nyuuzyou/wikis huggingface dataset:
First-Order Reaction/Proof 1

== Theorem ==
== Proof ==
From Differential Equation governing First-Order Reaction, the differential equation governing this reaction is given by:
:$-\dfrac {\d x} {\d t} = k x$
for $k \in \R_{>0}$.
At time $t = 0$ we have that $x = x_0$.
So:
:$x_0 = e^c e^0 = e^c$
and hence the result
:$x = x_0 e^{-k t}$
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $1$: How Differential Equations Originate
* : $1$: The Nature of Differential Equations: $\S 4$: Growth, Decay and Chemical Reactions
Category:First-Order Reactions
First-Order Reaction/Proof 2

== Theorem ==
== Proof ==
From Differential Equation governing First-Order Reaction, the differential equation governing this reaction is given by:
:$-\dfrac {\d x} {\d t} = k x$
for $k \in \R_{>0}$.
This is an instance of the Decay Equation, and has the solution:
:$x = x_0 e^{-k t}$
== Also see ==
* Half-Life of Radioactive Substance
Category:First-Order Reactions
First Apotome/Example

== Example of First Apotome ==
Let $a, b \in \set {x \in \R_{>0} : x^2 \in \Q}$ be two rationally expressible numbers such that $a - b$ is an apotome.
By definition, $a - b$ is a first apotome :
:$(1): \quad a \in \Q$
:$(2): \quad \dfrac {\sqrt {a^2 - b^2}} a \in \Q$
where $\Q$ denotes the set of rational numbers.
Let $a = 9$ and $b = \sqrt {17}$.
Then:
Therefore $9 - \sqrt {17}$ is a first apotome.
Category:Examples of Apotome
First Apotome of Medial is Irrational

== Theorem ==
== Proof ==
:
Let $AB$ be a medial straight line.
Let a medial straight line $BC$ such that:
:$BC$ is commensurable in square only with $AB$
:the rectangle contained by $AB$ and $BC$ is rational
be cut off from $AB$.
We have that $AB$ and $BC$ are medial.
So by definition $AB^2$ and $BC^2$ are both medial.
But $2 \cdot AB \cdot BC$ is rational.
Therefore:
:$AB^2$ and $BC^2$ are incommensurable with $2 \cdot AB \cdot BC$.
From :
:if $AB$ is incommensurable with either $AC$ or $CB$, $AC$ and $CB$ are incommensurable with each other.
Therefore by :
:$2 \cdot AB \cdot BC$ is incommensurable with $AC^2$.
Therefore $AC^2$ is irrational.
Therefore by definition $AC$ is irrational.
Such a straight line is known as a first apotome of a medial.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
First Bimedial Straight Line is Divisible Uniquely

== Theorem ==
== Proof ==
Let $AB$ be a first bimedial straight line.
Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that:
:$AC$ and $CB$ are medial straight lines
:$AC$ and $CB$ are commensurable in square only
:$AC$ and $CB$ contain a rational rectangle.
Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$.
From :
:$AB^2 = \paren {AC + CB}^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
and:
:$AB^2 = \paren {AD + DB}^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$
and so:
:$\paren {AC^2 + CB^2} - \paren {AD^2 + DB^2} = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$
By definition, $AC \cdot CB$ and $AD \cdot DB$ are both rational areas.
So by :
:$2 AC \cdot CB$ and $2 AD \cdot DB$ are both rational areas.
But $AB$
Thus $2 AC \cdot CB - 2 AD \cdot DB$ is also a rational area.
As:
:$AC$ and $CB$ are medial straight lines
and
:$AD$ and $DB$ are medial straight lines
it follows by definition that $AC^2$, $CB^2$, $AD^2$ and $DB^2$ are all medial areas.
But from this cannot be the case.
From this contradiction it follows that $D$ cannot divide $AB$ into medial straight lines commensurable in square only and containing a rational rectangle.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
First Bimedial is Irrational

