nyuuzyou/wikis huggingface dataset:
Power of Product in Abelian Group

== Theorem ==
Let $\struct {G, \circ}$ be an abelian group.
Then:
:$\forall x, y \in G: \forall k \in \Z: \paren {x \circ y}^k = x^k \circ y^k$
=== Additive Notation ===
This can also be written in additive notation as:
== Proof ==
By definition of abelian group, $x$ and $y$ commute.
That is:
:$x \circ y = y \circ x$
The result follows from Power of Product of Commutative Elements in Group.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.7)$
* : Chapter $14$: The classification of finite abelian groups: Proposition $14.2$
Category:Abelian Groups
Power of Product in Abelian Group/Additive Notation

== Theorem ==
Let $\struct {G, +}$ be an abelian group.
Then:
:$\forall x, y \in G: \forall k \in \Z: k \cdot \paren {x + y} = \paren {k \cdot x} + \paren {k \cdot y}$
== Proof ==
By definition of abelian group, $x$ and $y$ commute.
That is:
:$x + y = y + x$
The result follows from Power of Product of Commutative Elements in Group.
== Sources ==
* : $\S 2.1$: Subrings: Notation $2$
Category:Abelian Groups
Power of Product of Commutative Elements in Group

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $a, b \in G$.
Then:
:$a \circ b = b \circ a \iff \forall n \in \Z: \paren {a \circ b}^n = a^n \circ b^n$
That is:
:$a$ and $b$ commute
:
:the product of their powers equals the power of their product:
This can be expressed in additive notation in the group $\struct {G, +}$ as:
:$a + b = b + a \iff \forall n \in \Z: n \cdot \paren {a + b} = \paren {n \cdot a} + \paren {n \cdot b}$
== Proof ==
=== Necessary Condition ===
Let $a \circ b = b \circ a$.
By definition, all elements of a group are invertible.
Therefore the results in Power of Product of Commutative Elements in Monoid can be applied directly.
=== Sufficient Condition ===
If $\paren {a \circ b}^n = a^n \circ b^n$ for all $n$, then it certainly holds for $n = 2$:
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: $(2)$ Groups of order $21$: Proposition $12.2$
Category:Group Theory
Category:Commutativity
Power of Product of Commutative Elements in Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $a, b \in S$ be invertible elements for $\circ$ that also commute.
Then:
:$\forall n \in \Z: \paren {a \circ b}^n = a^n \circ b^n$
== Proof ==
From Power of Product of Commutative Elements in Semigroup, this result holds if $n \ge 0$.
Since $a$ and $b$ commute, then so do $a^{-1}$ and $b^{-1}$ by Commutation of Inverses in Monoid.
Hence, if $n > 0$:
The result follows.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.11 \ (4)$
* : $\S 3.1$: Monoids
Category:Monoids
Category:Commutativity
Power of Product of Commutative Elements in Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $x, y \in S$ both be cancellable elements of $S$.
Then:
:$\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n \iff x \circ y = y \circ x$
== Proof ==
=== Necessary Condition ===
Let $x \circ y = y \circ x$.
Then by Power of Product of Commuting Elements in Semigroup equals Product of Powers:
:$\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n$
=== Sufficient Condition ===
Suppose $\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n$.
In particular, when $n = 2$,
== Examples ==
== Sources ==
* : Chapter $4$. Groups: Exercise $8$
Category:Semigroups
Category:Commutativity
Category:Power of Product of Commutative Elements in Semigroup
Power of Product of Commutative Elements in Semigroup/Examples/Elements of 3rd Symmetric Group

== Examples of Use of Power of Product of Commutative Elements in Semigroup ==
Let $S = \set {1, 2, 3}$.
Let $S_3$ denote the symmetric group on $3$ letters.
Let $\rho, \sigma \in S_3$ defined in two-row notation as:
Then:
:$\rho^2 \sigma^2 \ne \paren {\rho \sigma}^2$
== Proof ==
== Sources ==
* : Chapter $4$. Groups: Exercise $9$
Category:Power of Product of Commutative Elements in Semigroup
Power of Product of Commuting Elements in Monoid equals Product of Powers

