the rank function on $M$:
:$\size X \le \map \rho A$
=== Case 1 : $\size X = \map \rho A$ ===
Let $\size X = \map \rho A$.
Let $Y = X$ and the result follows.
 
=== Case 2 : $\size X < \map \rho A$ ===
 
Let $\size X < \map \rho A$.
By definition of the rank function on $M$:
:$\map \rho A = \max \set{\size I : I \subseteq A \land I \in \mathscr I}$
From Max Operation Equals an Operand:
:$\exists Z \in \mathscr I : Z \subseteq A$ and $\size Z = \map \rho A$
From Independent Set can be Augmented by Larger Independent Set:
:$\exists Y' \subseteq Z \setminus X: \size {X \cup Y'} = \size Z$ and $X \cup Y' \in \mathscr I$
We have:
From Union of Subsets is Subset:
:$X \cup Y' \subseteq A$
Let $Y = X \cup Y'$ and the result follows.
In either case, the result follows.
Category:Matroid Independent Subsets
Index Laws/Common Index/Field

== Theorem ==
Let $\struct {F, +, \circ}$ be a field with zero $0_F$ and unity $1_F$.
Let $F^* = F \setminus {0_F}$ denote the set of elements of $F$ without the zero $0_F$.
Then:
:$(a):\quad \forall a, b \in \F^* : \forall n \in \Z : a^n \circ b^n = \paren{ab}^n$
:$(b):\quad \forall a, b \in \F : \forall n \in \Z_{\ge 0} : a^n \circ b^n = \paren{ab}^n$
== Proof ==
=== Statement $(a)$ ===
By :
:$\struct{F^*, \circ}$ is an Abelian group
By :
:For all $a \in F^*$ and $n \in \Z$, $a^n$ is defined as the $n$th power of $a$ with respect to the Abelian group $\struct {F^*, \circ}$
From Sum of Powers of Group Elements:
:$\forall a \in \F^* : \forall n, m \in \Z : a^m \circ a^n = a^\paren{m + n}$
=== Statement $(b)$ ===
Let $m,n \in \Z_{\ge 0}$ be arbitrary elements of $\Z_{\ge 0}$.
For $a \in F^*$, $(b)$ follows from $(a)$.
It remains to show that $(b)$ holds for $0_F$.
==== Case 1: $m = 0$ ====
Let $m = 0$.
We have:
==== Case 2: $m \ne 0$ ====
Let $m \ne 0$.
Hence:
:$m + n \ne 0$
We have:
In both cases:
:$\paren{0_F}^m \circ \paren{0_F}^n = \paren{0_F}^\paren{m + n}$
Since $m,n$ were arbitrary elements of $\Z_{\ge 0}$:
:$\forall n, m \in \Z_{\ge 0} : \paren{0_F}^m \circ \paren{0_F}^n = \paren{0_F}^\paren{m + n}$
Category:Index Laws
Index Laws/Negative Index/Field

== Theorem ==
Let $\struct {F, +, \circ}$ be a field with zero $0_F$ and unity $1_F$.
Let $F^* = F \setminus {0_F}$ denote the set of elements of $F$ without the zero $0_F$.
Then:
:$\forall a \in \F^* : \forall n \in \Z : \paren{a^n}^{-1} = a^{-n} = \paren{a^{-1}}^n$
== Proof ==
By :
:$\struct{F^*, \circ}$ is an Abelian group
By :
:For all $a \in F^*$ and $n \in \Z$, $a^n$ is defined as the $n$th power of $a$ with respect to the Abelian group $\struct {F^*, \circ}$
From Negative Powers of Group Elements:
:$\forall a \in \F^* : \forall n \in \Z : \paren{a^n}^{-1} = a^{-n} = \paren{a^{-1}}^n$
Category:Index Laws
Index Laws/Product of Indices/Field

