nyuuzyou/wikis huggingface dataset:
Power of Absolute Value is Convex Real Function

== Theorem ==
Let $p \ge 1$ be a real number.
Define $f : \R \to \R$ by: 
:$\map f x = {\size x}^p$
for each $x \in \R$. 
Then $f$ is a convex function.
== Proof ==
From Absolute Value Function is Convex:
:$x \mapsto \size x$ is a convex function.
Note now that: 
:$x \mapsto x^p$ is increasing on $\hointr 0 \infty$ 
From Power Function is Convex Real Function, we also have: 
:$x \mapsto x^p$ is convex.
Since $f$ is the composition of the maps $x \mapsto \size x$ and $x \mapsto x^p$, we have:
:$f$ is convex
by Convex Real Function Composed with Increasing Convex Real Function is Convex.
Category:Convex Real Functions
Power of Base minus 1 is Repdigit Base minus 1

== Theorem ==
Let $b \in \Z_{>1}$ be an integer greater than $1$.
Let $n = b^k - 1$ for some integer $k$ such that $k \ge 1$.
Let $n$ be expressed in base $b$ representation.
Then $n$ is a repdigit number consisting of $k$ instances of digit $b - 1$.
== Proof ==
Thus, by the definition of base $b$ representation, $n$ consists of $k$ occurrences of the digit $b - 1$.
Category:Repdigit Numbers
Category:Number Bases
Power of Complex Conjugate is Complex Conjugate of Power

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\overline z$ denote the complex conjugate of $z$.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\overline {z^n} = \left({\overline z}\right)^n$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
:$\overline {z^n} = \left({\overline z}\right)^n$
$P \left({0}\right)$ is the case:
Thus $P \left({0}\right)$ is seen to hold.
=== Basis for the Induction ===
$P \left({1}\right)$ is the case:
Thus $P \left({1}\right)$ is seen to hold.
This is the basis for the induction.
=== Basis for the Induction ===
$P \left({2}\right)$ is the case:
:$\overline {z^2} = \left({\overline z}\right)^2$
which is demonstrated in Square of Complex Conjugate is Complex Conjugate of Square.
Thus $P \left({2}\right)$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is the induction hypothesis:
:$\overline {z^k} = \left({\overline z}\right)^k$
from which it is to be shown that:
:$\overline {z^{k + 1} } = \left({\overline z}\right)^{k + 1}$
=== Induction Step ===
This is the induction step:
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \ Z_{\ge 0}: \overline {z^n} = \left({\overline z}\right)^n$
Category:Complex Conjugates
Category:Complex Powers
Power of Complex Modulus equals Complex Modulus of Power

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\left\vert{z}\right\vert$ be the modulus of $z$.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\left\vert{z^n}\right\vert = \left\vert{z}\right\vert^n$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
:$\left\vert{z^n}\right\vert = \left\vert{z}\right\vert^n$
$P \left({0}\right)$ is the case:
Thus $P \left({0}\right)$ is seen to hold.
=== Basis for the Induction ===
$P \left({1}\right)$ is the case:
Thus $P \left({1}\right)$ is seen to hold.
This is the basis for the induction.
=== Basis for the Induction ===
$P \left({2}\right)$ is the case:
: $\left\vert{z^2}\right\vert = \left\vert{z}\right\vert^2$
which is demonstrated in Square of Complex Modulus equals Complex Modulus of Square.
Thus $P \left({2}\right)$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is the induction hypothesis:
:$\left\vert{z^k}\right\vert = \left\vert{z}\right\vert^k$
from which it is to be shown that:
:$\left\vert{z^{k + 1} }\right\vert = \left\vert{z}\right\vert^{k + 1}$
=== Induction Step ===
This is the induction step:
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \ Z_{\ge 0}: \left\vert{z^n}\right\vert = \left\vert{z}\right\vert^n$
== Examples ==
== Sources ==
* : $\S 1.2$. The Algebraic Theory
Category:Complex Modulus
Category:Complex Powers
Power of Complex Modulus equals Complex Modulus of Power/Examples/(1+i)^4

