nyuuzyou/wikis huggingface dataset:
Positive Rational Numbers under Addition fulfil Naturally Ordered Semigroup Axioms 2 to 4

== Theorem ==
Let $\Q_{\ge 0}$ denote the set of positive rational numbers.
Consider the naturally ordered semigroup axioms:
The algebraic structure:
:$\struct {\Q_{\ge 0}, +, \le}$
is an ordered semigroup which fulfils the axioms:
:
:
:
but:
:does not fulfil 
:$\struct {\Q_{\ge 0}, +}$is not isomorphic to $\struct {\N, +}$.
== Proof ==
First we note that from Positive Rational Numbers under Addition form Ordered Semigroup:
:$\struct {\Q_{\ge 0}, +, \le}$ is an ordered semigroup.
=== ===
From Rational Numbers under Addition form Infinite Abelian Group, $\struct {\Q, +}$ is a group.
By the Cancellation Laws, all elements of $\struct {\Q, +}$ are cancellable.
From Cancellable Element is Cancellable in Subset, it follows that all elements of $\struct {\Q_{\ge 0}, +}$ are likewise cancellable.
Hence holds.
=== ===
Let $a, b \in \Q_{\ge 0}$ such that $a \le b$.
Then as $\struct {\Q, +}$ is a group it follows that:
:$\exists c \in \Q: a + c = b$
and so:
:$c = b + \paren {-a}$
But as $a \le b$ it follows that $c \ge 0$.
That is:
:$\exists c \in \Q_{\ge 0}: a + c = b$
Hence holds.
=== ===
We have that:
:$0 \in \Q_{\ge 0}$
and:
:$1 \in \Q_{\ge 0}$
and trivially holds.
=== ===
We consider the subset $S$ of $\struct {\Q_{\ge 0}, +, \le}$ defined as:
:$S = \set {x \in \Q_{\ge 0}: x > 0}$
From Smallest Strictly Positive Rational Number does not Exist:
:there exists no smallest element of $S$.
Hence by definition $\struct {\Q_{\ge 0}, +, \le}$ is not well-ordered by $\le$.
That is, $\struct {\Q_{\ge 0}, +, \le}$ does not fulfil .
From Positive Rational Numbers under Addition not Isomorphic to Natural Numbers:
:$\struct {\Q_{\ge 0}, +}$is not isomorphic to $\struct {\N, +}$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.1$
Category:Naturally Ordered Semigroup
Positive Rational Numbers under Addition not Isomorphic to Natural Numbers

== Theorem ==
The positive rational numbers $\Q_{\ge 0}$ under addition:
:$\struct {\Q_{\ge 0}, +}$
is not isomorphic to the natural numbers under addition:
:$\struct {\N, +}$
== Proof ==
From:
:Positive Rational Numbers under Addition form Commutative Monoid
:Natural Numbers under Addition form Commutative Monoid
both $\struct {\Q_{\ge 0}, +}$ and $\struct {\N, +}$ form commutative monoids.
 there exists an semigroup isomorphism $\phi$ from $\struct {\Q_{\ge 0}, +}$ to $\struct {\N, +}$.
By definition of isomorphism:
:$\phi$ is a homomorphism
:$\phi$ is a bijection.
Let $n \in \N$ be odd.
Let $q \in \Q_{\ge 0}$ such that $\map \phi q = n$.
Such a $q$ exists and is unique by definition of bijection.
But then we have:
But this contradicts the assertion that $n$ is odd.
So by Proof by Contradiction there can be no such isomorphism from $\struct {\Q_{\ge 0}, +}$ to $\struct {\N, +}$.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.1$
Category:Natural Number Addition
Category:Rational Addition
Positive Rational Numbers under Division do not form Group

