nyuuzyou/wikis huggingface dataset:
Universal Property of Quotient Ring

== Theorem ==
Let $R, S$ be commutative rings.
Let $I \trianglelefteq R$ be an ideal of $R$.
Let $\pi : R \to R / I$ be the quotient epimorphism.
Let $f: R \to S$ be a ring homomorphism with $\map f I = \set 0$.
Then there exists a unique ring homomorphism $\overline f: R / I \to S$ such that $f = \overline f \circ \pi$.
:$\xymatrix {
R \ar[d]_\pi \ar[r]^{\forall f} & S \\
R/I \ar[ru]_{\exists ! \bar f} }$
== Proof ==
 
Define $\overline f: R / I \to S$ by:
:$\forall r \in R: \map {\overline f} {r + I} = \map f r$
Since $f$ is a ring homomorphism, $f$ is well-defined.
Suppose for some $r_1, r_2 \in R$ that: 
:$r_1 + I = r_2 + I$
Since $I$ is an ideal, it contains the zero $0_R$ of $R$.
Then for some $i \in I$, we have that: 
Then: 
so $\overline f$ is seen to be well-defined.
Next, for all $r_1, r_2 \in R$:
Therefore $\overline f$ is a homomorphism.
It remains to demonstrate uniqueness.
Suppose there were another homomorphism $g: R / I \to S$ with $f = g \circ \pi$.
Then we would require that, for all $r\in R$:
That is:
:$g = \overline f$
and so $\overline f$ is unique.
== Also see ==
* First Ring Isomorphism Theorem
Category:Quotient Rings
Category:Universal Properties
Universal Property of Quotient Set

== Theorem ==
Let $X$ and $Y$ be sets.
Let $\sim$ be an equivalence relation on $X$.
Let $\pi: X \to X / {\sim}$ be the quotient mapping to the quotient set.
Let $f: X \to Y$ be $\sim$-invariant.
Then there exists a unique mapping $\overline f : X / {\sim} \to Y$ such that $f = \overline f \circ \pi$.
:$\xymatrix{
X \ar[d]_\pi \ar[r]^f & Y\\
X/\sim \ar@{.>}[ru]_{\overline f} }$
== Proof ==
=== Existence ===
For $x \in X$, let $\eqclass x {} \in X / {\sim}$ denote its equivalence class under $\sim$.
Define the relation $\overline f: X / {\sim} \to Y$ by:
:$\tuple {\eqclass x {}, y} \in \overline f \iff \map f x = y$
We check that this is a well-defined mapping.
By Quotient Mapping is Surjection, $f$ is left-total.
Let $\eqclass x {} \in X / {\sim}$ and $y, z \in Y$ with:
:$\tuple {\eqclass x {}, y} \in \overline f$
:$\tuple {\eqclass x {}, z} \in \overline f$
We show that $y = z$.
By definition of $\overline f$, there exist $u, v \in X$ with $\eqclass u {} = \eqclass v {} = \eqclass x {}$ and $\map f u = y$ and $\map f v = z$.
By definition of equivalence class, $u \sim v$.
Because $f$ is $\sim$-invariant, $y = z$.
Thus $\overline f$ is many-to-one.
Thus $\overline f$ is a well-defined mapping.
=== Uniqueness ===
Follows from:
* Quotient Mapping is Surjection
* Surjection iff Epimorphism in Category of Sets
Category:Equivalence Relations
Category:Universal Properties
Universal Property of Quotient Space

