nyuuzyou/wikis huggingface dataset:
Restriction of Reflexive Relation is Reflexive

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a reflexive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a reflexive relation on $T$.
== Proof ==
Suppose $\RR$ is reflexive on $S$.
So:
:$\forall x \in S: \tuple {x, x} \in \RR$
We are given $T$ is a subset of $S$, so:
:$\forall x \in T: \tuple {x, x} \in \RR$
By definition of restriction:
:$\forall x \in T: \tuple {x, x} \in \RR {\restriction_T}$
Hence $\RR {\restriction_T}$ is by definition reflexive on $T$.
== Also see ==
* Properties of Restriction of Relation‎ for other similar properties of the restriction of a relation.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Reflexive Relations
Category:Restrictions
Restriction of Restriction is Restriction

== Theorem ==
Let $\RR$ be a relation on $S \times T$.
Let $X \subseteq S$, $Y \subseteq T$.
Let $W \subseteq X$, $V \subseteq Y$.
Then:
:$\paren{\RR {\restriction_{X \times Y} } } {\restriction_{W \times V} } = \RR {\restriction_{W \times V} }$
That is, the restriction of $\RR {\restriction_{X \times Y} }$ to $W \times V$ is the restriction of $\RR$ to $W \times V$
== Proof ==
From Cartesian Product of Subsets:
:$W \times V \subseteq X \times Y$
We have:
Category:Restrictions
Restriction of Ringed Space to Open Set is Ringed Space

== Theorem ==
Let $\struct {X, \OO_X}$ be a ringed space.
Let $U \subset X$ be an open subset of $X$.
Let $\struct {U, \OO_X {\restriction_U}}$ denote the restriction of $\struct {X, \OO_X}$ to $U$.
Then $\struct {U, \OO_X {\restriction_U}}$ is a ringed space.
== Proof ==
By Restriction of Sheaf to Open Set is Sheaf $\OO_X {\restriction_U}$ is a sheaf of commutative rings on $U$.
It follows, that $\struct {U, \OO_X {\restriction_U}}$ is a ringed space.
Category:Ringed Spaces
Restriction of Serial Relation is Not Necessarily Serial

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a serial relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is not necessarily a serial relation on $T$.
== Proof ==
Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, b}, \tuple {b, b} }$.
$\RR$ is a serial relation, as can be seen by definition.
Now let $T = \set a$.
Then:
:$\RR {\restriction_T} = \O$
So:
:$\not \exists y \in T: \tuple {a, y} \in \RR {\restriction_T}$
That is, $\RR {\restriction_T}$ is not a serial relation on $T$.
== Also see ==
* Properties of Relation Not Preserved by Restriction‎ for other similar results.
Category:Serial Relations
Category:Restrictions
Restriction of Sheaf to Open Set is Sheaf

== Theorem ==
Let $X$ be a topological space.
Let $\FF$ be a sheaf of sets on $X$.
Let $U \subset X$ be an open subset.
Then the restriction $\FF {\restriction_U}$ of $\FF$ to $U$ is a sheaf.
== Proof ==
Category:Algebraic Geometry
Category:Sheaf Theory
Restriction of Strict Total Ordering is Strict Total Ordering

== Theorem ==
Let $\struct {S, \prec}$ be a strict total ordering.
Let $T \subseteq S$.
Let $\prec \restriction_T$ be the restriction of $\prec$ to $T$.
Then $\prec \restriction_T$ is a strict total ordering of $T$.
== Proof ==
By definition of strict total ordering, $\prec$ is:
:$(1): \quad$ a relation which is transitive and antireflexive 
:$(2): \quad$ a relation which is connected.
By Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering:
:$\prec \restriction_T$ is a strict ordering.
It follows from Restriction of Connected Relation is Connected that:
:$\prec \restriction_T$ is connected.
Thus $\prec \restriction_T$ is a strict ordering which is connected.
So by definition $\prec \restriction_T$ is a strict total ordering of $T$.
== Sources ==
* : $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.7$
Category:Total Orderings
Category:Strict Orderings
Restriction of Strict Well-Ordering is Strict Well-Ordering