== Theorem ==
Let $\struct {\Q_{>0}, /}$ denote the algebraic structure consisting of the set of (strictly) positive rational numbers $\Q_{>0}$ under the operation $/$ of division.
We have that $\struct {\Q_{>0}, /}$ is not a group.
== Proof ==
In order to be a group, it is necessary that $\struct {\Q_{>0}, /}$ be an associative structure.
But consider the elements $2, 6, 12$ of $\Q_{>0}$.
We have:
whereas:
That is:
:$\paren {12 / 6} / 2 \ne 12 / \paren {6 / 2}$
So it is not generally the case that for $a, b, c \in \Q_{>0}$:
:$\paren {a / b} / c = a / \paren {b / c}$
and so $/$ is not associative on $\Q_{>0}$.
Hence $\struct {\Q_{>0}, /}$ is not a group.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: Examples: $(2)$
Category:Rational Numbers
Category:Rational Division
Positive Real Axis forms Subgroup of Complex Numbers under Multiplication

== Theorem ==
Let $S$ be the subset of the set of complex numbers $\C$ defined as:
:$S = \set {z \in \C: z = x + 0 i, x > 0}$
That is, let $S$ be the positive real axis of the complex plane.
Then the algebraic structure $\struct {S, \times}$ is a subgroup of the multiplicative group of complex numbers $\struct {\C_{\ne 0}, \times}$.
== Proof ==
We have that $S$ is the same thing as $\R_{>0}$, the set of strictly positive real numbers:
:$\R_{>0} = \set {x \in \R: x > 0}$
From Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group, $\struct {S, \times}$ is a group.
Hence as $S$ is a group which is a subset of $\struct {\C_{\ne 0}, \times}$, it follows that $\struct {S, \times}$ is a subgroup of $\struct {\C_{\ne 0}, \times}$.
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{E v}$
Category:Real Multiplication
Category:Multiplicative Group of Complex Numbers
Category:Examples of Subgroups
Positive Real Complex Root of Unity

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n = \set {z \in \C: z^n = 1}$ be the set of complex $n$th roots of unity.
The only $x \in U_n$ such that $x \in \R_{>0}$ is:
:$x = 1$
That is, $1$ is the only complex $n$th root of unity which is a positive real number.
== Proof ==
We have that $1$ is a positive real number.
The result follows from Existence and Uniqueness of Positive Root of Positive Real Number.
== Sources ==
* : $\S 3$. Roots of Unity
Category:Complex Roots of Unity
Positive Real Number Inequalities can be Multiplied

== Theorem ==
Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.
Let $b > 0$ and $d > 0$.
Then $a c > b d$.
If $b < 0$ or $d < 0$ the inequality does not hold.
== Proof ==
Note that as $d > 0$ it follows that $c > d > 0$ and so $c > 0$.
Finally:
=== Disproof for Negative Parameters ===
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.8 \ (3)$
Category:Positive Real Number Inequalities can be Multiplied
Category:Real Numbers
Category:Multiplication
Category:Inequalities
Positive Real Number Inequalities can be Multiplied/Disproof for Negative Parameters

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $a = c = -1, b = d = -2$.
Then $a c = 1$ but $b d = 2$.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.8 \ (4) \ \text{(iii)}$
Category:Positive Real Number Inequalities can be Multiplied
Positive Real Number has Simple Continued Fraction Expansion

== Theorem ==
Let $x \in \R_{>0}$ be a (strictly) positive real number.
Then $x$ can be expressed as a simple continued fraction.
== Proof ==
We have that $x$ is either rational or irrational.
;$x$ rational
Let $x$ be rational.
Then from Rational Number can be Expressed as Simple Finite Continued Fraction, $x$ has a simple continued fraction expansion.
;$x$ irrational
Let $x$ be irrational.
The result follows from Correspondence between Irrational Numbers and Simple Infinite Continued Fractions.
== Sources ==
* 
Category:Simple Continued Fractions
Positive Real Numbers under Max Operation form Monoid

== Theorem ==
Let $\R_{\ge 0}$ be the set of positive (that is, non-negative) real numbers.
Let $\max: \R_{\ge 0}^2 \to \R_{\ge 0}$ be the max operation on $\R_{\ge 0}$.
Then $\struct {\R_{\ge 0}, \max}$ is a monoid whose identity is $0$.
== Proof ==
From Real Numbers are Totally Ordered, $\R$ is a totally ordered set.
From Max Operation on Toset forms Semigroup, $\struct {\R_{\ge 0}, \max}$ is a semigroup.
By definition of $\R_{\ge 0}$:
:$\forall x \in \R_{\ge 0}: 0 \le x$
Thus by definition of the max operation:
:$\forall x \in \R_{\ge 0}: \map \max {0, x} = x = \map \max {x, 0}$
So $0$ is the identity of $\struct {\R_{\ge 0}, \max}$ by definition.
The result follows by definition of monoid.
== Sources ==
* : $\S 3.1$: Monoids
Category:Real Numbers
Category:Max Operation
Category:Examples of Monoids
Positive Real Numbers whose Reciprocals Sum to 1

