hcju/proofwikips huggingface dataset:
Taylor's Theorem/One Variable/Proof by Rolle's Theorem
== Theorem == {{:Taylor's Theorem/One Variable/Statement of Theorem}} == Proof ==  Let the function $g$ be defined as: :$\map g t = \map {R_n} t - \dfrac {\paren {t - a}^{n + 1} } {\paren {x - a}^{n + 1} } \map {R_n} x$ Then: :$\map {g^{\paren k} } a = 0$ for $k = 0, \dotsc, n$, and $\map g x = 0$. Apply Rolle's Theorem successively to $g, g', \dotsc, g^{\paren n}$. Then there exist: :$\xi_1, \ldots, \xi_{n + 1}$ between $a$ and $x$ such that: :$\map {g'} {\xi_1} = 0, \map {g''} {\xi_2} = 0, \ldots, \map {g^{\paren {n + 1} } } {\xi_{n + 1} } = 0$ Let $\xi = \xi_{n + 1}$. Then: :$0 = \map {g^{\paren {n + 1} } } \xi = \map {f^{\paren {n + 1} } } \xi - \dfrac {\paren {n + 1}!} {\paren {x - a}^{n + 1} } \map {R_n} x$ and the formula for $\map {R_n} x$ follows.   {{Namedfor|Brook Taylor}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Taylor's Theorem/One Variable/Statement of Theorem/Also presented as|next = Definition:Increasing Real Function}}: $\S 11.10$
Matrix Inverse Algorithm
== Algorithm ==  The '''matrix inverse algorithm''' is an algorithm which either: :$(1): \quad$ converts a matrix into its inverse, if it exists or: :$(2): \quad$ determines that such an inverse does not exist. Let $\mathbf A$ be the $n \times n$ square matrix in question. Let $\mathbf I$ be the unit matrix of order $n$. :'''Step $0$''': Create the augmented matrix $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$. :'''Step $1$''': Perform elementary row operations until $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$ is in reduced echelon form. By Matrix is Row Equivalent to Reduced Echelon Matrix, this is possible. By Reduced Echelon Matrix is Unique, this process is well-defined. Call this new augmented matrix $\begin {bmatrix} \mathbf H & \mathbf C \end {bmatrix}$. :'''Step $2$''':  ::If $\mathbf H = \mathbf I$, then take $\mathbf C = \mathbf A^{-1}$. ::If $\mathbf H \ne \mathbf I$, $\mathbf A$ is not invertible.  == Proof == Suppose $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$ can be transformed into an upper triangular matrix with no zeroes on its main diagonal. Then from Identity Matrix from Upper Triangular Matrix, it can further be transformed into $\begin {bmatrix} \mathbf H & \mathbf C \end {bmatrix}$ where $\mathbf H = \mathbf I$. From Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, the row operation to convert $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$ to $\begin {bmatrix} \mathbf H & \mathbf C \end {bmatrix}$ is equivalent to the matrix product of the elementary row matrices corresponding to the sequence of elementary row operations that compose that row operation. Let $\mathbf R$ be that matrix corresponding to that row operation. Because $\mathbf H = \mathbf I$, it follows that: :$\mathbf R \mathbf A = \mathbf I$ and so $\mathbf R$ is the inverse of $\mathbf A$. That is: :$\mathbf R = \mathbf A^{-1}$ We also have, from the action of $\mathbf R$ on the  of the augmented matrix $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$ that: :$\mathbf R \mathbf I = \mathbf C$ and so: :$\mathbf C = \mathbf R = \mathbf A^{-1}$ {{qed|lemma}} Now suppose that $\mathbf H \ne \mathbf I$. {{finish|It remains to be shown that in such a case $\mathbf A$ is not invertible.}} == Examples == {{:Matrix Inverse Algorithm/Examples}} == Sources == * {{BookReference|Linear Algebra|1998|Richard Kaye|author2 = Robert Wilson|prev = Definition:Augmented Matrix|next = Matrix Inverse Algorithm/Examples/Arbitrary Matrix 1}}: Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations
Convergence of Generalized Sum of Complex Numbers/Corollary
