nyuuzyou/wikis huggingface dataset:
Two-Person Zero-Sum Game/Examples/Abstract 4

== Example of Two-Person Zero-Sum Game ==
The two players are $A$ and $B$.
Player $A$ has $m$ strategies: $A_1, A_2, \ldots, A_m$.
Player $B$ has $n$ strategies: $B_1, B_2, \ldots, B_n$.
The game is zero-sum in that a payoff of $p$ to $A$ corresponds to a payoff of $-p$ to $B$.
The following table specifies the payoff to $A$ for each strategy of $A$ and $B$.
This has an equilibrium point at $\tuple {A_2, B_3}$ with a payoff to $A$ of $-1$.
== Proof ==
By inspection, strategy $B_3$ dominates strategy $B_1$.
Therefore by Elimination by Domination, $B_1$ can be eliminated.
$B_1$ having been eliminated, it is seen by inspection that $A_2$ dominates $A_1$.
$A_1$ having been eliminated, it is seen by inspection that $B_3$ dominates $B_2$.
$B_2$ having been eliminated, it is seen by inspection that $A_2$ dominates $A_3$.
The optimum strategies are therefore $A_2$ and $B_3$, and the payoff to $A$ is $-1$.
Given $B_3$, every other strategy adopted by $A$ will result in a smaller payoff to $A$.
Given $A_2$, every other strategy adopted by $B$ will result in a larger payoff to $A$.
Hence, by definition, $\tuple {A_2, B_3}$ is an equilibrium point.
== Sources ==
* : $\S 2$: The Two-Person, Zero-Sum Game with Equilibrium Points: Introductory Problems: Figure $2.4$
Category:Examples of Two-Person Zero-Sum Games
Two-Person Zero-Sum Game/Examples/Abstract 5

== Example of Two-Person Zero-Sum Game ==
The two players are $A$ and $B$.
Player $A$ has $m$ strategies: $A_1, A_2, \ldots, A_m$.
Player $B$ has $n$ strategies: $B_1, B_2, \ldots, B_n$.
The game is zero-sum in that a payoff of $p$ to $A$ corresponds to a payoff of $-p$ to $B$.
The following table partially specifies the payoff to $A$ for each strategy of $A$ and $B$.
This has an equilibrium point at $\tuple {A_3, B_3}$ with a payoff to $A$ of $5$.
== Proof ==
Given $B_3$, every other strategy adopted by $A$ will result in a smaller payoff to $A$ than $5$.
Given $A_3$, every other strategy adopted by $B$ will result in a larger payoff to $A$ than $5$.
Hence, by definition, $\tuple {A_3, B_3}$ is an equilibrium point.
== Sources ==
* : $\S 2$: The Two-Person, Zero-Sum Game with Equilibrium Points: Introductory Problems: Figure $2.5$
Category:Examples of Two-Person Zero-Sum Games
Two-Person Zero-Sum Game/Examples/Political Example 1

== Example of Two-Person Zero-Sum Game ==
The two players are political candidates: $\text A$ and $\text B$.
Each may choose between the following three strategies:
:supporting policy $\text X$
:supporting policy $\text Y$
:taking no position on the matter.
It is assumed that voting will be restricted to either $\text A$ or $\text B$. That is, no abstentions will take place.
The expected share of the vote for either candidate depends on which strategy they adopt.
The following table gives the expected share of the vote for $\text A$ based upon the combined strategies of $\text A$ and $\text B$:
This is a two-person zero-sum game with an equilibrium point.
== Proof ==
=== Two-Person ===
The two players are $A$ and $B$.
=== Zero-Sum ===
The assumption is made at the start that the votes are divided solely between $A$ and $B$.
There is a fixed total share of the vote ($100\%$).
What is not gained by $A$ will be gained by $B$, whatever their strategies.
=== Equilibrium Point ===
It can be seen by inspection that whatever $B$'s platform, $A$'s share of the vote is a maximum when $A$ supports policy $Y$.
It can also be seen by inspection that whatever $A$'s platform, $A$'s share of the vote is a minimum (and therefore that of $B$ is a maximum) when $B$ takes no position on the matter.
If either party deviates from this strategy, their share of the vote will decrease.
Thus, by definition, this is an equilibrium point.
== Sources ==
* : $\S 2$: The Two-Person, Zero-Sum Game with Equilibrium Points: A Political Example: Figure $2.7$
Category:Examples of Two-Person Zero-Sum Games
Two-Person Zero-Sum Game/Examples/Political Example 2

