avewright/proofwiki-math huggingface dataset:
Theorem: Image of Interval by Derivative

Let $f$ be a real function that is everywhere differentiable.
Let $I \subseteq \Dom f$ be a real interval.
Then:
:$f' \sqbrk I$ is a real interval
where $f'$ denotes the derivative of $f$

Proof:
Let $x_1, x_2 \in f' \sqbrk I: x_1 < x_2$.
Let $\xi \in \openint {x_1} {x_2}$.
We need to show that $\xi \in f' \sqbrk I$.
Let $a, b \in I : \map {f'} a = x_1 \land \map {f'} b = x_2$.
{{WLOG}}, assume $a < b$. The case $b < a$ is handled similarly.
Let $\map g x = \map f x - \xi x$.
Then:
:$\map {g'} x = \map {f'} x - \xi$
By Differentiable Function is Continuous, $g$ is continuous.
By Restriction of Continuous Mapping is Continuous, $g \restriction_{\closedint a b}$ is continuous.
From {{Corollary|Continuous Image of Compact Space is Compact|3}}:
:$g \restriction_{\closedint a b}$ attains a minimum.
By hypothesis, $\map {g'} a < 0$ and $\map {g'} b > 0$.
Let $\map {N^*_\epsilon} x$ denote the deleted $\epsilon$-neighborhood  of $x$.
Then from Behaviour of Function Near Limit:
:$\exists \map {N^*_\epsilon} a: \forall x \in \map {N^*_\epsilon} a: \dfrac {\map g x - \map g a} {x - a} < 0$
and:
:$\exists \map {N^*_\delta} b: \forall x \in \map {N^*_\delta} b: \dfrac {\map g x - \map g b} {x - b} > 0$
Thus:
:$\exists x \in \map {N^*_\epsilon} a \cap \closedint a b: \map g x < \map g a$
and
:$\exists x \in \map {N^*_\delta} b \cap \closedint a b: \map g x < \map g b$
Hence $\map g a$ and $\map g b$ are not minima of $g \restriction_{\closedint a b}$.
So $g \restriction_{\closedint a b}$ must attain its minimum at some $m \in \openint a b$.
By the Interior Extremum Theorem:
:$\map {g'} m = 0$
Hence:
:$\map {f'} m = \xi$
The result follows.
{{qed}}
Theorem: Intermediate Value Theorem for Derivatives

Let $I$ be an open interval.
Let $f : I \to \R$ be everywhere differentiable.
Then $f'$ satisfies the Intermediate Value Property.

Proof:
Since $\forall \set {a, b \in I: a < b}: \openint a b \subseteq I$, the result follows from Image of Interval by Derivative.
{{explain|Now I'm looking at this, I think we need to flesh this out a bit. Anyone up for it?}}
Theorem: Sine of Half-Integer Multiple of Pi

:$\forall n \in \Z: \map \sin {n + \dfrac 1 2} \pi = \paren {-1}^n$

Proof:
From the discussion of Sine and Cosine are Periodic on Reals:
:$\map \sin {x + \dfrac \pi 2} = \cos x$
The result then follows directly from the Cosine of Multiple of Pi.
{{qed}}
Category:Sine Function
mpm9bskwo8z0lvs9quo23dcsouumuhz
Theorem: Sine of Integer Multiple of Pi

:$\forall n \in \Z: \sin n \pi = 0$

Proof:
This is established in Zeroes of Sine and Cosine.
{{qed}}
Theorem: Cosine of Half-Integer Multiple of Pi

:$\forall n \in \Z: \map \cos {n + \dfrac 1 2} \pi = 0$

Proof:
This is established in Zeroes of Sine and Cosine.
{{qed}}
Category:Cosine Function
rfuhggw0gmxdy0pt571mvoo0kbkotq4
Theorem: Rational Square Root of Integer is Integer

Let $n$ be an integer.
Suppose that $\sqrt n$ is a rational number.
Then $\sqrt n$ is an integer.

