nyuuzyou/wikis huggingface dataset:
Power Set with Intersection and Subset Relation is Ordered Semigroup

== Theorem ==
Let $S$ be a set and let $\powerset S$ be its power set.
Let $\struct {\powerset S, \cap, \subseteq}$ be the ordered structure formed from the set intersection operation and subset relation.
Then $\struct {\powerset S, \cap, \subseteq}$ is an ordered semigroup.
== Proof ==
From Power Set with Intersection is Commutative Monoid, $\struct {\powerset S, \cap}$ is a semigroup.
From Subset Relation is Ordering, $\struct {\powerset S, \subseteq}$ is an ordered set.
It remains to be shown that $\subseteq$ is compatible with $\cap$.
Let $A, B \subseteq S$ be arbitrary such that $A \subseteq B$.
Thus:
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.8$
Category:Examples of Ordered Semigroups
Category:Power Set
Category:Set Intersection
Category:Subsets
Power Set with Intersection and Superset Relation is Ordered Semigroup

== Theorem ==
Let $S$ be a set and let $\powerset S$ be its power set.
Let $\struct {\powerset S, \cap, \supseteq}$ be the ordered structure formed from the set intersection operation and superset relation.
Then $\struct {\powerset S, \cap, \supseteq}$ is an ordered semigroup.
== Proof ==
From Power Set with Intersection is Commutative Monoid, $\struct {\powerset S, \cap}$ is a semigroup.
From Subset Relation is Ordering, $\struct {\powerset S, \subseteq}$ is an ordered set.
We have that $\supseteq$ is the dual to $\subseteq$.
Hence $\struct {\powerset S, \supseteq}$ is an ordered set.
It remains to be shown that $\supseteq$ is compatible with $\cap$.
Let $A, B \subseteq S$ be arbitrary such that $A \supseteq B$.
Thus:
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.8$
Category:Examples of Ordered Semigroups
Category:Power Set
Category:Set Intersection
Category:Subsets
Power Set with Intersection is Commutative Monoid

== Theorem ==
Let $S$ be a set and let $\powerset S$ be its power set.
Then $\struct {\powerset S, \cap}$ is a commutative monoid whose identity is $S$.
The only invertible element of this structure is $S$.
Thus (except in the degenerate case $S = \O$) $\struct {\powerset S, \cap}$ cannot be a group.
== Proof ==
From Power Set is Closed under Intersection:
:$\forall A, B \in \powerset S: A \cap B \in \powerset S$
From Set System Closed under Intersection is Commutative Semigroup, $\struct {\powerset S, \cap}$ is a commutative semigroup.
From Identity of Power Set with Intersection, we have that $S$ acts as the identity.
It remains to be shown that only $S$ has an inverse:
For $T \subseteq S$ to have an inverse under $\cap$, we require $T^{-1} \cap T = S$.
From this it follows that $T = S = T^{-1}$.
The result follows by definition of commutative monoid.
== Also see ==
* Power Set with Union is Commutative Monoid
== Sources ==
* : Chapter $\text{I}$: Semi-Groups and Groups: $1$: Definition and examples of semigroups: Example $7$
* : $\S 4.4$. Gruppoids, semigroups and groups: Example $77$
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Exercise $4.4$
* : Chapter $\text{II}$: Groups: Exercise $\text{T}$
* : $\S 29$. Semigroups: definition and examples: $(3)$
* : Chapter $5$: Semigroups: Exercise $2$
Category:Examples of Commutative Monoids
Category:Power Set
Category:Set Intersection
Power Set with Union and Intersection forms Boolean Algebra