== Example of Power of Complex Modulus equals Complex Modulus of Power ==
:$\left\vert{\left({1 + i}\right)^4}\right\rvert = 4$
== Proof ==
== Sources ==
* : $\S 1.2$. The Algebraic Theory: Examples
Category:Examples of Complex Modulus
Power of Complex Modulus equals Complex Modulus of Power/Examples/(2-i)^6

== Example of Power of Complex Modulus equals Complex Modulus of Power ==
:$\cmod {\paren {2 - i}^6} = 125$
== Proof ==
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $4 \ \text{(ii)}$
Category:Examples of Complex Modulus
Power of Complex Number as Summation of Stirling Numbers of Second Kind

== Theorem ==
Let $z \in \C$ be a complex number whose real part is positive.
Then:
:$z^r = \ds \sum_{k \mathop \in \Z} {r \brace r - k} z^{\underline {r - k} }$
where:
:$\ds {r \brace r - k}$ denotes the extension of the Stirling numbers of the second kind to the complex plane
:$z^{\underline {r - k} }$ denotes $z$ to the $r - k$ falling.
== Proof ==
== Historical Note ==
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: $(59)$
Category:Stirling Numbers
Category:Power of Complex Number as Summation of Stirling Numbers of Second Kind
Power of Complex Number minus 1

== Theorem ==
Let $z \in \C$ be a complex number.
Then:
:$z^n - 1 = \ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \alpha^k}$
where $\alpha$ is a primitive complex $n$th root of unity.
=== Corollary ===
== Proof ==
Follows directly from the corollary to the Polynomial Factor Theorem:
If $\zeta_1, \zeta_2, \ldots, \zeta_n \in \C$ such that all are different, and $\map P {\zeta_1} = \map P {\zeta_2} = \ldots = \map P {\zeta_n} = 0$, then:
:$\ds \map P z = k \prod_{j \mathop = 1}^n \paren {z - \zeta_j}$
where $k \in \C$.
In this context, each of $\alpha^k$ is a primitive complex $n$th root of unity:
:$\paren {\alpha^k}^n = 1$
and the result follows.
== Sources ==
* : $\S 3$. Roots of Unity: $(3.10)$
Category:Complex Powers
Power of Complex Number minus 1/Corollary

== Theorem ==
Let $z \in \C$ be a complex number.
Then:
:$\ds \sum_{k \mathop = 0}^{n - 1} z^k = \prod_{k \mathop = 1}^{n - 1} \paren {z - \alpha^k}$
where $\alpha$ is a primitive complex $n$th root of unity.
== Proof ==
== Sources ==
* : $\S 3$. Roots of Unity: $(3.11)$
Category:Complex Powers
Power of Conjugate equals Conjugate of Power

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $x, y \in G$ such that $\exists a \in G: x \circ a = a \circ y$.
That is, let $x$ and $y$ be conjugate.
Then:
: $\forall n \in \Z: y^n = \paren {a^{-1} \circ x \circ a}^n = a^{-1} \circ x^n \circ a$
It follows directly that:
: $\exists b \in G: \forall n \in \Z: y^n = b \circ x^n \circ b^{-1}$
In particular:
: $y^{-1} = \paren {a^{-1} \circ x \circ a}^{-1} = a^{-1} \circ x^{-1} \circ a$
== Proof ==
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition $y^n = a^{-1} \circ x^n \circ a$.
$\map P 0$ is true, as this just says $e = a^{-1} \circ e \circ a$.
=== Basis for the Induction ===
$\map P 1$ is the case $y = a^{-1} \circ x \circ a$, which is how conjugacy is defined for $x$ and $y$.
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:
:$y^k = a^{-1} \circ x^k \circ a$
Then we need to show:
:$y^{k + 1} = a^{-1} \circ x^{k + 1} \circ a$
=== 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:
:$\forall n \in \N: y^n = a^{-1} \circ x^n \circ a$
Now we need to show that if $\map P n$ holds, then $\map P {-n}$ holds.
That is:
:$y^{-n} = a^{-1} \circ x^{-n} \circ a$
Let $n \in \N$.
Then:
Thus $\map P n$ has been shown to hold for all $n \in \Z$.
Hence the result.
== Sources ==
* : $\S 6.6$. Normal subgroups: Example $124 \ \text{(ii)}$
* : Chapter $6$: An Introduction to Groups: Exercise $12$
* : Chapter $3$: Elementary consequences of the definitions: Exercise $7$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.18$
Category:Conjugacy
Power of Coset Product is Coset of Power

