Rule of Implication

Rule of Implication 0 273 493647 354568 2020-10-10T12:39:50Z Prime.mover 59 wikitext text/x-wiki == Proof Rule == The '''rule of implication''' is a valid deduction sequent in propositional logic. === Proof Rule === {{:Rule of Implication/Proof Rule}} === Sequent Form === {{:Rule of Implication/Sequent Form}} == Explanation == {{:Rule of Implication/Explanation}} == Also known as == {{:Rule of Implication/Also known as}} == Sources == * {{BookReference|Symbolic Logic|1973|Irving M. Copi|ed = 4th|edpage = Fourth Edition|prev = Implication Equivalent to Negation of Conjunction with Negative/Formulation 1|next = Rule of Transposition/Formulation 2}}: $2.4$: Statement Forms Category:Implication Category:Rule of Implication oaxbv4hmjqs1uazzych5wwy0o0b6lvx


Proof by Contradiction

Proof by Contradiction 0 275 452511 409742 2020-03-05T07:34:23Z Prime.mover 59 wikitext text/x-wiki == Proof Rule == '''Proof by contradiction''' is a valid deduction sequent in propositional logic. === Proof Rule === {{:Proof by Contradiction/Proof Rule}} === Sequent Form === {{:Proof by Contradiction/Sequent Form}} == Explanation == {{:Proof by Contradiction/Explanation}} == Also known as == {{:Proof by Contradiction/Also known as}} == Variants == The following forms can be used as variants of this theorem: === Variant 1 === {{:Proof by Contradiction/Variant 1}} === Variant 2 === {{:Proof by Contradiction/Variant 2}} === Variant 3 === {{:Proof by Contradiction/Variant 3}} == Also see == * Reductio ad Absurdum * Law of Excluded Middle == Sources == * {{BookReference|Adventures in Group Theory|2008|David Joyner|ed = 2nd|edpage = Second Edition|prev = Definition:Contrapositive Statement|next = Reductio ad Absurdum}}: Chapter $1$: Elementary, my dear Watson: $\S 1.1$: You have a logical mind if... * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Proof|next = Definition:Proper Divisor|entry = proof by contradiction}} Category:Proof Techniques Category:Negation Category:Contradiction Category:Proof by Contradiction 93pb8g7tez7ee41xr05zohr36av3p2t


Rule of Explosion

Rule of Explosion 0 278 323536 245018 2017-10-28T16:11:09Z Lord Farin 560 wikitext text/x-wiki == Proof Rule == The '''Rule of Explosion''' is a valid deduction sequent in propositional logic. === Proof Rule === {{:Rule of Explosion/Proof Rule}} === Sequent Form === {{:Rule of Explosion/Sequent Form}} == Explanation == {{:Rule of Explosion/Explanation}} == Variants == The following can be used as variants of this theorem: === Variant 1 === {{:Rule of Explosion/Variant 1}} === Variant 2 === {{:Rule of Explosion/Variant 2}} === Variant 3 === {{:Rule of Explosion/Variant 3}} == Also known as == {{:Rule of Explosion/Also known as}} == Also see == * False Statement implies Every Statement Category:Contradiction Category:Rule of Explosion 76pn95w4d19csj56l3czdm6kli77sae


Law of Excluded Middle

Law of Excluded Middle 0 279 490603 448675 2020-09-25T06:27:59Z Prime.mover 59 wikitext text/x-wiki == Proof Rule == The '''law of (the) excluded middle''' is a valid deduction sequent in propositional logic. === Proof Rule === {{:Law of Excluded Middle/Proof Rule}} === Sequent Form === {{:Law of Excluded Middle/Sequent Form}} == Explanation == {{:Law of Excluded Middle/Explanation}} == Also known as == {{:Law of Excluded Middle/Also known as}} == Also see == * Principle of Non-Contradiction == Sources == * {{BookReference|Algebra Volume 1|1982|P.M. Cohn|edpage = Second Edition|ed = 2nd|prev = Definition:Biconditional|next = Definition:Tautology (Boolean Interpretations)}}: Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic * {{BookReference|Logic for Mathematicians|1988|Alan G. Hamilton|ed = 2nd|edpage = Second Edition|prev = Definition:Unsatisfiable (Boolean Interpretations)|next = Principle of Non-Contradiction}}: $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Example $1.6 \ \text{(a)}$ * {{BookReference|Mathematical Logic for Computer Science|1993|M. Ben-Ari|prev = Definition:Intuitionistic Logic|next = Definition:Temporal Logic}}: Chapter $1$: Introduction: $\S 1.4$: Non-standard logics * {{BookReference|Introduction to Graph Theory|1993|Richard J. Trudeau|prev = Russell's Paradox|next = Definition:Theory of Types}}: $2$. Graphs: Paradox Category:Proof Rules Category:Philosophical Positions Category:Disjunction Category:Tautology Category:Law of Excluded Middle i05cf2wzoalh8whklj6dam1xq18gyg7


Modus Tollendo Tollens

Modus Tollendo Tollens 0 304 493614 466200 2020-10-10T12:25:25Z Prime.mover 59 wikitext text/x-wiki == Proof Rule == The '''modus tollendo tollens''' is a valid deduction sequent in propositional logic. === Proof Rule === {{:Modus Tollendo Tollens/Proof Rule}} === Sequent Form === {{:Modus Tollendo Tollens/Sequent Form}} == Explanation == {{:Modus Tollendo Tollens/Explanation}} == Also known as == {{:Modus Tollendo Tollens/Also known as}} == Also see == The following are related argument forms: * Modus Ponendo Ponens * Modus Ponendo Tollens * Modus Tollendo Ponens The Rule of Transposition is conceptually similar, and can be derived from the MTT by a simple application of the Rule of Implication. These are classic fallacies: * Affirming the Consequent * Denying the Antecedent == Linguistic Note == {{:Modus Tollendo Tollens/Linguistic Note}} == Sources == * {{BookReference|Symbolic Logic|1973|Irving M. Copi|ed = 4th|edpage = Fourth Edition|prev = Modus Ponendo Ponens|next = Hypothetical Syllogism/Formulation 1}}: $2.3$: Argument Forms and Truth Tables * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Modus Tollendo Ponens|next = Definition:Mole|entry = ''modus tollens''}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Modus Tollendo Ponens|next = Definition:Mole|entry = ''modus tollens''}} Category:Implication Category:Negation Category:Modus Tollendo Tollens e6awiocdjyyx9qiw71r61kc22yic59x


Tautology and Contradiction

Tautology and Contradiction 0 310 241510 238677 2015-12-24T17:23:17Z Prime.mover 59 wikitext text/x-wiki == Theorems == === Contradiction is Negation of Tautology === {{:Contradiction is Negation of Tautology}} === Tautology is Negation of Contradiction === {{:Tautology is Negation of Contradiction}} === Conjunction with Tautology === {{:Conjunction with Tautology}} === Disjunction with Tautology === {{:Disjunction with Tautology}} === Conjunction with Contradiction === {{:Conjunction with Contradiction}} === Disjunction with Contradiction === {{:Disjunction with Contradiction}} == Comment == Note that the proofs of: * $\neg \bot \vdash \top$ * $\neg \top \vdash \bot$ * $p \vdash p \land \top$ * $p \lor \top \vdash \top$ rely (directly or indirectly) upon Law of Excluded Middle - and it can be seen that they are just another way of stating that truth. The propositions: :''If it's not false, it must be true'' and :''If it's not true, it must be false'' are indeed valid ''only'' in the context where there are only two truth values. From the intuitionistic perspective, these results do not hold. == Sources == * {{BookReference|Elementary Logic|1980|D.J. O'Connor|author2 = Betty Powell|prev = Definition:Contingent Statement|next = Definition:Truth Value/Aristotelian Logic}}: $\S \text{I}: 3$: Logical Constants $(2)$ Category:Tautology Category:Contradiction eesax7f146b3vge7b2n6rksqx6ogou5