Proof:
Suppose that $\sqrt n = \dfrac a b$, with $a, b$ coprime integers and $b > 0$.
Then we would have:
:$n = \dfrac {a^2} {b^2}$
That is:
:$n b^2 = a^2$
From Number divides Number iff Square divides Square:
:$b^2 \divides a^2 \implies b \divides a$
However, since $a \perp b$ and $b \divides a$, this means that necessarily $b = 1$.
That is, $\sqrt n = a$, an integer.
Hence the result.
{{qed}}
Theorem: Definite Integral of Even Function

Let $f$ be an even function with a primitive on the closedinterval $\closedint {-a} a$, where $a > 0$.
Then:
:$\ds \int_{-a}^a \map f x \rd x = 2 \int_0^a \map f x \rd x$

Proof:
Let $F$ be a primitive for $f$ on the interval $\closedint {-a} a$.
Then, by Sum of Integrals on Adjacent Intervals for Integrable Functions, we have:
{{begin-eqn}}
{{eqn | l = \int_{-a}^a \map f x \rd x
      | r = \int_{-a}^0 \map f x \rd x + \int_0^a \map f x \rd x
      | c =
}}
{{end-eqn}}
Therefore, it suffices to prove that:
:$\ds \int_{-a}^0 \map f x \rd x = \int_0^a \map f x \rd x$
To this end, let $\phi: \R \to \R$ be defined by $x \mapsto -x$. 
From Derivative of Identity Function and Derivative of Constant Multiple, for all $x \in \R$, we have $\map {\phi'} x = -1$.
Then, by means of Integration by Substitution, we compute:
{{begin-eqn}}
{{eqn | l = \int_{\map \phi a}^{\map \phi 0} \map f x \rd x
      | r = \int_a^0 \map f {-u} \paren {-1} \rd u
}}
{{eqn | r = \int_0^a \map f {-u} \rd u
      | c = {{Defof|Definite Integral}}
}}
{{eqn | r = \int_0^a \map f {-x} \rd x
      | c = renaming integration variable
}}
{{eqn | r = \int_0^a \map f x \rd x
      | c = $f$ is an even function
}}
{{end-eqn}}
This concludes the proof.
{{qed}}
Theorem: Definite Integral of Odd Function

Let $f$ be an odd function with a primitive on the open interval $\closedint {-a} a$, where $a > 0$.
Then:
:$\ds \int_{-a}^a \map f x \rd x = 0$

Proof:
Let $F$ be a primitive for $f$ on the interval $\closedint {-a} a$.
Then, by Sum of Integrals on Adjacent Intervals for Integrable Functions, we have:
{{begin-eqn}}
{{eqn | l = \int_{-a}^a \map f x \rd x
      | r = \int_{-a}^0 \map f x \rd x + \int_0^a \map f x \rd x
      | c =
}}
{{end-eqn}}
Therefore, it suffices to prove that:
:$\ds \int_{-a}^0 \map f x \rd x = -\int_0^a \map f x \rd x$
To this end, let $\phi: \R \to \R$ be defined by $x \mapsto -x$. 
From Derivative of Identity Function and Derivative of Constant Multiple, for all $x \in \R$, we have $\map {\phi'} x = -1$.
Then, by means of Integration by Substitution, we compute:
{{begin-eqn}}
{{eqn | l = \int_{\map \phi a}^{\map \phi 0} \map f x \rd x
      | r = \int_a^0 \map f {-u} \paren {-1} \rd u
}}
{{eqn | r = \int_0^a \map f {-u} \rd u
      | c = {{Defof|Definite Integral}}
}}
{{eqn | r = \int_0^a \map f {-x} \rd x
      | c = Renaming integration variable
}}
{{eqn | r = -\int_0^a \map f x \rd x
      | c = $f$ is an odd function
}}
{{end-eqn}}
This concludes the proof.
{{qed}}
Theorem: Union of Equivalence Classes is Whole Set

Let $\RR \subseteq S \times S$ be an equivalence on a set $S$.
Then the set of $\RR$-classes constitutes the whole of $S$.