== Theorem ==
== Proof ==
Let $AB$ and $BC$ be medial straight lines which are commensurable in square only.
Let $AB$ and $BC$ contain a rational rectangle.
By definition, $AB$ and $BC$ are incommensurable in length.
We have:
:$AB : BC = AB \cdot BC : BC^2$
:$AB : BC = AB^2 : AB \cdot BC$
Therefore from :
: $AB \cdot BC$ is incommensurable with $AB^2$
and
: $AB \cdot BC$ is incommensurable with $BC^2$.
But by :
:$2 AB \cdot BC$ is commensurable with $AB \cdot BC$.
We have that $AB$ and $BC$ are commensurable in square.
So from :
:$AB^2 + BC^2$ is commensurable with $BC^2$.
So from :
:$2 AB \cdot BC$ is incommensurable with $AB^2 + BC^2$.
Thus from :
:$2 AB \cdot BC + AB^2 + BC^2$ is incommensurable with $AB \cdot BC$.
We have that $AB$ and $BC$ contain a rational rectangle.
Thus $AB \cdot BC$ is rational.
From :
:$AC^2 = \left({AB + BC}\right)^2 = 2 AB \cdot BC + AB^2 + BC^2$
Thus from :
:$AC$ is irrational.
Such a straight line is called first bimedial.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
First Complex Root of Unity is Primitive

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n$ denote the complex $n$th roots of unity:
:$U_n = \set {z \in \C: z^n = 1}$
Let $\alpha_1 = \exp \paren {\dfrac {2 \pi i} n}$ denote the first complex root of unity.
Then $\alpha_1$ is a primitive complex root of unity.
== Proof ==
From Condition for Complex Root of Unity to be Primitive:
:$\gcd \set {n, k} = 1$
where:
:$\alpha_k = \exp \paren {\dfrac {2 k \pi i} n}$
Here we have:
:$k = 1$
and:
:$\gcd \set {n, 1} = 1$
Hence the result.
== Sources ==
* : $\S 3$. Roots of Unity
Category:Complex Roots of Unity
First Cyclotomic Polynomial

== Theorem ==
The first cyclotomic polynomial is:
:$\map {\Phi_1} x = x - 1$
== Proof ==
By definition:
:$\ds \map {\Phi_1} x = \prod_\zeta \paren {x - \zeta}$
where the product runs over all primitive complex first roots of unity.
A root of unity has order $1$ it equals $1$.
Hence the only factor is $x - 1$.
Category:Examples of Cyclotomic Polynomials
First Derivative of PGF of Negative Binomial Distribution/Type 1

== Theorem ==
Let $X$ be a discrete random variable with the type 1 negative binomial distribution (second form) with parameters $n$ and $p$.
Then the first derivative of the PGF of $X$ $s$ is:
:$\dfrac \d {\d s} \map {\Pi_X} s = n p \paren {\dfrac {\paren {p s}^{n - 1} } {\paren {1 - q s}^{n + 1} } }$
== Proof ==
The Probability Generating Function of Negative Binomial Distribution (Type 1) is:
:$\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
We have that for a given type 1 negative binomial distribution, $n, p$ and $q$ are constant.
Thus we have:
Hence the result.
Category:First Derivative of PGF of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 1)
Category:Derivatives of PGFs
First Derivative of PGF of Negative Binomial Distribution/Type 2

== Theorem ==
Let $X$ be a discrete random variable with the type 2 negative binomial distribution with parameters $n$ and $p$.
Then the first derivative of the PGF of $X$ $s$ is:
:$\dfrac \d {\d s} \map {\Pi_X} s = \dfrac {n p} q \paren {\dfrac q {1 - p s} }^{n + 1}$
== Proof ==
The Probability Generating Function of Negative Binomial Distribution (Type 2) is:
:$\map {\Pi_X} s = \paren {\dfrac q {1 - p s} }^n$
We have that for a given type 2 negative binomial distribution, $n, p$ and $q$ are constant.
Thus we have:
Hence the result.
Category:First Derivative of PGF of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 2)
Category:Derivatives of PGFs
First Element of Geometric Sequence not dividing Second

== Theorem ==
Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers of length $n$.
Let $a_0$ not be a divisor of $a_1$.
Then:
:$\forall j, k \in \set {0, 1, \ldots, n}, j \ne k: a_j \nmid a_k$
That is, if the initial term of $P$ does not divide the second, no term of $P$ divides any other term of $P$.
== Proof 1 ==
== Proof 2 ==
Category:Geometric Sequences of Integers
Category:First Element of Geometric Sequence not dividing Second
First Element of Geometric Sequence not dividing Second/Proof 1

