nyuuzyou/wikis huggingface dataset:
Axiom of Subsets Equivalents

== Theorem ==
The Axiom of Specification states that:
:$\forall z: \forall \map P y: \exists x: \forall y: \paren {y \in x \iff \paren {y \in z \land \map P y} }$
We will prove that this statement is equivalent to the following statements:
:$\forall z: \forall A: \paren {\paren {z \cap A} \in U}$
:$\forall z: \forall A: \paren {A \subseteq z \implies A \in U}$
In the above statements, the universe is $U$.
== Proof of the First Statement ==
The Axiom of Specification states:
:$\forall z: \forall \map P y: \exists x: \forall y: \paren {y \in x \iff \paren {y \in z \land \map P y} }$
$y \in A$ is substituted for the propositional function $\map P y$.
This leads to the statement:
:$\forall z: \forall A: \exists x: \forall y: \paren {y \in x \iff \paren {y \in z \land y \in A} }$
By definition of intersection:
:$\forall z: \forall A: \exists x: \forall y: \paren {y \in x \iff y \in \paren {z \cap A} }$
By definition of class equality:
:$\forall z: \forall A: \exists x = \paren {z \cap A}$
This is equivalent to:
:$\forall z: \forall A: \paren {z \cap A} \in U$
because $A \in U \iff \exists x = A$.
== Re-derivation of the Axiom of Specification ==
Only bi-conditional ($\iff$) statements were used to prove the first result, so it is possible to reverse the step order and arrive at the original Axiom of Specification by Biconditional is Commutative.
Although this statement is shorter, it uses defined terms, and is thus unsuitable as an axiom.
== Proof of the Second Statement ==
We will take the result of the first statement:
:$\forall z: \forall A: \paren {\paren {z \cap A} \in U}$
We will now take the definition of subset:
:$A \subseteq B \iff \forall x: \paren {x \in A \implies x \in B}$
From Intersection with Subset is Subset:
:$A \subseteq B \iff \paren {A \cap B} = A$
Thus:
:$A \subseteq B \implies \paren {\paren {A \cap B} \in U \implies A \in U}$
We will take the result of the first statement:
:$\forall z: \forall A: \paren {\paren {z \cap A} \in U}$
Using the above two statements, substituting $z$ for $B$:
:$\forall z: \forall A: \paren {A \subseteq z \implies A \in U}$
== Re-derivation of the Axiom of Specification ==
Because $\paren {A \cap z} \subseteq z$, the antecedent of $\forall z: \forall A: \paren {A \subseteq z \implies A \in U}$ is satisfied.
We now arrive at the first statement (above), which in turn can prove the Axiom of Specification:
:$\forall z: \forall A: \paren {A \cap z} \in U$
== Sources ==
* : $\S 5.12$, $\S 5.13$
Category:Subsets
Category:Class Theory
Axiom of Swelledness is implied by Axiom of Replacement

== Theorem ==
Let the (in the context of class theory) be accepted.
Then the holds.
== Proof ==
Recall the Axiom of Replacement:
Recall the :
That is:
:Every subclass of a set is a set.
Let $x$ be a set.
Let $A$ be a class such that $A \subseteq x$.
Suppose $A$ is the empty class.
Then by the , $A$ is a set.
Suppose $A$ is a non-empty class.
Let $c \in A$ be arbitrary.
Let $f$ be:
:the class of all ordered pairs $\tuple {a, a}$ for $a \in A$
along with:
:all ordered pairs $\tuple {y, c}$ where $y \in x \setminus A$.
Thus:
:$f$ is a mapping whose domain is $x$
and:
:$\forall y \in x: \map f y = \begin {cases} y & : y \in A \\ c & : y \notin A \end {cases}$
Thus:
:$\forall y \in x: \map f y \in A$
and in fact:
:$f \sqbrk x = A$
We have that $x$ is a set.
Then by the :
:$A$ is therefore also a set.
Hence the result.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 3$ The axiom of substitution
Category:Axiom of Swelledness
Category:Axiom of Replacement
Axiomatize all of Physics