== Example of Two-Person Zero-Sum Game ==
The two players are political candidates: $\text A$ and $\text B$.
Each may choose between the following three strategies:
:supporting policy $\text X$
:supporting policy $\text Y$
:taking no position on the matter.
It is assumed that voting will be restricted to either $\text A$ or $\text B$. That is, no abstentions will take place.
The expected share of the vote for either candidate depends on which strategy they adopt.
The following table gives the expected share of the vote for $\text A$ based upon the combined strategies of $\text A$ and $\text B$:
This is a two-person zero-sum game with an equilibrium point.
== Proof ==
=== Two-Person ===
The two players are $A$ and $B$.
=== Zero-Sum ===
The assumption is made at the start that the votes are divided solely between $A$ and $B$.
There is a fixed total share of the vote ($100\%$).
What is not gained by $A$ will be gained by $B$, whatever their strategies.
=== Equilibrium Point ===
It can be seen by inspection that whatever $B$'s platform, $A$'s share of the vote is a maximum when $A$ supports policy $Y$.
It is therefore clear to $B$ that this will be $A$'s decision.
So, while there may be a slim chance that $A$ will make the ridiculous decision to either support $X$ or take no position, this is sufficiently unlikely as to make $B$'s decision clear: to take no position.
If either party deviates from this strategy, their share of the vote will decrease.
Thus, by definition, this is an equilibrium point.
== Sources ==
* : $\S 2$: The Two-Person, Zero-Sum Game with Equilibrium Points: A Political Example: Figure $2.8$
Category:Examples of Two-Person Zero-Sum Games
Two-Person Zero-Sum Game/Examples/Political Example 3

== Example of Two-Person Zero-Sum Game ==
The two players are political candidates: $\text A$ and $\text B$.
Each may choose between the following three strategies:
:supporting policy $\text X$
:supporting policy $\text Y$
:taking no position on the matter.
It is assumed that voting will be restricted to either $\text A$ or $\text B$. That is, no abstentions will take place.
The expected share of the vote for either candidate depends on which strategy they adopt.
The following table gives the expected share of the vote for $\text A$ based upon the combined strategies of $\text A$ and $\text B$:
This is a two-person zero-sum game with an equilibrium point.
== Proof ==
=== Two-Person ===
The two players are $A$ and $B$.
=== Zero-Sum ===
The assumption is made at the start that the votes are divided solely between $A$ and $B$.
There is a fixed total share of the vote ($100\%$).
What is not gained by $A$ will be gained by $B$, whatever their strategies.
=== Equilibrium Point ===
It can be seen by inspection that whatever $A$'s platform, $B$ should not take no position, because that is always inferior to supporting $X$.
Once it has been established that $B$ is either supporting $X$ or supporting $Y$, it can be seen that $A$ always does best when supporting $Y$.
It then follows that $B$ should support $X$, leaving $A$ with $45 \%$ of the vote.
If either party deviates from this strategy, their share of the vote will decrease.
Thus, by definition, this is an equilibrium point.
== Sources ==
* : $\S 2$: The Two-Person, Zero-Sum Game with Equilibrium Points: A Political Example: Figure $2.9$
Category:Examples of Two-Person Zero-Sum Games
Two-Person Zero-Sum Game is Non-Cooperative

== Theorem ==
A two-person zero-sum game necessarily has to be non-cooperative.
== Proof ==
A cooperative game is one where players form coalitions against the other players.
If the players in a two-person zero-sum game were to form a coalition, there would be no other players to form it against.
Further, as the total payoff is zero, there would be no benefit in collaborating on using one strategy over another, as when they pool their payoffs they are back where they started.
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $2$
Category:Two-Person Zero-Sum Games
Category:Cooperative Games
Two-Person Zero-Sum Game which is Not Completely Mixed

== Example of Two-Person Zero-Sum Game which is not Completely Mixed ==
Consider the two-person zero-sum game with the following payoff table:
It is seen by inspection that:
:$A$'s optimum strategy is $\left({1/4, 3/4}\right)$
:$B$'s optimum strategy is $\left({1/2, 1/2, 0}\right)$.
It is also noted, in passing, that $B_2$ dominates $B_3$.
Hence the result, by definition of completely mixed game.
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $3$
Category:Examples of Two-Person Zero-Sum Games
Two-Person Zero-Sum Game with Finite Strategies has Solution