Reductio ad Absurdum

Reductio ad Absurdum 0 312 415595 410772 2019-07-31T22:20:05Z Prime.mover 59 wikitext text/x-wiki == Proof Rule == The '''reductio ad absurdum''' is a valid deduction sequent in propositional logic: === Proof Rule === {{:Reductio ad Absurdum/Proof Rule}} === Sequent Form === {{:Reductio ad Absurdum/Sequent Form}} == Explanation == {{:Reductio ad Absurdum/Explanation}} == Variants == The following forms can be used as variants of this theorem: === Variant 1 === {{:Reductio ad Absurdum/Variant 1}} === Variant 2 === {{:Reductio ad Absurdum/Variant 2}} {{LEM}} == Also see == * Clavius's Law * Proof by Contradiction, often treated as another aspect of the same thing. From the point of view of purely classical logic, this is acceptable. However, in the context of intuitionistic logic, it is '''essential''' to bear in mind that only the Proof by Contradiction is valid. This is because Proof by Contradiction starts with a positive assumption $\phi$. As a result, it does not depend on the Law of Excluded Middle. == Linguistic Note == {{:Reductio ad Absurdum/Linguistic Note}} == Sources == * {{BookReference|A Handbook of Terms used in Algebra and Analysis|1972|A.G. Howson|prev = Definition:Consistent Proof System|next = Definition:Independent Proof System}}: $\S 1$: Some mathematical language: Axiom systems * {{BookReference|Elementary Logic|1980|D.J. O'Connor|author2 = Betty Powell|prev = Biconditional as Disjunction of Conjunctions/Formulation 2/Proof 2|next = Definition:Natural Deduction}}: $\S \text{I}: 13$: A Short Cut to Truth-tables * {{BookReference|Algebra Volume 1|1982|P.M. Cohn|edpage = Second Edition|ed = 2nd|prev = Definition:Converse Statement|next = Square Root of 2 is Irrational/Classic Proof}}: Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic * {{BookReference|Adventures in Group Theory|2008|David Joyner|ed = 2nd|edpage = Second Edition|prev = Proof by Contradiction|next = Definition:Conjunction/Truth Table}}: Chapter $1$: Elementary, my dear Watson: $\S 1.1$: You have a logical mind if... Category:Proof Rules Category:Negation Category:Reductio ad Absurdum 600liiy6mll0k1hdz49i5h42u7nz15y


Conjunction and Implication

Conjunction and Implication 0 314 163059 163057 2013-09-28T10:33:29Z Lord Farin 560 wikitext text/x-wiki == Theorems == === Conjunction Equivalent to Negation of Implication of Negative === {{:Conjunction Equivalent to Negation of Implication of Negative}} === Implication Equivalent to Negation of Conjunction with Negative === {{:Implication Equivalent to Negation of Conjunction with Negative}} === Conjunction with Negative Equivalent to Negation of Implication === {{:Conjunction with Negative Equivalent to Negation of Implication}} === Modus Ponendo Tollens === {{:Modus Ponendo Tollens/Variant}} == Law of Excluded Middle == Note that the Modus Ponendo Tollens: :$\neg \left({p \land q}\right) \dashv \vdash p \implies \neg q$ can be proved in both directions without resorting to Law of Excluded Middle. All the others: * $p \land q \vdash \neg \left({p \implies \neg q}\right)$ * $p \implies q \vdash \neg \left({p \land \neg q}\right)$ * $p \land \neg q \vdash \neg \left({p \implies q}\right)$ are not reversible in intuitionistic logic. == Also see == * Negation of Conditional implies Antecedent * Negation of Conditional implies Negation of Consequent Category:Conjunction Category:Implication dm3bild2uri1op8gxw84msinlx7bzb6


Disjunction and Implication

Disjunction and Implication 0 315 244713 230588 2016-01-16T22:21:32Z Prime.mover 59 wikitext text/x-wiki {{refactor|in progress}} == Theorems == === Modus Tollendo Ponens === {{:Modus Tollendo Ponens/Variant}} === Rule of Material Implication === {{:Rule of Material Implication}} Both of the above come in negative forms: {{begin-eqn}} {{eqn | l = \neg \left({p \implies q}\right)       | o = \dashv \vdash       | r = \neg \left({\neg p \lor q}\right)       | c =  }} {{eqn | l = \neg \left({\neg p \implies q}\right)       | o = \dashv \vdash       | r = \neg \left({p \lor q}\right)       | c =  }} {{end-eqn}} Disjunction is definable through implication: : $p \lor q \dashv \vdash \left({p \implies q}\right) \implies q$ === Alternative rendition === They can alternatively be rendered as: {{begin-eqn}} {{eqn | ll= \vdash       | l = \left({\neg \left({p \implies q}\right)}\right)       | o = \iff       | r = \left({\neg \left({\neg p \lor q}\right)}\right)       | c =  }} {{eqn | ll= \vdash       | l = \left({\neg \left({\neg p \implies q}\right)}\right)       | o = \iff       | r = \left({\neg \left({p \lor q}\right)}\right)       | c =  }} {{eqn | ll= \vdash       | l = \left({p \lor q}\right)       | o = \iff       | r = \left({\left({p \implies q}\right) \implies q}\right)       | c =  }} {{end-eqn}} They can be seen to be logically equivalent to the forms above. == Proof == {{BeginTableau|\neg \left({p \implies q}\right) \vdash \neg \left({\neg p \lor q}\right)}} {{Premise|1|\neg \left({p \implies q}\right)}} {{Assumption|2|\neg p \lor q|Assume the opposite of what is to be proved ...}} {{SequentIntro|3|2|p \implies q|2|Rule of Material Implication}} {{NonContradiction|4|1, 2|3|1| ... demonstrating a contradiction}} {{Contradiction|5|1|\neg \left({\neg p \lor q}\right)|2|4}} {{EndTableau}} {{qed}} {{BeginTableau|\neg \left({\neg p \lor q}\right) \vdash \neg \left({p \implies q}\right)}} {{Premise|1|\neg \left({\neg p \lor q}\right)}} {{Assumption|2|p \implies q|Assume the opposite of what is to be proved ...}} {{SequentIntro|3|2|\neg p \lor q|2|Rule of Material Implication}} {{NonContradiction|4|1, 2|3|1| ... demonstrating a contradiction}} {{Contradiction|5|1|\neg \left({p \implies q}\right)|2|4}} {{EndTableau}} {{qed}} {{LEM|Rule of Material Implication}} {{BeginTableau|\neg \left({\neg p \implies q}\right) \vdash \neg \left({p \lor q}\right)}} {{Premise|1|\neg \left({\neg p \implies q}\right)}} {{Assumption|2|p \lor q|Assume the opposite of what is to be proved ...}} {{SequentIntro|3|2|\neg p \implies q|2|Modus Tollendo Ponens}} {{NonContradiction|4|1, 2|3|1| ... demonstrating a contradiction}} {{Contradiction|5|1|\neg \left({p \lor q}\right)|2|4}} {{EndTableau}} {{qed}} {{BeginTableau|\neg \left({p \lor q}\right) \vdash \neg \left({\neg p \implies q}\right)}} {{Premise|1|\neg \left({p \lor q}\right)}} {{Assumption|2|\neg p \implies q|Assume the opposite of what is to be proved ...}} {{DeMorgan|3|1|\neg p \land \neg q|1|Conjunction of Negations}} {{Simplification|4|1|\neg p|3|1}} {{ModusPonens|5|1, 2|q|2|4|... from the assumption}} {{Simplification|6|1|\neg q|3|2}} {{NonContradiction|7|1, 2|5|6| ... demonstrating a contradiction}} {{Contradiction|8|1|\neg \left({\neg p \implies q}\right)|2|7}} {{EndTableau}} {{qed}} {{BeginTableau|p \lor q \vdash (p \implies q) \implies q}} {{Premise|1|p \lor q}} {{Assumption|2|p \implies q}} {{Assumption|3|p}} {{ModusPonens|4|2, 3|q|2|3}} {{Assumption|5|q}} {{ProofByCases|6|2|r|1|3|4|5|5}} {{Implication|7|1|\left({p \implies q}\right) \implies q|2|6}} {{EndTableau}} {{qed}} {{Proofread}} === Comment === Note that this: * $\neg \left({\neg p \implies q}\right) \dashv \vdash \neg \left({p \lor q}\right)$ can be proved in both directions without resorting to the Law of Excluded Middle. All the others: * $p \lor q \vdash \neg p \implies q$ * $\neg p \lor q \vdash p \implies q$ * $\neg \left({p \implies q}\right) \vdash \neg \left({\neg p \lor q}\right)$ are not reversible in intuitionistic logic. == Proof by Truth Table == We apply the Method of Truth Tables to the propositions in turn. As can be seen by inspection, in all cases the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|cccc||ccccc|} \hline \neg & (p & \lor & q) & \neg & (\neg & p & \implies & q) \\ \hline T & F & F & F & T & T & F & F & F \\ F & F & T & T & F & T & F & T & T \\ F & T & T & F & F & F & T & T & F \\ F & T & T & T & F & F & T & T & T \\ \hline \end{array}$ {{qed}} $\begin{array}{|cccc||ccccc|} \hline \neg & (p & \implies & q) & \neg & (\neg & p & \lor & q) \\ \hline F & F & T & F & F & T & F & T & F \\ F & F & T & T & F & T & F & T & T \\ T & T & F & F & T & F & T & F & F \\ F & T & T & T & F & F & T & T & T \\ \hline \end{array}$ {{qed}} Category:Disjunction Category:Implication g1zern52jt3d4o747h8wjo1ii9qzyfs


