nyuuzyou/wikis huggingface dataset:
Value of Adjugate of Determinant

== Theorem ==
Let $D$ be the determinant of order $n$.
Let $D^*$ be the adjugate of $D$.
Then $D^* = D^{n - 1}$.
== Proof ==
Let $\mathbf A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\
 a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
 a_{n1} & a_{n2} & \cdots & a_{nn}\end{bmatrix}$ and $\mathbf A^* = \begin{bmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\
 A_{21} & A_{22} & \cdots & A_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
 A_{n1} & A_{n2} & \cdots & A_{nn}\end{bmatrix}$.
Thus:
:$\paren {\mathbf A^*}^\intercal = \begin{bmatrix} A_{11} & A_{21} & \cdots & A_{n1} \\
 A_{12} & A_{22} & \cdots & A_{n2} \\
\vdots & \vdots & \ddots & \vdots \\
 A_{1n} & A_{2n} & \cdots & A_{nn}\end{bmatrix}$
is the transpose of $\mathbf A^*$.
Let $c_{ij}$ be the typical element of $\mathbf A \paren {\mathbf A^*}^\intercal$.
Then by definition of matrix product:
:$\ds c_{ij} = \sum_{k \mathop = 1}^n a_{ik} A_{jk}$
Thus by the corollary of the Expansion Theorem for Determinants:
:$c_{ij} = \delta_{ij} D$
So by Determinant of Diagonal Matrix:
:$\map \det {\mathbf A \paren {\mathbf A^*}^\intercal} = \begin{vmatrix} D & 0 & \cdots & 0 \\
 0 & D & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
 0 & 0 & \cdots & D\end{vmatrix} = D^n$
From Determinant of Matrix Product:
:$\map \det {\mathbf A} \map \det {\paren {\mathbf A^*}^\intercal} = \map \det {\mathbf A \paren {\mathbf A^*}^\intercal}$
From Determinant of Transpose:
:$\map \det {\paren {\mathbf A^*}^\intercal} = \map \det {\mathbf A^*}$
Thus as $D = \map \det {\mathbf A}$ and $D^* = \map \det {\mathbf A^*}$ it follows that:
:$DD^* = D^n$
Now if $D \ne 0$, the result follows.
However, if $D = 0$ we need to show that $D^* = 0$.
Let $D^* = \begin{vmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\
 A_{21} & A_{22} & \cdots & A_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
 A_{n1} & A_{n2} & \cdots & A_{nn}\end{vmatrix}$.
Suppose that at least one element of $\mathbf A$, say $a_{rs}$, is non-zero (otherwise the result follows immediately).
By Expansion Theorem for Determinants and its corollary, we can expand $D$ by row $r$, and get:
:$\ds D = 0 = \sum_{j \mathop = 1}^n A_{ij} t_j, \forall i = 1, 2, \ldots, n$
for all $t_1 = a_{r1}, t_2 = a_{r2}, \ldots, t_n = a_{rn}$.
But $t_s = a_{rs} \ne 0$.
So, by (work in progress):
:$D^* = \begin{vmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\
 A_{21} & A_{22} & \cdots & A_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
 A_{n1} & A_{n2} & \cdots & A_{nn}\end{vmatrix} = 0$
Category:Determinants
Value of Apéry's Constant

== Unsolved Problem ==
Apéry's constant:
:$\ds \map \zeta 3 = \sum_{n \mathop = 1}^\infty \frac 1 {n^3}$
has not been determined in closed form in terms of other familiar constants. 
It is of particular interest whether, similar to Riemann Zeta Function at Even Integers: 
:$\dfrac {\map \zeta 3} {\pi^3}$
is rational.
By using the techniques of Riemann Zeta Function as a Multiple Integral it can be established that:
:$\ds \int_0^1 \int_0^1 \int_0^1 \frac {\rd x \rd y \rd z} {1 - x y z} = \sum_{n \mathop = 1}^\infty \frac 1 {n^3}$
== Also see ==
* Apéry's Theorem, in which Apéry's constant is shown to be irrational.
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.24$: Euler's Formula $\ds \sum \limits_1^\infty \frac 1 {n^2} = \frac {\pi^2} 6$ by Double Integration
Category:Riemann Zeta Function
Category:Unsolved Problems
Category:Apéry's Constant
Value of Apéry's Constant/Historical Note