== Theorem ==
Let $R$ be a strict well-ordering of $A$.
Let $B \subseteq A$.
Then $R$ is a strict well-ordering of $B$.
== Proof ==
By Restriction of Strictly Well-Founded Relation is Strictly Well-Founded, $R$ is a strictly well-founded relation on $B$.
By Restriction of Total Ordering is Total Ordering, $R$ is a total ordering on $B$.
By the above two statements, $R$ is a strict well-ordering of $B$.
== Also see ==
* Restriction of Strictly Well-Founded Relation is Strictly Well-Founded
Category:Well-Orderings
Restriction of Strictly Well-Founded Relation is Strictly Well-Founded

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure.
Let $\RR \subseteq S \times S$ be a strictly well-founded relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a strictly well-founded relation on $T$.
== Proof ==
By definition of strictly well-founded relation, every non-empty subset of $S$ has a minimal element.
By Subset Relation is Transitive, every subset of $T$ is also a subset of $S$.
Therefore every non-empty subset of $T$ has a minimal element.
Thus by definition, $\RR$ is a strictly well-founded relation on $T$.
== Also see ==
* Definition:Strictly Well-Founded Relation
* Axiom of Foundation
Category:Well-Founded Relations
Restriction of Symmetric Relation is Symmetric

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a symmetric relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a symmetric relation on $T$.
== Proof ==
Suppose $\RR$ is symmetric on $S$.
Then:
and so $\RR {\restriction_T}$ is symmetric on $T$.
== Also see ==
* Properties of Restriction of Relation‎ for other similar properties of the restriction of a relation.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Symmetric Relations
Category:Restrictions
Restriction of Total Ordering is Total Ordering

== Theorem ==
Let $\struct {S, \preceq}$ be a total ordering.
Let $T \subseteq S$.
Let $\preceq \restriction_T$ be the restriction of $\preceq$ to $T$.
Then $\preceq \restriction_T$ is a total ordering of $T$.
== Proof ==
By Restriction of Ordering is Ordering, $\preceq \restriction_T$ is an ordering.
Let $x, y \in T$.
As $T \subseteq S$ it follows by definition of subset that:
:$x, y \in S$
As $\preceq$ is a total ordering:
:$\tuple {x, y} \in {\preceq}$
or:
:$\tuple {y, x} \in {\preceq}$
Suppose $\tuple {x, y} \in {\preceq}$.
As $x, y \in T$, it follows by definition of cartesian product that:
:$\tuple {x, y} \in T \times T$
Thus:
:$\tuple {x, y} \in \paren {T \times T} \cap {\preceq}$
By definition of the restriction of $\preceq$ to $T$:
:$\paren {T \times T} \cap {\preceq} = {\preceq \restriction_T}$
That is:
:$\tuple {x, y} \in {\preceq \restriction_T}$
A similar argument shows that:
:$\tuple {y, x} \in {\preceq} \implies \tuple {y, x} \in {\preceq \restriction_T}$
Thus $\preceq \restriction_T$ is a total ordering of $T$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
* : $\text I.1$: Theorem $1$
Category:Total Orderings
Restriction of Transitive Relation is Transitive

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a transitive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a transitive relation on $T$.
== Proof ==
Suppose $\RR$ is transitive on $S$.
Then by definition:
:$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR \implies \tuple {x, z} \in \RR$
So:
Therefore, if $x, y, z \in T$, it follows that:
:$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR {\restriction_T} \implies \tuple {x, z} \in \RR {\restriction_T}$
and so by definition $\RR {\restriction_T}$ is a transitive relation on $T$.
== Also see ==
* Properties of Restriction of Relation‎ for other similar properties of the restriction of a relation.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Transitive Relations
Category:Restrictions
Restriction of Union of Mappings which Agree Equals Mapping