== Theorem ==
== Proof ==
Let $P_a = \tuple {a_0, a_1, \ldots, a_n}$ be a geometric sequence of natural numbers such that $a_0 \nmid a_1$.
 $a_0 \divides a_k$ for some $k: 2 \le k \le n$.
Let $b_0, b_1, \ldots, b_k$ be the least natural numbers which have the same common ratio as $a_0, a_1, \ldots, a_k$.
These can be found by means of .
From 
:$a_0 : a_k = b_0 : b_k$
Also:
:$a_0 : a_1 = b_0 : b_1$
and so as $a_0 \nmid a_1$ it follows by :
:$b_0 \nmid b_1$
From One Divides all Integers it follows that:
:$b_0 \ne 1$
From :
:$b_0 \perp b_k$
But as:
:$a_0 : a_k = b_0 : b_k$
it follows that:
:$a_0 \nmid a_k$
Now suppose $a_j \divides a_k$ such that $0 < j < k$.
Let $b_j, \ldots, b_k$ be the least natural numbers which have the same common ratio as $a_j, \ldots, a_k$.
These can be found by means of .
From :
:$a_j : a_k = b_j : b_k$
The other cases follow similarly.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:First Element of Geometric Sequence not dividing Second
First Element of Geometric Sequence not dividing Second/Proof 2

== Theorem ==
== Proof ==
From Form of Geometric Sequence of Integers, the terms of $P$ are in the form:
:$\paren 1: \quad: a_j = k q^j p^{n - j}$
where $p \perp q$.
Let $a_0 \nmid a_1$.
That is:
: $\exists i, j \in \set {0, 1, \ldots, n}, i \ne j: a_i \divides a_j$
 let $i < j$.
But we have that $p \perp q$.
From Powers of Coprime Numbers are Coprime:
:$q^{j - i} \perp p^{j - i}$
This can happen only when $q = 1$.
This is incompatible with $q \nmid p$.
From this contradiction, the result follows.
Category:First Element of Geometric Sequence not dividing Second
First Element of Geometric Sequence that divides Last also divides Second

== Theorem ==
Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers of length $n$.
Let $a_0$ be a divisor of $a_n$.
Then $a_0$ is a divisor of $a_2$.
== Proof ==
By hypothesis, let $a_0$ be a divisor of $a_n$.
 $a_0$ is not a divisor of $a_2$.
From First Element of Geometric Sequence not dividing Second it would follow that $a_0$ does not divide $a_n$.
From this contradiction follows the result.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Geometric Sequences of Integers
First Fundamental Group of 1-Sphere

== Theorem ==
Let $\mathbb S^1$ be the $1$-sphere.
Let $\struct {\map {\pi _1} {\mathbb S^1}, \ast}$ be the first fundamental group of $\mathbb S^1$.
Let $\struct {\Z, +}$ be the additive group of integers.
Then $\struct {\map {\pi _1} {\mathbb S^1}, \ast}$ is isomorphic to $\struct {\Z, +}$.
== Proof ==
We are given that $\struct {\map {\pi _1} {\mathbb S^1}, \ast}$ is the first fundamental group of $\mathbb S^1$.
Let $x_0 \in \mathbb S^1$.
Let $\struct{\mathbb S^1, x_0}$ be the pointed topological space for $\mathbb S^1$.
Since Fundamental Group is Independent of Base Point for Path-Connected Space, what to be proved is:
:$\struct {\map {\pi _1} {\mathbb S^1, x_0}, \ast} \cong \struct {\Z, +}$ 
where the map:
:$\ast : \map {\pi_1} {\mathbb S^1, x_0} \times \map {\pi_1} {\mathbb S^1, x_0} \to \map {\pi_n} {\mathbb S^1, x_0}$ 
denotes the concatenation of homotopy classes of paths.
Hence $\struct {\map {\pi _1} {\mathbb S^1}, \ast}$ is isomorphic to $\struct {\Z, +}$.
Category:Homotopy Theory
First Harmonic Number to exceed 10

== Theorem ==
The first harmonic number that is greater than $10$ is $H_{12 \, 367}$.
That is, the number of terms of the harmonic series required for its partial sum to exceed $10$ is $12 \, 367$.
== Proof ==
We have:
:$H_{12 \, 366} = \ds \sum_{k \mathop = 1}^{12 \, 366} \frac 1 k \approx 9 \cdotp 99996 \, 214$
and:
:$H_{12 \, 367} = \ds \sum_{k \mathop = 1}^{12 \, 367} \frac 1 k \approx 10 \cdotp 00004 \, 30083$
== Sources ==
* : $272,400,600$
Category:Harmonic Numbers
Category:12,367
First Harmonic Number to exceed 100

== Theorem ==
The first harmonic number that is greater than $100$ is $H_n$ where $n \approx 1.5 \times 10^{43}$.
That is, it takes approximately $1.5 \times 10^{43}$ terms of the harmonic series required for its partial sum to exceed $100$.
== Proof ==
== Sources ==
* 
* : $272,400,600$
* : $272,400,600$
Category:Harmonic Numbers
First Harmonic Number to exceed 20