== Historical Note on Value of Apéry's Constant ==
The value of Apéry's constant remains unsolved since raised the question in $1736$.
== Sources ==
* : Chapter $\text {B}.24$: Euler's Formula $\ds \sum \limits_1^\infty \frac 1 {n^2} = \frac {\pi^2} 6$ by Double Integration
Category:Historical Notes
Value of Cauchy Determinant

== Theorem ==
Let $D_n$ be a Cauchy determinant of order $n$:
:$\begin{vmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_2} & \cdots & \dfrac 1 {x_n + y_n} \\
\end{vmatrix}$
Then the value of $D_n$ is given by:
:$D_n = \dfrac {\ds \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i} \paren {y_j - y_i} } {\ds \prod_{1 \mathop \le i, \, j \mathop \le n} \paren {x_i + y_j} }$
Let $D_n$ be given by:
:$\begin {vmatrix}
\dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\
\dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\
\end {vmatrix}$
Then its determinant is given by:
:$D_n = \dfrac {\ds \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i} \paren {y_i - y_j} } {\ds \prod_{1 \mathop \le i, \, j \mathop \le n} \paren {x_i - y_j} }$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $38$
Category:Value of Cauchy Determinant
Category:Cauchy Matrices
Category:Determinants
Value of Cauchy Determinant/Proof 1

== Theorem ==
== Proof ==
Take the version of the Cauchy matrix defined such that $a_{i j} = \dfrac 1 {x_i + y_j}$.
Subtract column $1$ from each of columns $2$ to $n$.
Thus:
From Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant.
Now:
:$1$: extract the factor $\dfrac 1 {x_i + y_1}$ from each row $1 \le i \le n$
:$2$: extract the factor $y_1 - y_j$ from each column $2 \le j \le n$.
Thus from Determinant with Row Multiplied by Constant we have the following:
:$\ds D_n = \paren {\prod_{i \mathop = 1}^n \frac 1 {x_i + y_1} } \paren {\prod_{j \mathop = 2}^n y_1 - y_j} \begin {vmatrix}
1 & \dfrac 1 {x_1 + y_2} & \dfrac 1 {x_1 + y_3} & \cdots & \dfrac 1 {x_1 + y_n} \\
1 & \dfrac 1 {x_2 + y_2} & \dfrac 1 {x_2 + y_3} & \cdots & \dfrac 1 {x_2 + y_n} \\
1 & \dfrac 1 {x_3 + y_2} & \dfrac 1 {x_3 + y_3} & \cdots & \dfrac 1 {x_3 + y_n} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \dfrac 1 {x_n + y_2} & \dfrac 1 {x_n + y_3} & \cdots & \dfrac 1 {x_n + y_n} \\
\end{vmatrix}$
Now subtract row $1$ from each of rows $2$ to $n$.
Column $1$ will go to $0$ for all but the first row.
Columns $2$ to $n$ will become:
From Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant.
Now:
:$1$: extract the factor $x_1 - x_i$ from each row $2 \le i \le n$
:$2$: extract the factor $\dfrac 1 {x_1 + y_j}$ from each column $2 \le j \le n$.
Thus from Determinant with Row Multiplied by Constant we have the following:
:$\ds D_n = \paren {\prod_{i \mathop = 1}^n \frac 1 {x_i + y_1} } \paren {\prod_{j \mathop = 1}^n \frac 1 {x_1 + y_j} } \paren {\prod_{i \mathop = 2}^n x_1 - x_i} \paren {\prod_{j \mathop = 2}^n y_1 - y_j} \begin {vmatrix}
1 & 1 & 1 & \cdots & 1 \\
0 & \dfrac 1 {x_2 + y_2} & \dfrac 1 {x_2 + y_3} & \cdots & \dfrac 1 {x_2 + y_n} \\
0 & \dfrac 1 {x_3 + y_2} & \dfrac 1 {x_3 + y_3} & \cdots & \dfrac 1 {x_3 + y_n} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & \dfrac 1 {x_n + y_2} & \dfrac 1 {x_n + y_3} & \cdots & \dfrac 1 {x_n + y_n} \\
\end {vmatrix}$
From Determinant with Unit Element in Otherwise Zero Row, and tidying up the products, we get:
:$D_n = \frac {\ds \prod_{i \mathop = 2}^n \paren {x_i - x_1} \paren {y_i - y_1} } {\ds \prod_{1 \mathop \le i, j \mathop \le n} \paren {x_i + y_1} \paren {x_1 + y_j} }
\begin {vmatrix}
\dfrac 1 {x_2 + y_2} & \dfrac 1 {x_2 + y_3} & \cdots & \dfrac 1 {x_2 + y_n} \\
\dfrac 1 {x_3 + y_2} & \dfrac 1 {x_3 + y_3} & \cdots & \dfrac 1 {x_3 + y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n + y_2} & \dfrac 1 {x_n + y_3} & \cdots & \dfrac 1 {x_n + y_n} \\
\end {vmatrix}$
Repeat the process for the remaining rows and columns $2$ to $n$.
The result follows.
A similar process obtains the result for the $a_{i j} = \dfrac 1 {x_i - y_j}$ form.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $38$
Category:Value of Cauchy Determinant
Value of Cauchy Determinant/Proof 2

