id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-12000 | Right Regular Representation wrt Right Cancellable Element on Finite Semigroup is Bijection | Let $\struct {S, \circ}$ be a finite semigroup.
Let $a \in S$ be right cancellable.
Then the right regular representation $\rho_a$ of $\struct {S, \circ}$ with respect to $a$ is a bijection. | By Right Cancellable iff Right Regular Representation Injective, $\rho_a$ is an injection.
By hypothesis, $S$ is finite.
From Injection from Finite Set to Itself is Surjection, $\rho_a$ is a surjection.
Thus $\rho_a$ is injective and surjective, and therefore a bijection.
{{qed}} | Let $\struct {S, \circ}$ be a [[Definition:Finite|finite]] [[Definition:Semigroup|semigroup]].
Let $a \in S$ be [[Definition:Right Cancellable Element|right cancellable]].
Then the [[Definition:Right Regular Representation|right regular representation]] $\rho_a$ of $\struct {S, \circ}$ with respect to $a$ is a [[Def... | By [[Right Cancellable iff Right Regular Representation Injective]], $\rho_a$ is an [[Definition:Injection|injection]].
[[Definition:By Hypothesis|By hypothesis]], $S$ is [[Definition:Finite Set|finite]].
From [[Injection from Finite Set to Itself is Surjection]], $\rho_a$ is a [[Definition:Surjection|surjection]].
... | Right Regular Representation wrt Right Cancellable Element on Finite Semigroup is Bijection | https://proofwiki.org/wiki/Right_Regular_Representation_wrt_Right_Cancellable_Element_on_Finite_Semigroup_is_Bijection | https://proofwiki.org/wiki/Right_Regular_Representation_wrt_Right_Cancellable_Element_on_Finite_Semigroup_is_Bijection | [
"Semigroups",
"Regular Representations",
"Cancellability"
] | [
"Definition:Finite",
"Definition:Semigroup",
"Definition:Cancellable Element/Right Cancellable",
"Definition:Regular Representations/Right Regular Representation",
"Definition:Bijection"
] | [
"Right Cancellable iff Right Regular Representation Injective",
"Definition:Injection",
"Definition:By Hypothesis",
"Definition:Finite Set",
"Injection from Finite Set to Itself is Surjection",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Bijection"
] |
proofwiki-12001 | Singleton is Chain | Let $\struct {S, \preceq}$ be an ordered set.
Let $x \in S$.
Then $\set x$ is a chain of $\struct {S, \preceq}$. | It suffices to prove that
:$\set x$ is connected
Let $y, z \in \set x$.
By definition of singleton:
:$y = x$ and $z = x$
By definition of reflexivity;
:$y \preceq z$
Thus
:$y \preceq z$ or $z \preceq y$
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $x \in S$.
Then $\set x$ is a [[Definition:Chain (Order Theory)|chain]] of $\struct {S, \preceq}$. | It suffices to prove that
:$\set x$ is [[Definition:Connected Relation|connected]]
Let $y, z \in \set x$.
By definition of [[Definition:Singleton|singleton]]:
:$y = x$ and $z = x$
By definition of [[Definition:Reflexivity|reflexivity]];
:$y \preceq z$
Thus
:$y \preceq z$ or $z \preceq y$
{{qed}} | Singleton is Chain | https://proofwiki.org/wiki/Singleton_is_Chain | https://proofwiki.org/wiki/Singleton_is_Chain | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Chain (Order Theory)"
] | [
"Definition:Connected Relation",
"Definition:Singleton",
"Definition:Reflexivity"
] |
proofwiki-12002 | Identity Elements occupy Diagonal of Cayley Table in Inverse Row Form | Let $\struct {G, \circ}$ be a finite group.
Let $\CC$ be a Cayley table for $\struct {G, \circ}$ presented in inverse row form.
Then all the entries in the main diagonal of $\CC$
are instances of the identity element. | By definition of inverse row form, the rows of $\CC$ are headed by the inverse elements of the elements which head the corresponding columns.
The entries in the main diagonal of $\CC$ have the same column number as row number.
Let $\sqbrk c_{k k}$ denote the entry of $\CC$ corresponding to the element where the $k$th r... | Let $\struct {G, \circ}$ be a [[Definition:Finite Group|finite group]].
Let $\CC$ be a [[Definition:Cayley Table|Cayley table]] for $\struct {G, \circ}$ presented in [[Definition:Inverse Row form of Cayley Table for Group|inverse row form]].
Then all the [[Definition:Entry of Cayley Table|entries]] in the [[Definitio... | By definition of [[Definition:Inverse Row form of Cayley Table for Group|inverse row form]], the rows of $\CC$ are headed by the [[Definition:Inverse Element|inverse elements]] of the [[Definition:Element|elements]] which head the corresponding columns.
The [[Definition:Entry of Cayley Table|entries]] in the [[Definit... | Identity Elements occupy Diagonal of Cayley Table in Inverse Row Form | https://proofwiki.org/wiki/Identity_Elements_occupy_Diagonal_of_Cayley_Table_in_Inverse_Row_Form | https://proofwiki.org/wiki/Identity_Elements_occupy_Diagonal_of_Cayley_Table_in_Inverse_Row_Form | [
"Cayley Tables"
] | [
"Definition:Finite Group",
"Definition:Cayley Table",
"Definition:Inverse Row form of Cayley Table for Group",
"Definition:Cayley Table/Entry",
"Definition:Matrix/Diagonal/Main",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Inverse Row form of Cayley Table for Group",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Element",
"Definition:Cayley Table/Entry",
"Definition:Matrix/Diagonal/Main",
"Definition:Cayley Table/Entry",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"De... |
proofwiki-12003 | Latin Square is not necessarily Cayley Table of Group | While it is true that the Cayley table of a (finite) group is in the form of a Latin square it is not necessarily the case that a Latin square is the Cayley table of a group. | Proof by Counterexample:
Let $\struct {S, \circ}$ be the algebraic structure defined by the following Cayley table:
:$\begin{array}{c|ccccc}
\circ & e & a & b & c & d \\
\hline
e & e & a & b & c & d \\
a & a & e & d & b & c \\
b & b & c & e & d & a \\
c & c & d & a & e & b \\
d & d & b & c & a & e \\
\end{array}$
... | While it is true that the [[Definition:Cayley Table|Cayley table]] of a [[Definition:Finite Group|(finite) group]] is in the form of a [[Definition:Latin Square|Latin square]] it is not necessarily the case that a [[Definition:Latin Square|Latin square]] is the [[Definition:Cayley Table|Cayley table]] of a [[Definition... | [[Proof by Counterexample]]:
Let $\struct {S, \circ}$ be the [[Definition:Algebraic Structure with One Operation|algebraic structure]] defined by the following [[Definition:Cayley Table|Cayley table]]:
:$\begin{array}{c|ccccc}
\circ & e & a & b & c & d \\
\hline
e & e & a & b & c & d \\
a & a & e & d & b & c \\
b &... | Latin Square is not necessarily Cayley Table of Group | https://proofwiki.org/wiki/Latin_Square_is_not_necessarily_Cayley_Table_of_Group | https://proofwiki.org/wiki/Latin_Square_is_not_necessarily_Cayley_Table_of_Group | [
"Cayley Tables"
] | [
"Definition:Cayley Table",
"Definition:Finite Group",
"Definition:Latin Square",
"Definition:Latin Square",
"Definition:Cayley Table",
"Definition:Group"
] | [
"Proof by Counterexample",
"Definition:Algebraic Structure/One Operation",
"Definition:Cayley Table",
"Definition:Cayley Table",
"Definition:Latin Square",
"Definition:Associative Operation",
"Definition:Group"
] |
proofwiki-12004 | Scaling preserves Modulo Addition | Let $m \in \Z_{> 0}$.
Let $x, y, c \in \Z$.
Let $x \equiv y \pmod m$.
Then:
:$c x \equiv c y \pmod m$ | Let $x \equiv y \pmod m$.
Then by definition of congruence:
:$\exists k \in Z: x - y = k m$
Hence:
:$c x - c y = c k m$
and so by definition of congruence:
:$c x \equiv c y \pmod m$
{{qed}}
Category:Modulo Addition
5ro93s15864yankv4wl8fdte2utm0ix | Let $m \in \Z_{> 0}$.
Let $x, y, c \in \Z$.
Let $x \equiv y \pmod m$.
Then:
:$c x \equiv c y \pmod m$ | Let $x \equiv y \pmod m$.
Then by definition of [[Definition:Congruence Modulo Integer|congruence]]:
:$\exists k \in Z: x - y = k m$
Hence:
:$c x - c y = c k m$
and so by definition of [[Definition:Congruence Modulo Integer|congruence]]:
:$c x \equiv c y \pmod m$
{{qed}}
[[Category:Modulo Addition]]
5ro93s15864yank... | Scaling preserves Modulo Addition | https://proofwiki.org/wiki/Scaling_preserves_Modulo_Addition | https://proofwiki.org/wiki/Scaling_preserves_Modulo_Addition | [
"Modulo Addition"
] | [] | [
"Definition:Congruence (Number Theory)/Integers",
"Definition:Congruence (Number Theory)/Integers",
"Category:Modulo Addition"
] |
proofwiki-12005 | Modulo Addition is Linear | Let $m \in \Z_{> 0}$.
Let $x_1, x_2, y_1, y_2, c_1, c_2 \in \Z$.
Let:
:$x_1 \equiv y_1 \pmod m$
:$x_2 \equiv y_2 \pmod m$
Then:
:$c_1 x_1 + c_2 x_2 \equiv c_1 y_1 + c_2 y_2 \pmod m$ | By Scaling preserves Modulo Addition:
:$c_1 x_1 \equiv c_1 y_1 \pmod m$
:$c_2 x_2 \equiv c_2 y_2 \pmod m$
and so by Modulo Addition is Well-Defined:
:$c_1 x_1 + c_2 x_2 \equiv c_1 y_1 + c_2 y_2 \pmod m$
{{qed}} | Let $m \in \Z_{> 0}$.
Let $x_1, x_2, y_1, y_2, c_1, c_2 \in \Z$.
Let:
:$x_1 \equiv y_1 \pmod m$
:$x_2 \equiv y_2 \pmod m$
Then:
:$c_1 x_1 + c_2 x_2 \equiv c_1 y_1 + c_2 y_2 \pmod m$ | By [[Scaling preserves Modulo Addition]]:
:$c_1 x_1 \equiv c_1 y_1 \pmod m$
:$c_2 x_2 \equiv c_2 y_2 \pmod m$
and so by [[Modulo Addition is Well-Defined]]:
:$c_1 x_1 + c_2 x_2 \equiv c_1 y_1 + c_2 y_2 \pmod m$
{{qed}} | Modulo Addition is Linear | https://proofwiki.org/wiki/Modulo_Addition_is_Linear | https://proofwiki.org/wiki/Modulo_Addition_is_Linear | [
"Modulo Addition"
] | [] | [
"Scaling preserves Modulo Addition",
"Modulo Addition is Well-Defined"
] |
proofwiki-12006 | Euler Phi Function of Prime | Let $p$ be a prime number.
Then:
:$\map \phi p = p - 1$
where $\phi: \Z_{>0} \to \Z_{>0}$ is the Euler $\phi$ function. | From the definition of a prime number, the only (strictly) positive integer less than or equal to a prime $p$ which is ''not'' prime to $p$ is $p$ itself.
Thus it follows directly that:
:$\map \phi p = p - 1$
{{qed}} | Let $p$ be a [[Definition:Prime Number|prime number]].
Then:
:$\map \phi p = p - 1$
where $\phi: \Z_{>0} \to \Z_{>0}$ is the [[Definition:Euler Phi Function|Euler $\phi$ function]]. | From the definition of a [[Definition:Prime Number|prime number]], the only [[Definition:Strictly Positive Integer|(strictly) positive integer]] less than or equal to a [[Definition:Prime Number|prime]] $p$ which is ''not'' [[Definition:Coprime Integers|prime to]] $p$ is $p$ itself.
Thus it follows directly that:
:$\m... | Euler Phi Function of Prime | https://proofwiki.org/wiki/Euler_Phi_Function_of_Prime | https://proofwiki.org/wiki/Euler_Phi_Function_of_Prime | [
"Euler Phi Function",
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Euler Phi Function"
] | [
"Definition:Prime Number",
"Definition:Strictly Positive/Integer",
"Definition:Prime Number",
"Definition:Coprime/Integers"
] |
proofwiki-12007 | Chain is Directed | Let $\struct {S, \preceq}$ be an ordered set.
Let $C$ be a non-empty chain of $S$.
Then $C$ is directed. | Let $x, y \in C$.
By definition of connected relation:
:$x \preceq y$ or $y \preceq x$
{{WLOG}}, suppose that
:$x \preceq y$
Define $z = y$.
Thus by definition of reflexivity
:$x \preceq z$ and $y \preceq z$
Hence $C$ is directed.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $C$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Chain (Order Theory)|chain]] of $S$.
Then $C$ is [[Definition:Directed Subset|directed]]. | Let $x, y \in C$.
By definition of [[Definition:Connected Relation|connected relation]]:
:$x \preceq y$ or $y \preceq x$
{{WLOG}}, suppose that
:$x \preceq y$
Define $z = y$.
Thus by definition of [[Definition:Reflexivity|reflexivity]]
:$x \preceq z$ and $y \preceq z$
Hence $C$ is [[Definition:Directed Subset|dire... | Chain is Directed | https://proofwiki.org/wiki/Chain_is_Directed | https://proofwiki.org/wiki/Chain_is_Directed | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Non-Empty Set",
"Definition:Chain (Order Theory)",
"Definition:Directed Subset"
] | [
"Definition:Connected Relation",
"Definition:Reflexivity",
"Definition:Directed Subset"
] |
proofwiki-12008 | Reduced Residue System Modulo Prime | Let $p$ be a prime number.
The reduced residue system modulo $p$ contains $p - 1$ elements:
:$Z'_p = \set {\eqclass 1 m, \eqclass 2 m, \ldots, \eqclass {p - 1} m}$
and so can be defined as:
:$Z'_p = Z_p \setminus \set {\eqclass 0 m}$ | From Prime not Divisor implies Coprime, each of $1, 2, \ldots, p - 1$ is coprime to $p$.
The result follows by definition of reduced residue system modulo $p$.
{{qed}} | Let $p$ be a [[Definition:Prime Number|prime number]].
The [[Definition:Reduced Residue System|reduced residue system modulo $p$]] contains $p - 1$ [[Definition:Element|elements]]:
:$Z'_p = \set {\eqclass 1 m, \eqclass 2 m, \ldots, \eqclass {p - 1} m}$
and so can be defined as:
:$Z'_p = Z_p \setminus \set {\eqclass ... | From [[Prime not Divisor implies Coprime]], each of $1, 2, \ldots, p - 1$ is [[Definition:Coprime Integers|coprime]] to $p$.
The result follows by definition of [[Definition:Reduced Residue System|reduced residue system modulo $p$]].
{{qed}} | Reduced Residue System Modulo Prime | https://proofwiki.org/wiki/Reduced_Residue_System_Modulo_Prime | https://proofwiki.org/wiki/Reduced_Residue_System_Modulo_Prime | [
"Reduced Residue Systems",
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Reduced Residue System",
"Definition:Element"
] | [
"Prime not Divisor implies Coprime",
"Definition:Coprime/Integers",
"Definition:Reduced Residue System"
] |
proofwiki-12009 | Modulo Multiplication on Reduced Residue System is Closed | Let $m \in \Z_{> 0}$ be a (strictly) positive integer.
Let $\Z'_m$ be the reduced residue system modulo $m$:
:$\Z'_m = \set {\eqclass k m \in \Z_m: k \perp m}$
Let $S = \struct {\Z'_m, \times_m}$ be the algebraic structure consisting of $\Z'_m$ under the modulo multiplication.
Then $S$ is closed, in the sense that:
:$\... | Let $\eqclass r m, \eqclass s m \in \Z'_m$.
Then by definition of reduced residue system:
:$r, s \perp m$
By Bézout's Identity:
:$\exists u_1, v_1 \in \Z: u_1 r + v_1 m = 1$
:$\exists u_2, v_2 \in \Z: u_2 s + v_2 m = 1$
Then:
{{begin-eqn}}
{{eqn | l = \paren {u_1 r + v_1 m} \paren {u_2 s + v_2 m}
| r = u_1 u_2 r ... | Let $m \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $\Z'_m$ be the [[Definition:Reduced Residue System|reduced residue system modulo $m$]]:
:$\Z'_m = \set {\eqclass k m \in \Z_m: k \perp m}$
Let $S = \struct {\Z'_m, \times_m}$ be the [[Definition:Algebraic Structure wi... | Let $\eqclass r m, \eqclass s m \in \Z'_m$.
Then by definition of [[Definition:Reduced Residue System|reduced residue system]]:
:$r, s \perp m$
By [[Bézout's Identity]]:
:$\exists u_1, v_1 \in \Z: u_1 r + v_1 m = 1$
:$\exists u_2, v_2 \in \Z: u_2 s + v_2 m = 1$
Then:
{{begin-eqn}}
{{eqn | l = \paren {u_1 r + v_1 m} ... | Modulo Multiplication on Reduced Residue System is Closed | https://proofwiki.org/wiki/Modulo_Multiplication_on_Reduced_Residue_System_is_Closed | https://proofwiki.org/wiki/Modulo_Multiplication_on_Reduced_Residue_System_is_Closed | [
"Modulo Multiplication",
"Reduced Residue Systems"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Reduced Residue System",
"Definition:Algebraic Structure/One Operation",
"Definition:Modulo Multiplication",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Definition:Reduced Residue System",
"Bézout's Identity",
"Bézout's Identity",
"Definition:Coprime/Integers",
"Definition:Group Product/Product Element",
"Definition:Element",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-12010 | Modulo Multiplication on Reduced Residue System is Cancellable | Let $m \in \Z_{> 0}$ be a (strictly) positive integer.
Let $\Z'_m$ be the reduced residue system modulo $m$:
:$\Z'_m = \set {\eqclass k m \in \Z_m: k \perp m}$
Let $S = \struct {\Z'_m, \times_m}$ be the algebraic structure consisting of $\Z'_m$ under modulo multiplication.
Then $\times_m$ is cancellable, in the sense t... | Let $a, b, c \in \Z'_m$ such that $a \times_m c = b \times_m c$
Let $p, q, r$ be integers such that:
:$p \in a$
:$q \in b$
:$r \in c$
By definition of residue class, this means:
{{begin-eqn}}
{{eqn | l = p r
| o = \equiv
| r = q r
| rr= \pmod m
| c =
}}
{{eqn | ll= \leadsto
| l = p
... | Let $m \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $\Z'_m$ be the [[Definition:Reduced Residue System|reduced residue system modulo $m$]]:
:$\Z'_m = \set {\eqclass k m \in \Z_m: k \perp m}$
Let $S = \struct {\Z'_m, \times_m}$ be the [[Definition:Algebraic Structure wi... | Let $a, b, c \in \Z'_m$ such that $a \times_m c = b \times_m c$
Let $p, q, r$ be [[Definition:Integer|integers]] such that:
:$p \in a$
:$q \in b$
:$r \in c$
By definition of [[Definition:Residue Class|residue class]], this means:
{{begin-eqn}}
{{eqn | l = p r
| o = \equiv
| r = q r
| rr= \pmod m
... | Modulo Multiplication on Reduced Residue System is Cancellable | https://proofwiki.org/wiki/Modulo_Multiplication_on_Reduced_Residue_System_is_Cancellable | https://proofwiki.org/wiki/Modulo_Multiplication_on_Reduced_Residue_System_is_Cancellable | [
"Modulo Multiplication",
"Reduced Residue Systems"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Reduced Residue System",
"Definition:Algebraic Structure/One Operation",
"Definition:Modulo Multiplication",
"Definition:Cancellable Operation"
] | [
"Definition:Integer",
"Definition:Residue Class",
"Cancellability of Congruences/Corollary 1"
] |
proofwiki-12011 | Order Generating iff Every Element is Infimum | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $X$ be a subset of $S$.
Then
:$X$ is order generating
{{iff}}
:$\forall x \in S: \exists Y \subseteq X: x = \inf Y$ | === Sufficient Condition ===
Let $X$ be order generating.
Let $x \in S$.
By definition of order generating:
:$x = \map \inf {x^\succeq \cap X}$
Thus
:$\exists Y \subseteq X: x = \inf Y$
{{qed|lemma}} | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $X$ be a [[Definition:Subset|subset]] of $S$.
Then
:$X$ is [[Definition:Order Generating Subset|order generating]]
{{iff}}
:$\forall x \in S: \exists Y \subseteq X: x = \inf Y$ | === Sufficient Condition ===
Let $X$ be [[Definition:Order Generating Subset|order generating]].
Let $x \in S$.
By definition of [[Definition:Order Generating Subset|order generating]]:
:$x = \map \inf {x^\succeq \cap X}$
Thus
:$\exists Y \subseteq X: x = \inf Y$
{{qed|lemma}} | Order Generating iff Every Element is Infimum | https://proofwiki.org/wiki/Order_Generating_iff_Every_Element_is_Infimum | https://proofwiki.org/wiki/Order_Generating_iff_Every_Element_is_Infimum | [
"Complete Lattices",
"Order Generating Subsets"
] | [
"Definition:Complete Lattice",
"Definition:Subset",
"Definition:Order Generating Subset"
] | [
"Definition:Order Generating Subset",
"Definition:Order Generating Subset"
] |
proofwiki-12012 | Weierstrass Factorization Theorem | Let $f$ be an entire function.
Let $0$ be a zero of $f$ of multiplicity $m \ge 0$.
Let the sequence $\sequence {a_n}$ consist of the nonzero zeroes of $f$, repeated according to multiplicity. | From Weierstrass Product Theorem, the function:
:$\ds \map h z = z^m \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$
defines an entire function that has the same zeroes as $f$ counting multiplicity.
Thus $f / h$ is both an entire function and non-vanishing.
As $f / h$ is both holomorphic and nowhere zer... | Let $f$ be an [[Definition:Entire Function|entire function]].
Let $0$ be a zero of $f$ of [[Definition:Multiplicity (Analytic Function)|multiplicity]] $m \ge 0$.