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a \in G$.
Then:
:$\forall n \in \Z: \paren {a \circ N}^n = \paren {a^n} \circ N$
== Proof ==
From Quotient Group is Group, the operation:
:$\forall a, b \in G: \paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is the group operation in the quotient group $\struct {G / N, \circ}$.
The result follows directly by definition of power of group element.
== Sources ==
* : Chapter $6$: Cosets: Exercise $13$
Category:Coset Product
Power of Element/Semigroup

== Theorem ==
Let $\struct {S, \oplus}$ be a magma.
Let $a \in S$.
Let $n \in \N_{>0}$.
Let $\tuple {a_1, a_2, \ldots, a_n}$ be the ordered $n$-tuple defined by $a_k = a$ for each $k \in \N_n$.
Then:
:$\ds \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a$
where:
:$\ds \bigoplus_{k \mathop = 1}^n a_k$ is the composite of $\tuple {a_1, a_2, \ldots, a_n}$ for $\oplus$
:$\oplus^n a$ is the $n$th power of $a$ under $\oplus$.
== Proof ==
The proof will proceed by the Principle of Mathematical Induction on $\N$.
Let $T$ be the set defined as:
:$\ds T := \set {n \in \N: \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a}$
First, recall the definition of the composite of $\tuple {a_1, a_2, \ldots, a_n}$ for $\oplus$:
:$\ds \bigoplus_{k \mathop = 1}^n a_k = \begin{cases}
a: & n = 1 \\
\map {\oplus_m} {a_1, \ldots, a_m} \oplus a_{m + 1}: & n = m + 1
\end{cases}$
Secondly, recall the definition of the $n$th power of $a$ under $\oplus$:
:$\forall n \in \N_{>0}: \oplus^n a = \begin{cases}
a & : n = 1 \\
\paren {\oplus^m a} \oplus a & : n = m + 1
\end{cases}$
=== Basis for the Induction ===
We have that:
So $1 \in T$.
This is our basis for the induction.
=== Induction Hypothesis ===
It is to be shown that, if $j \in T$ where $j \ge 1$, then it follows that $j + 1 \in T$.
This is the induction hypothesis:
:$\ds \bigoplus_{k \mathop = 1}^j a_k = \oplus^j a$
It is to be demonstrated that it follows that:
:$\ds \bigoplus_{k \mathop = 1}^{j + 1} a_k = \oplus^{j + 1} a$
=== Induction Step ===
This is our induction step:
So $k \in T \implies k + 1 \in T$ and the result follows by the Principle of Mathematical Induction:
:$\ds \forall n \in \N: \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a$
== Notation ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.2$
Category:Semigroups
Category:Powers (Abstract Algebra)
Power of Element in Subgroup

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\struct {H, \circ}$ be a subgroup of $\struct {G, \circ}$.
Let $x \in H$.
Then:
:$\forall n \in \Z: x^n \in H$
== Proof ==
Proof by induction:
For all $n \in \N^*$, let $\map P n$ be the compound proposition:
:$x^n \in H \text{ and } x^{-n} \in H$.
$\map P 0$ is true, as this just says $x^0 \in H$.
By Powers of Group Elements, $x^0 = e$.
This follows by Identity of Subgroup‎.
=== Basis for the Induction ===
$\map P 1$ is true, as this just says $x^1 = x \in H$.
By the Two-Step Subgroup Test, we also have that $x^{-1} \in H$.
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:
:$x^k \in H \text{ and } x^{-k} \in H$.
Then we need to show:
:$x^{k + 1} \in H \text{ and } x^{-\paren {k + 1} } \in H$.
=== Induction Step ===
This is our induction step:
:By Powers of Group Elements, $x^{k + 1} = x \circ x^k$.
:By the base case, $x \in H$.
:By the induction hypothesis, $x^k \in H$.
:By the closure axiom, $x \circ x^k \in H$.
:By the Two-Step Subgroup Test, it follows that $x^{-\paren {k + 1} } \in H$.
So $\map P k \implies \map P {k + 1}$.
The result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: x^n \in H$
:$\forall n \in \N: x^{-n} \in H$
Hence the result.
== Sources ==
* : $\S 5.2$. Subgroups: Example $89$
Category:Subgroups
Power of Elements is Subgroup