== Theorem ==
Let $G$ be a two-person zero-sum game.
Let each player of $G$ have a finite set of strategies available.
Then $G$ has at least one solution.
== Proof ==
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $3$
Category:Two-Person Zero-Sum Games
Two-Person Zero-Sum Game with Multiple Solutions

== Theorem ==
There exists a two-person zero-sum game with more than one solution.
== Proof ==
Consider the game defined by the following payoff table:
This has two solutions:
:$(1): \quad A: \left({1/3, 2/3}\right), B: \left({0, 1, 0}\right)$
:$(2): \quad A: \left({2/3, 1/3}\right), B: \left({0, 1, 0}\right)$
It follows that the strategies:
:$\left({\dfrac t 3, \dfrac {2 \left({1 - t}\right)} 3}\right), B: \left({0, 1, 0}\right)$
for all $0 \le t \le 1$ are also solutions.
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $3$
Category:Examples of Two-Person Zero-Sum Games
Two-Point Form of Equation of Straight Line in Plane/Examples

== Examples of Use of Two-Point Form of Equation of Straight Line in Plane ==
=== $\tuple {1, 3}$ and $\tuple {-2, 4}$ ===
=== $\tuple {2, 1}$ and $\tuple {-6, 7}$ ===
Category:Two-Point Form of Equation of Straight Line in Plane
Two-Point Form of Equation of Straight Line in Plane/Examples/(2, 1) and (-6, 7)

== Example of Use of Two-Point Form of Equation of Straight Line in Plane ==
Let $\LL$ be the straight line passing through the points $\tuple {2, 1}$ and $\tuple {-6, 7}$.
Then $\LL$ has the equation:
:$\dfrac {x - 2} {-6 - 2} = \dfrac {y - 1} {7 - 1}$
which after rearranging, can be written in the general form:
:$4 y = -3 x + 10$
== Sources ==
* 
* 
Category:Two-Point Form of Equation of Straight Line in Plane
Two-Row Notation/Examples/Permutation in S4

== Example of Two-Row Notation ==
The permutation on the symmetric group on $4$ letters $S_4$ defined as:
:$1 \mapsto 3, 2 \mapsto 2, 3 \mapsto 4, 4 \mapsto 1$
can be depicted in two-row notation as:
:$\begin {pmatrix} 1 & 2 & 3 & 4 \\ 3 & 2 & 4 & 1 \end {pmatrix}$
== Sources ==
* : $\S 34$. Examples of groups: $(4) \ \text {(c)}$
Category:Examples of Two-Row Notation
Two-Row Notation/Examples/Permutations in S3

== Examples of Two-Row Notation ==
The permutations on the symmetric group on $3$ letters $S_3$ can be depicted in two-row notation as:
:$\begin{pmatrix}
1 & 2 & 3 \\
1 & 2 & 3
\end{pmatrix}
\qquad
\begin{pmatrix}
1 & 2 & 3 \\
1 & 3 & 2
\end{pmatrix}
\qquad
\begin{pmatrix}
1 & 2 & 3 \\
2 & 1 & 3
\end{pmatrix}$
:$\begin{pmatrix}
1 & 2 & 3 \\
2 & 3 & 1
\end{pmatrix}
\qquad
\begin{pmatrix}
1 & 2 & 3 \\
3 & 1 & 2
\end{pmatrix}
\qquad
\begin{pmatrix}
1 & 2 & 3 \\
3 & 2 & 1
\end{pmatrix}$
== Sources ==
* : Chapter $9$: Permutations
Category:Examples of Two-Row Notation
Two-Step Subgroup Test using Subset Product