Let the [[Definition:Sequence|sequence]] $\sequence {a_n}$ consist of the nonzero [[Definition:Zero of Function|zeroes]] of $f$, repeated according to [[Def... | From [[Weierstrass Product Theorem]], the [[Definition:Complex Function|function]]:
:$\ds \map h z = z^m \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$
defines an [[Definition:Entire Function|entire function]] that has the same [[Definition:Zero of Function|zeroes]] as $f$ counting [[Definition:Multip... | Weierstrass Factorization Theorem | https://proofwiki.org/wiki/Weierstrass_Factorization_Theorem | https://proofwiki.org/wiki/Weierstrass_Factorization_Theorem | [
"Complex Analysis",
"Infinite Products",
"Entire Functions"
] | [
"Definition:Entire Function",
"Definition:Multiplicity (Complex Analysis)",
"Definition:Sequence",
"Definition:Root of Mapping",
"Definition:Multiplicity (Complex Analysis)",
"Definition:Sequence",
"Definition:Entire Function",
"Definition:Sequence",
"Definition:Entire Function"
] | [
"Weierstrass Product Theorem",
"Definition:Complex Function",
"Definition:Entire Function",
"Definition:Root of Mapping",
"Definition:Multiplicity (Complex Analysis)",
"Definition:Entire Function",
"Definition:Holomorphic Function",
"Definition:Holomorphic Function"
] |
proofwiki-12013 | Equivalence of Axiom Schemata for Finite Group | The following axiom schemata for the definition of a finite group are logically equivalent: | Let $S$ be an algebraic structure of finite order which fulfils the group axioms.
It is to be shown that $S$ also fulfils all the finite group axioms.
We have that $\text {FG} 0$ and $\text {FG} 1$ are the same as $\text G 0$ and $\text G 1$ and so {{apriori}} $\text {FG} 0$ and $\text {FG} 1$ are fulfilled.
We also ha... | The following [[Definition:Axiom Schema|axiom schemata]] for the definition of a [[Definition:Finite Group|finite group]] are [[Definition:Logical Equivalence|logically equivalent]]: | Let $S$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] of [[Definition:Finite Order (Structure)|finite order]] which fulfils the [[Axiom:Group Axioms|group axioms]].
It is to be shown that $S$ also fulfils all the [[Axiom:Finite Group Axioms|finite group axioms]].
We have that $\text ... | Equivalence of Axiom Schemata for Finite Group | https://proofwiki.org/wiki/Equivalence_of_Axiom_Schemata_for_Finite_Group | https://proofwiki.org/wiki/Equivalence_of_Axiom_Schemata_for_Finite_Group | [
"Finite Groups"
] | [
"Definition:Axiom/Formal Systems/Axiom Schema",
"Definition:Finite Group",
"Definition:Logical Equivalence"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Order of Structure/Finite Structure",
"Axiom:Group Axioms",
"Axiom:Finite Group/Axioms",
"Definition:Order of Structure/Finite Structure",
"Definition:Strictly Positive/Integer",
"Definition:Element",
"Axiom:Group Axioms",
"Definition:Group... |
proofwiki-12014 | Order Generating iff Every Superset Closed on Infima is Whole Space | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $X$ be a subset of $S$.
Then
:$X$ is order generating
{{iff}}
:$\forall Y \subseteq S: Y \supseteq X \land \paren {\forall Z \subseteq Y: \inf Z \in Y} \implies Y = S$ | === Sufficient Condition ===
Let $X$ be order generating.
Let $Y \subseteq S$ such that
:$Y \supseteq X \land \paren {\forall Z \subseteq Y: \inf Z \in Y}$
We will prove that
:$S \subseteq Y$
Let $s \in S$.
By Order Generating iff Every Element is Infimum:
:$\exists Z \subseteq X: s = \inf Z$
By Subset Relation is Tran... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $X$ be a [[Definition:Subset|subset]] of $S$.
Then
:$X$ is [[Definition:Order Generating Subset|order generating]]
{{iff}}
:$\forall Y \subseteq S: Y \supseteq X \land \paren {\forall Z \subseteq Y: \inf Z \in Y} ... | === Sufficient Condition ===
Let $X$ be [[Definition:Order Generating Subset|order generating]].
Let $Y \subseteq S$ such that
:$Y \supseteq X \land \paren {\forall Z \subseteq Y: \inf Z \in Y}$
We will prove that
:$S \subseteq Y$
Let $s \in S$.
By [[Order Generating iff Every Element is Infimum]]:
:$\exists Z \su... | Order Generating iff Every Superset Closed on Infima is Whole Space | https://proofwiki.org/wiki/Order_Generating_iff_Every_Superset_Closed_on_Infima_is_Whole_Space | https://proofwiki.org/wiki/Order_Generating_iff_Every_Superset_Closed_on_Infima_is_Whole_Space | [
"Complete Lattices",
"Order Generating Subsets"
] | [
"Definition:Complete Lattice",
"Definition:Subset",
"Definition:Order Generating Subset"
] | [
"Definition:Order Generating Subset",
"Order Generating iff Every Element is Infimum",
"Subset Relation is Transitive",
"Definition:Set Equality",
"Definition:Set Equality",
"Definition:Order Generating Subset",
"Order Generating iff Every Element is Infimum"
] |
proofwiki-12015 | Group Isomorphism Preserves Inverses | Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group isomorphism.
Let:
:$e_G$ be the identity of $\struct {G, \circ}$
:$e_H$ be the identity of $\struct {H, *}$.
Then:
: $\forall g \in G: \map \phi {g^{-1} } = \paren {\map \phi g}^{-1}$ | An group isomorphism is by definition a group epimorphism.
The result follows from Epimorphism Preserves Inverses.
{{Qed}} | Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Group Isomorphism|group isomorphism]].
Let:
:$e_G$ be the [[Definition:Identity Element|identity]] of $\struct {G, \circ}$
:$e_H$ be the [[Definition:Identity Element|identity]] of $\struct {H, *}$.
Then:
: $\forall g \in G: \map \phi {g^{-1} } = \p... | An [[Definition:Group Isomorphism|group isomorphism]] is by definition a [[Definition:Group Epimorphism|group epimorphism]].
The result follows from [[Epimorphism Preserves Inverses]].
{{Qed}} | Group Isomorphism Preserves Inverses/Proof 1 | https://proofwiki.org/wiki/Group_Isomorphism_Preserves_Inverses | https://proofwiki.org/wiki/Group_Isomorphism_Preserves_Inverses/Proof_1 | [
"Group Isomorphisms",
"Inverse Elements",
"Group Isomorphism Preserves Inverses"
] | [
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Group Epimorphism",
"Epimorphism Preserves Inverses"
] |
proofwiki-12016 | Group Isomorphism Preserves Inverses | Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group isomorphism.
Let:
:$e_G$ be the identity of $\struct {G, \circ}$
:$e_H$ be the identity of $\struct {H, *}$.
Then:
: $\forall g \in G: \map \phi {g^{-1} } = \paren {\map \phi g}^{-1}$ | Let $g \in G$.
{{begin-eqn}}
{{eqn | l = \map \phi g * \map \phi {g^{-1} }
| r = \map \phi {g \circ g^{-1} }
| c = {{Defof|Group Isomorphism}}
}}
{{eqn | r = \map \phi {e_G}
| c = {{Defof|Inverse Element}}
}}
{{eqn | r = e_H
| c = Group Isomorphism Preserves Identity
}}
{{end-eqn}}
It follows fr... | Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Group Isomorphism|group isomorphism]].
Let:
:$e_G$ be the [[Definition:Identity Element|identity]] of $\struct {G, \circ}$
:$e_H$ be the [[Definition:Identity Element|identity]] of $\struct {H, *}$.
Then:
: $\forall g \in G: \map \phi {g^{-1} } = \p... | Let $g \in G$.
{{begin-eqn}}
{{eqn | l = \map \phi g * \map \phi {g^{-1} }
| r = \map \phi {g \circ g^{-1} }
| c = {{Defof|Group Isomorphism}}
}}
{{eqn | r = \map \phi {e_G}
| c = {{Defof|Inverse Element}}
}}
{{eqn | r = e_H
| c = [[Group Isomorphism Preserves Identity]]
}}
{{end-eqn}}
It foll... | Group Isomorphism Preserves Inverses/Proof 2 | https://proofwiki.org/wiki/Group_Isomorphism_Preserves_Inverses | https://proofwiki.org/wiki/Group_Isomorphism_Preserves_Inverses/Proof_2 | [
"Group Isomorphisms",
"Inverse Elements",
"Group Isomorphism Preserves Inverses"
] | [
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Group Isomorphism Preserves Identity",
"Inverse in Group is Unique",
"Definition:Unique",
"Definition:Inverse (Abstract Algebra)/Inverse"
] |
proofwiki-12017 | Preimage of Composite Relation | Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be relations.
Let $\RR_2 \circ \RR_1 \subseteq S_1 \times T_2$ be the composition of $\RR_1$ and $\RR_2$.
Then the preimage of $\RR_2 \circ \RR_1$ is given by:
:$\Preimg {\RR_2 \circ \RR_1} = \Preimg {\Img {\RR_1} \cap \Preimg {\RR_2} }$ | {{proof wanted}}
Category:Composite Relations
6lh1ebouq13pwncjkma96kpqrizpu1t | Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be [[Definition:Relation|relations]].
Let $\RR_2 \circ \RR_1 \subseteq S_1 \times T_2$ be the [[Definition:Composition of Relations|composition of $\RR_1$ and $\RR_2$]].
Then the [[Definition:Preimage of Relation|preimage]] of $\RR_2 \circ \RR... | {{proof wanted}}
[[Category:Composite Relations]]
6lh1ebouq13pwncjkma96kpqrizpu1t | Preimage of Composite Relation | https://proofwiki.org/wiki/Preimage_of_Composite_Relation | https://proofwiki.org/wiki/Preimage_of_Composite_Relation | [
"Composite Relations"
] | [
"Definition:Relation",
"Definition:Composition of Relations",
"Definition:Preimage/Relation/Relation"
] | [
"Category:Composite Relations"
] |
proofwiki-12018 | Image of Composite Relation | Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be relations.
Let $\RR_2 \circ \RR_1 \subseteq S_1 \times T_2$ be the composition of $\RR_1$ and $\RR_2$.
Then the image of $\RR_2 \circ \RR_1$ is given by:
:$\Img {\RR_2 \circ \RR_1} = \Img {\Img {\RR_1} \cap \Preimg {\RR_2} }$ | We have by definition of composition of $\RR_1$ and $\RR_2$:
:$\RR_2 \circ \RR_1 := \set {\tuple {x, z} \in S_1 \times T_2: \exists y \in S_2 \cap T_1: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$
Let $\left({x, z}\right) \in \RR_2 \circ \RR_1$.
By definition of image:
:$z \in \Img {\RR_2 \circ \RR_1}$
By de... | Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be [[Definition:Relation|relations]].
Let $\RR_2 \circ \RR_1 \subseteq S_1 \times T_2$ be the [[Definition:Composition of Relations|composition of $\RR_1$ and $\RR_2$]].
Then the [[Definition:Image of Relation|image]] of $\RR_2 \circ \RR_1$ is... | We have by definition of [[Definition:Composition of Relations|composition of $\RR_1$ and $\RR_2$]]:
:$\RR_2 \circ \RR_1 := \set {\tuple {x, z} \in S_1 \times T_2: \exists y \in S_2 \cap T_1: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$
Let $\left({x, z}\right) \in \RR_2 \circ \RR_1$.
By definition of [[D... | Image of Composite Relation | https://proofwiki.org/wiki/Image_of_Composite_Relation | https://proofwiki.org/wiki/Image_of_Composite_Relation | [
"Composite Relations"
] | [
"Definition:Relation",
"Definition:Composition of Relations",
"Definition:Image (Set Theory)/Relation/Relation"
] | [
"Definition:Composition of Relations",
"Definition:Image (Set Theory)/Relation/Relation",
"Definition:Composition of Relations",
"Category:Composite Relations"
] |
proofwiki-12019 | Weierstrass Product Theorem | Let $\sequence {a_k}$ be a sequence of non-zero complex numbers such that:
:$\cmod {a_n} \to \infty$ as $n \to \infty$
Let $\sequence {p_n}$ be a sequence of non-negative integers for which the series:
:$\ds \sum_{n \mathop = 1}^\infty \size {\dfrac r {a_n} }^{1 + p_n}$
converges for every $r \in \R_{> 0}$.
Let:
:$\ds ... | By:
:Locally Uniformly Absolutely Convergent Product is Locally Uniformly Convergent
:Infinite Product of Analytic Functions is Analytic
:Zeroes of Infinite Product of Analytic Functions
it suffices to show that the product $\ds \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$ converges locally uniformly ... | Let $\sequence {a_k}$ be a [[Definition:Sequence|sequence]] of non-zero [[Definition:Complex Number|complex numbers]] such that:
:$\cmod {a_n} \to \infty$ as $n \to \infty$
Let $\sequence {p_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Non-Negative Integer|non-negative integers]] for which the [[Definitio... | By:
:[[Locally Uniformly Absolutely Convergent Product is Locally Uniformly Convergent]]
:[[Infinite Product of Analytic Functions is Analytic]]
:[[Zeroes of Infinite Product of Analytic Functions]]
it suffices to show that the product $\ds \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$ [[Definition:Lo... | Weierstrass Product Theorem | https://proofwiki.org/wiki/Weierstrass_Product_Theorem | https://proofwiki.org/wiki/Weierstrass_Product_Theorem | [
"Infinite Products"
] | [
"Definition:Sequence",
"Definition:Complex Number",
"Definition:Sequence",
"Definition:Positive/Integer",
"Definition:Series",
"Definition:Convergent Series",
"Definition:Weierstrass Elementary Factor",
"Definition:Entire Function",
"Definition:Root of Mapping",
"Definition:Multiplicity (Complex A... | [
"Locally Uniformly Absolutely Convergent Product is Locally Uniformly Convergent",
"Infinite Product of Analytic Functions is Analytic",
"Zeroes of Infinite Product of Analytic Functions",
"Definition:Locally Uniform Absolute Convergence of Product",
"Bounds for Weierstrass Elementary Factors",
"Weierstra... |
proofwiki-12020 | Isomorphism between Gaussian Integer Units and Rotation Matrices Order 4 | Let $\struct {U_\C, \times}$ be the group of Gaussian integer units under complex multiplication.
Let $\struct {R_4, \times}$ be the group of rotation matrices of order $4$ under modulo addition.
Then $\struct {U_\C, \times}$ and $\struct {R_4, \times}$ are isomorphic algebraic structures. | Establish the mapping $f: U_C \to R_4$ as follows:
{{begin-eqn}}
{{eqn | l = \map f 1
| r = r_0
}}
{{eqn | l = \map f i
| r = r_1
}}
{{eqn | l = \map f {-1}
| r = r_2
}}
{{eqn | l = \map f {-i}
| r = r_3
}}
{{end-eqn}}
From Isomorphism by Cayley Table, the two Cayley tables can be compared by ey... | Let $\struct {U_\C, \times}$ be the [[Definition:Group of Gaussian Integer Units|group of Gaussian integer units]] under [[Definition:Complex Multiplication|complex multiplication]].
Let $\struct {R_4, \times}$ be the [[Definition:Group of Rotation Matrices Order 4|group of rotation matrices of order $4$]] under [[Def... | Establish the [[Definition:Mapping|mapping]] $f: U_C \to R_4$ as follows:
{{begin-eqn}}
{{eqn | l = \map f 1
| r = r_0
}}
{{eqn | l = \map f i
| r = r_1
}}
{{eqn | l = \map f {-1}
| r = r_2
}}
{{eqn | l = \map f {-i}
| r = r_3
}}
{{end-eqn}}
From [[Isomorphism by Cayley Table]], the two [[Def... | Isomorphism between Gaussian Integer Units and Rotation Matrices Order 4 | https://proofwiki.org/wiki/Isomorphism_between_Gaussian_Integer_Units_and_Rotation_Matrices_Order_4 | https://proofwiki.org/wiki/Isomorphism_between_Gaussian_Integer_Units_and_Rotation_Matrices_Order_4 | [
"Examples of Group Isomorphisms/Order 4",
"Group of Rotation Matrices Order 4",
"Group of Gaussian Integer Units"
] | [
"Definition:Group of Gaussian Integer Units",
"Definition:Multiplication/Complex Numbers",
"Definition:Group of Rotation Matrices Order 4",
"Definition:Modulo Addition",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Mapping",
"Isomorphism by Cayley Table",
"Definition:Cayley Table",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Cayley Table",
"Definition:Cayley Table"
] |
proofwiki-12021 | Isomorphism between Gaussian Integer Units and Reduced Residue System Modulo 5 under Multiplication | Let $\struct {U_\C, \times}$ be the group of Gaussian integer units under complex multiplication.
Let $\struct {\Z'_5, \times_5}$ be the multiplicative group of reduced residues modulo $5$.
Then $\struct {U_\C, \times}$ and $\struct {\Z'_5, \times_5}$ are isomorphic algebraic structures. | Establish the mapping $f: U_C \to \Z'_5$ as follows:
{{begin-eqn}}
{{eqn | l = \map f 1
| r = \eqclass 1 5
}}
{{eqn | l = \map f i
| r = \eqclass 2 5
}}
{{eqn | l = \map f {-1}
| r = \eqclass 4 5
}}
{{eqn | l = \map f {-i}
| r = \eqclass 3 5
}}
{{end-eqn}}
From Isomorphism by Cayley Table, the t... | Let $\struct {U_\C, \times}$ be the [[Definition:Group of Gaussian Integer Units|group of Gaussian integer units]] under [[Definition:Complex Multiplication|complex multiplication]].
Let $\struct {\Z'_5, \times_5}$ be the [[Multiplicative Group of Reduced Residues Modulo 5|multiplicative group of reduced residues modu... | Establish the [[Definition:Mapping|mapping]] $f: U_C \to \Z'_5$ as follows:
{{begin-eqn}}
{{eqn | l = \map f 1
| r = \eqclass 1 5
}}
{{eqn | l = \map f i
| r = \eqclass 2 5
}}
{{eqn | l = \map f {-1}
| r = \eqclass 4 5
}}
{{eqn | l = \map f {-i}
| r = \eqclass 3 5
}}
{{end-eqn}}
From [[Isomor... | Isomorphism between Gaussian Integer Units and Reduced Residue System Modulo 5 under Multiplication | https://proofwiki.org/wiki/Isomorphism_between_Gaussian_Integer_Units_and_Reduced_Residue_System_Modulo_5_under_Multiplication | https://proofwiki.org/wiki/Isomorphism_between_Gaussian_Integer_Units_and_Reduced_Residue_System_Modulo_5_under_Multiplication | [
"Examples of Group Isomorphisms/Order 4",
"Multiplicative Group of Reduced Residues Modulo 5",
"Group of Gaussian Integer Units"
] | [
"Definition:Group of Gaussian Integer Units",
"Definition:Multiplication/Complex Numbers",
"Multiplicative Group of Reduced Residues/Examples/Modulo 5",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Mapping",
"Isomorphism by Cayley Table",
"Definition:Cayley Table",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Cayley Table",
"Definition:Cayley Table"
] |
proofwiki-12022 | Group Generated by Reciprocal of z and 1 minus z | Let $\struct {S, \circ}$ denote the '''group generated by $\dfrac 1 z$ and $1 - z$'''.
Then $\struct {S, \circ}$ is a finite group of order $6$. | By definition:
:$S = \set {f_1, f_2, f_3, f_4, f_5, f_6}$
where $f_1, f_2, \ldots, f_6$ are complex functions defined for all $z \in \C \setminus \set {0, 1}$ as:
{{begin-eqn}}
{{eqn | l = \map {f_1} z
| r = z
}}
{{eqn | l = \map {f_2} z
| r = \dfrac 1 {1 - z}
}}
{{eqn | l = \map {f_3} z
| r = \dfrac ... | Let $\struct {S, \circ}$ denote the '''[[Definition:Group Generated by Reciprocal of z and 1 minus z|group generated by $\dfrac 1 z$ and $1 - z$]]'''.
Then $\struct {S, \circ}$ is a [[Definition:Finite Group|finite group]] of [[Definition:Order of Structure|order $6$]]. | By definition:
:$S = \set {f_1, f_2, f_3, f_4, f_5, f_6}$
where $f_1, f_2, \ldots, f_6$ are [[Definition:Complex Function|complex functions]] defined for all $z \in \C \setminus \set {0, 1}$ as:
{{begin-eqn}}
{{eqn | l = \map {f_1} z
| r = z
}}
{{eqn | l = \map {f_2} z
| r = \dfrac 1 {1 - z}
}}
{{eqn | l =... | Group Generated by Reciprocal of z and 1 minus z | https://proofwiki.org/wiki/Group_Generated_by_Reciprocal_of_z_and_1_minus_z | https://proofwiki.org/wiki/Group_Generated_by_Reciprocal_of_z_and_1_minus_z | [
"Groups of Order 6",
"Group Generated by Reciprocal of z and 1 minus z"
] | [
"Definition:Group Generated by Reciprocal of z and 1 minus z",
"Definition:Finite Group",
"Definition:Order of Structure"
] | [
"Definition:Complex Function",
"Definition:Composition of Mappings",
"Definition:Cayley Table",
"Definition:Composition of Mappings",
"Definition:Composition of Mappings",
"Group Generated by Reciprocal of z and 1 minus z/Cayley Table",
"Axiom:Group Axioms",
"Axiom:Group Axioms"
] |
proofwiki-12023 | Group Generated by Reciprocal of z and Minus z is Klein Four-Group | Let $K_4$ denote the Klein $4$-group.
Let $S$ be the group generated by $1 / z$ and $-z$.
Then $K_4$ and $S$ are isomorphic algebraic structures. | Establish the mapping $\phi: K_4 \to S$ as follows:
{{begin-eqn}}
{{eqn | l = \map \phi e
| r = z
}}
{{eqn | l = \map \phi a
| r = -z
}}
{{eqn | l = \map \phi b
| r = \dfrac 1 z
}}
{{eqn | l = \map \phi c
| r = -\dfrac 1 z
}}
{{end-eqn}}
From Isomorphism by Cayley Table, the two Cayley tables ca... | Let $K_4$ denote the [[Definition:Klein Four-Group|Klein $4$-group]].
Let $S$ be the [[Group Generated by Reciprocal of z and Minus z|group generated by $1 / z$ and $-z$]].
Then $K_4$ and $S$ are [[Definition:Isomorphic Algebraic Structures|isomorphic algebraic structures]]. | Establish the [[Definition:Mapping|mapping]] $\phi: K_4 \to S$ as follows:
{{begin-eqn}}
{{eqn | l = \map \phi e
| r = z
}}
{{eqn | l = \map \phi a
| r = -z
}}
{{eqn | l = \map \phi b
| r = \dfrac 1 z
}}
{{eqn | l = \map \phi c
| r = -\dfrac 1 z
}}
{{end-eqn}}
From [[Isomorphism by Cayley Tab... | Group Generated by Reciprocal of z and Minus z is Klein Four-Group | https://proofwiki.org/wiki/Group_Generated_by_Reciprocal_of_z_and_Minus_z_is_Klein_Four-Group | https://proofwiki.org/wiki/Group_Generated_by_Reciprocal_of_z_and_Minus_z_is_Klein_Four-Group | [
"Examples of Group Isomorphisms/Order 4",
"Klein Four-Group",
"Group Generated by Reciprocal of z and Minus z"
] | [
"Definition:Klein Four-Group",
"Group Generated by Reciprocal of z and Minus z",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Mapping",
"Isomorphism by Cayley Table",
"Definition:Cayley Table",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Cayley Table"
] |
proofwiki-12024 | Group of Reflection Matrices Order 4 is Klein Four-Group | Let $K_4$ denote the Klein $4$-group.
