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-7500 | Polynomial has Integer Coefficients iff Content is Integer | $f$ has integer coefficients {{iff}} $\cont f$ is an integer. | If $f \in \Z \sqbrk X$ then $\cont f \in \Z$ by definition of content.
Conversely, suppose that:
:$f = a_d X^d + \cdots + a_1 X + a_0 \notin \Z \sqbrk X$
Let $m = \min \set {n \in \N : n f \in \Z \sqbrk X}$.
Then, by definition of content:
:$\cont f = \dfrac 1 m \gcd \set {m a_d, \ldots, m a_0}$
So $\cont f \in \Z$ wou... | $f$ has [[Definition:Integer|integer]] [[Definition:Polynomial Coefficient|coefficients]] {{iff}} $\cont f$ is an [[Definition:Integer|integer]]. | If $f \in \Z \sqbrk X$ then $\cont f \in \Z$ by definition of [[Definition:Content of Rational Polynomial|content]].
Conversely, suppose that:
:$f = a_d X^d + \cdots + a_1 X + a_0 \notin \Z \sqbrk X$
Let $m = \min \set {n \in \N : n f \in \Z \sqbrk X}$.
Then, by definition of [[Definition:Content of Rational Polyno... | Polynomial has Integer Coefficients iff Content is Integer | https://proofwiki.org/wiki/Polynomial_has_Integer_Coefficients_iff_Content_is_Integer | https://proofwiki.org/wiki/Polynomial_has_Integer_Coefficients_iff_Content_is_Integer | [
"Content of Polynomial"
] | [
"Definition:Integer",
"Definition:Coefficient of Polynomial",
"Definition:Integer"
] | [
"Definition:Content of Polynomial/Rational",
"Definition:Content of Polynomial/Rational",
"Definition:Greatest Common Divisor/Integers",
"Definition:Multiple",
"Definition:Integer",
"Category:Content of Polynomial"
] |
proofwiki-7501 | Content of Monic Polynomial | If $f$ is monic, then $\cont f = \dfrac 1 n$ for some integer $n$. | Since $f$ is monic, it can be written as:
:$f = X^r + \cdots + a_1 X + a_0$
Let $n = \inf \set {n \in \N : n f \in \Z \sqbrk X}$.
Let $d = \cont {n f}$.
Then by definition of content:
:$d = \gcd \set {n, n a_{r - 1}, \ldots, n a_1, n a_0}$
Therefore, by definition of GCD, $d$ divides $n$.
So say $n = k d$ with $k \in \... | If $f$ is [[Definition:Monic Polynomial|monic]], then $\cont f = \dfrac 1 n$ for some [[Definition:Integer|integer]] $n$. | Since $f$ is [[Definition:Monic Polynomial|monic]], it can be written as:
:$f = X^r + \cdots + a_1 X + a_0$
Let $n = \inf \set {n \in \N : n f \in \Z \sqbrk X}$.
Let $d = \cont {n f}$.
Then by definition of [[Definition:Content of Rational Polynomial|content]]:
:$d = \gcd \set {n, n a_{r - 1}, \ldots, n a_1, n a_0... | Content of Monic Polynomial | https://proofwiki.org/wiki/Content_of_Monic_Polynomial | https://proofwiki.org/wiki/Content_of_Monic_Polynomial | [
"Content of Polynomial",
"Monic Polynomials"
] | [
"Definition:Monic Polynomial",
"Definition:Integer"
] | [
"Definition:Monic Polynomial",
"Definition:Content of Polynomial/Rational",
"Definition:Greatest Common Divisor/Integers",
"Definition:Divisor (Algebra)/Integer",
"Category:Content of Polynomial",
"Category:Monic Polynomials"
] |
proofwiki-7502 | Content of Scalar Multiple | :$\cont {q f} = q \cont f$ | Let $q = \dfrac a b$ with $a, b \in \Z$.
Let $\Z \sqbrk X$ denote the ring of polynomials over $\Z$.
Let $n \in \Z$ such that $n f \in \Z \sqbrk X$.
Then we have:
:$b n \paren {q f} = a n f \in \Z \sqbrk X$
By the definition of content, and using that $a \in \Z$:
:$\cont {b n q f} = a \cont {n f}$
{{handwaving|why is ... | :$\cont {q f} = q \cont f$ | Let $q = \dfrac a b$ with $a, b \in \Z$.
Let $\Z \sqbrk X$ denote the [[Definition:Ring of Polynomials|ring of polynomials]] over $\Z$.
Let $n \in \Z$ such that $n f \in \Z \sqbrk X$.
Then we have:
:$b n \paren {q f} = a n f \in \Z \sqbrk X$
By the definition of [[Definition:Content of Rational Polynomial|content]... | Content of Scalar Multiple | https://proofwiki.org/wiki/Content_of_Scalar_Multiple | https://proofwiki.org/wiki/Content_of_Scalar_Multiple | [
"Polynomial Theory",
"Content of Polynomial"
] | [] | [
"Definition:Polynomial Ring",
"Definition:Content of Polynomial/Rational",
"Definition:Content of Polynomial/Rational",
"Category:Polynomial Theory",
"Category:Content of Polynomial"
] |
proofwiki-7503 | Fixed Point of Composition of Inflationary Mappings | Let $\struct {S, \preceq}$ be an ordered set.
Let $f, g: S \to S$ be inflationary mappings.
Let $x \in S$.
Then:
:$x$ is a fixed point of $f \circ g$
{{iff}}:
:$x$ is a fixed point of both $f$ and $g$. | === Necessary Condition ===
Follows from Fixed Point of Mappings is Fixed Point of Composition.
{{qed|lemma}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $f, g: S \to S$ be [[Definition:Inflationary Mapping|inflationary mappings]].
Let $x \in S$.
Then:
:$x$ is a [[Definition:Fixed Point|fixed point]] of $f \circ g$
{{iff}}:
:$x$ is a [[Definition:Fixed Point|fixed point]] of both $f$ and $g... | === Necessary Condition ===
Follows from [[Fixed Point of Mappings is Fixed Point of Composition]].
{{qed|lemma}} | Fixed Point of Composition of Inflationary Mappings | https://proofwiki.org/wiki/Fixed_Point_of_Composition_of_Inflationary_Mappings | https://proofwiki.org/wiki/Fixed_Point_of_Composition_of_Inflationary_Mappings | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Inflationary Mapping",
"Definition:Fixed Point",
"Definition:Fixed Point"
] | [
"Fixed Point of Mappings is Fixed Point of Composition"
] |
proofwiki-7504 | Group of Units is Group | Let $\struct {R, +, \circ}$ be a ring with unity.
Then the set of units of $\struct {R, +, \circ}$ forms a group under $\circ$.
Hence the justification for referring to the group of units of $\struct {R, +, \circ}$. | Follows directly from Invertible Elements of Monoid form Subgroup of Cancellable Elements.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
Then the [[Definition:Set|set]] of [[Definition:Unit of Ring|units]] of $\struct {R, +, \circ}$ forms a [[Definition:Group|group]] under $\circ$.
Hence the justification for referring to the [[Definition:Group of Units of Ring|group of ... | Follows directly from [[Invertible Elements of Monoid form Subgroup of Cancellable Elements]].
{{qed}} | Group of Units is Group | https://proofwiki.org/wiki/Group_of_Units_is_Group | https://proofwiki.org/wiki/Group_of_Units_is_Group | [
"Rings with Unity"
] | [
"Definition:Ring with Unity",
"Definition:Set",
"Definition:Unit of Ring",
"Definition:Group",
"Definition:Group of Units/Ring"
] | [
"Invertible Elements of Monoid form Subgroup of Cancellable Elements"
] |
proofwiki-7505 | Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication | Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $p$ be an irreducible element of $D$.
Let $\ideal p$ be the principal ideal of $D$ generated by $p$.
Then $\ideal p$ is a maximal ideal of $D$. | Let $p$ be irreducible in $D$.
Let $U_D$ be the group of units of $D$.
By definition, an irreducible element is not a unit.
So from Principal Ideals in Integral Domain:
:$\ideal p \subset D$
Suppose the principal ideal $\ideal p$ is not maximal.
Then there exists an ideal $K$ of $D$ such that:
:$\ideal p \subset K \sub... | Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
Let $p$ be an [[Definition:Irreducible Element of Ring|irreducible element]] of $D$.
Let $\ideal p$ be the [[Definition:Principal Ideal of Ring|principal ideal of $D$ generated by $p$]].
Then $\ideal p$ is a [[Definition:... | Let $p$ be [[Definition:Irreducible Element of Ring|irreducible]] in $D$.
Let $U_D$ be the [[Definition:Group of Units of Ring|group of units]] of $D$.
By definition, an [[Definition:Irreducible Element of Ring|irreducible element]] is not a [[Definition:Unit of Ring|unit]].
So from [[Principal Ideals in Integral D... | Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication | https://proofwiki.org/wiki/Principal_Ideal_of_Principal_Ideal_Domain_is_of_Irreducible_Element_iff_Maximal/Forward_Implication | https://proofwiki.org/wiki/Principal_Ideal_of_Principal_Ideal_Domain_is_of_Irreducible_Element_iff_Maximal/Forward_Implication | [
"Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal"
] | [
"Definition:Principal Ideal Domain",
"Definition:Irreducible Element of Ring",
"Definition:Principal Ideal of Ring",
"Definition:Maximal Ideal of Ring"
] | [
"Definition:Irreducible Element of Ring",
"Definition:Group of Units/Ring",
"Definition:Irreducible Element of Ring",
"Definition:Unit of Ring",
"Principal Ideals in Integral Domain",
"Definition:Principal Ideal of Ring",
"Definition:Maximal Ideal of Ring",
"Definition:Ideal of Ring",
"Definition:Pr... |
proofwiki-7506 | Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Reverse Implication | Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $\ideal p$ be the principal ideal of $D$ generated by $p$.
Let $\ideal p$ be a maximal ideal of $D$.
Then $p$ is irreducible. | Let $\ideal p$ be a maximal ideal of $D$.
Let $p = f g$ be any factorization of $p$.
We must show that one of $f, g$ is a unit.
{{AimForCont}} neither of $f, g$ is a unit.
First it will be shown that:
:$\ideal p \subsetneqq \ideal f$
Let $x \in \ideal p$.
That is:
:$\exists q \in D: x = p q$
Then:
:$x = f g q \in \idea... | Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
Let $\ideal p$ be the [[Definition:Principal Ideal of Ring|principal ideal of $D$ generated by $p$]].
Let $\ideal p$ be a [[Definition:Maximal Ideal of Ring|maximal ideal]] of $D$.
Then $p$ is [[Definition:Irreducible Ele... | Let $\ideal p$ be a [[Definition:Maximal Ideal of Ring|maximal ideal]] of $D$.
Let $p = f g$ be any [[Definition:Factorization|factorization]] of $p$.
We must show that one of $f, g$ is a [[Definition:Unit of Ring|unit]].
{{AimForCont}} neither of $f, g$ is a [[Definition:Unit of Ring|unit]].
First it will be show... | Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Reverse Implication | https://proofwiki.org/wiki/Principal_Ideal_of_Principal_Ideal_Domain_is_of_Irreducible_Element_iff_Maximal/Reverse_Implication | https://proofwiki.org/wiki/Principal_Ideal_of_Principal_Ideal_Domain_is_of_Irreducible_Element_iff_Maximal/Reverse_Implication | [
"Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal"
] | [
"Definition:Principal Ideal Domain",
"Definition:Principal Ideal of Ring",
"Definition:Maximal Ideal of Ring",
"Definition:Irreducible Element of Ring"
] | [
"Definition:Maximal Ideal of Ring",
"Definition:Divisor (Algebra)/Factorization",
"Definition:Unit of Ring",
"Definition:Unit of Ring",
"Definition:Unit of Ring",
"Definition:Maximal Ideal of Ring",
"Definition:Unit of Ring",
"Definition:Unit of Ring"
] |
proofwiki-7507 | Subring of Polynomials over Integral Domain Contains that Domain | Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Let $x \in R$.
Let $D \sqbrk x$ denote the ring of polynomials in $x$ over $D$.
Then $D \sqbrk x$ contains $D$ as a subring and $x$ as an element. | We have that $\ds \sum_{k \mathop = 0}^m a_k \circ x^k$ is a polynomial for all $m \in \Z_{\ge 0}$.
Set $m = 0$:
:$\ds \sum_{k \mathop = 0}^0 a_k \circ x^k = a_k \circ x^0 = a_k \circ 1_D = a_k$
Thus:
:$\ds \forall a_k \in D: \sum_{k \mathop = 0}^0 a_k \circ x^k \in D$
It follows directly that $D$ is a subring of $D \s... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]].
Let $\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$.
Let $x \in R$.
Let $D \sqbrk x$ denote the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $x$ over $D$.
Then $D \sqbrk ... | We have that $\ds \sum_{k \mathop = 0}^m a_k \circ x^k$ is a [[Definition:Polynomial over Integral Domain|polynomial]] for all $m \in \Z_{\ge 0}$.
Set $m = 0$:
:$\ds \sum_{k \mathop = 0}^0 a_k \circ x^k = a_k \circ x^0 = a_k \circ 1_D = a_k$
Thus:
:$\ds \forall a_k \in D: \sum_{k \mathop = 0}^0 a_k \circ x^k \in D$... | Subring of Polynomials over Integral Domain Contains that Domain | https://proofwiki.org/wiki/Subring_of_Polynomials_over_Integral_Domain_Contains_that_Domain | https://proofwiki.org/wiki/Subring_of_Polynomials_over_Integral_Domain_Contains_that_Domain | [
"Polynomial Theory",
"Subrings"
] | [
"Definition:Commutative Ring",
"Definition:Subdomain",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Subring",
"Definition:Element"
] | [
"Definition:Polynomial over Ring",
"Subring Test"
] |
proofwiki-7508 | Subring of Polynomials over Integral Domain is Smallest Subring containing Element and Domain | Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Let $x \in R$.
Let $D \sqbrk x$ denote the ring of polynomials in $x$ over $D$.
Then $D \sqbrk x$ is the smallest subring of $R$ which contains $D$ as a subring and $x$ as an element. | {{proof wanted|Whitelaw says "fairly obviously", so should be more or less straightforward.}} | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]].
Let $\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$.
Let $x \in R$.
Let $D \sqbrk x$ denote the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $x$ over $D$.
Then $D \sqbrk ... | {{proof wanted|Whitelaw says "fairly obviously", so should be more or less straightforward.}} | Subring of Polynomials over Integral Domain is Smallest Subring containing Element and Domain | https://proofwiki.org/wiki/Subring_of_Polynomials_over_Integral_Domain_is_Smallest_Subring_containing_Element_and_Domain | https://proofwiki.org/wiki/Subring_of_Polynomials_over_Integral_Domain_is_Smallest_Subring_containing_Element_and_Domain | [
"Polynomial Theory",
"Subrings"
] | [
"Definition:Commutative Ring",
"Definition:Subdomain",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Smallest Set by Set Inclusion",
"Definition:Subring",
"Definition:Subring",
"Definition:Element"
] | [] |
proofwiki-7509 | Cantor-Bernstein-Schröder Theorem/Lemma | Let $S$ be a set.
Let $T \subseteq S$.
Suppose that $f: S \to T$ is an injection.
Then there is a bijection $g: S \to T$. | Recursively define a sequence $\langle C_n \rangle$ in the power set of $S$ as follows:
:$C_0 = S \setminus T$, the difference between $S$ and $T$.
:$C_{n + 1} = f \sqbrk {C_n}$, the image of $C_n$ under $f$.
Let $\ds C := \bigcup_{n \mathop \in \N} C_n$.
Define a mapping $h: S \to T$ as follows:
:$\map h x = \begin {c... | Let $S$ be a [[Definition:Set|set]].
Let $T \subseteq S$.
Suppose that $f: S \to T$ is an [[Definition:Injection|injection]].
Then there is a [[Definition:Bijection|bijection]] $g: S \to T$. | [[Principle of Recursive Definition|Recursively]] define a [[Definition:Sequence|sequence]] $\langle C_n \rangle$ in the [[Definition:Power Set|power set]] of $S$ as follows:
:$C_0 = S \setminus T$, the [[Definition:Set Difference|difference]] between $S$ and $T$.
:$C_{n + 1} = f \sqbrk {C_n}$, the [[Definition:Image ... | Cantor-Bernstein-Schröder Theorem/Lemma/Proof 1 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Lemma | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Lemma/Proof_1 | [
"Cantor-Bernstein-Schröder Theorem"
] | [
"Definition:Set",
"Definition:Injection",
"Definition:Bijection"
] | [
"Principle of Recursive Definition",
"Definition:Sequence",
"Definition:Power Set",
"Definition:Set Difference",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Mapping",
"Definition:Mapping",
"Law of Excluded Middle",
"Definition:Mapping",
"Set is Subset of Union/Family of Sets",
"... |
proofwiki-7510 | Cantor-Bernstein-Schröder Theorem/Lemma | Let $S$ be a set.
Let $T \subseteq S$.
Suppose that $f: S \to T$ is an injection.
Then there is a bijection $g: S \to T$. | Define a mapping $E: \powerset S \to \powerset S$ as:
:$\map E K = S \setminus \paren {T \setminus f \sqbrk K}$
where $f \sqbrk K$ is the image of $K$ under $f$.
Then:
:$\map E K = \paren {S \setminus T} \cup f \sqbrk K$
By Image of Subset under Relation is Subset of Image and {{Corollary|Set Union Preserves Subsets}},... | Let $S$ be a [[Definition:Set|set]].
Let $T \subseteq S$.
Suppose that $f: S \to T$ is an [[Definition:Injection|injection]].
Then there is a [[Definition:Bijection|bijection]] $g: S \to T$. | Define a mapping $E: \powerset S \to \powerset S$ as:
:$\map E K = S \setminus \paren {T \setminus f \sqbrk K}$
where $f \sqbrk K$ is the [[Definition:Image of Subset under Mapping|image of $K$ under $f$]].
Then:
:$\map E K = \paren {S \setminus T} \cup f \sqbrk K$
By [[Image of Subset under Relation is Subset of I... | Cantor-Bernstein-Schröder Theorem/Lemma/Proof 2 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Lemma | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Lemma/Proof_2 | [
"Cantor-Bernstein-Schröder Theorem"
] | [
"Definition:Set",
"Definition:Injection",
"Definition:Bijection"
] | [
"Definition:Image (Set Theory)/Mapping/Subset",
"Image of Subset under Relation is Subset of Image",
"Definition:Increasing/Mapping",
"Knaster-Tarski Lemma/Power Set",
"Definition:Fixed Point",
"Definition:Fixed Point",
"Definition:Set Difference",
"Definition:Restriction/Mapping",
"Injection to Ima... |
proofwiki-7511 | Cantor-Bernstein-Schröder Theorem/Lemma | Let $S$ be a set.
Let $T \subseteq S$.
Suppose that $f: S \to T$ is an injection.
Then there is a bijection $g: S \to T$. | Define $C = \{f^k(x) | k \in \N, x \in S \setminus T \}$.
Clearly, $C = C_0 \cup C_1$, where:
$C_0 = S \setminus T$, the difference between $S$ and $T$,
$C_1 = \{f^k(x) | k \in \N_{> 0}, x \in S \setminus T \}$.
Note, that $S \setminus C_0 = S \setminus (S \setminus T) = S \cap T = T$ (use this theorem and $T \subsete... | Let $S$ be a [[Definition:Set|set]].
Let $T \subseteq S$.
Suppose that $f: S \to T$ is an [[Definition:Injection|injection]].
Then there is a [[Definition:Bijection|bijection]] $g: S \to T$. | Define $C = \{f^k(x) | k \in \N, x \in S \setminus T \}$.
Clearly, $C = C_0 \cup C_1$, where:
$C_0 = S \setminus T$, the [[Definition:Set Difference|difference]] between $S$ and $T$,
$C_1 = \{f^k(x) | k \in \N_{> 0}, x \in S \setminus T \}$.
Note, that $S \setminus C_0 = S \setminus (S \setminus T) = S \cap T = T$... | Cantor-Bernstein-Schröder Theorem/Lemma/Proof 3 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Lemma | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Lemma/Proof_3 | [
"Cantor-Bernstein-Schröder Theorem"
] | [
"Definition:Set",
"Definition:Injection",
"Definition:Bijection"
] | [
"Definition:Set Difference",
"Set Difference with Set Difference",
"Definition:Identity Mapping",
"Restriction of Injection is Injection"
] |
proofwiki-7512 | Cantor-Bernstein-Schröder Theorem/Proof 5 | Let $S$ and $T$ be sets.
Let $f: S \to T$ and $g: T \to S$ be injections.
Then there exists a bijection $\phi: S \to T$. | By Injection to Image is Bijection:
:$g: T \to g \sqbrk T$ is a bijection.
Thus $T$ is equivalent to $g \sqbrk T$.
By Composite of Injections is Injection $g \circ f: S \to g \sqbrk T \subset S$ is also an injection (to a subset of the domain of $g \circ f$).
Then by Cantor-Bernstein-Schröder Theorem: Lemma:
:There exi... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ and $g: T \to S$ be [[Definition:Injection|injections]].
Then there exists a [[Definition:Bijection|bijection]] $\phi: S \to T$. | By [[Injection to Image is Bijection]]:
:$g: T \to g \sqbrk T$ is a [[Definition:Bijection|bijection]].
Thus $T$ is [[Definition:Set Equivalence|equivalent]] to $g \sqbrk T$.
By [[Composite of Injections is Injection]] $g \circ f: S \to g \sqbrk T \subset S$ is also an [[Definition:Injection|injection]] (to a [[Defi... | Cantor-Bernstein-Schröder Theorem/Proof 5 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_5 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_5 | [
"Cantor-Bernstein-Schröder Theorem"
] | [
"Definition:Set",
"Definition:Injection",
"Definition:Bijection"
] | [
"Injection to Image is Bijection",
"Definition:Bijection",
"Definition:Set Equivalence",
"Composite of Injections is Injection",
"Definition:Injection",
"Definition:Subset",
"Definition:Domain (Set Theory)/Mapping",
"Cantor-Bernstein-Schröder Theorem/Lemma",
"Definition:Bijection",
"Definition:Set... |
proofwiki-7513 | Quotient Space of Real Line may be Indiscrete | Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Let $\Q$ denote the set of rational numbers.
Let $\mathbb I$ denote the set of irrational numbers.
Then $\set {\Q, \mathbb I}$ is a partition of $\R$.
Let $\sim$ be the equivalence relation induced on $\R$ by $\set {\Q, \mathbb I}... | Let $\phi: \R \to \R / {\sim}$ be the quotient mapping.
Then:
:$\forall x \in \Q: \map \phi x = \Q$
:$\forall x \in \mathbb I: \map \phi x = \mathbb I$
{{AimForCont}} $\set {\mathbb I} \in \tau_\sim$.
Then by the definition of the quotient topology:
:$\O \subsetneqq \mathbb I = \phi^{-1} \sqbrk {\set {\mathbb I} } \in ... | Let $\struct {\R, \tau}$ denote the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]].
Let $\Q$ denote the [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]].
Let $\mathbb I$ denote the [[Definition:Set|set]] of [[Definition:Irrational ... | Let $\phi: \R \to \R / {\sim}$ be the [[Definition:Quotient Mapping|quotient mapping]].