== Theorem ==
Let $\struct {F, +, \circ}$ be a field with zero $0_F$ and unity $1_F$.
Let $F^* = F \setminus {0_F}$ denote the set of elements of $F$ without the zero $0_F$.
Then:
:$(a):\quad \forall a \in \F^* : \forall n, m \in \Z : \paren{a^m}^n = a^\paren{mn}$
:$(b):\quad \forall a \in \F : \forall n, m \in \Z_{\ge 0} : \paren{a^m}^n = a^\paren{mn}$
== Proof ==
=== Statement $(a)$ ===
By :
:$\struct{F^*, \circ}$ is an Abelian group
By :
:For all $a \in F^*$ and $n \in \Z$, $a^n$ is defined as the $n$th power of $a$ with respect to the Abelian group $\struct {F^*, \circ}$
From Product of Powers of Group Elements:
:$\forall a \in \F^* : \forall n, m \in \Z : \paren{a^m}^n = a^\paren{mn}$
=== Statement $(b)$ ===
Let $m,n \in \Z_{\ge 0}$ be arbitrary elements of $\Z_{\ge 0}$.
For $a \in F^*$, $(b)$ follows from $(a)$.
It remains to show that $(b)$ holds for $0_F$.
==== Case 1: $m = 0$ ====
Let $m = 0$.
We have:
==== Case 2: $m \ne 0, n = 0$ ====
Let $m \ne 0$ and $n = 0$.
Hence:
:$m n = 0$
We have:
==== Case 3: $m \ne 0, n = 0$ ====
Let $m \ne 0$ and $n \ne 0$.
Hence:
:$m n \ne 0$
We have:
In all cases:
:$\paren{\paren{0_F}^m}^n = \paren{0_F}^\paren{mn}$
Since $m,n$ were arbitrary elements of $\Z_{\ge 0}$:
:$\forall n, m \in \Z_{\ge 0} : \paren{\paren{0_F}^m}^n = \paren{0_F}^\paren{mn}$
Category:Index Laws
Index Laws/Product of Indices/Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.
For $a \in S$, let $\circ^n a = a^n$ be the $n$th power of $a$.
Then:
:$\forall m, n \in \N: a^{n m} = \paren {a^n}^m = \paren {a^m}^n$
That is:
:$\forall m, n \in \N: \circ^{n m} a = \circ^m \paren {\circ^n a} = \circ^n \paren {\circ^m a}$
== Proof ==
Because $\struct {S, \circ}$ is a monoid, it is a fortiori a semigroup.
Hence, from Index Laws for Semigroup: Product of Indices:
:$\forall m, n \in \N_{>0}: \circ^{n m} a = \circ^m \paren {\circ^n a} = \circ^n \paren {\circ^m a}$
That is:
:$\forall m, n \in \N_{>0}: a^{n m} = \paren {a^n}^m = \paren {a^m}^n$
It remains to be shown that the result holds for the cases where $m = 0$ and $n = 0$.
so the condition holds for either $n = 0$ or $m = 0$.
Finally, we also have:
:$\circ^n \paren {\circ^0 a} = e = \circ^0 \paren {\circ^m a}$
:$\circ^0 \paren {\circ^n a} = e = \circ^m \paren {\circ^0 a}$
== Notation ==
== Source of Name ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.11$
Category:Index Laws
Category:Monoids
Index Laws/Product of Indices/Notation