== Theorem ==
== Proof ==
Let:
To be proved:
Assume hereafter that set $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ consists of distinct values, because otherwise $\map \det C$ is undefined or zero.
Preliminaries:
Vandermonde Matrix Identity for Cauchy Matrix supplies matrix equation
:$(1): \quad - C = P V_x^{-1} V_y Q^{-1}$
:Definitions of symbols:
::$V_x = \begin {pmatrix}
1 & 1 & \cdots & 1 \\
x_1 & x_2 & \cdots & x_n \\
\vdots & \vdots & \ddots & \vdots \\
{x_1}^{n - 1} & {x_2}^{n - 1} & \cdots & {x_n}^{n - 1} \\
\end {pmatrix}$
::$V_y = \begin {pmatrix}
1 & 1 & \cdots & 1 \\
y_1 & y_2 & \cdots & y_n \\
\vdots & \vdots & \ddots & \vdots \\
{y_1}^{n - 1} & {y_2}^{n - 1} & \cdots & {y_n}^{n - 1} \\
\end {pmatrix}$ 
::$P = \begin {pmatrix}
\map {p_1} {x_1} & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & \map {p_n} {x_n} \\
\end {pmatrix}$
::$Q = \begin {pmatrix}
\map p {y_1} & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & \map p {y_n)} \\
\end {pmatrix}$ 
:$1 \mathop \le k \mathop \le n$ Polynomials
::$\ds \map p x = \prod_{i \mathop = 1}^n \paren {x - x_i}$
::$\ds \map {p_k} x = \prod_{i \mathop = 1, i \mathop \ne k}^n \paren {x - x_i}$
Determinant of $C$ Calculation:
Lemma: $\map \det P = \paren {-1}^m \paren {\map \det {V_x} }^2$ where $m = \frac 1 2 n \paren {n - 1}$
:Details: Determinant $\map \det P$ expands to:
:Pair factors $\paren {x_r - x_s}$ and $\paren {x_s - x_r}$ into factor $-\paren {x_s - x_r}^2$, then:
Apply the Lemma to equation (2):
Category:Value of Cauchy Determinant
Value of Compactly Supported Function outside its Support