== Theorem ==
Let $X$ and $Y$ be topological spaces.
Let $\sim$ be an equivalence relation on $X$.
Let $\pi : X \to X / {\sim}$ be the quotient mapping.
Let $f : X \to Y$ be continuous and $\sim$-invariant.
Then there exists a unique continuous map $\overline f : X / {\sim} \to Y$ such that $f = \overline f \circ \pi$.
== Proof ==
=== Existence ===
For $x \in X$, let $\eqclass x {} \in X / \sim$ denote its equivalence class under $\sim$.
Define $\map {\overline f} {\eqclass x {} } = \map f x$; note that this does not depend on the choice of $x$.
To see that $\overline f$ is continuous, simply observe that for any open set $U \subseteq Y$, the preimage $\inv {\overline f} U$ is open in $X$ and is a union of equivalence classes, hence is also open in $X / {\sim}$, as desired.
=== Uniqueness ===
Follows from:
* Quotient Mapping is Surjection
* Surjection iff Epimorphism in Category of Topological Spaces
* Condition for Existence of Epimorphism from Quotient Structure to Epimorphic Image
== Also see ==
* Universal Property of Quotient of Topological Group
* Surjection iff Epimorphism in Category of Sets
Category:Quotient Spaces
Category:Universal Properties
Universal Property of Quotient of Topological Group

== Theorem ==
Let $G$ and $H$ be topological groups.
Let $N$ be a normal subgroup of $G$.
Let $\pi : G \to G/N$ be the quotient mapping.
Let $f : G \to H$ be a continuous group homomorphism whose kernel contains $N$.
Then there exists a unique continuous group homomorphism $\overline f: G / N \to H$ such that:
:$f = \overline f \circ \pi$
== Proof ==
Because $N \subset \ker f$, $f$ is constant on the cosets of $N$.
Thus $f$ is invariant under congruence modulo $N$.
By Universal Property of Quotient Set, there exists a unique mapping $\overline f: G / N \to H$ such that:
:$f = \overline f \circ \pi$
It suffices to verify that it is a continuous group homomorphism.
By Universal Property of Quotient Space, there exists a continuous mapping $\overline g: G / N \to H$ such that:
:$f = \overline g \circ \pi$
By uniqueness of $f$:
:$g = f$
By Universal Property of Quotient Group, there exists a group homomorphism $\overline h: G / N \to H$ such that:
:$f = \overline h \circ \pi$
By uniqueness of $f$:
:$h = f$
Category:Topological Groups
Category:Universal Properties
Universal Quantifier Distributes over Conjunction

== Theorem ==
Let $\map \phi x, \map \psi x$ be WFFs of the free variable $x$.
Then:
:$\forall x: \paren {\map \phi x \land \map \psi x} \dashv \vdash \paren {\forall x: \map \phi x} \land \paren {\forall x: \map \psi x}$
== Proof ==
=== Forward Implication ===
=== Reverse Implication ===
Category:Universal Quantifier
Universal Quantifier Distributes over Disjunction

== Theorem ==
Let $\map \phi x, \map \psi x$ be WFFs of the free variable $x$.
Then:
:$\paren {\forall x: \map \phi x} \lor \paren {\forall x: \map \psi x} \vdash \forall x: \paren {\map \phi x \lor \map \psi x}$
== Proof ==
Category:Universal Quantifier
Universal Statement/Examples/All Men are Mortal

== Examples of Universal Statements ==
The classical example of a universal statement is the statement:
:All men are mortal.
It has the logical form:
:$\forall x: \map {\operatorname {Man} } x \implies \map {\operatorname {Mortal} } x$
where $\operatorname {Man}$ and $\operatorname {Mortal}$ are the predicates is a man and is mortal respectively.
== Sources ==
* 
* 
Category:Examples of Universal Statements
Universal Statement has no Existential Import

== Theorem ==
A universal statement of the form:
:All $A$ are $B$
has no existential import.
== Proof ==
If there exist no $A$, then:
:All $A$ are $B$
is vacuously true, and hence remains true.
If there exist no $B$, then:
:All $A$ are $B$
is vacuously true when there exist no $A$.
Hence the result by definition of existential import.
== Examples ==
== Sources ==
* 
* 
Category:Universal Statement has no Existential Import
Category:Universal Quantifier
Category:Existential Import
Universal URM Computable Functions