== Theorem ==
Let $G$ be a group.
Let $\O\subset H \subseteq G$ be a non-empty subset of $G$.
Then $H$ is a subgroup of $G$ :
:$H H \subseteq H$
:$H^{-1} \subseteq H$
where:
:$H H$ is the product of $H$ with itself
:$H^{-1}$ is the inverse of $H$.
== Proof ==
This is a reformulation of the Two-Step Subgroup Test in terms of subset product.
=== Necessary Condition ===
Let $H$ is a subgroup of $G$.
Then $H$ is closed.
It follows from Magma Subset Product with Self:
:$H H \subseteq H$
Then:
:$g \in H^{-1} \implies \exists h \in H: g = h^{-1} \implies g \in H$
so:
:$H^{-1} \subseteq H$
=== Sufficient Condition ===
Let:
:$H H \subseteq H$
:$H^{-1} \subseteq H$
From the definition of subset product:
:$\forall x, y \in H: x y \in H$
:$\forall x \in H^{-1}: x^{-1} \in H$
So by the Two-Step Subgroup Test, $H$ is a subgroup of $G$.
== Also see ==
* One-Step Subgroup Test using Subset Product
== Linguistic Note ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Theorem $9.1: \ 2^\circ$
* : Chapter $\text{II}$: Groups: Subgroups
* : $\S 3.2$: Groups; the axioms
Category:Subgroups
Category:Subset Products
Two-Step Vector Subspace Test

== Theorem ==
Let $V$ be a vector space over a division ring $K$.
Let $U \subseteq V$ be a non-empty subset of $V$ such that:
Then $U$ is a subspace of $V$.
== Proof ==
Suppose that $(1)$ and $(2)$ hold.
From $(1)$, we obtain for every $\lambda \in K$ and $u \in U$ that $\lambda u \in U$.
An application of $(2)$ yields the condition of the One-Step Vector Subspace Test.
Hence $U$ is a vector subspace of $V$.
== Also see ==
* One-Step Vector Subspace Test
Category:Linear Algebra
Category:Vector Subspaces
Two-Valued Functions form Boolean Algebra

== Theorem ==
Let $\mathbf 2$ be the Boolean algebra two, and let $X$ be a set.
Let $\mathbf 2^X$ be the set of all mappings $p: X \to \mathbf 2$.
Define the operations $\vee$, $\wedge$ and $\neg$ on $\mathbf 2^X$ in pointwise fashion thus:
:$\vee: \mathbf 2^X \times \mathbf 2^X \to \mathbf 2^X, \left({p \vee q}\right) (x) := p (x) \vee q (x)$
:$\wedge: \mathbf 2^X \times \mathbf 2^X \to \mathbf 2^X, \left({p \wedge q}\right) (x) := p (x) \wedge q (x)$
:$\neg: \mathbf 2^X \to \mathbf 2^X, \left({\neg p}\right) (x) := \neg p (x)$
Furthermore, write $\bot$ and $\top$ for the constant mappings with these values, viz:
:$\bot: X \to \mathbf 2, \bot (x) := \bot$
:$\top: X \to \mathbf 2, \top (x) := \top$
Then $\left({\mathbf 2^X, \vee, \wedge, \neg}\right)$ is a Boolean algebra, with $\bot$ and $\top$ as identities for $\vee$ and $\wedge$, respectively.
== Proof ==
Let us verify the axioms for a Boolean algebra in turn.
=== $(BA \ 0)$: Closure ===
Follows from Induced Operations Preserve Closure.
=== $(BA \ 1)$: Commutativity ===
Follows from Induced Operations Preserve Commutativity.
=== $(BA \ 2)$: Distributivity ===
Follows from Induced Operations Preserve Distributivity.
=== $(BA \ 3)$: Identities ===
Follows from Identity for Induced Operation.
=== $(BA \ 4)$: Complements ===
Follows from Induced Operations Preserve Identities.
Having verified all the axioms, we conclude $\mathbf 2^X$ is a Boolean algebra.
== Sources ==
* : $\S 2$: Exercise $4$
Category:Boolean Algebras
Two-Valued Functions form Boolean Ring

== Theorem ==
Let $S$ be a set, and let $2$ be the two ring.
Let $2^S$ be the set of all $2$-valued functions on $S$.
Denote with $+$ and $\cdot$ the pointwise operations induced on $2^S$ by $+_2$ and $\times_2$, respectively.
Then $\struct {2^S, +, \cdot}$ is a Boolean ring.
== Proof ==
By Structure Induced by Ring Operations is Ring, $\struct {2^S, +, \cdot}$ is a ring.
By Unity of Induced Structure, $\struct {2^S, +, \cdot}$ also has a unity.
By Induced Structure is Idempotent, $\cdot$ is an idempotent operation.
Hence $\struct {2^S, +, \cdot}$ is a Boolean ring.
== Sources ==
* : $\S 1$: Exercise $3$
Category:Boolean Rings
Two Angles making Two Right Angles make Straight Line