== Theorem ==
The first harmonic number that is greater than $20$ is $H_{272 \, 400 \, 600}$.
That is, the number of terms of the harmonic series required for its partial sum to exceed $20$ is $272 \, 400 \, 600$.
== Proof ==
We have:
:$H_{272 \, 400 \, 599} = \ds \sum_{k \mathop = 1}^{272 \, 400 \, 599} \frac 1 k \approx 19 \cdotp 99999 \, 99979$
and:
:$H_{272 \, 400 \, 600} = \ds \sum_{k \mathop = 1}^{272 \, 400 \, 600} \frac 1 k \approx 20 \cdotp 00000 \, 00016$
== Sources ==
* : $272,400,600$
* : $272,400,600$
Category:Harmonic Numbers
Category:272,400,600
First Hyperoperation is Addition Operation

== Theorem ==
The $1$st hyperoperation is the addition operation restricted to the positive integers:
:$\forall x, y \in \Z_{\ge 0}: H_1 \left({x, y}\right) = x + y$
== Proof ==
By definition of the hyperoperation sequence:
:$\forall n, x, y \in \Z_{\ge 0}: H_n \left({x, y}\right) = \begin{cases}
y + 1 & : n = 0 \\
x & : n = 1, y = 0 \\
0 & : n = 2, y = 0 \\
1 & : n > 2, y = 0 \\
H_{n - 1} \left({x, H_n \left({x, y - 1}\right)}\right) & : n > 0, y > 0 \end{cases}$
Thus the $1$st hyperoperation is defined as:
:$\forall x, y \in \Z_{\ge 0}: H_1 \left({x, y}\right) = \begin{cases}
x & : y = 0 \\
H_0 \left({x, H_1 \left({x, y - 1}\right)}\right) & : y > 0 \end{cases}$
From Zeroth Hyperoperation is Successor Function:
:$(1): \quad \forall x, y \in \Z_{\ge 0}: H_1 \left({x, y}\right) = \begin{cases}
x & : y = 0 \\
H_1 \left({x, y - 1}\right) + 1 & : y > 0 \end{cases}$
The proof proceeds by induction.
For all $y \in \Z_{\ge 0}$, let $P \left({y}\right)$ be the proposition:
:$\forall x \in \Z_{\ge 0}: H_1 \left({x, y}\right) = x + y$
=== Basis for the Induction ===
$P \left({0}\right)$ is the case:
Thus $P \left({0}\right)$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is the induction hypothesis:
:$\forall x \in \Z_{\ge 0}: H_1 \left({x, k}\right) = x + k$
from which it is to be shown that:
:$\forall x \in \Z_{\ge 0}: H_1 \left({x, k + 1}\right) = x + k + 1$
=== Induction Step ===
This is the induction step:
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall x, y \in \Z_{\ge 0}: H_1 \left({x, y}\right) = x + y$
Category:Hyperoperation
First Inversion Formula for Stirling Numbers

== Theorem ==
For all $m, n \in \Z_{\ge 0}$:
:$\ds \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$
where:
:$\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
:$\ds {k \brace m}$ denotes a Stirling number of the second kind
:$\delta_{m n}$ denotes the Kronecker delta.
== Proof ==
The proof proceeds by strong induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$
=== Basis for the Induction ===
$\map P 0$ is the case:
Thus $\map P 0$ has been shown to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P j$ is true for all $j$ such $0 \le j \le r$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \forall j: 0 \le j \le r: \forall m \in \Z_{\ge 0}: \sum_k {j \brack k} {k \brace m} \paren {-1}^{j - k} = \delta_{m j}$
from which it is to be shown that:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{r + 1 - k} = \delta_{m \paren {r + 1} }$
=== 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:
:$\ds \forall m, n \in \Z_{\ge 0}: \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$
== Also see ==
* Second Inversion Formula for Stirling Numbers
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(47)$
Category:Stirling Numbers
First Isomorphism Theorem

== Context ==
This theorem applies for Groups, Rings, Modules, Algebras, and any other algebraic structure where you see the word homomorphism.
It is a categorical result; that is, it is independent of the structure used.
== Theorem ==
=== Groups ===
=== Rings ===
=== Vector Spaces ===
=== Topological Vector Spaces ===
== Also known as ==
There is no standard numbering for the Isomorphism Theorems. Different authors use different labellings.
This particular result, for example, is also known as the Homomorphism Theorem.
== Also see ==
* Isomorphism Theorems
Category:Named Theorems
Category:Isomorphism Theorems
Category:First Isomorphism Theorem
First Isomorphism Theorem/Groups