== Theorem ==
Let $S$ be a set, and let $\powerset S$ be its power set.
Denote with $\cup$, $\cap$ and $\complement$ the operations of union, intersection and complement on $\powerset S$, respectively.
Then $\struct {\powerset S, \cup, \cap, \complement}$ is a Boolean algebra.
== Proof ==
Taking the criteria for definition 1 of a Boolean algebra in turn:
=== $(\text {BA} 0):$ Closure ===
$\powerset S$ is closed under both $\cup$ and $\cap$:
:Power Set is Closed under Intersection
:Power Set is Closed under Union
:Power Set is Closed under Complement
=== $(\text {BA} 1):$ Commutativity ===
Both $\cup$ and $\cap$ are commutative from Intersection is Commutative and Union is Commutative.
=== $(\text {BA} 2):$ Distributivity ===
Both $\cup$ and $\cap$ distribute over the other, from Union Distributes over Intersection and Intersection Distributes over Union.
=== $(\text {BA} 3):$ Identity Elements ===
Both $\cup$ and $\cap$ have identities:
From Power Set with Intersection is Monoid‎, $S$ is the identity for $\cap$. 
From Power Set with Union is Monoid, $\O$ is the identity for $\cup$.
=== $(\text {BA} 4):$ Complements ===
From Union with Complement:
:$\forall A \in S: A \cup \map \complement A = S$
which is the identity for $\cap$.
From Intersection with Complement:
:$\forall A \in S: A \cap \map \complement A = \O$
which is the identity for $\cup$.
All the criteria for a Boolean algebra are therefore fulfilled.
== Sources ==
* : $\S 1.5$
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : $\S 2$
* : $\S 2$: Exercise $1$
Category:Power Set
Category:Boolean Algebras
Power Set with Union and Subset Relation is Ordered Semigroup

== Theorem ==
Let $S$ be a set and let $\powerset S$ be its power set.
Let $\struct {\powerset S, \cup, \subseteq}$ be the ordered structure formed from the set union operation and subset relation.
Then $\struct {\powerset S, \cup, \subseteq}$ is an ordered semigroup.
== Proof ==
From Power Set with Union is Commutative Monoid, $\struct {\powerset S, \cup}$ is a semigroup.
From Subset Relation is Ordering, $\struct {\powerset S, \subseteq}$ is an ordered set.
It remains to be shown that $\subseteq$ is compatible with $\cup$.
Let $A, B \subseteq S$ be arbitrary such that $A \subseteq B$.
Thus:
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.8$
Category:Examples of Ordered Semigroups
Category:Power Set
Category:Set Union
Category:Subsets
Power Set with Union and Superset Relation is Ordered Semigroup

== Theorem ==
Let $S$ be a set and let $\powerset S$ be its power set.
Let $\struct {\powerset S, \cup, \supseteq}$ be the ordered structure formed from the set union operation and superset relation.
Then $\struct {\powerset S, \cup, \supseteq}$ is an ordered semigroup.
== Proof ==
From Power Set with Union is Commutative Monoid, $\struct {\powerset S, \cup}$ is a semigroup.
From Subset Relation is Ordering, $\struct {\powerset S, \subseteq}$ is an ordered set.
We have that $\supseteq$ is the dual to $\subseteq$.
Hence $\struct {\powerset S, \supseteq}$ is an ordered set.
It remains to be shown that $\supseteq$ is compatible with $\cup$.
Let $A, B \subseteq S$ be arbitrary such that $A \supseteq B$.
Thus:
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.8$
Category:Examples of Ordered Semigroups
Category:Power Set
Category:Set Union
Category:Subsets
Power Set with Union is Commutative Monoid

== Theorem ==
Let $S$ be a set and let $\powerset S$ be its power set.
Then $\struct {\powerset S, \cup}$ is a commutative monoid whose identity is $\O$.
The only invertible element of this structure is $\O$.
Thus (except in the degenerate case $S = \O$) $\struct {\powerset S, \cup}$ cannot be a group.
== Proof ==
From Power Set is Closed under Union:
:$\forall A, B \in \powerset S: A \cup B \in \powerset S$
From Set System Closed under Union is Commutative Semigroup, $\struct {\powerset S, \cup}$ is a commutative semigroup.
From Identity of Power Set with Union, $\O$ acts as the identity of $\struct {\powerset S, \cup}$.
It remains to be shown that only $\O$ has an inverse:
For $T \subseteq S$ to have an inverse under $\cup$, we require $T^{-1} \cup T = \O$.
From this it follows that $T = \O = T^{-1}$.
The result follows by definition of commutative monoid.
== Also see ==
* Power Set with Intersection is Commutative Monoid
== Sources ==
* : Chapter $\text{I}$: Semi-Groups and Groups: $1$: Definition and examples of semigroups: Example $6$
* : $\S 4.4$. Gruppoids, semigroups and groups: Example $77$
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Exercise $4.4$
* : Chapter $\text{II}$: Groups: Exercise $\text{T}$
* : $\S 29$. Semigroups: definition and examples: $(3)$
* : Chapter $5$: Semigroups: Exercise $2$
Category:Examples of Commutative Monoids
Category:Set Union
Category:Power Set
Power Sets of Equinumerous Sets are Equinumerous