== Theorem ==
Let $f : \R \to \R$ be a continuous real function.
Let $K \subseteq \R$ be a compact subset.
Let $K$ be the support of $f$:
:$\map \supp f = K$.
Then:
:$\forall x \notin K : \map f x = 0$
== Proof ==
We have that:
:$\R = K \cup \paren {\R \setminus K}$.
By definition of the support:
:$x \in \map \supp f \iff \map f x \ne 0$
By Biconditional Equivalent to Biconditional of Negations:
:$\neg \paren {x \in \map \supp f} \iff \neg \paren {\map f x \ne 0}$
That is:
:$x \notin K \iff \map f x = 0$
or
:$x \in \R \setminus K \iff \map f x = 0$
Hence:
:$\forall x \notin K : \map f x = 0$
Category:Real Analysis
Value of Curvilinear Coordinate Metric

== Definition ==
Let $A$ and $B$ be two points in space.
Let a curvilinear $3$-Space coordinate system $\QQ$ be applied on top of a Cartesian $3$-space.
Let $h_{i j}$ be the metric of $\QQ$.
Then:
:$\forall i, j \in \set {1, 2, 3}: {h_{i j} }^2 = \dfrac {\partial x} {\partial q_i} \dfrac {\partial x} {\partial q_j} + \dfrac {\partial y} {\partial q_i} \dfrac {\partial y} {\partial q_j} + \dfrac {\partial z} {\partial q_i} \dfrac {\partial z} {\partial q_j}$
where $\partial$ denotes partial differentiation.
== Proof ==
From the Cartesian representation of $\QQ$:
where:
:$\tuple {x, y, z}$ denotes the Cartesian coordinates
:$\tuple {q_1, q_2, q_3}$ denotes the corresponding curvilinear coordinates.
Then:
By definition of the metric of $\QQ$:
Thus we can extract the appropriate terms with $\d q_i \rd q_j$:
Hence the result.
== Sources ==
* : Chapter $2$ Coordinate Systems $2.1$ Curvilinear Coordinates: $(2.6)$
Category:Definitions/Curvilinear Coordinates
Value of Degree in Radians

== Theorem ==
The value of a degree in radians is given by:
:$1 \degrees = \dfrac {\pi} {180} \radians \approx 0 \cdotp 01745 \, 32925 \, 19943 \, 29576 \, 9236 \ldots \radians$
== Proof ==
By Measurement of Full Angle, a full angle measures $2 \pi$ radians.
By definition of degree of angle, a full angle measures $360$ degrees.
Thus $1$ degree of angle is given by:
:$1 \degrees = \dfrac {2 \pi} {360} = \dfrac \pi {180}$
== Also see ==
* Value of Radian in Degrees
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.27$
* : $\S 5$: Trigonometric Functions: $5.14$
* : $1$. Functions: $1.5$ Trigonometric or Circular Functions: $1.5.1$ Unit Circle
* 
* 
* : $\S 1$: Special Constants: $1.10.$
Category:Degrees of Angle
Category:Radians
Category:Angles
Category:Units of Measurement
Value of Discriminant of Conic Section

== Theorem ==
Let $K$ be a conic section embedded in a Cartesian plane with the general equation:
:$a x^2 + 2 h x y + b y^2 + 2 g x + 2 f y + c = 0$
where $a, b, c, f, g, h \in \R$.
The value of the discriminant of $K$ is:
:$\Delta = a b c + 2 f g h - a f^2 - b g^2 - c h^2$
== Proof ==
By definition, the discriminant of $K$ is:
== Sources ==
* 
* 
Category:Discriminants of Conic Sections
Value of Field Norm on 5th Cyclotomic Ring is Integer

== Theorem ==
Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
Let $\alpha = a + i b \sqrt 5$ be an arbitrary element of $\Z \sqbrk {i \sqrt 5}$.
Let $\map N \alpha$ denoted the field norm of $\alpha$.
Then $\map N \alpha$ is an integer.
== Proof ==
From Field Norm on 5th Cyclotomic Ring:
:$\map N \alpha = a^2 + 5 b^2$
From the definition of the $5$th cyclotomic ring:
:$\Z \sqbrk {i \sqrt 5} = \set {a + i \sqrt 5 b: a, b \in \Z}$
That is, both $a$ and $b$ are integers.
Hence $a^2 + 5 b^2$ is also an integer.
== Sources ==
* : Chapter $9$: Rings: Exercise $19$
Category:Cyclotomic Rings
Value of Finite Continued Fraction equals Numerator Divided by Denominator