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.
For $a \in S$, let $\circ^n a = a^n$ denote the $n$th power of $a$.
Let $a, b \in S$ such that $a$ commutes with $b$:
:$a \circ b = b \circ a$
Then:
:$\forall n \in \N: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
That is:
:$\forall n \in \N: \paren {a \circ b}^n = a^n \circ b^n$ 
== Proof ==
Because $\struct {S, \circ}$ is a monoid, it is a fortiori also a semigroup.
From Power of Product of Commuting Elements in Semigroup equals Product of Powers:
:$\forall n \in \N_{>0}: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
That is:
:$\forall n \in \N_{>0}: \paren {a \circ b}^n = a^n \circ b^n$
It remains to be shown that the result holds for the cases where $n = 0$.
Thus:
Thus:
:$\paren {a \circ b}^n = a^n \circ b^n$
holds for $n = 0$.
Thus:
:$\forall n \in \N: \paren {a \circ b}^n = a^n \circ b^n$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.8$
Category:Monoids
Category:Commutativity
Category:Powers (Abstract Algebra)
Power of Product of Commuting Elements in Semigroup equals Product of Powers

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
For $a \in S$, let $\circ^n a = a^n$ denote the $n$th power of $a$.
Let $a, b \in S$ such that $a$ commutes with $b$:
:$a \circ b = b \circ a$
Then:
:$\forall n \in \N_{>0}: \map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
That is:
:$\forall n \in \N_{>0}: \paren {a \circ b}^n = a^n \circ b^n$ 
== Proof ==
The proof proceeds by the Principle of Mathematical Induction:
Let $\map P n$ be the proposition:
:$\map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
=== Basis of the Induction ===
So $\map P 1$ holds.
This is the basis for the induction.
=== Induction Hypothesis ===
Suppose that $\map P k$ holds:
:$\map {\circ^k} {a \circ b} = \paren {\circ^k a} \circ \paren {\circ^k b}$
This is the induction hypothesis.
It remains to be shown that:
:$\map P k \implies \map P {k + 1}$
That is, that:
:$\map {\circ^{k + 1} } {a \circ b} = \paren {\circ^{k + 1} a} \circ \paren {\circ^{k + 1} b}$
=== Induction Step ===
This is the induction step:
So $\map P {k + 1}$ holds.
Thus by the Principle of Mathematical Induction, the result holds for all $n \in \N_{>0}$:
:$\forall n \in \N_{>0}: \map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
== Sources ==
* : $\S 4.2$. Commutative and associative operations: Example $70$
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.8 \ (3)$
Category:Semigroups
Category:Commutativity
Category:Powers (Abstract Algebra)
Power of Product with Inverse

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $a, b \in G: a b = b a^{-1}$.
Then:
: $\forall n \in \Z: a^n b = b a^{-n}$
== Proof ==
Proof by induction:
For all $n \in \Z$, let $\map P n$ be the proposition $a^n b = b a^{-n}$.
$\map P 0$ is trivially true, as $a^0 b = e b = b = b e = b a^{-0}$.
=== Basis for the Induction ===
$\map P 1$ is true, as this is the given relation between $a$ and $b$:
:$a b = b a^{-1}$
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:
:$a^k b = b a^{-k}$
Then we need to show:
:$a^{k + 1} b = b a^{-\paren {k + 1} }$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore $\forall n \in \N: a^n b = b a^{-n}$.
Now we show that $\map P {-1}$ holds, that is, $a^{-1} b = b a$.
thus showing that $\map P {-1}$ holds.
The proof that $\map P n$ holds for all $n \in \Z: n < 0$ then follows by induction, similarly to the proof for $n > 0$.
Category:Group Theory
Power of Random Variable with Continuous Uniform Distribution has Beta Distribution

== Theorem ==
Let $X \sim \ContinuousUniform 0 1$ where $\ContinuousUniform 0 1$ is the continuous uniform distribution on $\closedint 0 1$.
Let $n$ be a positive real number. 
Then:
:$X^n \sim \BetaDist {\dfrac 1 n} 1$
where $\operatorname {Beta}$ is the beta distribution.
== Proof ==
Let:
:$Y \sim \BetaDist {\dfrac 1 n} 1$
We aim to show that: 
:$\map \Pr {Y < x^n} = \map \Pr {X < x}$
for all $x \in \closedint 0 1$. 
We have:
Category:Continuous Uniform Distribution
Category:Beta Distribution
Power of Real Number between Zero and One is Bounded