== Theorem ==
== Proof ==
Let $AB$ be the given straight line and $B$ the given point on it.
Let $BC$ and $BD$ be the two straight lines not lying on the same side of it such that $\angle ABC + \angle ABD$ equal two right angles.
Suppose $BD$ is not in a straight line with $BC$.
Now some line must be, so let $BE$ be in a straight line with $BC$ instead.
Then since $AB$ stands on the (supposed) straight line $CBE$, $\angle ABC + \angle ABE$ equal two right angles.
But we already know that $\angle ABC + \angle ABD$ equal two right angles.
Therefore, by Common Notion 1 and the fact that all right angles are congruent, $\angle ABC + \angle ABD = \angle ABC + \angle ABE$.
So let $\angle ABC$ be subtracted from each.
Then by Common Notion 3, the remaining angle $\angle ABE$ equals remaining angle $\angle ABD$.
But $\angle ABD$ is greater than $\angle ABE$, so this is impossible.
So $BE$ can not be in a straight line with $BC$.
Similarly we can show that any straight line which is not $BD$ can not be in a straight line with $BC$.
Therefore $BC$ is in a straight line with $BD$, hence the result.
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.2$
Category:Angles
Two Angles of Triangle are Less than Two Right Angles

== Theorem ==
== Proof ==
:
Let $\triangle ABC$ be a triangle.
Let the side $BC$ be extended to $D$.
Since the angle $\angle ACD$ is an external angle of $\triangle ABC$, it follows that it is greater than both $\angle BAC$ and $\angle ABC$.
We add $\angle ACB$ to both, so that $\angle ACD + \angle ACB$ is greater than $\angle ABC + \angle ACB$.
But $\angle ACD + \angle ACB$ is equal to two right angles.
Therefore $\angle ABC + \angle ACB$ is less than two right angles.
In a similar manner we show that the same applies to the other two pairs of internal angles of $\triangle ABC$.
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.8$: Corollary $3$
Category:Triangles
Two Angles on Straight Line make Two Right Angles

== Theorem ==
== Proof ==
:
Let the line $AB$ be set up on the line $CD$.
If $\angle ABC = \angle ABD$ then from they are two right angles.
If not, we draw $BE$ at right angles to $CD$.
Then:
:$\angle CBE = \angle EBD$
which from are both right angles. 
Since $\angle CBE = \angle ABC + \angle ABE$, we add $\angle EBD$ to each.
Therefore by :
:$\angle CBE + \angle EBD = \angle ABC + \angle ABE + \angle EBD$
Therefore:
:$\angle ABC + \angle ABD = \angle ABC + \angle ABE + \angle EBD$
Again, since $\angle ABD = \angle EBD + \angle ABE$, we add $\angle ABC$ to each.
Therefore by :
:$\angle ABD + \angle ABC = \angle ABC + \angle ABE + \angle EBD$
But $\angle CBE + \angle EBD$ equals the same three angles.
Therefore by :
:$\angle CBE + \angle EBD = \angle ABC + \angle ABD$
But $\angle CBE$ and $\angle EBD$ are both right angles.
Hence the result.
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.1$
Category:Angles
Two Circles have at most Two Points of Intersection

== Theorem ==
== Proof ==
Suppose the opposite, and let circle $ABC$ cut the circle $DEF$ at more points than two, that is at $B, G, F, H$.
:
Clearly, in the diagram, circle $DEF$ is not actually a circle.
 it is.
Join $BH$ and $BG$, and bisect them at $K$ and $L$.
From $K$ and $L$, draw perpendiculars to $BH$ and $BG$, through to $A$ and $E$.
Since $AC$ cuts $BH$ into two equal parts at right angles, from the porism to Finding Center of Circle, the center of circle $ABC$ is on $AC$.
For the same reason, the center of circle $ABC$ is also on $NO$.
But as the only point on both $AC$ and $NO$ is $P$, it follows that $P$ is the center of circle $ABC$.
Similarly, we can show that $P$ is also the center of circle $DEF$.
So we have two circles that intersect which have the same center.
From Intersecting Circles have Different Centers, this statement is contradictory.
Hence such a pair of circles, that is, those that intersect at more points than two, can not exist.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Two Consecutive Integers each Product of Four Distinct Primes