== Theorem ==
Let $p, q \in \R_{\ge 0}$ be positive real numbers such that:
:$\dfrac 1 p + \dfrac 1 q = 1$
Then $p > 1$ and $q > 1$.
== Proof ==
From Division by Zero it is immediate that $p > 0$ and $q > 0$.
 either $0 < p \le 1$ or $0 < q \le 1$.
, suppose $0 < p \le 1$.
Note we have that:
:$\dfrac 1 q = 1 - \dfrac 1 p$
First suppose $p = 1$.
Then:
:$\dfrac 1 q = 0$
But there exists no $q \in \R$ such that $\dfrac 1 q = 0$.
Hence it cannot be the case that $p = 1$.
Now suppose $0 < p < 1$.
Then:
But this contradicts our assertion that $q > 0$.
Hence by Proof by Contradiction it follows that $p > 1$.
Similarly we reach a contradiction by assuming $0 < q \le 1$.
Hence $p > 1$ and $q > 1$ as required.
Category:Real Analysis
Positive Real has Real Square Root

== Theorem ==
Let $x \in \R_{>0}$ be a (strictly) positive real number.
Then:
:$\exists y \in \R: x = y^2$
== Proof ==
Let $f: \R \to \R$ be defined as:
:$\forall x \in \R: \map f x = x^2$
We have that $f$ is the pointwise product of the identity mapping with itself.
By Product Rule for Continuous Real Functions and Identity Mapping is Continuous, $f$ is continuous.
By Power Function is Unbounded Above:
:$\exists q \in \R: \map f q > x$
Then:
:$0^2 = 0 \le x$
By the Intermediate Value Theorem:
:$\exists y \in \R: 0 < y < q: y^2 = x$
Category:Real Numbers
Positive Scalar Multiple of Norm on Vector Space is Norm

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$. 
Let $\alpha > 0$ be a real number. 
Define $\norm {\, \cdot \,}' : X \to \R_{\ge 0}$ by: 
:$\norm x' = \alpha \norm x$
for each $x \in X$.
Then $\norm {\, \cdot \,}'$ is a norm on $X$. 
== Proof ==
=== ===
Suppose that $x \in X$ is such that: 
:$\norm x' = 0$
Then we have: 
:$\alpha \norm x = 0$
Since $\alpha > 0$, it follows that: 
:$\norm x = 0$ 
From , we have $x = {\mathbf 0}_X$.
=== ===
Let $\lambda \in \GF$ and $x \in X$.
Then we have: 
=== ===
Let $x, y \in X$.
Then we have: 
Category:Normed Vector Spaces
Positive and Negative Parts of Signed Measure are Mutually Singular

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space. 
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then $\mu^+$ and $\mu^-$ are mutually singular.
== Proof ==
From the Hahn Decomposition Theorem, there exists $\mu$-positive set and a $\mu$-negative set such that:
:$X = P \cup N$
and:
:$P \cap N = \O$ 
From the Jordan Decomposition Theorem, we have:
:$\map {\mu^+} A = \map \mu {A \cap P}$
and:
:$\map {\mu^-} A = -\map \mu {A \cap N}$
for each $A \in \Sigma$.
From the definition of a signed measure we have $\map \mu \O = 0$. 
Hence $\map {\mu^+} N = 0$ and $\map {\mu^-} {X \setminus N} = 0$.
So $\mu^+$ is concentrated on $N^c$ and $\mu%-$ is concentrated on $N$. 
So $\mu^+$ and $\mu^-$ are mutually singular.
Category:Mutually Singular Measures
Category:Signed Measures
Postfix Notation/Examples