== Theorem ==
For each integer $k \ge 1$, there exists a URM computable function:
:$\Phi_k: \N^{k+1} \to \N$
such that for each URM computable function $f: \N^k \to \N$ there exists a natural number $e$ such that:
:$\forall \left({n_1, n_2, \ldots, n_k}\right) \in \N^k: f \left({n_1, n_2, \ldots, n_k}\right) \approx \Phi_k \left({e, n_1, n_2, \ldots, n_k}\right)$.
This function $\Phi_k$ is universal for URM computable functions of $k$ variables.
== Proof ==
Let $\Phi_k: \N^{k+1} \to \N$ be given by:
:$\Phi_k \left({e, n_1, n_2, \ldots, n_k}\right) = U \left({\mu z \ T_k \left({e, n_1, n_2, \ldots, n_k, z}\right)}\right)$
where $T_k$ and $U$ are as in Kleene's Normal Form Theorem.
Thus we have reinterpreted Kleene's Normal Form Theorem as being about URM computable functions.
This is legitimate, as a URM Computable Function is Recursive and vice versa.
== Comment ==
Thus we can obtain all URM computable functions of $k$ variables by letting the value of the first variable of $\Phi_k$ to range through all the natural numbers.
So we can think of a corresponding URM program $P$ which computes $\Phi_k$ as being a universal machine which computes all URM computable functions of $k$ variables.
So we now have a recipe for constructing a suitable URM program $P$ for this universal machine.
This recipe will be fairly complicated.
Category:URM Programs
Universal URM Programs

== Theorem ==
For each integer $k \ge 1$, there exists a URM program $P_k$ such that:
For each URM program $P$ there exists a natural number $e$ such that:
For all $\left({n_1, n_2, \ldots, n_k}\right) \in \N^k$, the computation using the program $P_k$ with input $\left({e, n_1, n_2, \ldots, n_k}\right)$
has the same output as the computation using the program $P$ with input $\left({n_1, n_2, \ldots, n_k}\right)$.
This function $P_k$ is a universal program for URM computations with $k$ inputs.
== Proof ==
This follows directly from:
* Kleene's Normal Form Theorem‎;
* Universal URM Computable Functions‎.
Category:URM Programs
Universal Upper Bound greater than Supremum Operator Norm

== Theorem ==
Let $\map {CL} {X, Y}$ be the continuous linear transformation space.
Let $T \in \map {CL} {X, Y}$.
Let $\norm {\, \cdot \,}$ be an operator norm on $\map {CL} {X, Y}$ defined as:
:$\norm T := \sup \set {\norm {Tx}_Y : x \in X : \norm x_X \le 1}$
Suppose:
:$\exists M \in \R_{> 0} : \forall x \in X : \norm {T x}_Y \le M \norm x_X$
Then:
:$\norm {T} \le M$
== Proof ==
Let $x \in X$ be such that $\norm x_X \le 1$.
Then:
Hence, $M$ is an upper bound for $S = \set {\norm {T x}_Y : x \in X : \norm x_X \le 1}$.
By definition of the supremum of subset of real numbers:
:$\sup S \le M$.
That is:
:$\norm T \le M$.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$
Category:Linear Transformations
Category:Continuous Linear Transformations
Category:Continuous Mappings
Category:Supremum Norm
Unlike Electric Charges Attract