Let $R_4$ be the Group of Reflection Matrices Order $4$.
Then $K_4$ and $R_4$ are isomorphic algebraic structures. | Establish the mapping $\phi: K_4 \to R_4$ as follows:
{{begin-eqn}}
{{eqn | l = \map \phi e
| r = r_0
}}
{{eqn | l = \map \phi a
| r = r_1
}}
{{eqn | l = \map \phi b
| r = r_2
}}
{{eqn | l = \map \phi c
| r = r_3
}}
{{end-eqn}}
From Isomorphism by Cayley Table, the two Cayley tables can be compa... | Let $K_4$ denote the [[Definition:Klein Four-Group|Klein $4$-group]].
Let $R_4$ be the [[Group of Reflection Matrices Order 4|Group of Reflection Matrices Order $4$]].
Then $K_4$ and $R_4$ are [[Definition:Isomorphic Algebraic Structures|isomorphic algebraic structures]]. | Establish the [[Definition:Mapping|mapping]] $\phi: K_4 \to R_4$ as follows:
{{begin-eqn}}
{{eqn | l = \map \phi e
| r = r_0
}}
{{eqn | l = \map \phi a
| r = r_1
}}
{{eqn | l = \map \phi b
| r = r_2
}}
{{eqn | l = \map \phi c
| r = r_3
}}
{{end-eqn}}
From [[Isomorphism by Cayley Table]], the ... | Group of Reflection Matrices Order 4 is Klein Four-Group | https://proofwiki.org/wiki/Group_of_Reflection_Matrices_Order_4_is_Klein_Four-Group | https://proofwiki.org/wiki/Group_of_Reflection_Matrices_Order_4_is_Klein_Four-Group | [
"Examples of Group Isomorphisms/Order 4",
"Klein Four-Group",
"Group of Reflection Matrices Order 4"
] | [
"Definition:Klein Four-Group",
"Group of Reflection Matrices Order 4",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Mapping",
"Isomorphism by Cayley Table",
"Definition:Cayley Table",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Cayley Table"
] |
proofwiki-12025 | Multiplicative Group of Reduced Residues Modulo 8 is Klein Four-Group | Let $K_4$ denote the Klein $4$-group.
Let $R_4$ be the multiplicative group of reduced residues Modulo $8$.
Then $K_4$ and $R_4$ are isomorphic algebraic structures. | Establish the mapping $\phi: K_4 \to R_4$ as follows:
{{begin-eqn}}
{{eqn | l = \map \phi e
| r = \eqclass 1 8
}}
{{eqn | l = \map \phi a
| r = \eqclass 3 8
}}
{{eqn | l = \map \phi b
| r = \eqclass 5 8
}}
{{eqn | l = \map \phi c
| r = \eqclass 7 8
}}
{{end-eqn}}
From Isomorphism by Cayley Table... | Let $K_4$ denote the [[Definition:Klein Four-Group|Klein $4$-group]].
Let $R_4$ be the [[Multiplicative Group of Reduced Residues Modulo 8|multiplicative group of reduced residues Modulo $8$]].
Then $K_4$ and $R_4$ are [[Definition:Isomorphic Algebraic Structures|isomorphic algebraic structures]]. | Establish the [[Definition:Mapping|mapping]] $\phi: K_4 \to R_4$ as follows:
{{begin-eqn}}
{{eqn | l = \map \phi e
| r = \eqclass 1 8
}}
{{eqn | l = \map \phi a
| r = \eqclass 3 8
}}
{{eqn | l = \map \phi b
| r = \eqclass 5 8
}}
{{eqn | l = \map \phi c
| r = \eqclass 7 8
}}
{{end-eqn}}
From [... | Multiplicative Group of Reduced Residues Modulo 8 is Klein Four-Group | https://proofwiki.org/wiki/Multiplicative_Group_of_Reduced_Residues_Modulo_8_is_Klein_Four-Group | https://proofwiki.org/wiki/Multiplicative_Group_of_Reduced_Residues_Modulo_8_is_Klein_Four-Group | [
"Examples of Group Isomorphisms/Order 4",
"Klein Four-Group",
"Multiplicative Group of Reduced Residues Modulo 8"
] | [
"Definition:Klein Four-Group",
"Multiplicative Group of Reduced Residues/Examples/Modulo 8",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Mapping",
"Isomorphism by Cayley Table",
"Definition:Cayley Table",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Cayley Table"
] |
proofwiki-12026 | Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic | The Klein $4$-group $K_4$ and the cyclic group of order $4$ $C_4$ are not isomorphic. | Recall the Cayley table for $K_4$:
{{:Klein Four-Group/Cayley Table}}
From Finite Cyclic Group is Isomorphic to Integers under Modulo Addition, $C_4$ can be exemplified using the additive group of integers modulo $4$.
Recall the Cayley table for $\struct {\Z_4, +_4}$:
{{:Modulo Addition/Cayley Table/Modulo 4}}
Note tha... | The [[Definition:Klein Four-Group|Klein $4$-group]] $K_4$ and the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order $4$]] $C_4$ are not [[Definition:Group Isomorphism|isomorphic]]. | Recall the [[Klein Four-Group/Cayley Table|Cayley table]] for $K_4$:
{{:Klein Four-Group/Cayley Table}}
From [[Finite Cyclic Group is Isomorphic to Integers under Modulo Addition]], $C_4$ can be exemplified using the [[Definition:Additive Group of Integers Modulo 4|additive group of integers modulo $4$]].
Recall the ... | Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic/Proof 1 | https://proofwiki.org/wiki/Klein_Four-Group_and_Group_of_Cyclic_Group_of_Order_4_are_not_Isomorphic | https://proofwiki.org/wiki/Klein_Four-Group_and_Group_of_Cyclic_Group_of_Order_4_are_not_Isomorphic/Proof_1 | [
"Examples of Group Isomorphisms/Order 4",
"Klein Four-Group",
"Cyclic Group of Order 4",
"Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic"
] | [
"Definition:Klein Four-Group",
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Klein Four-Group/Cayley Table",
"Finite Cyclic Group is Isomorphic to Integers under Modulo Addition",
"Definition:Additive Group of Integers Modulo 4",
"Modulo Addition/Cayley Table/Modulo 4",
"Definition:Element",
"Definition:Self-Inverse Element",
"Definition:Self-Inverse Element",
"Definition:Bij... |
proofwiki-12027 | Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic | The Klein $4$-group $K_4$ and the cyclic group of order $4$ $C_4$ are not isomorphic. | Note that both $C_4$ and $K_4$ are of order $4$.
Also note that both $C_4$ and $K_4$ are abelian.
By definition, $C_4$ has elements of order $4$.
From Group Isomorphism Preserves Order of Group Element, the image of an element of $C_4$ under an isomorphism to $K_4$ would also be of order $4$.
But $K_4$ has no elements ... | The [[Definition:Klein Four-Group|Klein $4$-group]] $K_4$ and the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order $4$]] $C_4$ are not [[Definition:Group Isomorphism|isomorphic]]. | Note that both $C_4$ and $K_4$ are of [[Definition:Order of Group|order $4$]].
Also note that both $C_4$ and $K_4$ are [[Definition:Abelian Group|abelian]].
By definition, $C_4$ has [[Definition:Element|elements]] of [[Definition:Order of Group Element|order $4$]].
From [[Group Isomorphism Preserves Order of Group E... | Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic/Proof 2 | https://proofwiki.org/wiki/Klein_Four-Group_and_Group_of_Cyclic_Group_of_Order_4_are_not_Isomorphic | https://proofwiki.org/wiki/Klein_Four-Group_and_Group_of_Cyclic_Group_of_Order_4_are_not_Isomorphic/Proof_2 | [
"Examples of Group Isomorphisms/Order 4",
"Klein Four-Group",
"Cyclic Group of Order 4",
"Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic"
] | [
"Definition:Klein Four-Group",
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Order of Structure",
"Definition:Abelian Group",
"Definition:Element",
"Definition:Order of Group Element",
"Group Isomorphism Preserves Order of Group Element",
"Definition:Image (Set Theory)/Mapping/Element",
"Definition:Element",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorph... |
proofwiki-12028 | Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $X$ be a subset of $S$.
Then
:$X$ is order generating
{{iff}}
:$\forall x, y \in S: \paren {y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p}$ | === Sufficient Condition ===
Let $X$ be order generating.
Let $x, y \in S$ such that:
:$y \npreceq x$
By Order Generating iff Every Element is Infimum:
:$\exists P \subseteq X: x = \inf P$
By definition of infimum:
:$y$ is lower bound for $P \implies y \preceq x$
By definition of lower bound:
:$\exists p \in P: y \npre... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $X$ be a [[Definition:Subset|subset]] of $S$.
Then
:$X$ is [[Definition:Order Generating Subset|order generating]]
{{iff}}
:$\forall x, y \in S: \paren {y \npreceq x \implies \exists p \in X: x \preceq p \land y \... | === Sufficient Condition ===
Let $X$ be [[Definition:Order Generating Subset|order generating]].
Let $x, y \in S$ such that:
:$y \npreceq x$
By [[Order Generating iff Every Element is Infimum]]:
:$\exists P \subseteq X: x = \inf P$
By definition of [[Definition:Infimum of Set|infimum]]:
:$y$ is [[Definition:Lower B... | Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding | https://proofwiki.org/wiki/Order_Generating_iff_Not_Preceding_implies_There_Exists_Element_Preceding_and_Not_Preceding | https://proofwiki.org/wiki/Order_Generating_iff_Not_Preceding_implies_There_Exists_Element_Preceding_and_Not_Preceding | [
"Complete Lattices",
"Order Generating Subsets"
] | [
"Definition:Complete Lattice",
"Definition:Subset",
"Definition:Order Generating Subset"
] | [
"Definition:Order Generating Subset",
"Order Generating iff Every Element is Infimum",
"Definition:Infimum of Set",
"Definition:Lower Bound of Set",
"Definition:Lower Bound of Set",
"Definition:Subset",
"Definition:Infimum of Set",
"Definition:Lower Bound of Set",
"Definition:Lower Bound of Set",
... |
proofwiki-12029 | Permutation Induces Equivalence Relation/Corollary | :$i \mathrel {\RR_\pi} j$ {{iff}} $i$ and $j$ are in the same cycle of $\pi$. | We have that Permutation Induces Equivalence Relation.
The equivalence classes of that equivalence relation are the cycles of $\pi$.
Hence the result by definition of equivalence class.
{{qed}}
Category:Symmetric Groups
Category:Examples of Equivalence Relations
o3tk79kr9ipf4kqz1tdnyyic21qg3h0 | :$i \mathrel {\RR_\pi} j$ {{iff}} $i$ and $j$ are in the same [[Definition:Cyclic Permutation|cycle]] of $\pi$. | We have that [[Permutation Induces Equivalence Relation]].
The [[Definition:Equivalence Class|equivalence classes]] of that [[Definition:Equivalence Relation|equivalence relation]] are the [[Definition:Cyclic Permutation|cycles]] of $\pi$.
Hence the result by definition of [[Definition:Equivalence Class|equivalence c... | Permutation Induces Equivalence Relation/Corollary | https://proofwiki.org/wiki/Permutation_Induces_Equivalence_Relation/Corollary | https://proofwiki.org/wiki/Permutation_Induces_Equivalence_Relation/Corollary | [
"Symmetric Groups",
"Examples of Equivalence Relations"
] | [
"Definition:Cyclic Permutation"
] | [
"Permutation Induces Equivalence Relation",
"Definition:Equivalence Class",
"Definition:Equivalence Relation",
"Definition:Cyclic Permutation",
"Definition:Equivalence Class",
"Category:Symmetric Groups",
"Category:Examples of Equivalence Relations"
] |
proofwiki-12030 | Order of Group Element not less than Order of Power | Let $\left({G, \circ}\right)$ be a group whose identity is $e$.
Let $g \in G$ be an element of $g$.
Let $\left\lvert{g}\right\rvert$ denote the order of $g$ in $G$.
Then:
:$\forall m \in \Z: \left\lvert{g^m}\right\rvert \le \left\lvert{g}\right\rvert$
where $g^m$ denotes the $m$th power of $g$ in $G$. | Let $\left\lvert{g}\right\rvert = n$.
Then from Order of Power of Group Element:
:$\forall m \in \Z: \left\lvert{g^m}\right\rvert = \dfrac n {\gcd \left\{ {m, n}\right\} }$
where $\gcd \left\{ {m, n}\right\}$ denotes the greatest common divisor of $m$ and $n$.
The result follows from Greatest Common Divisor is at least... | Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $g \in G$ be an [[Definition:Element|element]] of $g$.
Let $\left\lvert{g}\right\rvert$ denote the [[Definition:Order of Group Element|order]] of $g$ in $G$.
Then:
:$\forall m \in \Z: \left\lvert... | Let $\left\lvert{g}\right\rvert = n$.
Then from [[Order of Power of Group Element]]:
:$\forall m \in \Z: \left\lvert{g^m}\right\rvert = \dfrac n {\gcd \left\{ {m, n}\right\} }$
where $\gcd \left\{ {m, n}\right\}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $m$ and $n$.
T... | Order of Group Element not less than Order of Power/Proof 1 | https://proofwiki.org/wiki/Order_of_Group_Element_not_less_than_Order_of_Power | https://proofwiki.org/wiki/Order_of_Group_Element_not_less_than_Order_of_Power/Proof_1 | [
"Order of Group Elements",
"Order of Group Element not less than Order of Power"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Element",
"Definition:Order of Group Element",
"Definition:Power of Element/Group"
] | [
"Order of Power of Group Element",
"Definition:Greatest Common Divisor/Integers",
"Greatest Common Divisor is at least 1"
] |
proofwiki-12031 | Order of Group Element not less than Order of Power | Let $\left({G, \circ}\right)$ be a group whose identity is $e$.
Let $g \in G$ be an element of $g$.
Let $\left\lvert{g}\right\rvert$ denote the order of $g$ in $G$.
Then:
:$\forall m \in \Z: \left\lvert{g^m}\right\rvert \le \left\lvert{g}\right\rvert$
where $g^m$ denotes the $m$th power of $g$ in $G$. | Let $g^n = e$.
Let $h = g^m$.
Then:
:$h^n = g^{m n} = \paren {g^n}^m = e^m = e$
Hence by definition of order of group element:
:$\order h \le n$
Hence the result.
{{qed}} | Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $g \in G$ be an [[Definition:Element|element]] of $g$.
Let $\left\lvert{g}\right\rvert$ denote the [[Definition:Order of Group Element|order]] of $g$ in $G$.
Then:
:$\forall m \in \Z: \left\lvert... | Let $g^n = e$.
Let $h = g^m$.
Then:
:$h^n = g^{m n} = \paren {g^n}^m = e^m = e$
Hence by definition of [[Definition:Order of Group Element|order of group element]]:
:$\order h \le n$
Hence the result.
{{qed}} | Order of Group Element not less than Order of Power/Proof 2 | https://proofwiki.org/wiki/Order_of_Group_Element_not_less_than_Order_of_Power | https://proofwiki.org/wiki/Order_of_Group_Element_not_less_than_Order_of_Power/Proof_2 | [
"Order of Group Elements",
"Order of Group Element not less than Order of Power"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Element",
"Definition:Order of Group Element",
"Definition:Power of Element/Group"
] | [
"Definition:Order of Group Element"
] |
proofwiki-12032 | Order of Group Element equals Order of Coprime Power | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $g \in G$ be an element of $g$.
Let $\order g$ denote the order of $g$ in $G$.
Then:
:$\forall m \in \Z: \order {g^m} = \order g \iff m \perp \order g$
where:
:$g^m$ denotes the $m$th power of $g$ in $G$
:$\perp$ denotes coprimality. | Let $\order g = n$.
Then from Order of Power of Group Element:
:$\forall m \in \Z: \order {g^m} = \dfrac n {\gcd \set {m, n}}$
where $\gcd \set {m, n}$ denotes the greatest common divisor of $m$ and $n$.
Thus:
:$\order {g^m} = \order g \iff \gcd \set {m, n} = 1$
The result follows by definition of coprime integers.
{{q... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $g \in G$ be an [[Definition:Element|element]] of $g$.
Let $\order g$ denote the [[Definition:Order of Group Element|order]] of $g$ in $G$.
Then:
:$\forall m \in \Z: \order {g^m} = \order g \iff m \pe... | Let $\order g = n$.
Then from [[Order of Power of Group Element]]:
:$\forall m \in \Z: \order {g^m} = \dfrac n {\gcd \set {m, n}}$
where $\gcd \set {m, n}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $m$ and $n$.
Thus:
:$\order {g^m} = \order g \iff \gcd \set {m, n} = 1$... | Order of Group Element equals Order of Coprime Power/Proof 1 | https://proofwiki.org/wiki/Order_of_Group_Element_equals_Order_of_Coprime_Power | https://proofwiki.org/wiki/Order_of_Group_Element_equals_Order_of_Coprime_Power/Proof_1 | [
"Order of Group Elements",
"Order of Group Element equals Order of Coprime Power"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Element",
"Definition:Order of Group Element",
"Definition:Power of Element/Group",
"Definition:Coprime/Integers"
] | [
"Order of Power of Group Element",
"Definition:Greatest Common Divisor/Integers",
"Definition:Coprime/Integers"
] |
proofwiki-12033 | Order of Group Element equals Order of Coprime Power | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $g \in G$ be an element of $g$.
Let $\order g$ denote the order of $g$ in $G$.
Then:
:$\forall m \in \Z: \order {g^m} = \order g \iff m \perp \order g$
where:
:$g^m$ denotes the $m$th power of $g$ in $G$
:$\perp$ denotes coprimality. | Let $\order g = n$.
=== Necessary Condition ===
Let $m \perp n$.
Then by Bézout's Identity:
:$\exists x, y \in \Z: x m + y n = 1$
Let $h = g^m$.
Then:
:$h^x = g^{m x} = g^{1 - y n} = g g^{- y n} = g e = g$
and so $g$ is also a power of $h$.
Hence from Order of Group Element not less than Order of Power:
:$\order g \le ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $g \in G$ be an [[Definition:Element|element]] of $g$.
Let $\order g$ denote the [[Definition:Order of Group Element|order]] of $g$ in $G$.
Then:
:$\forall m \in \Z: \order {g^m} = \order g \iff m \pe... | Let $\order g = n$.
=== Necessary Condition ===
Let $m \perp n$.
Then by [[Bézout's Identity]]:
:$\exists x, y \in \Z: x m + y n = 1$
Let $h = g^m$.
Then:
:$h^x = g^{m x} = g^{1 - y n} = g g^{- y n} = g e = g$
and so $g$ is also a [[Definition:Power of Group Element|power]] of $h$.
Hence from [[Order of Group E... | Order of Group Element equals Order of Coprime Power/Proof 2 | https://proofwiki.org/wiki/Order_of_Group_Element_equals_Order_of_Coprime_Power | https://proofwiki.org/wiki/Order_of_Group_Element_equals_Order_of_Coprime_Power/Proof_2 | [
"Order of Group Elements",
"Order of Group Element equals Order of Coprime Power"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Element",
"Definition:Order of Group Element",
"Definition:Power of Element/Group",
"Definition:Coprime/Integers"
] | [
"Bézout's Identity",
"Definition:Power of Element/Group",
"Order of Group Element not less than Order of Power",
"Definition:Contrapositive Statement"
] |
proofwiki-12034 | Unique Composition of Group Element whose Order is Product of Coprime Integers | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $g \in G$ be an element of $g$.
Let:
:$\order g = m n$
where:
:$\order g$ denotes the order of $g$ in $G$
:$m$ and $n$ are coprime integers.
Then $g$ can be expressed uniquely as the product of two commuting elements $a$ and $b$ of order $m$ and $n$ respect... | Let $g_1 = g^n$ and $g_2 = g^m$.
By Powers of Group Element Commute, $g_1$ and $g_2$ commute.
We have:
:${g_1}^m = g^{n m} = e$
:${g_2}^n = g^{m n} = e$
It follows that:
:the order of $g_1$ is $m$
:the order of $g_2$ is $n$
otherwise if either $g_1$ or $g_2$ were of a smaller order then $g$ would also be of a smaller o... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $g \in G$ be an [[Definition:Element|element]] of $g$.
Let:
:$\order g = m n$
where:
:$\order g$ denotes the [[Definition:Order of Group Element|order]] of $g$ in $G$
:$m$ and $n$ are [[Definition:Copri... | Let $g_1 = g^n$ and $g_2 = g^m$.
By [[Powers of Group Element Commute]], $g_1$ and $g_2$ [[Definition:Commuting Elements|commute]].
We have:
:${g_1}^m = g^{n m} = e$
:${g_2}^n = g^{m n} = e$
It follows that:
:the [[Definition:Order of Group Element|order]] of $g_1$ is $m$
:the [[Definition:Order of Group Element|ord... | Unique Composition of Group Element whose Order is Product of Coprime Integers | https://proofwiki.org/wiki/Unique_Composition_of_Group_Element_whose_Order_is_Product_of_Coprime_Integers | https://proofwiki.org/wiki/Unique_Composition_of_Group_Element_whose_Order_is_Product_of_Coprime_Integers | [
"Order of Group Elements"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Element",
"Definition:Order of Group Element",
"Definition:Coprime/Integers",
"Definition:Unique",
"Definition:Group Product/Product Element",
"Definition:Commutative/Elements",
"Definition:Order of Group El... | [
"Powers of Group Element Commute",
"Definition:Commutative/Elements",
"Definition:Order of Group Element",
"Definition:Order of Group Element",
"Definition:Order of Group Element",
"Definition:Order of Group Element",
"Bézout's Identity",
"Bézout's Identity",
"Order of Group Element equals Order of ... |
proofwiki-12035 | Order of Cyclic Group equals Order of Generator | Let $G$ be a finite cyclic group which is generated by $a \in G$.
Then:
:$\order a = \order G$
where:
:$\order a$ denotes the order of $a$ in $G$
:$\order G$ denotes the order of $G$. | Let $\order a = n$.
From List of Elements in Finite Cyclic Group:
:$G = \set {a_0, a_1, \ldots, a_{n - 1} }$
Hence the result.
{{qed}} | Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Cyclic Group|cyclic group]] which is [[Definition:Generator of Cyclic Group|generated]] by $a \in G$.
Then:
:$\order a = \order G$
where:
:$\order a$ denotes the [[Definition:Order of Group Element|order]] of $a$ in $G$
:$\order G$ denotes the [[Definition:... | Let $\order a = n$.
From [[List of Elements in Finite Cyclic Group]]:
:$G = \set {a_0, a_1, \ldots, a_{n - 1} }$
Hence the result.