Then:
:$\forall x \in \Q: \map \phi x = \Q$
:$\forall x \in \mathbb I: \map \phi x = \mathbb I$
{{AimForCont}} $\set {\mathbb I} \in \tau_\sim$.
Then by the definition of the [[Definition:Quotient Topology|quotient topology]]:
... | Quotient Space of Real Line may be Indiscrete | https://proofwiki.org/wiki/Quotient_Space_of_Real_Line_may_be_Indiscrete | https://proofwiki.org/wiki/Quotient_Space_of_Real_Line_may_be_Indiscrete | [
"Quotient Spaces (Topology)",
"Real Number Line with Euclidean Topology"
] | [
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Set",
"Definition:Rational Number",
"Definition:Set",
"Definition:Irrational Number",
"Definition:Set Partition",
"Definition:Equivalence Relation Induced by Partition",
"Definition:Quotient Topology/Quotient Space",
"Defi... | [
"Definition:Quotient Mapping",
"Definition:Quotient Topology",
"Rational Numbers are Everywhere Dense in Set of Real Numbers/Topology",
"Definition:Rational Number",
"Definition:Contradiction",
"Irrationals are Everywhere Dense in Reals/Topology",
"Definition:Irrational Number",
"Definition:Contradict... |
proofwiki-7514 | Quotient Space of Real Line may not be T0 | Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Then there exists an equivalence relation $\sim$ on $\R$ such that the quotient space $\struct {\R / {\sim}, \tau_\sim}$ is not a $T_0$ space. | By Quotient Space of Real Line may be Indiscrete, there is an equivalence relation $\sim$ on $\R$ such that the quotient space $\struct {\R / {\sim}, \tau_\sim}$ has two points and is indiscrete.
It follows directly from the definition of $T_0$ space that $\struct {\R / {\sim}, \tau_\sim}$ is not a $T_0$ space.
{{qed}}... | Let $\struct {\R, \tau}$ denote the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]].
Then there exists an [[Definition:Equivalence Relation|equivalence relation]] $\sim$ on $\R$ such that the [[Definition:Quotient Space (Topology)|quotient space]] $\struct ... | By [[Quotient Space of Real Line may be Indiscrete]], there is an [[Definition:Equivalence Relation|equivalence relation]] $\sim$ on $\R$ such that the [[Definition:Quotient Space (Topology)|quotient space]] $\struct {\R / {\sim}, \tau_\sim}$ has two [[Definition:Point of Set|points]] and is [[Definition:Indiscrete Spa... | Quotient Space of Real Line may not be T0 | https://proofwiki.org/wiki/Quotient_Space_of_Real_Line_may_not_be_T0 | https://proofwiki.org/wiki/Quotient_Space_of_Real_Line_may_not_be_T0 | [
"Real Number Line with Euclidean Topology",
"Quotient Spaces (Topology)",
"Examples of T0 Spaces"
] | [
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Equivalence Relation",
"Definition:Quotient Topology/Quotient Space",
"Definition:T0 Space"
] | [
"Quotient Space of Real Line may be Indiscrete",
"Definition:Equivalence Relation",
"Definition:Quotient Topology/Quotient Space",
"Definition:Element",
"Definition:Indiscrete Topology",
"Definition:T0 Space",
"Definition:T0 Space",
"Category:Real Number Line with Euclidean Topology",
"Category:Quot... |
proofwiki-7515 | Unique Representation in Polynomial Forms/General Result | Let $f$ be a polynomial form in the indeterminates $\set {X_j: j \in J}$ such that $f: \mathbf X^k \mapsto a_k$.
For $r \in \R$, $\mathbf X^k \in M$, let $r \mathbf X^k$ denote the polynomial form that takes the value $r$ on $\mathbf X^k$ and zero on all other monomials.
Let $Z$ denote the set of all multiindices index... | Suppose that the sum has infinitely many non-zero terms.
Then infinitely many $a_k$ are non-zero, which contradicts the definition of a polynomial.
Therefore the sum consists of finitely many non-zero terms.
Let $\mathbf X^m \in M$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map {\hat f} {\mathbf X^m}
| r = \p... | Let $f$ be a [[Definition:Polynomial Form|polynomial form]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\set {X_j: j \in J}$ such that $f: \mathbf X^k \mapsto a_k$.
For $r \in \R$, $\mathbf X^k \in M$, let $r \mathbf X^k$ denote the [[Definition:Polynomial Form|polynomial form]] that takes ... | Suppose that the sum has infinitely many non-zero terms.
Then infinitely many $a_k$ are non-zero, which contradicts the definition of a polynomial.
Therefore the sum consists of finitely many non-zero terms.
Let $\mathbf X^m \in M$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map {\hat f} {\mathbf X^m}
| r... | Unique Representation in Polynomial Forms/General Result | https://proofwiki.org/wiki/Unique_Representation_in_Polynomial_Forms/General_Result | https://proofwiki.org/wiki/Unique_Representation_in_Polynomial_Forms/General_Result | [
"Polynomial Theory"
] | [
"Definition:Polynomial over Ring as Function on Free Monoid on Set",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Polynomial over Ring as Function on Free Monoid on Set",
"Definition:Monomial",
"Definition:Multiindex",
"Definition:Finite",
"Definition:Polynomial/Term"
] | [
"Category:Polynomial Theory"
] |
proofwiki-7516 | Unique Representation in Polynomial Forms/General Result/Corollary | Dropping the zero terms from the sum we can write the polynomial $f$ as
:$f = a_{k_1} \mathbf X^{k_1} + \cdots + a_{k_r} \mathbf X^{k_r}$
for some $a_{k_i}\in R$, $i = 1, \ldots, r$. | Follows directly from Unique Representation in Polynomial Forms: General Result
{{qed}}
Category:Polynomial Theory
lhg13al7b3sgikdeoaiida9jri8zfp3 | Dropping the zero terms from the sum we can write the polynomial $f$ as
:$f = a_{k_1} \mathbf X^{k_1} + \cdots + a_{k_r} \mathbf X^{k_r}$
for some $a_{k_i}\in R$, $i = 1, \ldots, r$. | Follows directly from [[Unique Representation in Polynomial Forms/General Result|Unique Representation in Polynomial Forms: General Result]]
{{qed}}
[[Category:Polynomial Theory]]
lhg13al7b3sgikdeoaiida9jri8zfp3 | Unique Representation in Polynomial Forms/General Result/Corollary | https://proofwiki.org/wiki/Unique_Representation_in_Polynomial_Forms/General_Result/Corollary | https://proofwiki.org/wiki/Unique_Representation_in_Polynomial_Forms/General_Result/Corollary | [
"Polynomial Theory"
] | [] | [
"Unique Representation in Polynomial Forms/General Result",
"Category:Polynomial Theory"
] |
proofwiki-7517 | Closed Element of Composite Closure Operator | Let $\struct {S, \preceq}$ be an ordered set.
Let $f, g: S \to S$ be closure operators.
Let $h = f \circ g$, where $\circ$ represents composition.
Suppose that $h$ is also a closure operator.
Then an element $x \in S$ is closed {{WRT}} $h$ {{iff}} it is closed {{WRT}} $f$ and {{WRT}} $g$. | An element is closed with respect to a closure operator {{iff}} it is a fixed point of that operator.
Since $f$ and $g$ are closure operators, they are inflationary.
Thus the result follows from Fixed Point of Composition of Inflationary Mappings.
{{qed}}
Category:Closure Operators
Category:Closed Elements
d5skj832j0qf... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $f, g: S \to S$ be [[Definition:Closure Operator|closure operators]].
Let $h = f \circ g$, where $\circ$ represents [[Definition:Composition of Mappings|composition]].
Suppose that $h$ is also a [[Definition:Closure Operator|closure operato... | An [[Definition:Element|element]] is [[Definition:Closed Element|closed]] with respect to a [[Definition:Closure Operator|closure operator]] {{iff}} it is a [[Definition:Fixed Point|fixed point]] of that [[Definition:Closure Operator|operator]].
Since $f$ and $g$ are [[Definition:Closure Operator|closure operators]], ... | Closed Element of Composite Closure Operator | https://proofwiki.org/wiki/Closed_Element_of_Composite_Closure_Operator | https://proofwiki.org/wiki/Closed_Element_of_Composite_Closure_Operator | [
"Closure Operators",
"Closed Elements"
] | [
"Definition:Ordered Set",
"Definition:Closure Operator",
"Definition:Composition of Mappings",
"Definition:Closure Operator",
"Definition:Element",
"Definition:Closed Element",
"Definition:Closed Element"
] | [
"Definition:Element",
"Definition:Closed Element",
"Definition:Closure Operator",
"Definition:Fixed Point",
"Definition:Closure Operator",
"Definition:Closure Operator",
"Definition:Inflationary Mapping",
"Fixed Point of Composition of Inflationary Mappings",
"Category:Closure Operators",
"Categor... |
proofwiki-7518 | Law of Excluded Middle for Two Variables | :$\vdash (p \land q) \lor (\lnot p \land q) \lor (p \land \lnot q) \lor (\lnot p \land \lnot q)$ | {{BeginTableau| \vdash ((p \land q) \lor (\lnot p \land q)) \lor ( (p \land \lnot q) \lor (\lnot p \land \lnot q)) }}
{{ExcludedMiddle|1|p \lor \lnot p}}
{{ExcludedMiddle|2|q \lor \lnot q}}
{{Conjunction|3||(p \lor \lnot p) \land (q \lor \lnot q)|1|2}}
{{SequentIntro|4||((p \lor \lnot p) \land q) \lor ((p \lor \lnot p)... | :$\vdash (p \land q) \lor (\lnot p \land q) \lor (p \land \lnot q) \lor (\lnot p \land \lnot q)$ | {{BeginTableau| \vdash ((p \land q) \lor (\lnot p \land q)) \lor ( (p \land \lnot q) \lor (\lnot p \land \lnot q)) }}
{{ExcludedMiddle|1|p \lor \lnot p}}
{{ExcludedMiddle|2|q \lor \lnot q}}
{{Conjunction|3||(p \lor \lnot p) \land (q \lor \lnot q)|1|2}}
{{SequentIntro|4||((p \lor \lnot p) \land q) \lor ((p \lor \lnot p)... | Law of Excluded Middle for Two Variables | https://proofwiki.org/wiki/Law_of_Excluded_Middle_for_Two_Variables | https://proofwiki.org/wiki/Law_of_Excluded_Middle_for_Two_Variables | [
"Propositional Logic"
] | [] | [
"Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1",
"Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1",
"Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1",
"Constructive Dil... |
proofwiki-7519 | Cardinality of Finite Set is Well-Defined | Let $S$ be a finite set.
Then there is a unique natural number $n$ such that $S \sim \N_n$, where:
:$\sim$ represents set equivalence
and:
:$\N_n = \set {0, 1, \dotsc, n - 1}$ is the initial segment of $\N$ determined by $n$. | By the definition of finite set, there is an $n \in \N$ such that $S \sim \N_n$.
Suppose $m \in \N$ and $S \sim \N_m$.
It follows from Set Equivalence behaves like Equivalence Relation that $\N_n \sim \N_m$.
Thus by Equality of Natural Numbers, $n = m$.
Therefore the cardinality of a finite set is well-defined.
{{qed... | Let $S$ be a [[Definition:Finite Set|finite set]].
Then there is a unique [[Definition:Natural Number|natural number]] $n$ such that $S \sim \N_n$, where:
:$\sim$ represents [[Definition:Set Equivalence|set equivalence]]
and:
:$\N_n = \set {0, 1, \dotsc, n - 1}$ is the [[Definition:Initial Segment of Natural Numbers|i... | By the definition of [[Definition:Finite Set|finite set]], there is an $n \in \N$ such that $S \sim \N_n$.
Suppose $m \in \N$ and $S \sim \N_m$.
It follows from [[Set Equivalence behaves like Equivalence Relation]] that $\N_n \sim \N_m$.
Thus by [[Equality of Natural Numbers]], $n = m$.
Therefore the [[Definition... | Cardinality of Finite Set is Well-Defined | https://proofwiki.org/wiki/Cardinality_of_Finite_Set_is_Well-Defined | https://proofwiki.org/wiki/Cardinality_of_Finite_Set_is_Well-Defined | [
"Set Theory",
"Cardinality"
] | [
"Definition:Finite Set",
"Definition:Natural Numbers",
"Definition:Set Equivalence",
"Definition:Initial Segment of Natural Numbers"
] | [
"Definition:Finite Set",
"Set Equivalence behaves like Equivalence Relation",
"Equality of Natural Numbers",
"Definition:Cardinality",
"Definition:Finite",
"Definition:Well-Defined/Mapping"
] |
proofwiki-7520 | Ring of Polynomial Forms is Commutative Ring with Unity | Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $A = R \sqbrk {\set {X_j: j \in J} }$ be the set of all polynomial forms over $R$ in the indeterminates $\set {X_j: j \in J}$.
Then $\struct {A, +, \circ}$ is a commutative ring with unity. | We must show that the commutative and unitary ring axioms are satisfied:
{{:Axiom:Commutative and Unitary Ring Axioms}} | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $A = R \sqbrk {\set {X_j: j \in J} }$ be the set of all [[Definition:Polynomial Form|polynomial forms]] over $R$ in the indeterminates $\set {X_j: j \in J}$.
Then $\struct {A, +, \circ}$ is a [[Definition:Co... | We must show that the [[Axiom:Commutative and Unitary Ring Axioms|commutative and unitary ring axioms]] are satisfied:
{{:Axiom:Commutative and Unitary Ring Axioms}} | Ring of Polynomial Forms is Commutative Ring with Unity | https://proofwiki.org/wiki/Ring_of_Polynomial_Forms_is_Commutative_Ring_with_Unity | https://proofwiki.org/wiki/Ring_of_Polynomial_Forms_is_Commutative_Ring_with_Unity | [
"Polynomial Rings"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Polynomial over Ring as Function on Free Monoid on Set",
"Definition:Commutative and Unitary Ring"
] | [
"Axiom:Commutative and Unitary Ring Axioms"
] |
proofwiki-7521 | Idempotent Elements form Subsemigroup of Commutative Semigroup | Let $\struct {S, \circ}$ be a semigroup such that $\circ$ is commutative.
Let $I$ be the set of all elements of $S$ that are idempotent under $\circ$.
That is:
:$I = \set {x \in S: x \circ x = x}$
Then $\struct {I, \circ}$ is a subsemigroup of $\struct {S, \circ}$. | By Subsemigroup Closure Test we need only show that:
:For all $x, y \in I$: $x \circ y \in I$.
That is:
:$\paren {x \circ y} \circ \paren {x \circ y} = x \circ y$
We reason as follows:
{{begin-eqn}}
{{eqn | l = \paren {x \circ y} \circ \paren {x \circ y}
| r = \paren {x \circ y} \circ \paren {y \circ x}
| c... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] such that $\circ$ is [[Definition:Commutative Operation|commutative]].
Let $I$ be the [[Definition:Set|set]] of all [[Definition:Element|elements]] of $S$ that are [[Definition:Idempotent Element|idempotent]] under $\circ$.
That is:
:$I = \set {x \in S... | By [[Subsemigroup Closure Test]] we need only show that:
:For all $x, y \in I$: $x \circ y \in I$.
That is:
:$\paren {x \circ y} \circ \paren {x \circ y} = x \circ y$
We reason as follows:
{{begin-eqn}}
{{eqn | l = \paren {x \circ y} \circ \paren {x \circ y}
| r = \paren {x \circ y} \circ \paren {y \circ x}
... | Idempotent Elements form Subsemigroup of Commutative Semigroup/Proof 1 | https://proofwiki.org/wiki/Idempotent_Elements_form_Subsemigroup_of_Commutative_Semigroup | https://proofwiki.org/wiki/Idempotent_Elements_form_Subsemigroup_of_Commutative_Semigroup/Proof_1 | [
"Subsemigroups",
"Idempotence",
"Idempotent Elements form Subsemigroup of Commutative Semigroup"
] | [
"Definition:Semigroup",
"Definition:Commutative/Operation",
"Definition:Set",
"Definition:Element",
"Definition:Idempotence/Element",
"Definition:Subsemigroup"
] | [
"Subsemigroup Closure Test",
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Definition:Idempotence/Element"
] |
proofwiki-7522 | Idempotent Elements form Subsemigroup of Commutative Semigroup | Let $\struct {S, \circ}$ be a semigroup such that $\circ$ is commutative.
Let $I$ be the set of all elements of $S$ that are idempotent under $\circ$.
That is:
:$I = \set {x \in S: x \circ x = x}$
Then $\struct {I, \circ}$ is a subsemigroup of $\struct {S, \circ}$. | By Subsemigroup Closure Test we need only show that:
: For all $x, y \in I$: $x \circ y \in I$.
As $x, y \in I$, they are idempotent.
We have that $\circ$ is commutative.
Thus, by definition, $x$ and $y$ commute.
From Product of Commuting Idempotent Elements is Idempotent, $\left({x \circ y}\right)$ is idempotent.
That... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] such that $\circ$ is [[Definition:Commutative Operation|commutative]].
Let $I$ be the [[Definition:Set|set]] of all [[Definition:Element|elements]] of $S$ that are [[Definition:Idempotent Element|idempotent]] under $\circ$.
That is:
:$I = \set {x \in S... | By [[Subsemigroup Closure Test]] we need only show that:
: For all $x, y \in I$: $x \circ y \in I$.
As $x, y \in I$, they are [[Definition:Idempotent Element|idempotent]].
We have that $\circ$ is [[Definition:Commutative Operation|commutative]].
Thus, by definition, $x$ and $y$ [[Definition:Commute|commute]].
Fro... | Idempotent Elements form Subsemigroup of Commutative Semigroup/Proof 2 | https://proofwiki.org/wiki/Idempotent_Elements_form_Subsemigroup_of_Commutative_Semigroup | https://proofwiki.org/wiki/Idempotent_Elements_form_Subsemigroup_of_Commutative_Semigroup/Proof_2 | [
"Subsemigroups",
"Idempotence",
"Idempotent Elements form Subsemigroup of Commutative Semigroup"
] | [
"Definition:Semigroup",
"Definition:Commutative/Operation",
"Definition:Set",
"Definition:Element",
"Definition:Idempotence/Element",
"Definition:Subsemigroup"
] | [
"Subsemigroup Closure Test",
"Definition:Idempotence/Element",
"Definition:Commutative/Operation",
"Definition:Commutative/Elements",
"Product of Commuting Idempotent Elements is Idempotent",
"Definition:Idempotence/Element"
] |
proofwiki-7523 | Product of Commuting Idempotent Elements is Idempotent | Let $\struct {S, \circ}$ be a semigroup.
Let $a, b \in S$ be idempotent elements of $S$.
Let $a$ and $b$ commute:
:$a \circ b = b \circ a$
Then $a \circ b$ is idempotent. | From {{Semigroup-axiom|0}} we take it for granted that $\struct {S, \circ}$ is closed under $\circ$.
Hence:
{{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {a \circ b}
| r = \paren {a \circ \paren {b \circ a} } \circ b
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = \paren {a \circ \paren {a \circ b} }... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $a, b \in S$ be [[Definition:Idempotent Element|idempotent elements]] of $S$.
Let $a$ and $b$ [[Definition:Commute|commute]]:
:$a \circ b = b \circ a$
Then $a \circ b$ is [[Definition:Idempotent Element|idempotent]]. | From {{Semigroup-axiom|0}} we take it for granted that $\struct {S, \circ}$ is [[Definition:Closure (Abstract Algebra)|closed]] under $\circ$.
Hence:
{{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {a \circ b}
| r = \paren {a \circ \paren {b \circ a} } \circ b
| c = {{Semigroup-axiom|1}}
}}
{{eq... | Product of Commuting Idempotent Elements is Idempotent | https://proofwiki.org/wiki/Product_of_Commuting_Idempotent_Elements_is_Idempotent | https://proofwiki.org/wiki/Product_of_Commuting_Idempotent_Elements_is_Idempotent | [
"Semigroups",
"Idempotence"
] | [
"Definition:Semigroup",
"Definition:Idempotence/Element",
"Definition:Commutative/Elements",
"Definition:Idempotence/Element"
] | [
"Definition:Closure (Abstract Algebra)",
"Definition:Commutative/Elements",
"Definition:Idempotence/Element",
"Definition:Idempotence/Element",
"Category:Semigroups",
"Category:Idempotence"
] |
proofwiki-7524 | Set of all Self-Maps under Composition forms Semigroup | Let $S$ be a set.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let the operation $\circ$ represent composition of mappings.
Then the algebraic structure $\struct {S^S, \circ}$ is a semigroup. | Let $f, g \in S^S$.
As the domain of $g$ and codomain of $f$ are the same, the composition $f \circ g$ is defined.
By the definition of composition, $f \circ g$ is a mapping from the domain of $g$ to the codomain of $f$.
Thus $f \circ g: S \to S$, so $f \circ g \in S^S$.
Since this holds for all $f, g \in S^S$, $\stru... | Let $S$ be a [[Definition:Set|set]].
Let $S^S$ be the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to itself.
Let the [[Definition:Binary Operation|operation]] $\circ$ represent [[Definition:Composition of Mappings|composition of mappings]].
Then the [[Definition:Algebraic Structure|algebraic str... | Let $f, g \in S^S$.
As the [[Definition:Domain of Mapping|domain]] of $g$ and [[Definition:Codomain of Mapping|codomain]] of $f$ are the same, the [[Definition:Composition of Mappings|composition]] $f \circ g$ is defined.
By the definition of composition, $f \circ g$ is a [[Definition:Mapping|mapping]] from the domai... | Set of all Self-Maps under Composition forms Semigroup | https://proofwiki.org/wiki/Set_of_all_Self-Maps_under_Composition_forms_Semigroup | https://proofwiki.org/wiki/Set_of_all_Self-Maps_under_Composition_forms_Semigroup | [
"Composite Mappings",
"Examples of Semigroups"
] | [
"Definition:Set",
"Definition:Set of All Mappings",
"Definition:Operation/Binary Operation",
"Definition:Composition of Mappings",
"Definition:Algebraic Structure",
"Definition:Semigroup"
] | [
"Definition:Domain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Composition of Mappings",
"Definition:Mapping",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Composition of Mappings is Associative",
"Definition:Associative Operation",
"Definition:Closu... |
proofwiki-7525 | Integers form Subdomain of Reals | The integral domain of integers $\struct {\Z, +, \times}$ forms a subdomain of the field of real numbers. | We have that Integers form Subdomain of Rationals.
We have that Rational Numbers form Subfield of Real Numbers.
Hence the result, from the definition of subdomain.
{{qed}}
Category:Integers
Category:Real Numbers
k7dkktib40yaper07azh5367u9i60v8 | The [[Integers form Integral Domain|integral domain of integers]] $\struct {\Z, +, \times}$ forms a [[Definition:Subdomain|subdomain]] of the [[Definition:Field of Real Numbers|field of real numbers]]. | We have that [[Integers form Subdomain of Rationals]].
We have that [[Rational Numbers form Subfield of Real Numbers]].
Hence the result, from the definition of [[Definition:Subdomain|subdomain]].