== Theorem ==
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.
Category:Abelian Groups
Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order

== Theorem ==
Let $C_n$ be the cyclic group of order $n$.
Let $C_n = \gen a$, that is, that $C_n$ is generated by $a$.
Then:
:$C_n = \gen {a^k} \iff k \perp n$
That is, $C_n$ is also generated by $a^k$ $k$ is coprime to $n$.
== Proof ==
=== Necessary Condition ===
Let $k \perp n$.
Then by Integer Combination of Coprime Integers:
:$\exists u, v \in \Z: 1 = u k + v n$
So $\forall m \in \Z$, we have:
Thus $a^k$ generate $C_n$.
=== Sufficient Condition ===
Let $C_n = \gen {a^k}$.
That is, let $a^k$ generate $C_n$.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 9$: Cyclic Groups
* : Chapter $2$: Subgroups and Cosets: $\S 44 \alpha$
* : Chapter $6$: An Introduction to Groups: Exercise $19$
Category:Cyclic Groups
Category:Coprime Integers
Category:Generators of Groups
Power of Golden Mean as Sum of Smaller Powers

== Theorem ==
Let $\phi$ denote the golden mean.
Then:
:$\forall z \in \C: \phi^z = \phi^{z - 1} + \phi^{z - 2}$
== Proof ==
Let $z \in \C$.
Let $w \in \C$ such that $w + 2 = z$.
Then:
== Sources ==
* 
Category:Golden Mean
Power of Group Element in Kernel of Homomorphism iff Power of Image is Identity

== Theorem ==
Let $G$ be a group whose identity is $e_G$.
Let $H$ be a group whose identity is $e_H$.
Let $\phi: G \to H$ be a (group) homomorphism.
Let $x^n \in \map \ker \phi$ for some integer $n$.
Then:
:$\paren {\map \phi x}^n = e_H$
== Proof ==
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $8$
Category:Kernels of Group Homomorphisms
Power of Ideal is Subset

== Theorem ==
Let $\struct {R, +, \cdot}$ be a ring.
Let $I \subseteq R$ be an ideal.
Let $n \in \Z$ be an integer such that $n \ge 1$.
Let $I^n$ denote the $n$th power of $I$.
Then $I^n$ is a subset of $I$:
:$I^n \subseteq I$
== Proof ==
Let $n = 1$.
Then by definition of Power of Ideal of Ring:
:$I^1 = I$
Hence by Set is Subset of Itself:
:$I^1 \subseteq I$.
Let $n > 1$.
By definition of Power of Ideal of Ring, $I^n$ is an ideal generated by elements of $R$ the form:
:$a_1 \cdots a_n$
where each $a_i \in I$.
From Ideal is Subring, $I$ is a subring of $R$.
Hence $I$ is closed under addition.
Hence it suffices to show that such generators of $I^n$ are elements of $I$.
Let $a_1 \cdots a_n \in I^n$ such that each $a_i \in I$.
Then:
:$a_1 \in I$
:$a_2 \cdots a_n \in I$
Hence in particular:
:$a_2 \cdots a_n \in R$
Because:
:$a_1 \in I$
:$a_2 \cdots a_n \in R$
it follows that:
:$a_1 \cdot \paren {a_2 \cdots a_n} \in I$
Thus:
:$a_1 \cdots a_n \in I$
and it follows by definition of subset that:
:$I^n \subseteq I$
Category:Ideal Theory
Category:Subsets
Category:Power Set
Power of Idempotent Element