== Hilbert $23$: Problem $6$ ==
Mathematical treatment of the axioms of physics:
:$(1): \quad$ An axiomatic treatment of probability with limit theorems for foundation of statistical physics
:$(2): \quad$ The rigorous theory of limiting processes "which lead from the atomistic view to the laws of motion of continua".
== Progress ==
This has been partially resolved.
This resolution depends on how the original statement is interpreted.
Items $(1)$ and $(2)$ were two specific problems given by in a later explanation.
The Kolmogorov's axioms are now accepted as a standard way of specifying probability.
There is also some success on the way from the "atomistic view to the laws of motion of continua."
Category:Physics
Category:Hilbert 23
B-Algebra/Examples/S3

== Example of $B$-Algebras==
This is a $B$-algebra for the finite set $\set {0, 1, 2, 3, 4, 5}$:
 
$\quad \begin{array}{c|cccccc} & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
0 & 0 & 2 & 1 & 3 & 4 & 5 \\
1 & 1 & 0 & 2 & 4 & 5 & 3 \\
2 & 2 & 1 & 0 & 5 & 3 & 4 \\
3 & 3 & 4 & 5 & 0 & 2 & 1 \\
4 & 4 & 5 & 3 & 1 & 0 & 2 \\
5 & 5 & 3 & 4 & 2 & 1 & 0 \\
\end{array}$
== Proof ==
== Sources ==
* 
Category:B-Algebras
B-Algebra Identity: x (y z) = (x (0 z)) y

== Theorem ==
Let $\struct {X, \circ}$ be a $B$-algebra.
Then:
:$\forall x, y, z \in X: x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$
== Proof ==
Let $x, y, z \in X$.
Then:
Hence the result.
== Also see ==
This identity is comparable to .
Category:B-Algebras
B-Algebra Identity: xy=x(0(0y))

== Theorem ==
Let $\struct {X, \circ}$ be a $B$-algebra.
Then:
:$\forall x, y \in X: x \circ y = x \circ \paren {0 \circ \paren {0 \circ y} }$
== Proof ==
Let $x, y \in X$.
Then:
Hence the result.
Category:B-Algebras
B-Algebra Identity: xy = 0 iff x = y

== Theorem ==
Let $\struct {X, \circ}$ be a $B$-algebra.
Then:
:$\forall x, y \in X: x \circ y = 0 \iff x = y$
== Proof ==
=== Sufficient Condition ===
Suppose that $x = y$.
Then by :
:$x \circ y = x \circ x = 0$
=== Necessary Condition ===
Let $x, y \in X$ such that $x \circ y = 0$.
Then:
Hence the result.
Category:B-Algebras
B-Algebra Induced by Group Induced by B-Algebra

== Theorem ==
Let $\struct {S, *}$ be a $B$-algebra.
Let $\struct {S, \circ}$ be the group described on $B$-Algebra Induces Group.
Let $\struct {S, *'}$ be the $B$-algebra described on Group Induces $B$-Algebra.
Then $\struct {S, *'} = \struct {S, *}$.
== Proof ==
Let $a, b \in S$.
It is required to show that:
:$a *' b = a * b$
To achieve this, recall that $*'$ is defined on Group Induces $B$-Algebra to satisfy:
:$a *' b = a \circ b^{-1}$
which, by the definition of $\circ$ on $B$-Algebra Induces Group comes down to:
Hence the result.
== Also see ==
* Group Induced by $B$-Algebra Induced by Group
Category:Group Theory
Category:B-Algebras
B-Algebra Induces Group

== Theorem ==
Let $\struct {X, \circ}$ be a $B$-algebra.
Let $*$ be the binary operation on $X$ defined as:
:$\forall a, b \in X: a * b := a \circ \paren {0 \circ b}$
Then the algebraic structure $\struct {X, *}$ is a group such that:
:$\forall x \in X: 0 \circ x$ is the inverse element of $x$ under $*$.
That is:
:$\forall a, b \in X: a * b^{-1} := a \circ b$
== Proof ==
Let $x, y, z \in X$:
We will show that $\struct {X, *}$ satisfies each of the group axioms in turn:
=== ===
By definition of $*$, we have:
:$x * y = x \circ \paren {0 \circ y}$
By Axiom $(AC)$ for $B$-algebras:
:$x \circ \paren {0 \circ y} \in X$
Whence $x * y \in X$, and so $\paren {X, *}$ is a closed structure.
=== ===
Thus it is seen that $*$ is associative.
=== ===
Let $e := 0$; we will show that it is an identity element of $\struct {X, *}$.
Hence $0$ is an identity for $*$.
=== ===
Let us prove that for all $x \in X$, $0 \circ x$ is an inverse element to $x$.
That is, each $x \in X$ has a unique inverse element $x^{-1}$ under $*$.
This inverse element is $0 \circ x$.
It follows that:
All the axioms have been shown to hold and the result follows.
== Also see ==
* Group Induces $B$-Algebra
Category:B-Algebras
Category:Group Theory
B-Algebra Power Law