== Notation for Index Laws: Product of Indices ==
Let $\struct {S, \circ}$ be a semigroup.
Let $a \in S$.
Let $a^n$ be defined as the power of an element of a magma:
:$a^n = \begin{cases}
a : & n = 1 \\
a^x \circ a : & n = x + 1
\end{cases}$
that is:
: $a^n = \underbrace {a \circ a \circ \cdots \circ a}_{n \text{ instances of } a} = \circ^n \paren a$
Recall the index law for product of indices:
:$\circ^{n m} a = \circ^m \paren {\circ^n a} = \circ^n \paren {\circ^m a}$
This result can be expressed:
:$a^{n m} = \paren {a^n}^m = \paren {a^m}^n$
When additive notation $\struct {S, +}$ is used, the following is a common convention:
:$\paren {n m} a = m \paren {n a} = n \paren {m a}$
or:
:$\forall m, n \in \N_{>0}: \paren {n m} \cdot a = m \cdot \paren {n \cdot a} = n \cdot \paren {m \cdot a}$
== Source of Name ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.11$
Category:Index Laws
Index Laws/Product of Indices/Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
For $a \in S$, let $\circ^n a = a^n$ be the $n$th power of $a$.
Then:
:$\forall m, n \in \N_{>0}: a^{n m} = \paren {a^n}^m = \paren {a^m}^n$
That is:
:$\forall m, n \in \N_{>0}: \circ^{n m} a = \circ^m \paren {\circ^n a} = \circ^n \paren {\circ^m a}$
== Proof ==
Let $b = \circ^m a$.
Let $h: \N_{>0} \to S$ be the mapping defined as:
:$\forall n \in \N_{>0}: \map h n = \circ^{n m} a$
Let the mapping $f_b: \N_{>0} \to S$ be recursively defined as:
:$\forall n \in \N_{>0}: \map {f_b} n = \circ^n b$
From the Principle of Recursive Definition:
:$f_b$ is the unique mapping which satisfies:
:$\forall n \in \N_{>0}: \map {f_b} n = \begin{cases}
b & : n = 1 \\
\map {f_b} r \circ b & : n = r \circ 1
\end{cases}$
But $\map h 1 = \circ^{1 \times m} a = \circ^m a = b$.
So:
Thus $h = f_b$, and so:
:$\forall n, m \in \N_{>0}: \circ^{n m} = \circ^n \paren {\circ^m a}$
From Natural Number Multiplication is Commutative:
:$\forall n, m \in \N_{>0}: \circ^m \paren {\circ^n a} = \circ^{m n} = \circ^{n m}$
== Notation ==
== Source of Name ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.11$
* : $2$: Examples of Groups and Homomorphisms: $2.1$ Definition
* : $\S 30.1$ Powers of an element in a semigroup: $\text{(ii)}$
* : Chapter $3$: Elementary consequences of the definitions: Remark
Category:Index Laws
Category:Semigroups
Index Laws/Sum of Indices/Field

== Theorem ==
Let $\struct {F, +, \circ}$ be a field with zero $0_F$ and unity $1_F$.
Let $F^* = F \setminus {0_F}$ denote the set of elements of $F$ without the zero $0_F$.
Then:
:$(a):\quad \forall a \in \F^* : \forall n, m \in \Z : a^m \circ a^n = a^\paren{m + n}$
:$(b):\quad \forall a \in \F : \forall n, m \in \Z_{\ge 0} : a^m \circ a^n = a^\paren{m + n}$
== Proof ==
=== Statement $(a)$ ===
By :
:$\struct{F^*, \circ}$ is an Abelian group
By :
:For all $a \in F^*$ and $n \in \Z$, $a^n$ is defined as the $n$th power of $a$ with respect to the Abelian group $\struct {F^*, \circ}$
From Sum of Powers of Group Elements:
:$\forall a \in \F^* : \forall n, m \in \Z : a^m \circ a^n = a^\paren{m + n}$
=== Statement $(b)$ ===
Let $m,n \in \Z_{\ge 0}$ be arbitrary elements of $\Z_{\ge 0}$.
For $a \in F^*$, $(b)$ follows from $(a)$.
It remains to show that $(b)$ holds for $0_F$.
==== Case 1: $m = 0$ ====
Let $m = 0$.
We have:
==== Case 2: $m \ne 0$ ====
Let $m \ne 0$.
Hence:
:$m + n \ne 0$
We have:
In both cases:
:$\paren{0_F}^m \circ \paren{0_F}^n = \paren{0_F}^\paren{m + n}$
Since $m,n$ were arbitrary elements of $\Z_{\ge 0}$:
:$\forall n, m \in \Z_{\ge 0} : \paren{0_F}^m \circ \paren{0_F}^n = \paren{0_F}^\paren{m + n}$
Category:Index Laws
Index Laws/Sum of Indices/Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.
For $a \in S$, let $\circ^n a = a^n$ be defined as the $n$th power of $a$:
:$a^n = \begin{cases}
e & : n = 0 \\
a^x \circ a & : n = x + 1
\end{cases}$
That is:
:$a^n = \underbrace {a \circ a \circ \cdots \circ a}_{n \text{ copies of } a} = \map {\circ^n} a$
while:
:$a^0 = e$
Then:
:$\forall m, n \in \N: a^{n + m} = a^n \circ a^m$
== Proof ==
Because $\struct {S, \circ}$ is a monoid, it is also a semigroup.
From Index Laws for Semigroup: Sum of Indices:
:$\forall m, n \in \N_{>0}: \circ^{n + m} a = \paren {\circ^n a} \circ \paren {\circ^m a}$
That is:
:$\forall m, n \in \N_{>0}: a^{n + m} = a^n \circ a^m$
It remains to be shown that the result holds for the cases where $m = 0$ and $n = 0$.
Let $n \in \N$:
Similarly, let $m \in \N$:
and:
Thus:
:$a^{n + m} = a^n \circ a^m$
holds for $n = 0$ and $m = 0$.
Thus:
:$\forall m, n \in \N: a^{n + m} = a^n \circ a^m$
== Also see ==
* Power of Identity is Identity
== Source of Name ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.8$
Category:Index Laws
Category:Monoids
Index Laws/Sum of Indices/Notation