== Examples of Postfix Notation ==
=== Example: $2 3 +$ ===
=== Example: $3 4 \times 2 +$ ===
=== Example: $2 3 + 4 \times$ ===
=== Example $1$ of Use of Łukasiewicz's Notation ===
=== Example $2$ of Use of Łukasiewicz's Notation ===
Category:Examples of Postfix Notation
Postfix Notation/Examples/Łukasiewicz's Notation Example 1

== Example of Postfix Notation ==
's Polish notation for propositional logic can also be implemented in its reverse Polish form.
Hence:
:$p q \operatorname K$
can be used to express the conjunction:
:$p \land q$
which would be expressed in Polish notation as:
:$\operatorname K p q$
== Sources ==
* 
Category:Examples of Postfix Notation
Postfix Notation/Examples/Łukasiewicz's Notation Example 2

== Example of Postfix Notation ==
's Polish notation for propositional logic can also be implemented in its reverse Polish form.
Hence the expression in postfix notation:
:$p q \operatorname K p q \operatorname N \operatorname A \operatorname C$
can be used to express the compound statement:
:$\paren {p \land q} \implies \paren {p \lor \lnot q}$
which would be expressed in prefix notation as:
:$\operatorname C \operatorname K p q \operatorname A p \operatorname N q$
== Sources ==
* 
Category:Examples of Postfix Notation
Potato Paradox

== Paradox ==
I dug my potatoes out of the ground after a particularly wet season when my garden got badly waterlogged.
When I finally got them up, I found that of my $100 \, \mathrm {kg}$ of taties, they were $99 \%$ water. Very soggy potatoes.
So I set them in the (finally arrived) sun to dry for a bit.
When I looked at them next, they were now $98 \%$ water.
How heavy are my taters now?
== Solution ==
$50 \, \mathrm {kg}$.
What, really?
== Resolution ==
Let $x \, \mathrm {kg}$ denote the dry stuff in the potato.
We have that $x$ forms $1 \%$ of the mass of the potatoes before we start drying them.
That is, $x = \dfrac 1 {100}$ of $100 \, \mathrm {kg}$:
:$x = 1 \, \mathrm {kg}$
After drying them, $x$ is $2 \%$, or $\dfrac 1 {50}$, of the mass of the potatoes.
But still $x = 1 \, \mathrm {kg}$.
So the weight of the potatoes after drying them is $50 \, \mathrm {kg}$.
Category:Veridical Paradoxes
Potential Energy/Examples/1 newton raised 1 metre

== Examples of Potential Energy ==
A body which requires $1$ newton to hold it against a gravitational force when moved against that force through a distance of $1$ metre has $1$ joule of potential energy transferred to it.
== Proof ==
== Sources ==
* : Chapter $1$: Electrical Quantities: Definitions and Laws: Definitions: Energy
Category:Examples of Potential Energy
Potential Energy of Position

== Physical Law ==
Let $B$ be a body of mass $m$ in a gravitational field.
Let the Acceleration Due to Gravity be of constant magnitude $g$.
Let $B$ be at a height $h$ above a reference level $h_0$.
Then the potential energy $P$ of $B$ due to the position of $B$ relative to $h_0$ is given by:
:$P = m g h$
== Proof ==
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems
Category:Physics
Category:Mechanics
Poulet Number/Examples/341

== Theorem ==
The smallest Poulet number is $341$:
:$2^{341} \equiv 2 \pmod {341}$
despite the fact that $341$ is not prime:
:$341 = 11 \times 31$
== Proof ==
We have that:
== Sources ==
* : $341$
* : $341$
* 
* 
Category:Poulet Numbers
Category:341
Poulet Numbers which are also Magic Constant for Magic Square

== Theorem ==
The sequence of Poulet numbers which are also the magic constant of a magic square begins:
:$1105, 2465, \ldots$
== Proof ==
From the sequence of Poulet numbers, these are Poulet numbers:
:$1105, 2465, \ldots$
Then we have:
Category:Poulet Numbers
Category:Magic Squares
Power (Algebra)/Examples