== Theorem ==
Let $a$ and $b$ be stationary particles, each carrying an electric charge of $q_a$ and $q_b$ respectively.
Let $q_a$ and $q_b$ be of the opposite polarity.
That is, let $q_a$ and $q_b$ be unlike charges.
Then the forces exerted by $a$ on $b$, and by $b$ on $a$, are such as to cause $a$ and $b$ to attract each other.
== Proof ==
By Coulomb's Law of Electrostatics:
:$\mathbf F_{a b} \propto \dfrac {q_a q_b {\mathbf r_{a b} } } {r^3}$
where:
:$\mathbf F_{a b}$ is the force exerted on $b$ by the electric charge on $a$
:$\mathbf r_{a b}$ is the displacement vector from $a$ to $b$
:$r$ is the distance between $a$ and $b$.
, let $q_a$ be positive and $q_b$ be negative.
Then $q_a q_b$ is a positive number multiplied by a negative number. 
Thus $q_a q_b$ is a negative number.
Hence $\mathbf F_{a b}$ is in the opposite direction to the displacement vector from $a$ to $b$.
That is, the force exerted on $b$ by the electric charge on $a$ is in the direction towards $a$.
The same applies to the force exerted on $a$ by the electric charge on $b$.
That is, the force exerted on $b$ by the electric charge on $a$ is in the direction towards $b$.
The effect of these forces is to cause $a$ and $b$ to pull together, that is, to attract each other.
== Also see ==
* Like Electric Charges Repel
== Sources ==
* : Chapter $\text {I}$: Preliminary Survey: $\S 1$. Electrostatics
* : Chapter $1$: Force and energy in electrostatics: $1.1$ Electric Charge
Category:Electric Charge
Unordered Pair is Finite

== Theorem ==
Let $x, y$ be arbitrary.
Then $\set {x, y}$ is finite.
== Proof ==
By Union of Disjoint Singletons is Doubleton:
:$\set {x, y} = \set x \cup \set y$
By Singleton is Finite:
:$\set x$ and $\set y$ are finite.
Thus by Union of Finite Sets is Finite:
:$\set {x, y}$ is finite.
== Sources ==
* 
Category:Doubletons
Category:Finite Sets
Unordered Pairs Exist

== Theorem ==
Let $A$ and $B$ be classes.
Then:
:$\forall A, B: \set {A, B} \in U$
where $U$ is the universal class.
== Proof ==
== Also see ==
* Axiom:Axiom of Pairing (Class Theory)
* Definition:Universal Class
* Definition:Doubleton
== Sources ==
* : $\S 7.10$
Category:Class Theory
Category:Doubletons
Unsatisfiable Set Union Formula is Unsatisfiable

== Theorem ==
Let $\LL$ be a logical language.
Let $\mathscr M$ be a formal semantics for $\LL$.
Let $\FF$ be an $\mathscr M$-unsatisfiable set of formulas from $\LL$.
Let $\phi$ be a logical formula.
Then $\FF \cup \set \phi$ is also $\mathscr M$-unsatisfiable.
== Proof ==
This is an immediate consequence of Superset of Unsatisfiable Set is Unsatisfiable.
== Also see ==
* Superset of Unsatisfiable Set is Unsatisfiable
== Sources ==
* : $\S 2.5.2$: Theorem $2.46$
* : $\S 2.10$: Exercise $2.15$
Category:Formal Semantics
Unsatisfiable Set minus Tautology is Unsatisfiable

== Theorem ==
Let $\LL$ be a logical language.
Let $\mathscr M$ be a formal semantics for $\LL$.
Let $\FF$ be an $\mathscr M$-unsatisfiable set of formulas from $\LL$.
Let $\phi \in \FF$ be a tautology.
Then $\FF \setminus \set {\phi}$ is also $\mathscr M$-unsatisfiable.
== Proof ==
Suppose $\FF \setminus \set {\phi}$ were satisfiable.
Then by Satisfiable Set Union Tautology is Satisfiable, so would $\FF$ be, because:
:$\FF = \paren {\FF \setminus \set {\phi} } \cup \set {\phi}$
by Set Difference Union Intersection and Intersection with Subset is Subset.
Therefore, $\FF \setminus \set {\phi}$ must be unsatisfiable.
== Sources ==
* : $\S 2.5.2$: Theorem $2.47$
* : $\S 2.10$: Exercise $2.15$
Category:Formal Semantics
Unsigned Stirling Number of the First Kind of 0