Proof:
We have that:
{{begin-eqn}}
{{eqn | q = \forall x \in S
      | l = x
      | o = \in 
      | r = \eqclass x \RR
      | c = {{Defof|Equivalence Class}}
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = x
      | o = \in 
      | r = \set {y \in S: x \mathrel \RR y}
      | c = {{Defof|Equivalence Class}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \eqclass x \RR
      | r = \set {y: x \mathrel \RR y}
      | c = {{Defof|Equivalence Class}}
}}
{{eqn | n = 2
      | o = \subseteq
      | r = S
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = S
      | r = \bigcup_{x \mathop \in S} \set x
      | c = {{Defof|Union of Set of Sets}}
}}
{{eqn | o = \subseteq
      | r = \bigcup_{x \mathop \in S} \eqclass x \RR
      | c = from $(1)$ and Set Union Preserves Subsets
}}
{{eqn | o = \subseteq
      | r = \bigcup_{x \mathop \in S} S
      | c = from $(2)$ and Set Union Preserves Subsets
}}
{{eqn | r = S
      | c = 
}}
{{end-eqn}}
Theorem: Ratio of Consecutive Fibonacci Numbers

For $n \in \N$, let $f_n$ be the $n$th Fibonacci number.
Then:
:$\ds \lim_{n \mathop \to \infty} \frac {f_{n + 1} } {f_n} = \phi$
where $\phi = \dfrac {1 + \sqrt 5} 2$ is the golden mean.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = \phi
      | o = :=
      | r = \dfrac {1 + \sqrt 5} 2
}}
{{eqn | l = \hat \phi
      | o = :=
      | r = \paren {1 - \phi}
      | rr= = \dfrac {1 - \sqrt 5} 2
}}
{{eqn | l = \alpha
      | o = :=
      | r = \dfrac {\phi} {\hat \phi}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \alpha
      | r = \dfrac {\phi} {\hat \phi}
}}
{{eqn | r = \dfrac {1 + \sqrt 5} {1 - \sqrt 5}
}}
{{eqn | r = \dfrac {1 + \sqrt 5} {1 - \sqrt 5} \paren {\dfrac {1 + \sqrt 5} {1 + \sqrt 5} }
      | c = multiplying top and bottom by $1 + \sqrt 5$
}}
{{eqn | r = \dfrac {6 + 2 \sqrt 5} {-4} 
      | c = 
}}
{{eqn | r = -\dfrac {3 + \sqrt 5} 2
}}
{{end-eqn}}
Recall the Euler-Binet Formula:
:$f_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
Let $n \ge 1$.
It immediately follows that:
{{begin-eqn}}
{{eqn | l = \frac {f_{n + 1} } {f_n} 
      | r = \dfrac {\phi^{n + 1} - \hat \phi^{n + 1} } {\phi^n - \hat \phi^n}
}}
{{eqn | r = \dfrac {\paren {\phi^{n + 1} - \phi \hat \phi^n} + \paren {\phi \hat \phi^n - \hat \phi^{n + 1} } } {\phi^n - \hat \phi^n}
      | c = adding and subtracting $\phi \hat \phi^n$ to the numerator
}}
{{eqn | r = \dfrac {\phi \paren {\phi^n - \hat \phi^n} + \hat \phi^n \paren {\phi - \hat \phi } } {\phi^n - \hat \phi^n}
      | c = factoring out $\phi$ and $\hat \phi^n$ in the numerator
}}
{{eqn | r = \phi + \dfrac {\hat \phi^n \paren {\phi - \hat \phi} } {\phi^n - \hat \phi^n}
      | c = 
}}
{{eqn | r = \phi + \dfrac {\hat \phi^n \paren {\phi - \hat \phi} } {\phi^n - \hat \phi^n} \dfrac {\dfrac 1 {\hat \phi^n } } {\dfrac 1 {\hat \phi^n } }
      | c = multiplying top and bottom by $\dfrac 1 {\hat \phi^n }$
}}
{{eqn | r = \phi + \dfrac {\paren {\phi - \hat \phi} } {\dfrac {\phi^n} {\hat \phi^n} - 1} 
      | c = 
}}
{{eqn | r = \phi + \dfrac {\sqrt 5} {\alpha^n - 1}
}}
{{end-eqn}}
From the definition of $\alpha$:
:$\size \alpha > 1$
Therefore:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \frac {f_{n + 1} } {f_n}
      | r = \lim_{n \mathop \to \infty}\ \phi + \dfrac {\sqrt 5} {\alpha^n - 1}
}}
{{eqn | r = \phi + 0
      | c = 
}}
{{eqn | r = \phi
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Ratio of Consecutive Fibonacci Numbers

For $n \in \N$, let $f_n$ be the $n$th Fibonacci number.
Then:
:$\ds \lim_{n \mathop \to \infty} \frac {f_{n + 1} } {f_n} = \phi$
where $\phi = \dfrac {1 + \sqrt 5} 2$ is the golden mean.