== Examples of Integer Powers ==
=== Sequence of Powers of 2 ===
=== Sequence of Powers of 3 ===
=== Sequence of Powers of 4 ===
=== Sequence of Powers of 5 ===
=== Sequence of Powers of 6 ===
=== Sequence of Powers of 7 ===
=== Sequence of Powers of 8 ===
=== Sequence of Powers of 9 ===
=== Sequence of Powers of 10 ===
=== Sequence of Powers of 11 ===
=== Sequence of Powers of 12 ===
Category:Examples of Real Powers
Power (Algebra)/Rational Number/Examples

== Examples of Rational Powers ==
=== Fractional Power: $8^{2/3}$ ===
=== Fractional Power: $27^{-4/3}$ ===
=== Fractional Power: $32^{6/5}$ ===
=== Fractional Power: $0 \cdotp 125^{-2/3}$ ===
Category:Examples of Real Powers
Power (Algebra)/Rational Number/Examples/27 to the Power of Minus Four Thirds

== Example of Rational Power ==
:$27^{-4/3} = \dfrac 1 {81}$
== Proof ==
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.12 \ (2) \, \text{(ii)}$
Category:Examples of Real Powers
Power (Algebra)/Real Number/Examples

== Examples of Power to Real Number ==
=== Euler's Number to Power of Itself ===
=== Euler's Number to Power of its Negative ===
=== Euler's Number to Power of its Reciprocal ===
=== Euler's Number to Power of Euler-Mascheroni Constant ===
=== Euler's Number to Power of Minus the Euler-Mascheroni Constant ===
Category:Examples of Real Powers
Power (Algebra)/Real Number/Examples/e to the Power of Minus gamma

== Example of Power to Real Number ==
Euler's Number $e$ to the power of the negative of the Euler-Mascheroni constant is approximately equal to:
:$e^{-\gamma} \approx 0 \cdotp 56145 \, 94835 \, 66885 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $0,56145 94835 66885 \ldots$
Category:Examples of Real Powers
Category:Euler's Number
Category:Euler-Mascheroni Constant
Power (Algebra)/Real Number/Examples/e to the Power of gamma

== Example of Power to Real Number ==
Euler's Number $e$ to the power of the Euler-Mascheroni constant is approximately equal to:
:$e^\gamma \approx 1 \cdotp 78107 \, 24179 \, 90197 \, 9852 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.21$
* : $\S 1$: Special Constants: $1.4.$
Category:Examples of Real Powers
Category:Euler's Number
Category:Euler-Mascheroni Constant
Power Dominates Logarithm

== Theorem ==
Let $\epsilon \in \R_{>0}$.
Let $B \in \N$ be arbitrary.
Then there exists $N \in \N$ such that:
:$\forall n > N: \paren {\ln n}^B < n^\epsilon$
== Proof ==
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
First we show that there exists $N$ such that $\ln n < n^\epsilon$ for all $n > N$,.
By Exponential is Strictly Increasing, the real exponential function is strictly increasing.
Therefore:
:$\ln n < n^\epsilon \iff n < \map \exp {n^\epsilon}$
Choose $k \in \N$ such that $k \epsilon > 2$ and $N = k!$.
By Taylor Series Expansion for Exponential Function:
:$\ds \map \exp {n^\epsilon} = \sum_{m \mathop \ge 0} \frac {n^{m \epsilon} } {m!} > \frac {n^{k \epsilon} } {k!}$
Then for all $n > N$:
This completes the proof when $B = 1$.
Now let $B \in \N$ be arbitrary.
By the above we can find $N \in \N$ such that:
:$\forall n > N: \ln n < n^{\epsilon / B}$
Then:
:$\forall n > N: \paren {\ln n}^B < \paren {n^{\epsilon / B} }^B = n^\epsilon$
Hence the result.
Category:Logarithms
Category:Powers
Power Function Preserves Ordering in Ordered Group

== Theorem ==
Let $\struct {S, \circ, \preccurlyeq}$ be an ordered group.
Let $n \in \N_{>0}$ be a strictly positive integer.
Let $\prec$ be the reflexive reduction of $\preccurlyeq$.
Then the following hold:
where $x^n$ denotes the $n$th power of $x$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
Category:Ordered Groups
Category:Power Function Preserves Ordering in Ordered Group
Power Function Preserves Ordering in Ordered Group/Corollary