== Theorem ==
Let $x$ and $y$ be sets such that $x \sim y$.
Then:
:$\powerset x \sim \powerset y$
== Proof 1 ==
Let $f: x \to y$ be a bijection.
Let $F$ send each $z \in \powerset x \mapsto \Img z$ where $\Img z$ denotes the image of the subset $z$ under $f$.
It follows that $F$ is a mapping whose domain is $\powerset x$.
Moreover, the image of $F$ is the collection of all images of $z$.
Each image is a subset of $y$ and thus:
:$F: \powerset x \to \powerset y$
Suppose that $\map F a = \map F b$.
Then if $z \in a$ then $\map f z \in \map F a$.
But then, $\map f z = \map f w$ for some $w \in y$.
Thus, $z = w$ and $w \in y$.
Therefore, $z = y$ and $F$ is injective.
Suppose that $a \in \powerset y$.
Let $b$ be the preimage of $a$ with respect to the $f$ mapping.
It follows that $b$ is a subset of $x$.
We have:
:$\map F b = a$
and so $a$ is in the image of $F$.
Generalizing for all $a \in \powerset y$, it follows that:
:$F: \powerset x \to \powerset y$ is a bijection.
== Proof using covariant power set functor ==
Let $\PP$ denote the covariant power set functor.
Let $f: x \to y$ be a bijection.
By Covariant Functors Preserve Isomorphisms, $\map \PP f: \powerset x \to \powerset y$ is a bijection.
== Proof using contravariant power set functor ==
Let $\overline \PP$ denote the contravariant power set functor.
Let $f: x \to y$ be a bijection.
By Contravariant Functors Preserve Isomorphisms, $\map {\overline \PP} f : \powerset y \to \powerset x$ is a bijection.
== Sources ==
* : $\S 10.6$
Category:Set Theory
Power Structure Operation on Set of Singleton Subsets is Closed

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $S$.
Let $S'$ denote the set of singleton elements of $\powerset S$.
Then the algebraic structure $\struct {S', \circ_\PP}$ is closed.
== Proof ==
Let $A, B \in S'$.
Then:
:$\exists a, b \in S: A = \set a, B = \set b$
Hence:
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.5$
Category:Power Structures
Category:Singletons
Power Structure Operation on Set of Singleton Subsets preserves Associativity

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Let $S'$ denote the set of singleton elements of $\powerset S$.
Then $\circ_\PP$ is associative $\circ$ is associative.
== Proof ==
From Operation is Isomorphic to Power Structure Operation on Set of Singleton Subsets:
:$\struct {S, \circ}$ is isomorphic to $\struct {S', \circ_\PP}$
The result follows from Isomorphism Preserves Associativity.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.5$
Category:Power Structures
Category:Associativity
Power Structure Operation on Set of Singleton Subsets preserves Commutativity

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Let $S'$ denote the set of singleton elements of $\powerset S$.
Then $\circ_\PP$ is commutative $\circ$ is commutative.
== Proof ==
From Operation is Isomorphic to Power Structure Operation on Set of Singleton Subsets:
:$\struct {S, \circ}$ is isomorphic to $\struct {S', \circ_\PP}$
The result follows from Isomorphism Preserves Commutativity.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.5$
Category:Power Structures
Category:Commutativity
Category:Singletons
Power Structure of Group is Monoid