== Theorem ==
Let $Y$ be a set.
Let $\family {A_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Let $\family {f_i: A_i \to Y}$ be a family of mappings indexed by $I$.
Let $X = \ds \bigcup_{i \mathop \in I} A_i$.
Let $f = \ds \bigcup_{i \mathop \in I} f_i : X \to Y$ where $\ds \bigcup_{i \mathop \in I} f_i$ is the union of relations.
Let for all $i, j \in I$, $f_i$ and $f_j$ agree on $A_i \cap A_j$.
Let $f \restriction_{A_i} : A_i \to Y$ denote the restriction of $f$ to $A_i$.
Then:
:$\forall i \in I: f \restriction_{A_i} = f_i: A_i \to Y$.
== Proof ==
From Union of Family of Mappings which Agree is Mapping:
:$f$ is a mapping from $X$ to $Y$.
Let $i \in I$ be an arbitrary index of $I$.
Consider the restriction of $f$ to $A_i$:
Let $j$ be an arbitrary index of $I$.
=== Case 1 : $i \ne j$ ===
Let $j \ne i$.
Let $\tuple {x, y} \in f_j \cap \paren{A_i \times Y}$.
Then:
:$x \in f_j \cap f_i$
We have:
:$f_i$ and $f_j$ agree on $A_i \cap A_j$
By definition of agreement of mappings:
:$y = \map {f_j} x = \map {f_i} x$
Hence:
:$\tuple {x, y} \in f_i$
By definition of subset:
:$f_j \cap \paren{A_i \times Y} \subseteq f_i$
=== Case 2 : $j = i$ ===
Let $j = i$.
By definition of mapping:
:$f_i \subseteq A_i \times Y$
From Intersection with Subset is Subset:
:$f_i = f_i \cap \paren{A_i \times Y}$
In either case:
:$f_j \cap \paren{A_i \times Y} \subseteq f_i$
Since $j \in I$ was arbitrary:
:$\forall j \in I : f_j \cap \paren{A_i \times Y} \subseteq f_i$
From Union of Family of Subsets is Subset:
:$\bigcup_{j \in I} \paren{f_j \cap \paren{A_i \times Y} } \subseteq f_i$
From Set is Subset of Union of Family:
:$f_i = f_i \cap \paren{A_i \times Y} \subseteq \bigcup_{j \in I} \paren{f_j \cap \paren{A_i \times Y} }$
By definition of set equality:
:$f_i = \bigcup_{j \in I} \paren{f_j \cap \paren{A_i \times Y} }$
Hence:
:$f \restriction _ { A_i } = f_i$
Since $i \in I$ was arbitrary:
:$\forall i \in I: f \restriction_{A_i} = f_i: A_i \to Y$.
Category:Mapping Theory
Category:Set Union
Category:Indexed Families
Category:Restrictions
Restriction of Well-Founded Ordering is Well-Founded

== Theorem ==
Let $S$ be a set or class.
Let $T$ be a subset or subclass of $S$.
Let $\preceq$ be a well-founded ordering of $A$.
Let $\preceq'$ be the restriction of $\preceq$ to $T$.
Then $\preceq'$ is a well-founded ordering of $T$.
== Proof ==
By Restriction of Ordering is Ordering, $\preceq'$ is an ordering.
By Restriction of Well-Founded Relation is Well-Founded, $\preceq'$ is a well-founded relation on $T$.
Hence the result by definition of well-founded ordering.
Category:Order Theory
Category:Well-Founded Relations
Restriction of Well-Founded Relation is Well-Founded

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure.
Let $\RR$ be a well-founded relation on $S$.
Let $T$ be a subset or subclass of $S$.
Let $\RR'$ be the restriction of $\RR$ to $T$.
Then $\preceq'$ is a well-founded relation on $T$.
== Proof ==
Let $A$ be a non-empty subset of $T$.
By Subset Relation is Transitive, $A$ is a non-empty subset of $S$.
Since $\RR$ is a well-founded relation on $S$, $A$ has a minimal element $m$ under $\RR$.
Let $x \in A$.
Let $x \mathrel {\RR'} m$.
By the definition of restriction:
:$x \mathrel \RR m$
Thus by the definition of minimal element:
:$x = m$
As $x$ is an arbitrary element of $A$, this holds for all $x \in A$.
Hence by definition $m$ is a minimal element of $A$ under $\RR'$.
As $A$ is an arbitrary non-empty subset of $T$, this holds for all non-empty subsets of $T$.
Hence $\RR'$ is a well-founded relation on $T$.
Category:Well-Founded Relations
Restriction of Well-Ordering is Well-Ordering