Proof:
From Continued Fraction Expansion of Golden Mean: Successive Convergents, the $n$th convergent of the continued fraction expansion of $\phi$ is:
:$C_n = \dfrac {f_{n + 1} } {f_n}$
The result follows from Continued Fraction Expansion of Irrational Number Converges to Number Itself.
{{qed}}
Theorem: Ratio of Consecutive Fibonacci Numbers

For $n \in \N$, let $f_n$ be the $n$th Fibonacci number.
Then:
:$\ds \lim_{n \mathop \to \infty} \frac {f_{n + 1} } {f_n} = \phi$
where $\phi = \dfrac {1 + \sqrt 5} 2$ is the golden mean.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = a_n
      | o = :=
      | r = \dfrac {f_{n + 1} } {f_n} 
}}
{{eqn | l = \map g x
      | o = :=
      | r = 1 + \dfrac 1 x
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \map g {a_n}
      | r = 1 + \dfrac {f_n} {f_{n + 1} }
      | c = 
}}
{{eqn | r = \dfrac {f_{n + 1} + f_n} {f_{n + 1} }
      | c = 
}}
{{eqn | r = \dfrac {f_{n + 2} } {f_{n + 1 } }
      | c = {{Defof|Fibonacci Number}}
}}
{{eqn | ll = \leadsto
      | l = \map g {a_n}
      | r = a_{n + 1}
      | c = 
}}
{{end-eqn}}
This sequence represents the iterated dynamical system $a_{n + 1} = \map g {a_n}$ with initial condition $a_1 = 1$.
From the Banach Fixed-Point Theorem, we have:
Let $\struct {M, d}$ be a complete metric space.
Let $f: M \to M$ be a contraction.
Then there exists a '''unique''' fixed point of $f$.
We wish to demonstrate that $\map g x$ is a contraction over the Real Interval $I := \closedint {\dfrac 3 2} 2$ and that $\phi$ is the unique fixed point of $g$ over this complete metric space.
The Banach Fixed-Point Theorem would then guarantee:
:$\ds \lim_{n \mathop \to \infty} \frac {f_{n + 1} } {f_n} = \phi$
We begin by demonstrating that $\phi$ is a fixed point of $g$.
From the {{Defof|Fixed Point}}, we have:
Let $f: S \to T$ be a mapping.
Then a '''fixed point of $S$ under $f$''' is an element $x \in S$ such that $\map f x = x$.
We have that:
{{begin-eqn}}
{{eqn | l = \map g \phi
      | r = 1 + \frac 1 \phi
      | c = 
}}
{{eqn | r = 1 + \dfrac 1 {\frac 1 {\phi - 1} }
      | c = {{Defof|Golden Mean|index = 3}}
}}
{{eqn | r = 1 + \paren {\phi - 1}
      | c = 
}}
{{eqn | r = \phi
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \map g \phi
      | r = \phi
      | c = 
}}
{{end-eqn}}
Hence the golden mean $\phi$ is a fixed point of $g$.
We now demonstrate that $\map g x$ is a contraction over the Real Interval $I := \closedint {\dfrac 3 2} 2$
From the {{Defof|Contraction Mapping (Metric Space)}}, we have:
$f$ is a '''contraction (mapping)''' {{iff}} there exists $\kappa \in \R: 0 \le \kappa < 1$ such that:
:$\forall x, y \in X: \map {d_2} {\map f x, \map f y} \le \kappa \map {d_1} {x, y}$
From Derivative of Reciprocal and Derivative of Constant, we have that:
:$\size {\map {g'} x} = \dfrac 1 {x^2} $
Therefore $\kappa$ of $g$ is bounded above on the Real Interval $I := \closedint {\dfrac 3 2} 2$ by $\dfrac 1 {\paren {3 / 2}^2} < 1$,
Therefore $g$ is a contraction mapping on $I$ and $\phi$ is the unique fixed point of $g$.
Therefore, by the Banach Fixed-Point Theorem:
:$\ds \lim_{n \mathop \to \infty} \dfrac {f_{n + 1} } {f_n} = \lim_{n \mathop \to \infty} a_n = \phi$
{{qed}}
Theorem: Cardinality of Image of Injection

Let $f: S \rightarrowtail T$ be an injection.
Let $A \subseteq S$ be a finite subset of $S$.
Then:
:$\card {f \paren A} = \card A$
where $\card A$ denotes the cardinality of $A$.