== Theorem ==
Let $\struct {G, \circ}$ be a group with identity $e$.
Let $\struct {\powerset G, \circ_\PP}$ be the power structure of $\struct {G, \circ}$.
Then $\struct {\powerset G, \circ_\PP}$ is a monoid with identity $\set e$.
== Proof ==
By definition of a group, $\struct {G, \circ}$ is a monoid.
The result follows from Power Structure of Monoid is Monoid.
Category:Group Theory
Category:Power Structures
Power Structure of Group is Semigroup

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\struct {\powerset G, \circ_\PP}$ be the power structure of $\struct {G, \circ}$.
Then $\struct {\powerset G, \circ_\PP}$ is a semigroup.
== Proof 1 ==
== Proof 2 ==
Category:Group Theory
Category:Semigroups
Category:Power Structures
Category:Power Structure of Group is Semigroup
Power Structure of Group is Semigroup/Proof 1

== Theorem ==
== Proof ==
We need to prove closure and associativity.
=== Closure ===
Let $\struct {G, \circ}$ be a group, and let $A, B \subseteq G$.
Thus $\struct {\powerset G, \circ_\PP}$ is closed.
=== Associativity ===
It follows from Subset Product within Semigroup is Associative that $\circ_\PP$ is associative in $\struct {\powerset G, \circ_\PP}$.
Thus $\struct {\powerset G, \circ_\PP}$ is a semigroup.
== Sources ==
* : $\S 41.1$ Multiplication of subsets of a group
Category:Power Structure of Group is Semigroup
Power Structure of Magma is Magma

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Then $\struct {\powerset S, \circ_\PP}$ is a magma.
That is, $\circ_\PP$ is closed in $\powerset S$.
== Proof ==
Let $\struct {S, \circ}$ be a magma.
Let $A, B \subseteq S$.
Thus $\struct {\powerset S, \circ_\PP}$ is a magma.
Category:Power Structures
Category:Magmas
Power Structure of Monoid is Monoid

== Theorem ==
Let $\struct {G, \circ}$ be a monoid with identity $e$.
Let $\struct {\powerset G, \circ_\PP}$ be the power structure of $\struct {G, \circ}$.
Then $\struct {\powerset G, \circ_\PP}$ is a monoid with identity $\set e$.
== Proof ==
By definition of a monoid, $\struct {G, \circ}$ is a semigroup.
By Power Structure of Semigroup is Semigroup, $\struct {\powerset G, \circ_\PP}$ is a semigroup.
By Subset Product by Identity Singleton, $\set e$ is an identity for $\struct {\powerset G, \circ_\PP}$.
Thus $\struct {\powerset G, \circ_\PP}$ is a monoid with identity $\set e$.
Category:Monoids
Category:Power Structures
Power Structure of Semigroup Ordered by Subsets is Ordered Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Let $\struct {\powerset S, \circ_\PP, \subseteq}$ be the ordered structure formed from $\struct {\powerset S, \circ_\PP}$ and the subset relation.
Then $\struct {\powerset S, \circ_\PP, \subseteq}$ is an ordered semigroup.
== Proof ==
From Power Structure of Semigroup is Semigroup, $\struct {\powerset S, \circ_\PP}$ is a semigroup.
From Subset Relation is Ordering, $\struct {\powerset S, \subseteq}$ is an ordered set.
It remains to be shown that $\subseteq$ is compatible with $\circ_\PP$.
This is demonstrated directly in Subset Relation is Compatible with Subset Product.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.8$
Category:Examples of Ordered Semigroups
Category:Power Structures
Power Structure of Semigroup Ordered by Supersets is Ordered Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Let $\struct {\powerset S, \circ_\PP, \supseteq}$ be the ordered structure formed from $\struct {\powerset S, \circ_\PP}$ and the superset relation.
Then $\struct {\powerset S, \circ_\PP, \supseteq}$ is an ordered semigroup.
== Proof ==
From Power Structure of Semigroup Ordered by Subsets is Ordered Semigroup, $\struct {\powerset S, \circ_\PP, \subseteq}$ is an ordered semigroup.
The result then follows from Dual of Ordered Semigroup is Ordered Semigroup.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.8$
Category:Examples of Ordered Semigroups
Category:Power Structures
Power Structure of Semigroup is Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Then $\struct {\powerset S, \circ_\PP}$ is a semigroup.
== Proof ==
From Power Structure of Magma is Magma we conclude that $\struct {\powerset S, \circ_\PP}$ is a magma.
It follows from Subset Product within Semigroup is Associative that $\circ_\PP$ is associative in $\struct {\powerset S, \circ_\PP}$.
Thus $\struct {\powerset S, \circ_\PP}$ is a semigroup.
Category:Semigroups
Category:Power Structures
Power Structure of Subset is Closed iff Subset is Closed