== Corollary to Convergence of Generalized Sum of Complex Numbers == Let $\left({z_i}\right)_{i \in I}$ be an $I$-indexed family of complex numbers. That is, let $z_i \in \C$ for all $i \in I$.  Suppose that $\displaystyle \sum \left\{{ z_i: i \in I }\right\}$ converges to $z \in \C$. Then $\displaystyle \sum \left\{{ \overline{z_i}: i \in I }\right\}$ converges to $\overline z$, where $\overline z$ denotes the complex conjugate of $z$.  Here, the $\sum$ denote generalized sums. == Proof == Using Convergence of Generalized Sum of Complex Numbers, one has: :$\displaystyle \sum \left\{{\operatorname{Re} z_i : i \in I}\right\} = \operatorname{Re} z$ :$\displaystyle \sum \left\{{\operatorname{Im} z_i : i \in I}\right\} = \operatorname{Im} z$ Now, observe that, from the definition of complex conjugate: {{begin-eqn}} {{eqn|l = \overline z      |r = \operatorname{Re} z - i \operatorname{Im} z }} {{eqn|r = \sum \left\{ {\operatorname{Re} z_i : i \in I}\right\} - i \sum \left\{ {\operatorname{Im} z_i : i \in I}\right\} }} {{eqn|r = \sum \left\{ {\operatorname{Re} z_j - i \operatorname{Im} z_j : j \in I}\right\}      |c = Generalized Sum is Linear }} {{eqn|r = \sum \left\{ {\overline{z_j}: j \in I}\right\} }} {{end-eqn}}
Bijection iff Left and Right Inverse/Corollary
== Corollary to Bijection iff Left and Right Inverse ==  Let $f: S \to T$ and $g: T \to S$ be mappings such that: : $g \circ f = I_S$ : $f \circ g = I_T$ Then both $f$ and $g$ are bijections.  == Proof  == Suppose we have such mappings $f$ and $g$ with the given properties. From Bijection iff Left and Right Inverse, we have that $f$ is a bijection, by considering $g = g_1$ and $g = g_2$. It directly follows that by setting $g = f, f = g_1, f = g_2$, the result Bijection iff Left and Right Inverse can be used the other way about.  == Sources == * {{BookReference|Lectures in Abstract Algebra|1951|Nathan Jacobson|volume = I|subtitle = Basic Concepts|prev = Composite of Bijection with Inverse is Identity Mapping|next = Definition:Binary Operation}}: Introduction $\S 2$: Product sets, mappings * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Composite of Bijection with Inverse is Identity Mapping|next = Composite of Bijections is Bijection/Proof 1}}: $\S 25.1$: Some further results and examples on mappings
Exponential of Sum/Real Numbers/Lemma
== Lemma ==  Let $x, y \in \R$. Let $n \in \N_{> 0}$ such that $n > -\paren {x + y}$. Then: :$1 + \dfrac {x + y} n + \dfrac {x y} {n^2} = \paren {1 + \dfrac {x + y} n} \paren {1 + \dfrac {\paren {\frac {x y} {n + x + y} } } n}$ {{qed|lemma}}  == Proof == As $n \in \N_{> 0}$ we have that $n \ne 0$ and so the fractions in the expressions are defined. {{begin-eqn}} {{eqn | l = 1 + \frac {x + y} n + \frac {x y} {n^2}       | r = \frac {\paren {1 + \frac {x + y} n} \paren {1 + \frac {x + y} n + \frac {x y} {n^2} } } {1 + \frac {x + y} n}        | c = multiplying and dividing by $1 + \dfrac {x + y} n$ }} {{eqn | r = \paren {1 + \frac {x + y} n} \frac {\paren {1 + \frac {x + y} n + \frac {x y} {n^2} } } {\paren {1 + \frac {x + y} n} }        | c = extracting a factor }} {{eqn | r = \paren {1 + \frac {x + y} n} \paren {\frac {n^2 + n x + n y + x y} {n^2 + n x + n y} }       | c = multiplying top and bottom by $n^2$ }} {{eqn | r = \paren {1 + \frac {x + y} n} \paren {1 + \frac {x y} {n^2 + n x + n y} }       | c = Polynomial Long Division }} {{eqn | r = \paren {1 + \frac {x + y} n} \paren {1 + \frac {\paren {\frac {x y} {n + x + y} } } n}       | c = dividing top and bottom by $n + x + y$ }} {{end-eqn}} That final step is justified, as we have that $n > -\paren {x + y}$ and so $n + x + y \ne 0$.  == Also see == This lemma is used in: :Proof 2 of (Real) Exponential of Sum :Proof 3 of (Real) Exponential of Sum :$\map \exp {x + y} = \paren {\exp x} \paren {\exp y}$ for real $x$ and $y$.
Natural Logarithm of 2 is Greater than One Half
== Lemma ==  :$\ln 2 \ge \dfrac 1 2$ where $\ln$ denotes the natural logarithm function.  == Proof 1 == {{:Natural Logarithm of 2 is Greater than One Half/Proof 1}} == Proof 2 == {{:Natural Logarithm of 2 is Greater than One Half/Proof 2}} == Also see == * Logarithm Tends to Infinity == Sources == * {{BookReference|Calculus|2005|Roland E. Larson|author2 = Robert P. Hostetler|author3 = Bruce H. Edwards|ed = 8th|edpage = Eighth Edition}}: Appendix $A$: ''Properties of the Natural Logarithmic Function''
Schur's Lemma (Representation Theory)/Corollary
 == Corollary to Schur's Lemma ==  Let $\struct {G, \cdot}$ be a finite group. Let $\struct {V, \phi}$ be a $G$-module. Let the underlying field $k$ of $V$ be algebraically closed. Let: :$\map {\mathrm {End}_G} V := \leftset {f: V \to V: f}$ is a homomorphism of $G$-modules$\rightset {}$ Then: :$\map {\mathrm {End}_G} V$ is a field, with the same structure as $k$.  == Proof == Denote the identity mapping on $V$ as $I_V: V \to V$. If $f = 0$, since $0\in k$ it can be written $f = 0 I_V$. {{explain|Not clear what the above line means.}} Let $f$ be an automorphism. We have that $k$ is algebraically closed. Therefore the characteristic polynomial of $f$ is complete reducible in $k \sqbrk x$. Hence $f$ has all eigenvalue in $k$. Let $\lambda \in k$ be an eigenvalue of $f$. Consider the endomorphism: :$f - \lambda I_V: V \to V$ Because $\lambda$ is an eigenvalue: :$\map \ker {f - \lambda I_V} \ne \set 0$ From Schur's Lemma: :$f = \lambda I_V$ :$\paren {\lambda I_V} \circ \paren {\mu I_V} = \paren {\lambda \mu} I_V$ :$\lambda I_V + \paren {-\mu I_V} = \paren {\lambda - \mu} I_V$ From Subring Test: :$\map {\mathrm {End}_G} V$ is a subring of the ring endomorphisms of $V$ as an abelian group. Let $\phi: \map {\mathrm {End}_G} V \to k$ be defined as: :$\map \phi {\lambda I_V} = \lambda$ Then: :$\map \phi {\lambda I_V + \mu I_V} = \lambda + \mu = \map \phi {\lambda I_V} + \map \phi {\mu I_V}$ :$\map \phi {\paren {\lambda I_V} \circ \paren {\mu I_V} } = \lambda \mu = \map \phi {\lambda I_V} \map \phi {\mu I_V}$ Hence $\phi$ is a ring isomorphism. But since $k$ is a field it is a field isomorphism.
De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection/Corollary
== Corollary to De Morgan's Laws: Difference with Intersection == Let $S, T_1, T_2$ be sets.  Suppose that $T_1 \subseteq S$. Then: :$S \setminus \left({T_1 \cap T_2}\right) = \left({S \setminus T_1}\right) \cup \left({T_1 \setminus T_2}\right)$  == Proof == {{begin-eqn}} {{eqn | l = S \setminus \left({T_1 \cap T_2}\right)       | r = \left({S \setminus T_1}\right) \cup \left({S \setminus T_2}\right)       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \left({S \setminus T_1}\right) \cup \left({\left({\left({S \setminus T_1}\right) \cup \left({S \cap T_1}\right)}\right) \setminus T_2}\right)       | c = Set Difference Union Intersection }} {{eqn | r = \left({S \setminus T_1}\right) \cup \left({\left({S \setminus T_1}\right) \setminus T_2}\right) \cup \left({\left({S \cap T_1}\right) \setminus T_2}\right)       | c = Set Difference is Right Distributive over Union }} {{eqn | r = \left({S \setminus T_1}\right) \cup \left({\left({S \setminus T_1}\right) \setminus T_2}\right) \cup \left({T_1 \setminus T_2}\right)       | c = Intersection with Subset is Subset: $T_1 \subseteq S$ }} {{eqn | r = \left({S \setminus T_1}\right) \cup \left({T_1 \setminus T_2}\right)       | c = Set Difference Union First Set is First Set }} {{end-eqn}}
Image of Preimage under Mapping/Corollary
== Corollary to Image of Preimage under Mapping == Let $f: S \to T$ be a mapping. Then:  :$B \subseteq \Img S \implies \paren {f \circ f^{-1} } \sqbrk B = B$  == Proof == From Image of Subset under Relation is Subset of Image/Corollary 3 we have: :$B \subseteq \Img S \implies f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {\Img S}$ and from Intersection with Subset is Subset we have: :$f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {\Img S} \implies f^{-1} \sqbrk B \cap f^{-1} \sqbrk {\Img S} = f^{-1} \sqbrk B$ Hence the result.  == Sources == * {{BookReference|Rings, Modules and Linear Algebra|1970|B. Hartley|author2 = T.O. Hawkes|prev = Definition:Direct Image Mapping of Mapping|next = Image of Subset under Relation is Subset of Image/Corollary 2}}: $\S 2.2$: Homomorphisms: $\text{(i)}$
Axiom:Axiom of Countable Choice
== Axiom ==  === Form 1 === {{:Axiom:Axiom of Countable Choice/Form 1}} === Form 2 === {{:Axiom:Axiom of Countable Choice/Form 2}}  == Also known as == This axiom can be abbreviated $\mathrm{ACC}$, $\mathrm{CC}$, $\mathrm{AC}_\omega$, or $\mathrm{AC}_\N$. == Also see == * Equivalence of Forms of Axiom of Countable Choice * This axiom is a weaker form of the axiom of dependent choice, as shown in Axiom of Dependent Choice Implies Axiom of Countable Choice. Countable Choice Countable Choice 4g8kezgx135n752bu4a8j7r5ij9wckk
Kernel of Linear Transformation contains Zero Vector
== Corollary to Linear Transformation Maps Zero Vector to Zero Vector == Let $\mathbf V$ be a vector space, with zero $\mathbf 0$. Likewise let $\mathbf V'$ be another vector space, with zero $\mathbf 0'$. Let $T: \mathbf V \to \mathbf V'$ be a linear transformation. Then:  :$\mathbf 0 \in \map \ker T$ where $\map \ker T$ is the kernel of $T$.  == Proof == Follows from Linear Transformation Maps Zero Vector to Zero Vector and the definition of kernel.
Fermat's Little Theorem/Corollary 1
== Corollary to Fermat's Little Theorem ==  If $p$ is a prime number, then $n^p \equiv n \pmod p$.  == Proof 1 == {{:Fermat's Little Theorem/Corollary 1/Proof 1}} == Proof 2 == {{:Fermat's Little Theorem/Corollary 1/Proof 2}} == Also known as == Some sources call this '''Fermat's Little Theorem''', and from it derive that theorem as a corollary. == Also reported as == This result can also be reported as: :If $p$ is a prime number, then $n^p - n$ is divisible by $p$. {{Namedfor|Pierre de Fermat}} == Historical Note == {{:Fermat's Little Theorem/Historical Note}} == Sources == * {{BookReference|Men of Mathematics|1937|Eric Temple Bell|prev = Construction of Regular Prime p-Gon Exists iff p is Fermat Prime/Historical Note|next = Fermat's Little Theorem/Historical Note}}: Chapter $\text{IV}$: The Prince of Amateurs * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Fermat's Little Theorem|next = Prime Group has no Proper Subgroups}}: Exercise $25.6$ * {{BookReference|Number Theory|1971|George E. Andrews|prev = Binomial Theorem/Integral Index|next = Alternating Summation of Binomial Coefficient of Summation of Binomial Coefficient of Sequence}}: $\text {3-1}$ Permutations and Combinations: Exercise $11$ * {{BookReference|Number Theory|1971|George E. Andrews|prev = Congruence Modulo Integer/Examples/11 equiv -1 mod 12|next = Wilson's Theorem}}: $\text {4-1}$ Basic Properties of Congruences: Example $\text {4-2}$ * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Fermat's Two Squares Theorem|next = Fermat's Last Theorem}}: $1$: The Nature of Differential Equations: $\S 6$: The Brachistochrone. Fermat and the Bernoullis * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Chinese Remainder Theorem|next = Fermat's Little Theorem/Proof 1}}: $\S 1.2.4$: Integer Functions and Elementary Number Theory: Theorem $\text{F}$ * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Fermat's Little Theorem|next = Fermat's Little Theorem/Historical Note|entry = Fermat's theorem}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Fermat's Little Theorem|next = Fermat's Little Theorem/Historical Note|entry = Fermat's theorem}} * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Lagrange's Four Square Theorem|next = Fermat's Last Theorem}}: Chapter $7$: Patterns in Numbers: Fermat * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Fermat's Little Theorem|next = Feuerbach's Theorem|entry = Fermat's Little Theorem}}
Generated Sigma-Algebra by Generated Monotone Class/Corollary
== Corollary to Generated Sigma-Algebra by Generated Monotone Class ==  Let $X$ be a set, and let $\mathcal G \subseteq \mathcal P \left({X}\right)$ be a nonempty collection of subsets of $X$. Define $\complement_X \left({\mathcal G}\right)$ by: :$\complement_X \left({\mathcal G}\right) := \left\{{\complement_X \left({A}\right): A \in \mathcal G}\right\}$ Then: :$\sigma \left({\mathcal G}\right) = \mathfrak m \left({\mathcal G \cup \complement_X \left({\mathcal G}\right)}\right)$  Here, $\mathfrak m$ denotes generated monotone class, and $\sigma$ denotes generated $\sigma$-algebra. == Proof == From Set is Subset of Union: : $\mathcal G \subseteq \mathcal G \cup \complement_X \left({\mathcal G}\right)$ Further, as $\sigma \left({\mathcal G}\right)$ is a $\sigma$-algebra: :$S \in \sigma \left({\mathcal G}\right) \implies \complement_X \left({X}\right) = X \setminus S \in \sigma \left({\mathcal G}\right)$ from Set Difference as Intersection with Relative Complement. Since $\mathcal G \subseteq \sigma \left({\mathcal G}\right)$: :$\mathcal G \cup \complement_X \left({\mathcal G}\right) \subseteq \sigma \left({\mathcal G}\right)$ By Condition on Equality of Generated Sigma-Algebras: :$\sigma \left({\mathcal G}\right) = \sigma \left({\mathcal G \cup \complement_X \left({\mathcal G}\right)}\right)$ Applying Generated Sigma-Algebra by Generated Monotone Class: :$\sigma \left({\mathcal G}\right) = \mathfrak m \left({\mathcal G \cup \complement_X \left({\mathcal G}\right)}\right)$
Nth Derivative of Nth Power
== Corollary to Nth Derivative of Mth Power == Let $n \in \Z$ be an integer such that $n \ge 0$. The $n$th derivative of $x^n$ {{WRT|Differentiation}} $x$ is:  :$\dfrac {\d^n} {\d x^n} x^n = n!$ where $n!$ denotes $n$ factorial.  == Proof == From Nth Derivative of Mth Power, we have: :$\dfrac {\d^n} {\d x^n} x^m = \begin {cases} m^\underline n \, x^{m - n} & : n \le m \\ 0 & : n > m \end {cases}$ where $m^\underline n$ denotes the falling factorial. Putting $m = n$: :$\dfrac {\d^n} {\d x^n} x^n = n^\underline n$ where from the definition of the falling factorial: :$n^\underline n = n!$ Hence the result.
Unique Linear Transformation Between Vector Spaces
Unique Linear Transformation Between Vector Spaces 0 16781 147340 93014 2013-05-21T00:14:28Z Dfeuer 1672 Add to vector spaces category wikitext text/x-wiki == Corollary to Unique Linear Transformation Between Modules ==  Let $G$ be a finite-dimensional $K$-vector space. Let $H$ be a $K$-vector space (not necessarily finite-dimensional). Let $\left \langle {a_n} \right \rangle$ be a linearly independent sequence of vectors of $G$. Let $\left \langle {b_n} \right \rangle$ be a sequence of vectors of $H$. Then there is a unique linear transformation $\phi: G \to H$ satisfying $\forall k \in \left[{1 \,.\,.\, n}\right]: \phi \left({a_k}\right) = b_k$  == Proof == From Generator of Vector Space Contains Basis, $\left\{{a_1, \ldots, a_m}\right\}$ is contained in a basis for $G$. The result then follows from Unique Linear Transformation Between Modules.  == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev=Unique Linear Transformation Between Modules|next=Definition:Rank of Linear Transformation}}: $\S 28$: Theorem $28.4$: Corollary
Linear Transformations Isomorphic to Matrix Space/Corollary
=== Corollary to Linear Transformations Isomorphic to Matrix Space ===  Let $R$ be a commutative ring with unity. Let $M: \struct {\map {\mathcal L_R} G, +, \circ} \to \struct {\map {\mathcal M_R} n, +, \times}$ be defined as: :$\forall u \in \map {\mathcal L_R} G: \map M u = \sqbrk {u; \sequence {a_n} }$ Then $M$ is an isomorphism.  == Proof == Follows directly from Linear Transformations Isomorphic to Matrix Space.   == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Definition:Order of Square Matrix|next = Ring of Square Matrices over Commutative Ring with Unity}}: $\S 29$: Theorem $29.2$
Fundamental Theorem of Calculus/First Part/Corollary
== Corollary to Fundamental Theorem of Calculus (First Part) == Let $f$ be a real function which is continuous on the closed interval $\closedint a b$. Let $F$ be a real function which is defined on $\closedint a b$ by: :$\displaystyle \map F x = \int_a^x \map f t \rd t$ Then:  :$\displaystyle \frac \d {\d x} \int_a^x \map f t \rd t = \map f x$  == Proof == Follows from the Fundamental Theorem of Calculus (First Part) and the definition of primitive.  == Sources == * {{BookReference|Calculus|2005|Roland E. Larson|author2 = Robert P. Hostetler|author3 = Bruce H. Edwards|ed = 8th|edpage = Eighth Edition}}: $\S 4.4$
Ring of Integers Modulo Prime is Integral Domain
== Corollary to Ring of Integers Modulo Prime is Field ==  Let $m \in \Z: m \ge 2$. Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$. Then: :$m$ is prime : :$\struct {\Z_m, +, \times}$ is an integral domain.  == Proof 1 == {{:Ring of Integers Modulo Prime is Integral Domain/Proof 1}} == Proof 2 == {{:Ring of Integers Modulo Prime is Integral Domain/Proof 2}}
Reduced Residue System under Multiplication forms Abelian Group/Corollary
