hcju/proofwikips huggingface dataset:
Natural Numbers Set Equivalent to Ideals of Integers
Let the mapping $\psi: \N \to$ the set of all ideals of $\Z$ be defined as: :$\forall b \in \N: \psi \left({b}\right) = \left({b}\right)$ where $\left({b}\right)$ is the principal ideal of $\Z$ generated by $b$. Then $\psi$ is a bijection.
Canonical Epimorphism from Integers by Principal Ideal
Let $m$ be a strictly positive integer. Let $\left({m}\right)$ be the principal ideal of $\Z$ generated by $m$. The restriction to $\N_m$ of the canonical epimorphism $q_m$ from the ring $\left({\Z, +, \times}\right)$ onto $\left({\Z, +, \times}\right) / \left({m}\right)$ is an isomorphism from the ring $\left({\N_m, +_m, \times_m}\right)$ of integers modulo $m$ onto the quotient ring $\left({\Z, +, \times}\right) / \left({m}\right)$. In particular, $\left({\Z, +, \times}\right) / \left({m}\right)$ has $m$ elements.
Integer Divisor is Equivalent to Subset of Ideal
Let $\Z$ be the set of all integers. Let $\Z_{>0}$ be the set of strictly positive integers. Let $m \in \Z_{>0}$ and let $n \in \Z$. Let $\ideal m$ be the principal ideal of $\Z$ generated by $m$. Then: :$m \divides n \iff \ideal n \subseteq \ideal m$
Principal Ideals in Integral Domain
Let $\struct {D, +, \circ}$ be an integral domain. Let $U_D$ be the group of units of $D$. Let $\ideal x$ be the principal ideal of $D$ generated by $x$. Let $x, y \in \struct {D, +, \circ}$. Then:
Principal Ideal Domain is Unique Factorization Domain
Every principal ideal domain is a unique factorization domain.
Maximal Ideal iff Quotient Ring is Field
Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$. Let $J$ be an ideal of $R$. The following are equivalent: :$(1): \quad$ $J$ is a maximal ideal. :$(2): \quad$ The quotient ring $R / J$ is a field.
Prime Number iff Generates Principal Maximal Ideal
Let $\Z_{>0}$ be the set of strictly positive integers. Let $p \in \Z_{>0}$. Let $\ideal p$ be the principal ideal of $\Z$ generated by $p$.  Then $p$ is prime  $\ideal p$ is a maximal ideal of $\Z$.
Integral Domain of Prime Order is Field
Let $\left({\Z_p, +_p, \times_p}\right)$ be the ring of integers modulo $p$. The following statements are equivalent: : $(1): \quad p$ is a prime. : $(2): \quad \left({\Z_p, +_p, \times_p}\right)$ is an integral domain. : $(3): \quad \left({\Z_p, +_p, \times_p}\right)$ is a field.
Quotient Ring of Integers and Zero
Let $\struct {\Z, +, \times}$ be the integral domain of integers. Let $\ideal 0$ be the principal ideal of $\struct {\Z, +, \times}$ generated by $0$. The quotient ring $\struct {\Z / \ideal 0, +, \times}$ is isomorphic to $\struct {\Z, +, \times}$.
Quotient Ring of Integers and Principal Ideal from Unity
Let $\left({\Z, +, \times}\right)$ be the integral domain of integers. Let $\left({1}\right)$ be the principal ideal of $\left({\Z, +, \times}\right)$ generated by $1$. The quotient ring $\left({\Z, +, \times}\right) / \left({1}\right)$ is isomorphic to the null ring.
Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal
Let $\struct {D, +, \circ}$ be a principal ideal domain. Let $\ideal p$ be the principal ideal of $D$ generated by $p$. Then $p$ is irreducible  $\ideal p$ is a maximal ideal of $D$.