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\circ_\PP$ be the operation induced on $\powerset S$, the power set of $S$.
Let $T \subseteq S$.
Then:
:the algebraic structure $\struct {\powerset T, \circ_\PP}$ is closed
:
:the algebraic structure $\struct {T, \circ}$ is closed.
== Proof ==
=== Sufficient Condition ===
Let $\struct {\powerset T, \circ_\PP}$ be closed.
Then:
That is, $\struct {T, \circ}$ is closed.
=== Necessary Condition ===
Let $\struct {T, \circ}$ be closed.
Then by definition $\struct {T, \circ}$ is a magma.
From Power Structure of Magma is Magma it follows that $\struct {\powerset T, \circ_\PP}$ is likewise a magma.
That is, the algebraic structure $\struct {\powerset T, \circ_\PP}$ is closed.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.8$
Category:Power Set
Category:Subset Products
Category:Power Structures
Power in Prime Ideal

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $\mathfrak p \subseteq A$ be a prime ideal.
Let $n > 0$ be a natural number.
Then for all elements $a \in A$, if $a^n \in \mathfrak p$ then $a \in \mathfrak p$.
== Proof ==
Proof by induction:
=== Basis for the Induction ===
Let $n = 1$.
By definition of Power of Element of Ring:
:$a^1 = a$.
Hence $a \in \mathfrak p$.
=== Induction Hypothesis ===
This is the induction hypothesis:
Let $n > 1$.
$\forall k = 1, \ldots, n - 1, a^k \in \mathfrak p \implies a \in \mathfrak p$
=== Induction Step ===
This is the induction step:
By definition of Power of Element of Ring:
:$a^n = a^{n - 1} \cdot a$.
Hence $a^{n - 1} \cdot a \in \mathfrak p$.
By definition of prime ideal:
: $a^{n - 1} \cdot a \in \mathfrak p \implies a^{n - 1} \in \mathfrak p$ or $a \in \mathfrak p$.
If $a \in \mathfrak{p}$ then our claim is satisfied.
Otherwise, suppose $a^{n - 1} \in \mathfrak p$.
Then by the induction hypothesis:
: $a^{n - 1} \in \mathfrak p \implies a \in \mathfrak p$.
By the Principle of Mathematical Induction, the theorem holds for all $n$.
Category:Ideal Theory
Category:Prime Ideals of Rings
Power is Well-Defined/Integer

== Theorem ==
Let $x$ be a non-zero real number.
Let $k$ be an integer.
Then $x^k$ is well-defined.
== Proof ==
Fix $x \in \R \setminus \set 0$.
=== Positive Integers ===
We first prove the theorem for positive integers by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
: $x^n$ is well-defined
==== Basis for the Induction ====
$\map P 1$ is true, since $x^1 = 1$ by definition.
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:
:$x^k$ is well-defined
Then we need to show:
:$x^{k + 1}$ is well-defined
==== Induction Step ====
This is our induction step:
By definition of integer power:
:$x^{k + 1} = x \cdot x^k$
From Real Multiplication is Well-Defined and the Induction Hypothesis:
:$x \cdot x^k$ is well-defined
So $x^{k + 1}$ is well-defined.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N: x^n$ is well-defined
=== Non-Positive Integers ===
Fix $k \in \Z_{< 0}$
First:
:$k \in \Z_{< 0} \implies -k \in \N$
From the development above, $x^{-k}$ is well-defined.
From Powers of Group Elements/Negative Index, $\dfrac 1 {x^k}$ is well-defined.
From Reciprocal of Real Number is Unique, $x^k$ is well-defined.
Finally:
:$\forall x \in \R \setminus \set 0: x^0 = 1$
So $x^k$ is well-defined for $k = 0$.
Hence the result.
Category:Powers
Power is Well-Defined/Rational