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $s \in S$ be an idempotent element with respect to $\circ$.
Then:
:$\forall n \in \Z_{> 0}: s^n = s$
where $s^n$ is defined as:
:$s^n = \begin{cases} s & : n = 1 \\
s^{n - 1} \circ s & : n > 1 \end{cases}$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$s^n = s$
$\map P 1$ is the case:
:$s^1 = s$
which holds by definition.
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
$\map P 2$ is the case:
Thus $\map P 2$ 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 $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$s^k = s$
from which it is to be shown that:
:$s^{k + 1} = s$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{> 0}: s^n = s$
</onlyinclude>
Category:Idempotence
Power of Identity is Identity

== Theorem ==
Let $\struct {M, \circ}$ be a monoid whose identity element is $e$.
Then:
:$\forall n \in \Z: e^n = e$
== Proof ==
Since $e$ is invertible, the power of $e$ is defined for all $n \in \Z$.
We prove the case $n \ge 0$ by induction.
=== Basis for the Induction ===
By definition of power of monoid element:
:$e^0 = e$
so the theorem holds for $n = 0$. 
This is our basis for the induction.
=== Induction Hypothesis ===
Our induction hypothesis is that the theorem is true for $n = k$:
:$e^k = e$
=== Induction Step ===
In the induction step, we prove that the theorem is true for $n = k + 1$.
We have:
Therefore, by Principle of Mathematical Induction:
:$\forall n \in \Z_{\ge 0} : e^n = e$
Now we prove the case $n < 0$.
We have:
Thus:
:$\forall n \in \Z : e^n = e$
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.8)$
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.8 \ (4)$
Category:Monoids
Category:Powers (Abstract Algebra)
Category:Proofs by Induction
Power of Maximum is not Greater than Sum of Powers

== Theorem ==
Let $a_1, a_2, \ldots, a_r$ be non-negative real numbers.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $a = \max \set {a_1, a_2, \ldots, a_r}$.
Then:
:$a^n \le a_1^n + a_2^n + \cdots + a_r^n$
== Proof ==
By definition of the $\max$ operation:
:$\exists k \in \set {1, 2, \ldots, r}: a_k = a$
Hence:
Hence the result.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 11$
Category:Inequalities
Power of Moved Element is Moved

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\sigma \in S_n$.
Then for all $m \in \Z$:
:$i \notin \Fix \sigma \implies \map {\sigma^m} i \notin \Fix \sigma$
where $\Fix \sigma$ denotes the set of fixed elements of $\sigma$.
== Proof ==
 that there exists some $i \notin \Fix \sigma$ and some $m \in \Z$ such that $\map {\sigma^m} i \in \Fix \sigma$.
Then:
But it was previously established that $i$ was moved by $\sigma$.
This is a contradiction.
Therefore $\map {\sigma^m} i$ is moved by $\sigma$.
Category:Symmetric Groups
Power of One plus x in terms of Gaussian Hypergeometric Function

== Theorem ==
:$\map F {-p, 1; 1; -x} = \paren {1 + x}^p$
where:
:$x$ and $p$ are real numbers with $\size x < 1$
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
== Proof ==
== Sources ==
* : $\S 31$: Hypergeometric Functions: Special Cases: $31.3$
Category:Gaussian Hypergeometric Function
Category:Hypergeometric Functions
Power of Positive Real Number is Positive/Integer

== Theorem ==
Let $x \in \R_{>0}$ be a (strictly) positive real number.
Let $n \in \Z$ be an integer.
Then:
:$x^n > 0$
where $x^n$ denotes the $n$th power of $x$.
== Proof ==
By Power of Positive Real Number is Positive: Natural Number, the theorem is already proven for non-negative integers.
Suppose $n \in \Z_{< 0}$.
When $n < 0$, by Real Number Ordering is Compatible with Multiplication: Negative Factor:
:$-n > 0$
Then, by Power of Positive Real Number is Positive: Natural Number:
:$x^{-n} > 0$
Therefore, by Reciprocal of Strictly Positive Real Number is Strictly Positive:
:$x^n = \dfrac 1 {x^{-n} } > 0$
Category:Power of Positive Real Number is Positive
Power of Positive Real Number is Positive/Rational Number