== Theorem ==
Let $x \in \R$ be a real number.
Let $0 < x < 1$.
Let set $S = \set {x^n: n \in \N}$.
Then:
:$\inf S = 0$
and:
:$\sup S = 1$
where $\inf S$ and $\sup S$ are the infimum and supremum of $S$ respectively.
== Proof ==
When $n = 0$ it follows that $x^n = 1$ and so $\sup S \ge 1$.
Let $x^k \in S$.
As $x < 1$, it follows from Real Number Ordering is Compatible with Multiplication that:
:$x^{k + 1} < x^k$
So:
:$\forall x \in S: x \le 1$
Hence it follows that $\sup S = 1$.
As $x > 0$, it follows by Real Number Ordering is Compatible with Multiplication that:
:$\forall x \in S: x \ge 0$
Therefore $x$ is a lower bound of $S$.
Now suppose $h > 0$ is also a lower bound of $S$.
Then:
:$\forall n \in \N: x^n \ge h$
and:
:$\forall n \in \N: \paren {\dfrac 1 x}^n \le \dfrac 1 h$
But as $0 < x < 1$ it follows that:
:$\dfrac 1 x > 1$
Thus $\dfrac 1 h$ is an upper bound for $\set {\paren {\dfrac 1 x}^n: n \in \N}$.
From Power of Real Number greater than One is Unbounded Above, this has been shown to be unbounded above.
Therefore there can be no such lower bound $h > 0$ of $S$.
Hence $\inf S = 0$.
== Sources ==
* : $\S 3$: Natural Numbers: Exercise $\S 3.6 \ (2)$
Category:Powers
Power of Real Number greater than One is Unbounded Above

== Theorem ==
Let $x \in \R$ be a real number such that $x > 1$.
Let set $S = \set {x^n: n \in \N}$.
Then $S$ is unbounded above.
== Proof ==
 $S$ were bounded above.
Then $S$ has a supremum $B$.
As $x > 1$, it follows that $\dfrac B x < B$ and so therefore $\dfrac B x$ can not be an upper bound.
Therefore:
:$\exists n \in \N: x^n > \dfrac B x \implies x^{n + 1} > B$
So $B$ can not be an upper bound.
From that contradiction it can be concluded that $S$ can not have an upper bound.
So by Proof by Contradiction it follows that $S$ is unbounded above.
== Sources ==
* : $\S 3$: Natural Numbers: Exercise $\S 3.6 \ (2)$
Category:Powers
Power of Ring Negative

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $n \in \N_{>0}$ be a strictly positive integer.
Let $x \in R$.
Then:
:If $n$ is even:
:::$\map {\circ^n} {-x} = \map {\circ^n} x$
:If $n$ is odd:
:::$\map {\circ^n} {-x} = -\map {\circ^n} x$
== Proof ==
First, suppose that $n$ is even.
Then for some $m \in \N_{>0}$:
:$n = 2 m = m + m$
Thus since $\circ$ is associative:
:$\ds \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \paren {-x} \circ \paren {-x}$
By Product of Ring Negatives:
:$\paren {-x} \circ \paren {-x} = x \circ x = \map {\circ^2} x$
Thus:
:$\ds \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \map {\circ^2} x$
By associativity:
:$\map {\circ^n} {-x} = \map {\circ^{2 m} } x = \map {\circ^n} x$
Now suppose instead that $n$ is odd.
If $n = 1$, then:
:$\map {\circ^n} {-x} = -x = -\map {\circ^n} x$
Otherwise:
Category:Ring Theory
Category:Powers
Power of Root of Unity Equals Power of Remainder

== Theorem ==
Let $n \in \Z_{> 0}$ be a strictly positive integer.
Let $F$ be a field.
Let $\alpha$ be an $n$-th root of unity.
Let $k \in \Z$.
Then:
:$\alpha^k = \alpha^r$
where $0 \le r < n$ is the remainder of $k$ on division by $n$.
== Proof ==
From Division Theorem:
:$\exists r, c \in \Z : 0 \le r < n : k = r + c n$
By :
:$k \equiv r \pmod n$
From Congruent Powers of Root of Unity are Equal
:$\alpha^k = \alpha^r$
Category:Roots of Unity
Power of Strictly Positive Real Number is Strictly Positive/Positive Integer