{{qed}}
[[Category:Integers]]
[[Category:Real Numbers]]
k7dkktib40yaper07azh5367u9i60v8 | Integers form Subdomain of Reals | https://proofwiki.org/wiki/Integers_form_Subdomain_of_Reals | https://proofwiki.org/wiki/Integers_form_Subdomain_of_Reals | [
"Integers",
"Real Numbers"
] | [
"Integers form Integral Domain",
"Definition:Subdomain",
"Definition:Field of Real Numbers"
] | [
"Integers form Subdomain of Rationals",
"Rational Numbers form Subfield of Real Numbers",
"Definition:Subdomain",
"Category:Integers",
"Category:Real Numbers"
] |
proofwiki-7526 | Identity of Algebraic Structure is Preserved in Substructure | Let $\struct {S, \circ}$ be an algebraic structure with identity $e$.
Let $\struct {T, \circ}$ be a algebraic substructure of $\struct {S, \circ}$.
That is, let $T \subseteq S$.
Let $e \in T$.
Then $e$ is an identity of $\struct {T, \circ}$. | Let $x \in T$.
By the definition of subset, $x \in S$.
Since $e$ is an identity of $\struct {S, \circ}$:
:$e \circ x = x \circ e = x$
Since this holds for all $x \in T$, $e$ is an identity of $\struct {T, \circ}$.
{{qed}}
Category:Identity Elements
n2ie5wgzy61jxufp2t0pdsatrdnts8o | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] with [[Definition:Identity Element|identity]] $e$.
Let $\struct {T, \circ}$ be a [[Definition:Algebraic Substructure|algebraic substructure]] of $\struct {S, \circ}$.
That is, let $T \subseteq S$.
Let $e \in T$.
Then $e$ is an [[... | Let $x \in T$.
By the definition of [[Definition:Subset|subset]], $x \in S$.
Since $e$ is an [[Definition:Identity Element|identity]] of $\struct {S, \circ}$:
:$e \circ x = x \circ e = x$
Since this holds for all $x \in T$, $e$ is an [[Definition:Identity Element|identity]] of $\struct {T, \circ}$.
{{qed}}
[[Categ... | Identity of Algebraic Structure is Preserved in Substructure | https://proofwiki.org/wiki/Identity_of_Algebraic_Structure_is_Preserved_in_Substructure | https://proofwiki.org/wiki/Identity_of_Algebraic_Structure_is_Preserved_in_Substructure | [
"Identity Elements"
] | [
"Definition:Algebraic Structure",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Algebraic Substructure",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Subset",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Category:Identity Elements"
] |
proofwiki-7527 | Idempotent Elements form Submonoid of Commutative Monoid | Let $\struct {S, \circ}$ be a commutative monoid.
Let $e \in S$ be the identity element of $\struct {S, \circ}$.
Let $I$ be the set of all elements of $S$ that are idempotent under $\circ$.
That is:
:$I = \set {x \in S: x \circ x = x}$
Then $\struct {I, \circ}$ is a submonoid of $\struct {S, \circ}$ with identity $e$. | By Idempotent Elements form Subsemigroup of Commutative Semigroup, $\struct {I, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
By Identity Element is Idempotent:
:$e \in I$
By Identity of Algebraic Structure is Preserved in Substructure, $e$ is an identity of $\struct {I, \circ}$.
Since $\struct {T, \circ}$ is a se... | Let $\struct {S, \circ}$ be a [[Definition:Commutative Monoid|commutative monoid]].
Let $e \in S$ be the [[Definition:Identity Element|identity element]] of $\struct {S, \circ}$.
Let $I$ be the [[Definition:Set|set]] of all [[Definition:Element|elements]] of $S$ that are [[Definition:Idempotent Element|idempotent]] u... | By [[Idempotent Elements form Subsemigroup of Commutative Semigroup]], $\struct {I, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$.
By [[Identity Element is Idempotent]]:
:$e \in I$
By [[Identity of Algebraic Structure is Preserved in Substructure]], $e$ is an [[Definition:Identity Elem... | Idempotent Elements form Submonoid of Commutative Monoid | https://proofwiki.org/wiki/Idempotent_Elements_form_Submonoid_of_Commutative_Monoid | https://proofwiki.org/wiki/Idempotent_Elements_form_Submonoid_of_Commutative_Monoid | [
"Commutative Monoids"
] | [
"Definition:Commutative Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Set",
"Definition:Element",
"Definition:Idempotence/Element",
"Definition:Submonoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Idempotent Elements form Subsemigroup of Commutative Semigroup",
"Definition:Subsemigroup",
"Identity Element is Idempotent",
"Identity of Algebraic Structure is Preserved in Substructure",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Semigroup",
"Definition:Identity (Abstra... |
proofwiki-7528 | Inverse Image under Order Embedding of Strict Upper Closure of Image of Point | Let $\struct {S, \preceq}$ and $\struct {T, \preceq'}$ be ordered sets.
Let $\phi: S \to T$ be an order embedding of $\struct {S, \preceq}$ into $\struct {T, \preceq'}$
Let $p \in S$.
Then:
:$\map {\phi^{-1} } {\map \phi p^{\succ'} } = p^\succ$
where $\cdot^\succ$ and $\cdot^{\succ'}$ represent strict upper closure wit... | Let $x \in \map {\phi^{-1} } {\map \phi p^{\succ'} }$.
By the definition of inverse image:
:$\map \phi x \in \map \phi p^{\succ'}$
By the definition of strict upper closure:
:$\map \phi p \prec' \map \phi x$
Since $\phi$ is an order embedding:
:$p \prec x$
Thus by the definition of strict upper closure:
:$x \in p^\succ... | Let $\struct {S, \preceq}$ and $\struct {T, \preceq'}$ be [[Definition:Ordered Set|ordered sets]].
Let $\phi: S \to T$ be an [[Definition:Order Embedding|order embedding]] of $\struct {S, \preceq}$ into $\struct {T, \preceq'}$
Let $p \in S$.
Then:
:$\map {\phi^{-1} } {\map \phi p^{\succ'} } = p^\succ$
where $\cdot... | Let $x \in \map {\phi^{-1} } {\map \phi p^{\succ'} }$.
By the definition of [[Definition:Inverse Image|inverse image]]:
:$\map \phi x \in \map \phi p^{\succ'}$
By the definition of [[Definition:Strict Upper Closure of Element|strict upper closure]]:
:$\map \phi p \prec' \map \phi x$
Since $\phi$ is an [[Definition... | Inverse Image under Order Embedding of Strict Upper Closure of Image of Point | https://proofwiki.org/wiki/Inverse_Image_under_Order_Embedding_of_Strict_Upper_Closure_of_Image_of_Point | https://proofwiki.org/wiki/Inverse_Image_under_Order_Embedding_of_Strict_Upper_Closure_of_Image_of_Point | [
"Order Embeddings",
"Upper Closures"
] | [
"Definition:Ordered Set",
"Definition:Order Embedding",
"Definition:Strict Upper Closure/Element"
] | [
"Definition:Inverse Image",
"Definition:Strict Upper Closure/Element",
"Definition:Order Embedding",
"Definition:Strict Upper Closure/Element",
"Definition:Strict Upper Closure/Element",
"Definition:Order Embedding",
"Definition:Strict Upper Closure/Element",
"Definition:Inverse Image",
"Definition:... |
proofwiki-7529 | Equivalence of Definitions of Order Embedding/Definition 1 implies Definition 3 | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: S \to T$ be a mapping.
Let $\phi: S \to T$ be an order embedding by Definition 1:
{{:Definition:Order Embedding/Definition 1}}
Then $\phi: S \to T$ is an order embedding by Definition 3:
{{:Definition:Order Embedding/Definition 3}} | Let $\phi$ be an order embedding by definition 1.
Then by definition:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
$\phi$ is injective by Order Embedding is Injection.
It remains to be shown that:
: $x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$
Suppose first that $x \prec_1 y$.
Then ... | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\phi: S \to T$ be a [[Definition:Mapping|mapping]].
Let $\phi: S \to T$ be an [[Definition:Order Embedding|order embedding]] by [[Definition:Order Embedding/Definition 1|Definition 1]]:
{{:Definition:Order Embe... | Let $\phi$ be an [[Definition:Order Embedding/Definition 1|order embedding by definition 1]].
Then by definition:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
$\phi$ is [[Definition:Injection|injective]] by [[Order Embedding is Injection]].
It remains to be shown that:
: $x \prec_1 y ... | Equivalence of Definitions of Order Embedding/Definition 1 implies Definition 3 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Order_Embedding/Definition_1_implies_Definition_3 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Order_Embedding/Definition_1_implies_Definition_3 | [
"Equivalence of Definitions of Order Embedding"
] | [
"Definition:Ordered Set",
"Definition:Mapping",
"Definition:Order Embedding",
"Definition:Order Embedding/Definition 1",
"Definition:Order Embedding",
"Definition:Order Embedding/Definition 3"
] | [
"Definition:Order Embedding/Definition 1",
"Definition:Injection",
"Order Embedding is Injection",
"Definition:Injection",
"Definition:Order Embedding/Definition 3"
] |
proofwiki-7530 | Equivalence of Definitions of Order Embedding/Definition 3 implies Definition 1 | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: S \to T$ be a mapping.
Let $\phi: S \to T$ be an order embedding by Definition 3:
{{:Definition:Order Embedding/Definition 3}}
Then $\phi: S \to T$ is an order embedding by Definition 1:
{{:Definition:Order Embedding/Definition 1}} | Let $\phi$ be an order embedding by definition 3.
Then by definition:
:$(1): \quad \phi$ is injective
:$(2): \quad \forall x, y, \in S: x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$
Let $x \preceq_1 y$.
Then $x \prec_1 y$ or $x = y$.
If $x \prec_1 y$, then by hypothesis:
:$\map \phi x \prec_2 \map \phi y$
Thus:
:$\... | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\phi: S \to T$ be a [[Definition:Mapping|mapping]].
Let $\phi: S \to T$ be an [[Definition:Order Embedding|order embedding]] by [[Definition:Order Embedding/Definition 3|Definition 3]]:
{{:Definition:Order Embe... | Let $\phi$ be an [[Definition:Order Embedding/Definition 3|order embedding by definition 3]].
Then by definition:
:$(1): \quad \phi$ is [[Definition:Injection|injective]]
:$(2): \quad \forall x, y, \in S: x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$
Let $x \preceq_1 y$.
Then $x \prec_1 y$ or $x = y$.
If $x ... | Equivalence of Definitions of Order Embedding/Definition 3 implies Definition 1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Order_Embedding/Definition_3_implies_Definition_1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Order_Embedding/Definition_3_implies_Definition_1 | [
"Equivalence of Definitions of Order Embedding"
] | [
"Definition:Ordered Set",
"Definition:Mapping",
"Definition:Order Embedding",
"Definition:Order Embedding/Definition 3",
"Definition:Order Embedding",
"Definition:Order Embedding/Definition 1"
] | [
"Definition:Order Embedding/Definition 3",
"Definition:Injection",
"Definition:Injection",
"Proof by Cases",
"Definition:Order Embedding/Definition 1"
] |
proofwiki-7531 | Inverse Image under Embedding of Image under Relation of Image of Point | Let $S$ and $T$ be sets.
Let $\RR_S$ and $\RR_t$ be relations on $S$ and $T$, respectively.
Let $\phi: S \to T$ be a mapping with the property that:
:$\forall p, q \in S: \paren {p \mathrel {\RR_S} q \iff \map \phi p \mathrel {\RR_T} \map \phi q}$
Then for each $p \in S$:
:$\map {\RR_S} p = \phi^{-1} \sqbrk {\map {\RR_... | Let $p \in S$.
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \map {\RR_S} p
}}
{{eqn | ll= \leadstoandfrom
| l = p
| o = \mathrel {\RR_S}
| r = x
| c = {{Defof|Image of Element under Relation|Image}} of $p$ under $\RR_S$
}}
{{eqn | ll= \leadstoandfrom
| l = \map \phi p
| o ... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR_S$ and $\RR_t$ be [[Definition:Endorelation|relations]] on $S$ and $T$, respectively.
Let $\phi: S \to T$ be a [[Definition:Mapping|mapping]] with the property that:
:$\forall p, q \in S: \paren {p \mathrel {\RR_S} q \iff \map \phi p \mathrel {\RR_T} \map \phi q}$... | Let $p \in S$.
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \map {\RR_S} p
}}
{{eqn | ll= \leadstoandfrom
| l = p
| o = \mathrel {\RR_S}
| r = x
| c = {{Defof|Image of Element under Relation|Image}} of $p$ under $\RR_S$
}}
{{eqn | ll= \leadstoandfrom
| l = \map \phi p
| o... | Inverse Image under Embedding of Image under Relation of Image of Point | https://proofwiki.org/wiki/Inverse_Image_under_Embedding_of_Image_under_Relation_of_Image_of_Point | https://proofwiki.org/wiki/Inverse_Image_under_Embedding_of_Image_under_Relation_of_Image_of_Point | [
"Relation Theory"
] | [
"Definition:Set",
"Definition:Endorelation",
"Definition:Mapping"
] | [
"Definition:Set Equality",
"Category:Relation Theory"
] |
proofwiki-7532 | Path as Parameterization of Contour | Let $\closedint a b$ be a closed real interval.
Let $\gamma: \closedint a b \to \C$ be a path.
Let there exist $n \in \N$ and a subdivision $\set {a_0, a_1, \ldots, a_n}$ of $\closedint a b$ such that:
:$\gamma {\restriction_{\closedint {a_{k - 1} } {a_k} } }$ is a smooth path for all $k \in \set {1, \ldots, n}$
where ... | Put $\gamma_k = \gamma {\restriction_{\closedint {a_{k - 1} } {a_k} } } : \closedint {a_{k - 1} } {a_k} \to \C$.
By definition, it follows that there exists a directed smooth curve $C_k$ with parameterization $\gamma_k$.
For all $k \in \set {1, \ldots, n - 1}$, we have:
:$\map {\gamma_k} {a_k} = \map {\gamma_{k + 1} } ... | Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
Let $\gamma: \closedint a b \to \C$ be a [[Definition:Path (Topology)|path]].
Let there exist $n \in \N$ and a [[Definition:Subdivision of Interval|subdivision]] $\set {a_0, a_1, \ldots, a_n}$ of $\closedint a b$ such that:
:$\gamma {... | Put $\gamma_k = \gamma {\restriction_{\closedint {a_{k - 1} } {a_k} } } : \closedint {a_{k - 1} } {a_k} \to \C$.
By definition, it follows that there exists a [[Definition:Directed Smooth Curve|directed smooth curve]] $C_k$ with [[Definition:Parameterization of Directed Smooth Curve|parameterization]] $\gamma_k$.
For... | Path as Parameterization of Contour | https://proofwiki.org/wiki/Path_as_Parameterization_of_Contour | https://proofwiki.org/wiki/Path_as_Parameterization_of_Contour | [
"Path as Parameterization of Contour",
"Contour Integrals",
"Complex Contour Integrals"
] | [
"Definition:Real Interval/Closed",
"Definition:Path (Topology)",
"Definition:Subdivision of Interval",
"Definition:Smooth Path/Complex",
"Definition:Restriction/Mapping",
"Definition:Contour/Complex Plane",
"Definition:Contour/Parameterization/Complex Plane"
] | [
"Definition:Directed Smooth Curve",
"Definition:Directed Smooth Curve/Parameterization",
"Definition:Contour/Complex Plane",
"Definition:Concatenation of Contours",
"Definition:Contour/Parameterization/Complex Plane",
"Category:Path as Parameterization of Contour",
"Category:Contour Integrals",
"Categ... |
proofwiki-7533 | Kernel of Induced Homomorphism of Polynomial Forms | Let $R$ and $S$ be commutative rings with unity.
Let $\phi: R \to S$ be a ring homomorphism.
Let $K = \ker \phi$.
Let $R \sqbrk X$ and $S \sqbrk X$ be the rings of polynomial forms over $R$ and $S$ respectively in the indeterminate $X$.
Let $\bar \phi: R \sqbrk X \to S \sqbrk X$ be the induced morphism of polynomial ri... | Let $\map P X = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$.
Suppose first that $\map \phi {a_i} = 0$ for $i = 0, \ldots, n$.
We have by definition that:
:$\map {\bar \phi} {a_0 + a_1 X + \cdots + a_n X^n} = \map \phi {a_0} + \map \phi {a_1} X + \cdots + \map \phi {a_n} X^n = 0$
That is to say, $\map P X \in \ker \b... | Let $R$ and $S$ be [[Definition:Commutative and Unitary Ring|commutative rings with unity]].
Let $\phi: R \to S$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Let $K = \ker \phi$.
Let $R \sqbrk X$ and $S \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|rings of polynomial forms]] over $R$ and $S$ re... | Let $\map P X = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$.
Suppose first that $\map \phi {a_i} = 0$ for $i = 0, \ldots, n$.
We have by definition that:
:$\map {\bar \phi} {a_0 + a_1 X + \cdots + a_n X^n} = \map \phi {a_0} + \map \phi {a_1} X + \cdots + \map \phi {a_n} X^n = 0$
That is to say, $\map P X \in \k... | Kernel of Induced Homomorphism of Polynomial Forms | https://proofwiki.org/wiki/Kernel_of_Induced_Homomorphism_of_Polynomial_Forms | https://proofwiki.org/wiki/Kernel_of_Induced_Homomorphism_of_Polynomial_Forms | [
"Polynomial Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring Homomorphism",
"Definition:Ring of Polynomial Forms",
"Induced Homomorphism of Polynomial Forms",
"Definition:Kernel of Ring Homomorphism"
] | [
"Definition:Null Polynomial/Polynomial Form",
"Category:Polynomial Theory"
] |
proofwiki-7534 | Boundary of Polygon as Contour | Let $P$ be a polygon embedded in the complex plane $\C$.
Denote the boundary of $P$ as $\partial P$.
Then there exists a simple closed contour $C$ such that:
:$\Img C = \partial P$
where $\Img C$ denotes the image of $C$. | Let $n \in \N$ be the number of sides of $P$.
Denote the vertices of $P$ as $A_1, \ldots, A_n$.
From Complex Plane is Metric Space, it follows that $\C$ is homeomorphic to $\R^2$.
Then, we can consider $\partial P$ as a subset of $\R^2$.
From Boundary of Polygon is Jordan Curve, it follows that there exists a Jordan cu... | Let $P$ be a [[Definition:Polygon|polygon]] embedded in the [[Definition:Complex Plane|complex plane]] $\C$.
Denote the [[Definition:Boundary (Geometry)|boundary]] of $P$ as $\partial P$.
Then there exists a [[Definition:Simple Contour (Complex Plane)|simple]] [[Definition:Closed Contour (Complex Plane)|closed conto... | Let $n \in \N$ be the number of [[Definition:Side of Polygon|sides]] of $P$.
Denote the [[Definition:Vertex of Polygon|vertices]] of $P$ as $A_1, \ldots, A_n$.
From [[Complex Plane is Metric Space]], it follows that $\C$ is [[Definition:Homeomorphic Metric Spaces|homeomorphic]] to $\R^2$.
Then, we can consider $\par... | Boundary of Polygon as Contour | https://proofwiki.org/wiki/Boundary_of_Polygon_as_Contour | https://proofwiki.org/wiki/Boundary_of_Polygon_as_Contour | [
"Complex Contour Integrals"
] | [
"Definition:Polygon",
"Definition:Complex Number/Complex Plane",
"Definition:Boundary (Geometry)",
"Definition:Contour/Simple/Complex Plane",
"Definition:Contour/Closed/Complex Plane",
"Definition:Contour/Image/Complex Plane"
] | [
"Definition:Polygon/Side",
"Definition:Polygon/Vertex",
"Complex Plane is Metric Space",
"Definition:Homeomorphism/Metric Spaces",
"Definition:Subset",
"Boundary of Polygon is Jordan Curve",
"Definition:Jordan Curve",
"Definition:Concatenation (Topology)",
"Definition:Convex Set (Vector Space)/Line ... |
proofwiki-7535 | Zero Simple Staircase Integral Condition for Primitive | Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.
Let $\ds \oint_C \map f z \rd z = 0$ for all simple closed staircase contours $C$ in $D$.
Then $f$ has a primitive $F: D \to \C$. | Let $C$ be a closed staircase contour in $D$, not necessarily simple.
If we show that $\ds \oint_C \map f z \rd z = 0$, then the result follows from Zero Staircase Integral Condition for Primitive.
The staircase contour $C$ is a concatenation of $C_1, \ldots, C_n$, where the image of each $C_k$ is a line segment parall... | Let $f: D \to \C$ be a [[Definition:Continuous Complex Function|continuous complex function]], where $D$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]].
Let $\ds \oint_C \map f z \rd z = 0$ for all [[Definition:Simple Contour (Complex Plane)|simple]] [[Definition:Closed Contour (Complex Plane... | Let $C$ be a [[Definition:Closed Contour (Complex Plane)|closed]] [[Definition:Staircase Contour|staircase contour]] in $D$, not necessarily [[Definition:Simple Contour (Complex Plane)|simple]].
If we show that $\ds \oint_C \map f z \rd z = 0$, then the result follows from [[Zero Staircase Integral Condition for Primi... | Zero Simple Staircase Integral Condition for Primitive | https://proofwiki.org/wiki/Zero_Simple_Staircase_Integral_Condition_for_Primitive | https://proofwiki.org/wiki/Zero_Simple_Staircase_Integral_Condition_for_Primitive | [
"Complex Contour Integrals",
"Zero Simple Staircase Integral Condition for Primitive"
] | [
"Definition:Continuous Complex Function",
"Definition:Connected Domain (Complex Analysis)",
"Definition:Contour/Simple/Complex Plane",
"Definition:Contour/Closed/Complex Plane",
"Definition:Staircase Contour",
"Definition:Primitive (Calculus)/Complex"
] | [
"Definition:Contour/Closed/Complex Plane",
"Definition:Staircase Contour",
"Definition:Contour/Simple/Complex Plane",
"Zero Staircase Integral Condition for Primitive",
"Definition:Concatenation of Contours/Complex Plane",
"Definition:Contour/Image/Complex Plane",
"Definition:Convex Set (Vector Space)/L... |
proofwiki-7536 | Preordering induces Equivalence Relation | Let $\struct {S, \RR}$ be a relational structure such that $\RR$ is a preordering.
Let $\sim$ be the relation induced by $\RR$:
:$x \sim y$ {{iff}} $x \precsim y$ and $y \precsim x$.
Then $\sim$ is an equivalence relation. | To show that $\sim$ is an equivalence relation, we must show that it is reflexive, transitive, and symmetric.
By the definition of preordering, $\precsim$ is transitive and reflexive. | Let $\struct {S, \RR}$ be a [[Definition:Relational Structure|relational structure]] such that $\RR$ is a [[Definition:Preordering|preordering]].
Let $\sim$ be the [[Definition:Equivalence Relation Induced by Preordering|relation induced by $\RR$]]:
:$x \sim y$ {{iff}} $x \precsim y$ and $y \precsim x$.
Then $\sim... | To show that $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]], we must show that it is [[Definition:Reflexive Relation|reflexive]], [[Definition:Transitive Relation|transitive]], and [[Definition:Symmetric Relation|symmetric]].