== Theorem ==
Let $\phi: G_1 \to G_2$ be a group homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Then:
:$\Img \phi \cong G_1 / \map \ker \phi$
where $\cong$ denotes group isomorphism.
== Proof ==
Let $K = \map \ker \phi$.
By Kernel is Normal Subgroup of Domain, $G_1 / K$ exists.
We need to establish that the mapping $\theta: G_1 / K \to G_2$ defined as:
:$\forall x \in G_1: \map \theta {x K} = \map \phi x$
is well-defined.
That is, we need to ensure that:
:$\forall x, y \in G_1: x K = y K \implies \map \theta {x K} = \map \theta {y K}$
Let $x, y \in G_1: x K = y K$.
Then:
Thus we see that $\theta$ is well-defined.
Since we also have that:
:$\map \phi x = \map \phi y \implies x K = y K$
it follows that:
:$\map \theta {x K} = \map \theta {y K} \implies x K = y K$
So $\theta$ is injective.
We also note that:
:$\Img \theta = \set {\map \theta {x K}: x \in G}$
So:
We also note that $\theta$ is a homomorphism:
Thus $\theta$ is a monomorphism whose image equals $\Img \phi$.
The result follows.
== Also known as ==
Some sources call this the homomorphism theorem.
Others combine this result with Group Homomorphism Preserves Subgroups, Kernel of Group Homomorphism is Subgroup and Kernel is Normal Subgroup of Domain.
Still others do not assign a special name to this theorem at all.
== Also see ==
* Isomorphism Theorems
== Sources ==
* : $\S 7.4$. Kernel and image: The Homomorphism Theorem
* : $\S 1.10$: Theorem $24$: Corollary
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 66$
* : $\S 52.1$ The first isomorphism theorem
* : Chapter $8$: The Homomorphism Theorem: Theorem $8.13: \ (3)$
Category:Group Homomorphisms
Category:Group Isomorphisms
Category:First Isomorphism Theorem
First Isomorphism Theorem/Rings

== Theorem ==
Let $\phi: R \to S$ be a ring homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Then:
:$\Img \phi \cong R / \map \ker \phi$
where $\cong$ denotes ring isomorphism.
== Proof ==
From Ring Homomorphism whose Kernel contains Ideal, let $J = \map \ker \phi$.
This gives the ring homomorphism $\mu: R / \map \ker \phi \to S$ as follows:
::$\begin {xy} \xymatrix@L + 2mu@ + 1em {
 R \ar[rr]^*{\nu}
 \ar[rdrd]_*{\phi}
& & R / \map \ker \phi \ar[dd]_*{\mu} \\
\\
& & S 
} \end {xy}$
That is:
:$\phi = \mu \circ \nu$
Then we have:
:$\map \ker \mu = \map \ker \phi / \map \ker \phi$
This is the null subring of $R / \map \ker \phi$ by Quotient Ring Defined by Ring Itself is Null Ring.
Then from Kernel is Trivial iff Monomorphism it follows that $\mu$ is a monomorphism.
From $\phi = \mu \circ \nu$, we have:
:$\Img \mu = \Img \phi$
It follows that $\mu$ is an isomorphism.
== Also known as ==
== Also see ==
* Isomorphism Theorems
* Universal Property of Quotient Ring
== Sources ==
* : Chapter $5$: Rings: $\S 24$. Homomorphisms: Theorem $47$
* : $\S 2.2$: Homomorphisms: Theorem $2.9$
* : $\S 60.3$ Factor rings
Category:First Ring Isomorphism Theorem
Category:Ring Homomorphisms
Category:Ring Isomorphisms
Category:First Isomorphism Theorem
First Isomorphism Theorem/Topological Vector Spaces