== Theorem ==
Let $S$ be a set or class.
Let $\preceq$ be a well-ordering of $S$.
Let $T$ be a subset or subclass of $S$.
Let $\preceq'$ be the restriction of $\preceq$ to $T$.
Then $\preceq'$ is a well-ordering of $T$.
== Proof ==
By the definition of well-ordering, $\preceq$ is a well-founded total ordering.
By Restriction of Total Ordering is Total Ordering, $\preceq'$ is a total ordering.
By Restriction of Well-Founded Ordering is Well-Founded, $\preceq'$ is a well-founded ordering.
Thus $\preceq'$ is a well-ordering.
Category:Well-Orderings
Restriction to Subset of Strict Total Ordering is Strict Total Ordering

== Theorem ==
Let $S$ be a set or class.
Let $\prec$ be a strict total ordering on $A$.
Let $T$ be a subset or subclass of $A$.
Then the restriction of $\prec$ to $B$ is a strict total ordering of $B$.
== Proof ==
Follows from:
:Restriction of Transitive Relation is Transitive
:Restriction of Antireflexive Relation is Antireflexive
:Restriction of Connected Relation is Connected
Category:Total Orderings
Resultant in Terms of Dot Product

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Let their resultant be $\mathbf v$:
:$\mathbf v = \mathbf a + \mathbf b$
Then:
:$\mathbf v^2 = \mathbf a^2 + 2 \mathbf a \cdot \mathbf b + \mathbf b^2$
where:
:$\mathbf v^2$ denotes the square of $\mathbf v$
:$\mathbf a \cdot \mathbf b$ denotes the dot product of $\mathbf a$ and $\mathbf b$
== Proof ==
== Also presented as ==
This result can also be presented as:
:$v^2 = a^2 + 2 a b \cos \theta + b^2$
where:
:$v$, $a$ and $b$ are the magnitudes of $\mathbf v$, $\mathbf a$ and $\mathbf b$ respectively
:$\theta$ is the angle between the directions of $\mathbf a$ and $\mathbf b$.
== Also see ==
* Law of Cosines
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product
Category:Vector Addition
Category:Dot Product
Results Concerning Annihilator of Vector Subspace

== Theorem ==
Let $G$ be an $n$-dimensional vector space over a field.
Let $G^*$ be the algebraic dual of $G$.
Let $G^{**}$ be the algebraic dual of $G^*$.
Let $J: G \to G^{**}$ be the evaluation isomorphism.
Let $M$ be an $m$-dimensional subspace of $G$.
Let $N$ be a $p$-dimensional subspace of $G^*$.
Let $M^\circ$ be the annihilator of $M$.
Let $J^\gets: \powerset {G^{**} } \to \powerset G$ be the inverse image mapping of $J$.
Then the following results hold:
=== Dimension of Annihilator on Algebraic Dual ===
=== Annihilator of Annihilator on Algebraic Dual of Subspace is Image under Evaluation Isomorphism ===
=== Dimension of Preimage under Evaluation Isomorphism of Annihilator on Algebraic Dual ===
=== Mapping to Annihilator on Algebraic Dual is Bijection ===
=== Inverse of Mapping to Annihilator on Algebraic Dual is Bijection ===
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.10$
Category:Annihilators
Category:Vector Subspaces
Results Concerning Limits of Sequences of Intervals

== Decreasing Sequences of Sets ==
=== Limit of Decreasing Sequence of Unbounded Below Closed Intervals with Endpoint Tending to Negative Infinity ===
=== Limit of Decreasing Sequence of Unbounded Below Closed Intervals ===
=== Limit of Decreasing Sequence of Left Half-Open Intervals with Lower Bound Converging to Upper Bound === 
Category:Decreasing Sequences of Sets
Category:Increasing Sequences of Sets
Results Concerning Set Difference with Intersection

== Theorem ==
Let:
:$S \setminus T$ denote set difference
:$S \cap T$ denote set intersection.
=== Intersection with Set Difference is Set Difference with Intersection ===
=== Set Difference is Right Distributive over Set Intersection ===
=== Set Intersection Distributes over Set Difference ===
== Also see ==
* De Morgan's Laws: Difference with Intersection, in which:
:$R \setminus \paren {S \cap T} = \paren {R \setminus S} \cup \paren {R \setminus T}$
shows that set difference is not left distributive over set intersection.
Category:Set Difference
Category:Set Intersection
Results Concerning Set Difference with Union