Proof:
Proof by induction:
For all $n \in \N_{>0}$, let $P_n$ be the proposition:
:$\card {f \paren A} = \card A$ when $\card A = n$
Suppose $\card A = 0$.
{{begin-eqn}}
{{eqn | l = \card A
      | r = 0
      | c =
}}
{{eqn | ll= \leadsto
      | l = A
      | r = \O
      | c = Cardinality of Empty Set
}}
{{eqn | ll= \leadsto
      | l = f \paren A
      | r = \O
      | c = Image of Empty Set is Empty Set
}}
{{eqn | ll= \leadsto
      | l = \card {f \paren A}
      | r = 0
      | c = Cardinality of Empty Set
}}
{{end-eqn}}
So $P_0$ holds.
Theorem: Composition of Mappings is not Commutative

The composition of mappings is '''not''' in general a commutative binary operation:
:$f_2 \circ f_1 \ne f_1 \circ f_2$

Proof:
;Proof by Counterexample:
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings.
First note that unless $S_1 = S_3$ then $f_2 \circ f_1$ is not even defined.
So in that case $f_2 \circ f_1$ is definitely not the same thing as $f_1 \circ f_2$.
So, let us suppose $S_1 = S_3$ and so we define $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_1$.
If $S_1 \ne S_2$ then:
:$f_2 \circ f_1: S_1 \to S_1$
:$f_1 \circ f_2: S_2 \to S_2$
and so by Equality of Mappings they are unequal because their domains and codomains are different.
Finally, suppose $S_1 = S_2$, and consider the following.
:$S_1 = S_2 = \set {a, b}$
:$f_1 = \set {\tuple {a, a}, \tuple {b, a} }$
:$f_2 = \set {\tuple {a, b}, \tuple {b, b} }$
It is straightforward to check that $f_1$ and $f_2$ are mappings, and that:
:$f_1 \circ f_2 = \set {\tuple {a, b}, \tuple {b, b} }$
:$f_2 \circ f_1 = \set {\tuple {a, a}, \tuple {b, a} }$
Thus, even in this limitingly simple case, we see that:
:$f_2 \circ f_1 \ne f_1 \circ f_2$
{{qed}}
Theorem: Intersection of Injective Image with Relative Complement

Let $f: S \to T$ be a mapping.
Then $f$ is an injection {{iff}}:
:$\forall A \subseteq S: f \sqbrk A \cap f \sqbrk {\relcomp S A} = \O$

Proof:
From Intersection with Relative Complement is Empty:
:$A \cap \relcomp S A = \O$
From Image of Intersection under Injection:
:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$
{{iff}} $f$ is an injection.
Hence the result.
{{qed}}
Theorem: Injection to Image is Bijection

Let $f: S \rightarrowtail T$ be an injection.
Let $X \subseteq T$ be the image of $f$.
Then the restriction $f {\restriction_{S \times X}}: S \to X$ of $f$ to the image of $f$ is a bijection of $S$ onto $X$.

Proof:
We have:
:Restriction of Injection is Injection
:Restriction of Mapping to Image is Surjection
Thus we have that:
:$f {\restriction_{S \times X}}: S \to X$ is an injection
and
:$f {\restriction_{S \times X}}: S \to X$ is a surjection
Hence the result by definition of bijection.
{{qed}}
Theorem: Cardinality of Set of Induced Equivalence Classes of Injection

Let $f: S \to T$ be a mapping.
Let $\RR_f \subseteq S \times S$ be the relation induced by $f$:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
Let $f$ be an injection.
Then there are $\card S$ different $\RR_f$-classes.