By the definition of [[Definition:Preordering|preordering]], $\precsim... | Preordering induces Equivalence Relation | https://proofwiki.org/wiki/Preordering_induces_Equivalence_Relation | https://proofwiki.org/wiki/Preordering_induces_Equivalence_Relation | [
"Preorderings",
"Examples of Equivalence Relations"
] | [
"Definition:Relational Structure",
"Definition:Preordering",
"Definition:Equivalence Relation Induced by Preordering",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Transitive Relation",
"Definition:Symmetric Relation",
"Definition:Preordering",
"Definition:Transitive Relation",
"Definition:Reflexive Relation",
"Definition:Transitive Relation",
"Definition:Transitive Relation",
"D... |
proofwiki-7537 | Content of Rational Polynomial is Multiplicative | Let $h \in \Q \sqbrk X$ be a polynomial with rational coefficients.
Let $\cont h$ denote the content of $h$.
Then for any polynomials $f, g \in \Q \sqbrk X$ with rational coefficients:
:$\cont {f g} = \cont f \cont g$ | From Rational Polynomial is Content Times Primitive Polynomial, let $\map f X$ and $\map g X$ be expressed as:
:$\map f X = \cont f \cdot \map {f^*} X$
:$\map g X = \cont g \cdot \map {g^*} X$
where:
:$\cont f, \cont g$ are the content of $f$ and $g$ respectively
:$f^*, g^*$ are primitive.
We have, by applications of R... | Let $h \in \Q \sqbrk X$ be a [[Definition:Polynomial over Ring in One Variable|polynomial]] with [[Definition:Rational Number|rational]] [[Definition:Polynomial Coefficient|coefficients]].
Let $\cont h$ denote the [[Definition:Content of Rational Polynomial|content]] of $h$.
Then for any polynomials $f, g \in \Q \sq... | From [[Rational Polynomial is Content Times Primitive Polynomial]], let $\map f X$ and $\map g X$ be expressed as:
:$\map f X = \cont f \cdot \map {f^*} X$
:$\map g X = \cont g \cdot \map {g^*} X$
where:
:$\cont f, \cont g$ are the [[Definition:Content of Rational Polynomial|content]] of $f$ and $g$ respectively
:$f... | Content of Rational Polynomial is Multiplicative/Proof 2 | https://proofwiki.org/wiki/Content_of_Rational_Polynomial_is_Multiplicative | https://proofwiki.org/wiki/Content_of_Rational_Polynomial_is_Multiplicative/Proof_2 | [
"Gauss's Lemma (Polynomial Theory)",
"Content of Polynomial",
"Content of Rational Polynomial is Multiplicative"
] | [
"Definition:Polynomial over Ring/One Variable",
"Definition:Rational Number",
"Definition:Coefficient of Polynomial",
"Definition:Content of Polynomial/Rational",
"Definition:Rational Number",
"Definition:Coefficient of Polynomial"
] | [
"Rational Polynomial is Content Times Primitive Polynomial",
"Definition:Content of Polynomial/Rational",
"Definition:Primitive Polynomial (Ring Theory)",
"Rational Polynomial is Content Times Primitive Polynomial",
"Gauss's Lemma on Primitive Rational Polynomials",
"Definition:Primitive Polynomial (Ring ... |
proofwiki-7538 | Gauss's Lemma on Irreducible Polynomials | Let $\Z$ be the ring of integers.
Let $\Z \sqbrk X$ be the ring of polynomials over $\Z$.
Let $h \in \Z \sqbrk X$ be a polynomial.
{{TFAE}}
:$(1): \quad h$ is irreducible in $\Q \sqbrk X$ and primitive
:$(2): \quad h$ is irreducible in $\Z \sqbrk X$. | {{explain|Needs to be made explicit as to exactly what is being assumed and what follows as a consequence. As it stands, the consequence of 1 implies 2, for example, still needs to be completed so as to explain exactly how that show $h$ is irreducible -- it relies upon the implicit understanding of what irreducible mea... | Let $\Z$ be the [[Definition:Ring of Integers|ring of integers]].
Let $\Z \sqbrk X$ be the [[Definition:Ring of Polynomials|ring of polynomials]] over $\Z$.
Let $h \in \Z \sqbrk X$ be a [[Definition:Polynomial over Ring|polynomial]].
{{TFAE}}
:$(1): \quad h$ is [[Definition:Irreducible Polynomial|irreducible]] in $... | {{explain|Needs to be made explicit as to exactly what is being assumed and what follows as a consequence. As it stands, the consequence of 1 implies 2, for example, still needs to be completed so as to explain exactly how that show $h$ is irreducible -- it relies upon the implicit understanding of what irreducible mea... | Gauss's Lemma on Irreducible Polynomials | https://proofwiki.org/wiki/Gauss's_Lemma_on_Irreducible_Polynomials | https://proofwiki.org/wiki/Gauss's_Lemma_on_Irreducible_Polynomials | [
"Gauss's Lemma (Polynomial Theory)"
] | [
"Definition:Ring of Integers",
"Definition:Polynomial Ring",
"Definition:Polynomial over Ring",
"Definition:Irreducible Polynomial",
"Definition:Primitive Polynomial (Ring Theory)",
"Definition:Irreducible Polynomial"
] | [] |
proofwiki-7539 | Antisymmetric Quotient of Preordered Set is Ordered Set | Let $\struct {S, \precsim}$ be a preordered set.
Let $\sim$ be the equivalence relation $S$ induced by $\precsim$.
Let $\struct {S / {\sim}, \preceq}$ be the antisymmetric quotient of $\struct {S, \precsim}$.
Then:
:$\struct {S / {\sim}, \preceq}$ is an ordered set.
:$\forall P, Q \in S / {\sim}: \paren {P \preceq Q} \... | By the definition of equivalence relation, $\sim$ is transitive, reflexive, and symmetric.
By the definition of preordering, $\precsim$ is transitive and reflexive.
From Preordering induces Ordering, $\preceq$ is an ordering.
Let $P, Q \in S / {\sim}$ with $P \preceq Q$.
Then by the definition of $\preceq$, there are $... | Let $\struct {S, \precsim}$ be a [[Definition:Preordered Set|preordered set]].
Let $\sim$ be the [[Definition:Equivalence Relation Induced by Preordering|equivalence relation $S$ induced by $\precsim$]].
Let $\struct {S / {\sim}, \preceq}$ be the [[Definition:Antisymmetric Quotient|antisymmetric quotient]] of $\struc... | By the definition of [[Definition:Equivalence Relation|equivalence relation]], $\sim$ is [[Definition:Transitive Relation|transitive]], [[Definition:Reflexive Relation|reflexive]], and [[Definition:Symmetric Relation|symmetric]].
By the definition of [[Definition:Preordering|preordering]], $\precsim$ is [[Definition:T... | Antisymmetric Quotient of Preordered Set is Ordered Set | https://proofwiki.org/wiki/Antisymmetric_Quotient_of_Preordered_Set_is_Ordered_Set | https://proofwiki.org/wiki/Antisymmetric_Quotient_of_Preordered_Set_is_Ordered_Set | [
"Order Theory",
"Preorder Theory",
"Quotient Sets"
] | [
"Definition:Preordering/Preordered Set",
"Definition:Equivalence Relation Induced by Preordering",
"Definition:Antisymmetric Quotient",
"Definition:Ordered Set"
] | [
"Definition:Equivalence Relation",
"Definition:Transitive Relation",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation",
"Definition:Preordering",
"Definition:Transitive Relation",
"Definition:Reflexive Relation",
"Preordering induces Ordering",
"Definition:Ordering",
"Definition:Quo... |
proofwiki-7540 | Ordering on Partition Determines Preordering | Let $S$ be a set.
Let $\PP$ be a partition of $S$.
Let $\phi: S \to \PP$ be the quotient mapping.
Let $\preceq$ be a ordering of $\PP$.
Define a relation $\precsim$ on $S$ by letting $p \precsim q$ {{iff}}:
:$\map \phi p \preceq \map \phi q$
Then:
:$\precsim$ is a preordering on $S$.
:$\precsim$ is the only preordering... | To show that $\precsim$ is a preordering we must show that it is reflexive and transitive. | Let $S$ be a [[Definition:Set|set]].
Let $\PP$ be a [[Definition:Partition (Set Theory)|partition]] of $S$.
Let $\phi: S \to \PP$ be the [[Definition:Quotient Mapping|quotient mapping]].
Let $\preceq$ be a [[Definition:ordering|ordering]] of $\PP$.
Define a [[Definition:Endorelation|relation]] $\precsim$ on $S$ by ... | To show that $\precsim$ is a [[Definition:preordering|preordering]] we must show that it is [[Definition:Reflexive Relation|reflexive]] and [[Definition:Transitive Relation|transitive]]. | Ordering on Partition Determines Preordering | https://proofwiki.org/wiki/Ordering_on_Partition_Determines_Preordering | https://proofwiki.org/wiki/Ordering_on_Partition_Determines_Preordering | [
"Orderings",
"Preorderings"
] | [
"Definition:Set",
"Definition:Set Partition",
"Definition:Quotient Mapping",
"Definition:ordering",
"Definition:Endorelation",
"Definition:Preordering",
"Antisymmetric Quotient of Preordered Set is Ordered Set"
] | [
"Definition:preordering",
"Definition:Reflexive Relation",
"Definition:Transitive Relation",
"Definition:Reflexive Relation",
"Definition:Transitive Relation",
"Definition:Transitive Relation",
"Definition:Reflexive Relation"
] |
proofwiki-7541 | Units of Ring of Polynomial Forms over Commutative Ring | Let $\struct {R, +, \circ}$ be a non-null commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $R \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $R$.
Let $\map P X = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$.
Then:
:$\map P X$ is a unit of $R \sqbrk X$
{{iff}}:
:$a_... | === Necessary condition ===
Let $a_0$ be a unit of $R$.
For $i = 1, \ldots, n$, let $a_i$ be nilpotent in $R$.
Because the nilradical is an ideal of $R$, it follows that:
:$Q = -a_1 X + \dotsb + -a_n X^n$
is nilpotent.
Moreover, multiplying through by $a_0^{-1}$ we may as well assume that $a_0 = 1_R$.
Then:
:$P = 1_R -... | Let $\struct {R, +, \circ}$ be a non-[[Definition:Null Ring|null]] [[Definition:Commutative Ring with Unity|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $R \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial ... | === Necessary condition ===
Let $a_0$ be a [[Definition:Unit of Ring|unit]] of $R$.
For $i = 1, \ldots, n$, let $a_i$ be [[Definition:Nilpotent Ring Element|nilpotent]] in $R$.
Because the [[Definition:Nilradical of Ring|nilradical]] is an [[Definition:Ideal of Ring|ideal]] of $R$, it follows that:
:$Q = -a_1 X + \d... | Units of Ring of Polynomial Forms over Commutative Ring | https://proofwiki.org/wiki/Units_of_Ring_of_Polynomial_Forms_over_Commutative_Ring | https://proofwiki.org/wiki/Units_of_Ring_of_Polynomial_Forms_over_Commutative_Ring | [
"Polynomial Theory"
] | [
"Definition:Null Ring",
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring of Polynomial Forms",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Unit of Ring",
"Definition:Unit of Ring",
"Definition:Nilpotent Ring E... | [
"Definition:Unit of Ring",
"Definition:Nilpotent Ring Element",
"Definition:Nilradical of Ring",
"Definition:Ideal of Ring",
"Definition:Nilpotent Ring Element",
"Unity plus Negative of Nilpotent Ring Element is Unit",
"Definition:Unit of Ring",
"Definition:Unit of Ring",
"Definition:Unit of Ring",
... |
proofwiki-7542 | Polynomials Closed under Addition/Polynomials over Integral Domain | Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Then $\forall x \in R$, the set $D \sqbrk x$ of polynomials in $x$ over $D$ is closed under the operation $+$. | Let $p, q$ be polynomials in $x$ over $D$.
We can express them as:
:$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k$
:$\ds q = \sum_{k \mathop = 0}^n b_k \circ x^k$
where:
:$(1): \quad a_k, b_k \in D$ for all $k$
:$(2): \quad m, n \in \Z_{\ge 0}$, that is, are non-negative integers.
Suppose $m = n$.
Then:
:$\ds p + q = \... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$.
Then $\forall x \in R$, the set $D \sqbrk x$ of [[Definition:Polynomial over Integral Domain|polynomials in $x$ over $D$]] is ... | Let $p, q$ be [[Definition:Polynomial over Integral Domain|polynomials in $x$ over $D$]].
We can express them as:
:$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k$
:$\ds q = \sum_{k \mathop = 0}^n b_k \circ x^k$
where:
:$(1): \quad a_k, b_k \in D$ for all $k$
:$(2): \quad m, n \in \Z_{\ge 0}$, that is, are [[Definition:... | Polynomials Closed under Addition/Polynomials over Integral Domain/Proof 1 | https://proofwiki.org/wiki/Polynomials_Closed_under_Addition/Polynomials_over_Integral_Domain | https://proofwiki.org/wiki/Polynomials_Closed_under_Addition/Polynomials_over_Integral_Domain/Proof_1 | [
"Polynomial Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Subdomain",
"Definition:Polynomial over Ring",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Definition:Polynomial over Ring",
"Definition:Positive/Integer",
"Definition:Commutative Ring",
"Definition:Polynomial over Ring",
"Definition:Polynomial over Ring",
"Definition:Polynomial over Ring",
"Definition:Polynomial over Ring"
] |
proofwiki-7543 | Polynomials Closed under Addition/Polynomials over Integral Domain | Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Then $\forall x \in R$, the set $D \sqbrk x$ of polynomials in $x$ over $D$ is closed under the operation $+$. | A commutative ring with unity is a ring.
An integral subdomain of a commutative ring with unity $R$ is a subring of $R$.
The result then follows as a special case of Polynomials Closed under Addition: Polynomials over Ring
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$.
Then $\forall x \in R$, the set $D \sqbrk x$ of [[Definition:Polynomial over Integral Domain|polynomials in $x$ over $D$]] is ... | A [[Definition:Commutative Ring with Unity|commutative ring with unity]] is a [[Definition:Ring (Abstract Algebra)|ring]].
An [[Definition:Subdomain|integral subdomain]] of a [[Definition:Commutative Ring with Unity|commutative ring with unity]] $R$ is a [[Definition:Subring|subring]] of $R$.
The result then follows... | Polynomials Closed under Addition/Polynomials over Integral Domain/Proof 2 | https://proofwiki.org/wiki/Polynomials_Closed_under_Addition/Polynomials_over_Integral_Domain | https://proofwiki.org/wiki/Polynomials_Closed_under_Addition/Polynomials_over_Integral_Domain/Proof_2 | [
"Polynomial Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Subdomain",
"Definition:Polynomial over Ring",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Subdomain",
"Definition:Commutative and Unitary Ring",
"Definition:Subring",
"Polynomials Closed under Addition/Polynomials over Ring"
] |
proofwiki-7544 | Polynomials Closed under Addition/Polynomials over Ring | Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +, \circ}$ be a subring of $R$.
Then $\forall x \in R$, the set $S \sqbrk x$ of polynomials in $x$ over $S$ is closed under the operation $+$. | Let $p, q$ be polynomials in $x$ over $S$.
We can express them as:
:$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k$
:$\ds q = \sum_{k \mathop = 0}^n b_k \circ x^k$
where:
:$(1): \quad a_k, b_k \in S$ for all $k$
:$(2): \quad m, n \in \Z_{\ge 0}$, that is, are non-negative integers.
Suppose $m = n$.
Then:
:$\ds p + q = \... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {S, +, \circ}$ be a [[Definition:Subring|subring]] of $R$.
Then $\forall x \in R$, the [[Definition:Set|set]] $S \sqbrk x$ of [[Definition:Polynomial over Ring|polynomials in $x$ over $S$]] is [[Definition:Closed Algebraic Str... | Let $p, q$ be [[Definition:Polynomial over Ring|polynomials in $x$ over $S$]].
We can express them as:
:$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k$
:$\ds q = \sum_{k \mathop = 0}^n b_k \circ x^k$
where:
:$(1): \quad a_k, b_k \in S$ for all $k$
:$(2): \quad m, n \in \Z_{\ge 0}$, that is, are [[Definition:Non-Negativ... | Polynomials Closed under Addition/Polynomials over Ring | https://proofwiki.org/wiki/Polynomials_Closed_under_Addition/Polynomials_over_Ring | https://proofwiki.org/wiki/Polynomials_Closed_under_Addition/Polynomials_over_Ring | [
"Polynomial Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Subring",
"Definition:Set",
"Definition:Polynomial over Ring",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Definition:Polynomial over Ring",
"Definition:Positive/Integer",
"Definition:Ring (Abstract Algebra)",
"Definition:Polynomial over Ring",
"Definition:Polynomial over Ring",
"Definition:Polynomial over Ring",
"Definition:Polynomial over Ring",
"Category:Polynomial Theory"
] |
proofwiki-7545 | Polynomials Closed under Addition/Polynomial Forms | Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
:$\ds g = \sum_{k \mathop \in Z} b_k \mathbf X^k$
be polynomials in the indeterminates $\set {X_j: j \in J}$ over the ring $R$.
Then the operation of polynomial addition on $f$ and $g$:
Define the sum:
:$\ds f \oplus g = \sum_{k \mathop \in Z} \paren {a_k + b_k} \m... | It is immediate that $f \oplus g$ is a map from the free commutative monoid to $R$, so we need only prove that $f \oplus g$ is nonzero on finitely many $\mathbf X^k$, $k \in Z$.
Suppose that for some $k \in Z$, $a_k + b_k \ne 0$
This forces at least one of $a_k$ and $b_k$ to be non-zero.
This can only be true for a f... | Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
:$\ds g = \sum_{k \mathop \in Z} b_k \mathbf X^k$
be [[Definition:Polynomial Form|polynomials]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\set {X_j: j \in J}$ over the [[Definition:Ring (Abstract Algebra)|ring]] $R$.
Then the operati... | It is immediate that $f \oplus g$ is a map from the [[Definition:Free Commutative Monoid|free commutative monoid]] to $R$, so we need only prove that $f \oplus g$ is nonzero on finitely many $\mathbf X^k$, $k \in Z$.
Suppose that for some $k \in Z$, $a_k + b_k \ne 0$
This forces at least one of $a_k$ and $b_k$ to b... | Polynomials Closed under Addition/Polynomial Forms | https://proofwiki.org/wiki/Polynomials_Closed_under_Addition/Polynomial_Forms | https://proofwiki.org/wiki/Polynomials_Closed_under_Addition/Polynomial_Forms | [
"Polynomial Theory"
] | [
"Definition:Polynomial over Ring as Function on Free Monoid on Set",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Ring (Abstract Algebra)",
"Definition:Polynomial Addition/Polynomial Forms",
"Definition:Polynomial over Ring as Function on Free Monoid on Set",
"Definition:Operation/Binary Operat... | [
"Definition:Free Commutative Monoid",
"Definition:Finite",
"Definition:Polynomial/Term",
"Definition:Polynomial over Ring as Function on Free Monoid on Set",
"Category:Polynomial Theory"
] |
proofwiki-7546 | Fuzzy Intersection is Commutative | Fuzzy intersection is commutative. | Let $\textbf A = \struct{A, \mu_A}$ and $\textbf B = \struct{B, \mu_B}$ be fuzzy sets. | [[Definition:Fuzzy Intersection|Fuzzy intersection]] is [[Definition:Commutative Operation|commutative]]. | Let $\textbf A = \struct{A, \mu_A}$ and $\textbf B = \struct{B, \mu_B}$ be [[Definition:Fuzzy Set|fuzzy sets]]. | Fuzzy Intersection is Commutative | https://proofwiki.org/wiki/Fuzzy_Intersection_is_Commutative | https://proofwiki.org/wiki/Fuzzy_Intersection_is_Commutative | [
"Fuzzy Set Theory"
] | [
"Definition:Fuzzy Intersection",
"Definition:Commutative/Operation"
] | [
"Definition:Fuzzy Set"
] |
proofwiki-7547 | Nilpotent Element is Zero Divisor | Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Suppose further that $R$ is not the null ring.
Let $x \in R$ be a nilpotent element of $R$.
Then $x$ is a zero divisor in $R$. | First note that when $R$ is the null ring the result is false.
This is because although $0_R$ is nilpotent element in the null ring, it is not actually a zero divisor.
Hence in this case $0_R$ is both nilpotent and a zero divisor.
So, let $R$ be a non-null ring.
By hypothesis, there exists $n \in \Z_{>0}$ such that $x^... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Suppose further that $R$ is not the [[Definition:Null Ring|null ring]].
Let $x \in R$ be a [[Definition:Nilpotent Ring Element|nilpotent element]] of $R$.
Then $x$ is a [[Definition:Zero Diviso... | First note that when $R$ is the [[Definition:Null Ring|null ring]] the result is false.
This is because although $0_R$ is [[Definition:Nilpotent Ring Element|nilpotent element]] in the [[Definition:Null Ring|null ring]], it is not actually a [[Definition:Zero Divisor of Ring|zero divisor]].
Hence in this case $0_R$ i... | Nilpotent Element is Zero Divisor | https://proofwiki.org/wiki/Nilpotent_Element_is_Zero_Divisor | https://proofwiki.org/wiki/Nilpotent_Element_is_Zero_Divisor | [
"Nilpotent Ring Elements",
"Zero Divisors"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Null Ring",
"Definition:Nilpotent Ring Element",
"Definition:Zero Divisor/Ring"
] | [
"Definition:Null Ring",
"Definition:Nilpotent Ring Element",
"Definition:Null Ring",
"Definition:Zero Divisor/Ring",
"Definition:Nilpotent Ring Element",
"Definition:Zero Divisor/Ring",
"Definition:Null Ring",
"Definition:Null Ring",
"Ring Product with Zero",
"Definition:Zero Divisor/Ring",
"Def... |
proofwiki-7548 | Integral Domain is Reduced Ring | Let $\left({D, +, \circ}\right)$ be an integral domain.
Then $D$ is reduced. | Let $x \in D$ be a nilpotent element.
Then by Nilpotent Element is Zero Divisor, $x$ is a zero divisor in $D$.
By the definition of an integral domain, this means that $x = 0$.
Therefore the only nilpotent element of $D$ is $0$.
That is, $D$ is reduced.
{{Qed}}
Category:Ring Theory
fm7xo02i9ov8o6hcq03u0qp2633wo8v | Let $\left({D, +, \circ}\right)$ be an [[Definition:Integral Domain|integral domain]].
Then $D$ is [[Definition:Reduced Ring|reduced]]. | Let $x \in D$ be a [[Definition:Nilpotent Ring Element|nilpotent element]].
Then by [[Nilpotent Element is Zero Divisor]], $x$ is a [[Definition:Zero Divisor|zero divisor]] in $D$.
By the definition of an [[Definition:Integral Domain|integral domain]], this means that $x = 0$.
Therefore the only [[Definition:Nilpote... | Integral Domain is Reduced Ring | https://proofwiki.org/wiki/Integral_Domain_is_Reduced_Ring | https://proofwiki.org/wiki/Integral_Domain_is_Reduced_Ring | [
"Ring Theory"
] | [
"Definition:Integral Domain",
"Definition:Reduced Ring"
] | [
"Definition:Nilpotent Ring Element",
"Nilpotent Element is Zero Divisor",
"Definition:Zero Divisor",
"Definition:Integral Domain",
"Definition:Nilpotent Ring Element",
"Definition:Reduced Ring",
"Category:Ring Theory"
] |
proofwiki-7549 | Units of Ring of Polynomial Forms over Integral Domain | Let $\struct {D, +, \circ}$ be an integral domain.