== Theorem ==
Let:
:$S \setminus T$ denote set difference
:$S \cup T$ denote set union
:$S \cap T$ denote set intersection.
=== Set Difference with Union ===
=== Set Difference is Right Distributive over Union ===
=== Set Difference with Set Difference is Union of Set Difference with Intersection ===
=== Set Difference is Subset of Union of Differences ===
Category:Set Difference
Category:Set Union
Results concerning Generators and Bases of Vector Spaces

== Theorem ==
Let $E$ be a vector space of $n$ dimensions.
Let $G$ be a generator for $E$.
Then $G$ has the following properties:
=== Cardinality of Generator of Vector Space is not Less than Dimension ===
=== Generator of Vector Space Contains Basis ===
=== Generator of Vector Space is Basis iff Cardinality equals Dimension ===
Let $H$ be a linearly independent subset of $E$.
Then $H$ has the following properties:
=== Cardinality of Linearly Independent Set is No Greater than Dimension ===
=== Linearly Independent Set is Contained in some Basis ===
=== Linearly Independent Set is Basis iff of Same Cardinality as Dimension ===
Category:Linear Algebra
Results concerning Order of Element

== Theorems ==
Let $G$ be a group whose identity is $e$.
Let $a \in G$ have finite order such that $\order a = k$.
Then the following results apply:
=== Equal Powers of Finite Order Element ===
=== Element to Power of Remainder ===
=== Element to Power of Multiple of Order is Identity ===
=== List of Elements in Finite Cyclic Group ===
Category:Order of Group Elements
Retract of Injective Space is Injective

== Theorem ==
Let $T = \struct {S, \tau}$ be an injective topological space.
Let $R = \struct {Z, \tau'}$ be a retract of $T$.
Then $R$ is injective.
== Proof ==
By definition of retract:
:there exists a continuous retraction $r: S \to Z$ of $T$.
Let $\YY = \struct {Y, \sigma}$ be a topological space.
Let $f: Y \to Z$ be a continuous mapping.
Let $\XX = \struct {X, \sigma'}$ such that
:$\YY$ is topological subspace of $\XX$.
By Inclusion Mapping is Continuous:
:$i_Z$ is continuous $\paren {R \to T}$
where $i_Z$ denotes the inclusion mapping from $Z$ in $S$.
By Composite of Continuous Mappings is Continuous:
:$i_Z \circ f: Y \to S$ is continuous.
By definition of injective space:
:there exists a continuous mapping $g: X \to S: g \restriction Y = i_Z \circ f$
Define $h := r \circ g$
By Composite of Continuous Mappings is Continuous:
:$h$ is continuous.
We will prove that
:$h \restriction Y = f$
By definition of topological subspace:
:$Y \subseteq X$ and $Z \subseteq S$
Thus by definitions of composition of mappings and restriction of mapping:
:$h \restriction Y: Y \to Z$ and $f: Y \to Z$
Let $y \in Y$.
By Restriction of Composition is Composition of Restriction:
:$h \restriction Y = r \circ \paren {g \restriction Y}$
By definition of mapping:
:$\map f y \in Z$
Thus
== Sources ==
* 
Category:Topology
Retraction Theorem

== Theorem ==
Let $M$ be a compact manifold with boundary $\partial M$.
Then there is no smooth mapping $f: M \to \partial M$ such that $\partial f: \partial M \to \partial M$ is the identity.
== Proof ==
 such a smooth mapping exists.
By the Morse-Sard Theorem, there exists a regular value $x \in \partial M$.
By the Preimage Theorem:
:$\map {f^{-1} } x$ is a submanifold of $M$ with boundary.
We have that the codimension of $\map {f^{-1} } x$ in $M$ equals the codimension of $x$ in $\partial M$, that is, $\map \dim M - 1$.
Then $\map {f^{-1} } x$ is one dimensional and compact.
Since $\partial f$ is the identity mapping:
:$\map {\partial f^{-1} } x = \map {f^{-1} } x \cap \partial M = \set x$
This contradicts the Classification of Compact One-Manifolds.
Category:Algebraic Topology
Category:Named Theorems
Retrograde Analysis/Examples/Example 1/Historical Note