== Theorem ==
Let $\struct {X, \circ}$ be a $B$-algebra.
Let $n, m \in \N$ such that $n \ge m$.
Then:
:$\forall x \in X: x^n \circ x^m = x^{n - m}$
where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$.
=== $B$-Algebra Power Law with Zero ===
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall m \in \N_{> 0}, m \le n: \forall x \in X: x^n \circ x^m = x^{n - m}$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
: $x \circ x = 0$
which follows from the definition of the zeroth power in $B$-algebra.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show 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 our induction hypothesis:
:$\forall m \in \N_{> 0}, m \le k: \forall x \in X: x^k \circ x^m = x^{k - m}$
Then we need to show:
:$\forall m \in \N_{> 0}, m \le k + 1: \forall x \in X: x^{k + 1} \circ x^m = x^{k + 1 - m}$
=== Induction Step ===
This is our induction step:
First we show that:
:$\forall x \in X: x^{k + 1} \circ x = x^k$
Thus:
By induction, it follows that:
:$\forall n \in \N_{>0}: \forall x \in X: x^n \circ x = x^{n - 1}$
Now let $1 \le m \le k$.
We have:
By induction, it follows that:
:$\forall x \in X: \forall n, m \in \N: n \ge m \implies x^n \circ x^m = x^{n - m}$
Category:B-Algebras
B-Algebra Power Law with Zero

== Theorem ==
Let $\struct {X, \circ}$ be a $B$-algebra.
Let $n, m \in \N_{>0}$ such that $n > m$.
Then:
:$\forall x \in X: n, m \in \N_{>0} \implies x^m \circ x^n = 0 \circ x^{n - m}$
where $x^k$ for $k \in \N_{>0}$ denotes the $k$th power of the element $x$.
== Proof ==
First we show that:
:$\forall x \in X: x \circ x^2 = 0 \circ x$
Now we show that:
:$\forall x \in X: m \in \N_{>0}: x \circ x^m = 0 \circ x^{m - 1}$
Now we prove the original proposition using a proof by induction.
The base case for $n = 1, m = 2$ was established in the first lemma. 
Let us assume $x^m \circ x^n = 0 \circ x^{n - m}$ for some $n, m \in \N_{>0}$.
Then: 
Hence the result.
Category:B-Algebras
B-Algebra is Commutative iff x(xy)=y

== Theorem ==
Let $\struct {X, \circ}$ be a $B$-algebra.
Then $\struct {X, \circ}$ is commutative :
:$\forall x, y \in X: x \circ \paren {x \circ y} = y$
== Proof ==
=== Necessary Condition ===
Let $x, y \in X$:
=== Sufficient Condition ===
Let $x, y \in X$:
Hence:
:$\paren {x \circ \paren {0 \circ y} } \circ y = \paren {y \circ \paren {0 \circ x} } \circ y$
From $B$-Algebra is Right Cancellable, we have:
:$x \circ \paren {0 \circ y} = y \circ \paren {0 \circ x}$
and hence commutativity.
Category: B-Algebras
B-Algebra is Quasigroup

== Theorem ==
Let $\struct {X, \circ}$ be a $B$-algebra.
Then $\struct {X, \circ}$ is a quasigroup.
== Proof ==
By the definition of a quasigroup it must be shown that $\forall x \in X$ the left and right regular representations $\lambda_x$ and $\rho_x$ are permutations on $X$.
As $\struct {X, \circ}$ is a magma:
:$\forall x \in X$ the codomain of $\lambda_x$ and $\rho_x$ is $X$.
Hence it is sufficient to prove that $\lambda_x$ and $\rho_x$ are bijections.
We have that:
:$B$-Algebras are left cancellable
:$B$-Algebras are right cancellable
:Cancellable elements have injective regular representations.
Therefore:
:$\forall x \in X$: $\lambda_x$ and $\rho_x$ are injective mappings.
We also have that regular representations of $B$-algebras are surjective.
Therefore: 
:$\forall x \in X$: $\lambda_x$ and $\rho_x$ are both injective and surjective mappings.
Hence by definition are $\lambda_x$ and $\rho_x$ bijections for all $x \in X$.
== Also see ==
* Quasigroup is not necessarily B-Algebra where it is demonstrated that the converse does not necessarily hold.
Category:B-Algebras
Category:Examples of Quasigroups
B-Algebra is Right Cancellable