== Corollary of Reduced Residue System under Multiplication forms Abelian Group ==  Let $p$ be a prime number. Let $\Z_p$ be the set of integers modulo $p$. Let $\struct {\Z'_p, \times}$ denote the multiplicative group of reduced residues modulo $p$. Then $\struct {\Z'_p, \times}$ is an abelian group.  == Proof == Suppose $p \in \Z$ be a prime number. From the definition of reduced residue system modulo $p$, as $p$ is prime, $\Z'_p$ becomes: :$\set {\eqclass 1 p, \eqclass 2 p, \ldots, \eqclass {p - 1} p}$ This is precisely $\Z_p \setminus \set {\eqclass 0 p}$ which is what we wanted to show. The result follows from Reduced Residue System under Multiplication forms Abelian Group.  == Sources == * {{BookReference|The Theory of Groups|1968|Ian D. Macdonald|prev = Modulo Multiplication is Well-Defined/Proof 1|next = Modulo Multiplication is Associative}}: $\S 1$: Some examples of groups: Example $1.11$ * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Multiplicative Inverse in Monoid of Integers Modulo m|next = Definition:Multiplicative Group of Reduced Residues}}: Chapter $2$: Examples of Group Structure: $\S 34$ * {{BookReference|Algebra|1974|Thomas W. Hungerford|prev = Reduced Residue System under Multiplication forms Abelian Group/Proof 2|next = Definition:Group of Rationals Modulo One}}: $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups * {{BookReference|Algebra|1974|Thomas W. Hungerford|prev = Klein Four-Group/Cayley Table|next = Modulo One is Congruence Relation on Rational Numbers}}: $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Exercise $7$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Integers Modulo m under Addition form Abelian Group|next = Definition:Symmetric Group}}: $\S 34$. Examples of groups: $(3)$
Product with Ring Negative/Corollary
== Corollary to Product with Ring Negative ==  Let $\struct {R, +, \circ}$ be a ring with unity $1_R$. Then: :$\forall x \in R: \paren {-1_R} \circ x = -x$  == Proof == {{begin-eqn}} {{eqn | l = \paren {-1_R} \circ x       | r = -\paren {1_R \circ x}       | c = Product with Ring Negative }} {{eqn | r = -x       | c = {{Defof|Unity of Ring}} }} {{end-eqn}}  == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Product of Ring Negatives|next = Cancellation Law of Ring Product of Integral Domain}}: Chapter $1$: Integral Domains: $\S 4$. Elementary Properties: Theorem $2 \ \text{(vi)}$
Ring Homomorphism Preserves Subrings/Corollary
== Corollary to Ring Homomorphism Preserves Subrings == Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be rings.  The image of a ring homomorphism $\phi: R_1 \to R_2$ is a subring of $R_2$.  == Proof == From Null Ring and Ring Itself Subrings, $R_1$ is a subring of itself. The result then follows from Ring Homomorphism Preserves Subrings.  == Sources == * {{BookReference|Rings, Modules and Linear Algebra|1970|B. Hartley|author2 = T.O. Hawkes|prev = Kernel is Trivial iff Monomorphism/Ring|next = Definition:Coset}}: $\S 2.2$: Homomorphisms: Lemma $2.6 \ \text{(ii)}$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Ring Homomorphism Preserves Subrings/Proof 3|next = Kernel of Ring Homomorphism is Subring}}: $\S 57.3$ Ring homomorphisms: $\text{(i)}$