== Theorem ==
Let $K$ be a topological field.
Let $\struct {X, \tau_X}$ and $\struct {Y, \tau_Y}$ be vector spaces over $K$.
Let $T : X \to Y$ be a continuous and open linear transformation.
Let $\ker T$ be the kernel of $T$.
Let $X/\ker T$ be the quotient topological vector space of $X$ modulo $\ker T$.
Then $X/\ker T$ is topologically isomorphic to $\Img T$.
== Proof ==
Let $\pi : X \to X/\ker T$ be the quotient mapping.
From First Isomorphism Theorem: Vector Spaces, there exists a linear isomorphism $\Lambda : X/\ker T \to \Img T$ with $T = \Lambda \circ \pi$. 
It remains to argue that $\Lambda$ is a homeomorphism.
Since $T$ is continuous, we obtain that $\Lambda$ is continuous from Factorization of Continuous Linear Transformation between Topological Vector Spaces.
Since $T$ is open, we obtain that $\Lambda$ is open from Factorization of Open Linear Transformation between Topological Vector Spaces.
From Bijection is Open iff Inverse is Continuous, $\Lambda^{-1}$ is continuous.
So $\Lambda$ is a homeomorphism, and hence a topological isomorphism.
Category:First Isomorphism Theorem
Category:Topological Vector Spaces
First Isomorphism Theorem/Vector Spaces

== Theorem ==
Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Let $\ker T$ be the kernel of $T$.
Let $X/\ker T$ be the quotient vector space of $X$ modulo $\ker T$.
Then $X/\ker T$ is isomorphic to $\Img T$ as a vector space.
== Proof ==
From Image of Linear Transformation is Submodule, we assure ourselves that $\Img T$ is indeed a vector space over $K$.
Let $\pi : X \to X/\ker T$ be the quotient mapping. 
From Condition for Mapping from Quotient Vector Space to be Well-Defined, there exists a linear transformation $\Lambda : X/\ker T \to \Img T$ such that:
:$\map \Lambda {\map \pi x} = T x$ for each $x \in X$.
To finish, we need to show that $\Lambda$ is injective and surjective.
From Linear Transformation is Injective iff Kernel Contains Only Zero, to show $\Lambda$ is injective it suffices to show that:
:$\ker \Lambda = \set { {\mathbf 0}_{X/\ker T} }$
Suppose that $\map \Lambda {\map \pi x} = {\mathbf 0}_Y$.
Then $T x = {\mathbf 0}_Y$.
So $x \in \ker T$. 
That is, by Kernel of Quotient Mapping, we have $\map \pi x = {\mathbf 0}_{X/\ker T}$.
So we have $\ker \Lambda = \set { {\mathbf 0}_{X/\ker T} }$, and so $\ker \Lambda$ is injective by Linear Transformation is Injective iff Kernel Contains Only Zero.
Now let $y \in \Img T$.
Then there exists $x \in X$ such that $y = T x$.
Then, we have $\map \Lambda {\map \pi x} = T x = y$. 
So $\Img T \subseteq \Img \Lambda$.
Since we already have $\Img \Lambda \subseteq \Img T$, we obtain $\Img \Lambda = \Img T$.
So $\Lambda : X/\ker T \to \Img T$ is a vector space isomorphism.
Category:First Isomorphism Theorem
Category:Vector Spaces
First Joiner's Problem

== Problem ==
Cut each of a $6 \times 6$ chessboard and an $8 \times 8$ chessboard into exactly $2$ pieces.
Then assemble those $4$ pieces into a $10 \times 10$ chessboard.
== Solution ==
:
== Historical Note ==
== Sources ==
* : $\text {IV}$: Measure Seven Times Before you Cut: $162$. A Gift for Grandmother
* : $\text {IV}$ Geometric Puzzles: $4.2$ The First Joiner's Problem
* : Chapter $1$: Elementary, my dear Watson: $\S 1.2$: Elements, my dear Watson: Ponderable $1.2.2$
Category:Dissections
Category:Chessboard Puzzles
First Occurrence of Prime Gap of 864

== Theorem ==
The first occurrence of a prime gap of $864$ is between $6 \, 505 \, 941 \, 701 \, 960 \, 039$ and $6 \, 505 \, 941 \, 701 \, 960 \, 903$.
== Proof ==
$6 \, 505 \, 941 \, 701 \, 960 \, 039$ is a prime number.
$6 \, 505 \, 941 \, 701 \, 960 \, 903$ is a prime number.
It can be checked that all numbers between these two are composite.
It can also be checked by computer that there is no prime gap of $864$ between smaller prime numbers.
== Sources ==
* 
* : $6,505,941,701,960,039$
Category:Prime Gaps
First Order ODE/(1 + x^2) y' + x y = 0

== Theorem ==
The first order ODE:
:$\paren {1 + x^2} y' + x y = 0$
has the general solution:
:$y = \dfrac C {\sqrt {1 + x^2} }$
== Proof ==
Category:Examples of Linear First Order ODEs
Category:Examples of Solutions to Separable Differential Equation