{{qed}} | Order of Cyclic Group equals Order of Generator | https://proofwiki.org/wiki/Order_of_Cyclic_Group_equals_Order_of_Generator | https://proofwiki.org/wiki/Order_of_Cyclic_Group_equals_Order_of_Generator | [
"Order of Group Elements",
"Order of Groups",
"Cyclic Groups"
] | [
"Definition:Finite Group",
"Definition:Cyclic Group",
"Definition:Cyclic Group/Generator",
"Definition:Order of Group Element",
"Definition:Order of Structure"
] | [
"List of Elements in Finite Cyclic Group"
] |
proofwiki-12036 | Group whose Order equals Order of Element is Cyclic | Let $G$ be a finite group of order $n$.
Let $g \in G$ have order $n$.
Then $G$ is a cyclic group which is generated by $g$. | The subgroup of $G$ generated by $g$ is:
:$\gen g = \set {g^0, g^1, g^2, \ldots, g^{n - 1} }$
This contains $n$ elements.
Thus:
:$\gen g = G$
and the result follows by definition of cyclic group.
{{qed}} | Let $G$ be a [[Definition:Finite Group|finite group]] of [[Definition:Order of Group Element|order]] $n$.
Let $g \in G$ have [[Definition:Order of Group Element|order]] $n$.
Then $G$ is a [[Definition:Cyclic Group|cyclic group]] which is [[Definition:Generator of Cyclic Group|generated]] by $g$. | The [[Definition:Subgroup|subgroup]] of $G$ [[Definition:Generator of Subgroup|generated]] by $g$ is:
:$\gen g = \set {g^0, g^1, g^2, \ldots, g^{n - 1} }$
This contains $n$ [[Definition:Element|elements]].
Thus:
:$\gen g = G$
and the result follows by definition of [[Definition:Cyclic Group|cyclic group]].
{{qed}} | Group whose Order equals Order of Element is Cyclic | https://proofwiki.org/wiki/Group_whose_Order_equals_Order_of_Element_is_Cyclic | https://proofwiki.org/wiki/Group_whose_Order_equals_Order_of_Element_is_Cyclic | [
"Order of Group Elements",
"Order of Groups",
"Cyclic Groups"
] | [
"Definition:Finite Group",
"Definition:Order of Group Element",
"Definition:Order of Group Element",
"Definition:Cyclic Group",
"Definition:Cyclic Group/Generator"
] | [
"Definition:Subgroup",
"Definition:Generator of Subgroup",
"Definition:Element",
"Definition:Cyclic Group"
] |
proofwiki-12037 | Integer Powers of 2 under Multiplication form Infinite Abelian Group | Let $S$ be the set of integers defined as:
:$S = \set {2^k: k \in \Z}$
Then $\struct {S, \times}$ is an infinite abelian group. | It can be seen by inspection that $S \subseteq \Q_{>0}$.
That is, all the elements of $S$ are strictly positive rational numbers.
From Strictly Positive Rational Numbers under Multiplication form Countably Infinite Abelian Group:
:$\struct {\Q_{>0}, \times}$ is an infinite abelian group.
It is noted that $S$ is an infi... | Let $S$ be the [[Definition:Set|set]] of [[Definition:Integer|integers]] defined as:
:$S = \set {2^k: k \in \Z}$
Then $\struct {S, \times}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]]. | It can be seen by inspection that $S \subseteq \Q_{>0}$.
That is, all the [[Definition:Element|elements]] of $S$ are [[Definition:Strictly Positive Rational Number|strictly positive rational numbers]].
From [[Strictly Positive Rational Numbers under Multiplication form Countably Infinite Abelian Group]]:
:$\struct {\... | Integer Powers of 2 under Multiplication form Infinite Abelian Group | https://proofwiki.org/wiki/Integer_Powers_of_2_under_Multiplication_form_Infinite_Abelian_Group | https://proofwiki.org/wiki/Integer_Powers_of_2_under_Multiplication_form_Infinite_Abelian_Group | [
"Examples of Abelian Groups",
"Examples of Infinite Groups"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Infinite Group",
"Definition:Abelian Group"
] | [
"Definition:Element",
"Definition:Strictly Positive/Rational Number",
"Strictly Positive Rational Numbers under Multiplication form Countably Infinite Abelian Group",
"Definition:Infinite Group",
"Definition:Abelian Group",
"Definition:Infinite Set",
"Definition:Inverse (Abstract Algebra)/Inverse",
"O... |
proofwiki-12038 | Numbers of form 1 + 2m over 1 + 2n form Infinite Abelian Group under Multiplication | Let $S$ be the set of integers defined as:
:$S = \set {\dfrac {1 + 2 m} {1 + 2 n}: m, n \in \Z}$
Then $\struct {S, \times}$ is an infinite abelian group. | Let $k \in \Z$.
Then $1 + 2 k \ne 0$.
Thus:
:$\forall x \in S: x \in \Q_{\ne 0}$
Thus by definition of subset:
:$S \subseteq \Q_{\ne 0}$
From Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group:
:$\struct {\Q_{\ne 0}, \times}$ is an infinite abelian group.
It is noted that $S$ is an infinite set ... | Let $S$ be the [[Definition:Set|set]] of [[Definition:Integer|integers]] defined as:
:$S = \set {\dfrac {1 + 2 m} {1 + 2 n}: m, n \in \Z}$
Then $\struct {S, \times}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]]. | Let $k \in \Z$.
Then $1 + 2 k \ne 0$.
Thus:
:$\forall x \in S: x \in \Q_{\ne 0}$
Thus by definition of [[Definition:Subset|subset]]:
:$S \subseteq \Q_{\ne 0}$
From [[Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group]]:
:$\struct {\Q_{\ne 0}, \times}$ is an [[Definition:Infinite Group|infin... | Numbers of form 1 + 2m over 1 + 2n form Infinite Abelian Group under Multiplication | https://proofwiki.org/wiki/Numbers_of_form_1_+_2m_over_1_+_2n_form_Infinite_Abelian_Group_under_Multiplication | https://proofwiki.org/wiki/Numbers_of_form_1_+_2m_over_1_+_2n_form_Infinite_Abelian_Group_under_Multiplication | [
"Examples of Abelian Groups",
"Examples of Infinite Groups"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Infinite Group",
"Definition:Abelian Group"
] | [
"Definition:Subset",
"Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group",
"Definition:Infinite Group",
"Definition:Abelian Group",
"Definition:Infinite Set",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Two-Step Subgroup Test",
"Definition:Subgroup",
"Definition:Infinit... |
proofwiki-12039 | Circle Group is Group | The circle group $\struct {K, \times}$ is a group. | First we note that $K \subseteq \C$.
So to show that $K$ is a group it is sufficient to show that $K$ is a subgroup of the multiplicative group of complex numbers $\struct {\C_{\ne 0}, \times}$.
From Complex Multiplication Identity is One, the identity element $1 + 0 i$ is in $K$.
Thus $K \ne \O$.
We now show that $z, ... | The [[Definition:Circle Group|circle group]] $\struct {K, \times}$ is a [[Definition:Group|group]]. | First we note that $K \subseteq \C$.
So to show that $K$ is a [[Definition:Group|group]] it is sufficient to show that $K$ is a [[Definition:Subgroup|subgroup]] of the [[Definition:Multiplicative Group of Complex Numbers|multiplicative group of complex numbers]] $\struct {\C_{\ne 0}, \times}$.
From [[Complex Multipl... | Circle Group is Group/Proof 1 | https://proofwiki.org/wiki/Circle_Group_is_Group | https://proofwiki.org/wiki/Circle_Group_is_Group/Proof_1 | [
"Circle Group",
"Complex Numbers",
"Circle Group is Group"
] | [
"Definition:Circle Group",
"Definition:Group"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Multiplicative Group of Complex Numbers",
"Complex Multiplication Identity is One",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Two-Step Subgroup Test",
"Definition:Group"
] |
proofwiki-12040 | Circle Group is Group | The circle group $\struct {K, \times}$ is a group. | We note that $K \ne \O$ as the identity element $1 + 0 i \in K$.
Since all $z \in K$ have modulus $1$, they have, for some $\theta \in \hointr 0 {2 \pi}$, the polar form:
:$z = \map \exp {i \theta} = \cos \theta + i \sin \theta$
Conversely, if a complex number has such a polar form, it has modulus $1$.
Observe the foll... | The [[Definition:Circle Group|circle group]] $\struct {K, \times}$ is a [[Definition:Group|group]]. | We note that $K \ne \O$ as the [[Complex Multiplication Identity is One|identity element $1 + 0 i \in K$]].
Since all $z \in K$ have [[Definition:Complex Modulus|modulus]] $1$, they have, for some $\theta \in \hointr 0 {2 \pi}$, the [[Definition:Polar Form of Complex Number|polar form]]:
:$z = \map \exp {i \theta} =... | Circle Group is Group/Proof 2 | https://proofwiki.org/wiki/Circle_Group_is_Group | https://proofwiki.org/wiki/Circle_Group_is_Group/Proof_2 | [
"Circle Group",
"Complex Numbers",
"Circle Group is Group"
] | [
"Definition:Circle Group",
"Definition:Group"
] | [
"Complex Multiplication Identity is One",
"Definition:Complex Modulus",
"Definition:Complex Number/Polar Form",
"Definition:Complex Number",
"Definition:Complex Number/Polar Form",
"Definition:Complex Modulus",
"Definition:Exponential Function/Complex",
"Two-Step Subgroup Test",
"Definition:Multipli... |
proofwiki-12041 | Circle Group is Group | The circle group $\struct {K, \times}$ is a group. | Taking the group axioms in turn:
=== {{Group-axiom|0|nolink}} ===
{{begin-eqn}}
{{eqn | l = z, w
| o = \in
| r = K
| c =
}}
{{eqn | ll= \leadsto
| l = \cmod z
| r = 1 = \cmod w
| c =
}}
{{eqn | ll= \leadsto
| l = \cmod {z w}
| r = \cmod z \cmod w
| c =
}}
{{eqn |... | The [[Definition:Circle Group|circle group]] $\struct {K, \times}$ is a [[Definition:Group|group]]. | Taking the [[Axiom:Group Axioms|group axioms]] in turn:
=== {{Group-axiom|0|nolink}} ===
{{begin-eqn}}
{{eqn | l = z, w
| o = \in
| r = K
| c =
}}
{{eqn | ll= \leadsto
| l = \cmod z
| r = 1 = \cmod w
| c =
}}
{{eqn | ll= \leadsto
| l = \cmod {z w}
| r = \cmod z \cmod... | Circle Group is Group/Proof 3 | https://proofwiki.org/wiki/Circle_Group_is_Group | https://proofwiki.org/wiki/Circle_Group_is_Group/Proof_3 | [
"Circle Group",
"Complex Numbers",
"Circle Group is Group"
] | [
"Definition:Circle Group",
"Definition:Group"
] | [
"Axiom:Group Axioms",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Complex Multiplication is Associative",
"Complex Multiplication Identity is One",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Axiom:Group Axioms"
] |
proofwiki-12042 | Circle Group is Group | The circle group $\struct {K, \times}$ is a group. | Consider the complex modulus function $\cmod {\, \cdot \,}: \C \to \R, z \mapsto \cmod z$.
By Complex Modulus is Norm, we have that $\cmod z \ge 0$ for all $z \in \C$, and:
:$\cmod z = 0 \iff z = 0$
Let $\C_{\ne 0} := \C \setminus \set 0$ denote the complex numbers without zero.
From Group of Units of Field and Complex... | The [[Definition:Circle Group|circle group]] $\struct {K, \times}$ is a [[Definition:Group|group]]. | Consider the [[Definition:Complex Modulus|complex modulus]] function $\cmod {\, \cdot \,}: \C \to \R, z \mapsto \cmod z$.
By [[Complex Modulus is Norm]], we have that $\cmod z \ge 0$ for all $z \in \C$, and:
:$\cmod z = 0 \iff z = 0$
Let $\C_{\ne 0} := \C \setminus \set 0$ denote the [[Definition:Complex Numbers|com... | Circle Group is Group/Proof 4 | https://proofwiki.org/wiki/Circle_Group_is_Group | https://proofwiki.org/wiki/Circle_Group_is_Group/Proof_4 | [
"Circle Group",
"Complex Numbers",
"Circle Group is Group"
] | [
"Definition:Circle Group",
"Definition:Group"
] | [
"Definition:Complex Modulus",
"Complex Modulus is Norm",
"Definition:Complex Number",
"Definition:Zero (Number)",
"Group of Units of Field",
"Complex Numbers form Field",
"Definition:Group",
"Definition:Complex Modulus",
"Definition:Restriction/Mapping",
"Complex Modulus of Product of Complex Numb... |
proofwiki-12043 | Set of Meet Irreducible Elements Excluded Top is Order Generating | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a continuous complete lattice.
Let $X = \map {\mathit {IRR} } L \setminus \set \top$
where:
:$\map {\mathit {IRR} } L$ denotes the set of all meet irreducible element of $S$
:$\top$ denotes the top of $L$.
Then $X$ is order generating. | We will prove that:
:$\forall x, y \in S: \paren {y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p}$
Let $x, y \in S$ such that:
:$y \npreceq x$
By Not Preceding implies There Exists Meet Irreducible Element Not Preceding:
:$\exists p \in S: p$ is meet irreducible and $x \preceq p$ and $y \npreceq ... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Continuous Ordered Set|continuous]] [[Definition:Complete Lattice|complete lattice]].
Let $X = \map {\mathit {IRR} } L \setminus \set \top$
where:
:$\map {\mathit {IRR} } L$ denotes the [[Definition:Set|set]] of all [[Definition:Meet Irreducible Element|m... | We will prove that:
:$\forall x, y \in S: \paren {y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p}$
Let $x, y \in S$ such that:
:$y \npreceq x$
By [[Not Preceding implies There Exists Meet Irreducible Element Not Preceding]]:
:$\exists p \in S: p$ is [[Definition:Meet Irreducible Element|meet ir... | Set of Meet Irreducible Elements Excluded Top is Order Generating | https://proofwiki.org/wiki/Set_of_Meet_Irreducible_Elements_Excluded_Top_is_Order_Generating | https://proofwiki.org/wiki/Set_of_Meet_Irreducible_Elements_Excluded_Top_is_Order_Generating | [
"Continuous Lattices",
"Order Generating Subsets",
"Meet Irreducible Elements"
] | [
"Definition:Continuous Ordered Set",
"Definition:Complete Lattice",
"Definition:Set",
"Definition:Meet Irreducible Element",
"Definition:Element",
"Definition:Top of Lattice",
"Definition:Order Generating Subset"
] | [
"Not Preceding implies There Exists Meet Irreducible Element Not Preceding",
"Definition:Meet Irreducible Element",
"Definition:Greatest Element",
"Definition:Set Difference",
"Definition:Singleton",
"Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding",
"Definiti... |
proofwiki-12044 | Superset of Order Generating is Order Generating | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $X, Y$ be subsets of $S$ such that
:$X$ is order generating
and
:$X \subseteq Y$
Then $Y$ is order generating. | Let $x \in S$.
Thus by definition of complete lattice:
:$x^\succeq \cap Y$ admits an infimum.
By definition of complete lattice:
:$x^\succeq$ admits an infimum
and
:$x^\succeq \cap X$ admits an infimum.
By Intersection is Subset:
:$x^\succeq \cap Y \subseteq x^\succeq$
By Infimum of Subset:
:$\map \inf {x^\succeq} \pre... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $X, Y$ be [[Definition:Subset|subsets]] of $S$ such that
:$X$ is [[Definition:Order Generating Subset|order generating]]
and
:$X \subseteq Y$
Then $Y$ is [[Definition:Order Generating Subset|order generating]]. | Let $x \in S$.
Thus by definition of [[Definition:Complete Lattice|complete lattice]]:
:$x^\succeq \cap Y$ admits an [[Definition:Infimum of Set|infimum]].
By definition of [[Definition:Complete Lattice|complete lattice]]:
:$x^\succeq$ admits an [[Definition:Infimum of Set|infimum]]
and
:$x^\succeq \cap X$ admits an ... | Superset of Order Generating is Order Generating | https://proofwiki.org/wiki/Superset_of_Order_Generating_is_Order_Generating | https://proofwiki.org/wiki/Superset_of_Order_Generating_is_Order_Generating | [
"Complete Lattices",
"Order Generating Subsets"
] | [
"Definition:Complete Lattice",
"Definition:Subset",
"Definition:Order Generating Subset",
"Definition:Order Generating Subset"
] | [
"Definition:Complete Lattice",
"Definition:Infimum of Set",
"Definition:Complete Lattice",
"Definition:Infimum of Set",
"Definition:Infimum of Set",
"Intersection is Subset",
"Infimum of Subset",
"Infimum of Upper Closure of Element",
"Set Intersection Preserves Subsets/Corollary",
"Infimum of Sub... |
proofwiki-12045 | Modulus 1 Rational Argument Complex Numbers under Multiplication form Infinite Abelian Group | Let $S$ be the set defined as:
:$S = \set {\cos \theta + i \sin \theta: \theta \in \Q}$
Then the algebraic structure $\struct {S, \times}$ is an infinite abelian group. | By definition of polar form of complex numbers, the elements of $S$ are also elements of the circle group $\struct {K, \times}$:
:$K = \set {z \in \C: \cmod z = 1}$
$S$ is infinite by construction.
Thus $S \subseteq C$ and trivially $S \ne \O$.
Let $a, b \in S$.
Then:
:$a = \cos \theta_1 + i \sin \theta_1$
and:
:$b = \... | Let $S$ be the [[Definition:Set|set]] defined as:
:$S = \set {\cos \theta + i \sin \theta: \theta \in \Q}$
Then the [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {S, \times}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]]. | By definition of [[Definition:Polar Form of Complex Number|polar form of complex numbers]], the [[Definition:Element|elements]] of $S$ are also [[Definition:Element|elements]] of the [[Definition:Circle Group|circle group]] $\struct {K, \times}$:
:$K = \set {z \in \C: \cmod z = 1}$
$S$ is [[Definition:Infinite Set|inf... | Modulus 1 Rational Argument Complex Numbers under Multiplication form Infinite Abelian Group | https://proofwiki.org/wiki/Modulus_1_Rational_Argument_Complex_Numbers_under_Multiplication_form_Infinite_Abelian_Group | https://proofwiki.org/wiki/Modulus_1_Rational_Argument_Complex_Numbers_under_Multiplication_form_Infinite_Abelian_Group | [
"Examples of Abelian Groups",
"Examples of Infinite Groups"
] | [
"Definition:Set",
"Definition:Algebraic Structure/One Operation",
"Definition:Infinite Group",
"Definition:Abelian Group"
] | [
"Definition:Complex Number/Polar Form",
"Definition:Element",
"Definition:Element",
"Definition:Circle Group",
"Definition:Infinite Set",
"Two-Step Subgroup Test",
"Definition:Subgroup",
"Definition:Infinite Set",
"Definition:Infinite Group",
"Subgroup of Abelian Group is Abelian",
"Definition:A... |
proofwiki-12046 | Inversion Mapping is Automorphism iff Group is Abelian | Let $\struct {G, \circ}$ be a group.
Let $\iota: G \to G$ be the inversion mapping on $G$, defined as:
:$\forall g \in G: \map \iota g = g^{-1}$
Then $\iota$ is an automorphism {{iff}} $G$ is abelian. | From Inversion Mapping is Permutation, $\iota$ is a permutation.
It remains to be shown that $\iota$ has the morphism property {{iff}} $G$ is abelian. | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\iota: G \to G$ be the [[Definition:Inversion Mapping|inversion mapping]] on $G$, defined as:
:$\forall g \in G: \map \iota g = g^{-1}$
Then $\iota$ is an [[Definition:Group Automorphism|automorphism]] {{iff}} $G$ is [[Definition:Abelian Group|abelian]]. | From [[Inversion Mapping is Permutation]], $\iota$ is a [[Definition:Permutation|permutation]].
It remains to be shown that $\iota$ has the [[Definition:Morphism Property|morphism property]] {{iff}} $G$ is [[Definition:Abelian Group|abelian]]. | Inversion Mapping is Automorphism iff Group is Abelian | https://proofwiki.org/wiki/Inversion_Mapping_is_Automorphism_iff_Group_is_Abelian | https://proofwiki.org/wiki/Inversion_Mapping_is_Automorphism_iff_Group_is_Abelian | [
"Inversion Mappings",
"Abelian Groups",
"Group Automorphisms"
] | [
"Definition:Group",
"Definition:Inversion Mapping",
"Definition:Group Automorphism",
"Definition:Abelian Group"
] | [
"Inversion Mapping is Permutation",
"Definition:Permutation",
"Definition:Morphism Property",
"Definition:Abelian Group",
"Definition:Abelian Group",
"Definition:Abelian Group",
"Definition:Morphism Property"
] |
proofwiki-12047 | Even Order Group has Odd Number of Order 2 Elements | Let $G$ be a group whose identity is $e$.
Let $G$ be of even order.
Then $G$ has an odd number of elements of order $2$. | Let $S = \set {x \in G: \order x > 2}$ be the set of all elements of $G$ whose order is strictly greater than $2$.
Let $h \in S$.
Then from Order of Group Element equals Order of Inverse:
:$h^{-1} \in S$.
Because $\order h > 2$ it follows that $h^{-1} \ne h$ by definition of self-inverse.
Thus every element in $S$ can ... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $G$ be of even [[Definition:Order of Structure|order]].
Then $G$ has an [[Definition:Odd Integer|odd number]] of [[Definition:Element|elements]] of [[Definition:Order of Group Element|order $2$]]. | Let $S = \set {x \in G: \order x > 2}$ be the [[Definition:Set|set]] of all [[Definition:Element|elements]] of $G$ whose [[Definition:Order of Group Element|order]] is strictly greater than $2$.
Let $h \in S$.
Then from [[Order of Group Element equals Order of Inverse]]:
:$h^{-1} \in S$.
Because $\order h > 2$ it fo... | Even Order Group has Odd Number of Order 2 Elements | https://proofwiki.org/wiki/Even_Order_Group_has_Odd_Number_of_Order_2_Elements | https://proofwiki.org/wiki/Even_Order_Group_has_Odd_Number_of_Order_2_Elements | [
"Finite Groups",
"Order of Group Elements"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Structure",
"Definition:Odd Integer",
"Definition:Element",
"Definition:Order of Group Element"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Order of Group Element",
"Order of Group Element equals Order of Inverse",
"Definition:Self-Inverse Element",
"Definition:Element",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Even Integer",
"Definition:Element",
"Definition:Even... |
proofwiki-12048 | Roots of Complex Number/Examples/Cube Roots | Let $z := \polar {r, \theta}$ be a complex number expressed in polar form, such that $z \ne 0$.
Then the complex cube roots of $z$ are given by:
:$z^{1 / 3} = \set {r^{1 / 3} \paren {\map \cos {\dfrac {\theta + 2 \pi k} 3} + i \, \map \sin {\dfrac {\theta + 2 \pi k} 3} }: k \in \set {0, 1, 2} }$
There are $3$ distinct ... | An example of Roots of Complex Number.