Subring Generated by Unity of Ring with Unity
Let $\left({R, +, \circ}\right)$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$. Let the mapping $g: \Z \to R$ be defined as $\forall n \in \Z: g \left({n}\right) = n 1_R$, where $n 1_R$ the $n$th power of $1_R$. Let $\left({x}\right)$ be the principal ideal of $\left({R, +, \circ}\right)$ generated by $x$. Then $g$ is an epimorphism from $\Z$ onto the subring $S$ of $R$ generated by $1_R$. If $R$ has no proper zero divisors, then $g$ is the only nonzero homomorphism from $\Z$ into $R$. The kernel of $g$ is either: : $(1): \quad \left({0_R}\right)$, in which case $g$ is an isomorphism from $\Z$ onto $S$ or: : $(2): \quad \left({p}\right)$ for some prime $p$, in which case $S$ is isomorphic to the field $\Z_p$.
Null Ring iff Characteristic is One
The only ring whose characteristic is $1$ is the null ring.
Characteristic of Finite Ring with No Zero Divisors
Let $\struct {R, +, \circ}$ be a finite ring with unity with no proper zero divisors whose zero is $0_R$ and whose unity is $1_R$. Let $n \ne 0$ be the characteristic of $R$. Then: :$(1): \quad n$ must be a prime number :$(2): \quad n$ is the order of all non-zero elements in $\struct {R, +}$. It follows that $\struct {R, +} \cong C_n$, where $C_n$ is the cyclic group of order $n$.
Integral Domain with Characteristic Zero
In an integral domain with characteristic zero, every non-zero element has infinite order under ring addition.
Characteristic of Field is Zero or Prime
Let $F$ be a field. Then the characteristic of $F$ is either zero or a prime number.
Field of Characteristic Zero has Unique Prime Subfield
Let $F$ be a field, whose zero is $0_F$ and whose unity is $1_F$, with characteristic zero. Then there exists a unique $P \subseteq F$ such that: :$(1): \quad P$ is a subfield of $F$ :$(2): \quad P$ is isomorphic to the field of rational numbers $\struct {\Q, +, \times}$. That is, $P \cong \Q$ is a unique minimal subfield of $F$, and all other subfields of $F$ contain $P$. This field $P$ is called the prime subfield of $F$.
Field of Prime Characteristic has Unique Prime Subfield
Let $F$ be a field whose characteristic is $p$. Then there exists a unique $P \subseteq F$ such that: :$(1): \quad P$ is a subfield of $F$ :$(2): \quad P \cong \Z_p$. That is, $P \cong \Z_p$ is a unique minimal subfield of $F$, and all other subfields of $F$ contain $P$. This field $P$ is called the prime subfield of $F$.
Intersection of All Division Subrings is Prime Subfield
Let $\struct {K, +, \circ}$ be a division ring. Let $P$ be the intersection of the set of all division subrings of $K$. Then $P$ is the prime subfield of $K$.
Characteristic of Ordered Integral Domain is Zero
Let $\left({D, +, \circ}\right)$ be an ordered integral domain whose zero is $0_D$ and whose unity is $1_D$. Let $\operatorname{Char} \left({D}\right)$ be the characteristic of $D$. Then $\operatorname{Char} \left({D}\right) = 0$. Let $g: \Z \to D$ be the mapping defined as: :$\forall n \in \Z: g \left({n}\right) = n \cdot 1_D$ where $n \cdot 1_D$ is defined as the $n$th power of $1_D$. Then $g$ is the only monomorphism from the ordered ring $\Z$ onto the ordered ring $D$.
Monomorphism from Rational Numbers to Totally Ordered Field
Let $\struct {F, +, \circ, \le}$ be a totally ordered field. There is one and only one (ring) monomorphism from the totally ordered field $\Q$ onto $F$. Its image is the prime subfield of $F$.
Set of Polynomials over Integral Domain is Subring
Let $\struct {R, +, \circ}$ be a commutative ring. Let $\struct {D, +, \circ}$ be an integral subdomain of $R$. Then $\forall x \in R$, the set $D \sqbrk x$ of polynomials in $x$ over $D$ is a subring of $R$.
Polynomials Closed under Ring Product
Let $\left({R, +, \circ}\right)$ be a commutative ring. Let $\displaystyle f = \sum_{k \in Z} a_k \mathbf X^k$, $\displaystyle g =  \sum_{k \in Z} b_k \mathbf X^k$ be polynomials in the indeterminates $\left\{{X_j: j \in J}\right\}$ over $R$, where $Z$ is the set of all multiindices indexed by $\left\{{X_j: j \in J}\right\}$. Define the product :$\displaystyle f \otimes g = \sum_{k \in Z} c_k \mathbf X^k$ where :$\displaystyle c_k = \sum_{\substack{p + q = k \\ p, q \in Z}} a_p b_q$ Then $f \otimes g$ is a polynomial.
Ring of Polynomial Forms is Integral Domain
Let $\struct {R, +, \circ}$ be a commutative ring with unity. Let $\struct {D, +, \circ}$ be an integral subdomain of $R$. Let $X \in R$ be transcendental over $D$. Let $D \sqbrk X$ be the ring of polynomials in $X$ over $D$. Then $D \sqbrk X$ is an integral domain.
Injection is Bijection iff Inverse is Injection
Let $\phi: S \to T$ be an injection. Then $\phi$ is a bijection  its inverse $\phi^{-1}$ is also an injection.
Epimorphism from Polynomial Forms to Polynomial Functions
Let $D$ be an integral domain. Let $D \sqbrk X$ be the ring of polynomial forms in $X$ over $D$. Let $\map P D$ be the ring of polynomial functions over $D$. The mapping $\kappa: D \sqbrk X \to \map P D$ given by: :$\displaystyle \map \kappa {\sum_{k \mathop = 0}^n {a_k \circ X^k} } = f$ where $\displaystyle f = \sum_{k \mathop = 0}^n {a_k \circ x^k}, x \in D$ is a ring epimorphism.
Division Theorem for Polynomial Forms over Field
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$. Let $X$ be transcendental over $F$. Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$. Let $d$ be an element of $F \sqbrk X$ of degree $n \ge 1$. Then $\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ such that either: :$(1): \quad r = 0_F$ or: :$(2): \quad r \ne 0_F$ and $r$ has degree that is less than $n$.
Polynomial Forms over Field form Principal Ideal Domain
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$. Let $X$ be transcendental over $F$. Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$. Then $F \sqbrk X$ is a principal ideal domain.
Equal Consecutive Prime Number Gaps are Multiples of Six
If you list the gaps between consecutive primes greater than $5$: :$2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, \ldots$ you will notice that consecutive gaps that are equal are of the form $6 x$. This is ''always'' the case. {{OEIS|A001223}}
Standard Discrete Metric is Metric
The standard discrete metric is a metric.
Derivative of Constant
Let $\map {f_c} x$ be the constant function on $\R$, where $c \in \R$. Then: :$\map {f_c'} x = 0$
Derivative of Identity Function
Let $X$ be either set of either the real numbers $\R$ or the complex numbers $\C$. Let $I_X: X \to X$ be the identity function. Then: :$\map {I_X'} x = 1$ where $\map {I_X'} x$ denotes the derivative of $I_X$ {{WRT|Differentiation}} $x$. This can be presented for each of $\R$ and $\C$:
Product Rule for Derivatives
Let $\map f x, \map j x, \map k x$ be real functions defined on the open interval $I$. Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable. Let $\map f x = \map j x \map k x$. Then: :$\map {f'} \xi = \map j \xi \map {k'} \xi + \map {j'} \xi \map k \xi$ It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then: :$\forall x \in I: \map {f'} x = \map j x \map {k'} x + \map {j'} x \map k x$ Using Leibniz's notation for derivatives, this can be written as: :$\map {\dfrac \d {\d x} } {y \, z} = y \dfrac {\d z} {\d x} + \dfrac {\d y} {\d x} z$ where $y$ and $z$ represent functions of $x$.
Derivative of Composite Function
Let $f, g, h$ be continuous real functions such that: :$\forall x \in \R: \map h x = \map {f \circ g} x = \map f {\map g x}$ Then: :$\map {h'} x = \map {f'} {\map g x} \map {g'} x$ where $h'$ denotes the derivative of $h$. Using the $D_x$ notation: :$\map {D_x} {\map f {\map g x} } = \map {D_{\map g x} } {\map f {\map g x} } \, \map {D_x} {\map g x}$ This is often informally referred to as the '''chain rule (for differentiation)'''.
Derivative of Inverse Function
Let $I = \closedint a b$ and $J = \closedint c d$ be closed real intervals. Let $I^o = \openint a b$ and $J^o = \openint c d$ be the corresponding open real intervals. Let $f: I \to J$ be a real function which is continuous on $I$ and differentiable on $I^o$ such that $J = f \sqbrk I$. Let either: :$\forall x \in I^o: D \map f x > 0$ or: :$\forall x \in I^o: D \map f x < 0$ Then: :$f^{-1}: J \to I$ exists and is continuous on $J$ :$f^{-1}$ is differentiable on $J^o$ :$\forall y \in J^o: D \map {f^{-1} } y = \dfrac 1 {D \map f x}$
Upper Sum Never Smaller than Lower Sum
Let $\closedint a b$ be a closed interval of the set $\R$ of real numbers. Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ be a finite subdivision of $\closedint a b$. Let $f: \R \to \R$ be a real function. Let $f$ be bounded on $\closedint a b$. Let $\map L P$ be the lower sum of $\map f x$ on $\closedint a b$ belonging to the subdivision $P$. Let $\map U P$ be the upper sum of $\map f x$ on $\closedint a b$ belonging to the subdivision $P$. Then $\map L P \le \map U P$.
Dedekind's Theorem
Let $\tuple {L, R}$ be a Dedekind cut of the set of real numbers $\R$. Then there exists a unique real number which is a producer of $\tuple {L, R}$.
Complex Addition is Closed
The set of complex numbers $\C$ is closed under addition: :$\forall z, w \in \C: z + w \in \C$
Complex Addition is Associative
The operation of addition on the set of complex numbers $\C$ is associative: :$\forall z_1, z_2, z_3 \in \C: z_1 + \paren {z_2 + z_3} = \paren {z_1 + z_2} + z_3$
Complex Addition is Commutative
The operation of addition on the set of complex numbers is commutative: :$\forall z, w \in \C: z + w = w + z$
Integers are Countably Infinite
The set $\Z$ of integers is countably infinite.
Permutation of Determinant Indices
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$ over a field. Let $\lambda: \N_{> 0} \to \N_{> 0}$ be any fixed permutation on $\N_{> 0}$. Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$. Let $\struct {S_n, \circ}$ be the symmetric group of $n$ letters.  Then: :$\displaystyle \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k, \map \mu k} }$ :$\displaystyle \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \mu k, \map \lambda k} }$ where: :the summation $\displaystyle \sum_{\mu \mathop \in S_n}$ goes over all the $n!$ permutations of $\set {1, 2, \ldots, n}$ :$\map \sgn \mu$ is the sign of the permutation $\mu$.
Determinant of Transpose
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$. Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$. Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$. Then: :$\map \det {\mathbf A} = \map \det {\mathbf A^\intercal}$
Determinant with Rows Transposed
If two rows of a matrix with determinant $D$ are transposed, its determinant becomes $-D$.
Square Matrix with Duplicate Rows has Zero Determinant
If two rows of a square matrix over a commutative ring $\struct {R, +, \circ}$ are the same, then its determinant is zero.
Determinant with Row Multiplied by Constant
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$. Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$. Let $\mathbf B$ be the matrix resulting from one row of $\mathbf A$ having been multiplied by a constant $c$. Then: :$\map \det {\mathbf B} = c \map \det {\mathbf A}$ That is, multiplying one row of a square matrix by a constant multiplies its determinant by that constant.
Determinant as Sum of Determinants
Let $\begin{vmatrix}   a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix}$ be a determinant. Then $\begin{vmatrix}   a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{r1} + a'_{r1} & \cdots & a_{rs} + a'_{rs} & \cdots & a_{rn} + a'_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix} = \begin{vmatrix}   a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix} + \begin{vmatrix}   a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a'_{r1} & \cdots & a'_{rs} & \cdots & a'_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix}$. Similarly: Then $\begin{vmatrix}   a_{11} & \cdots & a_{1s} + a'_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{r1} & \cdots & a_{rs} + a'_{rs} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{n1} & \cdots & a_{ns} + a'_{ns} & \cdots & a_{nn} \end{vmatrix} = \begin{vmatrix}   a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix} + \begin{vmatrix}   a_{11} & \cdots & a'_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{r1} & \cdots & a'_{rs} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{n1} & \cdots & a'_{ns} & \cdots & a_{nn} \end{vmatrix}$.
Multiple of Row Added to Row of Determinant
Let $\mathbf A = \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{r 1} & a_{r 2} & \cdots & a_{r n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{s 1} & a_{s 2} & \cdots & a_{s n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$ be a square matrix of order $n$. Let $\map \det {\mathbf A}$ denote the determinant of $\mathbf A$. Let $\mathbf B = \begin{bmatrix} a_{1 1}             & a_{1 2}             & \cdots &             a_{1 n} \\  \vdots             & \vdots              & \ddots &              \vdots \\ a_{r 1} + k a_{s 1} & a_{r 2} + k a_{s 2} & \cdots & a_{r n} + k a_{s n} \\  \vdots             & \vdots              & \ddots &              \vdots \\ a_{s 1}             & a_{s 2}             & \cdots &             a_{s n} \\  \vdots             & \vdots              & \ddots &              \vdots \\ a_{n 1}             & a_{n 2}             & \cdots &             a_{n n} \\ \end{bmatrix}$. Then $\map \det {\mathbf B} = \map \det {\mathbf A}$. That is, the value of a determinant remains unchanged if a constant multiple of any row is added to any other row.
Determinant of Matrix Product
Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be a square matrices of order $n$. Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$. Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$. Then: :$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$ That is, the determinant of the product is equal to the product of the determinants.
Expansion Theorem for Determinants
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$. Let $D = \map \det {\mathbf A}$ be the determinant of $\mathbf A$: :$\displaystyle \map \det {\mathbf A} := \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} } = \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots a_{n \map \lambda n}$ where: :the summation $\displaystyle \sum_\lambda$ goes over all the $n!$ permutations of $\set {1, 2, \ldots, n}$ :$\map \sgn \lambda$ is the sign of the permutation $\lambda$. Let $a_{p q}$ be an element of $\mathbf A$. Let $A_{p q}$ be the cofactor of $a_{p q}$ in $D$. Then: :$(1): \quad \displaystyle \forall r \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$ :$(2): \quad \displaystyle \forall r \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{k r} A_{k r}$ Thus the value of a determinant can be found either by: :multiplying all the elements in a row by their cofactors and adding up the products or: :multiplying all the elements in a column by their cofactors and adding up the products. The identity: :$\displaystyle D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$ is known as the '''expansion of $D$ in terms of row $r$''', while: :$\displaystyle D = \sum_{k \mathop = 1}^n a_{k r} A_{k r}$ is known as the '''expansion of $D$ in terms of column $r$'''.
Determinant with Unit Element in Otherwise Zero Row
Let $D$ be the determinant: :$D = \begin {vmatrix}       1 &       0 & \cdots &       0 \\ b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\  \vdots &  \vdots & \ddots &  \vdots \\ b_{n 1} & b_{n 2} & \cdots & b_{n n} \end {vmatrix}$ Then: :$D = \begin {vmatrix} b_{2 2} & \cdots & b_{2 n} \\  \vdots & \ddots &  \vdots \\ b_{n 2} & \cdots & b_{n n} \end {vmatrix}$
Vandermonde Determinant
The '''Vandermonde determinant of order $n$''' is the determinant defined as follows: :$V_n = \begin {vmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n - 2} & x_1^{n - 1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n - 2} & x_2^{n - 1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n - 2} & x_n^{n - 1} \end {vmatrix}$ Its value is given by: :$\displaystyle V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$
Equality of Polynomials
$f$ and $g$ are equal as polynomials  $f$ and $g$ are equal as functions. Thus we can say $f = g$ without ambiguity as to what it means. {{explain|In the exposition, the term was "equal as forms", but it has now morphed into "equal as polynomials". Needs to be resolved.}}
Value of Adjugate of Determinant
Let $D$ be the determinant of order $n$. Let $D^*$ be the adjugate of $D$. Then $D^* = D^{n-1}$.
Determinant of Diagonal Matrix
Let $\mathbf A = \begin{bmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \\ \end{bmatrix}$ be a diagonal matrix. Then the determinant of $\mathbf A$ is the product of the elements of $\mathbf A$. That is: :$\ds \map \det {\mathbf A} = \prod_{i \mathop = 1}^n a_{ii}$
Matrix is Row Equivalent to Reduced Echelon Matrix
Let $\mathbf A = \sqbrk a_{m n}$ be a matrix of order $m \times n$ over a field $F$. Then $A$ is row equivalent to a reduced echelon matrix of order $m \times n$.
Square Matrix is Row Equivalent to Triangular Matrix
Let $\mathbf A = \left[{a}\right]_n$ be a square matrix of order $n$ over a commutative ring $R$. Then $\mathbf A$ can be converted to an upper or lower triangular matrix by elementary row operations.
Product of Triangular Matrices
Let $\mathbf A = \sqbrk a_n, \mathbf B = \sqbrk b_n$ be upper triangular matrices of order $n$. Let $\mathbf C = \mathbf A \mathbf B$. Then :$(1): \quad$ the diagonal elements of $\mathbf C$ are given by: ::::$\forall j \in \closedint 1 n: c_{j j} = a_{j j} b_{j j}$ :::That is, the diagonal elements of $\mathbf C$ are those of the factor matrices multiplied together. :$(2): \quad$ The matrix $\mathbf C$ is itself upper triangular. The same applies if both $\mathbf A$ and $\mathbf B$ are lower triangular matrices.
Effect of Elementary Row Operations on Determinant
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$. Let $\map \det {\mathbf A}$ denote the determinant of $\mathbf A$. Take the elementary row operations: {{begin-axiom}} {{axiom | n = \text {ERO} 1         | t = For some $\lambda$, multiply row $i$ by $\lambda$         | m = r_i \to \lambda r_i }} {{axiom | n = \text {ERO} 2         | t = For some $\lambda$, add $\lambda$ times row $j$ to row $i$         | m = r_i \to r_i + \lambda r_j }} {{axiom | n = \text {ERO} 3         | t = Exchange rows $i$ and $j$         | m = r_i \leftrightarrow r_j }} {{end-axiom}} Applying $\text {ERO} 1$ has the effect of multiplying $\map \det {\mathbf A}$ by $\lambda$. Applying $\text {ERO} 2$ has no effect on $\map \det {\mathbf A}$. Applying $\text {ERO} 3$ has the effect of multiplying $\map \det {\mathbf A}$ by $-1$.
Modulus in Terms of Conjugate
Let $z = a + i b$ be a complex number. Let $\cmod z$ be the modulus of $z$. Let $\overline z$ be the conjugate of $z$. Then: :$\cmod z^2 = z \overline z$
Existence and Uniqueness of Positive Root of Positive Real Number
Let $x \in \R$ be a real number such that $x \ge 0$. Let $n \in \Z$ be an integer such that $n \ne 0$. Then there always exists a unique $y \in \R: \paren {y \ge 0} \land \paren {y^n = x}$. Hence the justification for the terminology '''the positive $n$th root of $x$''' and the notation $x^{1/n}$.
Triangle Inequality
=== Geometry === {{:Triangle Inequality/Geometry}} === Real Numbers === {{:Triangle Inequality/Real Numbers}} === Complex Numbers === {{:Triangle Inequality/Complex Numbers}}
Even Power is Non-Negative
Let $x \in \R$ be a real number. Let $n \in \Z$ be an even integer. Then $x^n \ge 0$. That is, all even powers are positive.