Let $D \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $D$.
Then the group of units of $D \sqbrk X$ is precisely the group of elements of $D \sqbrk X$ of degree zero that are units of $D$. | It is immediate that a unit of $D$ is also a unit of $D \sqbrk X$.
Let $P$ be a unit of $D \sqbrk X$.
Then there exists $Q \in D \sqbrk X$ such that $P Q = 1$.
By Corollary 2 to Degree of Product of Polynomials over Ring we have:
:$0 = \map \deg 1 = \map \deg P + \map \deg Q$
Therefore:
:$\map \deg P = \map \deg Q = 0$... | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]].
Let $D \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] in an [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$ over $D$.
Then the [[Definition:Group of Units of Ring|group of units]] of... | It is immediate that a [[Definition:Unit of Ring|unit]] of $D$ is also a [[Definition:Unit of Ring|unit]] of $D \sqbrk X$.
Let $P$ be a [[Definition:Unit of Ring|unit]] of $D \sqbrk X$.
Then there exists $Q \in D \sqbrk X$ such that $P Q = 1$.
By [[Degree of Product of Polynomials over Ring/Corollary 2|Corollary 2 t... | Units of Ring of Polynomial Forms over Integral Domain | https://proofwiki.org/wiki/Units_of_Ring_of_Polynomial_Forms_over_Integral_Domain | https://proofwiki.org/wiki/Units_of_Ring_of_Polynomial_Forms_over_Integral_Domain | [
"Polynomial Theory"
] | [
"Definition:Integral Domain",
"Definition:Ring of Polynomial Forms",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Group of Units/Ring",
"Definition:Group",
"Definition:Element",
"Definition:Degree of Polynomial/Zero",
"Definition:Unit of Ring"
] | [
"Definition:Unit of Ring",
"Definition:Unit of Ring",
"Definition:Unit of Ring",
"Degree of Product of Polynomials over Ring/Corollary 2",
"Definition:Unit of Ring",
"Category:Polynomial Theory"
] |
proofwiki-7550 | Kernel of Magma Homomorphism is Submagma | Let $\struct {S, *}$ be a magma.
Let $\struct {T, \circ}$ be an algebraic structure with an identity $e$.
Let $\phi: S \to T$ be a homomorphism.
Then the kernel of $\phi$ is a submagma of $\struct {S, *}$.
That is:
:$\struct {\map {\phi^{-1} } e, *}$ is a submagma of $\struct {S, *}$
where $\map {\phi^{-1} } e$ denote ... | Let $x, y \in \map {\phi^{-1} } e$.
By the definition of a magma, $S$ is closed under $*$.
That is:
:$\forall x, y \in S: x * y \in S$
Hence:
:$x * y \in \Dom \phi$
It is to be shown that:
:$x * y \in \map {\phi^{-1} } e$
Thus:
{{begin-eqn}}
{{eqn | l = x, y
| o = \in
| r = \map {\phi^{-1} } e
| c = {... | Let $\struct {S, *}$ be a [[Definition:Magma|magma]].
Let $\struct {T, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] with an [[Definition:Identity Element|identity]] $e$.
Let $\phi: S \to T$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
Then the [[Definition:Kernel of Magm... | Let $x, y \in \map {\phi^{-1} } e$.
By the definition of a [[Definition:Magma|magma]], $S$ is closed under $*$.
That is:
:$\forall x, y \in S: x * y \in S$
Hence:
:$x * y \in \Dom \phi$
It is to be shown that:
:$x * y \in \map {\phi^{-1} } e$
Thus:
{{begin-eqn}}
{{eqn | l = x, y
| o = \in
| r = \map {\... | Kernel of Magma Homomorphism is Submagma | https://proofwiki.org/wiki/Kernel_of_Magma_Homomorphism_is_Submagma | https://proofwiki.org/wiki/Kernel_of_Magma_Homomorphism_is_Submagma | [
"Abstract Algebra"
] | [
"Definition:Magma",
"Definition:Algebraic Structure",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Kernel of Magma Homomorphism",
"Definition:Submagma",
"Definition:Submagma",
"Definition:Preimage/Mapping/Element"
] | [
"Definition:Magma",
"Category:Abstract Algebra"
] |
proofwiki-7551 | Preimage of Zero of Homomorphism is Submagma | Let $\struct {S, *}$ be a magma.
Let $\struct {T, \circ}$ be a magma with a zero element $0$.
Let $\phi: S \to T$ be a magma homomorphism.
Then $\struct {\phi^{-1} \sqbrk 0, *}$ is a submagma of $\struct {S, *}$. | Let $x, y \in \phi^{-1} \sqbrk 0$.
It is to be shown that:
:$x * y \in \phi^{-1} \sqbrk 0$
Thus:
{{begin-eqn}}
{{eqn | l = x, y \in \phi^{-1} \sqbrk 0
| o = \leadstoandfrom
| r = \paren {\map \phi x = 0} \land \paren {\map \phi y = 0}
| c = {{Defof|Preimage of Element under Mapping}}
}}
{{eqn | o = \l... | Let $\struct {S, *}$ be a [[Definition:Magma|magma]].
Let $\struct {T, \circ}$ be a [[Definition:Magma|magma]] with a [[Definition:Zero Element|zero element]] $0$.
Let $\phi: S \to T$ be a [[Definition:Homomorphism (Abstract Algebra)|magma homomorphism]].
Then $\struct {\phi^{-1} \sqbrk 0, *}$ is a [[Definition:Sub... | Let $x, y \in \phi^{-1} \sqbrk 0$.
It is to be shown that:
:$x * y \in \phi^{-1} \sqbrk 0$
Thus:
{{begin-eqn}}
{{eqn | l = x, y \in \phi^{-1} \sqbrk 0
| o = \leadstoandfrom
| r = \paren {\map \phi x = 0} \land \paren {\map \phi y = 0}
| c = {{Defof|Preimage of Element under Mapping}}
}}
{{eqn | o = ... | Preimage of Zero of Homomorphism is Submagma | https://proofwiki.org/wiki/Preimage_of_Zero_of_Homomorphism_is_Submagma | https://proofwiki.org/wiki/Preimage_of_Zero_of_Homomorphism_is_Submagma | [
"Abstract Algebra"
] | [
"Definition:Magma",
"Definition:Magma",
"Definition:Zero Element",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Submagma"
] | [
"Category:Abstract Algebra"
] |
proofwiki-7552 | Polynomial over Field is Reducible iff Scalar Multiple is Reducible | Let $K$ be a field.
Let $K \sqbrk X$ be the ring of polynomial forms over $K$.
Let $P \in K \sqbrk X$.
Let $\lambda \in K \setminus \set 0$.
Then $P$ is irreducible in $K \sqbrk X$ {{iff}} $\lambda P$ is also irreducible in $K \sqbrk X$.
{{expand|Investigate whether this result also holds where $K$ is a general ring.}} | === Necessary Condition ===
Let $P$ be irreducible.
Suppose further that $ \lambda P$ has a non-trivial factorization:
:$\lambda P = Q_1 Q_2$
that is, such that $Q_1$ and $Q_2$ are not units of $K \sqbrk X$.
By Units of Ring of Polynomial Forms over Field it follows that $\deg Q_1 \ge 1$ and $\deg Q_2 \ge 1$.
Let $Q_1'... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $K \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over $K$.
Let $P \in K \sqbrk X$.
Let $\lambda \in K \setminus \set 0$.
Then $P$ is [[Definition:Irreducible Polynomial|irreducible]] in $K \sqbrk X$ {{iff}} $\lambd... | === Necessary Condition ===
Let $P$ be [[Definition:Irreducible Polynomial|irreducible]].
Suppose further that $ \lambda P$ has a [[Definition:Trivial Factorization|non-trivial factorization]]:
:$\lambda P = Q_1 Q_2$
that is, such that $Q_1$ and $Q_2$ are not [[Definition:Unit of Ring|units]] of $K \sqbrk X$.
By [[U... | Polynomial over Field is Reducible iff Scalar Multiple is Reducible | https://proofwiki.org/wiki/Polynomial_over_Field_is_Reducible_iff_Scalar_Multiple_is_Reducible | https://proofwiki.org/wiki/Polynomial_over_Field_is_Reducible_iff_Scalar_Multiple_is_Reducible | [
"Polynomial Theory"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Ring of Polynomial Forms",
"Definition:Irreducible Polynomial",
"Definition:Irreducible Polynomial"
] | [
"Definition:Irreducible Polynomial",
"Definition:Trivial Factorization",
"Definition:Unit of Ring",
"Units of Ring of Polynomial Forms over Field",
"Definition:Trivial Factorization",
"Definition:Irreducible Polynomial",
"Definition:Trivial Factorization",
"Definition:Irreducible Polynomial",
"Defin... |
proofwiki-7553 | Conjunction of Disjunctions Consequence | :$\paren {p \lor q} \land \paren {r \lor s} \vdash p \lor r \lor \paren {q \land s}$ | {{BeginTableau|\paren {p \lor q} \land \paren {r \lor s} \vdash \paren {p \lor r} \lor \paren {q \land s} }}
{{Premise|1|\paren {p \lor q} \land \paren {r \lor s} }}
{{SequentIntro|2|1|\paren {p \land \paren {r \lor s} } \lor \paren {q \land \paren {r \lor s} }|1|Conjunction Distributes over Disjunction}}
{{TheoremIntr... | :$\paren {p \lor q} \land \paren {r \lor s} \vdash p \lor r \lor \paren {q \land s}$ | {{BeginTableau|\paren {p \lor q} \land \paren {r \lor s} \vdash \paren {p \lor r} \lor \paren {q \land s} }}
{{Premise|1|\paren {p \lor q} \land \paren {r \lor s} }}
{{SequentIntro|2|1|\paren {p \land \paren {r \lor s} } \lor \paren {q \land \paren {r \lor s} }|1|[[Rule of Distribution/Conjunction Distributes over Disj... | Conjunction of Disjunctions Consequence | https://proofwiki.org/wiki/Conjunction_of_Disjunctions_Consequence | https://proofwiki.org/wiki/Conjunction_of_Disjunctions_Consequence | [
"Conjunction",
"Disjunction"
] | [] | [
"Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1",
"Rule of Simplification/Sequent Form/Formulation 2",
"Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Forward Implication",
"Rule of Simplification/Sequent Form/Formu... |
proofwiki-7554 | Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Forward Implication | :$\vdash \paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} }$ | {{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} } }}
{{Assumption|1|p \land \paren {q \lor r} }}
{{SequentIntro|2|1|\paren {\paren {p \land q} \lor \paren {p \land r} }|1|Conjunction is Left Distributive over Disjunction: Formulation 1}}
{{Implicatio... | :$\vdash \paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} }$ | {{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} } }}
{{Assumption|1|p \land \paren {q \lor r} }}
{{SequentIntro|2|1|\paren {\paren {p \land q} \lor \paren {p \land r} }|1|[[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributi... | Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Forward Implication | https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Forward_Implication | [
"Rule of Distribution"
] | [] | [
"Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Forward Implication",
"Category:Rule of Distribution"
] |
proofwiki-7555 | Existence of Ring of Polynomial Forms in Transcendental over Integral Domain | Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$ whose zero is $0_D$.
Let $X \in R$ be transcendental over $D$
Then the ring of polynomials $D \sqbrk X$ in $X$ over $D$ exists. | {{finish|The following is an outline only}}
Suppose that $D \sqbrk X$ exists.
Let $\ds \map P X = \sum_{k \mathop = 0}^n a_k X^k$, where $a_n \ne 0_D$, be an arbitrary element of $D \sqbrk X$.
Then $\map P X$ corresponds to, and is completely described by, the ordered tuple of coefficients $\tuple {a_0, a_1, \dotsc, a_... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$ whose [[Definition:Ring Zero|zero]] is $0_D$.
Let $X \in R$ be [[Definition:Transcendental over Integral Domain|transcendental o... | {{finish|The following is an outline only}}
Suppose that $D \sqbrk X$ exists.
Let $\ds \map P X = \sum_{k \mathop = 0}^n a_k X^k$, where $a_n \ne 0_D$, be an arbitrary [[Definition:Element|element]] of $D \sqbrk X$.
Then $\map P X$ corresponds to, and is completely described by, the [[Definition:Ordered Tuple|ordere... | Existence of Ring of Polynomial Forms in Transcendental over Integral Domain | https://proofwiki.org/wiki/Existence_of_Ring_of_Polynomial_Forms_in_Transcendental_over_Integral_Domain | https://proofwiki.org/wiki/Existence_of_Ring_of_Polynomial_Forms_in_Transcendental_over_Integral_Domain | [
"Polynomial Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Subdomain",
"Definition:Ring Zero",
"Definition:Transcendental (Abstract Algebra)/Ring",
"Definition:Ring of Polynomials in Ring Element"
] | [
"Definition:Element",
"Definition:Ordered Tuple",
"Definition:Coefficient",
"Definition:Sequence/Infinite Sequence",
"Definition:Element",
"Definition:Element",
"Definition:Polynomial Ring/Sequences",
"Polynomial Ring of Sequences is Ring",
"Definition:Ring (Abstract Algebra)"
] |
proofwiki-7556 | Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Reverse Implication | :$\vdash \paren {\paren {p \land q} \lor \paren {p \land r} } \implies \paren {p \land \paren {q \lor r} }$ | {{BeginTableau|\vdash \paren {\paren {p \land q} \lor \paren {p \land r} } \implies \paren {p \land \paren {q \lor r} } }}
{{Assumption|1|\paren {p \land q} \lor \paren {p \land r} }}
{{SequentIntro|2|1|p \land \paren {q \lor r}|1|Conjunction is Left Distributive over Disjunction: Formulation 1}}
{{Implication|3||\pare... | :$\vdash \paren {\paren {p \land q} \lor \paren {p \land r} } \implies \paren {p \land \paren {q \lor r} }$ | {{BeginTableau|\vdash \paren {\paren {p \land q} \lor \paren {p \land r} } \implies \paren {p \land \paren {q \lor r} } }}
{{Assumption|1|\paren {p \land q} \lor \paren {p \land r} }}
{{SequentIntro|2|1|p \land \paren {q \lor r}|1|[[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formula... | Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Reverse Implication | https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Reverse_Implication | [
"Rule of Distribution"
] | [] | [
"Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Reverse Implication",
"Category:Rule of Distribution"
] |
proofwiki-7557 | Eisenstein Integers form Integral Domain | The ring of Eisenstein integers $\struct {\Z \sqbrk \omega, +, \times}$ is an integral domain. | By Eisenstein Integers form Subring of Complex Numbers we know that $\struct {\Z \sqbrk \omega, +, \times}$ is a subring of the complex numbers $\C$.
Let $1_\C$ be the unity of $\C$.
Let $1_\omega$ be the unity of $\Z \sqbrk \omega$.
By the Subdomain Test it suffices to show that $1_\C = 1_\omega$.
By Unity of Ring is ... | The [[Definition:Ring of Eisenstein Integers|ring of Eisenstein integers]] $\struct {\Z \sqbrk \omega, +, \times}$ is an [[Definition:Integral Domain|integral domain]]. | By [[Eisenstein Integers form Subring of Complex Numbers]] we know that $\struct {\Z \sqbrk \omega, +, \times}$ is a [[Definition:Subring|subring]] of the [[Definition:Complex Number|complex numbers]] $\C$.
Let $1_\C$ be the [[Definition:Unity of Ring|unity]] of $\C$.
Let $1_\omega$ be the [[Definition:Unity of Ring|... | Eisenstein Integers form Integral Domain | https://proofwiki.org/wiki/Eisenstein_Integers_form_Integral_Domain | https://proofwiki.org/wiki/Eisenstein_Integers_form_Integral_Domain | [
"Number Theory",
"Examples of Integral Domains",
"Eisenstein Integers"
] | [
"Definition:Ring of Eisenstein Integers",
"Definition:Integral Domain"
] | [
"Eisenstein Integers form Subring of Complex Numbers",
"Definition:Subring",
"Definition:Complex Number",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Unity (Abstract Algebra)/Ring",
"Subdomain Test",
"Unity of Ring is Unique",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring ... |
proofwiki-7558 | Eisenstein Integers form Subring of Complex Numbers | The set of Eisenstein integers $\Z \sqbrk \omega$, under the operations of complex addition and complex multiplication, forms a subring of the set of complex numbers $\C$. | We will use the Subring Test.
This is valid, as the set of complex numbers $\C$ forms a field, which is by definition itself a ring.
We note that $\Z \sqbrk \omega$ is not empty, as (for example) $0 + 0 \omega \in \Z \sqbrk \omega$.
Let $a + b \omega, c + d \omega \in \Z \sqbrk \omega$.
Then we have $-\paren {c + d \om... | The set of [[Definition:Eisenstein Integer|Eisenstein integers]] $\Z \sqbrk \omega$, under the operations of [[Definition:Complex Addition|complex addition]] and [[Definition:Complex Multiplication|complex multiplication]], forms a [[Definition:Subring|subring]] of the set of [[Definition:Complex Number|complex numbers... | We will use the [[Subring Test]].
This is valid, as the [[Complex Numbers form Field|set of complex numbers $\C$ forms a field]], which is [[Definition:Field (Abstract Algebra)|by definition]] itself a [[Definition:Ring (Abstract Algebra)|ring]].
We note that $\Z \sqbrk \omega$ is not [[Definition:Empty Set|empty]],... | Eisenstein Integers form Subring of Complex Numbers | https://proofwiki.org/wiki/Eisenstein_Integers_form_Subring_of_Complex_Numbers | https://proofwiki.org/wiki/Eisenstein_Integers_form_Subring_of_Complex_Numbers | [
"Examples of Integral Domains",
"Subrings",
"Complex Numbers",
"Eisenstein Integers"
] | [
"Definition:Eisenstein Integer",
"Definition:Addition/Complex Numbers",
"Definition:Multiplication/Complex Numbers",
"Definition:Subring",
"Definition:Complex Number"
] | [
"Subring Test",
"Complex Numbers form Field",
"Definition:Field (Abstract Algebra)",
"Definition:Ring (Abstract Algebra)",
"Definition:Empty Set",
"Integers form Integral Domain",
"Definition:Integral Domain",
"Definition:Ring (Abstract Algebra)",
"Definition:Multiplication/Complex Numbers",
"Defi... |
proofwiki-7559 | Norm of Eisenstein Integer | Let $\alpha$ be an Eisenstein integer.
That is:
:$\alpha = a + b \omega$ for some $a, b \in \Z$
where $\omega := e^{2 \pi i /3}$ is a cube root of unity.
Then:
:$\cmod \alpha^2 = a^2 - a b + b^2$
where $\cmod {\, \cdot \,}$ denotes the modulus of a complex number. | We find that:
{{begin-eqn}}
{{eqn | l = \cmod \alpha^2
| r = \alpha \overline \alpha
| c = Modulus in Terms of Conjugate
}}
{{eqn | r = \paren {a + b \omega} \paren {\overline {a + b \omega} }
| c = Modulus in Terms of Conjugate
}}
{{eqn | r = \paren {a + b \omega} \paren {\overline a + \overline b \o... | Let $\alpha$ be an [[Definition:Eisenstein Integer|Eisenstein integer]].
That is:
:$\alpha = a + b \omega$ for some $a, b \in \Z$
where $\omega := e^{2 \pi i /3}$ is a [[Cube Roots of Unity|cube root of unity]].
Then:
:$\cmod \alpha^2 = a^2 - a b + b^2$
where $\cmod {\, \cdot \,}$ denotes the [[Definition:Complex Mo... | We find that:
{{begin-eqn}}
{{eqn | l = \cmod \alpha^2
| r = \alpha \overline \alpha
| c = [[Modulus in Terms of Conjugate]]
}}
{{eqn | r = \paren {a + b \omega} \paren {\overline {a + b \omega} }
| c = [[Modulus in Terms of Conjugate]]
}}
{{eqn | r = \paren {a + b \omega} \paren {\overline a + \overl... | Norm of Eisenstein Integer | https://proofwiki.org/wiki/Norm_of_Eisenstein_Integer | https://proofwiki.org/wiki/Norm_of_Eisenstein_Integer | [
"Algebraic Number Theory"
] | [
"Definition:Eisenstein Integer",
"Complex Roots of Unity/Examples/Cube Roots",
"Definition:Complex Modulus",
"Definition:Complex Number"
] | [
"Modulus in Terms of Conjugate",
"Modulus in Terms of Conjugate",
"Sum of Complex Conjugates",
"Product of Complex Conjugates",
"Complex Number equals Conjugate iff Wholly Real",
"Definition:Complex Number/Polar Form",
"Sum of Complex Number with Conjugate",
"Polar Form of Complex Conjugate",
"Expon... |
proofwiki-7560 | Polynomial Forms over Field form Integral Domain/Formulation 2 | Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $\GF$ be the set of all polynomials over $\struct {F, +, \circ}$ defined as sequences.
Let polynomial addition and polynomial multiplication be defined as:
:$\forall f = \sequence {a_k} = \tuple {a_0, a_1, a_2, \ldots}, g = \sequen... | As $\struct {F, +, \circ}$ is a field, it is also by definition a ring.
Thus from Polynomial Ring of Sequences is Ring we have that $\struct {\GF, \oplus, \otimes}$ is a ring.
{{explain|Use an analogous result to Ring of Polynomial Forms is Commutative Ring with Unity to get the CRU bit done}}
From Field is Integral Do... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $\GF$ be the [[Definition:Set|set]] of all [[Definition:Polynomial over Field as Sequence|polynomials over $\struct {F, +, \circ}$ def... | As $\struct {F, +, \circ}$ is a [[Definition:Field (Abstract Algebra)|field]], it is also by definition a [[Definition:Ring (Abstract Algebra)|ring]].
Thus from [[Polynomial Ring of Sequences is Ring]] we have that $\struct {\GF, \oplus, \otimes}$ is a [[Definition:Ring (Abstract Algebra)|ring]].
{{explain|Use an ana... | Polynomial Forms over Field form Integral Domain/Formulation 2 | https://proofwiki.org/wiki/Polynomial_Forms_over_Field_form_Integral_Domain/Formulation_2 | https://proofwiki.org/wiki/Polynomial_Forms_over_Field_form_Integral_Domain/Formulation_2 | [
"Polynomial Forms over Field form Integral Domain"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Set",
"Definition:Polynomial over Ring as Sequence",
"Definition:Polynomial Addition/Sequence",
"Definition:Multiplication of Polynomials/Sequence",
"Definition:Integral Domain"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Ring (Abstract Algebra)",
"Polynomial Ring of Sequences is Ring",
"Definition:Ring (Abstract Algebra)",
"Ring of Polynomial Forms is Commutative Ring with Unity",
"Field is Integral Domain",
"Definition:Field (Abstract Algebra)",
"Definition:Integral ... |
proofwiki-7561 | Maximal Spectrum of Ring is Nonempty | Let $A$ be a non-trivial commutative ring with unity.
Then its maximal spectrum is non-empty:
:$\operatorname {Max} \Spec A \ne \O$ | This is a reformulation of Krull's Theorem.
{{qed}} | Let $A$ be a [[Definition:Non-Trivial Ring|non-trivial]] [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Then its [[Definition:Maximal Spectrum of Ring|maximal spectrum]] is [[Definition:Non-Empty Set|non-empty]]:
:$\operatorname {Max} \Spec A \ne \O$ | This is a reformulation of [[Krull's Theorem]].