== Corollary to Power Function Preserves Ordering in Ordered Group ==
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group with identity $e$.
Let $\prec$ be the reflexive reduction of $\preceq$.
Let $x \in G$.
Let $n \in \N_{>0}$ be a strictly positive integer.
Then the following hold:
== Proof 1 ==
== Proof 2 ==
Category:Ordered Groups
Category:Power Function Preserves Ordering in Ordered Group
Power Function Preserves Ordering in Ordered Group/Corollary/Proof 1

== Corollary to Power Function Preserves Ordering in Ordered Group ==
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group with identity $e$.
Let $\prec$ be the reflexive reduction of $\preceq$.
Let $x \in G$.
Let $n \in \N_{>0}$ be a strictly positive integer.
Then the following hold:
== Proof ==
By Power Function Preserves Ordering in Ordered Group:
By Identity Element is Idempotent, $e$ is idempotent with respect to $\circ$.
Therefore by the definition of an idempotent element, $e^n = e$.
Thus the theorem holds.
Category:Power Function Preserves Ordering in Ordered Group
Power Function Preserves Ordering in Ordered Group/Corollary/Proof 2

== Corollary to Power Function Preserves Ordering in Ordered Group ==
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group with identity $e$.
Let $\prec$ be the reflexive reduction of $\preccurlyeq$.
Let $x \in G$.
Let $n \in \N_{>0}$ be a strictly positive integer.
Then the following hold:
== Proof ==
By the definition of an ordered group, $\preccurlyeq$ is a transitive relation compatible with $\circ$.
By Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements:
:$x \preccurlyeq e \implies x^n \preccurlyeq e^n$
:$e \preccurlyeq x \implies e^n \preccurlyeq x^n$
By Identity Element is Idempotent, $e$ is idempotent with respect to $\circ$.
Thus we obtain the first two results:
:$x \preccurlyeq e \implies x^n \preccurlyeq e$
:$e \preccurlyeq x \implies e \preccurlyeq x^n$
By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is compatible with $\circ$.
By Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering, $\prec$ is transitive.
Thus by the same method as above, we obtain the remaining results:
:$x \prec e \implies x^n \prec e$
:$e \prec x \implies e \prec x^n$
Category:Power Function Preserves Ordering in Ordered Group
Power Function Preserves Ordering in Ordered Group/Proof 1

== Theorem ==
== Proof ==
By definition of ordered group:
:$\preccurlyeq$ is compatible with $\circ$.
By definition of ordering:
:$\preccurlyeq$ is transitive.
From Reflexive Reduction of Relation Compatible with Group Operation is Compatible:
:$\prec$ is also compatible with $\circ$.
From Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering:
:$\prec$ is also transitive.
By definition of ordered group:
:$\struct {S, \circ}$ is a ordered group, and therefore a semigroup.
The result follows from Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements.
Category:Power Function Preserves Ordering in Ordered Group
Power Function Preserves Ordering in Ordered Group/Proof 2

== Theorem ==
== Proof ==
An ordered group is an ordered structure which is also a group.
Hence an ordered group is an ordered semigroup.
From Power Function Preserves Ordering in Ordered Semigroup:
:$\forall x, y \in S: x \preccurlyeq y \implies x^n \preccurlyeq y^n$
From the Cancellation Laws, every element of a group is cancellable.
Hence from Power Function with Cancellable Element Preserves Strict Ordering in Ordered Semigroup:
:$\forall x, y \in S: x \prec y \implies x^n \prec y^n$
Category:Power Function Preserves Ordering in Ordered Group
Power Function Preserves Ordering in Ordered Semigroup

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $x, y \in S$ such that $x \preceq y$.
Let $n \in \N_{>0}$ be a strictly positive integer.
Then:
:$x^n \preceq y^n$
where $x^n$ is the $n$th power of $x$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.4$
Category:Ordered Semigroups
Category:Power Function Preserves Ordering in Ordered Semigroup
Power Function Preserves Ordering in Ordered Semigroup/Proof 1

== Theorem ==
== Proof ==
By definition of ordered semigroup:
:$\preceq$ is compatible with $\circ$.
By definition of ordering:
:$\preceq$ is transitive.
Thus by Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements:
:$x^n \preceq y^n$
Category:Power Function Preserves Ordering in Ordered Semigroup