== Theorem ==
Let $F$ be a field.
Let $\tuple {a_0, a_1, \ldots, a_n}$ be a finite continued fraction of length $n \ge 0$.
Let $p_n$ and $q_n$ be its $n$th numerator and denominator.
Then the value $\sqbrk {a_0, a_1, \ldots, a_n}$ equals $\dfrac {p_n} {q_n}$.
== Proof ==
We will use a proof by induction on the length $n$.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\sqbrk {a_0, a_1, \ldots, a_n} = \dfrac {p_n} {q_n}$
=== Basis for the Induction ===
$\map P 0$ is the case: 
:$\sqbrk {a_0} = \dfrac {a_0} 1 = \dfrac {p_0} {q_0}$
This holds for any continued fraction.
$\map P 1$ is the case:
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show 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 our induction hypothesis:
:$\sqbrk {a_0, a_1, \ldots, a_k} = \dfrac {p_k} {q_k}$
Then we need to show:
:$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = \dfrac {p_{k + 1} } {q_{k + 1} }$
=== Induction Step ===
This is our induction step:
Consider the continued fraction:
:$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} }$
The numerators are:
:$p_0, p_1, \ldots, p_k, p_{k + 1}$
and the denominators are:
:$q_0, q_1, \ldots, q_k, q_{k + 1}$
By definition of value of a finite continued fraction:
:$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = \sqbrk {a_0, a_1, \ldots, a_{k - 1}, a_k'}$
where $a_k' = a_k + \dfrac 1 {a_{k + 1} }$
Consider the .
Take the continued fraction:
:$\sqbrk {a_0, a_1, \ldots, a_{k - 1}, a_k'}$
Its numerators are:
:$p_0, p_1, \ldots, p_{k-1}$ and $p_k'$
where $p_k' = \paren {a_k + \dfrac 1 {a_{k + 1} } } p_{k - 1} + p_{k - 2}$ by definition.
Its denominators are:
:$q_0, q_1, \ldots, p_{k - 1}$ and $q_k'$
where $q_k' = \paren {a_k + \dfrac 1 {a_{k + 1} } } q_{k - 1} + q_{k - 2}$ by definition.
As it has just $k$ partial denominators, the induction hypothesis tells us that its value is:
:$\dfrac {p_k'} {q_k'}$
So:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = \dfrac {p_{k + 1} } {q_{k + 1} }$
Category:Continued Fractions
Category:Proofs by Induction
Value of Finite Continued Fraction of Real Numbers is at Least First Term

== Theorem ==
Let $\sequence {a_0, \ldots, a_n}$ be a finite continued fraction in $\R$ of length $n \ge 0$.
Let the partial denominators $a_k > 0$ be strictly positive for $k>0$.
Let $x = [a_0, a_1, \ldots, a_n]$ be its value.
Then $x \ge a_0$, and $x > a_0$ if the length $n \ge 1$.
== Proof ==
If $n = 0$, we have $x = \sqbrk {a_0} = a_0$ by definition of value.
Let $n>0$.
By definition of value:
:$\sqbrk {a_0, a_1, \ldots, a_n} = a_0 + \dfrac 1 {\sqbrk {a_1, a_2, \ldots, a_n} }$
By Value of Finite Continued Fraction of Strictly Positive Real Numbers is Strictly Positive:
:$\sqbrk {a_1, a_2, \ldots, a_n} > 0$
Thus:
:$\sqbrk {a_0, a_1, \ldots, a_n} = a_0 + \dfrac 1 {\sqbrk {a_1, a_2, \ldots, a_n} } > a_0$
== Also see ==
* Properties of Value of Finite Continued Fraction
Category:Continued Fractions
Value of Finite Continued Fraction of Strictly Positive Real Numbers is Strictly Positive