{{qed}} | Maximal Spectrum of Ring is Nonempty | https://proofwiki.org/wiki/Maximal_Spectrum_of_Ring_is_Nonempty | https://proofwiki.org/wiki/Maximal_Spectrum_of_Ring_is_Nonempty | [
"Commutative Algebra"
] | [
"Definition:Non-Trivial Ring",
"Definition:Commutative and Unitary Ring",
"Definition:Maximal Spectrum of Ring",
"Definition:Non-Empty Set"
] | [
"Krull's Theorem"
] |
proofwiki-7562 | Ring of Polynomial Functions is Commutative Ring with Unity | Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $R \sqbrk {\set {X_j: j \in J} }$ be the ring of polynomial forms over $R$ in the indeterminates $\set {X_j: j \in J}$.
Let $R^J$ be the free module on $J$.
Let $A$ be the set of all polynomial functions $R^J \to R$.
Let $\struct {A, +, \circ}$ be the ri... | First we check that the operations of ring product and ring addition are closed in $A$.
Let $Z$ be the set of all multiindices indexed by $J$.
Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k, \ g = \sum_{k \mathop \in Z} b_k \mathbf X^k \in R \sqbrk {\set {X_j: j \in J} }$.
Under the evaluation homomorphism, $f$ ... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $R \sqbrk {\set {X_j: j \in J} }$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over $R$ in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\set {X_j: j \in J}$.
... | First we check that the operations of [[Definition:Ring Product|ring product]] and [[Definition:Ring Addition|ring addition]] are [[Definition:Closed Operation|closed in $A$]].
Let $Z$ be the set of all [[Definition:Multiindex|multiindices]] indexed by $J$.
Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k, \ g =... | Ring of Polynomial Functions is Commutative Ring with Unity | https://proofwiki.org/wiki/Ring_of_Polynomial_Functions_is_Commutative_Ring_with_Unity | https://proofwiki.org/wiki/Ring_of_Polynomial_Functions_is_Commutative_Ring_with_Unity | [
"Polynomial Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring of Polynomial Forms",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Free Module over Ring",
"Definition:Polynomial Function/General Definition",
"Definition:Ring of Polynomial Functions",
"Definition:Commutative and Unitary Ring"
] | [
"Definition:Ring (Abstract Algebra)/Product",
"Definition:Ring (Abstract Algebra)/Addition",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiindex",
"Equality of Polynomials",
"Definition:Pointwise Operation",
"Definition:Polynomial Function/Ring",
"Definition:Closure (Ab... |
proofwiki-7563 | Knaster-Tarski Lemma | Let $\struct {L, \preceq}$ be a complete lattice.
Let $f: L \to L$ be an increasing mapping.
Then $f$ has a smallest fixed point and a greatest fixed point. | Let $P = \set {x \in L: x \preceq \map f x}$.
Let $p = \bigvee P$, the supremum of $P$.
Let $x \in P$.
Then by the definition of supremum:
:$x \preceq p$
Since $f$ is increasing:
:$\map f x \preceq \map f p$
By the definition of $P$:
:$x \preceq \map f x$
Thus because $\preceq$ is an ordering, and therefore transitive:... | Let $\struct {L, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $f: L \to L$ be an [[Definition:Increasing Mapping|increasing mapping]].
Then $f$ has a [[Definition:Smallest Element|smallest]] [[Definition:Fixed Point|fixed point]] and a [[Definition:Greatest Element|greatest]] [[Definition:Fix... | Let $P = \set {x \in L: x \preceq \map f x}$.
Let $p = \bigvee P$, the [[Definition:Supremum of Set|supremum]] of $P$.
Let $x \in P$.
Then by the definition of [[Definition:Supremum of Set|supremum]]:
:$x \preceq p$
Since $f$ is [[Definition:Increasing Mapping|increasing]]:
:$\map f x \preceq \map f p$
By the defi... | Knaster-Tarski Lemma | https://proofwiki.org/wiki/Knaster-Tarski_Lemma | https://proofwiki.org/wiki/Knaster-Tarski_Lemma | [
"Complete Lattices",
"Knaster-Tarski Lemma"
] | [
"Definition:Complete Lattice",
"Definition:Increasing/Mapping",
"Definition:Smallest Element",
"Definition:Fixed Point",
"Definition:Greatest Element",
"Definition:Fixed Point"
] | [
"Definition:Supremum of Set",
"Definition:Supremum of Set",
"Definition:Increasing/Mapping",
"Definition:Ordering",
"Definition:Transitive Relation",
"Definition:Upper Bound of Set",
"Definition:Supremum of Set",
"Definition:Increasing/Mapping",
"Definition:Supremum of Set",
"Definition:Ordering",... |
proofwiki-7564 | Knaster-Tarski Lemma | Let $\struct {L, \preceq}$ be a complete lattice.
Let $f: L \to L$ be an increasing mapping.
Then $f$ has a smallest fixed point and a greatest fixed point. | By the Knaster-Tarski Lemma: Power Set, $f$ has a least fixed point.
Thus it has a fixed point.
{{qed}} | Let $\struct {L, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $f: L \to L$ be an [[Definition:Increasing Mapping|increasing mapping]].
Then $f$ has a [[Definition:Smallest Element|smallest]] [[Definition:Fixed Point|fixed point]] and a [[Definition:Greatest Element|greatest]] [[Definition:Fix... | By the [[Knaster-Tarski Lemma/Power Set|Knaster-Tarski Lemma: Power Set]], $f$ has a [[Definition:Smallest Set by Set Inclusion|least]] [[Definition:Fixed Point|fixed point]].
Thus it has a fixed point.
{{qed}} | Knaster-Tarski Lemma/Corollary/Power Set/Proof 1 | https://proofwiki.org/wiki/Knaster-Tarski_Lemma | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Corollary/Power_Set/Proof_1 | [
"Complete Lattices",
"Knaster-Tarski Lemma"
] | [
"Definition:Complete Lattice",
"Definition:Increasing/Mapping",
"Definition:Smallest Element",
"Definition:Fixed Point",
"Definition:Greatest Element",
"Definition:Fixed Point"
] | [
"Knaster-Tarski Lemma/Power Set",
"Definition:Smallest Set by Set Inclusion",
"Definition:Fixed Point"
] |
proofwiki-7565 | Knaster-Tarski Lemma | Let $\struct {L, \preceq}$ be a complete lattice.
Let $f: L \to L$ be an increasing mapping.
Then $f$ has a smallest fixed point and a greatest fixed point. | Let $P = \set {x \in \powerset S: x \subseteq \map f x}$.
Let $\ds \bigcup P$ be the union of $P$.
Let $x \in P$.
Then by Set is Subset of Union: General Result:
:$\ds x \subseteq \bigcup P$
Since $f$ is increasing, $\map f x \subseteq \map f {\bigcup P}$.
By definition of $P$, $x \subseteq \map f x$.
Thus $\ds x \subs... | Let $\struct {L, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $f: L \to L$ be an [[Definition:Increasing Mapping|increasing mapping]].
Then $f$ has a [[Definition:Smallest Element|smallest]] [[Definition:Fixed Point|fixed point]] and a [[Definition:Greatest Element|greatest]] [[Definition:Fix... | Let $P = \set {x \in \powerset S: x \subseteq \map f x}$.
Let $\ds \bigcup P$ be the [[Definition:Union of Set of Sets|union]] of $P$.
Let $x \in P$.
Then by [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]]:
:$\ds x \subseteq \bigcup P$
Since $f$ is [[Definition:Increasing Mapping|in... | Knaster-Tarski Lemma/Corollary/Power Set/Proof 2 | https://proofwiki.org/wiki/Knaster-Tarski_Lemma | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Corollary/Power_Set/Proof_2 | [
"Complete Lattices",
"Knaster-Tarski Lemma"
] | [
"Definition:Complete Lattice",
"Definition:Increasing/Mapping",
"Definition:Smallest Element",
"Definition:Fixed Point",
"Definition:Greatest Element",
"Definition:Fixed Point"
] | [
"Definition:Set Union/Set of Sets",
"Set is Subset of Union/General Result",
"Definition:Increasing/Mapping",
"Subset Relation is Transitive",
"Union is Smallest Superset/General Result",
"Definition:Increasing/Mapping",
"Set is Subset of Union/General Result",
"Definition:Set Equality/Definition 2",
... |
proofwiki-7566 | Knaster-Tarski Theorem | Let $\struct {L, \preceq}$ be a complete lattice.
Let $f: L \to L$ be an increasing mapping.
Let $F$ be the set (or class) of fixed points of $f$.
Then $\struct {F, \preceq}$ is a complete lattice. | Let $S \subseteq F$.
Let $s = \bigvee S$ be the supremum of $S$.
We wish to show that there is an element of $F$ that succeeds all elements of $S$ and is the smallest element of $F$ to do so.
By the definition of supremum, an element succeeds all elements of $S$ {{iff}} it succeeds $s$.
Let $U = s^\succeq$ be the upper... | Let $\struct {L, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $f: L \to L$ be an [[Definition:Increasing Mapping|increasing mapping]].
Let $F$ be the [[Definition:Set|set]] (or [[Definition:Class (Class Theory)|class]]) of [[Definition:Fixed Point|fixed points]] of $f$.
Then $\struct {F, \pr... | Let $S \subseteq F$.
Let $s = \bigvee S$ be the [[Definition:Supremum of Set|supremum]] of $S$.
We wish to show that there is an [[Definition:Element|element]] of $F$ that [[Definition:Succeed|succeeds]] all [[Definition:Element|elements]] of $S$ and is the [[Definition:Smallest Element|smallest element]] of $F$ to d... | Knaster-Tarski Theorem | https://proofwiki.org/wiki/Knaster-Tarski_Theorem | https://proofwiki.org/wiki/Knaster-Tarski_Theorem | [
"Complete Lattices",
"Increasing Mappings"
] | [
"Definition:Complete Lattice",
"Definition:Increasing/Mapping",
"Definition:Set",
"Definition:Class (Class Theory)",
"Definition:Fixed Point",
"Definition:Complete Lattice"
] | [
"Definition:Supremum of Set",
"Definition:Element",
"Definition:Succeed",
"Definition:Element",
"Definition:Smallest Element",
"Definition:Supremum of Set",
"Definition:Element",
"Definition:Succeed",
"Definition:Element",
"Definition:Succeed",
"Definition:Upper Closure/Element",
"Definition:S... |
proofwiki-7567 | Degree of Product of Polynomials over Ring/Corollary 2 | Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
Let $D \sqbrk X$ be the ring of polynomials over $D$ in the indeterminate $X$.
For $f \in D \sqbrk X$ let $\map \deg f$ denote the degree of $f$.
Then:
:$\forall f, g \in D \sqbrk X: \map \deg {f g} = \map \deg f + \map \deg g$ | An integral domain is a commutative and unitary ring with no proper zero divisors.
The result follows from Degree of Product of Polynomials over Ring: Corollary 1.
{{qed}} | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
Let $D \sqbrk X$ be the [[Definition:Ring of Polynomials|ring of polynomials]] over $D$ in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
For $f \in D \sqbrk X$ let $... | An [[Definition:Integral Domain|integral domain]] is a [[Definition:Commutative and Unitary Ring|commutative and unitary ring]] with no [[Definition:Proper Zero Divisor|proper zero divisors]].
The result follows from [[Degree of Product of Polynomials over Ring/Corollary 1|Degree of Product of Polynomials over Ring: C... | Degree of Product of Polynomials over Ring/Corollary 2 | https://proofwiki.org/wiki/Degree_of_Product_of_Polynomials_over_Ring/Corollary_2 | https://proofwiki.org/wiki/Degree_of_Product_of_Polynomials_over_Ring/Corollary_2 | [
"Degree of Product of Polynomials over Ring"
] | [
"Definition:Integral Domain",
"Definition:Ring Zero",
"Definition:Polynomial Ring",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Degree of Polynomial"
] | [
"Definition:Integral Domain",
"Definition:Commutative and Unitary Ring",
"Definition:Proper Zero Divisor",
"Degree of Product of Polynomials over Ring/Corollary 1"
] |
proofwiki-7568 | Knaster-Tarski Lemma/Power Set | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $f: \powerset S \to \powerset S$ be a $\subseteq$-increasing mapping.
That is, suppose that for all $T, U \in \powerset S$:
:$T \subseteq U \implies \map f T \subseteq \map f U$
Then $f$ has a greatest fixed point and a least fixed point. | By Power Set is Complete Lattice, $\struct {\powerset S, \cap, \cup, \subseteq}$ is a complete lattice.
Thus the theorem holds by the Knaster-Tarski Lemma.
{{qed}}
{{Namedfor|Bronisław Knaster|name2 = Alfred Tarski|cat = Knaster|cat2 = Tarski}} | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $f: \powerset S \to \powerset S$ be a $\subseteq$-[[Definition:Increasing Mapping|increasing mapping]].
That is, suppose that for all $T, U \in \powerset S$:
:$T \subseteq U \implies \map f T \subseteq \map ... | By [[Power Set is Complete Lattice]], $\struct {\powerset S, \cap, \cup, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
Thus the theorem holds by the [[Knaster-Tarski Lemma]].
{{qed}}
{{Namedfor|Bronisław Knaster|name2 = Alfred Tarski|cat = Knaster|cat2 = Tarski}} | Knaster-Tarski Lemma/Power Set | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Power_Set | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Power_Set | [
"Knaster-Tarski Lemma",
"Power Set"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Increasing/Mapping",
"Definition:Greatest Set by Set Inclusion",
"Definition:Fixed Point",
"Definition:Smallest Set by Set Inclusion",
"Definition:Fixed Point"
] | [
"Power Set is Complete Lattice",
"Definition:Complete Lattice",
"Knaster-Tarski Lemma"
] |
proofwiki-7569 | Cantor-Bernstein-Schröder Theorem/Proof 6 | Let $A$ and $B$ be sets.
Let $f: A \to B$ and $g: B \to A$ be injections.
Then there is a bijection $h: A \to B$; so that $A$ and $B$ are equivalent.
Furthermore:
:For all $x \in A$ and $y \in B$, if $y = \map h x$ then either $y = \map f x$ or $x = \map g y$. | Let $\powerset A$ be the power set of $A$.
Define a mapping $E: \powerset A \to \powerset A$ thus:
:$\map E S = A \setminus g \sqbrk {B \setminus f \sqbrk S}$
=== $E$ is increasing ===
Let $S, T \in \powerset A$ such that $S \subseteq T$.
Then:
{{begin-eqn}}
{{eqn | l = f \sqbrk S
| o = \subseteq
| r = f \s... | Let $A$ and $B$ be [[Definition:Set|sets]].
Let $f: A \to B$ and $g: B \to A$ be [[Definition:Injection|injections]].
Then there is a [[Definition:Bijection|bijection]] $h: A \to B$; so that $A$ and $B$ are [[Definition:Set Equivalence|equivalent]].
Furthermore:
:For all $x \in A$ and $y \in B$, if $y = \map h x$ ... | Let $\powerset A$ be the [[Definition:Power Set|power set]] of $A$.
Define a mapping $E: \powerset A \to \powerset A$ thus:
:$\map E S = A \setminus g \sqbrk {B \setminus f \sqbrk S}$
=== $E$ is increasing ===
Let $S, T \in \powerset A$ such that $S \subseteq T$.
Then:
{{begin-eqn}}
{{eqn | l = f \sqbrk S
... | Cantor-Bernstein-Schröder Theorem/Proof 6 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_6 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_6 | [
"Cantor-Bernstein-Schröder Theorem"
] | [
"Definition:Set",
"Definition:Injection",
"Definition:Bijection",
"Definition:Set Equivalence"
] | [
"Definition:Power Set",
"Image of Subset under Relation is Subset of Image",
"Set Difference with Subset is Superset of Set Difference",
"Image of Subset under Relation is Subset of Image",
"Set Difference with Subset is Superset of Set Difference",
"Knaster-Tarski Lemma/Power Set",
"Definition:Fixed Po... |
proofwiki-7570 | Ring of Polynomial Forms over Integral Domain is Integral Domain | Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
Let $\struct {D \sqbrk X, \oplus, \odot}$ be the ring of polynomial forms over $D$ in the indeterminate $X$.
Then $\struct {D \sqbrk X, \oplus, \odot}$ is an integral domain. | By definition an integral domain is a commutative ring with unity.
From Ring of Polynomial Forms is Commutative Ring with Unity it follows that $\struct {D \sqbrk X, +, \circ}$ is a commutative ring with unity.
Suppose $f, g \in D \sqbrk X$ such that neither $f$ nor $g$ are the null polynomial.
Let $\map \deg f = n$ a... | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
Let $\struct {D \sqbrk X, \oplus, \odot}$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over $D$ in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]]... | By definition an [[Definition:Integral Domain|integral domain]] is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
From [[Ring of Polynomial Forms is Commutative Ring with Unity]] it follows that $\struct {D \sqbrk X, +, \circ}$ is a [[Definition:Commutative and Unitary Ring|commutative rin... | Ring of Polynomial Forms over Integral Domain is Integral Domain | https://proofwiki.org/wiki/Ring_of_Polynomial_Forms_over_Integral_Domain_is_Integral_Domain | https://proofwiki.org/wiki/Ring_of_Polynomial_Forms_over_Integral_Domain_is_Integral_Domain | [
"Polynomial Rings",
"Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Ring Zero",
"Definition:Ring of Polynomial Forms",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Integral Domain"
] | [
"Definition:Integral Domain",
"Definition:Commutative and Unitary Ring",
"Ring of Polynomial Forms is Commutative Ring with Unity",
"Definition:Commutative and Unitary Ring",
"Definition:Null Polynomial",
"Degree of Product of Polynomials over Ring/Corollary 2",
"Definition:Degree of Polynomial",
"Def... |
proofwiki-7571 | Union of One-to-Many Relations with Disjoint Images is One-to-Many | Let $S_1, S_2, T_1, T_2$ be sets or classes.
Let $\RR_1$ be a one-to-many relation on $S_1 \times T_1$.
Let $\RR_2$ be a one-to-many relation on $S_2 \times T_2$.
Suppose that the images of $\RR_1$ and $\RR_2$ are disjoint.
Then $\RR_1 \cup \RR_2$ is a one-to-many relation on $\paren {S_1 \cup S_2} \times \paren {T_1 \... | Let $\QQ = \RR_1 \cup \RR_2$.
Then $\QQ \subseteq \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \subseteq \paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.
Thus $\QQ$ is a relation on $\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.
Let $T'_1$ and $T'_2$ be the images of $\RR_1$ and $\RR_2$, respectively... | Let $S_1, S_2, T_1, T_2$ be [[Definition:Set|sets]] or [[Definition:Class (Class Theory)|classes]].
Let $\RR_1$ be a [[Definition:One-to-Many Relation|one-to-many relation]] on $S_1 \times T_1$.
Let $\RR_2$ be a [[Definition:One-to-Many Relation|one-to-many relation]] on $S_2 \times T_2$.
Suppose that the [[Definiti... | Let $\QQ = \RR_1 \cup \RR_2$.
Then $\QQ \subseteq \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \subseteq \paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.
Thus $\QQ$ is a [[Definition:Relation|relation]] on $\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.
Let $T'_1$ and $T'_2$ be the [[Definition:Ima... | Union of One-to-Many Relations with Disjoint Images is One-to-Many | https://proofwiki.org/wiki/Union_of_One-to-Many_Relations_with_Disjoint_Images_is_One-to-Many | https://proofwiki.org/wiki/Union_of_One-to-Many_Relations_with_Disjoint_Images_is_One-to-Many | [
"Relation Theory"
] | [
"Definition:Set",
"Definition:Class (Class Theory)",
"Definition:One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:Image (Set Theory)/Relation/Relation",
"Definition:Disjoint Sets",
"Definition:One-to-Many Relation"
] | [
"Definition:Relation",
"Definition:Image (Set Theory)/Relation/Relation",
"Definition:One-to-Many Relation",
"Category:Relation Theory"
] |
proofwiki-7572 | Union of Many-to-One Relations with Disjoint Domains is Many-to-One | Let $S_1, S_2, T_1, T_2$ be sets or classes.
Let $\RR_1$ be a many-to-one relation on $S_1 \times T_1$.
Let $\RR_2$ be a many-to-one relation on $S_2 \times T_2$.
Suppose that the domains of $\RR_1$ and $\RR_2$ are disjoint.
Then $\RR_1 \cup \RR_2$ is a many-to-one relation on $\paren {S_1 \cup S_2} \times \paren {T_1 ... | Let $\RR = \RR_1 \cup \RR_2$.
Let $\tuple {x, y_1}, \tuple {x, y_2} \in \RR$.
By the definition of union, $\tuple {x, y_1}$ and $\tuple {x, y_2}$ are each in $\RR_1$ or $\RR_2$.
Suppose that both are in $\RR_1$.
Then since $\RR_1$ is a many-to-one relation, $y_1 = y_2$.
Suppose that $\tuple {x, y_1} \in \RR_1$ and $\tu... | Let $S_1, S_2, T_1, T_2$ be [[Definition:Set|sets]] or [[Definition:Class (Class Theory)|classes]].
Let $\RR_1$ be a [[Definition:Many-to-One Relation|many-to-one relation]] on $S_1 \times T_1$.
Let $\RR_2$ be a [[Definition:Many-to-One Relation|many-to-one relation]] on $S_2 \times T_2$.
Suppose that the [[Definiti... | Let $\RR = \RR_1 \cup \RR_2$.
Let $\tuple {x, y_1}, \tuple {x, y_2} \in \RR$.
By the definition of [[Definition:Set Union|union]], $\tuple {x, y_1}$ and $\tuple {x, y_2}$ are each in $\RR_1$ or $\RR_2$.
Suppose that both are in $\RR_1$.
Then since $\RR_1$ is a [[Definition:Many-to-One Relation|many-to-one relation... | Union of Many-to-One Relations with Disjoint Domains is Many-to-One | https://proofwiki.org/wiki/Union_of_Many-to-One_Relations_with_Disjoint_Domains_is_Many-to-One | https://proofwiki.org/wiki/Union_of_Many-to-One_Relations_with_Disjoint_Domains_is_Many-to-One | [
"Relation Theory"
] | [
"Definition:Set",
"Definition:Class (Class Theory)",
"Definition:Many-to-One Relation",
"Definition:Many-to-One Relation",
"Definition:Domain (Set Theory)/Relation",
"Definition:Disjoint Sets",
"Definition:Many-to-One Relation"
] | [
"Definition:Set Union",
"Definition:Many-to-One Relation",
"Definition:Domain (Set Theory)/Relation",
"Definition:Many-to-One Relation",
"Category:Relation Theory"
] |
proofwiki-7573 | Nth Root of Integer is Integer or Irrational | Let $n$ be a natural number.
Let $x$ be an integer.
If the $n$th root of $x$ is not an integer, it must be irrational. | We prove the contrapositive: if the $n$th root of $x$ is rational, it must be an integer.
By Existence of Canonical Form of Rational Number, there exist an integer $a$ and a natural number $b$ which are coprime such that:
{{begin-eqn}}
{{eqn | l = x^{1/n}
| r = \frac a b
}}
{{eqn | ll= \leadsto
| l = x
... | Let $n$ be a [[Definition:Natural Number|natural number]].