== Historical Note on Retrograde Analysis Example 1 ==
 reports in his of $2008$ that this problem was the work of , but he gives no further details.
== Sources ==
* : Chapter $1$: Elementary, my dear Watson: $\S 1.1.1$: 'You talking to me?': Ponderable $1.1.7$
Category:Historical Notes
Returning Explorer Puzzle

== Puzzle ==
An explorer walks:
:one mile due south, 
:one mile due east,
:and one mile due north.
He finds himself back where he started.
He shoots a .
What colour is that ?
=== Variant 1 ===
=== Variant 2 ===
== Solution ==
White.
== Proof ==
The explorer starts from the North Pole.
Every direction from the North Pole is due south.
Walking due east keeps him the same distance (one mile) from the North Pole.
Walking due north takes him back to the North Pole again.
The bear, as a consequence, must be a .
== Sources ==
* : Chapter Two: Paradoxes for Everyone
* : Nine Problems: $1$
* : Nine Problems: $1$
Category:Geography Puzzles
Category:Returning Explorer Puzzle
Returning Explorer Puzzle/Variant 1

== Puzzle ==
A "sportsman" who shoots things for fun decided to go out and shoot a .
Suddenly he saw a huge bear a hundred metres due east.
In a panic, in confusion, he ran due north for a hundred metres.
Then he caught hold of himself and stopped, and turned.
He saw the , which had not moved, a hundred metres due south.
Of course, he shot and killed the poor .
What colour is that ?
== Solution ==
White.
== Proof ==
The explorer starts a hundred metres from the North Pole.
The , being due east of the man, is also a hundred metres from the North Pole.
Running due north for a hundred metres places him exactly at the North Pole.
Every direction from the North Pole is due south.
The bear, as a consequence, must be a .
== Sources ==
* : Chapter Two: Paradoxes for Everyone
Category:Returning Explorer Puzzle
Returning Explorer Puzzle/Variant 2

== Puzzle ==
An explorer walks:
:one mile due south, 
:one mile due east,
:and one mile due north.
He finds himself back where he started.
Where are all the places in the world where this is possible?
== Solution ==
:$(1): \quad$ The North Pole
:$(2): \quad$ Any position $1 + \dfrac 1 {2 \pi n}$ miles from the South Pole, where $n$ can be any (strictly) positive integer.
== Proof ==
The North Pole solution is explored (no pun intended) in Returning Explorer Puzzle, where he is specifically located, as he then shoots a .
It remains to explore the region of the South Pole.
Let $S$ denote the South Pole.
Let the explorer starts some distance $x$ from $S$, where $x$ is greater than $1$ mile.
He walks due south to a point $P$ which is $x - 1$ miles from $S$.
He walks $1$ mile due east, in the process circumnavigating $S$ a total of $n$ times, arriving back at $P$.
He then walks due north again, to the place he started from.
Let us treat the location around $S$ as a plane surface.
Thus we have that:
where $n$ is a (strictly) positive integer.
== Sources ==
* : Nine Problems: $1$
* : Nine Problems: $1$
Category:Returning Explorer Puzzle
Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication

== Theorem ==
Let $b \in \Z_{>1}$ be an integer greater than $1$.
Let $n = b^k - 1$ for some integer $k$ such that $k \ge 1$.
Then:
: $n + n$ is the reversal of $\paren {b - 1} n$
when both are expressed in base $b$ representation.
== Proof ==
By Power of Base minus 1 is Repdigit Base minus 1, $n$ is a repdigit number consisting of $k$ occurrences of $b - 1$.
Let $a = b - 1$.
Thus $n$ can be expressed in base $b$ as:
:$n = {\overbrace {\sqbrk {aaa \cdots a} }^k}_b$
We have that:
where there are:
:$k - 1$ occurrences of $a$
:$c = b - 2$.
Then:
where there are:
:$k - 1$ occurrences of $a$
:$c = b - 2$.
Hence the result.
== Examples ==
== Historical Note ==
== Sources ==
* : $18$
* : $18$
Category:Number Bases
Category:Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication
Category:Repdigit Numbers
Category:Reversals
Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication/Examples

== Examples of Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication ==
=== Reversal of $9 + 9$ ===
=== Reversal of $99 + 99$ ===
=== Reversal of $7 + 7$ Base $8$ ===
Category:Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication