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-2300 | König's Tree Lemma | Let $T$ be a rooted tree with an infinite number of nodes, each with a finite number of children.
Then $T$ has a branch of infinite length. | Suppose $T$ is a labeled tree with root $r$.
Let $L_0 = \set r, L_1, L_2, \ldots$ be the levels of $T$.
We construct our path by starting with $r$.
Next we partition all vertices in levels higher then $L_0$ (that is, all vertices in levels $L_i$ where $i > 0$) into as many finite parts as is the cardinality of $L_1$.
I... | Let $T$ be a [[Definition:Rooted Tree|rooted tree]] with an [[Definition:Infinite Set|infinite number]] of [[Definition:Node of Tree|nodes]], each with a [[Definition:Finite Set|finite number]] of [[Definition:Child Node|children]].
Then $T$ has a [[Definition:Branch (Graph Theory)|branch]] of [[Definition:Infinite Br... | Suppose $T$ is a labeled tree with root $r$.
Let $L_0 = \set r, L_1, L_2, \ldots$ be the levels of $T$.
We construct our path by starting with $r$.
Next we partition all vertices in levels higher then $L_0$ (that is, all vertices in levels $L_i$ where $i > 0$) into as many finite parts as is the cardinality of $L_1$... | König's Tree Lemma/Proof 3 | https://proofwiki.org/wiki/König's_Tree_Lemma | https://proofwiki.org/wiki/König's_Tree_Lemma/Proof_3 | [
"Tree Theory",
"König's Tree Lemma"
] | [
"Definition:Rooted Tree",
"Definition:Infinite Set",
"Definition:Tree (Graph Theory)/Node",
"Definition:Finite Set",
"Definition:Rooted Tree/Child Node",
"Definition:Rooted Tree/Branch",
"Definition:Rooted Tree/Branch/Infinite",
"Definition:Rooted Tree/Branch/Length"
] | [
"Infinite Ramsey's Theorem"
] |
proofwiki-2301 | Compactness Theorem for Boolean Interpretations | Let $\mathbf H$ be a countable set of WFFs of propositional logic.
Suppose $\mathbf H$ is finitely satisfiable for boolean interpretations.
That is, suppose that every finite subset $\mathbf H' \subseteq \mathbf H$ is satisfiable for boolean interpretations.
Then $\mathbf H$ has a model. | Suppose $\mathbf H$ does '''not''' have a model.
By the Main Lemma of Propositional Tableaux, $\mathbf H$ has a tableau confutation $T$.
By Tableau Confutation contains Finite Tableau Confutation, $T$ may be assumed to be finite.
Hence the set $\mathbf H'$ of all WFFs in $\mathbf H$ used somewhere in $T$ is finite.
Now... | Let $\mathbf H$ be a [[Definition:Countable Set|countable set]] of [[Definition:WFF of Propositional Logic|WFFs of propositional logic]].
Suppose $\mathbf H$ is [[Definition:Finitely Satisfiable|finitely satisfiable]] for [[Definition:Boolean Interpretation|boolean interpretations]].
That is, suppose that every [[Def... | Suppose $\mathbf H$ does '''not''' have a [[Definition:Model (Boolean Interpretations)|model]].
By the [[Main Lemma of Propositional Tableaux]], $\mathbf H$ has a [[Definition:Tableau Confutation|tableau confutation]] $T$.
By [[Tableau Confutation contains Finite Tableau Confutation]], $T$ may be assumed to be [[Defi... | Compactness Theorem for Boolean Interpretations/Proof 1 | https://proofwiki.org/wiki/Compactness_Theorem_for_Boolean_Interpretations | https://proofwiki.org/wiki/Compactness_Theorem_for_Boolean_Interpretations/Proof_1 | [
"Compactness Theorem for Boolean Interpretations",
"Propositional Tableaux",
"Named Theorems"
] | [
"Definition:Countable Set",
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Finitely Satisfiable",
"Definition:Boolean Interpretation",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Satisfiable/Boolean Interpretations",
"Definition:Boolean Interpretation",
"D... | [
"Definition:Model (Boolean Interpretations)",
"Main Lemma of Propositional Tableaux",
"Definition:Tableau Confutation",
"Tableau Confutation contains Finite Tableau Confutation",
"Definition:Propositional Tableau/Construction/Finite",
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Def... |
proofwiki-2302 | Compactness Theorem for Boolean Interpretations | Let $\mathbf H$ be a countable set of WFFs of propositional logic.
Suppose $\mathbf H$ is finitely satisfiable for boolean interpretations.
That is, suppose that every finite subset $\mathbf H' \subseteq \mathbf H$ is satisfiable for boolean interpretations.
Then $\mathbf H$ has a model. | If $\mathbf H$ is finite, the result is trivial.
So let $\mathbf H = \set {\mathbf A_n: n \in \N}$ be an enumeration of $\mathbf H$.
Define $\mathbf H_m = \set {\mathbf A_n: n \le m}$.
Let $T_1$ be the propositional tableau consisting of only a root node with hypothesis set $\mathbf H_1$.
For each $m \in \N$, construct... | Let $\mathbf H$ be a [[Definition:Countable Set|countable set]] of [[Definition:WFF of Propositional Logic|WFFs of propositional logic]].
Suppose $\mathbf H$ is [[Definition:Finitely Satisfiable|finitely satisfiable]] for [[Definition:Boolean Interpretation|boolean interpretations]].
That is, suppose that every [[Def... | If $\mathbf H$ is [[Definition:Finite Set|finite]], the result is trivial.
So let $\mathbf H = \set {\mathbf A_n: n \in \N}$ be an [[Definition:Enumeration|enumeration]] of $\mathbf H$.
Define $\mathbf H_m = \set {\mathbf A_n: n \le m}$.
Let $T_1$ be the [[Definition:Propositional Tableau|propositional tableau]] co... | Compactness Theorem for Boolean Interpretations/Proof 2 | https://proofwiki.org/wiki/Compactness_Theorem_for_Boolean_Interpretations | https://proofwiki.org/wiki/Compactness_Theorem_for_Boolean_Interpretations/Proof_2 | [
"Compactness Theorem for Boolean Interpretations",
"Propositional Tableaux",
"Named Theorems"
] | [
"Definition:Countable Set",
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Finitely Satisfiable",
"Definition:Boolean Interpretation",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Satisfiable/Boolean Interpretations",
"Definition:Boolean Interpretation",
"D... | [
"Definition:Finite Set",
"Definition:Enumeration",
"Definition:Propositional Tableau",
"Definition:Rooted Tree/Root Node",
"Definition:Labeled Tree for Propositional Logic/Hypothesis Set",
"Tableau Extension Lemma",
"Definition:Sequence",
"Definition:Finished Propositional Tableau",
"Definition:Exha... |
proofwiki-2303 | Prefix of WFF of Predicate Logic is not WFF | Let $\mathbf A$ be a WFF of predicate logic.
Let $\mathbf S$ be a prefix of $\mathbf A$.
Then $\mathbf S$ is not a WFF of predicate logic. | Let $\map l {\mathbf Q}$ denote the length of a string $\mathbf Q$.
By definition, $\mathbf S$ is a prefix of $\mathbf A$ iff $\mathbf A = \mathbf{ST}$ for some non-null string $\mathbf T$.
Thus we note that $\map l {\mathbf S} < \map l {\mathbf A}$.
The proof proceeds by induction on $\map l {\mathbf A}$. | Let $\mathbf A$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
Let $\mathbf S$ be a [[Definition:Prefix|prefix]] of $\mathbf A$.
Then $\mathbf S$ is not a [[Definition:WFF of Predicate Logic|WFF of predicate logic]]. | Let $\map l {\mathbf Q}$ denote the [[Definition:Length of String|length]] of a [[Definition:String|string]] $\mathbf Q$.
By definition, $\mathbf S$ is a [[Definition:Prefix|prefix]] of $\mathbf A$ iff $\mathbf A = \mathbf{ST}$ for some [[Definition:Null String|non-null string]] $\mathbf T$.
Thus we note that $\map l... | Prefix of WFF of Predicate Logic is not WFF | https://proofwiki.org/wiki/Prefix_of_WFF_of_Predicate_Logic_is_not_WFF | https://proofwiki.org/wiki/Prefix_of_WFF_of_Predicate_Logic_is_not_WFF | [
"Predicate Logic"
] | [
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Prefix",
"Definition:Language of Predicate Logic/Formal Grammar"
] | [
"Definition:Length of String",
"Definition:String",
"Definition:Prefix",
"Definition:Null String",
"Second Principle of Mathematical Induction",
"Definition:Prefix",
"Definition:Null String",
"Definition:Prefix",
"Definition:Null String",
"Definition:Prefix",
"Definition:Prefix",
"Definition:P... |
proofwiki-2304 | Language of Predicate Logic has Unique Parsability | Each WFF of predicate logic which starts with a left bracket or a negation sign has exactly one main connective.
{{Explain|probably, in the correct formulation of 'main connective', the quantifiers ought to be connectives as well}} | We have the following cases:
# $\mathbf A = \neg \mathbf B$, where $\mathbf B$ is a WFF of length $k$.
# $\mathbf A = \paren {\mathbf B \circ \mathbf C}$ where $\circ$ is one of the binary connectives.
# $\mathbf A = \map p {t_1, t_2, \ldots, t_n}$, where $t_1, t_2, \ldots, t_n$ are terms, and $p \in \PP_n$.
# $\mathbf... | Each [[Definition:WFF of Predicate Logic|WFF of predicate logic]] which starts with a [[Definition:Parenthesis|left bracket]] or a [[Definition:Logical Not|negation sign]] has exactly one [[Definition:Main Connective (Propositional Logic)|main connective]].
{{Explain|probably, in the correct formulation of 'main connec... | We have the following cases:
# $\mathbf A = \neg \mathbf B$, where $\mathbf B$ is a WFF of length $k$.
# $\mathbf A = \paren {\mathbf B \circ \mathbf C}$ where $\circ$ is one of the binary connectives.
# $\mathbf A = \map p {t_1, t_2, \ldots, t_n}$, where $t_1, t_2, \ldots, t_n$ are [[Definition:Term (Predicate Logic)|... | Language of Predicate Logic has Unique Parsability | https://proofwiki.org/wiki/Language_of_Predicate_Logic_has_Unique_Parsability | https://proofwiki.org/wiki/Language_of_Predicate_Logic_has_Unique_Parsability | [
"Language of Predicate Logic",
"Named Theorems"
] | [
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Parenthesis",
"Definition:Logical Not",
"Definition:Main Connective/Propositional Logic"
] | [
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Quantifier",
"Definition:Variable",
"Language of Propositional Logic has Unique Parsability",
"Definition:Parenthesis",
"Definition:Logical Not"
] |
proofwiki-2305 | Quantifier has Unique Scope | Let $\mathbf A$ be a WFF of predicate logic.
Let $Q$ be an occurrence of a quantifier in $\mathbf A$.
Then there exists a unique well-formed part of $\mathbf A$ which (omitting outermost parentheses) begins with that occurrence $Q$.
This unique well-formed part of $\mathbf A$ is called the scope of the occurrence of $Q... | === Existence ===
First, from the rules of formation of predicate logic, we have that whenever a quantifier is included in a WFF, it appears in the form:
:$\paren {Q x: \mathbf B }$
where $\mathbf B$ is itself a WFF.
Hence it is clear that $\paren {Q x: \mathbf B}$ is a well-formed part of $\mathbf A$ which begins with... | Let $\mathbf A$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
Let $Q$ be an [[Definition:Occurrence (Predicate Logic)|occurrence]] of a [[Definition:Quantifier|quantifier]] in $\mathbf A$.
Then there exists a unique [[Definition:Well-Formed Part|well-formed part]] of $\mathbf A$ which (omitting ... | === Existence ===
First, from the [[Definition:Formal Grammar of Predicate Logic|rules of formation of predicate logic]], we have that whenever a [[Definition:Quantifier|quantifier]] is included in a [[Definition:Well-Formed Formula|WFF]], it appears in the form:
:$\paren {Q x: \mathbf B }$
where $\mathbf B$ is itse... | Quantifier has Unique Scope | https://proofwiki.org/wiki/Quantifier_has_Unique_Scope | https://proofwiki.org/wiki/Quantifier_has_Unique_Scope | [
"Predicate Logic"
] | [
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Occurrence (Predicate Logic)",
"Definition:Quantifier",
"Definition:Well-Formed Part",
"Definition:Parenthesis",
"Definition:Occurrence (Predicate Logic)",
"Definition:Well-Formed Part",
"Definition:Scope (Logic)/Quantifier",
"Defi... | [
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Quantifier",
"Definition:Well-Formed Formula",
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Well-Formed Part",
"Definition:Well-Formed Part",
"Definition:Well-Formed Part"
] |
proofwiki-2306 | Restriction of Operation Distributivity | Let $\struct {S, *, \circ}$ be an algebraic structure.
Let $T \subseteq S$.
If the operation $\circ$ is distributive over $*$ in $\struct {S, *, \circ}$, then it is also distributive over $*$ on a restriction $\struct {T, * \restriction_T, \circ \restriction_T}$. | {{begin-eqn}}
{{eqn | o =
| r = T \subseteq S
| c =
}}
{{eqn | ll= \leadsto
| o =
| r = \forall a, b, c \in T: a, b, c \in S
| c = {{Defof|Subset}}
}}
{{eqn | ll= \leadsto
| o =
| r = a \circ_T \paren {b *_T c}
| c =
}}
{{eqn | r = a \circ \paren {b * c}
| c = ... | Let $\struct {S, *, \circ}$ be an [[Definition:Algebraic Structure with Two Operations|algebraic structure]].
Let $T \subseteq S$.
If the [[Definition:Binary Operation|operation]] $\circ$ is [[Definition:Distributive Operation|distributive]] over $*$ in $\struct {S, *, \circ}$, then it is also [[Definition:Distribut... | {{begin-eqn}}
{{eqn | o =
| r = T \subseteq S
| c =
}}
{{eqn | ll= \leadsto
| o =
| r = \forall a, b, c \in T: a, b, c \in S
| c = {{Defof|Subset}}
}}
{{eqn | ll= \leadsto
| o =
| r = a \circ_T \paren {b *_T c}
| c =
}}
{{eqn | r = a \circ \paren {b * c}
| c = ... | Restriction of Operation Distributivity | https://proofwiki.org/wiki/Restriction_of_Operation_Distributivity | https://proofwiki.org/wiki/Restriction_of_Operation_Distributivity | [
"Abstract Algebra",
"Distributive Operations"
] | [
"Definition:Algebraic Structure/Two Operations",
"Definition:Operation/Binary Operation",
"Definition:Distributive Operation",
"Definition:Distributive Operation",
"Definition:Restriction/Operation"
] | [
"Definition:Distributive Operation",
"Category:Abstract Algebra",
"Category:Distributive Operations"
] |
proofwiki-2307 | Disjunction of Conjunctions | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | We apply the Method of Truth Tables.
As can be seen by inspection, in all cases the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin {array} {|ccc||ccccccccc|} \hline
p & \iff & q & (p & \land & q) & \lor & (\neg & p & \land & \neg & q) \\
\hline
\F & \T & \F & \F & \F & \... | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, in all cases the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin {array} {|ccc||c... | Biconditional as Disjunction of Conjunctions/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Disjunction_of_Conjunctions | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_1/Proof_by_Truth_Table | [
"Conjunction",
"Disjunction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-2308 | Disjunction of Conjunctions | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | {{BeginTableau|\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} } }}
{{Assumption |1|p \iff q}}
{{SequentIntro |2|1|\paren {p \land q} \lor \paren {\neg p \land \neg q}|1|Biconditional as Disjunction of Conjunctions: Formulation 1}}
{{Implication |3| |\paren {... | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | {{BeginTableau|\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} } }}
{{Assumption |1|p \iff q}}
{{SequentIntro |2|1|\paren {p \land q} \lor \paren {\neg p \land \neg q}|1|[[Biconditional as Disjunction of Conjunctions/Formulation 1|Biconditional as Disjunction of Co... | Biconditional as Disjunction of Conjunctions/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Disjunction_of_Conjunctions | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_2/Proof_1 | [
"Conjunction",
"Disjunction"
] | [] | [
"Biconditional as Disjunction of Conjunctions/Formulation 1",
"Biconditional as Disjunction of Conjunctions/Formulation 1"
] |
proofwiki-2309 | Disjunction of Conjunctions | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | We apply the Method of Truth Tables.
As can be seen by inspection, in all cases the truth values under the main connective is true for all boolean interpretations.
$\begin{array}{|ccc|c|ccccccccc|} \hline
(p & \iff & q) & \iff & ((p & \land & q) & \lor & (\neg & p & \land & \neg & q)) \\
\hline
\F & \T & \F & \T & \F &... | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, in all cases the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{arra... | Biconditional as Disjunction of Conjunctions/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Disjunction_of_Conjunctions | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_2/Proof_by_Truth_Table | [
"Conjunction",
"Disjunction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-2310 | Disjunction of Conjunctions | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | {{BeginTableau|\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r}}
{{Premise|1|\paren {p \land q} \lor \paren {r \land s} }}
{{Assumption|2|p \land q}}
{{Simplification|3|2|p|2|1}}
{{Addition|4|2|p \lor r|3|1}}
{{Assumption|5|r \land s}}
{{Simplification|6|5|r|5|1}}
{{Addition|7|5|p \lor r|6|2}}
{{ProofByCases... | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | {{BeginTableau|\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r}}
{{Premise|1|\paren {p \land q} \lor \paren {r \land s} }}
{{Assumption|2|p \land q}}
{{Simplification|3|2|p|2|1}}
{{Addition|4|2|p \lor r|3|1}}
{{Assumption|5|r \land s}}
{{Simplification|6|5|r|5|1}}
{{Addition|7|5|p \lor r|6|2}}
{{ProofByCases... | Disjunction of Conjunctions | https://proofwiki.org/wiki/Disjunction_of_Conjunctions | https://proofwiki.org/wiki/Disjunction_of_Conjunctions | [
"Conjunction",
"Disjunction"
] | [] | [
"Category:Conjunction",
"Category:Disjunction"
] |
proofwiki-2311 | Disjunction of Conjunctions | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | {{begin-eqn}}
{{eqn | l = p \oplus q
| o = \dashv \vdash
| r = \neg \left ({p \iff q}\right)
| c = Exclusive Or is Negation of Biconditional
}}
{{eqn | o = \dashv \vdash
| r = \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)
| c = Non-Equivalence as Disjunction of Conjuncti... | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | {{begin-eqn}}
{{eqn | l = p \oplus q
| o = \dashv \vdash
| r = \neg \left ({p \iff q}\right)
| c = [[Exclusive Or is Negation of Biconditional]]
}}
{{eqn | o = \dashv \vdash
| r = \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)
| c = [[Non-Equivalence as Disjunction of Con... | Exclusive Or as Disjunction of Conjunctions/Proof 1 | https://proofwiki.org/wiki/Disjunction_of_Conjunctions | https://proofwiki.org/wiki/Exclusive_Or_as_Disjunction_of_Conjunctions/Proof_1 | [
"Conjunction",
"Disjunction"
] | [] | [
"Exclusive Or is Negation of Biconditional",
"Non-Equivalence as Disjunction of Conjunctions/Formulation 1"
] |
proofwiki-2312 | Disjunction of Conjunctions | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin {array} {|ccc||ccccccccc|} \hline
p & \oplus & q & (\neg & p & \land & q) & \lor & (p & \land & \neg & q) \\
\hline
\F & \F & \F & \T & \F & \F & \F & \F... | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin {array} {|ccc||ccccccccc|} \h... | Exclusive Or as Disjunction of Conjunctions/Proof by Truth Table | https://proofwiki.org/wiki/Disjunction_of_Conjunctions | https://proofwiki.org/wiki/Exclusive_Or_as_Disjunction_of_Conjunctions/Proof_by_Truth_Table | [
"Conjunction",
"Disjunction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-2313 | Disjunction of Conjunctions | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q} }}
{{Premise|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }|1
|Rule of Material Equivalence}}
{{SequentIntro|3|1|\neg \paren {\paren {\neg p \lor q} \land ... | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q} }}
{{Premise|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }|1
|[[Rule of Material Equivalence]]}}
{{SequentIntro|3|1|\neg \paren {\paren {\neg p \lor q} \l... | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Disjunction_of_Conjunctions | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Forward_Implication/Proof | [
"Conjunction",
"Disjunction"
] | [] | [
"Rule of Material Equivalence",
"Rule of Material Implication"
] |
proofwiki-2314 | Disjunction of Conjunctions | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | === Forward Implication: Proof ===
{{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof}}
=== Reverse Implication: Proof ===
{{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof}} | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | === [[Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof|Forward Implication: Proof]] ===
{{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof}}
=== [[Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof|Reverse I... | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Disjunction_of_Conjunctions | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Proof_1 | [
"Conjunction",
"Disjunction"
] | [] | [
"Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof",
"Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof"
] |
proofwiki-2315 | Disjunction of Conjunctions | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|cccc||ccccccccc|} \hline
\neg & (p & \iff & q) & (\neg & p & \land & q) & \lor & (p & \land & \neg & q) \\
\hline
\F & \F & \T & \F & \T & \F & \... | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|cccc||ccccccccc|} \hl... | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Disjunction_of_Conjunctions | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Proof_by_Truth_Table | [
"Conjunction",
"Disjunction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-2316 | Disjunction of Conjunctions | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | {{BeginTableau|\paren {\neg p \land q} \lor \paren {p \land \neg q} \vdash \neg \paren {p \iff q} }}
{{Premise|1|\paren {\neg p \land q} \lor \paren {p \land \neg q} }}
{{Commutation|2|1|\paren {p \land \neg q} \lor \paren {\neg p \land q}|1|Disjunction}}
{{Commutation|3|1|\paren {p \land \neg q} \lor \paren {q \land \... | :$\paren {p \land q} \lor \paren {r \land s} \vdash p \lor r$ | {{BeginTableau|\paren {\neg p \land q} \lor \paren {p \land \neg q} \vdash \neg \paren {p \iff q} }}
{{Premise|1|\paren {\neg p \land q} \lor \paren {p \land \neg q} }}
{{Commutation|2|1|\paren {p \land \neg q} \lor \paren {\neg p \land q}|1|Disjunction}}
{{Commutation|3|1|\paren {p \land \neg q} \lor \paren {q \land \... | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Disjunction_of_Conjunctions | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Reverse_Implication/Proof | [
"Conjunction",
"Disjunction"
] | [] | [
"Rule of Material Implication",
"Rule of Material Equivalence"
] |
proofwiki-2317 | Union of Intersections | :$\paren {S_1 \cap S_2} \cup \paren {T_1 \cap T_2} \subseteq S_1 \cup T_1$ | {{begin-eqn}}
{{eqn | l = \paren {S_1 \cap S_2} \cup \paren {T_1 \cap T_2}
| r = \paren {\paren {S_1 \cap S_2} \cup T_1} \cap \paren {\paren {S_1 \cap S_2} \cup T_2}
| c = Union Distributes over Intersection
}}
{{eqn | r = \paren {S_1 \cup T_1} \cap \paren {S_2 \cup T_1} \cap \paren {\paren {S_1 \cap S_2} \... | :$\paren {S_1 \cap S_2} \cup \paren {T_1 \cap T_2} \subseteq S_1 \cup T_1$ | {{begin-eqn}}
{{eqn | l = \paren {S_1 \cap S_2} \cup \paren {T_1 \cap T_2}
| r = \paren {\paren {S_1 \cap S_2} \cup T_1} \cap \paren {\paren {S_1 \cap S_2} \cup T_2}
| c = [[Union Distributes over Intersection]]
}}
{{eqn | r = \paren {S_1 \cup T_1} \cap \paren {S_2 \cup T_1} \cap \paren {\paren {S_1 \cap S_... | Union of Intersections | https://proofwiki.org/wiki/Union_of_Intersections | https://proofwiki.org/wiki/Union_of_Intersections | [
"Set Union",
"Set Intersection",
"Union of Intersections"
] | [] | [
"Union Distributes over Intersection",
"Union Distributes over Intersection",
"Intersection is Subset"
] |
proofwiki-2318 | Union of Symmetric Differences | Let $R, S, T$ be sets.
Then:
:$\paren {R \symdif S} \cup \paren {S \symdif T} = \paren {R \cup S \cup T} \setminus \paren {R \cap S \cap T}$
where $R \symdif S$ denotes the symmetric difference between $R$ and $S$. | From the definition of symmetric difference, we have:
:$R \symdif S = \paren {R \setminus S} \cup \paren {S \setminus R}$
Thus, expanding:
{{begin-eqn}}
{{eqn | l = \paren {R \symdif S} \cup \paren {S \symdif T}
| r = \paren {R \setminus S} \cup \paren {S \setminus R} \cup \paren {S \setminus T} \cup \paren {T \s... | Let $R, S, T$ be [[Definition:Set|sets]].
Then:
:$\paren {R \symdif S} \cup \paren {S \symdif T} = \paren {R \cup S \cup T} \setminus \paren {R \cap S \cap T}$
where $R \symdif S$ denotes the [[Definition:Symmetric Difference|symmetric difference]] between $R$ and $S$. | From the definition of [[Definition:Symmetric Difference|symmetric difference]], we have:
:$R \symdif S = \paren {R \setminus S} \cup \paren {S \setminus R}$
Thus, expanding:
{{begin-eqn}}
{{eqn | l = \paren {R \symdif S} \cup \paren {S \symdif T}
| r = \paren {R \setminus S} \cup \paren {S \setminus R} \cup ... | Union of Symmetric Differences | https://proofwiki.org/wiki/Union_of_Symmetric_Differences | https://proofwiki.org/wiki/Union_of_Symmetric_Differences | [
"Symmetric Difference",
"Set Union"
] | [
"Definition:Set",
"Definition:Symmetric Difference"
] | [
"Definition:Symmetric Difference",
"Set Difference is Right Distributive over Union",
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection/Corollary",
"Set Difference with Union is Set Difference",
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection/Corollary"
] |
proofwiki-2319 | Difference of Unions is Subset of Union of Differences | Let $I$ be an indexing set.
Let $S_\alpha, T_\alpha$ be sets, for all $\alpha \in I$.
Then:
:$\ds \paren {\bigcup_{\alpha \mathop \in I} S_\alpha} \setminus \paren {\bigcup_{\alpha \mathop \in I} T_\alpha} \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$
where $S_\alpha \setminus T_\alpha$... | Let $\ds x \in \paren {\bigcup_{\alpha \mathop \in I} S_\alpha} \setminus \paren {\bigcup_{\alpha \mathop \in I} T_\alpha}$.
Then by definition of set difference:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup_{\alpha \mathop \in I} S_\alpha
| c =
}}
{{eqn | l = x
| o = \notin
| r = ... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $S_\alpha, T_\alpha$ be [[Definition:Set|sets]], for all $\alpha \in I$.
Then:
:$\ds \paren {\bigcup_{\alpha \mathop \in I} S_\alpha} \setminus \paren {\bigcup_{\alpha \mathop \in I} T_\alpha} \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setmi... | Let $\ds x \in \paren {\bigcup_{\alpha \mathop \in I} S_\alpha} \setminus \paren {\bigcup_{\alpha \mathop \in I} T_\alpha}$.
Then by definition of [[Definition:Set Difference|set difference]]:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup_{\alpha \mathop \in I} S_\alpha
| c =
}}
{{eqn | l = x... | Difference of Unions is Subset of Union of Differences | https://proofwiki.org/wiki/Difference_of_Unions_is_Subset_of_Union_of_Differences | https://proofwiki.org/wiki/Difference_of_Unions_is_Subset_of_Union_of_Differences | [
"Set Union",
"Set Difference"
] | [
"Definition:Indexing Set",
"Definition:Set",
"Definition:Set Difference"
] | [
"Definition:Set Difference",
"Definition:Set Union",
"De Morgan's Laws (Predicate Logic)",
"Definition:Set Union",
"Definition:Subset"
] |
proofwiki-2320 | Symmetric Difference of Unions is Subset of Union of Symmetric Differences | Let $I$ be an indexing set.
Let $S_\alpha, T_\alpha$ be sets, for all $\alpha \in I$.
Then:
:$\ds \bigcup_{\alpha \mathop \in I} S_\alpha \symdif \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \symdif T_\alpha}$
where $S \symdif T$ is the symmetric difference between $... | From Difference of Unions is Subset of Union of Differences, we have:
:$\ds \bigcup_{\alpha \mathop \in I} S_\alpha \setminus \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$
:$\ds \bigcup_{\alpha \mathop \in I} T_i \setminus \bigcup_{\alpha \mathop ... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $S_\alpha, T_\alpha$ be [[Definition:Set|sets]], for all $\alpha \in I$.
Then:
:$\ds \bigcup_{\alpha \mathop \in I} S_\alpha \symdif \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \symdif T_\alpha}$
where ... | From [[Difference of Unions is Subset of Union of Differences]], we have:
:$\ds \bigcup_{\alpha \mathop \in I} S_\alpha \setminus \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$
:$\ds \bigcup_{\alpha \mathop \in I} T_i \setminus \bigcup_{\alpha \m... | Symmetric Difference of Unions is Subset of Union of Symmetric Differences | https://proofwiki.org/wiki/Symmetric_Difference_of_Unions_is_Subset_of_Union_of_Symmetric_Differences | https://proofwiki.org/wiki/Symmetric_Difference_of_Unions_is_Subset_of_Union_of_Symmetric_Differences | [
"Set Union",
"Symmetric Difference"
] | [
"Definition:Indexing Set",
"Definition:Set",
"Definition:Symmetric Difference"
] | [
"Difference of Unions is Subset of Union of Differences",
"Definition:Set Difference",
"Difference of Unions is Subset of Union of Differences",
"Union is Associative",
"Union is Commutative",
"Category:Set Union",
"Category:Symmetric Difference"
] |
proofwiki-2321 | Pseudometric Defines an Equivalence Relation | Let $X$ be a set on which there is a pseudometric $d: X \times X \to \R$.
For any $x, y \in X$, let $x \sim y$ denote that $\map d {x, y} = 0$.
Then $\sim$ is an equivalence relation, and the equivalence classes consist of sets of elements of $X$ at zero distance from each other. | Checking in turn each of the criteria for equivalence: | Let $X$ be a [[Definition:Set|set]] on which there is a [[Definition:Pseudometric|pseudometric]] $d: X \times X \to \R$.
For any $x, y \in X$, let $x \sim y$ denote that $\map d {x, y} = 0$.
Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]], and the [[Definition:Equivalence Class|equivalence... | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Pseudometric Defines an Equivalence Relation | https://proofwiki.org/wiki/Pseudometric_Defines_an_Equivalence_Relation | https://proofwiki.org/wiki/Pseudometric_Defines_an_Equivalence_Relation | [
"Examples of Equivalence Relations",
"Pseudometric Spaces"
] | [
"Definition:Set",
"Definition:Pseudometric",
"Definition:Equivalence Relation",
"Definition:Equivalence Class",
"Definition:Distance Function"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-2322 | Metric Induced by a Pseudometric | Let $X$ be a set on which there is a pseudometric $d: X \times X \to \R$.
For any $x, y \in X$, let $x \sim y$ denote that $\map d {x, y} = 0$.
Let $\eqclass x \sim$ denote the equivalence class of $x$ under $\sim$.
Let $X^*$ be the quotient of $X$ by $\sim$.
Then the mapping $d^*: X^* \times X^* \to \R$ defined by:
:$... | From Pseudometric Defines an Equivalence Relation we have that $\sim$ is indeed an equivalence relation.
First we verify that $d^*$ is well-defined.
Let $z \in \eqclass x \sim$ and $w \in \eqclass y \sim$.
From Equivalence Class Equivalent Statements:
:$\eqclass z \sim = \eqclass x \sim$ and $\eqclass w \sim = \eqclass... | Let $X$ be a [[Definition:Set|set]] on which there is a [[Definition:Pseudometric|pseudometric]] $d: X \times X \to \R$.
For any $x, y \in X$, let $x \sim y$ denote that $\map d {x, y} = 0$.
Let $\eqclass x \sim$ denote the [[Definition:Equivalence Class|equivalence class of $x$ under $\sim$]].
Let $X^*$ be the [[De... | From [[Pseudometric Defines an Equivalence Relation]] we have that $\sim$ is indeed an [[Definition:Equivalence Relation|equivalence relation]].
First we verify that $d^*$ is [[Definition:Well-Defined Mapping|well-defined]].
Let $z \in \eqclass x \sim$ and $w \in \eqclass y \sim$.
From [[Equivalence Class Equivalen... | Metric Induced by a Pseudometric | https://proofwiki.org/wiki/Metric_Induced_by_a_Pseudometric | https://proofwiki.org/wiki/Metric_Induced_by_a_Pseudometric | [
"Pseudometric Spaces",
"Metric Spaces"
] | [
"Definition:Set",
"Definition:Pseudometric",
"Definition:Equivalence Class",
"Definition:Quotient Set",
"Definition:Mapping",
"Definition:Metric Space/Metric",
"Definition:Metric Space"
] | [
"Pseudometric Defines an Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Well-Defined/Mapping",
"Equivalence Class Equivalent Statements",
"Definition:Equivalence Class",
"Definition:Metric Space/Metric",
"Category:Pseudometric Spaces",
"Category:Metric Spaces"
] |
proofwiki-2323 | First-Order Reaction | Let a substance decompose spontaneously in a '''first-order reaction'''.
Let $x_0$ be a measure of the quantity of that substance at time $t = 0$.
Let the quantity of the substance that remains after time $t$ be $x$.
Then:
:$x = x_0 e^{-k t}$
where $k$ is the rate constant. | From Differential Equation governing First-Order Reaction, the differential equation governing this reaction is given by:
:$-\dfrac {\d x} {\d t} = k x$
for $k \in \R_{>0}$.
{{begin-eqn}}
{{eqn | l = \int \dfrac {\d x} x
| r = \int - k \rd t
| c = Solution to Separable Differential Equation
}}
{{eqn | ll= \... | Let a [[Definition:Substance|substance]] decompose spontaneously in a '''[[Definition:First-Order Reaction|first-order reaction]]'''.
Let $x_0$ be a measure of the quantity of that [[Definition:Substance|substance]] at time $t = 0$.
Let the quantity of the [[Definition:Substance|substance]] that remains after time $... | From [[Differential Equation governing First-Order Reaction]], the [[Definition:Differential Equation|differential equation]] governing this reaction is given by:
:$-\dfrac {\d x} {\d t} = k x$
for $k \in \R_{>0}$.
{{begin-eqn}}
{{eqn | l = \int \dfrac {\d x} x
| r = \int - k \rd t
| c = [[Solution to Sepa... | First-Order Reaction/Proof 1 | https://proofwiki.org/wiki/First-Order_Reaction | https://proofwiki.org/wiki/First-Order_Reaction/Proof_1 | [
"First-Order Reactions",
"Decay Equation"
] | [
"Definition:Substance",
"Definition:First-Order Reaction",
"Definition:Substance",
"Definition:Substance",
"Definition:First-Order Reaction/Rate Constant"
] | [
"Differential Equation governing First-Order Reaction",
"Definition:Differential Equation",
"Solution to Separable Differential Equation"
] |
proofwiki-2324 | First-Order Reaction | Let a substance decompose spontaneously in a '''first-order reaction'''.
Let $x_0$ be a measure of the quantity of that substance at time $t = 0$.
Let the quantity of the substance that remains after time $t$ be $x$.
Then:
:$x = x_0 e^{-k t}$
where $k$ is the rate constant. | From Differential Equation governing First-Order Reaction, the differential equation governing this reaction is given by:
:$-\dfrac {\d x} {\d t} = k x$
for $k \in \R_{>0}$.
This is an instance of the Decay Equation, and has the solution:
:$x = x_0 e^{-k t}$
{{qed}} | Let a [[Definition:Substance|substance]] decompose spontaneously in a '''[[Definition:First-Order Reaction|first-order reaction]]'''.
Let $x_0$ be a measure of the quantity of that [[Definition:Substance|substance]] at time $t = 0$.
Let the quantity of the [[Definition:Substance|substance]] that remains after time $... | From [[Differential Equation governing First-Order Reaction]], the [[Definition:Differential Equation|differential equation]] governing this reaction is given by:
:$-\dfrac {\d x} {\d t} = k x$
for $k \in \R_{>0}$.
This is an instance of the [[Decay Equation]], and has the solution:
:$x = x_0 e^{-k t}$
{{qed}} | First-Order Reaction/Proof 2 | https://proofwiki.org/wiki/First-Order_Reaction | https://proofwiki.org/wiki/First-Order_Reaction/Proof_2 | [
"First-Order Reactions",
"Decay Equation"
] | [
"Definition:Substance",
"Definition:First-Order Reaction",
"Definition:Substance",
"Definition:Substance",
"Definition:First-Order Reaction/Rate Constant"
] | [
"Differential Equation governing First-Order Reaction",
"Definition:Differential Equation",
"Decay Equation"
] |
proofwiki-2325 | Solution to First Order ODE | Let:
:$\Phi = \dfrac {\d y} {\d x} = \map f {x, y}$
be a first order ordinary differential equation.
Then $\Phi$ has a general solution which can be expressed in terms of an indefinite integral of $\map f x$:
:$\ds y = \int \map f {x, y} \rd x + C$
where $C$ is an arbitrary constant. | Integrating both sides with respect to $x$:
{{begin-eqn}}
{{eqn | l = \int \paren {\frac {\d y} {\d x} } \rd x
| r = \int \map f {x, y} \rd x
| c =
}}
{{eqn | ll= \leadsto
| l = y + C_1
| r = \int \map f {x, y} \rd x
| c = {{Defof|Indefinite Integral}}: $C_1$ is arbitrary
}}
{{eqn | ll= \... | Let:
:$\Phi = \dfrac {\d y} {\d x} = \map f {x, y}$
be a [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]].
Then $\Phi$ has a [[Definition:General Solution to Differential Equation|general solution]] which can be expressed in terms of an [[Definition:Indefinite Integ... | Integrating both sides with respect to $x$:
{{begin-eqn}}
{{eqn | l = \int \paren {\frac {\d y} {\d x} } \rd x
| r = \int \map f {x, y} \rd x
| c =
}}
{{eqn | ll= \leadsto
| l = y + C_1
| r = \int \map f {x, y} \rd x
| c = {{Defof|Indefinite Integral}}: $C_1$ is [[Definition:Arbitrary Co... | Solution to First Order ODE | https://proofwiki.org/wiki/Solution_to_First_Order_ODE | https://proofwiki.org/wiki/Solution_to_First_Order_ODE | [
"First Order ODEs"
] | [
"Definition:First Order Ordinary Differential Equation",
"Definition:Differential Equation/Solution/General Solution",
"Definition:Primitive (Calculus)",
"Definition:Primitive (Calculus)/Constant of Integration"
] | [
"Definition:Primitive (Calculus)/Constant of Integration",
"Picard's Existence Theorem",
"Category:First Order ODEs"
] |
proofwiki-2326 | Solution to First Order Initial Value Problem | Let $\map y x$ be a solution to the first order ordinary differential equation:
:$\dfrac {\d y} {\d x} = \map f {x, y}$
which is subject to an initial condition: $\tuple {a, b}$.
Then this problem is equivalent to the integral equation:
:$\ds y = b + \int_a^x \map f {t, \map y t} \rd t$ | From Solution to First Order ODE, the general solution of:
:$\dfrac {\d y} {\d x} = \map f {x, y}$
is:
:$\ds y = \int \map f {x, \map y x} \rd x + C$
When $x = a$, we have $y = b$.
Thus:
:$\ds b = \valueat {\int \map f {x, \map y x} \rd x + C} a$
{{MissingLinks|to the notation $[..]_a$}}
which gives:
:$\ds C = b - \val... | Let $\map y x$ be a [[Definition:Solution of Differential Equation|solution]] to the [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
:$\dfrac {\d y} {\d x} = \map f {x, y}$
which is subject to an [[Definition:Initial Condition|initial condition]]: $\tuple {a, b}$.
... | From [[Solution to First Order ODE]], the [[Definition:General Solution to Differential Equation|general solution]] of:
:$\dfrac {\d y} {\d x} = \map f {x, y}$
is:
:$\ds y = \int \map f {x, \map y x} \rd x + C$
When $x = a$, we have $y = b$.
Thus:
:$\ds b = \valueat {\int \map f {x, \map y x} \rd x + C} a$
{{Missing... | Solution to First Order Initial Value Problem | https://proofwiki.org/wiki/Solution_to_First_Order_Initial_Value_Problem | https://proofwiki.org/wiki/Solution_to_First_Order_Initial_Value_Problem | [
"First Order ODEs"
] | [
"Definition:Differential Equation/Solution",
"Definition:First Order Ordinary Differential Equation",
"Definition:Initial Condition",
"Definition:Integral Equation"
] | [
"Solution to First Order ODE",
"Definition:Differential Equation/Solution/General Solution",
"Fundamental Theorem of Calculus/Second Part",
"Category:First Order ODEs"
] |
proofwiki-2327 | Condition for Lipschitz Condition to be Satisfied | Let $\phi$ be a real function.
Then $\phi$ satisfies the Lipschitz condition on a closed real interval $\closedint a b$ if:
:$\forall y \in \closedint a b: \exists A \in \R: \size {\map {\phi'} y} \le A$ | Integrating both sides of $\size {\map {\phi'} y} \le A$ gives us:
{{begin-eqn}}
{{eqn | l = \size {\map {\phi'} y}
| o = \le
| r = A
| c =
}}
{{eqn | ll= \leadsto
| l = -A
| o = \le
| r = \map {\phi'} y \le A
| c =
}}
{{eqn | ll= \leadsto
| l = \int {-A} \rd y
... | Let $\phi$ be a [[Definition:Real Function|real function]].
Then $\phi$ satisfies the [[Definition:Lipschitz Condition (Real Function)|Lipschitz condition]] on a [[Definition:Closed Real Interval|closed real interval]] $\closedint a b$ if:
:$\forall y \in \closedint a b: \exists A \in \R: \size {\map {\phi'} y} \le A... | [[Definition:Integration|Integrating]] both sides of $\size {\map {\phi'} y} \le A$ gives us:
{{begin-eqn}}
{{eqn | l = \size {\map {\phi'} y}
| o = \le
| r = A
| c =
}}
{{eqn | ll= \leadsto
| l = -A
| o = \le
| r = \map {\phi'} y \le A
| c =
}}
{{eqn | ll= \leadsto
... | Condition for Lipschitz Condition to be Satisfied | https://proofwiki.org/wiki/Condition_for_Lipschitz_Condition_to_be_Satisfied | https://proofwiki.org/wiki/Condition_for_Lipschitz_Condition_to_be_Satisfied | [
"Calculus"
] | [
"Definition:Real Function",
"Definition:Lipschitz Continuity/Real Function",
"Definition:Real Interval/Closed"
] | [
"Definition:Primitive (Calculus)/Integration",
"Definition:Bounded Mapping/Real-Valued",
"Category:Calculus"
] |
proofwiki-2328 | Quotient of Homogeneous Functions | Let $\map M {x, y}$ and $\map N {x, y}$ be homogeneous functions of the same degree.
Then:
:$\dfrac {\map M {x, y} } {\map N {x, y} }$
is homogeneous of zero degree. | Let:
:$\map Q {x, y} = \dfrac {\map M {x, y} } {\map N {x, y} }$
where $M$ and $N$ are homogeneous functions of degree $n$.
Let $t \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = \map Q {t x, t y}
| r = \frac {\map M {t x, t y} } {\map N {t x, t y} }
| c =
}}
{{eqn | r = \frac {t^n \map M {x, y} } {t^n \map N {x,... | Let $\map M {x, y}$ and $\map N {x, y}$ be [[Definition:Homogeneous Real Function|homogeneous functions]] of the same [[Definition:Degree of Homogeneous Real Function|degree]].
Then:
:$\dfrac {\map M {x, y} } {\map N {x, y} }$
is [[Definition:Homogeneous Real Function|homogeneous]] of [[Definition:Homogeneous Real Fu... | Let:
:$\map Q {x, y} = \dfrac {\map M {x, y} } {\map N {x, y} }$
where $M$ and $N$ are [[Definition:Homogeneous Real Function|homogeneous functions]] of [[Definition:Degree of Homogeneous Real Function|degree]] $n$.
Let $t \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = \map Q {t x, t y}
| r = \frac {\map M {t x, t ... | Quotient of Homogeneous Functions | https://proofwiki.org/wiki/Quotient_of_Homogeneous_Functions | https://proofwiki.org/wiki/Quotient_of_Homogeneous_Functions | [
"Homogeneous Functions"
] | [
"Definition:Homogeneous Function/Real Space",
"Definition:Homogeneous Function/Real Space/Degree",
"Definition:Homogeneous Function/Real Space",
"Definition:Homogeneous Function/Real Space/Zero Degree"
] | [
"Definition:Homogeneous Function/Real Space",
"Definition:Homogeneous Function/Real Space/Degree",
"Definition:Homogeneous Function/Real Space",
"Definition:Homogeneous Function/Real Space/Degree",
"Definition:Homogeneous Function/Real Space"
] |
proofwiki-2329 | Solution to Homogeneous Differential Equation | Let:
:$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be a homogeneous differential equation.
It can be solved by making the substitution $z = \dfrac y x$.
Its solution is:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {x, y} = -\dfrac {\map M {x, y} } {\map N {x, y} }$ | From the original equation, we see:
:$\dfrac {\d y} {\d x} = \map f {x, y} = -\dfrac {\map M {x, y} } {\map N {x, y} }$
From Quotient of Homogeneous Functions it follows that $\map f {x, y}$ is homogeneous of degree zero.
Thus:
:$\map f {t x, t y} = t^0 \map f {x, y} = \map f {x, y}$
Set $t = \dfrac 1 x$ in this equati... | Let:
:$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]].
It can be solved by making the substitution $z = \dfrac y x$.
Its solution is:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {x, y} = -\... | From the original equation, we see:
:$\dfrac {\d y} {\d x} = \map f {x, y} = -\dfrac {\map M {x, y} } {\map N {x, y} }$
From [[Quotient of Homogeneous Functions]] it follows that $\map f {x, y}$ is [[Definition:Homogeneous Real Function of Zero Degree|homogeneous of degree zero]].
Thus:
:$\map f {t x, t y} = t^0 \ma... | Solution to Homogeneous Differential Equation | https://proofwiki.org/wiki/Solution_to_Homogeneous_Differential_Equation | https://proofwiki.org/wiki/Solution_to_Homogeneous_Differential_Equation | [
"Homogeneous Differential Equations"
] | [
"Definition:Homogeneous Differential Equation"
] | [
"Quotient of Homogeneous Functions",
"Definition:Homogeneous Function/Real Space/Zero Degree",
"Product Rule for Derivatives",
"Definition:Separable Differential Equation"
] |
proofwiki-2330 | Solution to Exact Differential Equation | The first order ordinary differential equation:
:$F = \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
is an exact differential equation {{iff}}:
:$\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$
The general solution of such an equation is:
:$\map f {x, y} = C$
where:
:$\dfrac {\partial f} {... | === Necessary Condition ===
Let $F$ be exact.
Then by definition there exists a function whose second partial derivatives of $f$ exist and are continuous:
:$\map f {x, y}$
where:
:$\dfrac {\partial f} {\partial x} = M$
:$\dfrac {\partial f} {\partial y} = N$
Differentiating $M$ and $N$ partially {{WRT|Differentiation}}... | The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
:$F = \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
is an [[Definition:Exact Differential Equation|exact differential equation]] {{iff}}:
:$\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial... | === Necessary Condition ===
Let $F$ be [[Definition:Exact Differential Equation|exact]].
Then by definition there exists a [[Definition:Real Function|function]] whose [[Definition:Partial Derivative|second partial derivatives]] of $f$ exist and are [[Definition:Continuous Real Function|continuous]]:
:$\map f {x, y}$
... | Solution to Exact Differential Equation | https://proofwiki.org/wiki/Solution_to_Exact_Differential_Equation | https://proofwiki.org/wiki/Solution_to_Exact_Differential_Equation | [
"Exact Differential Equations"
] | [
"Definition:First Order Ordinary Differential Equation",
"Definition:Exact Differential Equation",
"Definition:Differential Equation/Solution/General Solution",
"Definition:Exact Differential Equation"
] | [
"Definition:Exact Differential Equation",
"Definition:Real Function",
"Definition:Partial Derivative",
"Definition:Continuous Real Function",
"Mixed Second Partial Derivatives are Equal"
] |
proofwiki-2331 | Existence of Integrating Factor | Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be such that $M$ and $N$ are real functions of two variables which are ''not'' homogeneous functions of the same degree.
Suppose also that:
:$\dfrac {\partial M} {\partial y} \ne \dfrac {\partial N} ... | Let us for ease of manipulation express $(1)$ in the form of differentials:
:$(2): \quad \map M {x, y} \rd x + \map N {x, y} \rd y = 0$
Suppose that $(2)$ has a general solution:
:$(3): \quad \map f {x, y} = C$
where $C$ is some constant.
We can eliminate $C$ by differentiating:
:$\dfrac {\partial f} {\partial x} \rd x... | Let the [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be such that $M$ and $N$ are [[Definition:Real Function|real functions]] of two variables which are ''not'' [[Definition:Homogeneous Real Fun... | Let us for ease of manipulation express $(1)$ in the form of [[Definition:Differential of Real Function|differentials]]:
:$(2): \quad \map M {x, y} \rd x + \map N {x, y} \rd y = 0$
Suppose that $(2)$ has a [[Definition:General Solution to Differential Equation|general solution]]:
:$(3): \quad \map f {x, y} = C$
where... | Existence of Integrating Factor | https://proofwiki.org/wiki/Existence_of_Integrating_Factor | https://proofwiki.org/wiki/Existence_of_Integrating_Factor | [
"Integrating Factors"
] | [
"Definition:First Order Ordinary Differential Equation",
"Definition:Real Function",
"Definition:Homogeneous Function/Real Space",
"Definition:Homogeneous Function/Real Space/Degree",
"Definition:Exact Differential Equation",
"Definition:Differential Equation/Solution/General Solution",
"Definition:Inte... | [
"Definition:Differential of Mapping/Real Function",
"Definition:Differential Equation/Solution/General Solution",
"Definition:Constant",
"Definition:Partial Derivative",
"Definition:Exact Differential Equation",
"Definition:Differential Equation/Solution/General Solution",
"Definition:Integrating Factor... |
proofwiki-2332 | Integrating Factor for First Order ODE | Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be non-homogeneous and not exact.
By Existence of Integrating Factor, if $(1)$ has a general solution, there exists an integrating factor $\map \mu {x, y}$ such that:
:$\ds \map \mu {x, y} \paren {\m... | We have one of these:
: Integrating Factor for First Order ODE: Function of One Variable: $x$ or $y$ only
: Integrating Factor for First Order ODE: Function of $x + y$
: Integrating Factor for First Order ODE: Function of $x y$ | Let the [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be non-[[Definition:Homogeneous Differential Equation|homogeneous]] and not [[Definition:Exact Differential Equation|exact]].
By [[Existence... | We have one of these:
: [[Integrating Factor for First Order ODE/Function of One Variable|Integrating Factor for First Order ODE: Function of One Variable: $x$ or $y$ only]]
: [[Integrating Factor for First Order ODE/Function of Sum of Variables|Integrating Factor for First Order ODE: Function of $x + y$]]
: [[Integrat... | Integrating Factor for First Order ODE | https://proofwiki.org/wiki/Integrating_Factor_for_First_Order_ODE | https://proofwiki.org/wiki/Integrating_Factor_for_First_Order_ODE | [
"First Order ODEs",
"Integrating Factors"
] | [
"Definition:First Order Ordinary Differential Equation",
"Definition:Homogeneous Differential Equation",
"Definition:Exact Differential Equation",
"Existence of Integrating Factor",
"Definition:Differential Equation/Solution/General Solution",
"Definition:Integrating Factor",
"Definition:Exact Different... | [
"Integrating Factor for First Order ODE/Function of One Variable",
"Integrating Factor for First Order ODE/Function of Sum of Variables",
"Integrating Factor for First Order ODE/Function of Product of Variables"
] |
proofwiki-2333 | Solution to Linear First Order Ordinary Differential Equation | A linear first order ordinary differential equation in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
has the general solution:
:$\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x} \rd x + C}$ | Consider the first order ordinary differential equation:
:$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
We can put our equation:
:$(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x$
into this format by identifying:
:$\map M {x, y} \equiv \map P x y - \map Q x, \map N {x, y} \equiv 1$
We see that:
:$\df... | A [[Definition:Linear First Order Ordinary Differential Equation|linear first order ordinary differential equation]] in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
has the [[Definition:General Solution to Differential Equation|general solution]]:
:$\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x... | Consider the [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
:$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
We can put our equation:
:$(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x$
into this format by identifying:
:$\map M {x, y} \equiv \map ... | Solution to Linear First Order Ordinary Differential Equation/Proof 1 | https://proofwiki.org/wiki/Solution_to_Linear_First_Order_Ordinary_Differential_Equation | https://proofwiki.org/wiki/Solution_to_Linear_First_Order_Ordinary_Differential_Equation/Proof_1 | [
"Solution to Linear First Order Ordinary Differential Equation",
"Linear First Order ODEs"
] | [
"Definition:Linear First Order Ordinary Differential Equation",
"Definition:Differential Equation/Solution/General Solution"
] | [
"Definition:First Order Ordinary Differential Equation",
"Definition:Real Function",
"Integrating Factor for First Order ODE",
"Definition:Integrating Factor",
"Solution to Exact Differential Equation"
] |
proofwiki-2334 | Solution to Linear First Order Ordinary Differential Equation | A linear first order ordinary differential equation in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
has the general solution:
:$\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x} \rd x + C}$ | From the Product Rule for Derivatives:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} \paren {e^{\int \map P x \rd x} y}
| r = e^{\int \map P x \rd x} \dfrac {\d y} {\d x} + y e^{\int \map P x \rd x} \map P x
| c =
}}
{{eqn | r = e^{\int \map P x \rd x} \paren {\dfrac {\d y} {\d x} + \map P x y}
| c =
}}... | A [[Definition:Linear First Order Ordinary Differential Equation|linear first order ordinary differential equation]] in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
has the [[Definition:General Solution to Differential Equation|general solution]]:
:$\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x... | From the [[Product Rule for Derivatives]]:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} \paren {e^{\int \map P x \rd x} y}
| r = e^{\int \map P x \rd x} \dfrac {\d y} {\d x} + y e^{\int \map P x \rd x} \map P x
| c =
}}
{{eqn | r = e^{\int \map P x \rd x} \paren {\dfrac {\d y} {\d x} + \map P x y}
| c ... | Solution to Linear First Order Ordinary Differential Equation/Proof 2 | https://proofwiki.org/wiki/Solution_to_Linear_First_Order_Ordinary_Differential_Equation | https://proofwiki.org/wiki/Solution_to_Linear_First_Order_Ordinary_Differential_Equation/Proof_2 | [
"Solution to Linear First Order Ordinary Differential Equation",
"Linear First Order ODEs"
] | [
"Definition:Linear First Order Ordinary Differential Equation",
"Definition:Differential Equation/Solution/General Solution"
] | [
"Product Rule for Derivatives"
] |
proofwiki-2335 | Extremal Length of Composition | Let $\Gamma_1$ and $\Gamma_2$ be families of (unions of) rectifiable curves on a Riemann surface $X$.
Let $\Gamma_1$ and $\Gamma_2$ be disjoint in the sense that there exist disjoint Borel sets $E_1 \subseteq X$ and $E_2 \subseteq X$ with $\bigcup \Gamma_1 \subset E_1$ and $\bigcup \Gamma_2 \subset E_2$.
The extremal l... | By the Series Law for Extremal Length:
:$\map \lambda \Gamma \ge \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$
Hence it remains only to prove the opposite inequality.
Let $\rho$ be a metric as in the definition of extremal length, normalized such that $\map A \rho = 1$.
The claim to be proved is that:
:$\paren {\m... | Let $\Gamma_1$ and $\Gamma_2$ be [[Definition:Indexed Family|families]] of ([[Definition:Set Union|unions]] of) [[Definition:Rectifiable Curve|rectifiable curves]] on a [[Definition:Riemann Surface|Riemann surface]] $X$.
Let $\Gamma_1$ and $\Gamma_2$ be disjoint in the sense that there exist [[Definition:Disjoint Sets... | By the [[Series Law for Extremal Length]]:
:$\map \lambda \Gamma \ge \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$
Hence it remains only to prove the opposite inequality.
Let $\rho$ be a [[Definition:Metric|metric]] as in the definition of [[Definition:Extremal Length|extremal length]], normalized such that $\m... | Extremal Length of Composition | https://proofwiki.org/wiki/Extremal_Length_of_Composition | https://proofwiki.org/wiki/Extremal_Length_of_Composition | [
"Functional Analysis"
] | [
"Definition:Indexing Set/Family",
"Definition:Set Union",
"Definition:Rectifiable Curve",
"Definition:Riemann Surface",
"Definition:Disjoint Sets",
"Definition:Borel Sigma-Algebra/Borel Set",
"Definition:Extremal Length",
"Definition:Indexing Set/Family"
] | [
"Series Law for Extremal Length",
"Definition:Metric Space/Metric",
"Definition:Extremal Length",
"Definition:Positive/Real Number",
"Definition:Positive/Real Number",
"Definition:Disjoint Sets",
"Cauchy-Bunyakovsky-Schwarz Inequality",
"Category:Functional Analysis"
] |
proofwiki-2336 | Squares with All Odd Digits | The only squares whose digits (when written in base $10$ notation) are all odd are $1$ and $9$. | If $n$ is even, then at least the last digit of $n^2$ is even.
So for $n^2$ to have its digits all odd, $n$ itself must be odd.
We can see immediately that $1 = 1^2$ and $9 = 3^2$ fit the criterion.
Of the other $1$-digit odd integers, we have $5^2 = 25, 7^2 = 49, 9^2 = 81$, all of which have an even digit.
Now, let $n... | The only [[Definition:Square Number|squares]] whose [[Definition:Digit|digits]] (when written in [[Definition:Decimal Notation|base $10$ notation]]) are all [[Definition:Odd Integer|odd]] are $1$ and $9$. | If $n$ is [[Definition:Even Integer|even]], then at least the last [[Definition:Digit|digit]] of $n^2$ is [[Definition:Even Integer|even]].
So for $n^2$ to have its [[Definition:Digit|digits]] all [[Definition:Odd Integer|odd]], $n$ itself must be [[Definition:Odd Integer|odd]].
We can see immediately that $1 = 1^2$ ... | Squares with All Odd Digits | https://proofwiki.org/wiki/Squares_with_All_Odd_Digits | https://proofwiki.org/wiki/Squares_with_All_Odd_Digits | [
"Square Numbers"
] | [
"Definition:Square Number",
"Definition:Digit",
"Definition:Decimal Notation",
"Definition:Odd Integer"
] | [
"Definition:Even Integer",
"Definition:Digit",
"Definition:Even Integer",
"Definition:Digit",
"Definition:Odd Integer",
"Definition:Odd Integer",
"Definition:Digit",
"Definition:Odd Integer",
"Definition:Integer",
"Definition:Even Integer",
"Definition:Digit",
"Definition:Odd Integer",
"Defi... |
proofwiki-2337 | Parity of Integer equals Parity of its Square | Let $p \in \Z$ be an integer.
Then $p$ is even {{iff}} $p^2$ is even. | Let $p$ be an integer.
By the Division Theorem, there are unique integers $k$ and $r$ such that $p = 2k + r$ and $0 \le r < 2$.
That is, $r = 0$ or $r = 1$, where $r = 0$ corresponds to the case of $p$ being even and $r = 1$ corresponds to the case of $p$ being odd. | Let $p \in \Z$ be an [[Definition:Integer|integer]].
Then $p$ is [[Definition:Even Integer|even]] {{iff}} $p^2$ is [[Definition:Even Integer|even]]. | Let $p$ be an [[Definition:integer|integer]].
By the [[Division Theorem]], there are unique integers $k$ and $r$ such that $p = 2k + r$ and $0 \le r < 2$.
That is, $r = 0$ or $r = 1$, where $r = 0$ corresponds to the case of $p$ being [[Definition:Even Integer|even]] and $r = 1$ corresponds to the case of $p$ being [... | Parity of Integer equals Parity of its Square | https://proofwiki.org/wiki/Parity_of_Integer_equals_Parity_of_its_Square | https://proofwiki.org/wiki/Parity_of_Integer_equals_Parity_of_its_Square | [
"Parity of Integer equals Parity of its Square",
"Square Numbers"
] | [
"Definition:Integer",
"Definition:Even Integer",
"Definition:Even Integer"
] | [
"Definition:integer",
"Division Theorem",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Even Integer",
"Definition:Even Integer",
"Definition:Even Integer"
] |
proofwiki-2338 | Borel-Cantelli Lemma | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of $\Sigma$-measurable sets.
Let:
:$\ds \sum_{n \mathop = 1}^\infty \map \mu {E_n} < \infty$
Then:
:$\ds \map \mu {\limsup_{n \mathop \to \infty} E_n} = 0$
where $\limsup$ denotes limit superior of sets. | By definition of limit superior:
:$\ds \limsup_{n \mathop \to \infty} E_n = \bigcap_{i \mathop = 1}^\infty \bigcup_{j \mathop = i}^\infty E_j$
Thus, by Measure is Monotone and Intersection is Subset:
:$(1): \quad \ds \map \mu {\limsup_{n \mathop \to \infty} E_n} = \map \mu {\bigcap_{i \mathop = 1}^\infty \bigcup_{j \ma... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\sequence {E_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Measurable Set|$\Sigma$-measurable sets]].
Let:
:$\ds \sum_{n \mathop = 1}^\infty \map \mu {E_n} < \infty$
Then:
:$\ds \map \mu {\limsup_{n \... | By definition of [[Definition:Limit Superior of Sequence of Sets|limit superior]]:
:$\ds \limsup_{n \mathop \to \infty} E_n = \bigcap_{i \mathop = 1}^\infty \bigcup_{j \mathop = i}^\infty E_j$
Thus, by [[Measure is Monotone]] and [[Intersection is Subset]]:
:$(1): \quad \ds \map \mu {\limsup_{n \mathop \to \infty} E_n... | Borel-Cantelli Lemma | https://proofwiki.org/wiki/Borel-Cantelli_Lemma | https://proofwiki.org/wiki/Borel-Cantelli_Lemma | [
"Borel-Cantelli Lemma",
"Measure Theory",
"Probability Theory"
] | [
"Definition:Measure Space",
"Definition:Sequence",
"Definition:Measurable Set",
"Definition:Limit Superior of Sequence of Sets"
] | [
"Definition:Limit Superior of Sequence of Sets",
"Measure is Monotone",
"Intersection is Subset",
"Measure is Subadditive",
"Definition:Convergent Series",
"Tail of Convergent Series tends to Zero",
"Lower and Upper Bounds for Sequences",
"Definition:Measure (Measure Theory)"
] |
proofwiki-2339 | Weierstrass M-Test | Let $f_n$ be a sequence of real functions defined on a domain $D \subseteq \R$.
Let $\ds \sup_{x \mathop \in D} \size {\map {f_n} x} \le M_n$ for each integer $n$ and some constants $M_n$
Let $\ds \sum_{i \mathop = 1}^\infty M_i < \infty$.
Then $\ds \sum_{i \mathop = 1}^\infty f_i$ converges uniformly on $D$. | Let:
:$\ds S_n = \sum_{i \mathop = 1}^n f_i$
Let:
:$\ds f = \lim_{n \mathop \to \infty} S_n$
To show the sequence of partial sums converge uniformly to $f$, we must show that:
:$\ds \lim_{n \mathop \to \infty} \sup_{x \mathop \in D} \size {f - S_n} = 0$
But:
{{begin-eqn}}
{{eqn | l = \sup_{x \mathop \in D} \size {f - S... | Let $f_n$ be a [[Definition:Sequence|sequence]] of [[Definition:Real Function|real functions]] defined on a domain $D \subseteq \R$.
Let $\ds \sup_{x \mathop \in D} \size {\map {f_n} x} \le M_n$ for each [[Definition:Integer|integer]] $n$ and some constants $M_n$
Let $\ds \sum_{i \mathop = 1}^\infty M_i < \infty$.
... | Let:
:$\ds S_n = \sum_{i \mathop = 1}^n f_i$
Let:
:$\ds f = \lim_{n \mathop \to \infty} S_n$
To show the [[Definition:Sequence of Partial Sums|sequence of partial sums]] [[Definition:Uniformly Convergent Real Sequence|converge uniformly]] to $f$, we must show that:
:$\ds \lim_{n \mathop \to \infty} \sup_{x \mathop \i... | Weierstrass M-Test | https://proofwiki.org/wiki/Weierstrass_M-Test | https://proofwiki.org/wiki/Weierstrass_M-Test | [
"Real Analysis",
"Uniform Convergence"
] | [
"Definition:Sequence",
"Definition:Real Function",
"Definition:Integer",
"Definition:Uniform Convergence/Infinite Series"
] | [
"Definition:Series/Sequence of Partial Sums",
"Definition:Uniform Convergence/Real Sequence",
"Triangle Inequality",
"Tail of Convergent Series tends to Zero",
"Definition:Uniform Convergence/Infinite Series"
] |
proofwiki-2340 | Measurable Sets form Sigma-Algebra | Let $\mu^*$ be an outer measure on a set $X$.
Then the set $\map {\mathfrak M} {\mu^*}$ of all $\mu^*$-measurable subsets of $X$ is a $\sigma$-algebra. | First, note that $\map {\mathfrak M} {\mu^*}$ is an algebra (of sets).
It remains to be shown that $\map {\mathfrak M} {\mu^*}$ is closed under countable union.
Because $\map {\mathfrak M} {\mu^*}$ is an algebra (of sets), the union of any two $\mu^*$-measurable sets is $\mu^*$-measurable.
Using mathematical induction,... | Let $\mu^*$ be an [[Definition:Outer Measure|outer measure]] on a [[Definition:Set|set]] $X$.
Then the [[Definition:Set|set]] $\map {\mathfrak M} {\mu^*}$ of all [[Definition:Measurable Set of Arbitrary Outer Measure|$\mu^*$-measurable]] [[Definition:Subset|subsets]] of $X$ is a [[Definition:Sigma-Algebra|$\sigma$-alg... | First, note that [[Measurable Sets form Algebra of Sets|$\map {\mathfrak M} {\mu^*}$ is an algebra (of sets)]].
It remains to be shown that $\map {\mathfrak M} {\mu^*}$ is [[Definition:Closure (Abstract Algebra)|closed]] under [[Definition:Countable|countable]] [[Definition:Set Union|union]].
Because $\map {\mathfra... | Measurable Sets form Sigma-Algebra | https://proofwiki.org/wiki/Measurable_Sets_form_Sigma-Algebra | https://proofwiki.org/wiki/Measurable_Sets_form_Sigma-Algebra | [
"Analysis",
"Measurable Sets",
"Sigma-Algebras"
] | [
"Definition:Outer Measure",
"Definition:Set",
"Definition:Set",
"Definition:Measurable Set/Arbitrary Outer Measure",
"Definition:Subset",
"Definition:Sigma-Algebra"
] | [
"Measurable Sets form Algebra of Sets",
"Definition:Closure (Abstract Algebra)",
"Definition:Countable Set",
"Definition:Set Union",
"Definition:Algebra of Sets",
"Definition:Set Union",
"Definition:Measurable Set/Arbitrary Outer Measure",
"Definition:Set",
"Principle of Mathematical Induction",
"... |
proofwiki-2341 | Markov's Inequality | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $A \in \Sigma$.
Let $f : A \to \overline \R$ be an $A$-measurable function.
Then:
:$\ds \map \mu {\set {x \in A: \size {\map f x} \ge t} } \le \frac 1 t \int_A \size f \rd \mu$
for any positive $t \in \R$. | Let $t > 0$ and define:
:$B = \set {x \in A: \size {\map f x} \ge t}$
Let $\chi_B$ denote the indicator function of $B$ on $A$.
For any $x \in A$, either $x \in B$ or $x \notin B$.
In the first case:
:$t \map {\chi_B} x = t \cdot 1 = t \le \size {\map f x}$
In the second case:
:$t \map {\chi_B} x = t \cdot 0 = 0 \le \... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $A \in \Sigma$.
Let $f : A \to \overline \R$ be an [[Definition:Measurable Function|$A$-measurable function]].
Then:
:$\ds \map \mu {\set {x \in A: \size {\map f x} \ge t} } \le \frac 1 t \int_A \size f \rd \mu$
for any positive $t ... | Let $t > 0$ and define:
:$B = \set {x \in A: \size {\map f x} \ge t}$
Let $\chi_B$ denote the [[Definition:Characteristic Function of Set|indicator function]] of $B$ on $A$.
For any $x \in A$, either $x \in B$ or $x \notin B$.
In the first case:
:$t \map {\chi_B} x = t \cdot 1 = t \le \size {\map f x}$
In the se... | Markov's Inequality | https://proofwiki.org/wiki/Markov's_Inequality | https://proofwiki.org/wiki/Markov's_Inequality | [
"Measure Theory",
"Probability Theory",
"Markov's Inequality"
] | [
"Definition:Measure Space",
"Definition:Measurable Function"
] | [
"Definition:Characteristic Function (Set Theory)/Set",
"Integral of Integrable Function is Monotone",
"Integral of Integrable Function is Homogeneous"
] |
proofwiki-2342 | Limit Superior includes Limit Inferior | Let $\set {E_n : n \in \N}$ be a sequence of sets.
Then:
:$\ds \liminf_{n \mathop \to \infty} E_n \subseteq \limsup_{n \mathop \to \infty} E_n$ | Select any $\ds x \in \liminf_{n \mathop \to \infty} E_n$.
Then by definition 2 of Limit Inferior of Sequence of Sets, $x$ belongs to $E_i$ for all but finitely many values of $i$.
Therefore $x$ belongs to $E_i$ for infinitely many values of $i$.
So by the definition 2 of Limit Superior of Sequence of Sets:
:$\ds x \in... | Let $\set {E_n : n \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Set|sets]].
Then:
:$\ds \liminf_{n \mathop \to \infty} E_n \subseteq \limsup_{n \mathop \to \infty} E_n$ | Select any $\ds x \in \liminf_{n \mathop \to \infty} E_n$.
Then by [[Definition:Limit Inferior of Sequence of Sets/Definition 2|definition 2 of Limit Inferior of Sequence of Sets]], $x$ belongs to $E_i$ for all but [[Definition:Finite Set|finitely many]] values of $i$.
Therefore $x$ belongs to $E_i$ for [[Definition:... | Limit Superior includes Limit Inferior | https://proofwiki.org/wiki/Limit_Superior_includes_Limit_Inferior | https://proofwiki.org/wiki/Limit_Superior_includes_Limit_Inferior | [
"Limits Superior of Set Sequences",
"Limits Inferior of Set Sequences"
] | [
"Definition:Sequence",
"Definition:Set"
] | [
"Definition:Limit Inferior of Sequence of Sets/Definition 2",
"Definition:Finite Set",
"Definition:Infinite Set",
"Definition:Limit Superior of Sequence of Sets/Definition 2"
] |
proofwiki-2343 | Construction of Outer Measure | Let $X$ be a set.
Let $\powerset X$ be the power set of $X$.
Let $\AA$ be a subset of $\powerset X$ which contains the empty set.
Let $\overline \R_{\ge 0}$ denote the set of positive extended real numbers.
Let $\gamma: \AA \to \overline \R_{\ge 0}$ be a mapping such that $\map \gamma \O = 0$.
Let $\mu^*: \powerset X \... | We check each of the criteria for an outer measure.
From the assumption $\map \gamma \O = 0$:
:$\map {\mu^*} \O = 0$
Any cover of a set is also a cover of a subset of it.
Thus $\mu^*$ is monotone by Infimum of Subset.
It remains to be shown that $\mu^*$ is countably subadditive.
Let $\sequence {S_n}$ be a sequence of s... | Let $X$ be a [[Definition:Set|set]].
Let $\powerset X$ be the [[Definition:Power Set|power set]] of $X$.
Let $\AA$ be a [[Definition:Subset|subset]] of $\powerset X$ which contains the [[Definition:Empty Set|empty set]].
Let $\overline \R_{\ge 0}$ denote the [[Definition:Set|set]] of [[Definition:Positive Real Numb... | We check each of the criteria for an [[Definition:Outer Measure|outer measure]].
From the assumption $\map \gamma \O = 0$:
:$\map {\mu^*} \O = 0$
Any [[Definition:Cover of Set|cover]] of a [[Definition:Set|set]] is also a [[Definition:Cover of Set|cover]] of a [[Definition:Subset|subset]] of it.
Thus $\mu^*$ is [[De... | Construction of Outer Measure | https://proofwiki.org/wiki/Construction_of_Outer_Measure | https://proofwiki.org/wiki/Construction_of_Outer_Measure | [
"Outer Measures"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Subset",
"Definition:Empty Set",
"Definition:Set",
"Definition:Positive/Real Number",
"Definition:Extended Real Number Line",
"Definition:Mapping",
"Definition:Mapping",
"Infimum of Empty Set is Greatest Element",
"Definition:Outer Measure"
] | [
"Definition:Outer Measure",
"Definition:Cover of Set",
"Definition:Set",
"Definition:Cover of Set",
"Definition:Subset",
"Definition:Monotone (Measure Theory)",
"Infimum of Subset",
"Definition:Countably Subadditive Function",
"Definition:Sequence",
"Definition:Subset",
"Definition:Cover of Set/... |
proofwiki-2344 | Additive and Countably Subadditive Function is Countably Additive | Let $\Sigma$ be a $\sigma$-algebra over a set $X$.
Let $f: \Sigma \to \overline \R_{\ge 0}$ be an additive and countably subadditive function, where $\overline \R_{\ge 0}$ denotes the set of positive extended real numbers.
Then $f$ is countably additive. | {{rewrite|Eqn templates would be good, I think}}
Let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint elements of $\Sigma$.
Let $N \in \N$ be any natural number.
By Set is Subset of Union:
:$\ds \bigcup_{n \mathop = 0}^N S_n \subseteq \bigcup_{n \mathop = 0}^\infty S_n$
From Non-Negative Additive... | Let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] over a [[Definition:Set|set]] $X$.
Let $f: \Sigma \to \overline \R_{\ge 0}$ be an [[Definition:Additive Function (Measure Theory)|additive]] and [[Definition:Countably Subadditive Function|countably subadditive]] [[Definition:Mapping|function]], where $\o... | {{rewrite|Eqn templates would be good, I think}}
Let $\sequence {S_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Element|elements]] of $\Sigma$.
Let $N \in \N$ be any [[Definition:Natural Numbers|natural number]].
By [[Set is Subset of ... | Additive and Countably Subadditive Function is Countably Additive | https://proofwiki.org/wiki/Additive_and_Countably_Subadditive_Function_is_Countably_Additive | https://proofwiki.org/wiki/Additive_and_Countably_Subadditive_Function_is_Countably_Additive | [
"Measure Theory"
] | [
"Definition:Sigma-Algebra",
"Definition:Set",
"Definition:Additive Function (Measure Theory)",
"Definition:Countably Subadditive Function",
"Definition:Mapping",
"Definition:Positive/Real Number",
"Definition:Extended Real Number Line",
"Definition:Countably Additive Function"
] | [
"Definition:Sequence",
"Definition:Pairwise Disjoint",
"Definition:Element",
"Definition:Natural Numbers",
"Set is Subset of Union",
"Non-Negative Additive Function is Monotone",
"Finite Union of Sets in Additive Function",
"Lower and Upper Bounds for Sequences",
"Lower and Upper Bounds for Sequence... |
proofwiki-2345 | Measure Space from Outer Measure | Suppose $\mu^*$ is an outer measure on a set $X$.
Let $\map {\mathfrak M} {\mu^*}$ be the collection of $\mu^*$-measurable sets.
Let $\mu$ be the restriction of $\mu^*$ to $\map {\mathfrak M} {\mu^*}$.
Then $\struct {X, \map {\mathfrak M} {\mu^*}, \mu}$ is a measure space. | First, note that $\map {\mathfrak M} {\mu^*}$ is a $\sigma$-algebra over $X$.
Next, choose $E_1, E_2 \in \map {\mathfrak M} {\mu^*}$ such that $E_1 \cap E_2 = \O$.
Thus:
{{begin-eqn}}
{{eqn | l = \map \mu {E_1 \cup E_2}
| r = \map {\mu^*} {E_1 \cup E_2}
| c = as $E_1 \cup E_2 \in \map {\mathfrak M} {\mu^*}$... | Suppose $\mu^*$ is an [[Definition:Outer Measure|outer measure]] on a set $X$.
Let $\map {\mathfrak M} {\mu^*}$ be the collection of [[Definition:Measurable Set|$\mu^*$-measurable sets]].
Let $\mu$ be the [[Definition:Restriction of Mapping|restriction]] of $\mu^*$ to $\map {\mathfrak M} {\mu^*}$.
Then $\struct {X,... | First, note that [[Measurable Sets form Sigma-Algebra|$\map {\mathfrak M} {\mu^*}$ is a $\sigma$-algebra over $X$]].
Next, choose $E_1, E_2 \in \map {\mathfrak M} {\mu^*}$ such that $E_1 \cap E_2 = \O$.
Thus:
{{begin-eqn}}
{{eqn | l = \map \mu {E_1 \cup E_2}
| r = \map {\mu^*} {E_1 \cup E_2}
| c = as $E_... | Measure Space from Outer Measure | https://proofwiki.org/wiki/Measure_Space_from_Outer_Measure | https://proofwiki.org/wiki/Measure_Space_from_Outer_Measure | [
"Measure Theory",
"Outer Measures",
"Outer Measures"
] | [
"Definition:Outer Measure",
"Definition:Measurable Set",
"Definition:Restriction/Mapping",
"Definition:Measure Space"
] | [
"Measurable Sets form Sigma-Algebra",
"Definition:Disjoint Sets",
"Definition:Additive Function (Measure Theory)",
"Additive and Countably Subadditive Function is Countably Additive",
"Definition:Countably Additive Function"
] |
proofwiki-2346 | Banach Fixed-Point Theorem | Let $\struct {M, d}$ be a complete metric space.
Let $f: M \to M$ be a contraction.
That is, there exists $q \in \hointr 0 1$ such that for all $x, y \in M$:
:$\map d {\map f x, \map f y} \le q \, \map d {x, y}$
Then there exists a unique fixed point of $f$. | === Uniqueness ===
Let $f$ have two fixed points $p_1, p_2 \in M$.
It is to be proved that $p_1 = p_2$.
{{begin-eqn}}
{{eqn | l = \map d {p_1, p_2}
| r = \map d {\map f {p_1}, \map f {p_2} }
| c = {{Defof|Fixed Point}}
}}
{{eqn | o = \le
| r = q \, \map d {p_1, p_2}
| c = {{Defof|Contraction Map... | Let $\struct {M, d}$ be a [[Definition:Complete Metric Space|complete metric space]].
Let $f: M \to M$ be a [[Definition:Contraction Mapping (Metric Space)|contraction]].
That is, there exists $q \in \hointr 0 1$ such that for all $x, y \in M$:
:$\map d {\map f x, \map f y} \le q \, \map d {x, y}$
Then there exist... | === Uniqueness ===
Let $f$ have two [[Definition:Fixed Point|fixed points]] $p_1, p_2 \in M$.
It is to be proved that $p_1 = p_2$.
{{begin-eqn}}
{{eqn | l = \map d {p_1, p_2}
| r = \map d {\map f {p_1}, \map f {p_2} }
| c = {{Defof|Fixed Point}}
}}
{{eqn | o = \le
| r = q \, \map d {p_1, p_2}
... | Banach Fixed-Point Theorem | https://proofwiki.org/wiki/Banach_Fixed-Point_Theorem | https://proofwiki.org/wiki/Banach_Fixed-Point_Theorem | [
"Banach Fixed-Point Theorem",
"Metric Spaces",
"Functional Analysis",
"Fixed Point Theorems"
] | [
"Definition:Complete Metric Space",
"Definition:Contraction Mapping (Metric Space)",
"Definition:Fixed Point"
] | [
"Definition:Fixed Point",
"Definition:Fixed Point"
] |
proofwiki-2347 | Egorov's Theorem | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $D \in \Sigma$ be such that $\map \mu D < +\infty$.
Let $\sequence {f_n}_{n \mathop \in \N}, f_n: D \to \R$ be a sequence of $\Sigma$-measurable functions.
Suppose that $f_n$ converges a.e. to $f$, for some $\Sigma$-measurable function $f: D \to \R$.
Then $f_n$ con... | Let $\epsilon \in \R_{>0}$ be arbitrary.
By definition of convergence a.e., there is a set $E \in \Sigma$ such that $E \subseteq D$, $\map \mu E = 0$ and:
:$\ds \lim_{n \mathop \to \infty} \map {f_n} x = \map f x$
for each $x \in A$ where $A = D \setminus E$ is the set difference of $D$ with $E$.
For all $n, k \in \N$,... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $D \in \Sigma$ be such that $\map \mu D < +\infty$.
Let $\sequence {f_n}_{n \mathop \in \N}, f_n: D \to \R$ be a [[Definition:Sequence|sequence]] of [[Definition:Measurable Function|$\Sigma$-measurable functions]].
Suppose that $f_n$... | Let $\epsilon \in \R_{>0}$ be arbitrary.
By definition of [[Definition:Convergence Almost Everywhere|convergence a.e.]], there is a set $E \in \Sigma$ such that $E \subseteq D$, $\map \mu E = 0$ and:
:$\ds \lim_{n \mathop \to \infty} \map {f_n} x = \map f x$
for each $x \in A$ where $A = D \setminus E$ is the [[Defi... | Egorov's Theorem | https://proofwiki.org/wiki/Egorov's_Theorem | https://proofwiki.org/wiki/Egorov's_Theorem | [
"Measure Theory"
] | [
"Definition:Measure Space",
"Definition:Sequence",
"Definition:Measurable Function",
"Definition:Convergence Almost Everywhere",
"Definition:Measurable Function",
"Definition:Almost Uniform Convergence"
] | [
"Definition:Convergence Almost Everywhere",
"Definition:Set Difference",
"Definition:Pointwise Convergence",
"Definition:Pointwise Convergence",
"Definition:Limit Superior of Sequence of Sets/Definition 2",
"Measure of Empty Set is Zero",
"Definition:Limit of Sets",
"Continuity of Measure",
"Definit... |
proofwiki-2348 | Measure of Empty Set is Zero | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\map \mu \O = 0$.
That is, $\O$ is a $\mu$-null set. | By definition of measure, there exists at least one $E \in \Sigma$ such that $\map \mu E$ is finite.
So, suppose that $E \in \Sigma$ such that $\map \mu E$ is finite.
Let $\map \mu E = x$.
Consider the sequence $\sequence {S_n}_{n \mathop \in \N} \subseteq \Sigma$ defined as:
:$S_n = \begin {cases} E & : n = 1 \\ \O & ... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Then $\map \mu \O = 0$.
That is, $\O$ is a [[Definition:Null Set|$\mu$-null set]]. | By definition of [[Definition:Measure (Measure Theory)|measure]], there exists at least one $E \in \Sigma$ such that $\map \mu E$ is [[Definition:Finite|finite]].
So, suppose that $E \in \Sigma$ such that $\map \mu E$ is [[Definition:Finite|finite]].
Let $\map \mu E = x$.
Consider the [[Definition:Sequence|sequence... | Measure of Empty Set is Zero | https://proofwiki.org/wiki/Measure_of_Empty_Set_is_Zero | https://proofwiki.org/wiki/Measure_of_Empty_Set_is_Zero | [
"Measures",
"Empty Set"
] | [
"Definition:Measure Space",
"Definition:Null Set"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Finite",
"Definition:Finite",
"Definition:Sequence"
] |
proofwiki-2349 | Convergence a.u. Implies Convergence a.e. | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f_n: D \to \R$ be a sequence of $\Sigma$-measurable functions for $D \in \Sigma$.
Let $f_n$ converge a.u. to a function $f$ on $D$.
Then $f_n$ converges a.e. to $f$. | Assume $f_n$ converge a.u. to $f$ on $D$.
Then for each $\epsilon > 0$ there is a $B_\epsilon \subseteq D$ with $\map \mu {B_\epsilon} < \epsilon$ outside of which $f_n$ converges uniformly to $f$.
Thus, $f_n$ converges pointwise to $f$ outside of each $B_\epsilon$.
Next, define $\ds B \equiv \bigcap_{n \mathop \in \N}... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f_n: D \to \R$ be a sequence of [[Definition:Measurable Function|$\Sigma$-measurable functions]] for $D \in \Sigma$.
Let $f_n$ [[Definition:Almost Uniform Convergence|converge a.u.]] to a function $f$ on $D$.
Then $f_n$ [[Definitio... | Assume $f_n$ [[Definition:Almost Uniform Convergence|converge a.u.]] to $f$ on $D$.
Then for each $\epsilon > 0$ there is a $B_\epsilon \subseteq D$ with $\map \mu {B_\epsilon} < \epsilon$ outside of which $f_n$ [[Definition:Uniform Convergence|converges uniformly]] to $f$.
Thus, $f_n$ [[Definition:Pointwise Converge... | Convergence a.u. Implies Convergence a.e. | https://proofwiki.org/wiki/Convergence_a.u._Implies_Convergence_a.e. | https://proofwiki.org/wiki/Convergence_a.u._Implies_Convergence_a.e. | [
"Almost Uniform Convergence",
"Convergence Almost Everywhere"
] | [
"Definition:Measure Space",
"Definition:Measurable Function",
"Definition:Almost Uniform Convergence",
"Definition:Convergence Almost Everywhere"
] | [
"Definition:Almost Uniform Convergence",
"Definition:Uniform Convergence",
"Definition:Pointwise Convergence",
"Definition:Sigma-Algebra",
"Definition:Monotone (Measure Theory)",
"Definition:Convergence Almost Everywhere"
] |
proofwiki-2350 | Existence of Solution of 2nd Order Linear ODE | Let $\map P x$, $\map Q x$ and $\map R x$ be continuous real functions on a closed interval $I = \closedint a b$.
Let $x_0 \in I$, and let $y_0 \in \R$ and $y_0' \in \R$ be arbitrary.
Then the initial value problem:
:$\ds \frac {\d^2 y}{\d x^2} + \map P x \frac {\d y}{\d x} + \map Q x y = \map R x, \map y {x_0} = y_0, ... | Let us introduce the variable $z = \dfrac {\d y}{\d x}$.
Then the initial ODE can be written:
:<nowiki>$\begin{cases}
\dfrac {\d y}{\d x} = z & : \map y {x_0} = y_0 \\
& \\
\dfrac {\d z}{\d x} = - \map P x z - \map Q x y + \map R x & : \map z {x_0} = y_0'
\end{cases}$</nowiki>
The converse is also true.
The result foll... | Let $\map P x$, $\map Q x$ and $\map R x$ be [[Definition:Continuous Real Function|continuous real functions]] on a [[Definition:Closed Real Interval|closed interval]] $I = \closedint a b$.
Let $x_0 \in I$, and let $y_0 \in \R$ and $y_0' \in \R$ be arbitrary.
Then the [[Definition:Initial Value Problem|initial value ... | Let us introduce the variable $z = \dfrac {\d y}{\d x}$.
Then the initial ODE can be written:
:<nowiki>$\begin{cases}
\dfrac {\d y}{\d x} = z & : \map y {x_0} = y_0 \\
& \\
\dfrac {\d z}{\d x} = - \map P x z - \map Q x y + \map R x & : \map z {x_0} = y_0'
\end{cases}$</nowiki>
The converse is also true.
The result f... | Existence of Solution of 2nd Order Linear ODE | https://proofwiki.org/wiki/Existence_of_Solution_of_2nd_Order_Linear_ODE | https://proofwiki.org/wiki/Existence_of_Solution_of_2nd_Order_Linear_ODE | [
"Differential Equations"
] | [
"Definition:Continuous Real Function",
"Definition:Real Interval/Closed",
"Definition:Initial Value Problem",
"Definition:Unique"
] | [
"Existence of Solution to System of First Order ODEs",
"Category:Differential Equations"
] |
proofwiki-2351 | Existence of Solution to System of First Order ODEs | Consider the system of initial value problems:
:<nowiki>$\begin{cases}
\dfrac {\d y} {\d x} = \map f {x, y, z} & : \map y {x_0} = y_0 \\
& \\
\dfrac {\d z} {\d x} = \map g {x, y, z} & : \map z {x_0} = z_0 \\
\end{cases}$</nowiki>
where $\map f {x, y, z}$ and $\map g {x, y, z}$ are continuous real functions in some regi... | {{proof wanted|Needs more work so as to state the problem more precisely.}}
Category:Differential Equations
dufy90jx3cpjawt9yrvf4fd6m3hux5d | Consider the [[Definition:System of Differential Equations|system]] of [[Definition:Initial Value Problem|initial value problems]]:
:<nowiki>$\begin{cases}
\dfrac {\d y} {\d x} = \map f {x, y, z} & : \map y {x_0} = y_0 \\
& \\
\dfrac {\d z} {\d x} = \map g {x, y, z} & : \map z {x_0} = z_0 \\
\end{cases}$</nowiki>
wher... | {{proof wanted|Needs more work so as to state the problem more precisely.}}
[[Category:Differential Equations]]
dufy90jx3cpjawt9yrvf4fd6m3hux5d | Existence of Solution to System of First Order ODEs | https://proofwiki.org/wiki/Existence_of_Solution_to_System_of_First_Order_ODEs | https://proofwiki.org/wiki/Existence_of_Solution_to_System_of_First_Order_ODEs | [
"Differential Equations"
] | [
"Definition:Differential Equation/System",
"Definition:Initial Value Problem",
"Definition:Continuous Real Function",
"Definition:Differential Equation/System",
"Definition:Solution of System of Differential Equations",
"Definition:Real Interval/Closed"
] | [
"Category:Differential Equations"
] |
proofwiki-2352 | Cartesian Metric is Rotation Invariant | The cartesian metric does not change under rotation. | Let the cartesian metric be $\delta_{ij} = \innerprod {e_i} {e_j}$.
Also, let $\delta_{ij}'$ be the metric of the coordinate system of $\delta_{ij}$ rotated by a rotation matrix $A$.
Then, $\delta_{ij}' = \innerprod {A e_i} {A e_j}$.
$\innerprod {A x} y = \innerprod x {A^T y}$, see Factor Matrix in the Inner Product, s... | The cartesian [[Definition:Metric|metric]] does not change under rotation. | Let the cartesian metric be $\delta_{ij} = \innerprod {e_i} {e_j}$.
Also, let $\delta_{ij}'$ be the metric of the coordinate system of $\delta_{ij}$ rotated by a rotation matrix $A$.
Then, $\delta_{ij}' = \innerprod {A e_i} {A e_j}$.
$\innerprod {A x} y = \innerprod x {A^T y}$, see [[Factor Matrix in the Inner Produ... | Cartesian Metric is Rotation Invariant | https://proofwiki.org/wiki/Cartesian_Metric_is_Rotation_Invariant | https://proofwiki.org/wiki/Cartesian_Metric_is_Rotation_Invariant | [
"Metric Spaces"
] | [
"Definition:Metric Space/Metric"
] | [
"Factor Matrix in the Inner Product",
"Definition:Unit Matrix",
"Definition:Euclidean Space",
"Category:Metric Spaces"
] |
proofwiki-2353 | Little Bézout Theorem | Let $\map {P_n} x$ be a polynomial of degree $n$ in $x$.
Let $a$ be a constant.
Then the remainder of $\map {P_n} x$ when divided by $x - a$ is equal to $\map {P_n} a$. | By the process of Polynomial Long Division, we can express $\map {P_n} x$ as:
:$(1): \quad \map {P_n} x = \paren {x - a} \map {Q_{n - 1} } x + R$
where:
:$\map {Q_{n - 1} } x$ is a polynomial in $x$ of degree $n - 1$
:$R$ is a polynomial in $x$ of degree no greater than $0$; that is, a constant.
It follows that, by set... | Let $\map {P_n} x$ be a [[Definition:Polynomial|polynomial]] of [[Definition:Degree of Polynomial|degree]] $n$ in $x$.
Let $a$ be a [[Definition:Constant|constant]].
Then the [[Definition:Remainder (Polynomial Long Division)|remainder]] of $\map {P_n} x$ when divided by $x - a$ is equal to $\map {P_n} a$. | By the process of [[Polynomial Long Division]], we can express $\map {P_n} x$ as:
:$(1): \quad \map {P_n} x = \paren {x - a} \map {Q_{n - 1} } x + R$
where:
:$\map {Q_{n - 1} } x$ is a [[Definition:Polynomial|polynomial]] in $x$ of [[Definition:Degree of Polynomial|degree]] $n - 1$
:$R$ is a [[Definition:Polynomial|... | Little Bézout Theorem | https://proofwiki.org/wiki/Little_Bézout_Theorem | https://proofwiki.org/wiki/Little_Bézout_Theorem | [
"Little Bézout Theorem",
"Polynomial Theory",
"Algebra"
] | [
"Definition:Polynomial",
"Definition:Degree of Polynomial",
"Definition:Constant",
"Polynomial Long Division"
] | [
"Polynomial Long Division",
"Definition:Polynomial",
"Definition:Degree of Polynomial",
"Definition:Polynomial",
"Definition:Degree of Polynomial",
"Definition:Constant"
] |
proofwiki-2354 | Thales' Theorem | Let $A$ and $B$ be two points on opposite ends of the diameter of a circle.
Let $C$ be another point on the circle such that $C \ne A, B$.
Then the lines $AC$ and $BC$ are perpendicular to each other.
:400px | :400px
Let $O$ be the center of the circle, and define the vectors:
:$\mathbf u = \overrightarrow{OC}$
:$\mathbf v = \overrightarrow{OB}$
:$\mathbf w = \overrightarrow{OA}$
Thus:
:$\overrightarrow{AC} = \mathbf u - \mathbf w$
:$\overrightarrow{BC} = \mathbf u - \mathbf v$
Then:
{{begin-eqn}}
{{eqn | o =
| r = \ov... | Let $A$ and $B$ be two [[Definition:Point|points]] on opposite ends of the [[Definition:Diameter of Circle|diameter]] of a [[Definition:Circle|circle]].
Let $C$ be another [[Definition:Point|point]] on the [[Definition:Circle|circle]] such that $C \ne A, B$.
Then the lines $AC$ and $BC$ are [[Definition:Perpendicula... | :[[File:Thales theorem.jpg|400px]]
Let $O$ be the [[Definition:Center of Circle|center]] of the circle, and define the [[Definition:Plane Vector|vectors]]:
:$\mathbf u = \overrightarrow{OC}$
:$\mathbf v = \overrightarrow{OB}$
:$\mathbf w = \overrightarrow{OA}$
Thus:
:$\overrightarrow{AC} = \mathbf u - \mathbf w$
:$\o... | Thales' Theorem/Proof 1 | https://proofwiki.org/wiki/Thales'_Theorem | https://proofwiki.org/wiki/Thales'_Theorem/Proof_1 | [
"Circles",
"Thales' Theorem"
] | [
"Definition:Point",
"Definition:Circle/Diameter",
"Definition:Circle",
"Definition:Point",
"Definition:Circle",
"Definition:Right Angle/Perpendicular",
"File:Thales-Theorem.png"
] | [
"File:Thales theorem.jpg",
"Definition:Circle/Center",
"Definition:Vector/Real Euclidean Space/Plane Vector",
"Definition:Dot Product",
"Dot Product Distributes over Addition",
"Dot Product Operator is Commutative",
"Dot Product of Vector with Itself",
"Non-Zero Vectors are Orthogonal iff Perpendicula... |
proofwiki-2355 | Thales' Theorem | Let $A$ and $B$ be two points on opposite ends of the diameter of a circle.
Let $C$ be another point on the circle such that $C \ne A, B$.
Then the lines $AC$ and $BC$ are perpendicular to each other.
:400px | Let $O$ be the center of $ACB$.
From the Inscribed Angle Theorem, $\angle AOB = 2 \angle ACB$.
Then we have that $\angle AOB$ is a straight angle.
Hence the result.
{{qed}} | Let $A$ and $B$ be two [[Definition:Point|points]] on opposite ends of the [[Definition:Diameter of Circle|diameter]] of a [[Definition:Circle|circle]].
Let $C$ be another [[Definition:Point|point]] on the [[Definition:Circle|circle]] such that $C \ne A, B$.
Then the lines $AC$ and $BC$ are [[Definition:Perpendicula... | Let $O$ be the [[Definition:Center of Semicircle|center]] of $ACB$.
From the [[Inscribed Angle Theorem]], $\angle AOB = 2 \angle ACB$.
Then we have that $\angle AOB$ is a [[Definition:Straight Angle|straight angle]].
Hence the result.
{{qed}} | Thales' Theorem/Proof 2 | https://proofwiki.org/wiki/Thales'_Theorem | https://proofwiki.org/wiki/Thales'_Theorem/Proof_2 | [
"Circles",
"Thales' Theorem"
] | [
"Definition:Point",
"Definition:Circle/Diameter",
"Definition:Circle",
"Definition:Point",
"Definition:Circle",
"Definition:Right Angle/Perpendicular",
"File:Thales-Theorem.png"
] | [
"Definition:Circle/Semicircle/Center",
"Inscribed Angle Theorem",
"Definition:Straight Angle"
] |
proofwiki-2356 | Thales' Theorem | Let $A$ and $B$ be two points on opposite ends of the diameter of a circle.
Let $C$ be another point on the circle such that $C \ne A, B$.
Then the lines $AC$ and $BC$ are perpendicular to each other.
:400px | Let $O$ be the center of $ACB$.
We have that $AO$, $BO$ and $CO$ are radii of the same circle.
Thus by definition:
:$AO = BO = CO$
Thus $\triangle AOC$ and $\triangle BOC$ are isosceles triangles.
From External Angle of Triangle equals Sum of other Internal Angles:
:$\angle COB = \angle ACO + \angle CAO$
Hence:
{{begin... | Let $A$ and $B$ be two [[Definition:Point|points]] on opposite ends of the [[Definition:Diameter of Circle|diameter]] of a [[Definition:Circle|circle]].
Let $C$ be another [[Definition:Point|point]] on the [[Definition:Circle|circle]] such that $C \ne A, B$.
Then the lines $AC$ and $BC$ are [[Definition:Perpendicula... | Let $O$ be the [[Definition:Center of Semicircle|center]] of $ACB$.
We have that $AO$, $BO$ and $CO$ are [[Definition:Radius of Circle|radii]] of the same [[Definition:Circle|circle]].
Thus by definition:
:$AO = BO = CO$
Thus $\triangle AOC$ and $\triangle BOC$ are [[Definition:Isosceles Triangle|isosceles triangles... | Thales' Theorem/Proof 3 | https://proofwiki.org/wiki/Thales'_Theorem | https://proofwiki.org/wiki/Thales'_Theorem/Proof_3 | [
"Circles",
"Thales' Theorem"
] | [
"Definition:Point",
"Definition:Circle/Diameter",
"Definition:Circle",
"Definition:Point",
"Definition:Circle",
"Definition:Right Angle/Perpendicular",
"File:Thales-Theorem.png"
] | [
"Definition:Circle/Semicircle/Center",
"Definition:Circle/Radius",
"Definition:Circle",
"Definition:Triangle (Geometry)/Isosceles",
"External Angle of Triangle equals Sum of other Internal Angles",
"Isosceles Triangle has Two Equal Angles",
"External Angle of Triangle equals Sum of other Internal Angles... |
proofwiki-2357 | Edge of Polyhedron has no Curvature | The edge of a polyhedron has zero curvature. | File:Curvature.png
Let $X$ and $Y$ be two separate faces of a polyhedron separated by the edge $l$.
Let $P$ be a point on $X$ and let $Q$ be a point on $Y$.
The curvature inside an infinitesimal region $\delta a$ is given by the net angular displacement $\delta \theta$ a vector $v$ experiences as it is parallel transpo... | The [[Definition:Edge of Polyhedron|edge]] of a [[Definition:Polyhedron|polyhedron]] has zero [[Definition:Curvature|curvature]]. | [[File:Curvature.png]]
Let $X$ and $Y$ be two separate [[Definition:Face of Polyhedron|faces]] of a [[Definition:Polyhedron|polyhedron]] separated by the [[Definition:Edge of Polyhedron|edge]] $l$.
Let $P$ be a [[Definition:Point|point]] on $X$ and let $Q$ be a [[Definition:Point|point]] on $Y$.
The curvature inside... | Edge of Polyhedron has no Curvature | https://proofwiki.org/wiki/Edge_of_Polyhedron_has_no_Curvature | https://proofwiki.org/wiki/Edge_of_Polyhedron_has_no_Curvature | [
"Differential Geometry"
] | [
"Definition:Polyhedron/Edge",
"Definition:Polyhedron",
"Definition:Curvature"
] | [
"File:Curvature.png",
"Definition:Polyhedron/Face",
"Definition:Polyhedron",
"Definition:Polyhedron/Edge",
"Definition:Point",
"Definition:Point",
"Definition:Infinitesimal",
"Category:Differential Geometry"
] |
proofwiki-2358 | Dot Product is Inner Product | The dot product is an inner product. | Let $\mathbf u, \mathbf v \in \R^n$.
We will check the four defining properties of an inner product in turn. | The [[Definition:Dot Product|dot product]] is an [[Definition:Inner Product|inner product]]. | Let $\mathbf u, \mathbf v \in \R^n$.
We will check the four defining properties of an [[Definition:Inner Product|inner product]] in turn. | Dot Product is Inner Product | https://proofwiki.org/wiki/Dot_Product_is_Inner_Product | https://proofwiki.org/wiki/Dot_Product_is_Inner_Product | [
"Dot Product",
"Inner Products"
] | [
"Definition:Dot Product",
"Definition:Inner Product"
] | [
"Definition:Inner Product",
"Definition:Inner Product"
] |
proofwiki-2359 | Factor Matrix in the Inner Product | Let $\mathbf u$ and $\mathbf v$ be $1 \times n$ column vectors.
Then:
:$\innerprod {A \mathbf u} {\mathbf v} = \innerprod {\mathbf u} {A^\intercal \mathbf v}$ | {{begin-eqn}}
{{eqn|l =\innerprod {A \mathbf u} {\mathbf v}
|r =\paren {A \mathbf u}^\intercal \mathbf v
|c = {{Defof|Dot Product}}
}}
{{eqn|r = \mathbf u^\intercal A^\intercal \mathbf v
|c = Transpose of Matrix Product
}}
{{eqn|r = \innerprod {\mathbf u} {A^\intercal \mathbf v}
|c = {{Defof|Dot Pro... | Let $\mathbf u$ and $\mathbf v$ be $1 \times n$ [[Definition:Column Matrix|column vectors]].
Then:
:$\innerprod {A \mathbf u} {\mathbf v} = \innerprod {\mathbf u} {A^\intercal \mathbf v}$ | {{begin-eqn}}
{{eqn|l =\innerprod {A \mathbf u} {\mathbf v}
|r =\paren {A \mathbf u}^\intercal \mathbf v
|c = {{Defof|Dot Product}}
}}
{{eqn|r = \mathbf u^\intercal A^\intercal \mathbf v
|c = [[Transpose of Matrix Product]]
}}
{{eqn|r = \innerprod {\mathbf u} {A^\intercal \mathbf v}
|c = {{Defof|Dot... | Factor Matrix in the Inner Product | https://proofwiki.org/wiki/Factor_Matrix_in_the_Inner_Product | https://proofwiki.org/wiki/Factor_Matrix_in_the_Inner_Product | [
"Vector Algebra"
] | [
"Definition:Column Matrix"
] | [
"Transpose of Matrix Product",
"Category:Vector Algebra"
] |
proofwiki-2360 | Empty Set Disjoint with Itself | The empty set is disjoint with itself:
:$\O \cap \O = \O$ | From Intersection with Empty Set, for all sets $S$, $S \cap \O = \O$.
The result follows from the definition of disjoint sets.
{{qed}}
Category:Empty Set
2amrsj1on7qtxcgdc0dpnhjrv8ftuxe | The [[Definition:Empty Set|empty set]] is [[Definition:Disjoint Sets|disjoint]] with itself:
:$\O \cap \O = \O$ | From [[Intersection with Empty Set]], for all [[Definition:Set|sets]] $S$, $S \cap \O = \O$.
The result follows from the definition of [[Definition:Disjoint Sets|disjoint sets]].
{{qed}}
[[Category:Empty Set]]
2amrsj1on7qtxcgdc0dpnhjrv8ftuxe | Empty Set Disjoint with Itself | https://proofwiki.org/wiki/Empty_Set_Disjoint_with_Itself | https://proofwiki.org/wiki/Empty_Set_Disjoint_with_Itself | [
"Empty Set"
] | [
"Definition:Empty Set",
"Definition:Disjoint Sets"
] | [
"Intersection with Empty Set",
"Definition:Set",
"Definition:Disjoint Sets",
"Category:Empty Set"
] |
proofwiki-2361 | Overflow Theorem | Let $F$ be a set of first-order formulas which has finite models of arbitrarily large size.
Then $F$ has an infinite model. | For each $n$, let $\mathbf A_n$ be the formula:
:$\exists x_1 \exists x_2 \ldots \exists x_n: \paren {x_1 \ne x_2 \land x_1 \ne x_3 \land \ldots \land x_{n - 1} \ne x_n}$
Then $\mathbf A_i$ is true in a structure $\AA$ {{iff}} $\AA$ has at least $n$ elements.
Take:
:$\ds \Gamma := A \cup \bigcup_{i \mathop = 1}^\infty ... | Let $F$ be a set of [[Definition:First-Order Formula|first-order formulas]] which has [[Definition:Finite|finite]] [[Definition:Model (Predicate Logic)|models]] of arbitrarily large size.
Then $F$ has an [[Definition:Infinite|infinite]] [[Definition:Model (Predicate Logic)|model]]. | For each $n$, let $\mathbf A_n$ be the formula:
:$\exists x_1 \exists x_2 \ldots \exists x_n: \paren {x_1 \ne x_2 \land x_1 \ne x_3 \land \ldots \land x_{n - 1} \ne x_n}$
Then $\mathbf A_i$ is true in a [[Definition:First-Order Structure|structure]] $\AA$ {{iff}} $\AA$ has at least $n$ [[Definition:Element|elements]]... | Overflow Theorem | https://proofwiki.org/wiki/Overflow_Theorem | https://proofwiki.org/wiki/Overflow_Theorem | [
"Mathematical Logic",
"Named Theorems"
] | [
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Finite",
"Definition:Model (Predicate Logic)",
"Definition:Infinite",
"Definition:Model (Predicate Logic)"
] | [
"Definition:Structure for Predicate Logic",
"Definition:Element",
"Definition:Model (Predicate Logic)",
"Definition:Finite Subset",
"Definition:Satisfiable/Set of Formulas",
"Compactness Theorem",
"Definition:Satisfiable/Set of Formulas",
"Definition:Model (Predicate Logic)",
"Definition:Infinite",
... |
proofwiki-2362 | Existence of Non-Standard Models of Arithmetic | There exist non-standard models of arithmetic. | Let $P$ be the set of axioms of Peano arithmetic.
Let $Q = P \cup \left\{{\neg x = 0, \neg x = s0, \neg x = ss0, \ldots}\right\}$ where $x$ is a variable of the language.
Then each finite subset of $Q$ is satisfied by the standard model of arithmetic
Hence $Q$ is satisfiable by the Compactness theorem.
But any model sa... | There exist non-standard models of arithmetic. | Let $P$ be the set of axioms of [[Definition:Peano Arithmetic|Peano arithmetic]].
Let $Q = P \cup \left\{{\neg x = 0, \neg x = s0, \neg x = ss0, \ldots}\right\}$ where $x$ is a variable of the language.
Then each finite subset of $Q$ is satisfied by the standard model of arithmetic
Hence $Q$ is satisfiable by the [[... | Existence of Non-Standard Models of Arithmetic | https://proofwiki.org/wiki/Existence_of_Non-Standard_Models_of_Arithmetic | https://proofwiki.org/wiki/Existence_of_Non-Standard_Models_of_Arithmetic | [
"Mathematical Logic"
] | [] | [
"Definition:Peano Arithmetic",
"Compactness Theorem",
"Definition:Model (Logic)",
"Category:Mathematical Logic"
] |
proofwiki-2363 | Pi is Irrational | Pi ($\pi$) is irrational. | {{AimForCont}} $\pi$ is rational.
Then from Existence of Canonical Form of Rational Number:
:$\exists a \in \Z, b \in \Z_{>0}: \pi = \dfrac a b$
Let $n \in \Z_{>0}$.
We define the polynomial function:
:$\forall x \in \R: \map f x = \dfrac {x^n \paren {a - b x}^n} {n!}$
We differentiate this $2 n$ times, and then we bui... | [[Definition:Pi|Pi ($\pi$)]] is [[Definition:Irrational Number|irrational]]. | {{AimForCont}} $\pi$ is [[Definition:Rational Number|rational]].
Then from [[Existence of Canonical Form of Rational Number]]:
:$\exists a \in \Z, b \in \Z_{>0}: \pi = \dfrac a b$
Let $n \in \Z_{>0}$.
We define the [[Definition:Real Polynomial Function|polynomial function]]:
:$\forall x \in \R: \map f x = \dfrac {x... | Pi is Irrational/Proof 1 | https://proofwiki.org/wiki/Pi_is_Irrational | https://proofwiki.org/wiki/Pi_is_Irrational/Proof_1 | [
"Pi",
"Irrationality Proofs",
"Pi is Irrational"
] | [
"Definition:Pi",
"Definition:Irrational Number"
] | [
"Definition:Rational Number",
"Existence of Canonical Form of Rational Number",
"Definition:Polynomial Function/Real",
"Definition:Differentiation",
"Definition:Alternating Series",
"Definition:Even Integer",
"Definition:Derivative",
"Derivative of Composite Function",
"Principle of Mathematical Ind... |
proofwiki-2364 | Pi is Irrational | Pi ($\pi$) is irrational. | {{AimForCont}} $\pi$ is rational.
We establish a lemma:
=== Lemma ===
{{:Pi is Irrational/Proof 2/Lemma}}{{qed|lemma}}
We will use the definition of $A_n$ from the lemma.
Then we will deduce that $A_n$ is an integer for all $n$.
First we confirm by direct integration that $A_0$ and $A_1$ are integers:
{{begin-eqn}}
{{e... | [[Definition:Pi|Pi ($\pi$)]] is [[Definition:Irrational Number|irrational]]. | {{AimForCont}} $\pi$ is [[Definition:Rational Number|rational]].
We establish a [[Definition:Lemma|lemma]]:
=== [[Pi is Irrational/Proof 2/Lemma|Lemma]] ===
{{:Pi is Irrational/Proof 2/Lemma}}{{qed|lemma}}
We will use the definition of $A_n$ from the [[Pi is Irrational/Proof 2/Lemma|lemma]].
Then we will deduce t... | Pi is Irrational/Proof 2 | https://proofwiki.org/wiki/Pi_is_Irrational | https://proofwiki.org/wiki/Pi_is_Irrational/Proof_2 | [
"Pi",
"Irrationality Proofs",
"Pi is Irrational"
] | [
"Definition:Pi",
"Definition:Irrational Number"
] | [
"Definition:Rational Number",
"Definition:Lemma",
"Pi is Irrational/Proof 2/Lemma",
"Pi is Irrational/Proof 2/Lemma",
"Definition:Integer",
"Definition:Integer",
"Area under Arc of Sine Function",
"Linear Combination of Integrals",
"Primitive of Power of x by Sine of a x",
"Cosine of Straight Angl... |
proofwiki-2365 | Pi is Irrational | Pi ($\pi$) is irrational. | From Rational Points on Graph of Sine Function, the only rational point on the graph of the sine function in the real Cartesian plane $\R^2$:
:$f := \left\{ {\left({x, y}\right) \in \R^2: y = \sin x}\right\}$
is the point $\left({0, 0}\right)$.
But $\left({\pi, 0}\right)$ is also on $f$.
Hence $\pi$ cannot be rational.... | [[Definition:Pi|Pi ($\pi$)]] is [[Definition:Irrational Number|irrational]]. | From [[Rational Points on Graph of Sine Function]], the only [[Definition:Rational Point in Plane|rational point]] on the [[Definition:Graph of Mapping|graph]] of the [[Definition:Sine Function|sine function]] in the [[Definition:Cartesian Plane|real Cartesian plane]] $\R^2$:
:$f := \left\{ {\left({x, y}\right) \in \R^... | Pi is Irrational/Proof 3 | https://proofwiki.org/wiki/Pi_is_Irrational | https://proofwiki.org/wiki/Pi_is_Irrational/Proof_3 | [
"Pi",
"Irrationality Proofs",
"Pi is Irrational"
] | [
"Definition:Pi",
"Definition:Irrational Number"
] | [
"Rational Points on Graph of Sine Function",
"Definition:Rational Point in Plane",
"Definition:Graph of Mapping",
"Definition:Sine",
"Definition:Cartesian Plane",
"Definition:Rational Number"
] |
proofwiki-2366 | Properties of Hadamard Product | Let $\map {\MM_S} {m, n}$ be a $m \times n$ matrix space over $S$ over an algebraic structure $\struct {S, \cdot}$.
Let $\mathbf A, \mathbf B \in \map {\MM_S} {m, n}$.
Let $\mathbf A \circ \mathbf B$ be defined as the Hadamard product of $\mathbf A$ and $\mathbf B$.
The operation $\circ$ of Hadamard product satisfies t... | === Closure of Hadamard Product ===
{{:Closure of Hadamard Product}} | Let $\map {\MM_S} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $S$ over an [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, \cdot}$.
Let $\mathbf A, \mathbf B \in \map {\MM_S} {m, n}$.
Let $\mathbf A \circ \mathbf B$ be defined as the [[Definition:Hadamard Product|Hadamar... | === [[Closure of Hadamard Product]] ===
{{:Closure of Hadamard Product}} | Properties of Hadamard Product | https://proofwiki.org/wiki/Properties_of_Hadamard_Product | https://proofwiki.org/wiki/Properties_of_Hadamard_Product | [
"Hadamard Product"
] | [
"Definition:Matrix Space",
"Definition:Algebraic Structure",
"Definition:Hadamard Product",
"Definition:Hadamard Product",
"Definition:Closure (Abstract Algebra)",
"Definition:Closure (Abstract Algebra)",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:Commutative... | [
"Closure of Hadamard Product"
] |
proofwiki-2367 | Matrix Space is Module | Let $\struct {R, +, \circ}$ be a ring.
Let $\mathbf A = \sqbrk a_{m n}$ be an $m \times n$ matrix over $\struct {R, +, \circ}$.
Then the matrix space $\map {\MM_R} {m, n}$ of all $m \times n$ matrices over $R$ is a module. | This follows as $\map {\MM_R} {m, n}$ is a direct instance of the module given in the module of all mappings, where $\map {\MM_R} {m, n}$ is the $R$-module $R^{\closedint 1 m \times \closedint 1 n}$.
The $S$ of that example is the set $\closedint 1 m \times \closedint 1 n$, while the $G$ of that example is the $R$-modu... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\mathbf A = \sqbrk a_{m n}$ be an [[Definition:Matrix|$m \times n$ matrix]] over $\struct {R, +, \circ}$.
Then the [[Definition:Matrix Space|matrix space $\map {\MM_R} {m, n}$ of all $m \times n$ matrices]] over $R$ is a [[Definition:... | This follows as $\map {\MM_R} {m, n}$ is a direct instance of the [[Definition:Module over Ring|module]] given in the [[Definition:Module of All Mappings|module of all mappings]], where $\map {\MM_R} {m, n}$ is the [[Definition:Module on Cartesian Product|$R$-module $R^{\closedint 1 m \times \closedint 1 n}$]].
The $S... | Matrix Space is Module | https://proofwiki.org/wiki/Matrix_Space_is_Module | https://proofwiki.org/wiki/Matrix_Space_is_Module | [
"Matrix Algebra",
"Matrix Spaces"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Matrix",
"Definition:Matrix Space",
"Definition:Module over Ring"
] | [
"Definition:Module over Ring",
"Definition:Module of All Mappings",
"Definition:Module on Cartesian Product",
"Definition:Module on Cartesian Product/Special Case",
"Definition:Matrix/Element",
"Definition:Matrix",
"Definition:Element",
"Definition:Abuse of Notation"
] |
proofwiki-2368 | Graph Isomorphism is Equivalence Relation | Graph isomorphism is an equivalence relation. | From the formal definitions:
=== Simple Graph ===
A (simple) graph $G$ is a non-empty set $V$ together with an antireflexive, symmetric relation $\RR$ on $V$.
=== Digraph ===
A digraph $D$ is a non-empty set $V$ together with an antireflexive relation $\RR$ on $V$.
=== Loop-graph ===
A loop-graph $P$ is a non-empty set... | [[Definition:Isomorphism (Graph Theory)|Graph isomorphism]] is an [[Definition:Equivalence Relation|equivalence relation]]. | From the formal definitions:
=== Simple Graph ===
A [[Definition:Simple Graph/Formal Definition|(simple) graph]] $G$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]] $V$ together with an [[Definition:Antireflexive Relation|antireflexive]], [[Definition:Symmetric Relation|symmetric]] [[Definition:Re... | Graph Isomorphism is Equivalence Relation/Proof 1 | https://proofwiki.org/wiki/Graph_Isomorphism_is_Equivalence_Relation | https://proofwiki.org/wiki/Graph_Isomorphism_is_Equivalence_Relation/Proof_1 | [
"Graph Isomorphisms",
"Examples of Equivalence Relations",
"Graph Isomorphism is Equivalence Relation"
] | [
"Definition:Isomorphism (Graph Theory)",
"Definition:Equivalence Relation"
] | [
"Definition:Simple Graph/Formal Definition",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Antireflexive Relation",
"Definition:Symmetric Relation",
"Definition:Relation",
"Definition:Digraph/Formal Definition",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Antireflexive Re... |
proofwiki-2369 | Graph Isomorphism is Equivalence Relation | Graph isomorphism is an equivalence relation. | In the following, let:
{{begin-eqn}}
{{eqn | l = G_1
| r = \struct {\map V {G_1}, \map E {G_1} }
}}
{{eqn | l = G_2
| r = \struct {\map V {G_2}, \map E {G_2} }
}}
{{eqn | l = G_3
| r = \struct {\map V {G_3}, \map E {G_3} }
}}
{{end-eqn}}
be arbitrary graphs.
Checking in turn each of the criteria for e... | [[Definition:Isomorphism (Graph Theory)|Graph isomorphism]] is an [[Definition:Equivalence Relation|equivalence relation]]. | In the following, let:
{{begin-eqn}}
{{eqn | l = G_1
| r = \struct {\map V {G_1}, \map E {G_1} }
}}
{{eqn | l = G_2
| r = \struct {\map V {G_2}, \map E {G_2} }
}}
{{eqn | l = G_3
| r = \struct {\map V {G_3}, \map E {G_3} }
}}
{{end-eqn}}
be arbitrary [[Definition:Graph (Graph Theory)|graphs]].
Chec... | Graph Isomorphism is Equivalence Relation/Proof 2 | https://proofwiki.org/wiki/Graph_Isomorphism_is_Equivalence_Relation | https://proofwiki.org/wiki/Graph_Isomorphism_is_Equivalence_Relation/Proof_2 | [
"Graph Isomorphisms",
"Examples of Equivalence Relations",
"Graph Isomorphism is Equivalence Relation"
] | [
"Definition:Isomorphism (Graph Theory)",
"Definition:Equivalence Relation"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Equivalence Relation",
"Definition:Identity Mapping",
"Definition:Bijection",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Adjacent (Graph Theory)/Vertices",
"Definition:Isomorphism (Graph Theory)",
"Definition:Reflexive Relation",
"Definition:... |
proofwiki-2370 | Relation Isomorphism is Equivalence Relation | Relation isomorphism is an equivalence relation. | Let $\struct {S_1, \RR_1}$, $\struct {S_2, \RR_2}$ and $\struct {S_3, \RR_3}$ be relational structures.
Let $S \cong T$ denote the relation that $S$ is (relation) isomorphic to $T$.
Checking in turn each of the criteria for equivalence: | [[Definition:Relation Isomorphism|Relation isomorphism]] is an [[Definition:Equivalence Relation|equivalence relation]]. | Let $\struct {S_1, \RR_1}$, $\struct {S_2, \RR_2}$ and $\struct {S_3, \RR_3}$ be [[Definition:Relational Structure|relational structures]].
Let $S \cong T$ denote the relation that $S$ is [[Definition:Relation Isomorphism|(relation) isomorphic]] to $T$.
Checking in turn each of the criteria for [[Definition:Equivale... | Relation Isomorphism is Equivalence Relation | https://proofwiki.org/wiki/Relation_Isomorphism_is_Equivalence_Relation | https://proofwiki.org/wiki/Relation_Isomorphism_is_Equivalence_Relation | [
"Graph Theory",
"Equivalence Relations",
"Relation Isomorphisms",
"Examples of Equivalence Relations"
] | [
"Definition:Relation Isomorphism",
"Definition:Equivalence Relation"
] | [
"Definition:Relational Structure",
"Definition:Relation Isomorphism",
"Definition:Equivalence Relation",
"Definition:Relation Isomorphism",
"Definition:Relation Isomorphism",
"Definition:Relation Isomorphism",
"Definition:Relation Isomorphism",
"Definition:Relation Isomorphism",
"Definition:Relation... |
proofwiki-2371 | Vertex Condition for Isomorphic Graphs | Let $G_1$ and $G_2$ be isomorphic graphs.
Then the degrees of the vertices of $G_1$ are exactly the same as the degrees of the vertices of $G_2$. | Let $\phi: \map V {G_1} \to \map V {G_2}$ be an isomorphism.
Let $u \in \map V {G_1}$ be an arbitrary vertex of $G_1$ such that $\map \phi u = v \in \map V {G_2}$.
Let $\map {\deg_{G_1} } u = n$.
We need to show that $\map {\deg_{G_2} } v = n$.
As $\map {\deg_{G_1} } u = n$, there exist $u_1, u_2, \ldots, u_n \in \map ... | Let $G_1$ and $G_2$ be [[Definition:Isomorphism (Graph Theory)|isomorphic]] [[Definition:Graph (Graph Theory)|graphs]].
Then the [[Definition:Degree of Vertex|degrees]] of the [[Definition:Vertex of Graph|vertices]] of $G_1$ are exactly the same as the [[Definition:Degree of Vertex|degrees]] of the [[Definition:Vertex... | Let $\phi: \map V {G_1} \to \map V {G_2}$ be an [[Definition:Isomorphism (Graph Theory)|isomorphism]].
Let $u \in \map V {G_1}$ be an arbitrary [[Definition:Vertex of Graph|vertex]] of $G_1$ such that $\map \phi u = v \in \map V {G_2}$.
Let $\map {\deg_{G_1} } u = n$.
We need to show that $\map {\deg_{G_2} } v = n$.... | Vertex Condition for Isomorphic Graphs | https://proofwiki.org/wiki/Vertex_Condition_for_Isomorphic_Graphs | https://proofwiki.org/wiki/Vertex_Condition_for_Isomorphic_Graphs | [
"Graph Isomorphisms",
"Degrees of Vertices"
] | [
"Definition:Isomorphism (Graph Theory)",
"Definition:Graph (Graph Theory)",
"Definition:Degree of Vertex",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Degree of Vertex",
"Definition:Graph (Graph Theory)/Vertex"
] | [
"Definition:Isomorphism (Graph Theory)",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Adjacent (Graph Theory)/Vertices",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Adjacent (Graph Theory)/Vertices",
"Definition:Isomorphism (Graph Theory)",
"Definition:Graph (Graph Theory)/Vertex",
... |
proofwiki-2372 | Dirichlet's Box Principle | Let $S$ be a finite set whose cardinality is $n$.
Let $S_1, S_2, \ldots, S_k$ be a partition of $S$ into $k$ subsets.
Then:
:at least one subset $S_i$ of $S$ contains at least $\ceiling {\dfrac n k}$ elements
where $\ceiling {\, \cdot \,}$ denotes the ceiling function. | {{AimForCont}} no subset $S_i$ of $S$ has as many as $\ceiling {\dfrac n k}$ elements.
Then the maximum number of elements of any $S_i$ would be $\ceiling {\dfrac n k} - 1$.
So the total number of elements of $S$ would be no more than $k \paren {\ceiling {\dfrac n k} - 1} = k \ceiling {\dfrac n k} - k$.
There are two c... | Let $S$ be a [[Definition:Finite Set|finite set]] whose [[Definition:Cardinality|cardinality]] is $n$.
Let $S_1, S_2, \ldots, S_k$ be a [[Definition:Set Partition|partition]] of $S$ into $k$ [[Definition:Subset|subsets]].
Then:
:at least one [[Definition:Subset|subset]] $S_i$ of $S$ contains at least $\ceiling {\dfra... | {{AimForCont}} no [[Definition:Subset|subset]] $S_i$ of $S$ has as many as $\ceiling {\dfrac n k}$ [[Definition:Element|elements]].
Then the maximum number of [[Definition:Element|elements]] of any $S_i$ would be $\ceiling {\dfrac n k} - 1$.
So the total number of [[Definition:Element|elements]] of $S$ would be no mo... | Dirichlet's Box Principle | https://proofwiki.org/wiki/Dirichlet's_Box_Principle | https://proofwiki.org/wiki/Dirichlet's_Box_Principle | [
"Dirichlet's Box Principle",
"Combinatorics"
] | [
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Set Partition",
"Definition:Subset",
"Definition:Subset",
"Definition:Element",
"Definition:Ceiling Function"
] | [
"Definition:Subset",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Integer",
"Definition:Subset",
"Definition:Element",
"Definition:Subset",
"Definition:Element",
"Proof by Contradic... |
proofwiki-2373 | Graph Components are Equivalence Classes | The components of a graph are equivalence classes under the relation '''is connected to''' on the set of vertices. | We have that Graph Connectedness is Equivalence Relation.
The result follows directly from the definition of component.
{{qed}} | The [[Definition:Component of Graph|components]] of a [[Definition:Graph (Graph Theory)|graph]] are [[Definition:Equivalence Class|equivalence classes]] under the relation '''is [[Definition:Connected Vertices|connected]] to''' on the set of [[Definition:Vertex of Graph|vertices]]. | We have that [[Graph Connectedness is Equivalence Relation]].
The result follows directly from the definition of [[Definition:Component of Graph|component]].
{{qed}} | Graph Components are Equivalence Classes | https://proofwiki.org/wiki/Graph_Components_are_Equivalence_Classes | https://proofwiki.org/wiki/Graph_Components_are_Equivalence_Classes | [
"Graph Theory",
"Examples of Equivalence Classes"
] | [
"Definition:Component of Graph",
"Definition:Graph (Graph Theory)",
"Definition:Equivalence Class",
"Definition:Connected (Graph Theory)/Vertices",
"Definition:Graph (Graph Theory)/Vertex"
] | [
"Graph Connectedness is Equivalence Relation",
"Definition:Component of Graph"
] |
proofwiki-2374 | Graph Connectedness is Equivalence Relation | Let $G = \struct {V, E}$ be a graph.
Let $\to$ denote the relation '''is connected to''' on the set $V$.
Then $\to$ is an equivalence relation. | Let $u, v, w$ be arbitrary vertices of a graph $G$.
Checking in turn each of the criteria for equivalence: | Let $G = \struct {V, E}$ be a [[Definition:Graph (Graph Theory)|graph]].
Let $\to$ denote the [[Definition:Relation|relation]] '''is [[Definition:Connected Vertices|connected]] to''' on the set $V$.
Then $\to$ is an [[Definition:Equivalence Relation|equivalence relation]]. | Let $u, v, w$ be arbitrary [[Definition:Vertex of Graph|vertices]] of a [[Definition:Graph (Graph Theory)|graph]] $G$.
Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Graph Connectedness is Equivalence Relation | https://proofwiki.org/wiki/Graph_Connectedness_is_Equivalence_Relation | https://proofwiki.org/wiki/Graph_Connectedness_is_Equivalence_Relation | [
"Connectedness (Graph Theory)",
"Examples of Equivalence Relations"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Relation",
"Definition:Connected (Graph Theory)/Vertices",
"Definition:Equivalence Relation"
] | [
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)",
"Definition:Equivalence Relation",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Equivalence Relation"
] |
proofwiki-2375 | Condition for Edge to be Bridge | Let $G = \struct {V, E}$ be a connected graph.
Let $e \in E$ be an edge of $G$.
Then $e$ is a bridge {{iff}} $e$ does not lie on any circuit of $G$. | Let $G - e$ denote the edge deletion of $e$ from $G$. | Let $G = \struct {V, E}$ be a [[Definition:Connected Graph|connected graph]].
Let $e \in E$ be an [[Definition:Edge of Graph|edge]] of $G$.
Then $e$ is a [[Definition:Bridge (Graph Theory)|bridge]] {{iff}} $e$ does not lie on any [[Definition:Circuit (Graph Theory)|circuit]] of $G$. | Let $G - e$ denote the [[Definition:Edge Deletion|edge deletion]] of $e$ from $G$. | Condition for Edge to be Bridge | https://proofwiki.org/wiki/Condition_for_Edge_to_be_Bridge | https://proofwiki.org/wiki/Condition_for_Edge_to_be_Bridge | [
"Graph Theory",
"Proofs by Contraposition"
] | [
"Definition:Connected (Graph Theory)/Graph",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Bridge (Graph Theory)",
"Definition:Circuit (Graph Theory)"
] | [
"Definition:Edge Deletion"
] |
proofwiki-2376 | Characteristics of Traversable Graph | A finite loop-multigraph is '''traversable''' {{iff}} it is connected and no more than two vertices are odd.
An Eulerian trail which is not an Eulerian circuit must start and end at an odd vertex. | {{improve|this can be presented in a neater way}}
Let $G$ be a graph.
Suppose all the vertices are even, that is, there are no odd vertices.
Then $G$ is Eulerian, and the result holds.
Similarly, by the same result, if $G$ is Eulerian, it is by definition traversable.
So the question of graphs all of whose vertices are... | A [[Definition:Finite Graph|finite]] [[Definition:Loop-Multigraph|loop-multigraph]] is '''[[Definition:Traversable Graph|traversable]]''' {{iff}} it is [[Definition:Connected Graph|connected]] and no more than two [[Definition:Vertex of Graph|vertices]] are [[Definition:Odd Vertex of Graph|odd]].
An [[Definition:Eule... | {{improve|this can be presented in a neater way}}
Let $G$ be a [[Definition:Graph (Graph Theory)|graph]].
Suppose all the [[Definition:Vertex of Graph|vertices]] are [[Definition:Even Vertex of Graph|even]], that is, there are no [[Definition:Odd Vertex of Graph|odd vertices]].
Then [[Characteristics of Eulerian Grap... | Characteristics of Traversable Graph | https://proofwiki.org/wiki/Characteristics_of_Traversable_Graph | https://proofwiki.org/wiki/Characteristics_of_Traversable_Graph | [
"Characteristics of Traversable Graph",
"Semi-Eulerian Graphs"
] | [
"Definition:Finite Graph",
"Definition:Loop-Graph/Loop-Multigraph",
"Definition:Semi-Eulerian Graph",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Odd Vertex of Graph",
"Definition:Eulerian Trail",
"Definition:Eulerian Circuit",
"Definition:Odd V... | [
"Definition:Graph (Graph Theory)",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Even Vertex of Graph",
"Definition:Odd Vertex of Graph",
"Characteristics of Eulerian Graph",
"Definition:Eulerian Graph",
"Definition:Semi-Eulerian Graph",
"Definition:Undirected Graph",
"Definition:Graph (Grap... |
proofwiki-2377 | Odd Order Complete Graph is Eulerian | Let $K_n$ be the complete graph of $n$ vertices.
Then $K_n$ is Eulerian {{iff}} $n$ is odd.
If $n$ is even, then $K_n$ is traversable iff $n = 2$. | From the definition, the complete graph $K_n$ is $n-1$-regular.
That is, every vertex of $K_n$ is of degree $n-1$.
Suppose $n$ is odd. Then $n-1$ is even, and so $K_n$ is Eulerian.
Suppose $n$ is even. Then $n-1$ is odd.
Hence for $n \ge 4$, $K_n$ has more than $2$ odd vertices and so can not be traversable, let alone ... | Let $K_n$ be the [[Definition:Complete Graph|complete graph]] of $n$ [[Definition:Vertex of Graph|vertices]].
Then $K_n$ is [[Definition:Eulerian Graph|Eulerian]] {{iff}} $n$ is [[Definition:Odd Integer|odd]].
If $n$ is even, then $K_n$ is [[Definition:Traversable Graph|traversable]] iff $n = 2$. | From the definition, the [[Definition:Complete Graph|complete graph]] $K_n$ is [[Definition:Regular Graph|$n-1$-regular]].
That is, every [[Definition:Vertex of Graph|vertex]] of $K_n$ is of [[Definition:Degree of Vertex|degree]] $n-1$.
Suppose $n$ is odd. Then $n-1$ is even, and so $K_n$ is [[Characteristics of Eul... | Odd Order Complete Graph is Eulerian | https://proofwiki.org/wiki/Odd_Order_Complete_Graph_is_Eulerian | https://proofwiki.org/wiki/Odd_Order_Complete_Graph_is_Eulerian | [
"Eulerian Graphs",
"Complete Graphs"
] | [
"Definition:Complete Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Eulerian Graph",
"Definition:Odd Integer",
"Definition:Semi-Eulerian Graph"
] | [
"Definition:Complete Graph",
"Definition:Regular Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Degree of Vertex",
"Characteristics of Eulerian Graph",
"Definition:Odd Vertex of Graph",
"Definition:Semi-Eulerian Graph",
"Definition:Eulerian Graph",
"Definition:Odd Vertex of Graph",
... |
proofwiki-2378 | Odd Vertices Determines Edge-Disjoint Trails | Let $G$ be a loop-multigraph with $2 n$ odd vertices, $n > 0$.
Then $G$ has $n$ edge-disjoint trails such that every edge of $G$ is contained in one of these trails.
Each of these trails starts and ends on an odd vertex. | We prove this result by induction:
For all $n \in \N_{>0}$, let $\map P N$ be the proposition:
:if $G$ has $2 n$ odd vertices, it consists entirely of $n$ edge-disjoint trails, each starting and ending on an odd vertex.
First note that from {{Corollary|Handshake Lemma}}, no graph can have an odd number of odd vertices. | Let $G$ be a [[Definition:Loop-Multigraph|loop-multigraph]] with $2 n$ [[Definition:Odd Vertex of Graph|odd vertices]], $n > 0$.
Then $G$ has $n$ [[Definition:Edge-Disjoint Trails|edge-disjoint trails]] such that every [[Definition:Edge of Graph|edge]] of $G$ is contained in one of these [[Definition:Edge-Disjoint Tr... | We prove this result by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\map P N$ be the [[Definition:Proposition|proposition]]:
:if $G$ has $2 n$ [[Definition:Odd Vertex of Graph|odd vertices]], it consists entirely of $n$ [[Definition:Edge-Disjoint Trails|edge-disjoint trails]], each... | Odd Vertices Determines Edge-Disjoint Trails | https://proofwiki.org/wiki/Odd_Vertices_Determines_Edge-Disjoint_Trails | https://proofwiki.org/wiki/Odd_Vertices_Determines_Edge-Disjoint_Trails | [
"Degrees of Vertices",
"Proofs by Induction"
] | [
"Definition:Loop-Graph/Loop-Multigraph",
"Definition:Odd Vertex of Graph",
"Definition:Edge-Disjoint Trails",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Edge-Disjoint Trails",
"Definition:Odd Vertex of Graph"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Odd Vertex of Graph",
"Definition:Edge-Disjoint Trails",
"Definition:Odd Vertex of Graph",
"Definition:Odd Integer",
"Definition:Odd Vertex of Graph",
"Definition:Odd Vertex of Graph",
"Definition:Odd Vertex of Graph",
"D... |
proofwiki-2379 | Number of Edges of Regular Graph | An $r$-regular graph of order $n$ is of size $\dfrac {n r} 2$. | The size of a $r$-regular graph is its number of edges.
The order of a $r$-regular graph is its number of vertices.
The degree of each vertex of an $r$-regular graph is $r$.
Hence the total of all the degrees of an $r$-regular graph of order $n$ is $nr$.
The result follows directly from the Handshake Lemma.
{{qed}}
Cat... | An [[Definition:Regular Graph|$r$-regular graph]] of [[Definition:Order of Graph|order]] $n$ is of [[Definition:Size of Graph |size]] $\dfrac {n r} 2$. | The [[Definition:Size of Graph|size]] of a [[Definition:Regular Graph|$r$-regular graph]] is its number of [[Definition:Edge of Graph|edges]].
The [[Definition:Order of Graph|order]] of a [[Definition:Regular Graph|$r$-regular graph]] is its number of [[Definition:Vertex of Graph|vertices]].
The [[Definition:Degree o... | Number of Edges of Regular Graph | https://proofwiki.org/wiki/Number_of_Edges_of_Regular_Graph | https://proofwiki.org/wiki/Number_of_Edges_of_Regular_Graph | [
"Regular Graphs"
] | [
"Definition:Regular Graph",
"Definition:Graph (Graph Theory)/Order",
"Definition:Size of Graph "
] | [
"Definition:Graph (Graph Theory)/Size",
"Definition:Regular Graph",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Graph (Graph Theory)/Order",
"Definition:Regular Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Degree of Vertex",
"Definition:Graph (Graph Theory)/Vertex",
"Definit... |
proofwiki-2380 | Graph is Bipartite iff No Odd Cycles | Let $G$ be an undirected graph.
Then $G$ is bipartite {{iff}} it has no odd cycles. | === Sufficient Condition ===
Let $G = \struct {V, E}$ be bipartite.
So, let $V = A \cup B$ such that $A \cap B = \O$ and that all edges $e \in E$ are such that $e$ is of the form $\set {a, b}$ where $a \in A$ and $b \in B$.
(This is the definition of a bipartite graph.)
Suppose $G$ has (at least) one odd cycle $C$.
Let... | Let $G$ be an [[Definition:Undirected Graph|undirected graph]].
Then $G$ is [[Definition:Bipartite Graph|bipartite]] {{iff}} it has no [[Definition:Odd Cycle (Graph Theory)|odd cycles]]. | === Sufficient Condition ===
Let $G = \struct {V, E}$ be [[Definition:Bipartite Graph|bipartite]].
So, let $V = A \cup B$ such that $A \cap B = \O$ and that all [[Definition:Edge of Graph|edges]] $e \in E$ are such that $e$ is of the form $\set {a, b}$ where $a \in A$ and $b \in B$.
(This is the definition of a [[De... | Graph is Bipartite iff No Odd Cycles | https://proofwiki.org/wiki/Graph_is_Bipartite_iff_No_Odd_Cycles | https://proofwiki.org/wiki/Graph_is_Bipartite_iff_No_Odd_Cycles | [
"Bipartite Graphs"
] | [
"Definition:Undirected Graph",
"Definition:Bipartite Graph",
"Definition:Cycle (Graph Theory)/Odd"
] | [
"Definition:Bipartite Graph",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Bipartite Graph",
"Definition:Cycle (Graph Theory)/Odd",
"Definition:Walk (Graph Theory)/Length",
"Definition:Odd Integer",
"Definition:Bipartite Graph",
"Definition:Bipartite Graph",
"Definition:Cycle (Graph Theory)/O... |
proofwiki-2381 | Ore's Theorem | Let $G = \struct {V, E}$ be a simple graph of order $n \ge 3$.
Let $G$ be an Ore graph, that is:
:For each pair of non-adjacent vertices $u, v \in V$:
::$(1): \quad \deg u + \deg v \ge n$
Then $G$ is Hamiltonian. | From Ore Graph is Connected it is not necessary to demonstrate that $G$ is connected.
{{AimForCont}} it were possible to construct a graph that fulfils condition $(1)$ which is not Hamiltonian.
For a given $n \ge 3$, let $G$ be the graph with the most possible edges such that $G$ is non-Hamiltonian which satisfies $(1)... | Let $G = \struct {V, E}$ be a [[Definition:Simple Graph|simple graph]] of [[Definition:Order of Graph|order $n \ge 3$]].
Let $G$ be an [[Definition:Ore Graph|Ore graph]], that is:
:For each pair of [[Definition:Adjacent Vertices (Undirected Graph)|non-adjacent]] [[Definition:Vertex of Graph|vertices]] $u, v \in V$:
:... | From [[Ore Graph is Connected]] it is not necessary to demonstrate that $G$ is [[Definition:Connected Graph|connected]].
{{AimForCont}} it were possible to construct a graph that fulfils condition $(1)$ which is not [[Definition:Hamiltonian Graph|Hamiltonian]].
For a given $n \ge 3$, let $G$ be the [[Definition:Simp... | Ore's Theorem | https://proofwiki.org/wiki/Ore's_Theorem | https://proofwiki.org/wiki/Ore's_Theorem | [
"Hamiltonian Graphs",
"Ore Graphs"
] | [
"Definition:Simple Graph",
"Definition:Graph (Graph Theory)/Order",
"Definition:Ore Graph",
"Definition:Adjacent (Graph Theory)/Vertices/Undirected Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Hamiltonian Graph"
] | [
"Ore Graph is Connected",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Hamiltonian Graph",
"Definition:Simple Graph",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Hamiltonian Graph",
"Definition:Hamilton Cycle",
"Definition:Hamiltonian Path",
"Definition:Graph (Graph Theory)/Edge"... |
proofwiki-2382 | Path in Tree is Unique | Let $T$ be a graph.
Then $T$ is a tree {{iff}} there is exactly one path between any two distinct vertices. | === Necessary Condition ===
{{:Path in Tree is Unique/Necessary Condition/Proof 1}}{{qed|lemma}} | Let $T$ be a [[Definition:Graph (Graph Theory)|graph]].
Then $T$ is a [[Definition:Tree (Graph Theory)|tree]] {{iff}} there is [[Definition:Unique|exactly one]] [[Definition:Path (Graph Theory)|path]] between any two [[Definition:Distinct Elements|distinct]] [[Definition:Vertex of Graph|vertices]]. | === [[Path in Tree is Unique/Necessary Condition|Necessary Condition]] ===
{{:Path in Tree is Unique/Necessary Condition/Proof 1}}{{qed|lemma}} | Path in Tree is Unique | https://proofwiki.org/wiki/Path_in_Tree_is_Unique | https://proofwiki.org/wiki/Path_in_Tree_is_Unique | [
"Path in Tree is Unique",
"Paths (Graph Theory)",
"Tree Theory"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Tree (Graph Theory)",
"Definition:Unique",
"Definition:Path (Graph Theory)",
"Definition:Distinct/Plural",
"Definition:Graph (Graph Theory)/Vertex"
] | [
"Path in Tree is Unique/Necessary Condition"
] |
proofwiki-2383 | Path in Tree is Unique | Let $T$ be a graph.
Then $T$ is a tree {{iff}} there is exactly one path between any two distinct vertices. | Let $T$ be a tree.
{{AimForCont}} there exists a pair of distinct vertices $u$ and $v$ in $T$ such that there is not exactly one path between them.
If there is no path between $u$ and $v$, $T$ is not connected.
In this case, $T$ is certainly not a tree.
So, in keeping with our supposition, there is more than one path b... | Let $T$ be a [[Definition:Graph (Graph Theory)|graph]].
Then $T$ is a [[Definition:Tree (Graph Theory)|tree]] {{iff}} there is [[Definition:Unique|exactly one]] [[Definition:Path (Graph Theory)|path]] between any two [[Definition:Distinct Elements|distinct]] [[Definition:Vertex of Graph|vertices]]. | Let $T$ be a [[Definition:Tree (Graph Theory)|tree]].
{{AimForCont}} there exists a [[Definition:Doubleton|pair]] of [[Definition:Distinct Elements|distinct]] [[Definition:Vertex of Graph|vertices]] $u$ and $v$ in $T$ such that there is not [[Definition:Unique|exactly one]] [[Definition:Path (Graph Theory)|path]] betw... | Path in Tree is Unique/Necessary Condition/Proof 1 | https://proofwiki.org/wiki/Path_in_Tree_is_Unique | https://proofwiki.org/wiki/Path_in_Tree_is_Unique/Necessary_Condition/Proof_1 | [
"Path in Tree is Unique",
"Paths (Graph Theory)",
"Tree Theory"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Tree (Graph Theory)",
"Definition:Unique",
"Definition:Path (Graph Theory)",
"Definition:Distinct/Plural",
"Definition:Graph (Graph Theory)/Vertex"
] | [
"Definition:Tree (Graph Theory)",
"Definition:Doubleton",
"Definition:Distinct/Plural",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Unique",
"Definition:Path (Graph Theory)",
"Definition:Path (Graph Theory)",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Tree (Graph Theory)",
... |
proofwiki-2384 | Path in Tree is Unique | Let $T$ be a graph.
Then $T$ is a tree {{iff}} there is exactly one path between any two distinct vertices. | Let $T$ be such that between every pair of distinct vertices of $T$ there exists exactly one path.
Then {{apriori}} $T$ is connected.
Suppose $T$ had a circuit, say $\tuple {u, u_1, u_2, \ldots, u_n, v, u}$.
Then there are two paths from $u$ to $v$:
:$\tuple {u, u_1, u_2, \ldots, u_n, v}$
and
:$\tuple {u, v}$
Hence, by... | Let $T$ be a [[Definition:Graph (Graph Theory)|graph]].
Then $T$ is a [[Definition:Tree (Graph Theory)|tree]] {{iff}} there is [[Definition:Unique|exactly one]] [[Definition:Path (Graph Theory)|path]] between any two [[Definition:Distinct Elements|distinct]] [[Definition:Vertex of Graph|vertices]]. | Let $T$ be such that between every [[Definition:Doubleton|pair]] of [[Definition:Distinct Elements|distinct]] [[Definition:Vertex of Graph|vertices]] of $T$ there exists exactly one [[Definition:Path (Graph Theory)|path]].
Then {{apriori}} $T$ is [[Definition:Connected Graph|connected]].
Suppose $T$ had a [[Definiti... | Path in Tree is Unique/Sufficient Condition/Proof 1 | https://proofwiki.org/wiki/Path_in_Tree_is_Unique | https://proofwiki.org/wiki/Path_in_Tree_is_Unique/Sufficient_Condition/Proof_1 | [
"Path in Tree is Unique",
"Paths (Graph Theory)",
"Tree Theory"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Tree (Graph Theory)",
"Definition:Unique",
"Definition:Path (Graph Theory)",
"Definition:Distinct/Plural",
"Definition:Graph (Graph Theory)/Vertex"
] | [
"Definition:Doubleton",
"Definition:Distinct/Plural",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Path (Graph Theory)",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Circuit (Graph Theory)",
"Definition:Path (Graph Theory)",
"Modus Tollendo Tollens",
"Definition:Circuit (Graph T... |
proofwiki-2385 | Finite Connected Simple Graph is Tree iff Size is One Less than Order | Let $T$ be a finite connected simple graph of order $n$.
Then $T$ is a (finite) tree {{iff}} the size of $T$ is $n - 1$. | By definition:
:the order of a (finite) tree is how many nodes it has
and:
:the size of a (finite) tree is how many edges it has. | Let $T$ be a [[Definition:Finite Graph|finite]] [[Definition:Connected Graph|connected]] [[Definition:Simple Graph|simple graph]] of [[Definition:Order of Graph|order]] $n$.
Then $T$ is a [[Definition:Finite Tree|(finite) tree]] {{iff}} the [[Definition:Size of Graph|size]] of $T$ is $n - 1$. | By definition:
:the [[Definition:Order of Graph|order]] of a [[Definition:Finite Tree|(finite) tree]] is how many [[Definition:Node of Tree|nodes]] it has
and:
:the [[Definition:Size of Graph|size]] of a [[Definition:Finite Tree|(finite) tree]] is how many [[Definition:Edge of Graph|edges]] it has. | Finite Connected Simple Graph is Tree iff Size is One Less than Order | https://proofwiki.org/wiki/Finite_Connected_Simple_Graph_is_Tree_iff_Size_is_One_Less_than_Order | https://proofwiki.org/wiki/Finite_Connected_Simple_Graph_is_Tree_iff_Size_is_One_Less_than_Order | [
"Finite Connected Simple Graph is Tree iff Size is One Less than Order",
"Tree Theory"
] | [
"Definition:Finite Graph",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Simple Graph",
"Definition:Graph (Graph Theory)/Order",
"Definition:Tree (Graph Theory)/Finite",
"Definition:Graph (Graph Theory)/Size"
] | [
"Definition:Graph (Graph Theory)/Order",
"Definition:Tree (Graph Theory)/Finite",
"Definition:Tree (Graph Theory)/Node",
"Definition:Graph (Graph Theory)/Size",
"Definition:Tree (Graph Theory)/Finite",
"Definition:Graph (Graph Theory)/Edge"
] |
proofwiki-2386 | Finite Connected Simple Graph is Tree iff Size is One Less than Order | Let $T$ be a finite connected simple graph of order $n$.
Then $T$ is a (finite) tree {{iff}} the size of $T$ is $n - 1$. | Let $T_{k + 1}$ be an arbitrary tree with $k + 1$ nodes.
Take any node $v$ of $T_{k + 1}$ of degree $1$.
Such a node exists from Finite Tree has Leaf Nodes.
Consider $T_k$, the subgraph of $T_{k + 1}$ created by removing $v$ and the edge connecting it to the rest of the tree.
By Connected Subgraph of Tree is Tree, $T_k... | Let $T$ be a [[Definition:Finite Graph|finite]] [[Definition:Connected Graph|connected]] [[Definition:Simple Graph|simple graph]] of [[Definition:Order of Graph|order]] $n$.
Then $T$ is a [[Definition:Finite Tree|(finite) tree]] {{iff}} the [[Definition:Size of Graph|size]] of $T$ is $n - 1$. | Let $T_{k + 1}$ be an arbitrary [[Definition:Tree (Graph Theory)|tree]] with $k + 1$ [[Definition:Node of Tree|nodes]].
Take any [[Definition:Node of Tree|node]] $v$ of $T_{k + 1}$ of [[Definition:Degree of Vertex|degree]] $1$.
Such a node exists from [[Finite Tree has Leaf Nodes]].
Consider $T_k$, the [[Definition:... | Finite Connected Simple Graph is Tree iff Size is One Less than Order/Necessary Condition/Induction Step/Proof 1 | https://proofwiki.org/wiki/Finite_Connected_Simple_Graph_is_Tree_iff_Size_is_One_Less_than_Order | https://proofwiki.org/wiki/Finite_Connected_Simple_Graph_is_Tree_iff_Size_is_One_Less_than_Order/Necessary_Condition/Induction_Step/Proof_1 | [
"Finite Connected Simple Graph is Tree iff Size is One Less than Order",
"Tree Theory"
] | [
"Definition:Finite Graph",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Simple Graph",
"Definition:Graph (Graph Theory)/Order",
"Definition:Tree (Graph Theory)/Finite",
"Definition:Graph (Graph Theory)/Size"
] | [
"Definition:Tree (Graph Theory)",
"Definition:Tree (Graph Theory)/Node",
"Definition:Tree (Graph Theory)/Node",
"Definition:Degree of Vertex",
"Finite Tree has Leaf Nodes",
"Definition:Subgraph",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Tree (Graph Theory)",
"Connected Subgraph of Tree is... |
proofwiki-2387 | Finite Connected Simple Graph is Tree iff Size is One Less than Order | Let $T$ be a finite connected simple graph of order $n$.
Then $T$ is a (finite) tree {{iff}} the size of $T$ is $n - 1$. | Let $T_{k + 1}$ be an arbitrary tree with $k + 1$ nodes.
Remove any edge $e$ of $T_{k + 1}$.
By definition of tree $T_{k + 1}$ has no circuits.
Therefore from Condition for Edge to be Bridge it follows that $e$ must be a bridge.
So removing $e$ disconnects $T_{k + 1}$ into two trees $T_1$ and $T_2$, with $k_1$ and $k_2... | Let $T$ be a [[Definition:Finite Graph|finite]] [[Definition:Connected Graph|connected]] [[Definition:Simple Graph|simple graph]] of [[Definition:Order of Graph|order]] $n$.
Then $T$ is a [[Definition:Finite Tree|(finite) tree]] {{iff}} the [[Definition:Size of Graph|size]] of $T$ is $n - 1$. | Let $T_{k + 1}$ be an arbitrary [[Definition:Tree (Graph Theory)|tree]] with $k + 1$ [[Definition:Node of Tree|nodes]].
Remove any [[Definition:Edge of Graph|edge]] $e$ of $T_{k + 1}$.
By definition of [[Definition:Tree (Graph Theory)|tree]] $T_{k + 1}$ has no [[Definition:Circuit (Graph Theory)|circuits]].
Therefor... | Finite Connected Simple Graph is Tree iff Size is One Less than Order/Necessary Condition/Induction Step/Proof 2 | https://proofwiki.org/wiki/Finite_Connected_Simple_Graph_is_Tree_iff_Size_is_One_Less_than_Order | https://proofwiki.org/wiki/Finite_Connected_Simple_Graph_is_Tree_iff_Size_is_One_Less_than_Order/Necessary_Condition/Induction_Step/Proof_2 | [
"Finite Connected Simple Graph is Tree iff Size is One Less than Order",
"Tree Theory"
] | [
"Definition:Finite Graph",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Simple Graph",
"Definition:Graph (Graph Theory)/Order",
"Definition:Tree (Graph Theory)/Finite",
"Definition:Graph (Graph Theory)/Size"
] | [
"Definition:Tree (Graph Theory)",
"Definition:Tree (Graph Theory)/Node",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Tree (Graph Theory)",
"Definition:Circuit (Graph Theory)",
"Condition for Edge to be Bridge",
"Definition:Bridge (Graph Theory)",
"Definition:Connected (Graph Theory)/Graph/Disc... |
proofwiki-2388 | Connected Subgraph of Tree is Tree | Let $T$ be a tree.
Let $S$ be a subgraph of $T$ such that $S$ is connected.
Then $S$ is also a tree. | Follows directly from the fact that by definition, $T$ has no circuits.
As $T$ has no circuits, then nor can $S$ have.
Hence $S$ is a connected simple graph with no circuits.
Thus by definition, $S$ is a tree.
{{qed}}
Category:Tree Theory
1hwn2ak43lszy2qs63d7p3rp9490j5z | Let $T$ be a [[Definition:Tree (Graph Theory)|tree]].
Let $S$ be a [[Definition:Subgraph|subgraph]] of $T$ such that $S$ is [[Definition:Connected Graph|connected]].
Then $S$ is also a [[Definition:Tree (Graph Theory)|tree]]. | Follows directly from the fact that by definition, $T$ has no [[Definition:Circuit (Graph Theory)|circuits]].
As $T$ has no [[Definition:Circuit (Graph Theory)|circuits]], then nor can $S$ have.
Hence $S$ is a [[Definition:Connected Graph|connected]] [[Definition:Simple Graph|simple graph]] with no [[Definition:Circu... | Connected Subgraph of Tree is Tree | https://proofwiki.org/wiki/Connected_Subgraph_of_Tree_is_Tree | https://proofwiki.org/wiki/Connected_Subgraph_of_Tree_is_Tree | [
"Tree Theory"
] | [
"Definition:Tree (Graph Theory)",
"Definition:Subgraph",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Tree (Graph Theory)"
] | [
"Definition:Circuit (Graph Theory)",
"Definition:Circuit (Graph Theory)",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Simple Graph",
"Definition:Circuit (Graph Theory)",
"Definition:Tree (Graph Theory)",
"Category:Tree Theory"
] |
proofwiki-2389 | Finite Tree has Leaf Nodes | Every non-edgeless finite tree has at least two leaf nodes. | We use the Method of Infinite Descent.
Suppose $T$ is a tree which has no leaf nodes.
First note that no tree has all even nodes.
Indeed, because by Characteristics of Eulerian Graph, it would then be an Eulerian graph, and by definition, trees do not have circuits.
From the Handshake Lemma, we know that $T$ must there... | Every non-[[Definition:Edgeless Graph|edgeless]] [[Definition:Finite Tree|finite tree]] has at least two [[Definition:Leaf Node|leaf nodes]]. | We use the [[Method of Infinite Descent]].
Suppose $T$ is a [[Definition:Tree (Graph Theory)|tree]] which has no [[Definition:Leaf Node|leaf nodes]].
First note that no [[Definition:Tree (Graph Theory)|tree]] has all [[Definition:Even Vertex of Graph|even nodes]].
Indeed, because by [[Characteristics of Eulerian Gr... | Finite Tree has Leaf Nodes/Proof 1 | https://proofwiki.org/wiki/Finite_Tree_has_Leaf_Nodes | https://proofwiki.org/wiki/Finite_Tree_has_Leaf_Nodes/Proof_1 | [
"Finite Tree has Leaf Nodes",
"Finite Trees",
"Leaf Nodes"
] | [
"Definition:Edgeless Graph",
"Definition:Tree (Graph Theory)/Finite",
"Definition:Tree (Graph Theory)/Leaf Node"
] | [
"Method of Infinite Descent",
"Definition:Tree (Graph Theory)",
"Definition:Tree (Graph Theory)/Leaf Node",
"Definition:Tree (Graph Theory)",
"Definition:Even Vertex of Graph",
"Characteristics of Eulerian Graph",
"Definition:Eulerian Graph",
"Definition:Tree (Graph Theory)",
"Definition:Circuit (Gr... |
proofwiki-2390 | Finite Tree has Leaf Nodes | Every non-edgeless finite tree has at least two leaf nodes. | Let the proposition we are proving in the case of a tree of order $k$ be denoted $\map P k$.
=== Basis for the Induction ===
That $\map P 2$ is true follows as the Unique Tree of Order 2 contains exactly two leaves (viz. its two nodes).
This is our basis for the induction.
=== Induction Hypothesis ===
Suppose $\map P i... | Every non-[[Definition:Edgeless Graph|edgeless]] [[Definition:Finite Tree|finite tree]] has at least two [[Definition:Leaf Node|leaf nodes]]. | Let the proposition we are proving in the case of a [[Definition:Tree (Graph Theory)|tree]] of [[Definition:Order of Graph|order]] $k$ be denoted $\map P k$.
=== Basis for the Induction ===
That $\map P 2$ is true follows as the [[Unique Tree of Order 2]] contains exactly two [[Definition:Leaf Node|leaves]] (viz. it... | Finite Tree has Leaf Nodes/Proof 2 | https://proofwiki.org/wiki/Finite_Tree_has_Leaf_Nodes | https://proofwiki.org/wiki/Finite_Tree_has_Leaf_Nodes/Proof_2 | [
"Finite Tree has Leaf Nodes",
"Finite Trees",
"Leaf Nodes"
] | [
"Definition:Edgeless Graph",
"Definition:Tree (Graph Theory)/Finite",
"Definition:Tree (Graph Theory)/Leaf Node"
] | [
"Definition:Tree (Graph Theory)",
"Definition:Graph (Graph Theory)/Order",
"Unique Tree of Order 2",
"Definition:Tree (Graph Theory)/Leaf Node",
"Definition:Tree (Graph Theory)/Node",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Definition:Tr... |
proofwiki-2391 | Size of Cycle Graph equals Order | The size of a cycle graph equals its order. | Follows immediately from:
:The fact that a cycle graph is $2$-regular
:The result Number of Edges of Regular Graph which says that an $r$-regular graph with $n$ vertices has $\dfrac {n r} 2$ edges.
{{qed}}
Category:Cycle Graphs
7spwmoa9qtdscb9urh7gmg5tejimxf5 | The [[Definition:Size of Graph|size]] of a [[Definition:Cycle Graph|cycle graph]] equals its [[Definition:Order of Graph|order]]. | Follows immediately from:
:The fact that a [[Definition:Cycle Graph|cycle graph]] is [[Definition:Regular Graph|$2$-regular]]
:The result [[Number of Edges of Regular Graph]] which says that an [[Definition:Regular Graph|$r$-regular graph]] with $n$ [[Definition:Vertex of Graph|vertices]] has $\dfrac {n r} 2$ [[Defin... | Size of Cycle Graph equals Order | https://proofwiki.org/wiki/Size_of_Cycle_Graph_equals_Order | https://proofwiki.org/wiki/Size_of_Cycle_Graph_equals_Order | [
"Cycle Graphs"
] | [
"Definition:Graph (Graph Theory)/Size",
"Definition:Cycle Graph",
"Definition:Graph (Graph Theory)/Order"
] | [
"Definition:Cycle Graph",
"Definition:Regular Graph",
"Number of Edges of Regular Graph",
"Definition:Regular Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Edge",
"Category:Cycle Graphs"
] |
proofwiki-2392 | Path Graph from Cycle Graph | Removing one edge from a cycle graph leaves a path graph. | Let $G_n$ be the graph obtained by removing any edge from the cycle graph $C_n$.
As each of those edges lies on a cycle, by Condition for Edge to be Bridge, none of them is a bridge.
So removing any edge from $C_n$ leaves the resulting subgraph of $C_n$ connected.
From Size of Cycle Graph equals Order, the cycle graph ... | Removing one [[Definition:Edge of Graph|edge]] from a [[Definition:Cycle Graph|cycle graph]] leaves a [[Definition:Path Graph|path graph]]. | Let $G_n$ be the [[Definition:Graph (Graph Theory)|graph]] obtained by removing any [[Definition:Edge of Graph|edge]] from the [[Definition:Cycle Graph|cycle graph]] $C_n$.
As each of those [[Definition:Edge of Graph|edges]] lies on a [[Definition:Cycle (Graph Theory)|cycle]], by [[Condition for Edge to be Bridge]], n... | Path Graph from Cycle Graph | https://proofwiki.org/wiki/Path_Graph_from_Cycle_Graph | https://proofwiki.org/wiki/Path_Graph_from_Cycle_Graph | [
"Cycle Graphs",
"Path Graphs"
] | [
"Definition:Graph (Graph Theory)/Edge",
"Definition:Cycle Graph",
"Definition:Path Graph"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Cycle Graph",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Cycle (Graph Theory)",
"Condition for Edge to be Bridge",
"Definition:Bridge (Graph Theory)",
"Definition:Graph (Graph Theory)/Edge",
"Definition:S... |
proofwiki-2393 | Cycle in Balanced Signed Graph | Let $G$ be a balanced signed graph.
Let $C$ be a cycle in $G$.
Then $C$ has an even number of negative edges. | Let $G$ be a balanced signed graph.
Since $G$ is balanced, we can colour the vertices black and white, say, so that every edge has a black and a white end.
Consider any cycle $C$ in $G$.
As we proceed around $C$, there is a change of colour when we traverse a negative edge.
Since this is a cycle, the last colour is the... | Let $G$ be a [[Definition:Balanced Signed Graph|balanced signed graph]].
Let $C$ be a [[Definition:Cycle (Graph Theory)|cycle]] in $G$.
Then $C$ has an [[Definition:Even Integer|even number]] of [[Definition:Signed Graph|negative edges]]. | Let $G$ be a [[Definition:Balanced Signed Graph|balanced signed graph]].
Since $G$ is balanced, we can colour the [[Definition:Vertex of Graph|vertices]] black and white, say, so that every [[Definition:Edge of Graph|edge]] has a black and a white end.
Consider any [[Definition:Cycle (Graph Theory)|cycle]] $C$ in $G$... | Cycle in Balanced Signed Graph | https://proofwiki.org/wiki/Cycle_in_Balanced_Signed_Graph | https://proofwiki.org/wiki/Cycle_in_Balanced_Signed_Graph | [
"Cycles (Graph Theory)",
"Balanced Signed Graphs"
] | [
"Definition:Balanced Signed Graph",
"Definition:Cycle (Graph Theory)",
"Definition:Even Integer",
"Definition:Signed Graph"
] | [
"Definition:Balanced Signed Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Cycle (Graph Theory)",
"Definition:Signed Graph",
"Category:Cycles (Graph Theory)",
"Category:Balanced Signed Graphs"
] |
proofwiki-2394 | Graph with Even Vertices Partitions into Cycles | Let $G = \struct {V, E}$ be a graph whose vertices are all even.
Then its edge set $E$ can be partitioned into cycles, no two of which share an edge.
The converse also holds: a graph which can be partitioned into cycles must have all its vertices even. | Let $G = \struct {V, E}$ be a graph whose vertices are all even.
If there is more than one vertex in $G$, then each vertex must have degree greater than $0$.
Begin at any vertex $u$.
Since the graph is connected (if the graph is not connected then the argument will be applied to separate components), there must be an e... | Let $G = \struct {V, E}$ be a [[Definition:Graph (Graph Theory)|graph]] whose [[Definition:Vertex of Graph|vertices]] are all [[Definition:Even Vertex of Graph|even]].
Then its [[Definition:Edge Set|edge set]] $E$ can be [[Definition:Partition (Set Theory)|partitioned]] into [[Definition:Cycle (Graph Theory)|cycles]],... | Let $G = \struct {V, E}$ be a [[Definition:Graph (Graph Theory)|graph]] whose [[Definition:Vertex of Graph|vertices]] are all [[Definition:Even Vertex of Graph|even]].
If there is more than one [[Definition:Vertex of Graph|vertex]] in $G$, then each [[Definition:Vertex of Graph|vertex]] must have [[Definition:Degree ... | Graph with Even Vertices Partitions into Cycles | https://proofwiki.org/wiki/Graph_with_Even_Vertices_Partitions_into_Cycles | https://proofwiki.org/wiki/Graph_with_Even_Vertices_Partitions_into_Cycles | [
"Graph Theory"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Even Vertex of Graph",
"Definition:Edge Set",
"Definition:Set Partition",
"Definition:Cycle (Graph Theory)",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Converse Statement",
"Definition:Graph (Graph Theo... | [
"Definition:Graph (Graph Theory)",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Even Vertex of Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Degree of Vertex",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)"... |
proofwiki-2395 | Condition for Bipartite Graph to be Hamiltonian | Let $G = \struct {A \mid B, E}$ be a bipartite graph.
Let $G$ be Hamiltonian.
Then $\card A = \card B$.
That is, there is the same number of vertices in $A$ as there are in $B$. | Let $G = \struct {A \mid B, E}$ be a bipartite graph.
To be Hamiltonian, a graph $G$ needs to have a Hamilton cycle: that is, one which goes through all the vertices of $G$.
As each edge in $G$ connects a vertex in $A$ with a vertex in $B$, any cycle alternately passes through a vertex in $A$ then a vertex in $B$.
{{WL... | Let $G = \struct {A \mid B, E}$ be a [[Definition:Bipartite Graph|bipartite graph]].
Let $G$ be [[Definition:Hamiltonian Graph|Hamiltonian]].
Then $\card A = \card B$.
That is, there is the same number of [[Definition:Vertex of Graph|vertices]] in $A$ as there are in $B$. | Let $G = \struct {A \mid B, E}$ be a [[Definition:Bipartite Graph|bipartite graph]].
To be [[Definition:Hamiltonian Graph|Hamiltonian]], a graph $G$ needs to have a [[Definition:Hamilton Cycle|Hamilton cycle]]: that is, one which goes through all the [[Definition:Vertex of Graph|vertices]] of $G$.
As each [[Definitio... | Condition for Bipartite Graph to be Hamiltonian | https://proofwiki.org/wiki/Condition_for_Bipartite_Graph_to_be_Hamiltonian | https://proofwiki.org/wiki/Condition_for_Bipartite_Graph_to_be_Hamiltonian | [
"Hamiltonian Graphs",
"Bipartite Graphs"
] | [
"Definition:Bipartite Graph",
"Definition:Hamiltonian Graph",
"Definition:Graph (Graph Theory)/Vertex"
] | [
"Definition:Bipartite Graph",
"Definition:Hamiltonian Graph",
"Definition:Hamilton Cycle",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Cycle (Graph Theory)",
"Definit... |
proofwiki-2396 | Complete Hamiltonian Bipartite Graph | :$K_{m, n}$ is Hamiltonian {{iff}} $m = n > 1$. | Clearly $K_{1, 1}$ is not Hamiltonian, although it does have a Hamiltonian path.
From Condition for Bipartite Graph to be Hamiltonian, if $m \ne n$ then $K_{m, n}$ is not Hamiltonian.
So let $m = n$.
We note that the degree of any vertex in $K_{n, n}$ is $n$.
Then we can use Ore's Theorem, and the result follows direct... | :$K_{m, n}$ is [[Definition:Hamiltonian Graph|Hamiltonian]] {{iff}} $m = n > 1$. | Clearly $K_{1, 1}$ is not [[Definition:Hamiltonian Graph|Hamiltonian]], although it does have a [[Definition:Hamiltonian Path|Hamiltonian path]].
From [[Condition for Bipartite Graph to be Hamiltonian]], if $m \ne n$ then $K_{m, n}$ is not [[Definition:Hamiltonian Graph|Hamiltonian]].
So let $m = n$.
We note that t... | Complete Hamiltonian Bipartite Graph | https://proofwiki.org/wiki/Complete_Hamiltonian_Bipartite_Graph | https://proofwiki.org/wiki/Complete_Hamiltonian_Bipartite_Graph | [
"Hamiltonian Graphs",
"Complete Bipartite Graphs"
] | [
"Definition:Hamiltonian Graph"
] | [
"Definition:Hamiltonian Graph",
"Definition:Hamiltonian Path",
"Condition for Bipartite Graph to be Hamiltonian",
"Definition:Hamiltonian Graph",
"Definition:Degree of Vertex",
"Definition:Graph (Graph Theory)/Vertex",
"Ore's Theorem",
"Category:Hamiltonian Graphs",
"Category:Complete Bipartite Grap... |
proofwiki-2397 | Condition for Complete Bipartite Graph to be Semi-Hamiltonian | $K_{m, n}$ is semi-Hamiltonian {{iff}} either:
:$m = n = 1$
or:
:$m = n + 1$ (or $n = m + 1$). | Let $K_{m, n}$ be a complete bipartite graph.
If $m = n = 1$ then $K_{m, n} = K_{1, 1}$ is trivially semi-Hamiltonian:
:100px
{{qed|lemma}}
If $m = n + 1$, we can construct a Hamiltonian path as follows.
Let $K_{m, n} = \struct {A \mid B, E}$ such that $\card A = m, A = \set {u_1, u_2, \ldots, u_m}, \card B = n, B = \s... | $K_{m, n}$ is [[Definition:Semi-Hamiltonian Graph|semi-Hamiltonian]] {{iff}} either:
:$m = n = 1$
or:
:$m = n + 1$ (or $n = m + 1$). | Let $K_{m, n}$ be a [[Definition:Complete Bipartite Graph|complete bipartite graph]].
If $m = n = 1$ then $K_{m, n} = K_{1, 1}$ is trivially [[Definition:Semi-Hamiltonian Graph|semi-Hamiltonian]]:
:[[File:K1-1.png|100px]]
{{qed|lemma}}
If $m = n + 1$, we can construct a [[Definition:Hamiltonian Path|Hamiltonian pa... | Condition for Complete Bipartite Graph to be Semi-Hamiltonian | https://proofwiki.org/wiki/Condition_for_Complete_Bipartite_Graph_to_be_Semi-Hamiltonian | https://proofwiki.org/wiki/Condition_for_Complete_Bipartite_Graph_to_be_Semi-Hamiltonian | [
"Semi-Hamiltonian Graphs",
"Complete Bipartite Graphs"
] | [
"Definition:Semi-Hamiltonian Graph"
] | [
"Definition:Complete Bipartite Graph",
"Definition:Semi-Hamiltonian Graph",
"File:K1-1.png",
"Definition:Hamiltonian Path",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Hamiltonian Path",
"Definition:Hamiltonian... |
proofwiki-2398 | Number of Components after Removal of Bridge | Let $G = \struct {V, E}$ be a graph.
Let $e \in E$ be a bridge.
Let $m$ be the number of components of $G$.
Then when $e$ is removed from $G$, the number of components in the remaining graph is $m + 1$. | It is clear, by definition of a bridge, that removing $e$ increases the number of components.
So after $e$ is removed from $G$, the number of components in the remaining graph is at least $m + 1$.
Suppose that removing $e$ disconnects $G$ into more than $m + 1$ components.
Since $e$ joins only two vertices of $G$, it c... | Let $G = \struct {V, E}$ be a [[Definition:Graph (Graph Theory)|graph]].
Let $e \in E$ be a [[Definition:Bridge (Graph Theory)|bridge]].
Let $m$ be the number of [[Definition:Component of Graph|components]] of $G$.
Then when $e$ is removed from $G$, the number of [[Definition:Component of Graph|components]] in the ... | It is clear, by definition of a [[Definition:Bridge (Graph Theory)|bridge]], that removing $e$ increases the number of [[Definition:Component of Graph|components]].
So after $e$ is removed from $G$, the number of [[Definition:Component of Graph|components]] in the remaining [[Definition:Graph (Graph Theory)|graph]] is... | Number of Components after Removal of Bridge | https://proofwiki.org/wiki/Number_of_Components_after_Removal_of_Bridge | https://proofwiki.org/wiki/Number_of_Components_after_Removal_of_Bridge | [
"Graph Theory"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Bridge (Graph Theory)",
"Definition:Component of Graph",
"Definition:Component of Graph",
"Definition:Graph (Graph Theory)"
] | [
"Definition:Bridge (Graph Theory)",
"Definition:Component of Graph",
"Definition:Component of Graph",
"Definition:Graph (Graph Theory)",
"Definition:Connected (Graph Theory)/Graph/Disconnected",
"Definition:Component of Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Component of Graph",... |
proofwiki-2399 | Adding Edge to Tree Creates One Cycle | Adding a new edge to a tree can create no more than one cycle. | From Equivalent Definitions for Finite Tree, adding an edge creates at least one cycle.
Suppose that adding an edge $\tuple {u, v}$ to a tree $T$ creates two or more cycles.
Let two such cycles be $\tuple {u, v, \ldots, u_1, u_2, \ldots, u}$ and $\tuple {u, v, \ldots, v_1, v_2, \ldots, u}$.
By removing the edge $\tuple... | Adding a new [[Definition:Edge of Graph|edge]] to a [[Definition:Tree (Graph Theory)|tree]] can create no more than one [[Definition:Cycle (Graph Theory)|cycle]]. | From [[Equivalent Definitions for Finite Tree]], adding an edge creates at least one [[Definition:Cycle (Graph Theory)|cycle]].
Suppose that adding an [[Definition:Edge of Graph|edge]] $\tuple {u, v}$ to a [[Definition:Tree (Graph Theory)|tree]] $T$ creates two or more [[Definition:Cycle (Graph Theory)|cycles]].
Let ... | Adding Edge to Tree Creates One Cycle | https://proofwiki.org/wiki/Adding_Edge_to_Tree_Creates_One_Cycle | https://proofwiki.org/wiki/Adding_Edge_to_Tree_Creates_One_Cycle | [
"Tree Theory"
] | [
"Definition:Graph (Graph Theory)/Edge",
"Definition:Tree (Graph Theory)",
"Definition:Cycle (Graph Theory)"
] | [
"Characteristics of Finite Tree",
"Definition:Cycle (Graph Theory)",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Tree (Graph Theory)",
"Definition:Cycle (Graph Theory)",
"Definition:Cycle (Graph Theory)",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Cycle (Graph Theory)",
"Definition:... |
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