== Theorem ==
Let $x \in \R_{>0}$ be a (strictly) positive real number.
Let $q \in \Q$ be a rational number.
Then:
:$x^q > 0$
where $x^q$ denotes the $x$ to the power of $q$.
== Proof ==
Let $q = \dfrac r s$, where $r \in \Z$, $s \in \Z \setminus \set 0$.
Then:
Hence the result.
Category:Power of Positive Real Number is Positive
Power of Positive Real Number is Positive/Real Number

== Theorem ==
Let $x \in \R_{>0}$ be a (strictly) positive real number.
Let $r \in \R$ be a real number.
Then:
:$x^r > 0$
where $x^r$ denotes the $x$ to the power of $r$.
== Proof ==
From the definition of $x$ to the power of $r$:
:$x^r = \map \exp {r \ln x}$
The result follows from Exponential of Real Number is Strictly Positive.
Category:Power of Positive Real Number is Positive
Power of Power of Positive Element of Unital C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $a \in A$ be positive elements.
Let $\alpha, \beta > 0$ be real numbers.
Let $a^{\alpha \beta}$ be given by the continuous functional calculus for $a$.
Let $\paren {a^\alpha}^\beta$ be given by the continuous functional calculus for $a^\alpha$.
Then:
:$\paren {a^\alpha}^\beta = a^{\alpha \beta}$
== Proof ==
First take $A$ unital.
Let $\map {\sigma_A} a$ be the spectrum of $a$ in $A$.
Let $\map {\sigma_A} {a^\alpha}$ be the spectrum of $a^\alpha$ in $A$.
From Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative, $a^\alpha$ is positive.
Let $\Theta_{a^\alpha}$ be the continuous functional calculus for $a^\alpha$.
Let $\Theta_a$ be the continuous functional calculus for $a$.
Define $e_{\alpha \beta} : \map {\sigma_A} a \to \hointr 0 \infty$ and $e_\alpha : \map {\sigma_A} \to \hointr 0 \infty$ by:
:$\map {e_{\alpha \beta} } t = t^{\alpha \beta}$
and:
:$\map {e_\alpha} t = t^\alpha$
for each $t \in \map {\sigma_A} a$. 
Define $e_\beta : \map {\sigma_A} {a^\alpha} \to \hointr 0 \infty$ by:
:$\map {e_\beta} t = t^\beta$ 
for each $t \in \map {\sigma_A} {a^\alpha}$.
From the Spectral Mapping Theorem, we have:
:$\map {\sigma_A} {a^\alpha} = \set {t^\alpha : t \in \map {\sigma_A} a}$
Hence we have $e_{\alpha \beta} = e_\beta \circ e_\alpha$ by Power of Power.
So:
Now take $A$ to not be unital.
Let $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ be the unitization of $A$.
From the definition of the continuous functional for non-unital $\text C^\ast$-algebras, we have:
Hence:
:$a^{\alpha \beta} = \paren {a^\alpha}^\beta$
in this case also.
Category:Positive Elements of C*-Algebras
Power of Prime is Deficient

== Theorem ==
Let $n \in \Z_{>0}$ be a power of a prime number $p$:
:$n = p^k$
for some $k \in \Z_{>0}$.
Then $n$ is deficient.
== Proof ==
From Divisor Sum of Power of Prime:
:$\map {\sigma_1} n = \dfrac {p^{k + 1} - 1} {p - 1}$
Thus:
If $p = 2$ then $\dfrac {p^k} {p - 1} - p^k = 0$ and so:
:$\dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1} = - \dfrac 1 {p - 1} < 0$
If $p > 2$ then $\dfrac {p^k} {p - 1} < p^k$ and so:
:$\dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1} < 0$
Thus in all cases:
:$\map A n < 0$
and so $n = p^k$ is deficient by definition.
== Sources ==
* : $2$
* : $12$
* : $2$
* : $12$
* 
Category:Deficient Numbers