== Notation for Index Laws: Sum of Indices ==
Let $\struct {S, \circ}$ be a semigroup.
Let $a \in S$.
Let $a^n$ be defined as the power of an element of a magma:
:$a^n = \begin{cases}
a : & n = 1 \\
a^x \circ a : & n = x + 1
\end{cases}$
that is:
: $a^n = \underbrace {a \circ a \circ \cdots \circ a}_{n \text{ copies of } a} = \map {\circ^n} a$
Recall the index law for sum of indices:
:$\circ^{n + m} a = \paren {\circ^n a} \circ \paren {\circ^m a}$
This result can be expressed:
:$a^{n + m} = a^n \circ a^m$
When additive notation $\struct {S, +}$ is used, the following is a common convention:
:$\left({n + m}\right) a = n a + m a$
or:
:$\forall m, n \in \N_{>0}: \paren {n + m} \cdot a = n \cdot a + m \cdot a$
== Source of Name ==
== Sources ==
* : $\S 4.2$. Commutative and associative operations: Example $69$
Category:Index Laws
Index Laws/Sum of Indices/Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
For $a \in S$, let $\circ^n a = a^n$ be defined as the $n$th power of $a$:
:$a^n = \begin{cases}
a & : n = 1 \\
a^x \circ a & : n = x + 1
\end{cases}$
That is:
:$a^n = \underbrace {a \circ a \circ \cdots \circ a}_{n \text{ copies of } a} = \circ^n \paren a$
Then:
:$\forall m, n \in \N_{>0}: a^{n + m} = a^n \circ a^m$
== Proof ==
Let $a \in S$.
Because $\struct {S, \circ}$ is a semigroup, $\circ$ is associative on $S$.
The proof proceeds by the Principle of Mathematical Induction.
Let $\map P m$ be the proposition:
:$\forall n \in \N_{>0}: a^{n + m} = a^n \circ a^m$
that is:
:$\forall n \in \N_{>0}: \circ^{n + m} a = \paren {\circ^n a} \circ \paren {\circ^m a}$
=== Basis for the Induction ===
So $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\circ^{n + k} a = \paren {\circ^n a} \circ \paren {\circ^k a}$
It is then to be shown that:
:$\circ^{n + \paren {k + 1} } a = \paren {\circ^n a} \circ \paren {\circ^{k + 1} a}$
=== Induction Step ===
This is our induction step:
So $\map P {k + 1}$ is true.
So by the Principle of Mathematical Induction, this result is true for all $m, n \in \N_{>0}$:
:$\forall m, n \in \N_{>0}: \circ^{n + m} a = \paren {\circ^n a} \circ \paren {\circ^m a}$
or:
:$\forall m, n \in \N_{>0}: a^{n + m} = a^n \circ a^m$
== Notation ==
== Source of Name ==
== Sources ==
* : $\S 4.2$. Commutative and associative operations
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.8 \ (1)$
* : $2$: Examples of Groups and Homomorphisms: $2.1$ Definition
* : $\S 30.1$ Powers of an element in a semigroup: $\text{(i)}$
* : Chapter $3$: Elementary consequences of the definitions: Remark
Category:Index Laws
Category:Semigroups
Index Laws for Field