== Theorem ==
:$\ds {0 \brack n} = \delta_{0 n}$
where:
:$\ds {0 \brack n}$ denotes an unsigned Stirling number of the first kind
:$\delta_{0 n}$ denotes the Kronecker delta.
== Proof ==
By definition of unsigned Stirling number of the first kind:
$\ds x^{\underline 0} = \sum_k \paren {-1}^{0 - k} {0 \brack k} x^k$
Thus we have:
Thus, in the expression:
:$\ds x^{\underline 0} = \sum_k \paren {-1}^{-k} {0 \brack k} x^k$
we have:
:$\ds {0 \brack 0} = 1$
and for all $k \in \Z_{>0}$:
:$\ds {0 \brack k} = 0$
That is:
:$\ds {0 \brack k} = \delta_{0 k}$
== Also see ==
* Signed Stirling Number of the First Kind of 0
* Stirling Number of the Second Kind of 0
* Particular Values of Unsigned Stirling Numbers of the First Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(48)$
Category:Examples of Stirling Numbers of the First Kind
Unsigned Stirling Number of the First Kind of 1

== Theorem ==
:$\ds {1 \brack n} = \delta_{1 n}$
where:
:$\ds {1 \brack n}$ denotes an unsigned Stirling number of the first kind
:$\delta_{1 n}$ is the Kronecker delta.
== Proof ==
== Also see ==
* Signed Stirling Number of the First Kind of 1
* Stirling Number of the Second Kind of 1
* Particular Values of Unsigned Stirling Numbers of the First Kind
Category:Examples of Stirling Numbers of the First Kind
Unsigned Stirling Number of the First Kind of Number with Greater

== Theorem ==
Let $n, k \in \Z_{\ge 0}$.
Let $k > n$.
Let $\ds {n \brack k}$ denote an unsigned Stirling number of the first kind.
Then:
:$\ds {n \brack k} = 0$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Signed Stirling Number of the First Kind of Number with Greater
* Stirling Number of the Second Kind of Number with Greater
Category:Stirling Numbers
Category:Unsigned Stirling Number of the First Kind of Number with Greater
Unsigned Stirling Number of the First Kind of Number with Greater/Proof 1

== Theorem ==
Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
== Proof ==
By definition, unsigned Stirling number of the first kind are defined as the polynomial coefficients $\ds {n \brack k}$ which satisfy the equation:
:$\ds x^{\underline n} = \sum_k \paren {-1}^{n - k} {n \brack k} x^k$
where $x^{\underline n}$ denotes the $n$th falling factorial of $x$.
Both of the expressions on the and are polynomials in $x$ of degree $n$.
Hence the coefficient $\ds {n \brack k}$ of $x^k$ where $k > n$ is $0$.
Category:Unsigned Stirling Number of the First Kind of Number with Greater
Unsigned Stirling Number of the First Kind of Number with Greater/Proof 2

== Theorem ==
Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
== Proof ==
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P v$ be the proposition:
:$\ds k > n \implies {n \brack k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brack k} = \delta_{0 k}$
from Unsigned Stirling Number of the First Kind of 0.
So by definition of Kronecker delta:
:$\forall k \in \Z_{\ge 0}: k > 0 \implies \ds {0 \brack k} = 0$
and so $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $0 \le r$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds k > r \implies {r \brack k} = 0$
from which it is to be shown that:
:$\ds k > r + 1 \implies {r + 1 \brack k} = 0$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: k > n \implies {n \brack k} = 0$
Category:Unsigned Stirling Number of the First Kind of Number with Greater
Unsigned Stirling Number of the First Kind of Number with Self

== Theorem ==
:$\ds {n \brack n} = 1$
where $\ds {n \brack n}$ denotes an unsigned Stirling number of the first kind.
== Proof ==
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds {n \brack n} = 1$
=== Basis for the Induction ===
$\map P 0$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds {k \brack k} = 1$
from which it is to be shown that:
:$\ds {k + 1 \brack k + 1} = 1$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: {n \brack n} = 1$
== Also see ==
* Signed Stirling Number of the First Kind of Number with Self
* Stirling Number of the Second Kind of Number with Self
* Particular Values of Unsigned Stirling Numbers of the First Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(48)$
Category:Stirling Numbers
Unsigned Stirling Number of the First Kind of n+1 with 0