Set Difference and Intersection form Partition/Corollary 2 0 473 418202 418200 2019-08-15T06:02:43Z Prime.mover 59 wikitext text/x-wiki == Corollary to Set Difference and Intersection form Partition == <onlyinclude> Let $\O \subsetneqq T \subsetneqq S$. Then: :$\set {T, \relcomp S T}$ is a partition of $S$. </onlyinclude> == Proof == First we note that: :$\O \subsetneqq T \implies T \ne \O$ :$T \subsetneqq S \implies T \ne S$ from the definition of proper subset. From the definition of relative complement, we have: :$\relcomp S T = S \setminus T$. It follows from Set Difference with Self is Empty Set that: :$T \ne S \iff \relcomp S T \ne \O$ From Intersection with Relative Complement is Empty: :$T \cap \relcomp S T = \O$ That is, $T$ and $\relcomp S T$ are disjoint. From Union with Relative Complement: :$T \cup \relcomp S T = S$ That is, the union of $T$ and $\relcomp S T$ forms the whole set $S$. Thus all the conditions for a partition are satisfied. {{qed}} == Sources == * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Definition:Set Partition/Definition 1|next = Definition:Endorelation}}: $\S 6$. Indexed families; partitions; equivalence relations: Example $6.4$ Category:Relative Complement ec8skoyuz4jxzrs02ej5wqxkc325k68


Empty Intersection iff Subset of Complement 0 511 416793 414063 2019-08-05T21:57:18Z Prime.mover 59 wikitext text/x-wiki == Corollary to Intersection with Complement is Empty iff Subset == <onlyinclude> :$S \cap T = \O \iff S \subseteq \relcomp {} T$ </onlyinclude> where: :$S \cap T$ denotes the intersection of $S$ and $T$ :$\O$ denotes the empty set :$\complement$ denotes set complement :$\subseteq$ denotes subset. === Corollary === {{:Empty Intersection iff Subset of Complement/Corollary}} == Proof 1 == {{:Empty Intersection iff Subset of Complement/Proof 1}} == Proof 2 == {{:Empty Intersection iff Subset of Complement/Proof 2}} == Sources == * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Complement Union with Superset is Universe/Corollary|next = Intersection is Empty and Union is Universe if Sets are Complementary}}: $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $11 \ \text{(b)}$ * {{BookReference|Introduction to Topology|1975|Bert Mendelson|ed = 3rd|edpage = Third Edition|prev = Intersection with Subset is Subset|next = Complement Union with Superset is Universe/Corollary}}: Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $1 \ \text{(c)}$ Category:Set Complement Category:Set Intersection Category:Empty Set Category:Subsets Category:Empty Intersection iff Subset of Complement 6sgo51puejhk70xmyjzrunwdfh8skgf


Image is Subset of Codomain/Corollary 1

Image is Subset of Codomain/Corollary 1 0 548 397277 397276 2019-03-25T20:03:12Z Prime.mover 59 wikitext text/x-wiki == Corollary to Image is Subset of Codomain == <onlyinclude> Let $\mathcal R = S \times T$ be a relation. The image of $\mathcal R$ is a subset of the codomain of $\mathcal R$: :$\Img {\mathcal R} \subseteq T$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \Img {\mathcal R}       | r = \mathcal R \sqbrk {\Dom {\mathcal R} }       | c = {{Defof|Image of Relation}} }} {{eqn | l = \Dom {\mathcal R}       | o = \subseteq       | r = \Dom {\mathcal R}       | c = Set is Subset of Itself }} {{eqn | ll= \leadsto       | l = \Img {\mathcal R}       | o = \subseteq       | r = T       | c = Image is Subset of Codomain }} {{end-eqn}} {{qed}} Category:Relation Theory 29kiylqq2rnxxkgg2fzbgbqg50epl7y


Sundry Coset Results

Sundry Coset Results 0 1196 372738 372734 2018-10-23T07:57:00Z Prime.mover 59 wikitext text/x-wiki {{rename}} {{refactor|Once extension works with TOC, use it to fix headings}} == Theorems == Let $G$ be a group and let $H$ be a subgroup of $G$. Let $x, y \in G$. Let: : $x H$ denote the left coset of $H$ by $x$; : $H y$ denote the right coset of $H$ by $y$. Then the following results apply: === Element in Coset iff Product with Inverse in Subgroup === {{:Element in Coset iff Product with Inverse in Subgroup}} === Cosets are Equal iff Product with Inverse in Subgroup === {{:Cosets are Equal iff Product with Inverse in Subgroup}} === Cosets are Equal iff Element in Other Coset === {{:Cosets are Equal iff Element in Other Coset}} === Coset Equals Subgroup iff Element in Subgroup === {{:Coset Equals Subgroup iff Element in Subgroup}} === Elements in Same Coset iff Product with Inverse in Subgroup === {{:Elements in Same Coset iff Product with Inverse in Subgroup}} === Regular Representation on Subgroup is Bijection to Coset === {{:Regular Representation on Subgroup is Bijection to Coset}} Category:Cosets dw55ecqly0ukrqdzu7y7en29tx3o5fn


Strictly Increasing Sequence on Ordered Set 0 1374 366577 366574 2018-09-20T12:36:57Z KarlFrei 3474 added category wikitext text/x-wiki == Lemma == Let $\left({S, \preceq}\right)$ be a totally ordered set. Let $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ be a finite sequence of elements of $\left({S, \preceq}\right)$. Then $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ is strictly increasing {{iff}}: :$\forall k \in \left[{p + 1 \,.\,.\, q}\right]: r_{k - 1} \prec r_k$ == Proof == Let $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ be strictly increasing. Because $\forall k \in \N_{>0}: k - 1 < k$, it follows directly that: : $\forall k \in \left[{p + 1 \,.\,.\, q}\right]: r_{k - 1} \prec r_k$ For the other direction, we use a Proof by Contraposition. To that end, suppose $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ is ''not'' strictly increasing. Let $K$ be the set of all $k \in \left[{p \,.\,.\, q}\right]$ such that: : $\exists j \in \left[{p \,.\,.\, q}\right]$ such that $j < k$ and $r_k \preceq r_j$ The set $K$ is not empty because $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ is not strictly increasing. As $K \subset \N$ and the latter is well-ordered, then so is $K$. Thus $K$ has a minimal element $m$. Thus there exists $j \in \left[{p \,.\,.\, q}\right]$ such that: : $j < m$ and: : $r_m \preceq r_j$ Because $m - 1 < m$: :$j \le m - 1$ and so: :$m - 1 \notin K$ So: : $r_m \preceq r_j\prec r_{m-1}.$ Since orderings are transitive, it follows : $r_m \preceq r_{m-1}.$ From Rule of Transposition it follows that :$\forall k \in \left[{p + 1 \,.\,.\, q}\right]: r_{k - 1} \prec r_k\implies \left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ is strictly increasing. The result follows. {{Qed}} == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev=Strictly Increasing Sequence induces Partition|next=Fundamental Principle of Counting}}: $\S 18$: Lemma for Theorem $18.4$ Category:Sequences Category:Proofs by Contraposition mkquczmokcibfdu9m0vax0lkugixbos