== Theorem ==
Let $\sequence {a_0, \ldots, a_n}$ be a finite continued fraction in $\R$ of length $n \ge 0$.
Let all partial denominators $a_k > 0$ be strictly positive.
Let $x = \sqbrk {a_0, a_1, \ldots, a_n}$ be its value.
Then $x > 0$.
== Proof ==
== Also see ==
* Properties of Value of Finite Continued Fraction
Category:Continued Fractions
Value of Formula under Assignment Determined by Free Variables

== Theorem ==
Let $\mathbf A$ be a WFF of predicate logic.
Let $\AA$ be a structure for predicate logic.
Let $\sigma, \sigma'$ be assignments for $\mathbf A$ in $\AA$ such that:
:For each free variable $x$ of $\mathbf A$, $\map \sigma x = \map {\sigma'} x$
Then:
:$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma'}$
where $\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma$ is the value of $\mathbf A$ under $\sigma$.
== Proof ==
Proceed by the Principle of Structural Induction applied to the bottom-up specification of predicate logic.
If $\mathbf A = \map p {\tau_1, \ldots, \tau_n}$, then:
:$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \map {p_\AA} {\map {\operatorname{val}_\AA} {\tau_1} \sqbrk \sigma, \ldots, \map {\operatorname{val}_\AA} {\tau_n} \sqbrk \sigma}$
Because $\mathbf A$ contains no quantifiers, all its variables are free, and hence are in the domain of $\sigma, \sigma'$ as assignments.
Thus $\sigma, \sigma'$ are assignments for each $\tau_i$, and by Value of Term under Assignment Determined by Variables:
:$\map {\operatorname{val}_\AA} {\tau_i} \sqbrk \sigma = \map {\operatorname{val}_\AA} {\tau_i} \sqbrk {\sigma'}$
for each $\tau_i$.
It is immediate that:
:$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma'}$
If $\mathbf A = \neg \mathbf B$ and the induction hypothesis applies to $\mathbf B$, then:
If $\mathbf A = \mathbf B \circ \mathbf B'$ for $\circ$ one of $\land, \lor, \implies, \iff$ and the induction hypothesis applies to $\mathbf B, \mathbf B'$:
If $\mathbf A = \exists x: \mathbf B$ or $\mathbf A = \forall x : \mathbf B$, and the induction hypothesis applies to $\mathbf B$, then from the definition of value under $\sigma$:
:$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma$
is determined by the values:
:$\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma + \paren {x / a} }$
where $a$ ranges over $\AA$, and $\sigma + \paren {x / a}$ is the extension of $\sigma$ mapping $x$ to $a$.
Now, for a free variable $y$ of $\mathbf B$:
Hence, by the induction hypothesis:
:$\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma + \paren {x / a} } = \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma' + \paren {x / a} }$
It follows that:
:$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma'}$
The result follows from the Principle of Structural Induction.
Category:Predicate Logic
Value of Multiplicative Function at One

== Theorem ==
Let $f: \N \to \C$ be a multiplicative function.
If $f$ is not identically zero, then $\map f 1 = 1$.
== Proof ==
If $f$ is not identically zero, then:
:$\exists m \in \Z: \map f m \ne 0$
Then:
:$\map f m = \map f {1 \times m} = \map f 1 \, \map f m$
Hence $\map f 1 = 1$.
Category:Multiplicative Functions
Value of Multiplicative Function is Product of Values of Prime Power Factors

== Theorem ==
Let $f: \N \to \C$ be a multiplicative function.
Let $n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ be the prime decomposition of $n$.
Then:
:$\map f n = \map f {p_1^{k_1} } \, \map f {p_2^{k_2} } \dotsm \map f {p_r^{k_r} }$
== Proof ==
We have:
:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$
We also have:
:$\forall i, j \in \closedint 1 n: i \ne j \implies p_i^{k_i} \perp p_j^{k_j}$
So:
:$\map f {p_i^{k_i} p_j^{k_j} } = \map f {p_i^{k_i} } \, \map f {p_j^{k_j} }$
It is a simple inductive process to show that $\map f n = \map f {p_1^{k_1} } \, \map f {p_2^{k_2} } \dotsm \map f {p_r^{k_r} }$.
Category:Multiplicative Functions
Value of Plastic Constant