== Theorem ==
Let $n \in \Z_{\ge 0}$.
Then:
:$\ds {n + 1 \brack 0} = 0$
where $\ds {n + 1 \brack 0}$ denotes an unsigned Stirling number of the first kind.
== Proof ==
We are given that $k = 0$.
So by definition of unsigned Stirling number of the first kind:
:$\ds {n \brack k} = \delta_{n k}$
where $\delta_{n k}$ is the Kronecker delta.
Thus:
Hence the result.
== Also see ==
* Signed Stirling Number of the First Kind of n+1 with 0
* Stirling Number of the Second Kind of n+1 with 0
* Particular Values of Unsigned Stirling Numbers of the First Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(50)$
Category:Examples of Stirling Numbers of the First Kind
Unsigned Stirling Number of the First Kind of n+1 with 1

== Theorem ==
Let $n \in \Z_{\ge 0}$.
Then:
:$\ds {n + 1 \brack 1} = n!$
where:
:$\ds {n + 1 \brack 1}$ denotes an unsigned Stirling number of the first kind
:$n!$ denotes $n$ factorial.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {n + 1 \brack 1} = n!$
=== Basis for the Induction ===
$\map P 0$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds {k + 1 \brack 1} = k!$
from which it is to be shown that:
:$\ds {k + 2 \brack 1} = \paren {k + 1}!$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: {n + 1 \brack 1} = n!$
== Also see ==
* Signed Stirling Number of the First Kind of n+1 with 1
* Stirling Number of the Second Kind of n+1 with 1
* Particular Values of Unsigned Stirling Numbers of the First Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(50)$
Category:Examples of Stirling Numbers of the First Kind
Unsigned Stirling Number of the First Kind of n with n-1

== Theorem ==
Let $n \in \Z_{> 0}$ be an integer greater than $0$.
Then:
:$\ds {n \brack n - 1} = \binom n 2$
where:
:$\ds {n \brack n - 1}$ denotes an unsigned Stirling number of the first kind
:$\dbinom n 2$ denotes a binomial coefficient.
== Proof ==
The proof proceeds by induction.
=== Basis for the Induction ===
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds {n \brack n - 1} = \binom n 2$
$\map P 1$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds {k \brack k - 1} = \binom k 2$
from which it is to be shown that:
:$\ds {k + 1 \brack k} = \binom {k + 1} 2$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 1}: {n \brack n - 1} = \dbinom n 2$
== Also see ==
* Signed Stirling Number of the First Kind of n with n-1
* Stirling Number of the Second Kind of n with n-1
* Particular Values of Unsigned Stirling Numbers of the First Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(49)$
Category:Examples of Stirling Numbers of the First Kind
Unsigned Stirling Number of the First Kind of n with n-2

== Theorem ==
Let $n \in \Z_{\ge 2}$ be an integer greater than or equal to $2$.
Then:
:$\ds {n \brack n - 2} = \binom n 4 + 2 \binom {n + 1} 4$
where:
:$\ds {n \brack n - 2}$ denotes an unsigned Stirling number of the first kind
:$\dbinom n 4$ denotes a binomial coefficient.
== Proof ==
The proof proceeds by induction.
=== Basis for the Induction ===
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\ds {n \brack n - 2} = \binom n 4 + 2 \binom {n + 1} 4$
$\map P 2$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds {k \brack k - 2} = \binom k 4 + 2 \binom {k + 1} 4$
from which it is to be shown that:
:$\ds {k + 1 \brack k - 1} = \binom {k + 1} 4 + 2 \binom {k + 2} 4$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 2}: {n \brack n - 2} = \binom n 4 + 2 \binom {n + 1} 4$
== Also see ==
* Stirling Number of the Second Kind of n with n-2
* Particular Values of Unsigned Stirling Numbers of the First Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(57)$
Category:Examples of Stirling Numbers of the First Kind