Euclidean Algorithm

Euclidean Algorithm 0 1527 493847 445819 2020-10-11T06:57:17Z Prime.mover 59 wikitext text/x-wiki == Algorithm == <onlyinclude> The '''Euclidean algorithm''' is a method for finding the greatest common divisor (GCD) of two integers $a$ and $b$. Let $a, b \in \Z$ and $a \ne 0 \lor b \ne 0$. The steps are: :$(1): \quad$ Start with $\tuple {a, b}$ such that $\size a \ge \size b$. If $b = 0$ then the task is complete and the GCD is $a$. :$(2): \quad$ If $b \ne 0$ then you take the remainder $r$ of $\dfrac a b$. :$(3): \quad$ Set $a \gets b, b \gets r$ (and thus $\size a \ge \size b$ again). :$(4): \quad$ Repeat these steps until $b = 0$. Thus the GCD of $a$ and $b$ is the value of the variable $a$ after the termination of the algorithm. </onlyinclude> {{:Euclid:Proposition/VII/2}} == Proof 1 == {{:Euclidean Algorithm/Proof 1}} == Euclid's Proof == {{:Euclidean Algorithm/Euclid's Proof}} == Demonstration == {{:Euclidean Algorithm/Demonstration}} == Algorithmic Nature == {{:Euclidean Algorithm/Algorithmic Nature}} == Formal Implementation == {{:Euclidean Algorithm/Formal Implementation}} == Constructing an Integer Combination == {{:Euclidean Algorithm/Construction of Integer Combination}} == Also known as == The '''Euclidean algorithm''' is also known as '''Euclid's algorithm''' or the '''Euclidean division algorithm'''. == Examples == {{:Euclidean Algorithm/Examples}} == Also see == * Rational Numbers and SFCFs are Equivalent for an application of the '''Euclidean algorithm''' in a slightly different context. {{Euclid Note|2|VII}} {{Namedfor|Euclid|cat = Euclid}} == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Bézout's Lemma/Proof 2|next = Definition:Prime Number/Definition 2}}: $\S 3.11$ * {{BookReference|Number Theory|1971|George E. Andrews|prev = Euclidean Algorithm/Examples/341 and 527|next = Euclidean Algorithm/Demonstration}}: $\text {2-2}$ Divisibility: Theorem $\text {2-2}$ * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = GCD and LCM Distribute Over Each Other|next = Euclidean Algorithm/Euclid's Proof}}: Chapter $1$: Properties of the Natural Numbers: $\S 23 \zeta$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Prime not Divisor implies Coprime|next = Euclidean Algorithm/Demonstration}}: $\S 12$: Highest common factors and Euclid's algorithm * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Abel-Ruffini Theorem/Historical Note|next = Lamé's Theorem/Historical Note}}: $5$ * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Definition:Euclidean|next = Definition:Euclidean Construction|entry = Euclidean algorithm}} * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Definition:Perfect Number|next = Axiom:Euclid's Postulates|entry = Euclid's algorithm''' or '''Euclidean algorithm}} * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Definition:Greatest Common Divisor/Integers/Definition 2|next = Euclidean Algorithm/Demonstration}}: $\text{A}.1$: Number theory * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Definition:Algorithm/Linguistic Note|next = Definition:Greatest Common Divisor/Integers/Definition 1}}: $\S 1.1$: Algorithms: Algorithm $\text{E}$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Abel-Ruffini Theorem/Historical Note|next = Lamé's Theorem/Historical Note}}: $5$ * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Mathematician:Euclid|next = Definition:Euclidean Construction|entry = Euclidean algorithm}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Mathematician:Euclid|next = Lamé's Theorem|entry = Euclidean algorithm}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Mathematician:Euclid|next = Definition:Euclidean Construction|entry = Euclidean Algorithm}} Category:Euclidean Algorithm Category:Number Theory Category:Greatest Common Divisor 7sgt63ld7znycn4u25fcsh8r7s54l1w


Real Numbers are Close Packed

Real Numbers are Close Packed 0 1643 411495 410825 2019-07-03T20:54:04Z Prime.mover 59 wikitext text/x-wiki == Corollary to Between two Real Numbers exists Rational Number == <onlyinclude> :$\forall a, b \in \R: a < b \implies \paren {\exists c \in \R: a < c < b}$ </onlyinclude> That is, the set of real numbers is close packed. == Proof 1 == {{:Real Numbers are Close Packed/Proof 1}} == Proof 2 == {{:Real Numbers are Close Packed/Proof 2}} == Sources == * {{BookReference|Algebra Volume 1|1982|P.M. Cohn|edpage = Second Edition|ed = 2nd|prev = Quantifier/Examples/Existence of Multiplicative Identity|next = Equivocation of Nothing}}: Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(8)$ Category:Real Analysis Category:Real Numbers are Close Packed 6dpt6r61ivdbhad6ncibxrh7qi3z9gn


Group Action on Prime Power Order Subset 0 1995 404172 404159 2019-05-04T12:52:20Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $G$ be a finite group. Let $\mathbb S = \set {S \subseteq G: \card S = p^n}$ where $p$ is prime. That is, the set of all subsets of $G$ whose cardinality is the power of a prime number. Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements: :$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$. Then: === Stabilizer is $p$-Subgroup === {{:Group Action on Prime Power Order Subset/Stabilizer is p-Subgroup}} === Stabilizer of Maximal Power Order Subset === {{:Group Action on Prime Power Order Subset/Stabilizer of Maximal Power Order Subset}} Category:Examples of Group Actions Category:Group Action on Prime Power Order Subset Category:Group Action on Sets with k Elements kmqpneic98hd4d379weg4zt375fwfzw


Orbits of Group Action on Sets with Power of Prime Size 0 2003 404166 404165 2019-05-04T12:47:52Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $G$ be a finite group such that $\order G = k p^n$ where $p \nmid k$. Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$ Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements: :$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$ Then: === Orbit Length === {{:Orbits of Group Action on Sets with Power of Prime Size/Orbit Length}} === Orbit whose Length is not Divisible by $p$ === {{:Orbits of Group Action on Sets with Power of Prime Size/Orbit whose Length is not Divisible by p}} === Orbit whose Length is Divisible by $p$ === {{:Orbits of Group Action on Sets with Power of Prime Size/Orbit whose Length is Divisible by p}} Category:Group Action on Prime Power Order Subset Category:Orbits of Group Action on Sets with Power of Prime Size ljy7325x3umbtw6he4d6nyl1dh3jr9e


Axiom:Euclid's First Postulate

Axiom:Euclid's First Postulate 100 2125 439989 425102 2019-12-17T21:27:44Z Prime.mover 59 wikitext text/x-wiki == Postulate == <onlyinclude> A straight line segment can be drawn joining any two points. {{EuclidPostulateStatement|First|To draw a straight line from any point to any point.}} </onlyinclude> == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Definition:Euclid's Definitions - Book I/23 - Parallel Lines|next = Axiom:Euclid's Second Postulate}}: Book $\text{I}$. Postulates * {{BookReference|Probability Theory|1965|A.M. Arthurs|prev = Definition:Straight Line|next = Definition:Axiom}}: Chapter $2$: Probability and Discrete Sample Spaces: $2.1$ Introduction * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Axiom:Euclid's Postulates|next = Axiom:Euclid's Second Postulate|entry = Euclid's axioms}} Euclid Postulate 1 n70ocyqo261qg5q7e6kw8y1t8m51qtw