== Theorem ==
Let $\struct {X, \circ}$ be a $B$-algebra.
Then $\circ$ is right-cancellable for $X$. That is:
:$\forall x, y, z \in X: x \circ z = y \circ z \implies x = y$
== Proof ==
Let $x, y \in X$.
Then:
Now if for some $z \in X$ we have $x \circ z = y \circ z$, then also:
:$\paren {x \circ z} \circ \paren {0 \circ z} = \paren {y \circ z} \circ \paren {0 \circ z}$ 
which by the above implies $x = y$.
Hence the result.
Category:B-Algebras
Babbage's Congruence

== Theorem ==
Let $p$ be a prime number.
Let $a, b \in \Z$ be integers.
Then:
:$\dbinom {a p} {b p} \equiv \dbinom a b \pmod {p^2}$
where $\dbinom a b$ denotes a binomial coefficient.
== Proof ==
By Lucas' Theorem, for $a, b \in \mathbb Z_{\ge 0}$:
:$\dbinom {a p} {b p} \equiv \dbinom {\floor {a p / p}} {\floor {b p / p}} \dbinom {a \mod p} {b \mod p} \pmod p$
Proof by Mathematical Induction on $a$:
For all $a \in \Z$, let $\map P a$ be the proposition:
:$\forall b \in \Z: \dbinom {a p} {b p} \equiv \dbinom a b \pmod {p^2}$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\forall b \in \Z: \dbinom 0 {b p} \equiv \dbinom 0 b \pmod {p^2}$
For $b = 0$, we have $\dbinom 0 0 \equiv \dbinom 0 0 \pmod {p^2}$.
For $b \ne 0$, by Zero Choose n:
:$\dbinom 0 {b p} = 0 = \dbinom 0 b \equiv \dbinom 0 b \pmod {p^2}$
Hence $\map P 0$ is true.
$\map P {-1}$ is the case:
:$\forall b \in \Z: \dbinom {-p} {b p} \equiv \dbinom {-1} b \pmod {p^2}$
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, then it logically follows that $\map P {k + 1}$ and $\map P {k - 1}$ are true.
So this is our induction hypothesis:
:$\forall b \in \Z: \dbinom {k p} {b p} \equiv \dbinom k b \pmod {p^2}$
Then we need to show:
:$\forall b \in \Z: \dbinom {\paren {k + 1} p} {b p} \equiv \dbinom {\paren {k + 1} } b \pmod {p^2}$
:$\forall b \in \Z: \dbinom {\paren {k - 1} p} {b p} \equiv \dbinom {\paren {k - 1} } b \pmod {p^2}$
=== Induction Step ===
This is our induction step:
Let $r \in \Z$ and $0 < r < p$.
By Binomial Coefficient of Prime:
:$\dbinom p r \equiv 0 \pmod p$
By Lucas' Theorem and Zero Choose n:
:$\dbinom {k p} {b p - r} \equiv \dbinom {\floor {k p / p}} {\floor {\paren {b p - r} / p}} \dbinom {k p \bmod p} {b p - r \bmod p} \equiv \dbinom k {b - 1} \dbinom 0 {p - r} \equiv 0 \pmod p$
Therefore:
:$(1): \dbinom {k p} {b p - r} \dbinom p r \equiv 0 \pmod {p^2}$
On the other hand, by Lucas' Theorem and Negated Upper Index of Binomial Coefficient:
:$\dbinom {-p} r = \paren {-1}^r \dbinom {p + r - 1} r \equiv \paren {-1}^r \dbinom {\floor {\paren {p + r - 1} / p}} {\floor {r / p}} \dbinom {p + r - 1 \bmod p} {r \bmod p} \pmod p$
In particular, $\dbinom {p + r - 1 \bmod p} {r \bmod p} = \dbinom {r - 1 \bmod p} {r \bmod p}$.
Note that $r - 1 \bmod p < r \bmod p$ except when $r \bmod p = 0$.
Hence $\dbinom {-p} r \equiv 0 \pmod p$ except when $p \divides r$.
Therefore, for $p \nmid r$:
:$(2): \dbinom {k p} {b p - r} \dbinom {-p} r \equiv 0 \pmod {p^2}$
Now, for $b \ge 0$:
So $\map P k \implies \map P {k + 1} \land \map P {k - 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall a, b \in \Z: \dbinom {a p} {b p} \equiv \dbinom a b \pmod {p^2}$
== Also see ==
* Wolstenholme's Theorem
== Sources ==
* 
Category:Binomial Coefficients
Babczyński Theorem