{{qed}} | Let $z := \polar {r, \theta}$ be a [[Definition:Polar Form of Complex Number|complex number expressed in polar form]], such that $z \ne 0$.
Then the [[Definition:Complex Root|complex cube roots]] of $z$ are given by:
:$z^{1 / 3} = \set {r^{1 / 3} \paren {\map \cos {\dfrac {\theta + 2 \pi k} 3} + i \, \map \sin {\dfra... | An example of [[Roots of Complex Number]].
{{qed}} | Roots of Complex Number/Examples/Cube Roots | https://proofwiki.org/wiki/Roots_of_Complex_Number/Examples/Cube_Roots | https://proofwiki.org/wiki/Roots_of_Complex_Number/Examples/Cube_Roots | [
"Examples of Complex Roots"
] | [
"Definition:Complex Number/Polar Form",
"Definition:Complex Root",
"Definition:Distinct/Plural",
"Definition:Complex Root",
"Definition:Root of Unity/Complex/First"
] | [
"Roots of Complex Number"
] |
proofwiki-12049 | Roots of Complex Number/Corollary | Let $z := \polar {r, \theta}$ be a complex number expressed in polar form, such that $z \ne 0$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $w$ be one of the complex $n$th roots of $z$.
Then the $n$th roots of $z$ are given by:
:$z^{1 / n} = \set {w \epsilon^k: k \in \set {1, 2, \ldots, n - 1} }$
where $\... | By definition of primitive complex $n$th root of unity:
:$\omega = e^{2 m i \pi k}$
for some $m \in \Z: 1 \le m < n$.
Thus:
{{begin-eqn}}
{{eqn | l = \paren {w \omega^k}^n
| r = w^n \paren {e^{2 m i \pi k / n} }^n
| c =
}}
{{eqn | r = z e^{2 m i \pi k}
| c =
}}
{{eqn | r = z \paren {e^{2 i \pi} }^{m... | Let $z := \polar {r, \theta}$ be a [[Definition:Polar Form of Complex Number|complex number expressed in polar form]], such that $z \ne 0$.
Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $w$ be one of the [[Definition:Complex Root|complex $n$th roots]] of $z$.
Th... | By definition of [[Definition:Primitive Complex Root of Unity|primitive complex $n$th root of unity]]:
:$\omega = e^{2 m i \pi k}$
for some $m \in \Z: 1 \le m < n$.
Thus:
{{begin-eqn}}
{{eqn | l = \paren {w \omega^k}^n
| r = w^n \paren {e^{2 m i \pi k / n} }^n
| c =
}}
{{eqn | r = z e^{2 m i \pi k}
... | Roots of Complex Number/Corollary | https://proofwiki.org/wiki/Roots_of_Complex_Number/Corollary | https://proofwiki.org/wiki/Roots_of_Complex_Number/Corollary | [
"Complex Roots"
] | [
"Definition:Complex Number/Polar Form",
"Definition:Strictly Positive/Integer",
"Definition:Complex Root",
"Definition:Complex Root",
"Definition:Root of Unity/Primitive"
] | [
"Definition:Root of Unity/Complex/Primitive",
"Definition:Complex Root",
"Definition:Root of Unity/Complex",
"Roots of Complex Number",
"Definition:Complex Root"
] |
proofwiki-12050 | Top is Prime Element | Let $L = \struct {S, \wedge, \preceq}$ be a bounded above meet semilattice.
Then:
:$\top$ is a prime element
where $\top$ denotes the greatest element of $L$. | Let $x, y \in S$ such that
:$x \wedge y \preceq \top$
Thus by definition of greatest element:
:$x \preceq \top$ or $y \preceq \top$
{{qed}} | Let $L = \struct {S, \wedge, \preceq}$ be a [[Definition:Bounded Above Set|bounded above]] [[Definition:Meet Semilattice|meet semilattice]].
Then:
:$\top$ is a [[Definition:Prime Element (Order Theory)|prime element]]
where $\top$ denotes the [[Definition:Greatest Element|greatest element]] of $L$. | Let $x, y \in S$ such that
:$x \wedge y \preceq \top$
Thus by definition of [[Definition:Greatest Element|greatest element]]:
:$x \preceq \top$ or $y \preceq \top$
{{qed}} | Top is Prime Element | https://proofwiki.org/wiki/Top_is_Prime_Element | https://proofwiki.org/wiki/Top_is_Prime_Element | [
"Prime Elements"
] | [
"Definition:Bounded Above Set",
"Definition:Meet Semilattice",
"Definition:Prime Element (Order Theory)",
"Definition:Greatest Element"
] | [
"Definition:Greatest Element"
] |
proofwiki-12051 | Characterization of Prime Element in Meet Semilattice | Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $p \in S$,
Then:
:$p$ is prime element
{{iff}}:
:for all non-empty finite subsets $A$ of $S$:
::if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$. | === Sufficient Condition ===
Let $p$ be prime element.
Let $A$ be non-empty finite subsets of $S$.
Define
:$\map P X : \equiv X \ne \O \land \inf X \preceq p \implies \exists x \in X: x \preceq p$
where $X \subseteq S$.
We will prove that
:$\forall x \in A, B \subseteq A: \map P B \implies \map P {B \cup \set x}$
Let $... | Let $L = \struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
Let $p \in S$,
Then:
:$p$ is [[Definition:Prime Element (Order Theory)|prime element]]
{{iff}}:
:for all [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] $A$ of $S$:
::if... | === Sufficient Condition ===
Let $p$ be [[Definition:Prime Element (Order Theory)|prime element]].
Let $A$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $S$.
Define
:$\map P X : \equiv X \ne \O \land \inf X \preceq p \implies \exists x \in X: x \preceq p$... | Characterization of Prime Element in Meet Semilattice | https://proofwiki.org/wiki/Characterization_of_Prime_Element_in_Meet_Semilattice | https://proofwiki.org/wiki/Characterization_of_Prime_Element_in_Meet_Semilattice | [
"Prime Elements"
] | [
"Definition:Meet Semilattice",
"Definition:Prime Element (Order Theory)",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Element"
] | [
"Definition:Prime Element (Order Theory)",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Subset",
"Union with Empty Set",
"Infimum of Singleton",
"Definition:Singleton",
"Subset of Finite Set is Finite",
"Definition:Finite Set",
"Existence of Non-Empty Finite Infima in Meet Semi... |
proofwiki-12052 | Isomorphism between Group Generated by Reciprocal of z and 1 minus z and Symmetric Group on 3 Letters | Let $S_3$ denote the symmetric group on $3$ letters.
Let $G$ be the group generated by $1 / z$ and $1 - z$.
Then $S_3$ and $G$ are isomorphic algebraic structures. | Establish the mapping $\phi: S_3 \to G$ as follows:
{{begin-eqn}}
{{eqn | l = \map \phi e
| r = z
}}
{{eqn | l = \map \phi {\paren {123} }
| r = \dfrac 1 {1 - z}
}}
{{eqn | l = \map \phi {\paren {132} }
| r = \dfrac {z - 1} z
}}
{{eqn | l = \map \phi {\paren {23} }
| r = \dfrac 1 z
}}
{{eqn | l ... | Let $S_3$ denote the [[Symmetric Group on 3 Letters|symmetric group on $3$ letters]].
Let $G$ be the [[Definition:Group Generated by Reciprocal of z and 1 minus z|group generated by $1 / z$ and $1 - z$]].
Then $S_3$ and $G$ are [[Definition:Isomorphic Algebraic Structures|isomorphic algebraic structures]]. | Establish the [[Definition:Mapping|mapping]] $\phi: S_3 \to G$ as follows:
{{begin-eqn}}
{{eqn | l = \map \phi e
| r = z
}}
{{eqn | l = \map \phi {\paren {123} }
| r = \dfrac 1 {1 - z}
}}
{{eqn | l = \map \phi {\paren {132} }
| r = \dfrac {z - 1} z
}}
{{eqn | l = \map \phi {\paren {23} }
| r = ... | Isomorphism between Group Generated by Reciprocal of z and 1 minus z and Symmetric Group on 3 Letters | https://proofwiki.org/wiki/Isomorphism_between_Group_Generated_by_Reciprocal_of_z_and_1_minus_z_and_Symmetric_Group_on_3_Letters | https://proofwiki.org/wiki/Isomorphism_between_Group_Generated_by_Reciprocal_of_z_and_1_minus_z_and_Symmetric_Group_on_3_Letters | [
"Examples of Group Isomorphisms/Order 6",
"Symmetric Group on 3 Letters",
"Group Generated by Reciprocal of z and 1 minus z"
] | [
"Symmetric Group on 3 Letters",
"Definition:Group Generated by Reciprocal of z and 1 minus z",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Mapping",
"Isomorphism by Cayley Table",
"Definition:Cayley Table",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Cayley Table",
"Definition:Cayley Table"
] |
proofwiki-12053 | Prime Element is Meet Irreducible | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $p \in S$.
Let $p$ be a prime element of $L$.
Then $p$ is meet irreducible in $L$. | Let $p$ be a prime element.
Let $x, y \in S$ such that
:$p = x \wedge y$
By definition of reflexivity:
:$x \wedge y \preceq p$
By definition of prime element:
:$x \preceq p$ or $y \preceq p$
By Meet Precedes Operands:
:$p \preceq x$ and $p \preceq y$
Thus by definition of antisymmetry:
:$p = x$ or $p = y$
{{qed}} | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice (Order Theory)|lattice]].
Let $p \in S$.
Let $p$ be a [[Definition:Prime Element (Order Theory)|prime element]] of $L$.
Then $p$ is [[Definition:Meet Irreducible Element|meet irreducible]] in $L$. | Let $p$ be a [[Definition:Prime Element (Order Theory)|prime element]].
Let $x, y \in S$ such that
:$p = x \wedge y$
By definition of [[Definition:Reflexivity|reflexivity]]:
:$x \wedge y \preceq p$
By definition of [[Definition:Prime Element (Order Theory)|prime element]]:
:$x \preceq p$ or $y \preceq p$
By [[Meet ... | Prime Element is Meet Irreducible | https://proofwiki.org/wiki/Prime_Element_is_Meet_Irreducible | https://proofwiki.org/wiki/Prime_Element_is_Meet_Irreducible | [
"Prime Elements",
"Meet Irreducible Elements"
] | [
"Definition:Lattice (Order Theory)",
"Definition:Prime Element (Order Theory)",
"Definition:Meet Irreducible Element"
] | [
"Definition:Prime Element (Order Theory)",
"Definition:Reflexivity",
"Definition:Prime Element (Order Theory)",
"Meet Precedes Operands",
"Definition:Antisymmetric Relation"
] |
proofwiki-12054 | Prime Element iff Meet Irreducible in Distributive Lattice | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a distributive lattice.
Let $p \in S$.
Then $p$ is a prime element {{iff}} $p$ is meet irreducible. | By Prime Element is Meet Irreducible:
:if $p$ is a prime element, then $p$ is meet irreducible.
Assume that:
:$p$ is meet irreducible.
Let $x, y \in S$ such that:
:$x \wedge y \preceq p$
{{begin-eqn}}
{{eqn | l = p
| r = p \vee \paren {x \wedge y}
| c = Preceding iff Join equals Larger Operand
}}
{{eqn | r ... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Distributive Lattice|distributive lattice]].
Let $p \in S$.
Then $p$ is a [[Definition:Prime Element (Order Theory)|prime element]] {{iff}} $p$ is [[Definition:Meet Irreducible Element|meet irreducible]]. | By [[Prime Element is Meet Irreducible]]:
:if $p$ is a [[Definition:Prime Element (Order Theory)|prime element]], then $p$ is [[Definition:Meet Irreducible Element|meet irreducible]].
Assume that:
:$p$ is [[Definition:Meet Irreducible Element|meet irreducible]].
Let $x, y \in S$ such that:
:$x \wedge y \preceq p$
{{... | Prime Element iff Meet Irreducible in Distributive Lattice | https://proofwiki.org/wiki/Prime_Element_iff_Meet_Irreducible_in_Distributive_Lattice | https://proofwiki.org/wiki/Prime_Element_iff_Meet_Irreducible_in_Distributive_Lattice | [
"Prime Elements",
"Meet Irreducible Elements"
] | [
"Definition:Distributive Lattice",
"Definition:Prime Element (Order Theory)",
"Definition:Meet Irreducible Element"
] | [
"Prime Element is Meet Irreducible",
"Definition:Prime Element (Order Theory)",
"Definition:Meet Irreducible Element",
"Definition:Meet Irreducible Element",
"Preceding iff Join equals Larger Operand",
"Definition:Meet Irreducible Element",
"Preceding iff Join equals Larger Operand"
] |
proofwiki-12055 | Multiple Equilibrium Points all have Equal Payoffs | Let $G$ be a two-person zero-sum game.
Let $G$ have more than one equilibrium point.
Then every equilibrium point of $G$ has the same payoff. | {{proof wanted|Stated without proof, or even an explanatory comment on its scope of applicability, in Davis. This work is too vague to be useful. I am going to have to use a different work.}} | Let $G$ be a [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]].
Let $G$ have more than one [[Definition:Equilibrium Point|equilibrium point]].
Then every [[Definition:Equilibrium Point|equilibrium point]] of $G$ has the same [[Definition:Payoff|payoff]]. | {{proof wanted|Stated without proof, or even an explanatory comment on its scope of applicability, in Davis. This work is too vague to be useful. I am going to have to use a different work.}} | Multiple Equilibrium Points all have Equal Payoffs | https://proofwiki.org/wiki/Multiple_Equilibrium_Points_all_have_Equal_Payoffs | https://proofwiki.org/wiki/Multiple_Equilibrium_Points_all_have_Equal_Payoffs | [
"Game Theory"
] | [
"Definition:Two-Person Zero-Sum Game",
"Definition:Equilibrium Point",
"Definition:Equilibrium Point",
"Definition:Game/Payoff"
] | [] |
proofwiki-12056 | Non-Zero-Sum Game as Zero-Sum Game | Let $G$ be a non-zero-sum game for $n$ players.
Then $G$ can be modelled as a zero-sum game for $n + 1$ players. | At each outcome, the total payoff of $G$ will be an amount which will (for at least one outcome) not be zero
Let an $n + 1$th player be introduced to $G$ who has one move:
:$(1): \quad$ Select any player $m$.
:$(2): \quad$ If the total payoff of $G$ is $+k$, receive in payment $k$ from player $m$.
:$(3): \quad$ If the ... | Let $G$ be a [[Definition:Non-Zero-Sum Game|non-zero-sum game]] for $n$ [[Definition:Player|players]].
Then $G$ can be modelled as a [[Definition:Zero-Sum Game|zero-sum game]] for $n + 1$ [[Definition:Player|players]]. | At each [[Definition:Outcome of Game|outcome]], the total [[Definition:Payoff|payoff]] of $G$ will be an amount which will (for at least one [[Definition:Outcome of Game|outcome]]) not be [[Definition:Zero (Number)|zero]]
Let an $n + 1$th [[Definition:Player|player]] be introduced to $G$ who has one [[Definition:Move|... | Non-Zero-Sum Game as Zero-Sum Game | https://proofwiki.org/wiki/Non-Zero-Sum_Game_as_Zero-Sum_Game | https://proofwiki.org/wiki/Non-Zero-Sum_Game_as_Zero-Sum_Game | [
"Non-Zero-Sum Game as Zero-Sum Game",
"Zero-Sum Games"
] | [
"Definition:Zero-Sum Game/Non-Zero",
"Definition:Game/Player",
"Definition:Zero-Sum Game",
"Definition:Game/Player"
] | [
"Definition:Outcome of Game",
"Definition:Game/Payoff",
"Definition:Outcome of Game",
"Definition:Zero (Number)",
"Definition:Game/Player",
"Definition:Move",
"Definition:Game/Player",
"Definition:Game/Payoff",
"Definition:Game/Player",
"Definition:Game/Payoff",
"Definition:Pay",
"Definition:G... |
proofwiki-12057 | Two-Person Zero-Sum Game is Non-Cooperative | A two-person zero-sum game necessarily has to be non-cooperative. | A cooperative game is one where players form coalitions against the other players.
If the players in a two-person zero-sum game were to form a coalition, there would be no other players to form it against.
Further, as the total payoff is zero, there would be no benefit in collaborating on using one strategy over anothe... | A [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]] necessarily has to be [[Definition:Non-Cooperative Game|non-cooperative]]. | A [[Definition:Cooperative Game|cooperative game]] is one where [[Definition:Player|players]] form [[Definition:Coalition|coalitions]] against the other [[Definition:Player|players]].
If the [[Definition:Player|players]] in a [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]] were to form a [[Definition:... | Two-Person Zero-Sum Game is Non-Cooperative | https://proofwiki.org/wiki/Two-Person_Zero-Sum_Game_is_Non-Cooperative | https://proofwiki.org/wiki/Two-Person_Zero-Sum_Game_is_Non-Cooperative | [
"Two-Person Zero-Sum Games",
"Cooperative Games"
] | [
"Definition:Two-Person Zero-Sum Game",
"Definition:Non-Cooperative Game"
] | [
"Definition:Cooperative Game",
"Definition:Game/Player",
"Definition:Coalition",
"Definition:Game/Player",
"Definition:Game/Player",
"Definition:Two-Person Zero-Sum Game",
"Definition:Coalition",
"Definition:Game/Player",
"Definition:Game/Payoff",
"Definition:Zero (Number)",
"Definition:Game/Str... |
proofwiki-12058 | Simple Graph with Finite Vertex Set is Finite | Let $G$ be a simple graph.
Suppose that the vertex set of $G$ is finite.
Then $G$ is a finite graph.
That is to say, its edge set is also finite. | Since $G$ is simple, it can have at most one edge between each two vertices.
Therefore, the mapping:
:$v: \map E G \to \powerset {\map V G}, \map v e = \set {a_e, b_e}$
which assigns to each edge its endvertices $a_e, b_e$, is an injection.
From Power Set of Finite Set is Finite and Domain of Injection Not Larger than ... | Let $G$ be a [[Definition:Simple Graph|simple graph]].
Suppose that the [[Definition:Vertex Set|vertex set]] of $G$ is [[Definition:Finite Set|finite]].
Then $G$ is a [[Definition:Finite Graph|finite graph]].
That is to say, its [[Definition:Edge Set|edge set]] is also [[Definition:Finite Set|finite]]. | Since $G$ is [[Definition:Simple Graph|simple]], it can have at most one [[Definition:Edge of Graph|edge]] between each two [[Definition:Vertex of Graph|vertices]].
Therefore, the [[Definition:Mapping|mapping]]:
:$v: \map E G \to \powerset {\map V G}, \map v e = \set {a_e, b_e}$
which assigns to each [[Definition:Ed... | Simple Graph with Finite Vertex Set is Finite | https://proofwiki.org/wiki/Simple_Graph_with_Finite_Vertex_Set_is_Finite | https://proofwiki.org/wiki/Simple_Graph_with_Finite_Vertex_Set_is_Finite | [
"Graph Theory",
"Simple Graphs"
] | [
"Definition:Simple Graph",
"Definition:Vertex Set",
"Definition:Finite Set",
"Definition:Finite Graph",
"Definition:Edge Set",
"Definition:Finite Set"
] | [
"Definition:Simple Graph",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Mapping",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Graph (Graph Theory)/Edge/Endvertex",
"Definition:Injection",
"Power Set of Finite Set is Finite",
"Domain of Injecti... |
proofwiki-12059 | Multigraph with Finite Vertex Set may not be Finite | Let $G$ be a multigraph.
Let the vertex set of $G$ be finite.
Then it is not necessarily the case that $G$ is also finite. | {{finish|two vertices, $\N$ as edges}}
Category:Graph Theory
Category:Multigraphs
4694d4g1dnmmomeps1l1rkeh713gspp | Let $G$ be a [[Definition:Multigraph|multigraph]].
Let the [[Definition:Vertex Set|vertex set]] of $G$ be [[Definition:Finite Set|finite]].
Then it is not necessarily the case that $G$ is also [[Definition:Finite Graph|finite]]. | {{finish|two vertices, $\N$ as edges}}
[[Category:Graph Theory]]
[[Category:Multigraphs]]
4694d4g1dnmmomeps1l1rkeh713gspp | Multigraph with Finite Vertex Set may not be Finite | https://proofwiki.org/wiki/Multigraph_with_Finite_Vertex_Set_may_not_be_Finite | https://proofwiki.org/wiki/Multigraph_with_Finite_Vertex_Set_may_not_be_Finite | [
"Graph Theory",
"Multigraphs"
] | [
"Definition:Multigraph",
"Definition:Vertex Set",
"Definition:Finite Set",
"Definition:Finite Graph"
] | [
"Category:Graph Theory",
"Category:Multigraphs"
] |
proofwiki-12060 | Modulus Larger than Real Part | :$\cmod z \ge \size {\map \Re z}$ | By the definition of a complex number, we have:
:$z = \map \Re z + i \map \Im z$
Then:
{{begin-eqn}}
{{eqn | l = \cmod z
| r = \sqrt {\paren {\map \Re z}^2 + \paren {\map \Im z}^2}
| c = {{Defof|Complex Modulus}}
}}
{{eqn | o = \ge
| r = \sqrt {\paren {\map \Re z}^2 }
| c = Square of Real Number... | :$\cmod z \ge \size {\map \Re z}$ | By the [[Definition:Complex Number|definition of a complex number]], we have:
:$z = \map \Re z + i \map \Im z$
Then:
{{begin-eqn}}
{{eqn | l = \cmod z
| r = \sqrt {\paren {\map \Re z}^2 + \paren {\map \Im z}^2}
| c = {{Defof|Complex Modulus}}
}}
{{eqn | o = \ge
| r = \sqrt {\paren {\map \Re z}^2 }
... | Modulus Larger than Real Part | https://proofwiki.org/wiki/Modulus_Larger_than_Real_Part | https://proofwiki.org/wiki/Modulus_Larger_than_Real_Part | [
"Complex Modulus"
] | [] | [
"Definition:Complex Number",
"Square of Real Number is Non-Negative",
"Definition:Real Number",
"Square of Real Number is Non-Negative",
"Definition:Real Number"
] |
proofwiki-12061 | Modulus Larger than Imaginary Part | :$\cmod z \ge \size {\map \Im z}$ | By the definition of a complex number, we have:
:$z = \map \Re z + i \map \Im z$
{{begin-eqn}}
{{eqn | l = \cmod z
| r = \sqrt {\paren {\map \Re z}^2 + \paren {\map \Im z}^2}
| c = {{Defof|Complex Modulus}}
}}
{{eqn | o = \ge
| r = \sqrt {\paren {\map \Im z}^2 }
| c = Square of Real Number is No... | :$\cmod z \ge \size {\map \Im z}$ | By the [[Definition:Complex Number|definition of a complex number]], we have:
:$z = \map \Re z + i \map \Im z$
{{begin-eqn}}
{{eqn | l = \cmod z
| r = \sqrt {\paren {\map \Re z}^2 + \paren {\map \Im z}^2}
| c = {{Defof|Complex Modulus}}
}}
{{eqn | o = \ge
| r = \sqrt {\paren {\map \Im z}^2 }
| ... | Modulus Larger than Imaginary Part | https://proofwiki.org/wiki/Modulus_Larger_than_Imaginary_Part | https://proofwiki.org/wiki/Modulus_Larger_than_Imaginary_Part | [
"Complex Modulus"
] | [] | [
"Definition:Complex Number",
"Square of Real Number is Non-Negative",
"Definition:Real Number",
"Square of Real Number is Non-Negative",
"Definition:Real Number"
] |
proofwiki-12062 | Product Space is T0 iff Factor Spaces are T0 | :$T$ is a $T_0$ space {{iff}} each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_0$ space. | === Sufficient Condition ===
{{:Product Space is T0 iff Factor Spaces are T0/Sufficient Condition/Proof 1}}{{qed|lemma}} | :$T$ is a [[Definition:T0 Space|$T_0$ space]] {{iff}} each of $\struct {S_\alpha, \tau_\alpha}$ is a [[Definition:T0 Space|$T_0$ space]]. | === [[Product Space is T0 iff Factor Spaces are T0/Sufficient Condition|Sufficient Condition]] ===
{{:Product Space is T0 iff Factor Spaces are T0/Sufficient Condition/Proof 1}}{{qed|lemma}} | Product Space is T0 iff Factor Spaces are T0 | https://proofwiki.org/wiki/Product_Space_is_T0_iff_Factor_Spaces_are_T0 | https://proofwiki.org/wiki/Product_Space_is_T0_iff_Factor_Spaces_are_T0 | [
"Product Space is T0 iff Factor Spaces are T0",
"T0 Spaces",
"Separation Properties Preserved under Topological Product",
"Product Spaces"
] | [
"Definition:T0 Space",
"Definition:T0 Space"
] | [
"Product Space is T0 iff Factor Spaces are T0/Sufficient Condition"
] |
proofwiki-12063 | Product Space is T0 iff Factor Spaces are T0 | :$T$ is a $T_0$ space {{iff}} each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_0$ space. | Let $T$ be a $T_0$ space.