Axiom:Euclid's Second Postulate

Axiom:Euclid's Second Postulate 100 2126 425103 196201 2019-09-13T16:23:29Z Prime.mover 59 wikitext text/x-wiki == Postulate == <onlyinclude> A straight line segment can be extended indefinitely to form a straight line. {{EuclidPostulateStatement|Second|To produce a finite straight line continuously in a straight line.}} </onlyinclude> === Production === {{:Axiom:Euclid's Second Postulate/Production}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Axiom:Euclid's First Postulate|next = Axiom:Euclid's Third Postulate}}: Book $\text{I}$. Postulates * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Axiom:Euclid's First Postulate|next = Axiom:Euclid's Third Postulate|entry = Euclid's axioms}} Euclid Postulate 2 e9p10asae0xjolxihapf4tieho2npm3


Axiom:Euclid's Third Postulate

Axiom:Euclid's Third Postulate 100 2127 425104 196204 2019-09-13T16:24:18Z Prime.mover 59 wikitext text/x-wiki == Postulate == <onlyinclude> Given any line segment, a circle can be drawn using the segment as the radius with one endpoint as the center. {{EuclidPostulateStatement|Third|To describe a circle with any centre  and distance.}} </onlyinclude> == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Axiom:Euclid's Second Postulate|next = Axiom:Euclid's Fourth Postulate}}: Book $\text{I}$. Postulates * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Axiom:Euclid's Second Postulate|next = Axiom:Euclid's Fourth Postulate|entry = Euclid's axioms}} Euclid Postulate 3 6s9apddjmowqegm9i6175yvq7plt7fd


Axiom:Euclid's Fourth Postulate

Axiom:Euclid's Fourth Postulate 100 2128 430025 425105 2019-10-08T12:22:27Z Prime.mover 59 wikitext text/x-wiki == Postulate == <onlyinclude> All right angles are congruent. {{EuclidPostulateStatement|Fourth|That all right angles are equal to one another.}} </onlyinclude> == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Axiom:Euclid's Third Postulate|next = Axiom:Euclid's Fifth Postulate}}: Book $\text{I}$. Postulates * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Axiom:Euclid's Third Postulate|next = Axiom:Euclid's Fifth Postulate|entry = Euclid's axioms}} * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Axiom:Euclid's Common Notion 1|next = Definition:Axiom}}: Chapter $2$: The Logic of Shape: Euclid Euclid Postulate 4 gu9wlfhcdl178dovsnycj6mdlqgh6sq


Axiom:Euclid's Fifth Postulate

Axiom:Euclid's Fifth Postulate 100 2129 425109 425107 2019-09-13T16:40:36Z Prime.mover 59 wikitext text/x-wiki == {{AuthorRef|Euclid}}'s Statement == <onlyinclude> {{EuclidPostulateStatement|Fifth|If a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.}} </onlyinclude> There are many equivalent ways to state this postulate.  See below for a selection of them. == Parallel Postulate == {{:Axiom:Parallel Postulate}} == Playfair's Axiom == {{:Axiom:Playfair's Axiom}} == Other equivalent statements == Many further attempts have been made to formulate equivalent definitions of this axiom, often with a view to finding a proof which relies on the other four axioms. (Such attempts have been universally doomed to failure.) Here are a few examples, in approximate chronological order: === {{AuthorRef|Proclus Lycaeus|Proclus}} === :If a straight line intersects one of two parallels, it will intersect the other also. Straight lines parallel to the same straight line are parallel to one another. === {{AuthorRef|Posidonius of Apameia|Posidonius}} and {{AuthorRef|Geminus of Rhodes|Geminus}} === :There exist straight lines everywhere equidistant from one another. (This can be compared with {{AuthorRef|Proclus Lycaeus|Proclus}}' tacit assumption that "Parallels remain, throughout their length, at a finite distance from one another.") === {{AuthorRef|Adrien-Marie Legendre|Legendre}} === :There exists a triangle in which the sum of the three angles is equal to two right angles. === {{AuthorRef|John Wallis|Wallis}}, {{AuthorRef|Lazare Nicolas Marguerite Carnot|Carnot}}, {{AuthorRef|Pierre-Simon de Laplace|Laplace}} === :Given any figure, there exists a figure similar to it of any size we please. {{AuthorRef|Giovanni Girolamo Saccheri|Saccheri}} points out that it is necessary only to postulate that: :There exist two unequal triangles with equal angles. === {{AuthorRef|Adrien-Marie Legendre|Legendre}} (again) === :Through any point within an angle less than two-thirds of a right angle a straight line can always be drawn which meets both sides of the angle. === {{AuthorRef|Johann Friedrich Lorenz|Lorenz}} === :Every straight line through a point within an angle must meet one of the sides of the angle.<ref>{{AuthorRef|Johann Friedrich Lorenz}}: ''Grundriss der reinen und angewandten Mathematik'', 1791.</ref> === {{AuthorRef|Adrien-Marie Legendre|Legendre}} and {{AuthorRef|Farkas Wolfgang Bolyai|W. Bolyai}} === :Given any three points not in a straight line, there exists a circle passing through them. === {{AuthorRef|Carl Friedrich Gauss|Gauss}} === :If I could prove that a rectilineal triangle is possible the content of which is greater than any given area, I am in a position to prove perfectly rigorously the whole of geometry.<ref>{{AuthorRef|Carl Friedrich Gauss|Gauss}}, in a letter to {{AuthorRef|Farkas Wolfgang Bolyai|W. Bolyai}}, 1799.</ref> === {{AuthorRef|Julius Worpitzky|Worpitzky}} === :There exists no triangle in which every angle is as small as we please. === {{AuthorRef|Alexis Claude Clairaut|Clairault}} === :If in a quadrilateral three angles are right angles, the fourth angle is a right angle also. (1741) === {{AuthorRef|Giuseppe Veronese|Veronese}} === :If two straight lines are parallel, they are figures opposite to (or the reflex of) one another with respect to the middle points of all their transversal segments.<ref>{{AuthorRef|Giuseppe Veronese|Veronese}}: ''Elementi'', 1904.</ref> === {{AuthorRef|G. Ingrami|Ingrami}} === :Two parallel straight lines intercept, on every transversal which passes through the middle point of a segment included within them, another segment the middle point of which is the middle point of the first.<ref>{{AuthorRef|G. Ingrami|Ingrami}}: ''Elementi'', 1904.</ref> == Historical Note == {{:Axiom:Euclid's Fifth Postulate/Historical Note}} == Also see == * {{TarskiAxiomLink|axiom = Euclid's Axiom|pageName = Euclid's Fifth Postulate}} == References == <references/> == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Mathematician:Dionysius Lardner|next = Mathematician:Proclus Lycaeus}}: Preface to the Second Edition * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Axiom:Euclid's Fourth Postulate|next = Axiom:Euclid's Common Notion 1}}: Book $\text{I}$. Postulates * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Axiom:Euclid's Fourth Postulate|next = Axiom:Playfair's Axiom|entry = Euclid's axioms}} Euclid Postulate 5 cmj46mqk4dtm5m3ls5f1eonb6lckqmo