Let $x$ be an [[Definition:Integer|integer]].
If the [[Definition:Root of Number|$n$th root]] of $x$ is not an [[Definition:Integer|integer]], it must be [[Definition:Irrational Number|irrational]]. | We prove the [[Definition:Contrapositive Statement|contrapositive]]: if the [[Definition:Root of Number|$n$th root]] of $x$ is [[Definition:Rational Number|rational]], it must be an [[Definition:Integer|integer]].
By [[Existence of Canonical Form of Rational Number]], there exist an [[Definition:Integer|integer]] $a$... | Nth Root of Integer is Integer or Irrational | https://proofwiki.org/wiki/Nth_Root_of_Integer_is_Integer_or_Irrational | https://proofwiki.org/wiki/Nth_Root_of_Integer_is_Integer_or_Irrational | [
"Irrationality Proofs",
"Integers"
] | [
"Definition:Natural Numbers",
"Definition:Integer",
"Definition:Root of Number",
"Definition:Integer",
"Definition:Irrational Number"
] | [
"Definition:Contrapositive Statement",
"Definition:Root of Number",
"Definition:Rational Number",
"Definition:Integer",
"Existence of Canonical Form of Rational Number",
"Definition:Integer",
"Definition:Natural Numbers",
"Definition:Coprime/Integers",
"Definition:Coprime/Integers",
"Definition:Co... |
proofwiki-7574 | Union of Bijections with Disjoint Domains and Codomains is Bijection | Let $A$, $B$, $C$, and $D$ be sets or classes.
Let $A \cap B = C \cap D = \O$.
Let $f: A \to C$ and $g: B \to D$ be bijections.
Then $f \cup g: A \cup B \to C \cup D$ is also a bijection. | By the definition of bijection, $f$ and $g$ are many-to-one and one-to-many relations.
By Union of Many-to-One Relations with Disjoint Domains is Many-to-One and Union of One-to-Many Relations with Disjoint Images is One-to-Many:
:$f \cup g$ is many-to-one and one-to-many.
Thus to show $f \cup g$ is a bijection require... | Let $A$, $B$, $C$, and $D$ be [[Definition:Set|sets]] or [[Definition:Class (Class Theory)|classes]].
Let $A \cap B = C \cap D = \O$.
Let $f: A \to C$ and $g: B \to D$ be [[Definition:Bijection|bijections]].
Then $f \cup g: A \cup B \to C \cup D$ is also a [[Definition:Bijection|bijection]]. | By the definition of [[Definition:Bijection|bijection]], $f$ and $g$ are [[Definition:Many-to-One Relation|many-to-one]] and [[Definition:One-to-Many Relation|one-to-many relations]].
By [[Union of Many-to-One Relations with Disjoint Domains is Many-to-One]] and [[Union of One-to-Many Relations with Disjoint Images is... | Union of Bijections with Disjoint Domains and Codomains is Bijection | https://proofwiki.org/wiki/Union_of_Bijections_with_Disjoint_Domains_and_Codomains_is_Bijection | https://proofwiki.org/wiki/Union_of_Bijections_with_Disjoint_Domains_and_Codomains_is_Bijection | [
"Bijections",
"Set Union",
"Union of Bijections with Disjoint Domains and Codomains is Bijection"
] | [
"Definition:Set",
"Definition:Class (Class Theory)",
"Definition:Bijection",
"Definition:Bijection"
] | [
"Definition:Bijection",
"Definition:Many-to-One Relation",
"Definition:One-to-Many Relation",
"Union of Many-to-One Relations with Disjoint Domains is Many-to-One",
"Union of One-to-Many Relations with Disjoint Images is One-to-Many",
"Definition:Many-to-One Relation",
"Definition:One-to-Many Relation",... |
proofwiki-7575 | Closed Interval in Complete Lattice is Complete Lattice | Let $\struct {L, \preceq}$ be a complete lattice.
Let $a, b \in L$ with $a \preceq b$.
Let $I = \closedint a b$ be the closed interval between $a$ and $b$.
{{explain|Demonstrate that for each $a, b \in L$ that $\closedint a b$ exists and is unique.}}
Then $\struct {I, \preceq}$ is also a complete lattice. | Let $S \subseteq I$.
If $S = \O$, then it has a supremum in $I$ of $a$ and an infimum in $I$ of $b$.
Let $S \ne \O$.
Since $S \subseteq I$, $a$ is a lower bound of $S$ and $b$ is an upper bound of $S$.
Since $L$ is a complete lattice, $S$ has an infimum, $p$, and a supremum, $q$, in $L$.
Thus by the definitions of infi... | Let $\struct {L, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $a, b \in L$ with $a \preceq b$.
Let $I = \closedint a b$ be the [[Definition:Closed Interval|closed interval]] between $a$ and $b$.
{{explain|Demonstrate that for each $a, b \in L$ that $\closedint a b$ exists and is unique.}}
The... | Let $S \subseteq I$.
If $S = \O$, then it has a [[Definition:Supremum of Set|supremum]] in $I$ of $a$ and an [[Definition:Infimum of Set|infimum]] in $I$ of $b$.
Let $S \ne \O$.
Since $S \subseteq I$, $a$ is a [[Definition:Lower Bound of Set|lower bound]] of $S$ and $b$ is an [[Definition:Upper Bound of Set|upper b... | Closed Interval in Complete Lattice is Complete Lattice | https://proofwiki.org/wiki/Closed_Interval_in_Complete_Lattice_is_Complete_Lattice | https://proofwiki.org/wiki/Closed_Interval_in_Complete_Lattice_is_Complete_Lattice | [
"Complete Lattices"
] | [
"Definition:Complete Lattice",
"Definition:Interval/Ordered Set/Closed",
"Definition:Complete Lattice"
] | [
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Lower Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Complete Lattice",
"Definition:Infimum of Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Supremum of Set",
"Definition:Infimum of S... |
proofwiki-7576 | Dedekind-Complete Bounded Ordered Set is Complete Lattice | Let $\struct {L, \preceq}$ be an ordered set.
Let $L$ have a lower bound $\bot$ and an upper bound $\top$.
Let $\struct {L, \preceq}$ be Dedekind-complete.
Then $\struct {L, \preceq}$ is a complete lattice. | Let $S \subseteq L$.
If $S = \O$, then $S$ has a supremum of $\bot$ and an infimum of $\top$.
Let $S \ne \O$.
$S$ is bounded above by $\top$.
As $\struct {L, \preceq}$ is Dedekind complete, $S$ has a supremum.
$S$ is bounded below by $\bot$.
By Dedekind Completeness is Self-Dual, $S$ has an infimum.
Thus every subset o... | Let $\struct {L, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $L$ have a [[Definition:Lower Bound of Set|lower bound]] $\bot$ and an [[Definition:Upper Bound of Set|upper bound]] $\top$.
Let $\struct {L, \preceq}$ be [[Definition:Dedekind Complete|Dedekind-complete]].
Then $\struct {L, \preceq}$ is a... | Let $S \subseteq L$.
If $S = \O$, then $S$ has a [[Definition:Supremum of Set|supremum]] of $\bot$ and an [[Definition:Infimum of Set|infimum]] of $\top$.
Let $S \ne \O$.
$S$ is [[Definition:Bounded Above Set|bounded above]] by $\top$.
As $\struct {L, \preceq}$ is [[Definition:Dedekind Complete|Dedekind complete]],... | Dedekind-Complete Bounded Ordered Set is Complete Lattice | https://proofwiki.org/wiki/Dedekind-Complete_Bounded_Ordered_Set_is_Complete_Lattice | https://proofwiki.org/wiki/Dedekind-Complete_Bounded_Ordered_Set_is_Complete_Lattice | [
"Complete Lattices"
] | [
"Definition:Ordered Set",
"Definition:Lower Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Dedekind Completeness Property",
"Definition:Complete Lattice"
] | [
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Bounded Above Set",
"Definition:Dedekind Completeness Property",
"Definition:Supremum of Set",
"Definition:Bounded Below Set",
"Dedekind Completeness is Self-Dual",
"Definition:Infimum of Set",
"Definition:Subset",
"Definition:... |
proofwiki-7577 | Set Difference with Subset is Superset of Set Difference | Let $A, B, S$ be sets or classes.
Suppose that $A \subseteq B$.
Then:
:$S \setminus B \subseteq S \setminus A$
where $\setminus$ represents set difference. | Let $x \in S \setminus B$.
Then by definition of set difference:
:$x \in S$ and $x \notin B$
{{AimForCont}} $x \in A$.
Then since $A$ is a subset (or subclass) of $B$, $x \in B$, a contradiction.
Thus $x \notin A$.
Since $x \in S$ and $x \notin A$, we conclude that $x \in S \setminus A$.
As this holds for all $x \in S ... | Let $A, B, S$ be [[Definition:Set|sets]] or [[Definition:Class (Class Theory)|classes]].
Suppose that $A \subseteq B$.
Then:
:$S \setminus B \subseteq S \setminus A$
where $\setminus$ represents [[Definition:Set Difference|set difference]]. | Let $x \in S \setminus B$.
Then by definition of [[Definition:Set Difference|set difference]]:
:$x \in S$ and $x \notin B$
{{AimForCont}} $x \in A$.
Then since $A$ is a [[Definition:subset|subset]] (or [[Definition:subclass|subclass]]) of $B$, $x \in B$, a [[Definition:contradiction|contradiction]].
Thus $x \notin... | Set Difference with Subset is Superset of Set Difference | https://proofwiki.org/wiki/Set_Difference_with_Subset_is_Superset_of_Set_Difference | https://proofwiki.org/wiki/Set_Difference_with_Subset_is_Superset_of_Set_Difference | [
"Set Difference"
] | [
"Definition:Set",
"Definition:Class (Class Theory)",
"Definition:Set Difference"
] | [
"Definition:Set Difference",
"Definition:subset",
"Definition:subclass",
"Definition:contradiction",
"Category:Set Difference"
] |
proofwiki-7578 | Knaster-Tarski Lemma/Corollary | Let $\struct {L, \preceq}$ be a complete lattice.
Let $f: L \to L$ be an increasing mapping.
Then $f$ has a fixed point | By the Knaster-Tarski Lemma, $f$ has a least fixed point.
Thus it has a fixed point.
{{qed}}
Category:Knaster-Tarski Lemma
5rvzccftb6yjfu5kjc2201xq80xzx0b | Let $\struct {L, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $f: L \to L$ be an [[Definition:Increasing Mapping|increasing mapping]].
Then $f$ has a [[Definition:Fixed Point|fixed point]] | By the [[Knaster-Tarski Lemma]], $f$ has a [[Definition:Smallest Element|least]] [[Definition:Fixed Point|fixed point]].
Thus it has a [[Definition:Fixed Point|fixed point]].
{{qed}}
[[Category:Knaster-Tarski Lemma]]
5rvzccftb6yjfu5kjc2201xq80xzx0b | Knaster-Tarski Lemma/Corollary | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Corollary | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Corollary | [
"Knaster-Tarski Lemma"
] | [
"Definition:Complete Lattice",
"Definition:Increasing/Mapping",
"Definition:Fixed Point"
] | [
"Knaster-Tarski Lemma",
"Definition:Smallest Element",
"Definition:Fixed Point",
"Definition:Fixed Point",
"Category:Knaster-Tarski Lemma"
] |
proofwiki-7579 | Knaster-Tarski Lemma/Corollary | Let $\struct {L, \preceq}$ be a complete lattice.
Let $f: L \to L$ be an increasing mapping.
Then $f$ has a fixed point | By the Knaster-Tarski Lemma: Power Set, $f$ has a least fixed point.
Thus it has a fixed point.
{{qed}} | Let $\struct {L, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $f: L \to L$ be an [[Definition:Increasing Mapping|increasing mapping]].
Then $f$ has a [[Definition:Fixed Point|fixed point]] | By the [[Knaster-Tarski Lemma/Power Set|Knaster-Tarski Lemma: Power Set]], $f$ has a [[Definition:Smallest Set by Set Inclusion|least]] [[Definition:Fixed Point|fixed point]].
Thus it has a fixed point.
{{qed}} | Knaster-Tarski Lemma/Corollary/Power Set/Proof 1 | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Corollary | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Corollary/Power_Set/Proof_1 | [
"Knaster-Tarski Lemma"
] | [
"Definition:Complete Lattice",
"Definition:Increasing/Mapping",
"Definition:Fixed Point"
] | [
"Knaster-Tarski Lemma/Power Set",
"Definition:Smallest Set by Set Inclusion",
"Definition:Fixed Point"
] |
proofwiki-7580 | Knaster-Tarski Lemma/Corollary | Let $\struct {L, \preceq}$ be a complete lattice.
Let $f: L \to L$ be an increasing mapping.
Then $f$ has a fixed point | Let $P = \set {x \in \powerset S: x \subseteq \map f x}$.
Let $\ds \bigcup P$ be the union of $P$.
Let $x \in P$.
Then by Set is Subset of Union: General Result:
:$\ds x \subseteq \bigcup P$
Since $f$ is increasing, $\map f x \subseteq \map f {\bigcup P}$.
By definition of $P$, $x \subseteq \map f x$.
Thus $\ds x \subs... | Let $\struct {L, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $f: L \to L$ be an [[Definition:Increasing Mapping|increasing mapping]].
Then $f$ has a [[Definition:Fixed Point|fixed point]] | Let $P = \set {x \in \powerset S: x \subseteq \map f x}$.
Let $\ds \bigcup P$ be the [[Definition:Union of Set of Sets|union]] of $P$.
Let $x \in P$.
Then by [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]]:
:$\ds x \subseteq \bigcup P$
Since $f$ is [[Definition:Increasing Mapping|in... | Knaster-Tarski Lemma/Corollary/Power Set/Proof 2 | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Corollary | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Corollary/Power_Set/Proof_2 | [
"Knaster-Tarski Lemma"
] | [
"Definition:Complete Lattice",
"Definition:Increasing/Mapping",
"Definition:Fixed Point"
] | [
"Definition:Set Union/Set of Sets",
"Set is Subset of Union/General Result",
"Definition:Increasing/Mapping",
"Subset Relation is Transitive",
"Union is Smallest Superset/General Result",
"Definition:Increasing/Mapping",
"Set is Subset of Union/General Result",
"Definition:Set Equality/Definition 2",
... |
proofwiki-7581 | Knaster-Tarski Lemma/Corollary/Power Set | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $f: \powerset S \to \powerset S$ be a $\subseteq$-increasing mapping.
That is, suppose that for all $T, U \in \powerset S$:
:$T \subseteq U \implies \map f T \subseteq \map f U$
Then $f$ has a fixed point. | By the Knaster-Tarski Lemma: Power Set, $f$ has a least fixed point.
Thus it has a fixed point.
{{qed}} | Let $S$ be a [[Definition:set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $f: \powerset S \to \powerset S$ be a $\subseteq$-[[Definition:Increasing Mapping|increasing mapping]].
That is, suppose that for all $T, U \in \powerset S$:
:$T \subseteq U \implies \map f T \subseteq \map ... | By the [[Knaster-Tarski Lemma/Power Set|Knaster-Tarski Lemma: Power Set]], $f$ has a [[Definition:Smallest Set by Set Inclusion|least]] [[Definition:Fixed Point|fixed point]].
Thus it has a fixed point.
{{qed}} | Knaster-Tarski Lemma/Corollary/Power Set/Proof 1 | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Corollary/Power_Set | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Corollary/Power_Set/Proof_1 | [
"Knaster-Tarski Lemma"
] | [
"Definition:set",
"Definition:Power Set",
"Definition:Increasing/Mapping",
"Definition:Fixed Point"
] | [
"Knaster-Tarski Lemma/Power Set",
"Definition:Smallest Set by Set Inclusion",
"Definition:Fixed Point"
] |
proofwiki-7582 | Knaster-Tarski Lemma/Corollary/Power Set | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $f: \powerset S \to \powerset S$ be a $\subseteq$-increasing mapping.
That is, suppose that for all $T, U \in \powerset S$:
:$T \subseteq U \implies \map f T \subseteq \map f U$
Then $f$ has a fixed point. | Let $P = \set {x \in \powerset S: x \subseteq \map f x}$.
Let $\ds \bigcup P$ be the union of $P$.
Let $x \in P$.
Then by Set is Subset of Union: General Result:
:$\ds x \subseteq \bigcup P$
Since $f$ is increasing, $\map f x \subseteq \map f {\bigcup P}$.
By definition of $P$, $x \subseteq \map f x$.
Thus $\ds x \subs... | Let $S$ be a [[Definition:set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $f: \powerset S \to \powerset S$ be a $\subseteq$-[[Definition:Increasing Mapping|increasing mapping]].
That is, suppose that for all $T, U \in \powerset S$:
:$T \subseteq U \implies \map f T \subseteq \map ... | Let $P = \set {x \in \powerset S: x \subseteq \map f x}$.
Let $\ds \bigcup P$ be the [[Definition:Union of Set of Sets|union]] of $P$.
Let $x \in P$.
Then by [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]]:
:$\ds x \subseteq \bigcup P$
Since $f$ is [[Definition:Increasing Mapping|in... | Knaster-Tarski Lemma/Corollary/Power Set/Proof 2 | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Corollary/Power_Set | https://proofwiki.org/wiki/Knaster-Tarski_Lemma/Corollary/Power_Set/Proof_2 | [
"Knaster-Tarski Lemma"
] | [
"Definition:set",
"Definition:Power Set",
"Definition:Increasing/Mapping",
"Definition:Fixed Point"
] | [
"Definition:Set Union/Set of Sets",
"Set is Subset of Union/General Result",
"Definition:Increasing/Mapping",
"Subset Relation is Transitive",
"Union is Smallest Superset/General Result",
"Definition:Increasing/Mapping",
"Set is Subset of Union/General Result",
"Definition:Set Equality/Definition 2",
... |
proofwiki-7583 | Natural Number has Same Prime Factors as Integer Power | Let $x$ be a natural number such that $x > 1$.
Let $n \ge 1$ be a (strictly) positive integer.
The $n$th power of $x$ has the same prime factors as $x$. | Let $p$ be a prime number such that $p$ divides $x^n$.
This is possible because $x > 1$, so $x^n > 1$, hence $x^n$ has prime divisors due to Fundamental Theorem of Arithmetic.
To prove the statement, we need to show $p$ divides $x$.
We will prove this statement by the Principle of Mathematical Induction on $n$.
=== Bas... | Let $x$ be a [[Definition:Natural Number|natural number]] such that $x > 1$.
Let $n \ge 1$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
The [[Definition:Integer Power|$n$th power]] of $x$ has the same [[Definition:Prime Factor|prime factors]] as $x$. | Let $p$ be a [[Definition:Prime Number|prime number]] such that $p$ [[Definition:Divisor of Integer|divides]] $x^n$.
This is possible because $x > 1$, so $x^n > 1$, hence $x^n$ has [[Definition:Prime Divisor|prime divisors]] due to [[Fundamental Theorem of Arithmetic]].
To prove the statement, we need to show $p$ [[D... | Natural Number has Same Prime Factors as Integer Power/Proof 1 | https://proofwiki.org/wiki/Natural_Number_has_Same_Prime_Factors_as_Integer_Power | https://proofwiki.org/wiki/Natural_Number_has_Same_Prime_Factors_as_Integer_Power/Proof_1 | [
"Natural Number has Same Prime Factors as Integer Power",
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Strictly Positive/Integer",
"Definition:Power (Algebra)/Integer",
"Definition:Prime Factor"
] | [
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Factor",
"Fundamental Theorem of Arithmetic",
"Definition:Divisor (Algebra)/Integer",
"Principle of Mathematical Induction",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definitio... |
proofwiki-7584 | Alternative Definition of Ordinal in Well-Founded Theory | A set $S$ is an ordinal {{iff}} $S$ is transitive and $\forall x, y \in S: \paren {x \in y \lor x = y \lor y \in x}$. | === Forward Implication ===
Let $S$ be an ordinal.
By Alternative Definition of Ordinal, $S$ is transitive and strictly well-ordered by the epsilon relation.
By Strict Well-Ordering is Strict Total Ordering, $S$ is strictly totally ordered by $\in$.
Thus:
:$\forall x, y \in S: \paren {x \in y \lor x = y \lor y \in x}$
... | A [[Definition:Set|set]] $S$ is an [[Definition:Ordinal|ordinal]] {{iff}} $S$ is [[Definition:Transitive Set|transitive]] and $\forall x, y \in S: \paren {x \in y \lor x = y \lor y \in x}$. | === Forward Implication ===
Let $S$ be an [[Definition:Ordinal|ordinal]].
By [[Alternative Definition of Ordinal]], $S$ is [[Definition:Transitive Set|transitive]] and [[Definition:Strict Well-Ordering|strictly well-ordered]] by the [[Definition:Epsilon Relation|epsilon relation]].
By [[Strict Well-Ordering is Stric... | Alternative Definition of Ordinal in Well-Founded Theory | https://proofwiki.org/wiki/Alternative_Definition_of_Ordinal_in_Well-Founded_Theory | https://proofwiki.org/wiki/Alternative_Definition_of_Ordinal_in_Well-Founded_Theory | [
"Ordinals"
] | [
"Definition:Set",
"Definition:Ordinal",
"Definition:Transitive Class"
] | [
"Definition:Ordinal",
"Alternative Definition of Ordinal",
"Definition:Transitive Class",
"Definition:Strict Well-Ordering",
"Definition:Epsilon Relation",
"Strict Well-Ordering is Strict Total Ordering",
"Definition:Strict Total Ordering",
"Definition:Transitive Class",
"Definition:Strict Well-Orde... |
proofwiki-7585 | Strictly Well-Founded Relation is Antireflexive/Corollary | Let $\struct {S, \preceq}$ be an ordered set.
Suppose that $S$ is non-empty.
Then $\preceq$ is not a strictly well-founded relation. | Since $S$ is non-empty, it has an element $x$.
By the definition of ordering, $\preceq$ is a reflexive relation.
Thus $x \preceq x$.
By Strictly Well-Founded Relation is Antireflexive, $\preceq$ is not a strictly well-founded relation.
{{qed}}
Category:Well-Founded Relations
jonolnxmr636hgrl1f6j1fcs4elutvy | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Suppose that $S$ is [[Definition:Non-Empty Set|non-empty]].
Then $\preceq$ is not a [[Definition:Strictly Well-Founded Relation|strictly well-founded relation]]. | Since $S$ is [[Definition:Non-Empty Set|non-empty]], it has an [[Definition:Element|element]] $x$.
By the definition of [[Definition:Ordering|ordering]], $\preceq$ is a [[Definition:Reflexive Relation|reflexive relation]].
Thus $x \preceq x$.
By [[Strictly Well-Founded Relation is Antireflexive]], $\preceq$ is not a... | Strictly Well-Founded Relation is Antireflexive/Corollary | https://proofwiki.org/wiki/Strictly_Well-Founded_Relation_is_Antireflexive/Corollary | https://proofwiki.org/wiki/Strictly_Well-Founded_Relation_is_Antireflexive/Corollary | [
"Well-Founded Relations"
] | [
"Definition:Ordered Set",
"Definition:Non-Empty Set",
"Definition:Strictly Well-Founded Relation"
] | [
"Definition:Non-Empty Set",
"Definition:Element",
"Definition:Ordering",
"Definition:Reflexive Relation",
"Strictly Well-Founded Relation is Antireflexive",
"Definition:Strictly Well-Founded Relation",
"Category:Well-Founded Relations"
] |
proofwiki-7586 | Reflexive Reduction of Well-Founded Relation is Strictly Well-Founded Relation | Let $\struct {S, \RR}$ be a relational structure.