== Theorem ==
Let $\struct {F, +, \circ}$ be a field with zero $0_F$ and unity $1_F$.
Let $F^* = F \setminus {0_F}$ denote the set of elements of $F$ without the zero $0_F$.
Then the following hold:
=== Common Index ===
=== Negative Index ===
=== Sum of Indices ===
=== Product of Indices ===
== Also see ==
* Index Laws for Monoid
* Powers of Group Elements
== Source of Name ==
Category:Field Theory
Category:Index Laws
Index Laws for Monoids

== Theorem ==
These results are an extension of the results in Index Laws for Semigroup in which the domain of the indices is extended to include all integers.
Let $\struct {S, \circ}$ be a monoid whose identity is $e$.
Let $a \in S$ be invertible for $\circ$.
Let $n \in \N$.
Let $a^n$ be the $n$th power of $a$:
:$a^n = \begin{cases}
e : & n = 0 \\
a^{n - 1} \circ a : & n > 0 \\
\paren {a^{-n}}^{-1} : & n < 0
\end{cases}$
Then we have the following results:
=== Negative Index ===
=== Sum of Indices ===
=== Product of Indices ===
== Also see ==
* Definition:Integral Multiple
== Source of Name ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.11$
Category:Monoids
Category:Index Laws
Index Laws for Monoids/Negative Index

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $a \in S$ be invertible for $\circ$.
Let $n \in \N$.
Let $a^n = \map {\circ^n} a$ be defined as the power of an element of a monoid:
:$a^n = \begin{cases}
e_S : & n = 0 \\
a^x \circ a : & n = x + 1
\end{cases}$
that is:
:$a^n = \underbrace {a \circ a \circ \cdots \circ a}_{n \text { instances} } = \map {\circ^n} a$
For each $n \in \N$ we define:
:$a^{-n} = \paren {a^{-1} }^n$
Then:
:$\forall n \in \Z: \paren {a^n}^{-1} = a^{-n} = \paren {a^{-1} }^n$
== Proof ==
We have $a^0 = e$ so it follows trivially that $a^{-0} = \paren {a^{-1} }^0$.
From the general inverse of product, we have:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
where $a_1, a_2, \ldots, a_n \in S$ are all invertible for $\circ$.
Hence we have:
:$a_1, a_2, \ldots, a_n = a$
and we have that:
:$a^n$ is invertible for all $n \in \N$
:$\forall n \in \N: \paren {a^n}^{-1} = \paren {a^{-1} }^n$
From the above:
:$a^{-n} = \paren {a^{-1} }^n$
Thus:
Similarly, if $a$ is invertible then $a^{-1}$ is also invertible.
So we also have:
:$\circ^{-n} \paren {a^{-1} } = \circ^n \paren {\paren {a^{-1} }^{-1} }$
Thus:
Thus the result holds for all $n \in \Z$.
== Source of Name ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.11 \ (1)$
* : $\S 3.1$: Monoids: Exercise $(8)$
Category:Monoids
Category:Index Laws
Index Laws for Monoids/Product of Indices