== Theorem ==
The plastic constant $P$ is evaluated as:
== Proof ==
By definition, the plastic constant $P$ is the real root of the cubic:
:$x^3 - x - 1 = 0$
Recall Cardano's Formula:
Here we have:
Hence:
and so:
Then:
The number can then be calculated.
Since $S \ne T$, the other two roots $x_2, x_3$ has non-zero imaginary parts $\pm \dfrac {i \sqrt 3} 2 \paren {S - T}$.
Hence the root above is the only real root.
== Sources ==
* : $1,32471 795 \ldots$
* 
Category:Pisot-Vijayaraghavan Numbers
Value of Position-Ratio

== Theorem ==
Let $P$ and $Q$ be points in space.
Let $R$ be a point on the straight line passing through $P$ and $Q$.
Let $k$ denote the position-ratio of $R$.
Then:
:$k = \dfrac {PQ} {RQ} - 1$
== Proof ==
== Sources ==
* : Chapter $\text I$. Coordinates: $11$. Position-ratio of a point
Category:Position-Ratios
Value of Radian in Degrees

== Theorem ==
The value of a radian in degrees is given by:
:$1 \radians = \dfrac {180 \degrees} {\pi} \approx 57 \cdotp 29577 \, 95130 \ 82320 \, 87679 \, 8154 \ldots \degrees$ 
== Proof ==
By Measurement of Full Angle, a full angle measures $2 \pi$ radians.
By definition of degree of angle, a full angle measures $360$ degrees.
Thus $1$ radian is given by:
:$1 \radians = \dfrac {360 \degrees} {2 \pi} = \dfrac {180 \degrees} {\pi}$
== Also see ==
* Value of Degree in Radians
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.26$
* : $\S 5$: Trigonometric Functions: $5.13$
* : $1$. Functions: $1.5$ Trigonometric or Circular Functions: $1.5.1$ Unit Circle
* : $57 \cdotp 296 \ldots$
* : $57 \cdotp 296 \ldots$
* : $\S 1$: Special Constants: $1.9.$
Category:Degrees of Angle
Category:Radians
Category:Angles
Category:Units of Measurement
Value of Relation is Small

== Theorem ==
The value of a relation is always a small class.
== Proof ==
Let $\RR$ be an arbitrary relation.
Let $s$ be any set.
The value of a relation is either equal to some set $y$ or $\O$ by Uniqueness Condition for Relation Value.
If it is equal to some set $y$, then the value of $s$ under $\RR$ is a small class by the definition of small class.
If it is equal to $\O$, then the result follows from Empty Set is Small.
== Sources ==
* : $\S 6.13$
Category:Zermelo-Fraenkel Class Theory
Value of Skew-Symmetric Game is Zero

== Theorem ==
Let $G$ be a two-person zero-sum game.
Let $G$ be represented by a payoff table that is skew-symmetric.
Then the value of $G$ is zero.
== Proof ==
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $3$
Category:Two-Person Zero-Sum Games
Value of Term under Assignment Determined by Variables

== Theorem ==
Let $\tau$ be a term of predicate logic.
Let $\AA$ be a structure for predicate logic.
Let $\sigma, \sigma'$ be assignments for $\tau$ in $\AA$ such that:
:For each variable $x$ occurring in $\tau$, $\map \sigma x = \map {\sigma'} x$
Then:
:$\map {\operatorname{val}_\AA} \tau \sqbrk \sigma = \map {\operatorname{val}_\AA} \tau \sqbrk {\sigma'}$
where $\map {\operatorname{val}_\AA} \tau \sqbrk \sigma$ is the value of $\tau$ under $\sigma$.
== Proof ==
Proceed by the Principle of Structural Induction applied to the definition of a term.
If $\tau = x$, then:
as desired.
If $\tau = \map f {\tau_1, \ldots, \tau_n}$ and the induction hypothesis applies to each $\tau_i$, then:
The result follows from the Principle of Structural Induction.
Category:Predicate Logic