Axiom:Euclid's Common Notions

Axiom:Euclid's Common Notions 100 2634 445822 430022 2020-02-02T23:04:56Z Prime.mover 59 wikitext text/x-wiki == Common Notions == <onlyinclude> This is a set of axiomatic statements that appear at the start of Book $\text{I}$ of {{ElementsLink}}. === Common Notion 1 === {{:Axiom:Euclid's Common Notion 1}} === Common Notion 2 === {{:Axiom:Euclid's Common Notion 2}} === Common Notion 3 === {{:Axiom:Euclid's Common Notion 3}} === Common Notion 4 === {{:Axiom:Euclid's Common Notion 4}} === Common Notion 5 === {{:Axiom:Euclid's Common Notion 5}}</onlyinclude> == Historical Note == {{:Axiom:Euclid's Common Notions/Historical Note}} == References == <references/> == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Axiom:Euclid's Postulates|next = Axiom:Playfair's Axiom}}: Book $\text{I}$. Common Notions * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Axiom:Euclid's Postulates|next = Definition:Axiom}}: Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.) * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Definition:Euclidean Geometry|next = Axiom:Euclid's Postulates|entry = Euclidean geometry}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Definition:Euclidean Geometry|next = Axiom:Euclid's Postulates|entry = Euclidean geometry}} * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Definition:Euclid's Definitions|next = Axiom:Euclid's Postulates}}: Chapter $2$: The Logic of Shape: Euclid * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Axiom:Euclid's Postulates|next = Mathematician:Eudoxus of Cnidus|entry = Euclid's axioms}} Category:Euclid's Common Notions ey6aqgq3tenjaykk0290pw8jxo2pi4r


Closure of Subset of Metric Space by Convergent Sequence 0 3396 445080 334490 2020-01-28T21:42:27Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $M$ be a metric space. Let $C \subseteq M$. Let $x \in M$. Let $\map \cl C$ denote the closure of $C$ in $T$. Then $x \in \map \cl C$ {{iff}} there exists a sequence $\sequence {x_n}$ in $C$ which converges to $x$. == Proof == === Necessary Condition === Suppose there exists a sequence $\sequence {x_n}$ in $C$ which converges to $x$. Let $\epsilon > 0$. Then by definition: :$\exists N \in \N: \forall n > N: x_n \in \map {B_\epsilon} x$ where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$. Since $\forall n: x_n \in C$, it follows that: :$\forall \epsilon > 0: \map {B_\epsilon} x \cap C \ne \O$ Hence $x \in \map \cl C$. {{qed|lemma}} === Sufficient Condition === Now suppose $x \in \map \cl C$. By definition of closure: :$\forall n \in \N: \exists x_n \in C \cap \map {B_{1 / n} } x$ Thus clearly $\sequence {x_n}$ converges to $x$. {{handwaving|Lose the weasel word "clearly"}} {{qed}} == Sources == * {{BookReference|Introduction to Metric and Topological Spaces|1975|W.A. Sutherland|prev = Definition:Sequentially Compact Space|next = Definition:Cauchy Sequence (Metric Space)}}: $7.2$: Sequential compactness: Lemma $7.2.2$ Category:Metric Spaces Category:Sequences ipkattwqz1m3rifaregy8oby7lpkqwq


Series Law for Extremal Length/Rho is Well Defined 0 3435 435002 240506 2019-11-12T16:46:21Z Prime.mover 59 wikitext text/x-wiki {{MissingLinks|throughout}} == Lemma for Series Law for Extremal Length == {{explain|The details of what is to be proved needs to be stated here, in order that this page be appropriately self-contained.}} == Proof == To see that $\rho$ is a well-defined metric, we need to check that it transforms correctly when changing local coordinates. Let $z = \map z t$ and $w = \map w t$ be charts on the Riemann surface $X$. Let $\map {\rho_1^z} t$ and $\map {\rho_1^w} t$ be the coefficient functions when $\rho_1$ is expressed in the local coordinates $z$ and $w$, respectively. We use the analogous notation for $\rho_2$ and $\rho$. Since $\rho_j$ is a metric for $j \in \set {1, 2}$, we have: :$\map {\rho_j^w} t = \map {\rho_j^z} t \cdot \size {\dfrac {\d z} {\d w} }$ where $\dfrac {\d z} {\d w}$ denotes, the derivative of the coordinate change $z \circ w^{-1}$. Thus we have: {{begin-eqn}} {{eqn | l = \map {\rho^w} t       | r = \max \set {\map {\rho_1^w} t, \map {\rho_2^w} t}       | c =  }} {{eqn | r = \max \set {\map {\rho_1^z} t, \map {\rho_2^z} t} \cdot \size {\frac {\d z} {\d w} }       | c =  }} {{eqn | r = \map {\rho^z} t \cdot \size {\dfrac {\d z} {\d w} }       | c =  }} {{end-eqn}} This means that $\rho$ transforms correctly and is a metric, as desired. {{qed}} bp2yqxpiygxf9e7xgb07wg13fk5vska


Goldbach Conjecture

Goldbach Conjecture 0 3618 492809 474006 2020-10-06T06:23:02Z Prime.mover 59 wikitext text/x-wiki == Conjecture == <onlyinclude> Every even integer greater than $2$ is the sum of two primes. </onlyinclude> === Marginal Conjecture === {{:Goldbach Conjecture/Marginal}} {{Hilbert23|8b}} == Landau's Problems == This is the $1$st of Landau's problems. == Also see == * Goldbach's Weak Conjecture * Goldbach Conjecture implies Goldbach's Marginal Conjecture {{Namedfor|Christian Goldbach|cat = Goldbach}} == Historical Note == {{:Goldbach Conjecture/Historical Note}} == Sources == * {{BookReference|Algebra Volume 1|1982|P.M. Cohn|edpage = Second Edition|ed = 2nd|prev = Proof by Counterexample|next = Definition:Sufficient Condition}}: Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Tamref's Last Theorem/Mistake|next = Definition:Binary Notation}}: $2$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 5^x + 12^y equals 13^z has Unique Solution|next = Definition:Binary Notation}}: $2$ * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Gödel's Incompleteness Theorems/Historical Note|next = Goldbach Conjecture/Historical Note|entry = Goldbach's conjecture}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Gödel's Incompleteness Theorems/Historical Note|next = Goldbach Conjecture/Historical Note|entry = Goldbach's conjecture}} * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Goldbach Conjecture/Marginal/Historical Note|next = Goldbach Conjecture/Historical Note}}: Chapter $7$: Patterns in Numbers: Fermat Category:Unproven Hypotheses Category:Prime Numbers Category:Goldbach Conjecture Category:Landau's Problems huazswegt2nronbgpmj6n6cxmtr4p3g


Gauss's Lemma (Number Theory)