Proof:
From Cardinality of Image of Injection we have that $\card {f \sqbrk S} = \card S$.
From the nature of an injection, for all $s \in S$, the $\RR_f$-class of $s$ is a singleton.
Hence the result.
{{qed}}
Theorem: Cardinality of Set of Induced Equivalence Classes of Surjection

Let $f: S \to T$ be a mapping.
Let $\RR_f \subseteq S \times S$ be the relation induced by $f$:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
Let $f$ be a surjection.
Then there are $\card T$ different $\RR_f$-classes.

Proof:
From the definition of a surjection:
:$\forall t \in T: \exists s \in S: \map f s = t$
Thus there are as many $\RR_f$-classes of $f$ as there are elements of $T$.
Hence the result.
{{qed}}
Theorem: Power of Elements is Subgroup

Let $G$ be an abelian group whose identity is $e$.
Then for any $n \in \Z$, the set $G^n = \set {x^n: x \in G}$ is a subgroup of $G$.
Moreover, if:
:$(1): \quad G$ is finite
and:
:$(2): \quad n \ne 1$ is a divisor of the order of $G$
then $G^n$ is a proper subgroup of $G$.

Proof:
As $e^n = e \in G^n$, the set $G^n$ is not empty.
Let $x^n, y^n \in G^n$.
From Power of Product in Abelian Group, we have:
:$x^n \paren {y^n}^{-1} = x^n \paren {y^{-1} }^n = \paren {x y^{-1} }^n$
It follows that $x^n \paren {y^n}^{-1} \in G^n$.
From the One-Step Subgroup Test, we conclude that $G^n$ is a subgroup of $G$.
Now assume $G$ is finite.
Let $n \ne 1$ such that $n \divides \order G$.
For $G^n$ to be a proper subgroup, it is enough to show that the map $x \mapsto x^n$ is not injective.
Now as $n \ne 1$, from the Fundamental Theorem of Arithmetic it is possible to find a prime $p$ such that $p \divides n$.
By Cauchy's Lemma (Group Theory), there is an element $x$ with order $p$.
It then follows that $x^n = e$, but $x \ne e$.
Therefore, the map $x \mapsto x^n$ is not injective.
Hence the result.
{{qed}}
Category:Abelian Groups
5rybzdcslzin548decnogsxbwp6aj2g
Theorem: Index Laws/Sum of Indices/Semigroup

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 $a$:
:<nowiki>$a^n = \begin{cases}
a & : n = 1 \\
a^x \circ a & : n = x + 1
\end{cases}$</nowiki>
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:
:$\forall n \in \N_{>0}: \circ^{n + m} a = \paren {\circ^n a} \circ \paren {\circ^m a}$
Theorem: Index Laws/Product of Indices/Semigroup

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:
:$\forall m, n \in \N_{>0}: a^{n m} = \paren {a^n}^m = \paren {a^m}^n$

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:
{{begin-eqn}}
{{eqn | q = \forall n \in \N_{>0}
      | l = \map h {n + 1}
      | r = \circ^{\paren {n + 1} m} a
      | c = 
}}
{{eqn | r = \circ^{\paren {n m} + m} a
      | c = Natural Number Multiplication Distributes over Addition
}}
{{eqn | r = \paren {\circ ^{n \ast m} } \circ \paren {\circ^m a}
      | c = Index Laws for Semigroup: Sum of Indices
}}
{{eqn | r = \map h n \circ b
      | c = 
}}
{{end-eqn}}
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:
:$\forall n, m \in \N_{>0}: \circ^m \paren {\circ^n a} = \circ^{m n} = \circ^{n m}$
{{qed}}
Theorem: Semigroup is Subsemigroup of Itself

Let $\struct {S, \circ}$ be a semigroup.
Then $\struct {S, \circ}$ is a subsemigroup of itself.

Proof:
For all sets $S$, $S \subseteq S$, that is, $S$ is a subset of itself.
Thus $\struct {S, \circ}$ is a semigroup which is a subset of $\struct {S, \circ}$, and therefore a subsemigroup of $\struct {S, \circ}$.
{{Qed}}
Theorem: Identity of Monoid is Unique

Let $\struct {S, \circ}$ be a monoid.
Then $S$ has a unique identity.