== Theorem ==
Let $n$ be a $6$-digit integer of the form $\sqbrk {xyxyxy}$ for digits $x$ and $y$.
Then $n$ is divisible by $3$, $7$, $13$ and $37$.
== Proof ==
== Examples ==
== Source of Name ==
Despite the coincidence in name, the Babczyński Theorem is not named for , but for another Babczyński altogether.
Category:Babczyński Theorem
Category:3
Category:7
Category:13
Category:37
Category:Divisibility
Babylonian Mathematics/Examples/Division of Triangular Field

== Examples of Babylonian Mathematics ==
A triangular field is to be divided between $6$ brothers by equidistant lines parallel to one of the sides.
Expressed in Babylonian notation:
:the length of the marked side is $6; 30$
:the area of the triangle is $11, 22; 30$.
What is the difference between the brothers' shares?
== Solution ==
The difference between each successive share is:
:$37; 55$ in Babylonian notation
:$37 \frac {11} {12}$ in mixed fractions.
== Proof ==
:
Let $\triangle ABC$ be the triangular field in question.
Let $d$ be the marked side.
Let $a$ be the side which is parallel to the dividing lines.
Let $\AA$ be the total area of $ABC$.
Let $\AA_1, \AA_2, \ldots, \AA_6$ be the areas of each of the divisions of $ABC$ such that $\AA_1 > \AA_2 > \cdots > \AA_6$.
Let $a_1, a_2, \ldots, a_5$ denote the dividing lines such that $a_1 > a_2 > \cdots > a_5$.
From Area of Triangle, we have that:
:$\AA = \dfrac 1 2 k d a$
where $k = \sin \angle CAB$.
Then:
We have:
and so:
We have that:
Note that it is not necessary to know how long the marked side is.
== Sources ==
* 
* : Dividing a Field: $12$
Category:Examples of Babylonian Mathematics
Babylonian Mathematics/Examples/Division of Triangular Field/Mistake

== Source Work ==
: 
:The Puzzles:
::Dividing a Field: $12$
== Mistake ==
:A triangular field is to be divided between six brothers by equidistant lines parallel to one side. The length of the marked side is $6, 30$ and the area is $11, 22, 30$. What is the difference between the brothers' shares?
:Answer: The difference between each successive share is $37, 55$ or $37 \frac {11} {12}$.
== Correction ==
The author is lax here in his interpretation of where the radix point lies.
In order for the question both to make sense and to be consistent, it is necessary to interpret the numbers as:
:the area is $11, 22; 30$
and:
:the difference between each successive share is $37; 55$
where $;$ denotes the radix point.
Note that this is at odds with the interpretation as given in the explanatory section leading up to this question, where $11, 22, 30$ is interpreted as $11; 22, 30$, that is, $11 + \dfrac {22} {60} + \dfrac {30} {60 \times 60}$.
== Sources ==
* : Dividing a Field: $12$
Category:Mistakes/Curious and Interesting Puzzles
Babylonian Mathematics/Examples/Pythagorean Triangle whose Side Ratio is 1.54

== Example of Babylonian Mathematics ==
Consider a Pythagorean triangle whose hypotenuse and one leg are in the ratio $1.54 : 1$.
What are the lengths of that hypotenuse and that leg?
== Solution ==
The lengths in question are $829$ and $540$.
== Proof ==
Let $a$, $b$ and $c$ be positive integers such that $a^2 + b^2 = c^2$ and such that $1.54 \times a = c$.
, suppose $a$ is even.
From Solutions of Pythagorean Equation, there exist positive integers $p$ and $q$ such that:
Hence it follows that:
:$\dfrac c a = \dfrac 1 2 \paren {\dfrac p q + \dfrac q p}$
The Babylonians would then consult the various standard tables of reciprocals which they used for multiplication.
Without these tables, we set:
:$\dfrac p q = t$
and solve the quadratic equation:
We can discard $0.369$ because we are after $p > q$.
Hence:
which is what is found in the original Babylonian clay tablet.
== Historical Note ==
== Sources ==
* : pp. $38 - 41$
* : p. $37$
* : Pythagorean Triples: $15$
Category:Examples of Babylonian Mathematics
Babylonian Mathematics/Examples/Sextic Equation