== Theorem ==
Let $x \in \R_{>0}$ be a (strictly) positive real number.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$x^n > 0$
where $x^n$ denotes the $n$th power of $x$.
== Proof ==
Proof by Mathematical Induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\forall x \in \R_{>0}: x^n > 0$
$\map P 0$ is true, as this just says:
=== Basis for the Induction ===
$\map P 1$ true, as this just says:
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0 $, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\forall x \in \R_{>0}: x^k > 0$
Then we need to show:
:$\forall x \in \R_{>0}: x^{k + 1} > 0$
=== Inductive Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \forall x \in \R_{>0}: x^n > 0$
Category:Power of Positive Real Number is Positive
Power of Sum Modulo Prime

== Theorem ==
Let $p$ be a prime number.
Then:
:$\paren {a + b}^p \equiv a^p + b^p \pmod p$
=== Corollary ===
== Proof ==
From the Binomial Theorem:
:$\ds \paren {a + b}^p = \sum_{k \mathop = 0}^p \binom p k a^k b^{p - k}$
Also note that:
:$\ds \sum_{k \mathop = 0}^p \binom p k a^k b^{p-k} = a^p + \sum_{k \mathop = 1}^{p - 1} \binom p k a^k b^{p - k} + b^p$
So:
== Also see ==
* Freshman's Dream
* Prime Power of Sum Modulo Prime
== Sources ==
* : Chapter $1$: Properties of the Natural Numbers: $\S 23 \eta$
* : $\S 2.3$: Congruences: Exercise $8$
Category:Number Theory
Category:Combinatorics
Category:Prime Numbers
Power of Sum Modulo Prime/Corollary

== Corollary to Power of Sum Modulo Prime ==
Let $p$ be a prime number.
Then:
:$\paren {1 + b}^p \equiv 1 + b^p \pmod p$
== Proof ==
Follows immediately from Power of Sum Modulo Prime by putting $a = 1$.
== Sources ==
* : $\S 2.6$. Algebra of congruences: Example $42 \ (3)$
Category:Number Theory
Category:Combinatorics
Category:Prime Numbers
Power of Two is Even-Times Even Only

== Theorem ==
Let $a > 2$ be a power of $2$.
Then $a$ is even-times even only.
== Proof ==
As $a$ is a power of $2$ greater than $2$:
:$\exists k \in \Z_{>1}: a = 2^k$
Thus:
:$a = 2^2 2^{k - 2}$
and so has $2^2 = 4$ as a divisor.
Let $b$ be an odd number.
By definition:
:$b \nmid 2$
The result follows by Integer Coprime to Factors is Coprime to Whole.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Power of a Point Theorem

== Theorem ==
Let $C$ be a circle in the Euclidean plane whose center is $O$ and whose radius is $r$.
Let $P$ be an arbitrary point in the plane.
Let $p$ be the power of point $P$ $C$.
Let a directed line segment from $P$ be drawn either:
:intersecting $C$ at two points $A$ and $A'$
or:
:tangent to $C$ at $A = A'$.
Then:
:$PA \cdot PA' = p$
== Proof ==
Let $d$ be the distance from $P$ to $O$.
Let $t$ be the length of the tangent from $P$ to $C$.
We use the following several times.
;$P$ is Circle Center
By definition of center of circle:
:a straight line through $O$ is a diameter of $C$.
;$P$ is Interior Point of $C$
:
Let $P$ be in the interior of $C$ but not at $O$.
Let $BB'$ be the diameter of $C$ through $P$.
;$P$ is on circle $C$
;$P$ is Exterior Point and $PA$ tangent to $C$
Let $A$ be such that $PA$ is tangent to $C$ at $A$.
We have that the point $A'$ is the same point as $A$.
;$P$ is Exterior Point and $PAA'$ a secant of $C$
The result follows.
== Sources ==
* Intersecting Chords Theorem, of which the Power of a Point Theorem is a generalization
== Sources ==
* 
Category:Power of a Point Theorem
Category:Power of Point
Category:Circles
Category:Named Theorems
Power of n equalling (n - 1)! + 1