== Theorem ==
Let $x \in \R_{> 0}$ be a (strictly) positive real number.
Let $q$ be a rational number.
Then $x^q$ is well-defined.
== Proof ==
Let $x \in \R_{>0}$ be fixed.
Let $q \in \Q \setminus \set 0$.
Let $\dfrac r s$ and $\dfrac t u$ be two representations of $q$.
That is, $r, s, t, u$ are non-zero integers.
We now show that:
: $\dfrac r s = \dfrac t u \implies x^{r / s} = x^{t / u}$
So:
Hence the result.
Category:Powers
Power of 2^10 Minus Power of 10^3 is Divisible by 24

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a non-negative integer.
Then $2^{10 n} - 10^{3 n}$ is divisible by $24$.
That is:
:$2^{10 n} - 10^{3 n} \equiv 0 \pmod {24}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Example ==
Category:Modulo Arithmetic
Category:Power of 2^10 Minus Power of 10^3 is Divisible by 24
Power of 2^10 Minus Power of 10^3 is Divisible by 24/Example

== Example of Power of $2^{10}$ Minus Power of $10^3$ is Divisible by $24$ ==
Let us examine arguably the flagship example, the smallest number whose compliance is not obvious nor trivial:
:$1 \, 048 \, 576$
This is equal to $2^{20}$, which is equal to $2^{10 \times 2}$, thus one of the valid powers of $2$.
We then subtract $10^{3 \times 2}$, or $10^6$:
:$1 \, 048 \, 576 - 1 \, 000 \, 000 = 48 \, 576$
and we see that:
:$48 \, 576 = 2024 \times 24$
satisfying the theorem.
Category:Power of 2^10 Minus Power of 10^3 is Divisible by 24
Power of 2^10 Minus Power of 10^3 is Divisible by 24/Proof 2

== Theorem ==
== Proof ==
For $n = 0$ both powers are $1$, and $1 - 1 = 0$ is divisible by $24$.
For $n > 1$:
Category:Power of 2^10 Minus Power of 10^3 is Divisible by 24
Power of 2 containing no Digit 2

== Theorem ==
$2^{168}$ contains no $2$ anywhere in its decimal representation.
== Proof ==
:$2^{168} = 374 \, 144 \, 419 \, 156 \, 711 \, 147 \, 060 \, 143 \, 317 \, 175 \, 368 \, 453 \, 031 \, 918 \, 731 \, 001 \, 856$
== Historical Note ==
== Sources ==
* : $374,144,419,156,711,147,060,143,317,175,368,453,031,918,731,001,856$
Category:374,144,419,156,711,147,060,143,317,175,368,453,031,918,731,001,856
Category:Powers of 2
Power of 2 is Almost Perfect

== Theorem ==
Let $n \in \Z_{>0}$ be a power of $2$:
:$n = 2^k$
for some $k \in \Z_{>0}$.
Then $n$ is almost perfect.
== Proof ==
and so $n = p^k$ is almost perfect by definition.
== Sources ==
* : $2$
* : $12$
* : $16$
* : $2$
* : $12$
* : $16$
* : $32$
Category:Almost Perfect Numbers
Category:2
Power of 2 is Difference between Two Powers

== Theorem ==
Let $n \in \Z_{>0}$ be a power of $2$.
Then $n$ is the difference between powers of two positive integers greater than or equal to $2$.
== Proof ==
$2^k = 2^{k+1} - 2^k$
== Examples ==
== Sources ==
* 
* : $32$
Category:Power of 2 is Difference between Two Powers
Category:Powers of 2
Power of 2 is Difference between Two Powers/Examples

== Examples of Power of 2 is Difference between Two Powers ==
=== $2^0$ expressed as Difference between Two Powers ===
=== $2^1$ expressed as Difference between Two Powers ===
=== $2^2$ expressed as Difference between Two Powers ===
=== $2^4$ expressed as Difference between Two Powers ===
=== $2^5$ expressed as Difference between Two Powers ===
Category:Power of 2 is Difference between Two Powers