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $a \in S$ be invertible for $\circ$.
Let $n \in \N$.
Let $a^n = \map {\circ^n} a$ be defined as the power of an element of a monoid:
:$a^n = \begin{cases}
e_S : & n = 0 \\
a^x \circ a : & n = x + 1
\end{cases}$
that is:
:$a^n = \underbrace {a \circ a \circ \cdots \circ a}_{n \text { instances} } = \map {\circ^n} a$
For each $n \in \N$ we define:
:$a^{-n} = \paren {a^{-1} }^n$
Then:
:$\forall m, n \in \Z: a^{n m} = \paren {a^m}^n = \paren {a^n}^m$
== Proof ==
Let $m \in \N, c = a^m, d = \paren {a^{-1}}^m$.
We define the mapping $g_c: \Z \to S$ as:
:$\forall n \in \Z: \map {g_c} n = \map {\circ^n} c$
as defined in the proof of the Index Law for Sum of Indices.
Let $h: \Z \to \Z$ be the mapping defined as:
:$\forall z \in \Z: \map h z = z m$
Then:
By Index Law for Sum of Indices and Index Laws for Semigroup: Product of Indices, $g_a \circ h$ and $g_c$ are homomorphisms from $\Z$ to $S$ which coincide on $\N$.
So by the Extension Theorem for Homomorphisms:
:$g_a \circ h = g_c$
Therefore:
:$\forall n \in \Z, m \in \N: a^{n m} = \paren {a^m}^n$
Also:
and:
So, by the same reasoning as before:
:$g_{a^{-1} } \circ h = g_d$
Therefore:
:$\forall n \in \Z, m \in \N: a^{n \paren {-m}} = \paren {a^{-m}}^n$
Thus:
:$\forall n, m \in \Z: a^{n m} = \paren {a^m}^n$
As $n m = m n$, the result follows.
== Source of Name ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.11 \ (3)$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Theorem $1.9 \text{(ii)}$
* : $\S 3.1$: Monoids
Category:Monoids
Category:Index Laws
Index Laws for Monoids/Sum of Indices

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $a \in S$ be invertible for $\circ$.
Let $n \in \N$.
Let $a^n = \map {\circ^n} a$ be defined as the power of an element of a monoid:
:$a^n = \begin{cases}
e_S : & n = 0 \\
a^x \circ a : & n = x + 1
\end{cases}$
That is:
:$a^n = \underbrace {a \circ a \circ \cdots \circ a}_{\text {$n$ instances} } = \map {\circ^n} a$
For each $n \in \N$ we define:
:$a^{-n} = \paren {a^{-1} }^n$
Then:
:$\forall m, n \in \Z: a^{n + m} = a^n \circ a^m$
== Proof ==
For each $c \in S$ which is invertible for $\circ$, we define the mapping $g_c: \Z \to S$ as:
:$\forall n \in \Z: \map {g_c} n = \map {\circ^n} c$
By the Index Law for Monoids: Negative Index, $\map {g_a} n$ is invertible for all $n \in \Z$.
By definition of Power of Element of Monoid, the restriction of $g_a$ to $\N$ is a homomorphism from $\struct {\N, +}$ to $\struct {S, \circ}$.
From the definition of Power of Element of Monoid, $\map {g_a} 0$ is the identity for $\circ$.
Hence, by the Extension Theorem for Homomorphisms, there is a unique homomorphism $h_a: \paren {\N, +} \to \paren {S, \circ}$ which coincides in $\N$ with $g_c$.
But by Homomorphism with Identity Preserves Inverses:
Hence $h_a = g_a$ and so $g_a$ is a homomorphism and so the result follows.
== Source of Name ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.11 \ (2)$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Theorem $1.9 \text{(i)}$
* : $\S 3.1$: Monoids
Category:Monoids
Category:Index Laws
Index Number/Examples/Arbitrary Example 1

== Example of Index Number ==
Let the price of an item $I$ be $50$ in the base year.
Let the price of $I$ be $55$ and $70$ in the two subsequent years.
Then the index number of $I$ based on its price would be $100$, $110$ and $140$ for those successive years.
== Sources ==
* 
* 
Category:Examples of Index Numbers