Proof:
As $\struct {S, \circ}$ is an algebraic structure, the result Identity is Unique can be applied directly.
{{qed}}
Theorem: Identity of Group is Unique

Let $\struct {G, \circ}$ be a group which has an identity element $e \in G$.
Then $e$ is unique.

Proof:
By the definition of a group, $\struct {G, \circ}$ is also a monoid.
The result follows by applying the result Identity of Monoid is Unique.
{{qed}}
Theorem: Identity of Group is Unique

Let $\struct {G, \circ}$ be a group which has an identity element $e \in G$.
Then $e$ is unique.

Proof:
Let $e$ and $f$ both be identity elements of a group $\struct {G, \circ}$.
Then:
{{begin-eqn}}
{{eqn | l = e
      | r = e \circ f
      | c = $f$ is an identity
}}
{{eqn | r = f
      | c = $e$ is an identity
}}
{{end-eqn}}
So $e = f$ and there is only one identity after all.
{{qed}}
Theorem: Identity of Group is Unique

Let $\struct {G, \circ}$ be a group which has an identity element $e \in G$.
Then $e$ is unique.

Proof:
From Group has Latin Square Property, there exists a unique $x \in G$ such that:
:$a x = b$
and there exists a unique $y \in G$ such that:
:$y a = b$
Setting $b = a$, this becomes:
There exists a unique $x \in G$ such that:
:$a x = a$
and there exists a unique $y \in G$ such that:
:$y a = a$
These $x$ and $y$ are both $e$, by definition of identity element.
{{qed}}
Theorem: Unity of Integral Domain is Unique

Let $\struct {D, +, \times}$ be an integral domain.
Then the unity of $\struct {D, +, \times}$ is unique.

Proof:
From the definition of an integral domain, $\struct {D, +, \times}$ is a commutative ring such that $\struct {D^*, \circ}$ is a monoid.
The result follows from Identity of Monoid is Unique.
{{qed}}
Theorem: Zero of Integral Domain is Unique

Let $\struct {D, +, \times}$ be an integral domain.
Then the zero of $\struct {D, +, \times}$ is unique.

Proof:
By definition, an integral domain is a ring.
The result the follows from Ring Zero is Unique.
{{qed}}
Theorem: Negative in Integral Domain is Unique

Let $\struct {D, +, \times}$ be an integral domain.
Let $a \in R$.
Then the negative $-a$ of $a$ is unique.

Proof:
From the definition of an integral domain, $\struct {D, +, \times}$ is a ring.
The result follows from Ring Negative is Unique.
{{qed}}
Theorem: Inverse in Group is Unique

Let $\struct {G, \circ}$ be a group.
Then every element $x \in G$ has exactly one inverse:
:$\forall x \in G: \exists_1 x^{-1} \in G: x \circ x^{-1} = e^{-1} = x^{-1} \circ x$
where $e$ is the identity element of $\struct {G, \circ}$.

Proof:
By the definition of a group, $\struct {G, \circ}$ is a monoid each of whose elements has an inverse.
The result follows directly from Inverse in Monoid is Unique.
{{qed}}
Theorem: Inverse in Group is Unique

Let $\struct {G, \circ}$ be a group.
Then every element $x \in G$ has exactly one inverse:
:$\forall x \in G: \exists_1 x^{-1} \in G: x \circ x^{-1} = e^{-1} = x^{-1} \circ x$
where $e$ is the identity element of $\struct {G, \circ}$.

Proof:
Let $\struct {G, \circ}$ be a group whose identity element is $e$.
By {{Group-axiom|3}}, every element of $G$ has at least one inverse.
Suppose that:
:$\exists b, c \in G: a \circ b = e, a \circ c = e$
that is, that $b$ and $c$ are both inverse elements of $a$.
Then:
{{begin-eqn}}
{{eqn | l = b
      | r = b \circ e
      | c = as $e$ is the identity element
}}
{{eqn | r = b \circ \paren {a \circ c}
      | c = as $c$ is an inverse of $a$
}}
{{eqn | r = \paren {b \circ a} \circ c
      | c = {{Group-axiom|1}}
}}
{{eqn | r = e \circ c
      | c = as $b$ is an inverse of $a$
}}
{{eqn | r = c
      | c = as $e$ is the identity element
}}
{{end-eqn}}
So $b = c$ and hence the result.
{{qed}}