== Theorem ==
There is exactly one solution to the equation in the integers:
:$\paren {n - 1}! + 1 = n^k$
for $k > 1$, and that is:
:$n = 5$
:$k = 2$
== Theorem ==
We have that:
Let $n > 5$.
Suppose $n$ is composite.
By Divisibility of n-1 Factorial by Composite n:
:$n \divides \paren {n - 1}!$
and thus:
:$n \nmid \paren {n - 1}! + 1$
showing that $\paren {n - 1}! + 1$ is not a power of $n$.
Suppose $n$ is prime.
Further suppose:
:$\exists k \in \N: \paren {n - 1}! + 1 = n^k$
Then:
By Divisibility of n-1 Factorial by Composite n, since $n - 1 \ne 4$ and is composite:
:$n - 1 \divides \paren {n - 2}!$
thus $k$ must be a multiple of $n - 1$.
However:
which shows that no multiple of $n - 1$ can satisfy our equation.
Hence there is no solution for $n > 5$.
== Historical Note ==
== Sources ==
* : No. $113$
* : $25$
* : $25$
Category:Factorials
Power over Factorial

== Theorem ==
Let $x \in \R: x > 0$ be a positive real number.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = \dfrac {x^n} {n!}$.
Then $\sequence {x_n}$ converges to zero.
== Proof ==
We need to show that $x_n \to 0$ as $n \to \infty$.
Let $N \in \N$ be the smallest natural number which satisfies $N > x$.
(From the Axiom of Archimedes, such an $N$ always exists.)
First we show that:
: $\forall n > N: \dfrac {x^n} {n!} \le \dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{n - N + 1}$
Note that as $N > x$, $\dfrac x N < 1$.
Also:
:$m > n \implies \dfrac x m < \dfrac x n$
Thus:
As $\dfrac x N < 1$, it follows from Sequence of Powers of Number less than One that $\paren {\dfrac x N}^n \to 0$ as $n \to \infty$.
For a given $x$ and $N$, $\dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N}$ is constant.
Thus by the Multiple Rule for Real Sequences:
:$\dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N} \paren {\dfrac x N}^n \to 0$ as $n \to \infty$
As (from above):
:$\dfrac {x^n} {n!} \le \dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N} \paren {\dfrac x N}^n$
the result follows from the Squeeze Theorem for Real Sequences.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.20 \ (4)$
Category:Limits of Sequences
Power to Characteristic Power of Field is Monomorphism

== Theorem ==
Let $F$ be a field whose characteristic is $p$ where $p \ne 0$.
Let $n \in \Z_{\ge 0}$ be any positive integer.
Let $\phi_n: F \to F$ be the mapping on $F$ defined as:
:$\forall x \in F: \map {\phi_n} x = x^{p^n}$
Then $\phi_n$ is a (field) monomorphism.
== Proof ==
Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\phi_n$ is a (field) monomorphism.
$\map P 0$ is trivially true:
:$\map {\phi_0} x = x^{p^0} = x^1 = x$
and we see that $\phi_0$ is the identity automorphism.
This is not the zero homomorphism.
So from Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi_0$ is a ring monomorphism.
=== Basis for the Induction ===
First we need to show that $\map P 1$ is true:
:$\map {\phi_1} x = x^{p^1} = x^p$ is a (field) monomorphism.
This is demonstrated to be a monomorphism in Power to Characteristic of Field is Monomorphism.
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:
:$\map {\phi_k} x = x^{p^k}$ is a (field) monomorphism.
Then we need to show:
:$\map {\phi_{k + 1} } x = x^{p^{k + 1} }$ is a (field) monomorphism.
=== Induction Step ===
This is our induction step:
Multiplication is more straightforward:
and does not rely on the induction process.
Thus, $\phi_{k + 1}$ is a homomorphism.
$\phi_{k + 1}$ is not the zero homomorphism, since $\map {\phi_{k + 1} } 1 = 1^{p^{k + 1} } = 1 \ne 0$.
So from Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi_{k + 1}$ is a ring monomorphism.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \phi_n$ is a (field) monomorphism.
== Also see ==
* Prime Power of Sum Modulo Prime, where the same technique is used.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Theorem $3.3$ Corollary
Category:Characteristics of Fields
Category:Field Monomorphisms
Power to Characteristic of Field is Monomorphism