Gauss's Lemma (Number Theory) 0 3718 357332 267473 2018-05-27T08:56:10Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $p$ be an odd prime. Let $a \in \Z: a \not \equiv 0 \pmod p$. Let $S = \left\{{a, 2 a, 3 a, \ldots, \dfrac {p - 1} 2 a}\right\}$. Let $n$ denote the number of elements of $S$ whose least positive residue modulo $p$ is greater than $\dfrac p 2$. Then: :$\left({\dfrac a p}\right) = \left({-1}\right)^n$ where $\left({\dfrac a p}\right)$ is the Legendre symbol. == Proof == First note that no two elements of $s$ are congruent modulo $p$: Let $r a \equiv s a \pmod p$ for some $r, s$ such that $1 \le s \le r \le \dfrac {p-1} 2$. As $a \perp p$, we have that $r a \equiv s a \implies r \equiv s \pmod p$ from Cancellability of Congruences. Now $r \equiv s \pmod p$ can happen only when $r = s$ as $0 \le r - s \le \dfrac {p-1} 2$. Next, no element of $S$ is congruent modulo $p$ to $0$. This is because $r a \equiv 0 \pmod p$ when $a \not \equiv 0$ requires $r \equiv 0$ which doesn't happen. Now, we create $S'$ from $S$ by replacing each element of $S$ by its least positive residue modulo $p$. Arranging $S'$ into increasing order, we get: :$S' = \left\{{b_1, b_2, \ldots, b_m, c_1, c_2, \ldots, c_n}\right\}$ where $b_m < \dfrac p 2 < c_1$ and $m + n = \dfrac {p-1}2$. As $p$ is an odd prime, $\dfrac p 2$ is not an integer so neither $b_m$ nor $c_1$ can be equal to $\dfrac p 2$. Now we let $S'' = \left\{{b_1, b_2, \ldots, b_m, p-c_1, p-c_2, \ldots, p-c_n}\right\}$ We need to show that $S'' = \left\{{1, 2, 3, \ldots, \dfrac {p-1} 2}\right\}$. Now: * all the elements of $S''$ are positive, as all the $c_j < p$ * the largest one is either $b_m$ or $p - c_1$ * there are $\dfrac {p-1} 2$ of them * they are all less than $\dfrac p 2$. So $S''$ contains $\dfrac {p-1} 2$ positive integers all less than $\dfrac p 2$. So all we need to do is show that they are distinct. As all the elements of $S'$ are distinct, the $m$ elements $b_1, b_2, \ldots, b_m$ are distinct. Similarly, so are the $n$ elements $p-c_1, p-c_2, \ldots, p-c_n$. So, we suppose $b_i = p - c_j$ for some $i, j$ and try to derive a contradiction. So: :$p = b_i + c_j \equiv r a + s a \pmod p$ for some $r, s$ such that $1 \le r, s \le \dfrac {p-1} 2$. But from: :$p \equiv \left({r+s}\right) a \pmod p$ and $a \not \equiv 0 \pmod p$ we apply Euclid's Lemma to get $r + s \equiv 0 \pmod p$. But this is impossible since $2 \le r + s \le p-1$. This establishes that $S'' = \left\{{1, 2, 3, \ldots, \dfrac {p-1} 2}\right\}$. Now, we multiply all the elements of $S''$ together: {{begin-eqn}} {{eqn | l = \left({\frac {p - 1} 2}\right)!       | r = b_1 b_2 \cdots b_m \left({p - c_1}\right) \left({p - c_2}\right) \cdots \left({p - c_n}\right)       | c =  }} {{eqn | o = \equiv       | r = b_1 b_2 \cdots b_m \left({-c_1}\right) \left({-c_2}\right) \cdots \left({-c_n}\right)       | rr= \pmod p       | c =  }} {{eqn | o = \equiv       | r = \left({-1}\right)^n b_1 b_2 \cdots b_m c_1 c_2 \cdots c_n       | rr= \pmod p       | c =  }} {{eqn | o = \equiv       | r = \left({-1}\right)^n a \times 2 a \times 3 a \times \cdots \frac{p - 1} 2 a       | rr= \pmod p       | c =  }} {{eqn | o = \equiv       | r = \left({-1}\right)^n a^{\left({\frac {p - 1} 2}\right)} \times \left({\frac {p - 1} 2}\right)!       | rr=\pmod p       | c = as the elements in $S'$ are congruent, in some order, to those in $S$ }} {{end-eqn}} Now, from GCD with Prime: :$\gcd \left\{ {p, \left({\dfrac {p - 1} 2}\right)!}\right\} = 1$ So from Cancellability of Congruences the term $\left({\dfrac {p - 1} 2}\right)!$ can be cancelled from both sides of the above congruence: :$1 \equiv \left({-1}\right)^n a^{\left({\frac {p - 1} 2}\right)} \pmod p$ Finally, from the definition of the Legendre symbol: :$\left({\dfrac a p}\right) \equiv a^{\left({\frac {p - 1} 2}\right)} \pmod p$ Multiplying both sides of the congruence by $\left({-1}\right)^n$: :$\left({\dfrac a p}\right) = \left({-1}\right)^n$ {{qed}} {{Namedfor|Carl Friedrich Gauss|cat = Gauss}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Definition:Legendre Symbol/Definition 1|next = Second Supplement to Law of Quadratic Reciprocity}}: $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $47 \ \text {a)}$ Category:Legendre Symbol s2sbf32tuemp74ygoq89qclfms0pkzr


Euclid's Lemma for Prime Divisors

Euclid's Lemma for Prime Divisors 0 3733 462866 462838 2020-04-18T20:11:09Z Prime.mover 59 wikitext text/x-wiki {{Previous POTW|2 April 2009|10 April 2009}} == Lemma == <onlyinclude> Let $p$ be a prime number. Let $a$ and $b$ be integers such that: : $p \divides a b$ where $\divides$ means '''is a divisor of'''. Then $p \divides a$ or $p \divides b$. </onlyinclude> === General Result === {{:Euclid's Lemma for Prime Divisors/General Result}} === Corollary === {{:Euclid's Lemma for Prime Divisors/Corollary}} == Proof 1 == {{:Euclid's Lemma for Prime Divisors/Proof 1}} == Proof 2 == {{:Euclid's Lemma for Prime Divisors/Proof 2}} == Proof 3 == {{:Euclid's Lemma for Prime Divisors/Proof 3}} == Also presented as == Some sources present this as: Let $p$ be a prime number. Let $a$ and $b$ be integers such that: :$a b \equiv 0 \pmod p$ Then either $a \equiv 0 \pmod p$ or $b \equiv 0 \pmod p$. == Also known as == Some sources give the name of this as '''Euclid's first theorem'''. {{Namedfor|Euclid}} == Also see == Some sources use this property to '''define''' a prime number. == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Modulo Arithmetic/Examples/Residue of 2^512 Modulo 5|next = Converse of Euclid's Lemma does not Hold}}: Chapter $2$. Equivalence Relations: Exercise $7$ * {{BookReference |An Introduction to the Theory of Numbers|1979|G.H. Hardy|author2 = E.M. Wright|ed = 5th|edpage = Fifth Edition|prev = Fundamental Theorem of Arithmetic|next = Euclid's Lemma for Prime Divisors/General Result}}: $\text I$: The Series of Primes: $1.3$ Statement of the fundamental theorem of arithmetic: Theorem $3$ * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Expression for Integer as Product of Primes is Unique|next = Definition:Irrational Number/Historical Note}}: Chapter $\text {B}.16$: The Sequence of Primes: Problem $1 \ \text{(b)}$ Category:Prime Numbers Category:Euclid's Lemma Category:Euclid's Lemma for Prime Divisors pjrpdvhh81ve5wvdnyzmx4xl0yh6vyz


Continued Fraction Algorithm

Continued Fraction Algorithm 0 3829 470623 470622 2020-05-26T06:30:25Z Prime.mover 59 wikitext text/x-wiki == Algorithm == <onlyinclude> The '''Continued Fraction Algorithm''' is a method for finding the continued fraction expansion for any irrational number to as many partial quotients as desired. Let $x_0$ be the irrational number in question. The steps are: {{begin-eqn}} {{eqn | n = 1       | l = k       | r = 0       | c = initialise }} {{eqn | n = 2       | l = a_k       | r = \floor {x_k}       | c = the $k$th partial quotient (that is, $a_k$) is the integer part of $x_k$ }} {{eqn | n = 3       | l = x_{k + 1}       | r = \frac 1 {x_k - a_k}       | c = the subsequent term is the reciprocal of the fractional part of $x_k$ }} {{eqn | n = 4       | l = k       | r = k + 1       | c = increase $k$ by $1$ then go to step $(2)$ }} {{end-eqn}} Then $x_0 = \sqbrk {a_0, a_1, a_2, \ldots}$ is the required continued fraction expansion. </onlyinclude> == Examples == {{:Continued Fraction Algorithm/Examples}} == Proof 1 == {{:Continued Fraction Algorithm/Proof 1}} Category:Continued Fractions ijqk8sif36y0gd1bh8diivul5x2dpsa