== Example of Babylonian Mathematics ==
Simpify the system of simultaneous equations:
== Solution ==
:$x^3 = \dfrac a {2 b} \paren {-d \pm \sqrt {d^2 - 4 a b c} }$
== Proof ==
== Historical Note ==
== Sources ==
* : Dividing a Field
Category:Examples of Babylonian Mathematics
Babylonian Mathematics/Examples/Sliding Ladder

== Example of Babylonian Mathematics ==
A ladder of length $0; 30$ stands upright against a wall.
The upper end slides down a distance $0; 6$.
How far away will the lower end move out from the wall?
All lengths are expressed in Babylonian form.
== Solution ==
The lower end of the ladder will move $0; 18$ away from the wall.
== Proof ==
The ladder will be the hypotenuse of a right triangle whose legs are formed by the wall and the floor.
Let $x$ be the distance of the lower end from the wall after the ladder has finished moving.
We have:
== Sources ==
* 
* : Pythagorean Triples: $14$
Category:Examples of Babylonian Mathematics
Babylonian Mathematics/Examples/Sum of Squares

== Example of Babylonian Mathematics ==
An area $A$, consisting of the sum of $2$ squares, is $1000$.
The side of one square is $10$ less than $\dfrac 2 3$ of the other square.
What are the sides of the squares?
== Solution ==
The lengths of the sides of the $2$ squares are $10$ and $30$.
== Proof ==
Let $x$ be the length of the side of the smaller square.
Let $y$ be the length of the side of the larger square.
Thus:
We can dispose of the negative value, leaving us with: 
and we see that:
as required.
== Sources ==
* 
* : Dividing a Field: $13$
Category:Examples of Babylonian Mathematics
Babylonian Number System/Examples/12,59;57,17

== Example of Number represented in the Babylonian Number System ==
The number represented in the Babylonian number system as:
:$12,59;57,17$
is represented in decimal as:
:$12 \times 60 + 59 + \dfrac {57} {60} + \dfrac {17} {60 \times 60} \approx 779 \cdotp 955$
== Sources ==
* : Chapter $1$: Tokens, Tallies and Tablets: Symbols for small numbers
Category:Babylonian Number System
Babylonian Number System/Examples/25,267

== Example of Number represented in the Babylonian Number System ==
The number $25 \, 267$ is represented in the Babylonian number system as:
: 
That is:
:$7 \times 60 \times 60 + 7 \times 60 + 7 = 25 \, 200 + 420 + 7 = 25 \, 267$
== Sources ==
* : Chapter $1$: Tokens, Tallies and Tablets: The first numerals
Category:Babylonian Number System
Backward Path of Well-Ordering is Finite

== Theorem ==
Let $A$ be a class under a well-ordering $\preccurlyeq$.
Let $S$ be a backward path of $\preccurlyeq$ in $A$.
Then $S$ is finite.
== Proof ==
First suppose that $S$ has no terms.
Then $S$ is finite by default.
Hence let it be assumed that $S$ has at least $1$ term.
Let $T = \set {x: x \in \sequence {x_n} }$ be the range of $S$.
Then by definition $T$ is a subclass of $A$.
 $S$ is infinite.
By definition of well-ordering, $T$ has a smallest element, which can be referred to as $y$.
Hence:
:$\exists n \in \N: y = x_n$
But then by definition of backward path:
:$x_{n + 1} \prec x_n$
and so $y$ is not the smallest element of $T$ after all.
Hence $T$ can have no such smallest element.
From this contradiction we conclude that $S$ cannot be an infinite sequence.
Hence the result.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Exercise $1.4$
Category:Well-Orderings
Baer's Criterion

== Theorem ==
Let $R$ be a ring with unity.
Let $M$ be a left $R$-module.
Then $M$ is injective the following condition holds:
:For all left ideals $I$ of $R$ with inclusion map $\iota : I \to R$, and for all $R$-module homomorphisms $f : I \to M$, there exists an $R$-module homomorphism $\tilde f : R \to M$ such that:
::$\tilde f \circ \iota = f$
== Proof ==
Category:Homological Algebra
Category:Module Theory
Category:Ring Theory