== Theorem ==
Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let the characteristic of $F$ be $p$ where $p \ne 0$.
Let $\phi: F \to F$ be the mapping on $F$ defined as:
:$\forall x \in F: \map \phi x = x^p$
Then $\phi$ is a (field) monomorphism.
== Proof ==
Let $a, b \in F$.
First note that:
So:
Multiplication is more straightforward:
Thus, $\phi$ is a (field) homomorphism. 
It is clear that $\phi$ is not a zero homomorphism, since:
Hence, from Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi$ must be a monomorphism.
== Also see ==
* Prime Power of Sum Modulo Prime, where the same technique is used.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Theorem $3.3$
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 89 \gamma$
Category:Characteristics of Fields
Category:Field Monomorphisms
Power to Real Number by Decimal Expansion is Uniquely Defined

== Theorem ==
Let $r \in \R_{>1}$ be a real number greater than $1$, expressed by its decimal expansion:
:$r = n \cdotp d_1 d_2 d_3 \ldots$
The power $x^r$ of a (strictly) positive real number $x$ defined as:
:$(1): \quad \ds \lim_{k \mathop \to \infty} x^{\psi_1} \le \xi \le x^{\psi_2}$
where:
is unique.
== Proof ==
If $r$ is rational this has already been established.
Let $d$ denote the difference between $x^{\psi^1}$ and $x^{\psi^2}$:
It follows from Nth Root of 1 plus x not greater than 1 plus x over n that:
:$d < \dfrac {x^{n + 1} \paren {x - 1} } {10^k}$
Thus as $k \to \infty$, $d \to 0$.
The result follows from the Squeeze Theorem.
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: $(7)$
Category:Powers
Powers Drown Logarithms

== Theorem ==
Let $r \in \R_{>0}$ be a (strictly) positive real number.
Then:
:$\ds \lim_{x \mathop \to \infty} x^{-r} \ln x = 0$
=== Corollary ===
== Proof ==
From Upper Bound of Natural Logarithm:
When $x > 1$:
: $\forall s \in \R: s > 0: \ln x \le \dfrac {x^s} s$
Given that $r > 0$, we can plug $s = \dfrac r 2$ in:
From Sequence of Powers of Reciprocals is Null Sequence:
:$\ds \lim_{x \mathop \to \infty} x^{-r} \frac 1 {x^{r/2} } = 0$
and so:
:$\ds \lim_{x \mathop \to \infty} x^{-r} \ln x = 0$
by the Squeeze Theorem for Real Sequences.
== Sources ==
* : $\S 14.3 \ (2) \ \text{(i)}$
Category:Logarithms
Category:Powers
Powers Drown Logarithms/Corollary

== Theorem ==
Let $r \in \R_{>0}$ be a (strictly) positive real number.
Then:
:$\ds \lim_{y \mathop \to 0_+} y^r \ln y = 0$
== Proof ==
Put $y = \dfrac 1 x$ in the Powers Drown Logarithms.
== Sources ==
* : $\S 14.3 \ (2) \ \text{(ii)}$
Category:Logarithms
Category:Powers
Powers of 10 Expressible as Product of 2 Zero-Free Factors

== Theorem ==
The powers of $10$ which can be expressed as the product of $2$ factors neither of which has a zero in its decimal representation are:
== Proof ==
Let $p q = 10^n$ for some $n \in \Z_{>0}$.
Then $p q = 2^n 5^n$.
, suppose $p = 2^r 5^s$ for $r, s \ge 1$.
Then $2 \times 5 = 10$ is a divisor of $p$ and so $p$ ends with a zero.
Thus for $p$ and $q$ to be zero-free, it must be the case that $p = 2^n$ and $q = 5^n$ (or the other way around).
The result follows from Powers of 2 and 5 without Zeroes.
== Sources ==
* 
* : $10^{33}$
Category:Powers of 10
Powers of 16 Modulo 20/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$16^n \equiv 16 \pmod {20}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$16^1 \equiv 16 \pmod {20}$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown 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 the induction hypothesis:
:$16^k \equiv 16 \pmod {20}$
from which it is to be shown that:
:$16^{k + 1} \equiv 16 \pmod {20}$
=== Induction Step ===
This is the induction step:
We have:
So $\map P k \implies \map P {k + 1}$ and thus it follows by the Principle of Mathematical Induction that:
:$\forall n \in \Z_{> 0}: 16^n \equiv 16 \pmod {20}$
Category:Powers of 16 Modulo 20
Category:Proofs by Induction