== Theorem ==
The sequence of pairs of consecutive positive integers which are each the product of exactly $4$ distinct prime numbers begins:
:$\tuple {7314, 7315}, \tuple {8294, 8295}, \tuple {8645, 8646}, \tuple {9009, 9010}, \ldots$
== Proof ==
== Sources ==
* : $7314$
Category:Recreational Mathematics
Two Coprime Integers have no Third Integer Proportional

== Theorem ==
Let $a, b \in \Z_{>0}$ be integers such that $a$ and $b$ are coprime.
Then there is no integer $c \in \Z$ such that:
:$\dfrac a b = \dfrac b c$
== Proof ==
Suppose such a $c$ exists.
From Coprime Numbers form Fraction in Lowest Terms, $\dfrac a b$ is in canonical form.
From Ratios of Fractions in Lowest Terms:
:$a \divides b$
where $\divides$ denotes divisibility.
This contradicts the fact that $a$ and $b$ are coprime.
Hence such a $c$ cannot exist.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Coprime Integers
Category:Ratios
Two Ears Theorem

== Theorem ==
Let $P$ be a polygon that is not a triangle.
Then $P$ has at least two ears that do not overlap each other.
== Proof ==
Let $P$ have $n$ sides, where $n \in \N_{ \ge 3 }$.
Polygon Triangulation Theorem shows that there exists a triangulation $\family { \triangle_i }_{i \mathop = 1}^{n-2}$ of $P$, and the sides of each triangle $\triangle_i$ are either sides of $P$ or chords of $P$.
Suppose $\triangle_i$ has $0$ or $1$ sides in common with $P$.
In this case, $\triangle_i$ cannot be an ear of $P$.
Suppose $\triangle_i$ has $2$ sides in common with $P$.
In this case, $\triangle_i$ is an ear of $P$.
Suppose $\triangle_i$ has $3$ sides in common with $P$.
In this case, we have $\triangle_i = P$, which would contradict our assumption that $P$ is not a triangle.
Now, suppose that at most one of the triangles $\family { \triangle_i }_{i \mathop = 1}^{n-2}$ is an ear.
The maximal number of sides that the $n-2$ triangles could have in common with $P$ is:
:$2 + 1 \paren { n-2-1} = n-1$
As the triangulation covers all $n$ sides of $P$, this is a contradiction.
It follows that at least two of the triangles $\family { \triangle_i }_{i \mathop = 1}^{n-2}$ are ears.
As the two ears are part of a triangulation, they do not overlap.
== Sources ==
* : $\S 1.3.1$
Category:Triangles
Category:Polygons
Two Equal Straight Lines can be Constructed from Point to Straight Line

== Theorem ==
Let $AB$ be a straight line.
Let $C$ be a point which is not on $AB$.
Then exactly $2$ straight lines $CD$ and $CE$ can be drawn such that $CD = CE$ and $D, E$ on $AB$.
== Proof ==
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.21$: Corollary
Category:Lines
Category:Angles
Two Fathers and Two Sons Leaving Town

== Classic Problem ==
Two fathers and two sons leave town.
This reduces the population of the town by three.
== Solution ==
This is an instance of the riddle Two Fathers and Two Sons who are Three People.
The $3$ people involved are grandfather, father and son.
== Historical Note ==
== Sources ==
* : Chapter One: What is a Paradox?
Category:Two Fathers and Two Sons who are Three People
Two Fathers and Two Sons who are Three People

== Veridical Paradox ==
In some given place there are $2$ fathers, each accompanied by his son.
So there are $2$ fathers and $2$ sons.
Yet there are only $3$ people.
== Resolution ==
This is a veridical paradox.
This puzzle can have a solution only if one of the fathers and one of the sons is the same person.
So:
:there is a father;
:there is his son, who is also a father;
:there is his son.
The $3$ people are therefore: a grandfather, a father and a son.
== Variants ==
This problem is often seen dressed up with further details as a riddle:
=== Three Pheasants ===
=== Two Fathers and Two Sons Leaving Town ===
Category:Two Fathers and Two Sons who are Three People
Category:Classic Riddles
Category:Veridical Paradoxes
Two Fifths as Pandigital Fraction

== Theorem ==
There are $3$ ways $\dfrac 2 5$ can be expressed as a pandigital fraction:
:$\dfrac 2 5 = \dfrac {6894} {17235}$
:$\dfrac 2 5 = \dfrac {8694} {21735}$
:$\dfrac 2 5 = \dfrac {9486} {23715}$
== Proof ==
Can be verified by brute force.
Category:Pandigital Fractions