As $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice.
Let $\alpha \in I$.
From Subspace of Product Space is Homeomorphic to Factor Space, $\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$.
From $T_0$ Property is Hereditary, $T_\alpha$ is a $T_0$ s... | :$T$ is a [[Definition:T0 Space|$T_0$ space]] {{iff}} each of $\struct {S_\alpha, \tau_\alpha}$ is a [[Definition:T0 Space|$T_0$ space]]. | Let $T$ be a [[Definition:T0 Space|$T_0$ space]].
As $S_\alpha \ne \O$ we also have $S \ne \O$ by the [[Axiom:Axiom of Choice|axiom of choice]].
Let $\alpha \in I$.
From [[Subspace of Product Space is Homeomorphic to Factor Space]], $\struct {S_\alpha, \tau_\alpha}$ is [[Definition:Homeomorphism (Topological Spaces)... | Product Space is T0 iff Factor Spaces are T0/Sufficient Condition/Proof 1 | https://proofwiki.org/wiki/Product_Space_is_T0_iff_Factor_Spaces_are_T0 | https://proofwiki.org/wiki/Product_Space_is_T0_iff_Factor_Spaces_are_T0/Sufficient_Condition/Proof_1 | [
"Product Space is T0 iff Factor Spaces are T0",
"T0 Spaces",
"Separation Properties Preserved under Topological Product",
"Product Spaces"
] | [
"Definition:T0 Space",
"Definition:T0 Space"
] | [
"Definition:T0 Space",
"Axiom:Axiom of Choice",
"Subspace of Product Space is Homeomorphic to Factor Space",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Subspace",
"T0 Property is Hereditary",
"Definition:T0 Space",
"T0 Property is Preserved under Homeomorphism",
"Definition:T0 Space"... |
proofwiki-12064 | Product Space is T0 iff Factor Spaces are T0 | :$T$ is a $T_0$ space {{iff}} each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_0$ space. | Let $T$ be a $T_0$ space.
{{AimForCont}} $\exists \beta: \struct {S_\beta, \tau_\beta}$ is not a $T_0$ space.
Then $\exists a, b \in S_\beta$ such that $\forall U_\beta \in \tau_\beta$, either $a, b \in U_\beta$ or $a, b \notin U_\beta$.
Consider the elements $y, z \in S$ defined as:
:<nowiki>$y = \family {x_\alpha} :... | :$T$ is a [[Definition:T0 Space|$T_0$ space]] {{iff}} each of $\struct {S_\alpha, \tau_\alpha}$ is a [[Definition:T0 Space|$T_0$ space]]. | Let $T$ be a [[Definition:T0 Space|$T_0$ space]].
{{AimForCont}} $\exists \beta: \struct {S_\beta, \tau_\beta}$ is not a [[Definition:T0 Space|$T_0$ space]].
Then $\exists a, b \in S_\beta$ such that $\forall U_\beta \in \tau_\beta$, either $a, b \in U_\beta$ or $a, b \notin U_\beta$.
Consider the [[Definition:Eleme... | Product Space is T0 iff Factor Spaces are T0/Sufficient Condition/Proof 2 | https://proofwiki.org/wiki/Product_Space_is_T0_iff_Factor_Spaces_are_T0 | https://proofwiki.org/wiki/Product_Space_is_T0_iff_Factor_Spaces_are_T0/Sufficient_Condition/Proof_2 | [
"Product Space is T0 iff Factor Spaces are T0",
"T0 Spaces",
"Separation Properties Preserved under Topological Product",
"Product Spaces"
] | [
"Definition:T0 Space",
"Definition:T0 Space"
] | [
"Definition:T0 Space",
"Definition:T0 Space",
"Definition:Element",
"Definition:Cartesian Product/Coordinate",
"Definition:T0 Space",
"Proof by Contradiction",
"Definition:T0 Space"
] |
proofwiki-12065 | Equivalence of Definitions of Distance to Nearest Integer Function | The following definitions of the distance to nearest integer function $\norm \cdot: \R \to \closedint 0 {\dfrac 1 2}$ are equivalent: | Let $\alpha \in \R$, $n \in \Z$.
From Real Number is between Floor Functions:
:$\floor \alpha \le \alpha < \floor \alpha + 1$
For any $n < \floor \alpha \le \alpha$:
{{begin-eqn}}
{{eqn | l = \size {n - \alpha}
| r = \alpha - n
| c = {{Defof|Absolute Value}}
}}
{{eqn | o = >
| r = \alpha - \floor \alp... | The following definitions of the [[Definition:Distance to Nearest Integer Function|distance to nearest integer function]] $\norm \cdot: \R \to \closedint 0 {\dfrac 1 2}$ are [[Definition:Logical Equivalence|equivalent]]: | Let $\alpha \in \R$, $n \in \Z$.
From [[Real Number is between Floor Functions]]:
:$\floor \alpha \le \alpha < \floor \alpha + 1$
For any $n < \floor \alpha \le \alpha$:
{{begin-eqn}}
{{eqn | l = \size {n - \alpha}
| r = \alpha - n
| c = {{Defof|Absolute Value}}
}}
{{eqn | o = >
| r = \alpha - \flo... | Equivalence of Definitions of Distance to Nearest Integer Function | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Distance_to_Nearest_Integer_Function | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Distance_to_Nearest_Integer_Function | [
"Real Analysis"
] | [
"Definition:Distance to Nearest Integer Function",
"Definition:Logical Equivalence"
] | [
"Real Number is between Floor Functions",
"Category:Real Analysis"
] |
proofwiki-12066 | Quotient and Remainder to Number Base/General Result | :$\floor {\dfrac n {b^s} } = \sqbrk {r_m r_{m - 1} \ldots r_{s + 1} r_s}_b$
and:
:$\ds n \mod {b^s} = \sum_{j \mathop = 0}^{s - 1} {r_j b^j} = \sqbrk {r_{s - 1} r_{s - 2} \ldots r_1 r_0}_b$ | We have:
{{begin-eqn}}
{{eqn | l = \frac n {b^s}
| r = \frac {\sum_{j \mathop = 0}^m r_j b^j} {b^s}
}}
{{eqn | r = \frac {\sum_{j \mathop = 0}^{s - 1} r_j b^j} {b^s} + \sum_{j \mathop = 0}^{m - s} r_{s + j} b^j
}}
{{eqn | r = \frac {\sum_{j \mathop = 0}^{s - 1} r_j b^j} {b^s} + \sqbrk {r_m r_{m - 1} \ldots r_{s +... | :$\floor {\dfrac n {b^s} } = \sqbrk {r_m r_{m - 1} \ldots r_{s + 1} r_s}_b$
and:
:$\ds n \mod {b^s} = \sum_{j \mathop = 0}^{s - 1} {r_j b^j} = \sqbrk {r_{s - 1} r_{s - 2} \ldots r_1 r_0}_b$ | We have:
{{begin-eqn}}
{{eqn | l = \frac n {b^s}
| r = \frac {\sum_{j \mathop = 0}^m r_j b^j} {b^s}
}}
{{eqn | r = \frac {\sum_{j \mathop = 0}^{s - 1} r_j b^j} {b^s} + \sum_{j \mathop = 0}^{m - s} r_{s + j} b^j
}}
{{eqn | r = \frac {\sum_{j \mathop = 0}^{s - 1} r_j b^j} {b^s} + \sqbrk {r_m r_{m - 1} \ldots r_{s +... | Quotient and Remainder to Number Base/General Result | https://proofwiki.org/wiki/Quotient_and_Remainder_to_Number_Base/General_Result | https://proofwiki.org/wiki/Quotient_and_Remainder_to_Number_Base/General_Result | [
"Number Theory"
] | [] | [
"Sum of Geometric Sequence",
"Definition:Floor Function",
"Category:Number Theory"
] |
proofwiki-12067 | Remainder is Primitive Recursive | Let $m, n \in \N$ be natural numbers.
Let us define the function $\rem: \N^2 \to \N$:
:<nowiki>$\map \rem {n, m} = \begin{cases}
\text{the remainder when } n \text{ is divided by } m & : m \ne 0 \\
0 & : m = 0
\end{cases}$</nowiki>
where the $\text{remainder}$ is as defined in the Division Theorem:
:If $n = m q + r$, w... | Let $\map \rem {n, m} = r$.
We see that as $n$ increases by $1$, then so does $r$, except when $n = m-1$ in which case increasing $n$ by $1$ makes $r$ go to $0$.
We also take into account the case where $m = 0$:
So we can define $\rem$ by cases:
:<nowiki>$\map \rem {n, m} = \begin{cases}
0 & : \map \rem {n - 1, m} = m ... | Let $m, n \in \N$ be [[Definition:Natural Numbers|natural numbers]].
Let us define the [[Definition:Function|function]] $\rem: \N^2 \to \N$:
:<nowiki>$\map \rem {n, m} = \begin{cases}
\text{the remainder when } n \text{ is divided by } m & : m \ne 0 \\
0 & : m = 0
\end{cases}$</nowiki>
where the $\text{remainder}$ i... | Let $\map \rem {n, m} = r$.
We see that as $n$ increases by $1$, then so does $r$, except when $n = m-1$ in which case increasing $n$ by $1$ makes $r$ go to $0$.
We also take into account the case where $m = 0$:
So we can define $\rem$ by cases:
:<nowiki>$\map \rem {n, m} = \begin{cases}
0 & : \map \rem {n - 1, m} =... | Remainder is Primitive Recursive | https://proofwiki.org/wiki/Remainder_is_Primitive_Recursive | https://proofwiki.org/wiki/Remainder_is_Primitive_Recursive | [
"Remainders",
"Primitive Recursive Functions"
] | [
"Definition:Natural Numbers",
"Definition:Function",
"Division Theorem",
"Definition:Remainder",
"Definition:Primitive Recursive/Function"
] | [
"Definition:Primitive Recursive/Function",
"Signum Function is Primitive Recursive",
"Equality Relation is Primitive Recursive",
"Cut-Off Subtraction is Primitive Recursive",
"Definition:Primitive Recursion",
"Permutation of Variables of Primitive Recursive Function",
"Definition:Primitive Recursive/Fun... |
proofwiki-12068 | Quotient is Primitive Recursive | Let $m, n \in \N$ be natural numbers.
Let us define the function $\operatorname {quot}: \N^2 \to \N$:
:$\map {\operatorname {quot} } {n, m} = \begin{cases}
\text{the quotient when } n \text{ is divided by } m & : m \ne 0 \\
0 & : m = 0
\end{cases}$
where the $\text {quotient}$ and $\text {remainder}$ are as defined in ... | We note that if $m \ne 0$ and $n = m q + r$, we have:
:$\dfrac n m = q + \frac r m$
Also note that $\dfrac n m$ and $\dfrac r m$ are rational numbers and not necessarily natural numbers.
Indeed, we have:
:$0 \le \dfrac r m < 1$
So if $m > 0$ then $\map {\operatorname {quot} } {n, m}$ is the floor $\floor {\dfrac n m}$ ... | Let $m, n \in \N$ be [[Definition:Natural Numbers|natural numbers]].
Let us define the [[Definition:Function|function]] $\operatorname {quot}: \N^2 \to \N$:
:$\map {\operatorname {quot} } {n, m} = \begin{cases}
\text{the quotient when } n \text{ is divided by } m & : m \ne 0 \\
0 & : m = 0
\end{cases}$
where the $\... | We note that if $m \ne 0$ and $n = m q + r$, we have:
:$\dfrac n m = q + \frac r m$
Also note that $\dfrac n m$ and $\dfrac r m$ are [[Definition:Rational Number|rational numbers]] and not necessarily [[Definition:Natural Numbers|natural numbers]].
Indeed, we have:
:$0 \le \dfrac r m < 1$
So if $m > 0$ then $\map {\... | Quotient is Primitive Recursive | https://proofwiki.org/wiki/Quotient_is_Primitive_Recursive | https://proofwiki.org/wiki/Quotient_is_Primitive_Recursive | [
"Primitive Recursive Functions"
] | [
"Definition:Natural Numbers",
"Definition:Function",
"Division Theorem",
"Definition:Quotient (Integer Division)",
"Definition:Primitive Recursive/Function"
] | [
"Definition:Rational Number",
"Definition:Natural Numbers",
"Definition:Floor Function",
"Definition:Primitive Recursion",
"Permutation of Variables of Primitive Recursive Function",
"Definition:Primitive Recursive/Function",
"Signum Function is Primitive Recursive",
"Remainder is Primitive Recursive"... |
proofwiki-12069 | Tarski's Geometry is Complete/Corollary | Tarski's geometry does not contain minimal arithmetic. | Immediate from Tarski's Geometry is Complete and the {{Corollary|Gödel's First Incompleteness Theorem}}.
{{qed}}
Category:Tarski's Geometry
6npbwztu4wi4lm1c1hnsp8vwh6u33b4 | [[Definition:Tarski's Geometry|Tarski's geometry]] does not contain [[Definition:Minimal Arithmetic|minimal arithmetic]]. | Immediate from [[Tarski's Geometry is Complete]] and the {{Corollary|Gödel's First Incompleteness Theorem}}.
{{qed}}
[[Category:Tarski's Geometry]]
6npbwztu4wi4lm1c1hnsp8vwh6u33b4 | Tarski's Geometry is Complete/Corollary | https://proofwiki.org/wiki/Tarski's_Geometry_is_Complete/Corollary | https://proofwiki.org/wiki/Tarski's_Geometry_is_Complete/Corollary | [
"Tarski's Geometry"
] | [
"Definition:Tarski's Geometry",
"Definition:Minimal Arithmetic"
] | [
"Tarski's Geometry is Complete",
"Category:Tarski's Geometry"
] |
proofwiki-12070 | Symmetric Relation equals its Symmetric Closure | Let $\RR$ be a symmetric relation on a set $S$.
Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$.
Then:
:$\RR = \RR^\leftrightarrow$ | {{begin-eqn}}
{{eqn | l = \RR^\leftrightarrow
| r = \RR \cup \RR^{-1}
| c = {{Defof|Symmetric Closure}}
}}
{{eqn | r = \RR \cup \RR
| c = Inverse of Symmetric Relation is Symmetric
}}
{{eqn | r = \RR
| c = Set Union is Idempotent
}}
{{end-eqn}}
{{qed}}
Category:Symmetric Relations
Category:Symm... | Let $\RR$ be a [[Definition:Symmetric Relation|symmetric relation]] on a [[Definition:Set|set]] $S$.
Let $\RR^\leftrightarrow$ be the [[Definition:Symmetric Closure|symmetric closure]] of $\RR$.
Then:
:$\RR = \RR^\leftrightarrow$ | {{begin-eqn}}
{{eqn | l = \RR^\leftrightarrow
| r = \RR \cup \RR^{-1}
| c = {{Defof|Symmetric Closure}}
}}
{{eqn | r = \RR \cup \RR
| c = [[Inverse of Symmetric Relation is Symmetric]]
}}
{{eqn | r = \RR
| c = [[Set Union is Idempotent]]
}}
{{end-eqn}}
{{qed}}
[[Category:Symmetric Relations]]
... | Symmetric Relation equals its Symmetric Closure | https://proofwiki.org/wiki/Symmetric_Relation_equals_its_Symmetric_Closure | https://proofwiki.org/wiki/Symmetric_Relation_equals_its_Symmetric_Closure | [
"Symmetric Relations",
"Symmetric Closures"
] | [
"Definition:Symmetric Relation",
"Definition:Set",
"Definition:Symmetric Closure"
] | [
"Inverse of Symmetric Relation is Symmetric",
"Set Union is Idempotent",
"Category:Symmetric Relations",
"Category:Symmetric Closures"
] |
proofwiki-12071 | Not Every Two-Person Zero-Sum Game has Saddle Point | Not every two-person zero-sum game has a saddle point. | Consider the game of Matching Pennies.
Recall its payoff table:
{{:Definition:Matching Pennies/Payoff Table}}
Trivially, by inspection, this has no entry which is the smallest entry in its row and the largest entry in its column.
{{qed}} | Not every [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]] has a [[Definition:Saddle Point (Game Theory)|saddle point]]. | Consider the [[Definition:Matching Pennies|game of Matching Pennies]].
Recall its [[Definition:Matching Pennies/Payoff Table|payoff table]]:
{{:Definition:Matching Pennies/Payoff Table}}
Trivially, by inspection, this has no [[Definition:Entry in Payoff Table|entry]] which is the smallest [[Definition:Entry in Payoff... | Not Every Two-Person Zero-Sum Game has Saddle Point | https://proofwiki.org/wiki/Not_Every_Two-Person_Zero-Sum_Game_has_Saddle_Point | https://proofwiki.org/wiki/Not_Every_Two-Person_Zero-Sum_Game_has_Saddle_Point | [
"Two-Person Zero-Sum Games"
] | [
"Definition:Two-Person Zero-Sum Game",
"Definition:Saddle Point (Game Theory)"
] | [
"Definition:Matching Pennies",
"Definition:Matching Pennies/Payoff Table",
"Definition:Payoff Table/Entry",
"Definition:Payoff Table/Entry",
"Definition:Payoff Table/Entry"
] |
proofwiki-12072 | Eluding Game has no Saddle Point | The eluding game has no saddle point. | Recall the payoff table:
{{:Definition:Eluding Game/Payoff Table}}
Trivially, by inspection, this has no entry which is the smallest entry in its row and the largest entry in its column. | The [[Definition:Eluding Game|eluding game]] has no [[Definition:Saddle Point (Game Theory)|saddle point]]. | Recall the [[Definition:Eluding Game/Payoff Table|payoff table]]:
{{:Definition:Eluding Game/Payoff Table}}
Trivially, by inspection, this has no [[Definition:Entry in Payoff Table|entry]] which is the smallest [[Definition:Entry in Payoff Table|entry]] in its row and the largest [[Definition:Entry in Payoff Table|ent... | Eluding Game has no Saddle Point | https://proofwiki.org/wiki/Eluding_Game_has_no_Saddle_Point | https://proofwiki.org/wiki/Eluding_Game_has_no_Saddle_Point | [
"Two-Person Zero-Sum Games"
] | [
"Definition:Eluding Game",
"Definition:Saddle Point (Game Theory)"
] | [
"Definition:Eluding Game/Payoff Table",
"Definition:Payoff Table/Entry",
"Definition:Payoff Table/Entry",
"Definition:Payoff Table/Entry"
] |
proofwiki-12073 | Two-Person Zero-Sum Game with Multiple Solutions | There exists a two-person zero-sum game with more than one solution. | Consider the game defined by the following payoff table:
{{PayoffTable|table = <nowiki>$\begin{array} {r {{|}} c {{|}} c {{|}} c {{|}} }
& B_1 & B_2 & B_3 \\
\hline A_1 & 1 & 2 & 4 \\
\hline A_2 & 4 & 2 & 1 \\
\hline \end{array}$</nowiki>}}
This has two solutions:
:$(1): \quad A: \tuple {1/3, 2/3}, B: \tuple {0, 1, 0}$... | There exists a [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]] with more than one [[Definition:Solution of Game|solution]]. | Consider the [[Definition:Game|game]] defined by the following [[Definition:Payoff Table|payoff table]]:
{{PayoffTable|table = <nowiki>$\begin{array} {r {{|}} c {{|}} c {{|}} c {{|}} }
& B_1 & B_2 & B_3 \\
\hline A_1 & 1 & 2 & 4 \\
\hline A_2 & 4 & 2 & 1 \\
\hline \end{array}$</nowiki>}}
This has two [[Definition:Sol... | Two-Person Zero-Sum Game with Multiple Solutions | https://proofwiki.org/wiki/Two-Person_Zero-Sum_Game_with_Multiple_Solutions | https://proofwiki.org/wiki/Two-Person_Zero-Sum_Game_with_Multiple_Solutions | [
"Examples of Two-Person Zero-Sum Games"
] | [
"Definition:Two-Person Zero-Sum Game",
"Definition:Solution of Game"
] | [
"Definition:Game",
"Definition:Payoff Table",
"Definition:Solution of Game",
"Definition:Game/Strategy",
"Definition:Solution of Game"
] |
proofwiki-12074 | Two-Person Zero-Sum Game with Finite Strategies has Solution | Let $G$ be a two-person zero-sum game.
Let each player of $G$ have a finite set of strategies available.
Then $G$ has at least one solution. | {{ProofWanted|Proof given later in book. Chapter III gets technical.}} | Let $G$ be a [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]].
Let each [[Definition:Player|player]] of $G$ have a [[Definition:Finite Set|finite set]] of [[Definition:Strategy|strategies]] available.