Let $\RR$ be a well-founded relation on $S$.
Let $\RR^{\ne}$ be the reflexive reduction of $\RR$.
Then $\RR^{\ne}$ is a strictly well-founded relation. | Let $T$ be a non-empty subset of $S$.
Since $\RR$ is a well-founded relation, $T$ has a minimal element with respect to the relation $\RR$.
That is, there is an element $m \in T$ such that $\forall x \in T: \paren {\tuple {x, m} \notin \RR} \lor \paren {x = m}$.
Let $x \in T$.
Then $\tuple {x, m} \notin \RR$ or $x = m$... | Let $\struct {S, \RR}$ be a [[Definition:Relational Structure|relational structure]].
Let $\RR$ be a [[Definition:Well-Founded Relation|well-founded relation]] on $S$.
Let $\RR^{\ne}$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\RR$.
Then $\RR^{\ne}$ is a [[Definition:Strictly Well-Founded Rel... | Let $T$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
Since $\RR$ is a [[Definition:Well-Founded Relation|well-founded relation]], $T$ has a [[Definition:Minimal Element|minimal element]] with respect to the [[Definition:Relation|relation]] $\RR$.
That is, there is an [[Definition:E... | Reflexive Reduction of Well-Founded Relation is Strictly Well-Founded Relation | https://proofwiki.org/wiki/Reflexive_Reduction_of_Well-Founded_Relation_is_Strictly_Well-Founded_Relation | https://proofwiki.org/wiki/Reflexive_Reduction_of_Well-Founded_Relation_is_Strictly_Well-Founded_Relation | [
"Reflexive Reductions",
"Well-Founded Relations"
] | [
"Definition:Relational Structure",
"Definition:Well-Founded Relation",
"Definition:Reflexive Reduction",
"Definition:Strictly Well-Founded Relation"
] | [
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Well-Founded Relation",
"Definition:Minimal/Element",
"Definition:Relation",
"Definition:Element",
"Definition:Reflexive Reduction",
"Definition:Subset",
"Reflexive Reduction is Antireflexive",
"Definition:Strictly Minimal Element",
"D... |
proofwiki-7587 | Epsilon Relation is Proper | Let $\mathbb U$ be the universal class.
Let $\Epsilon$ be the epsilon relation.
Then $\struct {\mathbb U, \Epsilon}$ is a proper relational structure. | {{NotZFC}}
Let $x \in \mathbb U$.
Then by the Axiom of Extension:
:$x = \map {\Epsilon^{-1} } x$
where $\map {\Epsilon^{-1} } x$ denotes the preimage of $x$ under $\Epsilon$.
Since $x$ is a set, $\map {\prec^{-1} } x = x$ is a set.
As this holds for all $x \in \mathbb U$, $\struct {\mathbb U, \Epsilon}$ is a proper rel... | Let $\mathbb U$ be the [[Definition:Universal Class|universal class]].
Let $\Epsilon$ be the [[Definition:Epsilon Relation|epsilon relation]].
Then $\struct {\mathbb U, \Epsilon}$ is a [[Definition:Proper Relational Structure|proper relational structure]]. | {{NotZFC}}
Let $x \in \mathbb U$.
Then by the [[Axiom:Axiom of Extension (Classes)|Axiom of Extension]]:
:$x = \map {\Epsilon^{-1} } x$
where $\map {\Epsilon^{-1} } x$ denotes the [[Definition:Preimage of Element under Relation|preimage]] of $x$ under $\Epsilon$.
Since $x$ is a [[Definition:Set|set]], $\map {\prec^{... | Epsilon Relation is Proper | https://proofwiki.org/wiki/Epsilon_Relation_is_Proper | https://proofwiki.org/wiki/Epsilon_Relation_is_Proper | [
"Class Theory"
] | [
"Definition:Universal Class",
"Definition:Epsilon Relation",
"Definition:Proper Relational Structure"
] | [
"Axiom:Axiom of Extension/Class Theory",
"Definition:Preimage/Relation/Element",
"Definition:Set",
"Definition:Set",
"Definition:Proper Relational Structure",
"Category:Class Theory"
] |
proofwiki-7588 | Relationship between Transitive Closure Definitions | Let $x$ be a set.
Let $a$ be the smallest set such that $x \in a$ and $a$ is transitive.
Let $b$ be the smallest set such that $x \subseteq b$ and $b$ is transitive.
Then $a = b \cup \set x$. | We have that:
:$x \in a$
and $a$ is transitive.
So:
:$x \subseteq a$
Thus by the definition of $b$ and of smallest set:
:$b \subseteq a$
Since we also have $x \in a$:
:$b \cup \set x \subseteq a$
$x \in \set x$, so:
:$x \in b \cup \set x$
$b \cup \set x$ is transitive:
If $p \in b$ then:
:$p \subseteq b \subseteq b \cu... | Let $x$ be a [[Definition:Set|set]].
Let $a$ be the [[Definition:Smallest Set by Set Inclusion|smallest set]] such that $x \in a$ and $a$ is [[Definition:Transitive Set|transitive]].
Let $b$ be the [[Definition:Smallest Set by Set Inclusion|smallest set]] such that $x \subseteq b$ and $b$ is [[Definition:Transitive S... | We have that:
:$x \in a$
and $a$ is [[Definition:Transitive Set|transitive]].
So:
:$x \subseteq a$
Thus by the definition of $b$ and of [[Definition:Smallest Set by Set Inclusion|smallest set]]:
:$b \subseteq a$
Since we also have $x \in a$:
:$b \cup \set x \subseteq a$
$x \in \set x$, so:
:$x \in b \cup \set x$
... | Relationship between Transitive Closure Definitions | https://proofwiki.org/wiki/Relationship_between_Transitive_Closure_Definitions | https://proofwiki.org/wiki/Relationship_between_Transitive_Closure_Definitions | [
"Set Theory"
] | [
"Definition:Set",
"Definition:Smallest Set by Set Inclusion",
"Definition:Transitive Class",
"Definition:Smallest Set by Set Inclusion",
"Definition:Transitive Class"
] | [
"Definition:Transitive Class",
"Definition:Smallest Set by Set Inclusion",
"Definition:Transitive Class",
"Definition:Set Equality/Definition 2",
"Category:Set Theory"
] |
proofwiki-7589 | Ordinal is not Element of Itself | Let $x$ be an ordinal.
Then:
:$x \notin x$ | By Successor Set of Ordinal is Ordinal, the successor of $x$ is an ordinal.
That is, $x^+ = x \cup \set x$ is an ordinal.
By Set is Element of Successor, $x \in x^+$.
Because $x^+$ is an ordinal, it is strictly well-ordered by the epsilon restriction $\Epsilon {\restriction_{x^+} }$.
Because a strict ordering is antire... | Let $x$ be an [[Definition:Ordinal|ordinal]].
Then:
:$x \notin x$ | By [[Successor Set of Ordinal is Ordinal]], the [[Definition:Successor Set|successor]] of $x$ is an [[Definition:Ordinal|ordinal]].
That is, $x^+ = x \cup \set x$ is an [[Definition:Ordinal|ordinal]].
By [[Set is Element of Successor]], $x \in x^+$.
Because $x^+$ is an [[Definition:Ordinal|ordinal]], it is [[Definit... | Ordinal is not Element of Itself/Proof 1 | https://proofwiki.org/wiki/Ordinal_is_not_Element_of_Itself | https://proofwiki.org/wiki/Ordinal_is_not_Element_of_Itself/Proof_1 | [
"Ordinals",
"Ordinal is not Element of Itself"
] | [
"Definition:Ordinal"
] | [
"Successor Set of Ordinal is Ordinal",
"Definition:Successor Mapping/Successor Set",
"Definition:Ordinal",
"Definition:Ordinal",
"Set is Element of Successor",
"Definition:Ordinal",
"Definition:Strict Well-Ordering",
"Definition:Epsilon Relation/Restriction",
"Definition:Antireflexive Relation"
] |
proofwiki-7590 | Ordinal is not Element of Itself | Let $x$ be an ordinal.
Then:
:$x \notin x$ | This result follows immediately from Set is Not Element of Itself.
{{qed}} | Let $x$ be an [[Definition:Ordinal|ordinal]].
Then:
:$x \notin x$ | This result follows immediately from [[Set is Not Element of Itself]].
{{qed}} | Ordinal is not Element of Itself/Proof 2 | https://proofwiki.org/wiki/Ordinal_is_not_Element_of_Itself | https://proofwiki.org/wiki/Ordinal_is_not_Element_of_Itself/Proof_2 | [
"Ordinals",
"Ordinal is not Element of Itself"
] | [
"Definition:Ordinal"
] | [
"Set is Not Element of Itself"
] |
proofwiki-7591 | Ordinal is not Element of Itself | Let $x$ be an ordinal.
Then:
:$x \notin x$ | Let $\On$ denote the class of all ordinals.
From Class of All Ordinals is Minimally Superinductive over Successor Mapping, $\On$ is superinductive.
Hence we can use the Principle of Superinduction.
By Zero is Smallest Ordinal, $0$ is the smallest element of $\On$.
We identify the natural number $0$ via the von Neumann ... | Let $x$ be an [[Definition:Ordinal|ordinal]].
Then:
:$x \notin x$ | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
From [[Class of All Ordinals is Minimally Superinductive over Successor Mapping]], $\On$ is [[Definition:Superinductive Class|superinductive]].
Hence we can use the [[Principle of Superinduction]].
By [[Zero is Smallest Ordinal]], $0$ ... | Ordinal is not Element of Itself/Proof 3 | https://proofwiki.org/wiki/Ordinal_is_not_Element_of_Itself | https://proofwiki.org/wiki/Ordinal_is_not_Element_of_Itself/Proof_3 | [
"Ordinals",
"Ordinal is not Element of Itself"
] | [
"Definition:Ordinal"
] | [
"Definition:Class of All Ordinals",
"Class of All Ordinals is Minimally Superinductive over Successor Mapping",
"Definition:Superinductive Class",
"Principle of Superinduction",
"Zero is Smallest Ordinal",
"Definition:Smallest Element/Class Theory",
"Definition:Zero (Number)",
"Definition:Natural Numb... |
proofwiki-7592 | Ordinal is not Element of Itself | Let $x$ be an ordinal.
Then:
:$x \notin x$ | {{AimForCont}} $x \in x$.
From Equality is Reflexive, $x = x$.
From Ordinal Membership is Trichotomy, either $x = x$ or $x \in x$.
It follows from this contradiction that $x \notin x$.
{{qed}} | Let $x$ be an [[Definition:Ordinal|ordinal]].
Then:
:$x \notin x$ | {{AimForCont}} $x \in x$.
From [[Equality is Reflexive]], $x = x$.
From [[Ordinal Membership is Trichotomy]], either $x = x$ or $x \in x$.
It follows from this contradiction that $x \notin x$.
{{qed}} | Ordinal is not Element of Itself/Proof 4 | https://proofwiki.org/wiki/Ordinal_is_not_Element_of_Itself | https://proofwiki.org/wiki/Ordinal_is_not_Element_of_Itself/Proof_4 | [
"Ordinals",
"Ordinal is not Element of Itself"
] | [
"Definition:Ordinal"
] | [
"Equality is Reflexive",
"Ordinal Membership is Trichotomy"
] |
proofwiki-7593 | Set is Element of Successor | Let $x$ be a set.
Let $x^+$ be the successor of $x$.
Then $x \in x^+$. | By the definition of successor set:
: $x^+ = x \cup \{x\}$.
By the definition of singleton, $x \in \{x\}$.
Thus by the definition of union, $x \in x^+$.
{{qed}}
Category:Ordinals
7sif0tawwnckd2ghx1u58z7u5b04ugc | Let $x$ be a [[Definition:set|set]].
Let $x^+$ be the [[Definition:Successor Set|successor]] of $x$.
Then $x \in x^+$. | By the definition of [[Definition:Successor Set|successor set]]:
: $x^+ = x \cup \{x\}$.
By the definition of [[Definition:singleton|singleton]], $x \in \{x\}$.
Thus by the definition of [[Definition:Set Union|union]], $x \in x^+$.
{{qed}}
[[Category:Ordinals]]
7sif0tawwnckd2ghx1u58z7u5b04ugc | Set is Element of Successor | https://proofwiki.org/wiki/Set_is_Element_of_Successor | https://proofwiki.org/wiki/Set_is_Element_of_Successor | [
"Ordinals"
] | [
"Definition:set",
"Definition:Successor Mapping/Successor Set"
] | [
"Definition:Successor Mapping/Successor Set",
"Definition:singleton",
"Definition:Set Union",
"Category:Ordinals"
] |
proofwiki-7594 | Element of Ordinal is Ordinal | Let $n$ be an ordinal.
Let $m \in n$.
Then $m$ is also an ordinal.
That is, the class of all ordinals $\On$ is a transitive class. | Let $\On$ denote the class of all ordinals.
From Class of All Ordinals is Minimally Superinductive over Successor Mapping, $\On$ is superinductive.
By Zero is Smallest Ordinal, $0$ is the smallest element of $\On$.
We identify the natural number $0$ via the von Neumann construction of the natural numbers as:
:$0 := \O$... | Let $n$ be an [[Definition:Ordinal|ordinal]].
Let $m \in n$.
Then $m$ is also an [[Definition:Ordinal|ordinal]].
That is, the [[Definition:Class of All Ordinals|class of all ordinals]] $\On$ is a [[Definition:Transitive Class|transitive class]]. | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
From [[Class of All Ordinals is Minimally Superinductive over Successor Mapping]], $\On$ is [[Definition:Superinductive Class|superinductive]].
By [[Zero is Smallest Ordinal]], $0$ is the [[Definition:Smallest Element (Class Theory)|sma... | Element of Ordinal is Ordinal/Proof 1 | https://proofwiki.org/wiki/Element_of_Ordinal_is_Ordinal | https://proofwiki.org/wiki/Element_of_Ordinal_is_Ordinal/Proof_1 | [
"Element of Ordinal is Ordinal",
"Ordinals",
"Transitive Classes"
] | [
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Class of All Ordinals",
"Definition:Transitive Class"
] | [
"Definition:Class of All Ordinals",
"Class of All Ordinals is Minimally Superinductive over Successor Mapping",
"Definition:Superinductive Class",
"Zero is Smallest Ordinal",
"Definition:Smallest Element/Class Theory",
"Definition:Zero (Number)",
"Definition:Natural Numbers/Von Neumann Construction",
... |
proofwiki-7595 | Element of Ordinal is Ordinal | Let $n$ be an ordinal.
Let $m \in n$.
Then $m$ is also an ordinal.
That is, the class of all ordinals $\On$ is a transitive class. | By the definition of ordinal, $n$ is transitive.
Thus $m \subseteq n$.
By Restriction of Strict Well-Ordering is Strict Well-Ordering, it follows that $m$ is strictly well-ordered by the epsilon restriction $\Epsilon {\restriction_m}$.
It is now to be shown that $m$ is transitive.
If $m = \O$ then the result follows by... | Let $n$ be an [[Definition:Ordinal|ordinal]].
Let $m \in n$.
Then $m$ is also an [[Definition:Ordinal|ordinal]].
That is, the [[Definition:Class of All Ordinals|class of all ordinals]] $\On$ is a [[Definition:Transitive Class|transitive class]]. | By the definition of [[Definition:Ordinal|ordinal]], $n$ is [[Definition:Transitive Class|transitive]].
Thus $m \subseteq n$.
By [[Restriction of Strict Well-Ordering is Strict Well-Ordering]], it follows that $m$ is [[Definition:Strict Well-Ordering|strictly well-ordered]] by the [[Definition:Epsilon Restriction|eps... | Element of Ordinal is Ordinal/Proof 2 | https://proofwiki.org/wiki/Element_of_Ordinal_is_Ordinal | https://proofwiki.org/wiki/Element_of_Ordinal_is_Ordinal/Proof_2 | [
"Element of Ordinal is Ordinal",
"Ordinals",
"Transitive Classes"
] | [
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Class of All Ordinals",
"Definition:Transitive Class"
] | [
"Definition:Ordinal",
"Definition:Transitive Class",
"Restriction of Strict Well-Ordering is Strict Well-Ordering",
"Definition:Strict Well-Ordering",
"Definition:Epsilon Relation/Restriction",
"Definition:Transitive Class",
"Empty Class is Transitive",
"Empty Set is Subset of All Sets",
"Definition... |
proofwiki-7596 | Modulus of Exponential of Imaginary Number is One | Let $\cmod z$ denote the modulus of a complex number $z$.
Let $e^z$ be the complex exponential of $z$.
Let $x$ be wholly real.
Then:
:$\cmod {e^{i x} } = 1$ | {{begin-eqn}}
{{eqn | l = e^{i x}
| r = \cos x + i \sin x
| c = Euler's Formula
}}
{{eqn | ll= \leadsto
| l = \cmod {e^{i x} }
| r = \cmod {\cos x + i \sin x}
}}
{{eqn | r = \sqrt {\paren {\map \Re {\cos x + i \sin x} }^2 + \paren {\map \Im {\cos x + i \sin x} }^2}
| c = {{Defof|Complex Mo... | Let $\cmod z$ denote the [[Definition:Complex Modulus|modulus]] of a [[Definition:Complex Number|complex number]] $z$.
Let $e^z$ be the [[Definition:Complex Exponential Function|complex exponential]] of $z$.
Let $x$ be [[Definition:Wholly Real|wholly real]].
Then:
:$\cmod {e^{i x} } = 1$ | {{begin-eqn}}
{{eqn | l = e^{i x}
| r = \cos x + i \sin x
| c = [[Euler's Formula]]
}}
{{eqn | ll= \leadsto
| l = \cmod {e^{i x} }
| r = \cmod {\cos x + i \sin x}
}}
{{eqn | r = \sqrt {\paren {\map \Re {\cos x + i \sin x} }^2 + \paren {\map \Im {\cos x + i \sin x} }^2}
| c = {{Defof|Comple... | Modulus of Exponential of Imaginary Number is One | https://proofwiki.org/wiki/Modulus_of_Exponential_of_Imaginary_Number_is_One | https://proofwiki.org/wiki/Modulus_of_Exponential_of_Imaginary_Number_is_One | [
"Complex Modulus",
"Modulus of Exponential of Imaginary Number is One"
] | [
"Definition:Complex Modulus",
"Definition:Complex Number",
"Definition:Exponential Function/Complex",
"Definition:Complex Number/Wholly Real"
] | [
"Euler's Formula",
"Definition:Complex Number/Wholly Real",
"Sum of Squares of Sine and Cosine",
"Category:Complex Modulus",
"Category:Modulus of Exponential of Imaginary Number is One"
] |
proofwiki-7597 | Absolute Value of Power | Let $x$, $y$ be real numbers.
Let $x^y$, $x$ to the power of $y$, be real.
Then:
:$\size {x^y} = \size x^y$ | If $x = 0$, the theorem clearly holds, by the definition of powers of zero.
Suppose $x \ne 0$.
We use the interpretation of real numbers as wholly real complex numbers.
Likewise we interpret the absolute value of $x$ as the modulus of $x$.
Then $x$ can be expressed in polar form:
:$x = r e^{i\theta}$
where $r = \size x... | Let $x$, $y$ be [[Definition:Real Number|real numbers]].
Let $x^y$, [[Definition:Power (Algebra)|$x$ to the power of $y$]], be real.
Then:
:$\size {x^y} = \size x^y$ | If $x = 0$, the theorem [[Definition:Clearly|clearly]] holds, by the definition of [[Definition:Power of Zero|powers of zero]].
Suppose $x \ne 0$.
We use the interpretation of [[Definition:Real Number|real numbers]] as [[Definition:Wholly Real|wholly real complex numbers]].
Likewise we interpret the [[Definition:Ab... | Absolute Value of Power | https://proofwiki.org/wiki/Absolute_Value_of_Power | https://proofwiki.org/wiki/Absolute_Value_of_Power | [
"Absolute Value Function"
] | [
"Definition:Real Number",
"Definition:Power (Algebra)"
] | [
"Definition:Clearly",
"Definition:Power (Algebra)/Power of Zero",
"Definition:Real Number",
"Definition:Complex Number/Wholly Real",
"Definition:Absolute Value",
"Definition:Complex Modulus",
"Definition:Complex Number/Polar Form",
"Definition:Argument of Complex Number",
"Complex Modulus of Product... |
proofwiki-7598 | Count of Rows of Truth Table | Let $P$ be a WFF of propositional logic.
Suppose $\PP$ is of finite size such that it contains $n$ different letters.
Then a truth table constructed to express $P$ will contain $2^n$ rows. | In a truth table, one row is needed for each boolean interpretation of $P$.
Let $S$ be the set of different letters used in $P$.
The result then follows from applying Number of Boolean Interpretations for Finite Set of Variables to $S$.
{{qed}} | Let $P$ be a [[Definition:WFF of Propositional Logic|WFF of propositional logic]].
Suppose $\PP$ is of [[Definition:Finite Set|finite size]] such that it contains $n$ different [[Definition:Letter of Alphabet|letters]].
Then a [[Definition:Truth Table|truth table]] constructed to express $P$ will contain $2^n$ [[Def... | In a [[Definition:Truth Table|truth table]], one [[Definition:Row of Truth Table|row]] is needed for each [[Definition:Boolean Interpretation|boolean interpretation]] of $P$.
Let $S$ be the [[Definition:Set|set]] of different [[Definition:Letter of Alphabet|letters]] used in $P$.
The result then follows from applying... | Count of Rows of Truth Table | https://proofwiki.org/wiki/Count_of_Rows_of_Truth_Table | https://proofwiki.org/wiki/Count_of_Rows_of_Truth_Table | [
"Truth Tables"
] | [
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Finite Set",
"Definition:Letter of Alphabet",
"Definition:Truth Table",
"Definition:Truth Table/Row"
] | [
"Definition:Truth Table",
"Definition:Truth Table/Row",
"Definition:Boolean Interpretation",
"Definition:Set",
"Definition:Letter of Alphabet",
"Number of Boolean Interpretations for Finite Set of Variables"
] |
proofwiki-7599 | Equivalence of Definitions of Transitive Closure of Set | Let $x$ and $y$ be sets.
{{TFAE|def = Transitive Closure of Set}} | Let $x^t$ be the transitive closure of $x$ by Definition 2.
Let the mapping $G$ be defined as on that definition page. | Let $x$ and $y$ be [[Definition:Set|sets]].
{{TFAE|def = Transitive Closure of Set}} | Let $x^t$ be the [[Definition:Transitive Closure of Set/Definition 2|transitive closure of $x$ by Definition 2]].
Let the [[Definition:mapping|mapping]] $G$ be defined as on that definition page. | Equivalence of Definitions of Transitive Closure of Set | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Transitive_Closure_of_Set | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Transitive_Closure_of_Set | [
"Set Theory"
] | [
"Definition:Set"
] | [
"Definition:Transitive Closure of Set/Definition 2",
"Definition:mapping"
] |
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