Then $G$ has at least one [[Definition:Solution of Game|solution]]. | {{ProofWanted|Proof given later in book. Chapter III gets technical.}} | Two-Person Zero-Sum Game with Finite Strategies has Solution | https://proofwiki.org/wiki/Two-Person_Zero-Sum_Game_with_Finite_Strategies_has_Solution | https://proofwiki.org/wiki/Two-Person_Zero-Sum_Game_with_Finite_Strategies_has_Solution | [
"Two-Person Zero-Sum Games"
] | [
"Definition:Two-Person Zero-Sum Game",
"Definition:Game/Player",
"Definition:Finite Set",
"Definition:Game/Strategy",
"Definition:Solution of Game"
] | [] |
proofwiki-12075 | Card Game with Bluffing is Completely Mixed Game | The game of cards with bluffing is a completely mixed game. | From Solution to Card Game with Bluffing, the optimum strategy for the [[Definition:Card Game with Bluffing|card game with bluffing, is:
:For $A$, to use both pure strategies but favour $A_1$ by $2/3$ to $1/3$.
:For $B$, to use both pure strategies but favour $B_1$ by $2/3$ to $1/3$.
Hence the result by definition of c... | The [[Definition:Game|game]] of [[Definition:Card Game with Bluffing|cards with bluffing]] is a [[Definition:Completely Mixed Game|completely mixed game]]. | From [[Solution to Card Game with Bluffing]], the optimum [[Definition:Strategy|strategy]] for the [[Definition:Card Game with Bluffing|card game with bluffing, is:
:For $A$, to use both [[Definition:Pure Strategy|pure strategies]] but favour $A_1$ by $2/3$ to $1/3$.
:For $B$, to use both [[Definition:Pure Strategy|pur... | Card Game with Bluffing is Completely Mixed Game | https://proofwiki.org/wiki/Card_Game_with_Bluffing_is_Completely_Mixed_Game | https://proofwiki.org/wiki/Card_Game_with_Bluffing_is_Completely_Mixed_Game | [
"Examples of Two-Person Zero-Sum Games"
] | [
"Definition:Game",
"Definition:Card Game with Bluffing",
"Definition:Completely Mixed Game"
] | [
"Solution to Card Game with Bluffing",
"Definition:Game/Strategy",
"Definition:Pure Strategy",
"Definition:Pure Strategy",
"Definition:Completely Mixed Game"
] |
proofwiki-12076 | Matching Pennies is Completely Mixed Game | The game of matching pennies is a completely mixed game. | The optimum strategy for matching pennies, for both players, is to randomise both their pure strategies.
Hence the result by definition of completely mixed game. | The [[Definition:Game|game]] of [[Definition:Matching Pennies|matching pennies]] is a [[Definition:Completely Mixed Game|completely mixed game]]. | The optimum [[Definition:Strategy|strategy]] for [[Definition:Matching Pennies|matching pennies]], for both [[Definition:Player|players]], is to randomise both their [[Definition:Pure Strategy|pure strategies]].
Hence the result by definition of [[Definition:Completely Mixed Game|completely mixed game]]. | Matching Pennies is Completely Mixed Game | https://proofwiki.org/wiki/Matching_Pennies_is_Completely_Mixed_Game | https://proofwiki.org/wiki/Matching_Pennies_is_Completely_Mixed_Game | [
"Examples of Two-Person Zero-Sum Games"
] | [
"Definition:Game",
"Definition:Matching Pennies",
"Definition:Completely Mixed Game"
] | [
"Definition:Game/Strategy",
"Definition:Matching Pennies",
"Definition:Game/Player",
"Definition:Pure Strategy",
"Definition:Completely Mixed Game"
] |
proofwiki-12077 | Dominated Strategy may be Optimal | A dominated strategy of a game may be the optimal strategy for a player of that game. | Consider the game defined by the following payoff table:
{{PayoffTable|table = <nowiki>$\begin{array} {r {{|}} c {{|}} }
& B_1 & B_2 \\
\hline A_1 & 1 & 2 \\
\hline A_2 & 1 & 3 \\
\hline \end{array}$</nowiki>}}
This has two solutions:
:$(1): \quad A: \tuple {1, 0}, B: \tuple {1, 0}$
:$(2): \quad A: \tuple {0, 1}, B: \... | A [[Definition:Dominating Strategy|dominated strategy]] of a [[Definition:Game|game]] may be the [[Definition:Optimal Strategy|optimal strategy]] for a [[Definition:Player|player]] of that [[Definition:Game|game]]. | Consider the [[Definition:Game|game]] defined by the following [[Definition:Payoff Table|payoff table]]:
{{PayoffTable|table = <nowiki>$\begin{array} {r {{|}} c {{|}} }
& B_1 & B_2 \\
\hline A_1 & 1 & 2 \\
\hline A_2 & 1 & 3 \\
\hline \end{array}$</nowiki>}}
This has two [[Definition:Solution of Game|solutions]]:
:... | Dominated Strategy may be Optimal | https://proofwiki.org/wiki/Dominated_Strategy_may_be_Optimal | https://proofwiki.org/wiki/Dominated_Strategy_may_be_Optimal | [
"Dominating Strategies",
"Optimal Strategies"
] | [
"Definition:Dominating Strategy",
"Definition:Game",
"Definition:Optimal Strategy",
"Definition:Game/Player",
"Definition:Game"
] | [
"Definition:Game",
"Definition:Payoff Table",
"Definition:Solution of Game",
"Definition:Pure Strategy",
"Definition:Optimal Strategy",
"Definition:Dominating Strategy"
] |
proofwiki-12078 | Value of Skew-Symmetric Game is Zero | 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 wanted|Once we have a definition for skew-symmetric we can start thinking about how to present this.}} | Let $G$ be a [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]].
Let $G$ be represented by a [[Definition:Payoff Table|payoff table]] that is [[Definition:Skew-Symmetric|skew-symmetric]].
Then the [[Definition:Value|value]] of $G$ is zero. | {{proof wanted|Once we have a definition for skew-symmetric we can start thinking about how to present this.}} | Value of Skew-Symmetric Game is Zero | https://proofwiki.org/wiki/Value_of_Skew-Symmetric_Game_is_Zero | https://proofwiki.org/wiki/Value_of_Skew-Symmetric_Game_is_Zero | [
"Two-Person Zero-Sum Games"
] | [
"Definition:Two-Person Zero-Sum Game",
"Definition:Payoff Table",
"Definition:Skew-Symmetric",
"Definition:Value"
] | [] |
proofwiki-12079 | Prime Element iff Complement of Lower Closure is Filter | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded above lattice.
Let $p \in S$ such that
:$p \ne \top$
where $\top$ denotes the top of $L$.
Then:
:$p$ is a prime element
{{iff}}
:$\relcomp S {p^\preceq}$ is filter in $L$
where
:$p^\preceq$ denotes the lower closure of $p$
:$\relcomp S {p^\preceq}$ denotes the r... | === Sufficient Condition ===
Assume that $p$ is a prime element.
By definition of the greatest element:
:$p \preceq \top$
By definition of $\prec$:
:$p \prec \top$
By definition of antisymmetry:
:$\top \npreceq p$
By definition of lower closure of element:
:$\top \notin p^\preceq$
By definition of relative complement:
... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Above Set|bounded above]] [[Definition:Lattice (Order Theory)|lattice]].
Let $p \in S$ such that
:$p \ne \top$
where $\top$ denotes the [[Definition:Top of Lattice|top]] of $L$.
Then:
:$p$ is a [[Definition:Prime Element (Order Theory)|prime elem... | === Sufficient Condition ===
Assume that $p$ is a [[Definition:Prime Element (Order Theory)|prime element]].
By definition of the [[Definition:Greatest Element|greatest element]]:
:$p \preceq \top$
By definition of $\prec$:
:$p \prec \top$
By definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
:$\top ... | Prime Element iff Complement of Lower Closure is Filter | https://proofwiki.org/wiki/Prime_Element_iff_Complement_of_Lower_Closure_is_Filter | https://proofwiki.org/wiki/Prime_Element_iff_Complement_of_Lower_Closure_is_Filter | [
"Prime Elements"
] | [
"Definition:Bounded Above Set",
"Definition:Lattice (Order Theory)",
"Definition:Top of Lattice",
"Definition:Prime Element (Order Theory)",
"Definition:Filter in Ordered Set",
"Definition:Lower Closure/Element",
"Definition:Relative Complement"
] | [
"Definition:Prime Element (Order Theory)",
"Definition:Greatest Element",
"Definition:Antisymmetric Relation",
"Definition:Lower Closure/Element",
"Definition:Relative Complement",
"Definition:Non-Empty Set",
"Definition:Filtered Subset",
"Definition:Relative Complement",
"Definition:Lower Closure/E... |
proofwiki-12080 | Oscillation at Point (Infimum) equals Oscillation at Point (Epsilon-Neighborhood) | Let $f: D \to \R$ be a real function where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $N_x$ be the set of neighborhoods of $x$.
Let $E_x$ be the set of $\epsilon$-neighborhoods of $x$.
Let $\map {\omega_f} x$ be the oscillation of $f$ at $x$ based on $N_x$:
:$\map {\omega_f} x = \inf \set {\map {\omega_f} {I \cap... | === Lemma ===
Let $f: D \to \R$ be a real function where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $N_x$ be the set of neighborhoods of $x$.
Let $E_x$ be the set of $\epsilon$-neighborhoods of $x$.
Let $\map {\omega_f} x$ be the oscillation of $f$ at $x$ based on $N_x$:
:$\map {\omega_f} x = \inf \set {\map {\om... | Let $f: D \to \R$ be a [[Definition:Real Function|real function]] where $D \subseteq \R$.
Let $x$ be a [[Definition:Element|point]] in $D$.
Let $N_x$ be the [[Definition:Set|set]] of [[Definition:Neighborhood of Real Number|neighborhoods]] of $x$.
Let $E_x$ be the [[Definition:Set|set]] of [[Definition:Epsilon-Neigh... | === Lemma ===
Let $f: D \to \R$ be a [[Definition:Real Function|real function]] where $D \subseteq \R$.
Let $x$ be a [[Definition:Element|point]] in $D$.
Let $N_x$ be the [[Definition:Set|set]] of [[Definition:Neighborhood of Real Number|neighborhoods]] of $x$.
Let $E_x$ be the [[Definition:Set|set]] of [[Definitio... | Oscillation at Point (Infimum) equals Oscillation at Point (Epsilon-Neighborhood) | https://proofwiki.org/wiki/Oscillation_at_Point_(Infimum)_equals_Oscillation_at_Point_(Epsilon-Neighborhood) | https://proofwiki.org/wiki/Oscillation_at_Point_(Infimum)_equals_Oscillation_at_Point_(Epsilon-Neighborhood) | [
"Oscillation"
] | [
"Definition:Real Function",
"Definition:Element",
"Definition:Set",
"Definition:Neighborhood (Real Analysis)/Open Subset",
"Definition:Set",
"Definition:Neighborhood (Real Analysis)/Epsilon",
"Definition:Oscillation/Real Space/Oscillation at Point/Infimum",
"Definition:Oscillation/Real Space/Oscillati... | [
"Definition:Real Function",
"Definition:Element",
"Definition:Set",
"Definition:Neighborhood (Real Analysis)/Open Subset",
"Definition:Set",
"Definition:Neighborhood (Real Analysis)/Epsilon",
"Definition:Oscillation/Real Space/Oscillation at Point/Infimum",
"Definition:Oscillation/Real Space/Oscillati... |
proofwiki-12081 | Prime Element iff Element Greater is Top | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a Boolean lattice.
Let $p \in S$ such that:
:$p \ne \top$
Then:
:$p$ is prime element
{{iff}}:
:$\forall x \in S: \paren {p \prec x \implies x = \top}$ | === Sufficient Condition ===
Suppose that:
:$p$ is prime element
Let $x \in S$ such that:
:$p \prec x$
By definition of Boolean lattice:
:$L$ is complemented distributive lattice.
By definition of complemented:
:$\exists y \in S: y$ is complement of $x$.
By definition of complement:
:$x \wedge y = \bot$
By definition o... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Boolean Lattice|Boolean lattice]].
Let $p \in S$ such that:
:$p \ne \top$
Then:
:$p$ is [[Definition:Prime Element (Order Theory)|prime element]]
{{iff}}:
:$\forall x \in S: \paren {p \prec x \implies x = \top}$ | === Sufficient Condition ===
Suppose that:
:$p$ is [[Definition:Prime Element (Order Theory)|prime element]]
Let $x \in S$ such that:
:$p \prec x$
By definition of [[Definition:Boolean Lattice|Boolean lattice]]:
:$L$ is [[Definition:Complemented Lattice|complemented]] [[Definition:Distributive Lattice|distributive l... | Prime Element iff Element Greater is Top | https://proofwiki.org/wiki/Prime_Element_iff_Element_Greater_is_Top | https://proofwiki.org/wiki/Prime_Element_iff_Element_Greater_is_Top | [
"Prime Elements"
] | [
"Definition:Boolean Lattice",
"Definition:Prime Element (Order Theory)"
] | [
"Definition:Prime Element (Order Theory)",
"Definition:Boolean Lattice",
"Definition:Complemented Lattice",
"Definition:Distributive Lattice",
"Definition:Complemented Lattice",
"Definition:Complement (Lattice Theory)",
"Definition:Complement (Lattice Theory)",
"Definition:Smallest Element",
"Defini... |
proofwiki-12082 | Prime Element iff There Exists Way Below Open Filter which Complement has Maximum | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous distributive lattice.
Let $p \in S$ such that:
:$p \ne \top$
where $\top$ denotes the top of $L$.
Then
:$p$ is a prime element
{{iff}}:
:there exists a way below open filter $F$ in $L$: $p = \map \max {\relcomp S F}$ | === Sufficient Condition ===
Let $p$ be a prime element.
By definition of continuous:
:$\forall x \in S: x^\ll$ is directed
We will prove that:
:$\forall x \in S: \paren {x \in \relcomp S {p^\preceq} \implies \exists y \in S: y \in \relcomp S {p^\preceq} \land y \ll x}$
Let $x \in S$ such that:
:$x \in \relcomp S {p^\p... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Distributive Lattice|distributive lattice]].
Let $p \in S$ such that:
:$p \ne \top$
where $\top$ denotes the [[Definition:Top of Lattice|top]] of $L$.
Then
:$... | === Sufficient Condition ===
Let $p$ be a [[Definition:Prime Element (Order Theory)|prime element]].
By definition of [[Definition:Continuous Ordered Set|continuous]]:
:$\forall x \in S: x^\ll$ is [[Definition:Directed Subset|directed]]
We will prove that:
:$\forall x \in S: \paren {x \in \relcomp S {p^\preceq} \imp... | Prime Element iff There Exists Way Below Open Filter which Complement has Maximum | https://proofwiki.org/wiki/Prime_Element_iff_There_Exists_Way_Below_Open_Filter_which_Complement_has_Maximum | https://proofwiki.org/wiki/Prime_Element_iff_There_Exists_Way_Below_Open_Filter_which_Complement_has_Maximum | [
"Prime Elements"
] | [
"Definition:Bounded Below Set",
"Definition:Continuous Ordered Set",
"Definition:Distributive Lattice",
"Definition:Top of Lattice",
"Definition:Prime Element (Order Theory)",
"Definition:Way Below Open",
"Definition:Filter in Ordered Set"
] | [
"Definition:Prime Element (Order Theory)",
"Definition:Continuous Ordered Set",
"Definition:Directed Subset",
"Definition:Relative Complement",
"Definition:Lower Closure/Element",
"Axiom of Approximation in Up-Complete Semilattice",
"Definition:Lower Closure/Element",
"Definition:Relative Complement",... |
proofwiki-12083 | Characterization of Prime Ideal | Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $I$ be an proper ideal in $L$.
Then
:$I$ is a prime ideal
{{iff}}:
:$\forall x, y \in S: \paren {x \wedge y \in I \implies x \in I \lor y \in I}$ | === Sufficient Condition ===
Assume that
:$I$ is a prime ideal.
Let $x, y \in S$ such that
:$x \wedge y \in I$
By definition of relative complement:
:$x \wedge y \notin \relcomp S I$
By definition of prime ideal:
:$\relcomp S I$ is filter in $L$.
By Filtered in Meet Semilattice:
:$x \notin \relcomp S I$ or $y \notin \r... | Let $L = \struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
Let $I$ be an [[Definition:Proper Subset|proper]] [[Definition:Ideal (Order Theory)|ideal]] in $L$.
Then
:$I$ is a [[Definition:Prime Ideal (Order Theory)|prime ideal]]
{{iff}}:
:$\forall x, y \in S: \paren {x \wedge y \in ... | === Sufficient Condition ===
Assume that
:$I$ is a [[Definition:Prime Ideal (Order Theory)|prime ideal]].
Let $x, y \in S$ such that
:$x \wedge y \in I$
By definition of [[Definition:Relative Complement|relative complement]]:
:$x \wedge y \notin \relcomp S I$
By definition of [[Definition:Prime Ideal (Order Theory)... | Characterization of Prime Ideal | https://proofwiki.org/wiki/Characterization_of_Prime_Ideal | https://proofwiki.org/wiki/Characterization_of_Prime_Ideal | [
"Prime Ideals (Order Theory)"
] | [
"Definition:Meet Semilattice",
"Definition:Proper Subset",
"Definition:Ideal (Order Theory)",
"Definition:Prime Ideal (Order Theory)"
] | [
"Definition:Prime Ideal (Order Theory)",
"Definition:Relative Complement",
"Definition:Prime Ideal (Order Theory)",
"Definition:Filter",
"Filtered in Meet Semilattice",
"Definition:Relative Complement",
"Definition:Relative Complement",
"Definition:Relative Complement",
"Definition:Relative Compleme... |
proofwiki-12084 | Matching Pennies is Strictly Competitive | The game of Matching Pennies is strictly competitive. | By definition, a strictly competitive game is a game in which the interests of each player are diametrically opposed.
Recall the payoff table of Matching Pennies:
{{:Definition:Matching Pennies/Payoff Table}}
It can be seen by inspection that exchanging $\text A$ and $\text B$ results in exactly the same entries in the... | The [[Definition:Game|game]] of [[Definition:Matching Pennies|Matching Pennies]] is [[Definition:Strictly Competitive Game|strictly competitive]]. | By definition, a [[Definition:Strictly Competitive Game|strictly competitive game]] is a [[Definition:Game|game]] in which the interests of each [[Definition:Player|player]] are diametrically opposed.
Recall the [[Definition:Matching Pennies/Payoff Table|payoff table]] of [[Definition:Matching Pennies|Matching Pennies... | Matching Pennies is Strictly Competitive | https://proofwiki.org/wiki/Matching_Pennies_is_Strictly_Competitive | https://proofwiki.org/wiki/Matching_Pennies_is_Strictly_Competitive | [
"Examples of Two-Person Games"
] | [
"Definition:Game",
"Definition:Matching Pennies",
"Definition:Strictly Competitive Game"
] | [
"Definition:Strictly Competitive Game",
"Definition:Game",
"Definition:Game/Player",
"Definition:Matching Pennies/Payoff Table",
"Definition:Matching Pennies",
"Definition:Payoff Table/Entry",
"Definition:Matching Pennies/Payoff Table",
"Definition:Matching Pennies/Payoff Table",
"Definition:Game/Pl... |
proofwiki-12085 | Intersection Condition for Direct Sum of Subspaces | Let $U$ and $W$ be subspaces of a vector space $V$.
Then $U + W$ is a direct sum {{iff}} $U \cap W = 0$. | We must first prove that if $U + W$ is a direct sum, then:
:$U \cap W = 0$
Let $U + W$ be a direct sum.
Let $\mathbf v \in U \cap W$ be an arbitrary vector in $U \cap W$.
Then:
:$\mathbf 0 = \mathbf v + \paren {-\mathbf v}$
where:
:$\mathbf v \in U$
:$-\mathbf v \in W$
{{ProofWanted}} | Let $U$ and $W$ be subspaces of a [[Definition:Vector Space|vector space]] $V$.
Then $U + W$ is a [[Definition:Internal Direct Sum of Modules|direct sum]] {{iff}} $U \cap W = 0$. | We must first prove that if $U + W$ is a [[Definition:Internal Direct Sum of Modules|direct sum]], then:
:$U \cap W = 0$
Let $U + W$ be a [[Definition:Internal Direct Sum of Modules|direct sum]].
Let $\mathbf v \in U \cap W$ be an arbitrary [[Definition:Vector|vector]] in $U \cap W$.
Then:
:$\mathbf 0 = \mathbf v +... | Intersection Condition for Direct Sum of Subspaces | https://proofwiki.org/wiki/Intersection_Condition_for_Direct_Sum_of_Subspaces | https://proofwiki.org/wiki/Intersection_Condition_for_Direct_Sum_of_Subspaces | [
"Linear Algebra"
] | [
"Definition:Vector Space",
"Definition:Internal Direct Sum of Modules"
] | [
"Definition:Internal Direct Sum of Modules",
"Definition:Internal Direct Sum of Modules",
"Definition:Vector"
] |
proofwiki-12086 | Second Price Auction has Inefficient Equilibria | A second price auction has Nash equilibria which are inefficient. | {{ProofWanted|Awaiting a reliable definition of inefficient Nash equilibrium}} | A [[Definition:Second Price Auction|second price auction]] has [[Definition:Nash Equilibrium|Nash equilibria]] which are [[Definition:Inefficient Nash Equilibrium|inefficient]]. | {{ProofWanted|Awaiting a reliable definition of inefficient Nash equilibrium}} | Second Price Auction has Inefficient Equilibria | https://proofwiki.org/wiki/Second_Price_Auction_has_Inefficient_Equilibria | https://proofwiki.org/wiki/Second_Price_Auction_has_Inefficient_Equilibria | [
"Auctions"
] | [
"Definition:Sealed-Bid Auction/Second Price",
"Definition:Nash Equilibrium",
"Definition:Inefficient Nash Equilibrium"
] | [] |
proofwiki-12087 | Epic Equalizer is Isomorphism | Let $\mathbf C$ be a metacategory.
Let $e: E \to C$ be the equalizer of two morphisms $f, g: C \to D$.
Let $e$ be an epimorphism.
Then $e$ is an isomorphism. | As $e$ equalises $f$ and $g$, $f \circ e = g \circ e$.
Since $e$ is epic, it follows that $f = g$.
Then in the equaliser diagram:
:::<nowiki>$\begin{xy}\xymatrix{
E
\ar[r]^*{e}
&
A
\ar[r]<2pt>^*{f}
\ar[r]<-2pt>_*{g}
&
B
\\
A
\ar@{.>}[u]^*{k}
\ar[ur]_*{\operatorname{id}_A}
}\end{xy}</nowiki>$
We have th... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $e: E \to C$ be the [[Definition:Equalizer|equalizer]] of two [[Definition:Morphism|morphisms]] $f, g: C \to D$.
Let $e$ be an [[Definition:Epimorphism (Category Theory)|epimorphism]].
Then $e$ is an [[Definition:Isomorphism (Category Theory)|isomor... | As $e$ [[Definition:Equalizer|equalises]] $f$ and $g$, $f \circ e = g \circ e$.
Since $e$ is [[Definition:Epimorphism (Category Theory)|epic]], it follows that $f = g$.
Then in the equaliser diagram:
:::<nowiki>$\begin{xy}\xymatrix{
E
\ar[r]^*{e}
&
A
\ar[r]<2pt>^*{f}
\ar[r]<-2pt>_*{g}
&
B
\\
A
\ar@{... | Epic Equalizer is Isomorphism | https://proofwiki.org/wiki/Epic_Equalizer_is_Isomorphism | https://proofwiki.org/wiki/Epic_Equalizer_is_Isomorphism | [
"Epimorphisms",
"Morphisms"
] | [
"Definition:Metacategory",
"Definition:Equalizer",
"Definition:Morphism",
"Definition:Epimorphism (Category Theory)",
"Definition:Isomorphism (Category Theory)"
] | [
"Definition:Equalizer",
"Definition:Epimorphism (Category Theory)",
"Equalizer is Monomorphism",
"Definition:Inverse Morphism",
"Definition:Isomorphism (Category Theory)"
] |
proofwiki-12088 | Infimum of Subset of Extended Real Numbers is Arbitrarily Close | Let $A \subseteq \overline \R$ be a subset of the extended real numbers.
Let $b$ be an infimum (in $\R$) of $A$.
Let $\epsilon \in \R_{>0}$.
Then:
:$\exists x \in A \cap \R: x - b < \epsilon$ | We have that:
:$A$ is a a set of extended real numbers
:$A$ is bounded below (in $\R$) as the real number $b$ is a lower bound for $A$.
From this follows by Infimum of Real Subset:
:$\map \inf {A \cap \R} \in \R$ as $\inf A \in \R$
Let $\epsilon \in \R_{>0}$.
Then:
{{begin-eqn}}
{{eqn | l = \map \inf {A \cap \R}
... | Let $A \subseteq \overline \R$ be a [[Definition:Subset|subset]] of the [[Definition:Extended Real Number Line|extended real numbers]].
Let $b$ be an [[Definition:Infimum of Subset of Real Numbers|infimum]] (in $\R$) of $A$.
Let $\epsilon \in \R_{>0}$.
Then:
:$\exists x \in A \cap \R: x - b < \epsilon$ | We have that:
:$A$ is a a [[Definition:Set|set]] of [[Definition:Extended Real Number Line|extended real numbers]]
:$A$ is [[Definition:Bounded Below Subset of Real Numbers|bounded below]] (in $\R$) as the [[Definition:Real Number|real number]] $b$ is a [[Definition:Lower Bound of Subset of Real Numbers|lower bound]] f... | Infimum of Subset of Extended Real Numbers is Arbitrarily Close | https://proofwiki.org/wiki/Infimum_of_Subset_of_Extended_Real_Numbers_is_Arbitrarily_Close | https://proofwiki.org/wiki/Infimum_of_Subset_of_Extended_Real_Numbers_is_Arbitrarily_Close | [
"Extended Real Numbers"
] | [
"Definition:Subset",
"Definition:Extended Real Number Line",
"Definition:Infimum of Set/Real Numbers"
] | [
"Definition:Set",
"Definition:Extended Real Number Line",
"Definition:Bounded Below Set/Real Numbers",
"Definition:Real Number",
"Definition:Lower Bound of Set/Real Numbers",
"Infimum of Real Subset",
"Infimum of Subset of Real Numbers is Arbitrarily Close",
"Category:Extended Real Numbers"
] |
proofwiki-12089 | Oscillation on Set is an Extended Real Number | Let $f: D \to \R$ be a real function where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $I$ be a real set that contains (as an element) $x$.
Let $\map {\omega_f} {I \cap D}$ be the oscillation of $f$ on $I \cap D$:
:$\map {\omega_f} {I \cap D} = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
Then:
:$\m... | We observe that $\size {\map f y - \map f z} = 0$ for $y = z = x$.
Therefore:
:$0 \in \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ as $x \in I \cap D$
Accordingly:
:$\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is non-empty
We have that $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is a re... | Let $f: D \to \R$ be a [[Definition:Real Function|real function]] where $D \subseteq \R$.
Let $x$ be a [[Definition:Element|point]] in $D$.
Let $I$ be a [[Definition:Real Number|real set]] that contains (as an [[Definition:Element|element]]) $x$.
Let $\map {\omega_f} {I \cap D}$ be the [[Definition:Oscillation on Re... | We observe that $\size {\map f y - \map f z} = 0$ for $y = z = x$.
Therefore:
:$0 \in \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ as $x \in I \cap D$
Accordingly:
:$\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is [[Definition:Non-Empty Set|non-empty]]
We have that $\set {\size {\map f y - \map ... | Oscillation on Set is an Extended Real Number | https://proofwiki.org/wiki/Oscillation_on_Set_is_an_Extended_Real_Number | https://proofwiki.org/wiki/Oscillation_on_Set_is_an_Extended_Real_Number | [
"Real Analysis"
] | [
"Definition:Real Function",
"Definition:Element",
"Definition:Real Number",
"Definition:Element",
"Definition:Oscillation/Real Space/Oscillation on Set"
] | [
"Definition:Non-Empty Set",
"Definition:Real Number",
"Definition:Real Number",
"Definition:Bounded Above Set/Real Numbers",
"Definition:Bounded Above Set/Real Numbers",
"Definition:Bounded Above Set/Real Numbers",
"Definition:Bounded Above Set/Real Numbers",
"Definition:Non-Empty Set",
"Definition:... |
proofwiki-12090 | Oscillation on Subset | Let $f: D \to \R$ be a real function where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $S_x$ be a set of real sets that contain (as an element) $x$.
Let $I$ be an element of $S_x$.
Let $\map {\omega_f} {I \cap D}$ be the oscillation of $f$ on $I \cap D$:
:$\map {\omega_f} {I \cap D} = \sup \set {\size {\map f y - ... | Let:
:$I, J \in S_x$
:$J \subset I$
:$\map {\omega_f} {I \cap D} \in \R$
where:
:$\map {\omega_f} {I \cap D} = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
We need to prove:
:$\map {\omega_f} {J \cap D} \in \R$
:$\map {\omega_f} {J \cap D} \le \map {\omega_f} {I \cap D}$
We intend to prove that $\map {\o... | Let $f: D \to \R$ be a [[Definition:Real Function|real function]] where $D \subseteq \R$.
Let $x$ be a [[Definition:Element|point]] in $D$.
Let $S_x$ be a [[Definition:Set|set]] of [[Definition:Real Number|real sets]] that contain (as an [[Definition:Element|element]]) $x$.
Let $I$ be an [[Definition:Element|element... | Let:
:$I, J \in S_x$
:$J \subset I$
:$\map {\omega_f} {I \cap D} \in \R$
where:
:$\map {\omega_f} {I \cap D} = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
We need to prove:
:$\map {\omega_f} {J \cap D} \in \R$
:$\map {\omega_f} {J \cap D} \le \map {\omega_f} {I \cap D}$
We intend to prove that $\map... | Oscillation on Subset | https://proofwiki.org/wiki/Oscillation_on_Subset | https://proofwiki.org/wiki/Oscillation_on_Subset | [
"Real Analysis"
] | [
"Definition:Real Function",
"Definition:Element",
"Definition:Set",
"Definition:Real Number",
"Definition:Element",
"Definition:Element",
"Definition:Oscillation/Real Space/Oscillation on Set",
"Definition:Subset"
] | [
"Definition:Bounded Above Set/Real Numbers",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Subset",
"Definition:Supremum of Set/Real Numbers",
"Definition:Bounded Above Set/Real Numbers",
"Definition:Bounded Above Set/Real Numbers",
"Definition:Subset",
"Definition:Non-Empty Set",
"... |
proofwiki-12091 | Generating Fraction for Lucas Numbers | The fraction:
:$\dfrac {199} {9899}$
has a decimal expansion which contains within it the start of the Lucas sequence:
:$0 \cdotp 02010 \, 30407 \, 11 \ldots$ | By long division:
<pre>
0.0201030407111829
----------------------
9899 ) 199.0000000000000000
197 98
--- --
1 0200
9899
- ----
30100
29697
-----
40300
39596
-----
70400
... | The [[Definition:Fraction|fraction]]:
:$\dfrac {199} {9899}$
has a [[Definition:Decimal Expansion|decimal expansion]] which contains within it the start of the [[Definition:Lucas Number|Lucas sequence]]:
:$0 \cdotp 02010 \, 30407 \, 11 \ldots$ | By [[Definition:Long Division|long division]]:
<pre>
0.0201030407111829
----------------------
9899 ) 199.0000000000000000
197 98
--- --
1 0200
9899
- ----
30100
29697
-----
40300
39596
... | Generating Fraction for Lucas Numbers | https://proofwiki.org/wiki/Generating_Fraction_for_Lucas_Numbers | https://proofwiki.org/wiki/Generating_Fraction_for_Lucas_Numbers | [
"Lucas Numbers"
] | [
"Definition:Fraction",
"Definition:Decimal Expansion",
"Definition:Lucas Number"
] | [
"Definition:Classical Algorithm/Division",
"Generating Function for Lucas Numbers"
] |
proofwiki-12092 | Transcendental Numbers are Uncountable | The set of transcendental real numbers is uncountable. | By definition, a transcendental number (in this context) is a real number which is not an algebraic number.
Recall that the Real Numbers are Uncountable.
Also recall that the Algebraic Numbers are Countable.
The result follows from Uncountable Set less Countable Set is Uncountable.
{{qed}} | The [[Definition:Set|set]] of [[Definition:Transcendental Number|transcendental]] [[Definition:Real Number|real numbers]] is [[Definition:Uncountable Set|uncountable]]. | By definition, a [[Definition:Transcendental Number|transcendental number]] (in this context) is a [[Definition:Real Number|real number]] which is not an [[Definition:Algebraic Number|algebraic number]].
Recall that the [[Real Numbers are Uncountable]].
Also recall that the [[Algebraic Numbers are Countable]].
The r... | Transcendental Numbers are Uncountable | https://proofwiki.org/wiki/Transcendental_Numbers_are_Uncountable | https://proofwiki.org/wiki/Transcendental_Numbers_are_Uncountable | [
"Transcendental Numbers"
] | [
"Definition:Set",
"Definition:Transcendental Number",
"Definition:Real Number",
"Definition:Uncountable/Set"
] | [
"Definition:Transcendental Number",
"Definition:Real Number",
"Definition:Algebraic Number",
"Real Numbers are Uncountably Infinite",
"Algebraic Numbers are Countable",
"Uncountable Set less Countable Set is Uncountable"
] |
proofwiki-12093 | Almost All Real Numbers are Transcendental | Almost all real numbers are transcendental. | By definition, a transcendental number (in this context) is a real number which is not an algebraic number.
Recall that the Real Numbers are Uncountable.
Also recall that the Algebraic Numbers are Countable.
Thus the subset of all real numbers which are not transcendental is countable.
The result follows by the definit... | [[Definition:Almost All/Set Theory/Uncountable|Almost all]] [[Definition:Real Number|real numbers]] are [[Definition:Transcendental Number|transcendental]]. | By definition, a [[Definition:Transcendental Number|transcendental number]] (in this context) is a [[Definition:Real Number|real number]] which is not an [[Definition:Algebraic Number|algebraic number]].
Recall that the [[Real Numbers are Uncountable]].
Also recall that the [[Algebraic Numbers are Countable]].
Thus ... | Almost All Real Numbers are Transcendental | https://proofwiki.org/wiki/Almost_All_Real_Numbers_are_Transcendental | https://proofwiki.org/wiki/Almost_All_Real_Numbers_are_Transcendental | [
"Transcendental Numbers"
] | [
"Definition:Almost All/Set Theory/Uncountable",
"Definition:Real Number",
"Definition:Transcendental Number"
] | [
"Definition:Transcendental Number",
"Definition:Real Number",
"Definition:Algebraic Number",
"Real Numbers are Uncountably Infinite",
"Algebraic Numbers are Countable",
"Definition:Subset",
"Definition:Real Number",
"Definition:Transcendental Number",
"Definition:Countable Set",
"Definition:Almost... |
proofwiki-12094 | Anomalous Cancellation on 2-Digit Numbers | There are exactly four anomalously cancelling proper fractions having two-digit numerator and denominator when expressed in base $10$ notation:
{{:Anomalous Cancellation on 2-Digit Numbers/Examples}} | Let $\dfrac {\sqbrk {a x} } {\sqbrk {x b} }$ be an anomalously cancelling proper fraction.
Then we have:
{{begin-eqn}}
{{eqn | n = 1
| l = a, b, x \le 9
| o =
| c = as they are digits in base $10$
}}
{{eqn | n = 2
| l = a < b
| o =
| c = this fraction is a proper fraction
}}
{{eqn | ... | There are exactly four [[Definition:Anomalous Cancellation|anomalously cancelling]] [[Definition:Proper Fraction|proper fractions]] having two-digit [[Definition:Numerator|numerator]] and [[Definition:Denominator|denominator]] when expressed in [[Definition:Decimal Notation|base $10$ notation]]:
{{:Anomalous Cancellat... | Let $\dfrac {\sqbrk {a x} } {\sqbrk {x b} }$ be an [[Definition:Anomalous Cancellation|anomalously cancelling]] [[Definition:Proper Fraction|proper fraction]].
Then we have:
{{begin-eqn}}
{{eqn | n = 1
| l = a, b, x \le 9
| o =
| c = as they are digits in [[Definition:Decimal Notation|base $10$]]
}... | Anomalous Cancellation on 2-Digit Numbers | https://proofwiki.org/wiki/Anomalous_Cancellation_on_2-Digit_Numbers | https://proofwiki.org/wiki/Anomalous_Cancellation_on_2-Digit_Numbers | [
"Anomalous Cancellation"
] | [
"Definition:Anomalous Cancellation",
"Definition:Fraction/Proper",
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator",
"Definition:Decimal Notation"
] | [
"Definition:Anomalous Cancellation",
"Definition:Fraction/Proper",
"Definition:Decimal Notation",
"Definition:Fraction/Proper",
"Definition:Anomalous Cancellation",
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Euc... |
proofwiki-12095 | Anomalous Cancellation/Variants/37 + 13 over 37 + 24 | :$\dfrac {37 + 13} {37 + 24} = \dfrac {37^3 + 13^3} {37^3 + 24^3}$ | {{begin-eqn}}
{{eqn | l = 37^3 + 13^3
| r = \paren {37 + 13}^3 - \paren {3 \times 37^2 \times 13 + 3 \times 37 \times 13^2}
| c =
}}
{{eqn | r = \paren {37 + 13}^3 - 3 \times 13 \times 37 \paren {37 + 13}
| c =
}}
{{eqn | r = \paren {37 + 13} \paren {\paren {37 + 13}^2 - 3 \times 13 \times 37}
... | :$\dfrac {37 + 13} {37 + 24} = \dfrac {37^3 + 13^3} {37^3 + 24^3}$ | {{begin-eqn}}
{{eqn | l = 37^3 + 13^3
| r = \paren {37 + 13}^3 - \paren {3 \times 37^2 \times 13 + 3 \times 37 \times 13^2}
| c =
}}
{{eqn | r = \paren {37 + 13}^3 - 3 \times 13 \times 37 \paren {37 + 13}
| c =
}}
{{eqn | r = \paren {37 + 13} \paren {\paren {37 + 13}^2 - 3 \times 13 \times 37}
... | Anomalous Cancellation/Variants/37 + 13 over 37 + 24 | https://proofwiki.org/wiki/Anomalous_Cancellation/Variants/37_+_13_over_37_+_24 | https://proofwiki.org/wiki/Anomalous_Cancellation/Variants/37_+_13_over_37_+_24 | [
"Anomalous Cancellation"
] | [] | [] |
proofwiki-12096 | Anomalous Cancellation/Variants/3 + 25 + 38 over 7 + 20 + 39 | :$\dfrac {3 + 25 + 38} {7 + 20 + 39} = \dfrac {3^4 + 25^4 + 38^4} {7^4 + 20^4 + 39^4}$ | {{begin-eqn}}
{{eqn | l = \dfrac {3^4 + 25^4 + 38^4} {7^4 + 20^4 + 39^4}
| r = \dfrac {81 + 390625 + 2085136} {2401 + 160000 + 2313441}
| c =
}}
{{eqn | r = \dfrac {2475842} {2475842}
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \dfrac {3 + 25 + 38} {7 + 20 + 39}
| r = \dfrac {66} {66}
... | :$\dfrac {3 + 25 + 38} {7 + 20 + 39} = \dfrac {3^4 + 25^4 + 38^4} {7^4 + 20^4 + 39^4}$ | {{begin-eqn}}
{{eqn | l = \dfrac {3^4 + 25^4 + 38^4} {7^4 + 20^4 + 39^4}
| r = \dfrac {81 + 390625 + 2085136} {2401 + 160000 + 2313441}
| c =
}}
{{eqn | r = \dfrac {2475842} {2475842}
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \dfrac {3 + 25 + 38} {7 + 20 + 39}
| r = \dfrac {66} {66}
... | Anomalous Cancellation/Variants/3 + 25 + 38 over 7 + 20 + 39 | https://proofwiki.org/wiki/Anomalous_Cancellation/Variants/3_+_25_+_38_over_7_+_20_+_39 | https://proofwiki.org/wiki/Anomalous_Cancellation/Variants/3_+_25_+_38_over_7_+_20_+_39 | [
"Anomalous Cancellation"
] | [] | [] |
proofwiki-12097 | Anomalous Cancellation/Examples/3544 over 7531 | :$\dfrac {344} {731} = \dfrac {3544} {7531} = \dfrac {35544} {75531} = \cdots$ | {{begin-eqn}}
{{eqn | l = \frac {355 \cdots 544} {755 \cdots 531}
| r = \paren {3 \times 10^n + \paren {\sum_{i \mathop = 2}^{n - 1} 5 \times 10^i} + 44} \Big / \paren {7 \times 10^n + \paren {\sum_{i \mathop = 2}^{n - 1} 5 \times 10^i} + 31}
| c =
}}
{{eqn | r = \paren {3 \times 10^n + \paren {500 \times ... | :$\dfrac {344} {731} = \dfrac {3544} {7531} = \dfrac {35544} {75531} = \cdots$ | {{begin-eqn}}
{{eqn | l = \frac {355 \cdots 544} {755 \cdots 531}
| r = \paren {3 \times 10^n + \paren {\sum_{i \mathop = 2}^{n - 1} 5 \times 10^i} + 44} \Big / \paren {7 \times 10^n + \paren {\sum_{i \mathop = 2}^{n - 1} 5 \times 10^i} + 31}
| c =
}}
{{eqn | r = \paren {3 \times 10^n + \paren {500 \times ... | Anomalous Cancellation/Examples/3544 over 7531 | https://proofwiki.org/wiki/Anomalous_Cancellation/Examples/3544_over_7531 | https://proofwiki.org/wiki/Anomalous_Cancellation/Examples/3544_over_7531 | [
"Anomalous Cancellation"
] | [] | [
"Sum of Geometric Sequence"
] |
proofwiki-12098 | Anomalous Cancellation/Examples/143 185 over 17 018 560 | :$\dfrac {1435} {170 \, 560} = \dfrac {143 \, 185} {17 \, 018 \, 560} = \dfrac {14 \, 318 \, 185} {1 \, 701 \, 818 \, 560} = \cdots$ | Let $q = \dfrac r s = \dfrac {1431818 \cdots 185} {1701818 \cdots 18560}$.
Let the number of digits in $r$ be $n + 2$.
Then the number of digits in $s$ is $n + 4$.
By inspection, it is seen that $n$ is even.
Then it is seen that:
{{begin-eqn}}
{{eqn | l = r
| r = 5 + 18 \times 10 + 18 \times 1000 + \cdots + 18 \t... | :$\dfrac {1435} {170 \, 560} = \dfrac {143 \, 185} {17 \, 018 \, 560} = \dfrac {14 \, 318 \, 185} {1 \, 701 \, 818 \, 560} = \cdots$ | Let $q = \dfrac r s = \dfrac {1431818 \cdots 185} {1701818 \cdots 18560}$.
Let the number of [[Definition:Digit|digits]] in $r$ be $n + 2$.
Then the number of [[Definition:Digit|digits]] in $s$ is $n + 4$.
By inspection, it is seen that $n$ is [[Definition:Even Integer|even]].
Then it is seen that:
{{begin-eqn}}
... | Anomalous Cancellation/Examples/143 185 over 17 018 560 | https://proofwiki.org/wiki/Anomalous_Cancellation/Examples/143_185_over_17_018_560 | https://proofwiki.org/wiki/Anomalous_Cancellation/Examples/143_185_over_17_018_560 | [
"Anomalous Cancellation"
] | [] | [
"Definition:Digit",
"Definition:Digit",
"Definition:Even Integer",
"Sum of Geometric Sequence"
] |
proofwiki-12099 | Number of Digits in Power of 2 | Let $n$ be a positive integer.
Expressed in conventional decimal notation, the number of digits in the $n$th power of $2$:
:$2^n$
is equal to:
:$\ceiling {n \log_{10} 2}$
where $\ceiling x$ denotes the ceiling of $x$. | Let $2^n$ have $m$ digits when expressed in decimal notation.
By the Basis Representation Theorem and its implications, a positive integer $x$ has $m$ digits {{iff}}:
:$10^{m - 1} \le x < 10^m$
Thus:
{{begin-eqn}}
{{eqn | l = 10^{m - 1}
| o = \le
| m = 2^n
| mo= <
| r = 10^m
| c =
}}
{{eq... | Let $n$ be a [[Definition:Positive Integer|positive integer]].
Expressed in conventional [[Definition:Decimal Notation|decimal notation]], the number of [[Definition:Digit|digits]] in the [[Definition:Integer Power|$n$th power]] of $2$:
:$2^n$
is equal to:
:$\ceiling {n \log_{10} 2}$
where $\ceiling x$ denotes the [[D... | Let $2^n$ have $m$ [[Definition:Digit|digits]] when expressed in [[Definition:Decimal Notation|decimal notation]].
By the [[Basis Representation Theorem]] and its implications, a [[Definition:Positive Integer|positive integer]] $x$ has $m$ [[Definition:Digit|digits]] {{iff}}:
:$10^{m - 1} \le x < 10^m$
Thus:
{{begin-... | Number of Digits in Power of 2 | https://proofwiki.org/wiki/Number_of_Digits_in_Power_of_2 | https://proofwiki.org/wiki/Number_of_Digits_in_Power_of_2 | [
"Logarithms",
"Powers of 2",
"Number of Digits in Power of 2"
] | [
"Definition:Positive/Integer",
"Definition:Decimal Notation",
"Definition:Digit",
"Definition:Power (Algebra)/Integer",
"Definition:Ceiling Function"
] | [
"Definition:Digit",
"Definition:Decimal Notation",
"Basis Representation Theorem",
"Definition:Positive/Integer",
"Definition:Digit",
"Definition:Power (Algebra)/Integer",
"Definition:Power (Algebra)/Integer",
"Integer equals Ceiling iff Number between Integer and One Less"
] |
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