Pell's Equation

Pell's Equation 0 3848 446491 431415 2020-02-05T10:52:20Z Prime.mover 59 wikitext text/x-wiki == Definition == The Diophantine equation: :$x^2 - n y^2 = 1$ is known as '''Pell's equation'''. == Solution == Let the continued fraction of $\sqrt n$ have a cycle whose length is $s$: :$\sqrt n = \sqbrk {a_1 \sequence {a_2, a_3, \ldots, a_{s + 1} } }$ Let $a_n = \dfrac {p_n} {q_n}$ be a convergent of $\sqrt n$. Then: :${p_{r s} }^2 - n {q_{r s} }^2 = \paren {-1}^{r s}$ for $r = 1, 2, 3, \ldots$ and all solutions of: :$x^2 - n y^2 = \pm 1$ are given in this way. == Proof == First note that if $x = p, y = q$ is a positive solution of $x^2 - n y^2 = 1$ then $\dfrac p q$ is a convergent of $\sqrt n$. The continued fraction of $\sqrt n$ is periodic from Continued Fraction Expansion of Irrational Square Root and of the form: :$\sqbrk {a \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a} }$ or  :$\sqbrk {a \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a} }$ For each $r \ge 1$ we can write $\sqrt n$ as the (non-simple) finite continued fraction: :$\sqrt n = \sqbrk {a \sequence {b_1, b_2, \ldots, b_2, b_1, 2 a, b_1, b_2, \ldots, b_2, b_1, x} }$ which has a total of $r s + 1$ partial quotients. The last element $x$ is of course not an integer. What we do have, though, is: {{begin-eqn}} {{eqn | l = x       | r = \sqbrk {\sequence {2 a, b_1, b_2, \ldots, b_2, b_1} }       | c =  }} {{eqn | r = a + \sqbrk {a, \sequence {b_1, b_2, \ldots, b_2, b_1, 2 a} }       | c =  }} {{eqn | r = a + \sqrt n       | c =  }} {{end-eqn}} The final three convergents in the above FCF are: :$\dfrac {p_{r s - 1} } {q_{r s - 1} }, \quad \dfrac {p_{r s} } {q_{r s} }, \quad \dfrac {x p_{r s} + p_{r s - 1} } {x q_{r s} + q_{r s - 1} }$ The last one of these equals $\sqrt n$ itself. So: :$\sqrt n \paren {x q_{r s} + q_{r s - 1} } = \paren {x p_{r s} + p_{r s - 1} }$ Substituting $a + \sqrt n$ for $x$, we get: :$\sqrt n \paren {\paren {a + \sqrt n} q_{r s} + q_{r s - 1} } = \paren {\paren {a + \sqrt n} p_{r s} + p_{r s - 1} }$ This simplifies to: :$\sqrt n \paren {a q_{r s} + q_{r s - 1} - p_{r s} } = a p_{r s} + p_{r s - 1} - n q_{r s}$ The {{RHS}} of this is an integer while the {{LHS}} is $\sqrt n$ times an integer. Since $\sqrt n$ is irrational, the only way that can happen is if both sides equal zero. This gives us: {{begin-eqn}} {{eqn | n = 1       | l = a q_{r s} + q_{r s - 1}       | r = p_{r s} }} {{eqn | n = 2       | l = a p_{r s} + p_{r s - 1}       | r = n q_{r s} }} {{end-eqn}} Multiplying $(1)$ by $p_{r s}$, $(2)$ by $q_{r s}$ and then subtracting: :$p_{r s}^2 - n q_{r s}^2 = p_{r s} q_{r s - 1} - p_{r s - 1} q_{r s}$ By Difference between Adjacent Convergents of Simple Continued Fraction, the {{RHS}} of this is $\paren {-1}^{r s}$. When the cycle length $s$ of the continued fraction of $\sqrt n$ is even, we have $\paren {-1}^{r s} = 1$. Hence $x = p_{r s}, y = q_{r s}$ is a solution to Pell's Equation for each $r \ge 1$. When $s$ is odd, though: : $x = p_{r s}, y = q_{r s}$ is a solution of $x^2 - n y^2 = -1$ when $r$ is odd : $x = p_{r s}, y = q_{r s}$ is a solution of $x^2 - n y^2 = 1$ when $r$ is even. {{qed}} == Sequence == {{:Pell's Equation/Sequence}} == Examples == {{:Pell's Equation/Examples}} == Also known as == A particular instance of '''Pell's equation''' can also be seen referred to as '''a Pellian equation'''. {{Namedfor|John Pell|cat = Pell}} == Historical Note == {{:Pell's Equation/Historical Note}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Best Rational Approximations to Root 2 generate Pythagorean Triples|next = Sprague's Property of Root 2}}: $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$ * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 4,729,494|next = Pell's Equation/Historical Note}}: $4,729,494$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Best Rational Approximations to Root 2 generate Pythagorean Triples|next = Square Root of 2 as Sum of Egyptian Fractions}}: $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Definition:Emirp/Sequence|next = Pell's Equation/Examples/13}}: $13$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 4,729,494|next = Pell's Equation/Historical Note}}: $4,729,494$ * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Mathematician:Charles Sanders Peirce|next = Definition:Pencil|entry = Pell's equation}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Mathematician:Charles Sanders Peirce|next = Definition:Pencil|entry = Pell's equation}} * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Mathematician:Pierre de Fermat|next = Mathematician:Carl Friedrich Gauss}}: Chapter $7$: Patterns in Numbers: Fermat * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Mathematician:John Pell|next = Definition:Pendulum|entry = Pell's equation}} * {{MathWorld|Pell Equation|PellEquation}} Category:Diophantine Equations Category:Pell's Equation rh0e3rxucadso4h7vp3zw8sc7ymemco


Clear Registers Program

Clear Registers Program 0 3990 51940 14186 2011-04-13T21:48:52Z Prime.mover 59 wikitext text/x-wiki == URM Program == Let $a, b \in \N$ be natural numbers such that $0 < a$. Then we define the URM program $Z \left({a, b}\right)$ to be: {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $Z \left({a}\right)$ |- | align="right" | $2$ || | align="left" | $Z \left({a + 1}\right)$ |- | align="right" | $3$ || | align="left" | $Z \left({a + 2}\right)$ |- | align="right" | $\vdots$ || | align="left" | $\vdots$ |- | align="right" | $b - a + 1$ || | align="left" | $Z \left({b}\right)$ |} This program clears (that is, sets to $0$) all the registers from $R_a$ through to $R_b$. If $a > b$ then $Z \left({a, b}\right)$ is the null URM program. The length of $Z \left({a, b}\right)$ is: :$\lambda \left({Z \left({a, b}\right)}\right) = \begin{cases} 0 & : a > b \\ b - a + 1 & : a \le b \end{cases}$ == Proof == Self-evident. {{qed}} Category:URM Programs n08c640hq7dvmd9vt1l4oe1uij6w3bn


Biconditional Properties

Biconditional Properties 0 4483 126281 126278 2013-01-13T13:21:40Z Prime.mover 59 wikitext text/x-wiki == Theorems == === Biconditional is Commutative === {{:Biconditional is Commutative}} === Biconditional is Associative === {{:Biconditional is Associative}} === Biconditional is Reflexive === {{:Biconditional is Reflexive}} Category:Biconditional bt4u4kllobeob13f69cagqquif51ykv


Method of Truth Tables

Method of Truth Tables 0 4495 493612 492247 2020-10-10T12:24:57Z Prime.mover 59 wikitext text/x-wiki == Proof Technique == The '''method of truth tables''' is a technique for determining the validity of propositional formulas with respect to boolean interpretations. In particular, for discerning if a propositional formula is a tautology for boolean interpretations. To start with, we establish the characteristic truth tables of the various logical connectives. We write one row for each boolean interpretation of the set of variables that we are concerned with. From Count of Rows of Truth Table this amounts to $2^n$ rows if there are $n$ variables. There are therefore two rows in the truth table for the only non-trivial unary connective: $\begin{array}{|c||c|} \hline p & \neg p \\ \hline F & T \\ T & F \\ \hline \end{array}$ ... and four rows in the truth tables for the binary connectives (the usual subset of which is given below): $\begin{array}{|cc||c|c|c|c|c|c|c|c|} \hline p & q & p \land q  & p \lor q & p \implies q & p \iff q & p \impliedby q & p \oplus q & p \uparrow q & p \downarrow q \\ \hline F & F & F & F & T & T & T & F & T & T \\ F & T & F & T & T & F & F & T & T & F \\ T & F & F & T & F & F & T & T & T & F \\ T & T & T & T & T & T & T & F & F & F \\ \hline \end{array}$ There are various sorts of proof this technique can be put to, as follows. These will be illustrated by various examples. == Proof of Tautology == {{:Method of Truth Tables/Proof of Tautology}} == Proof of Interderivability == {{:Method of Truth Tables/Proof of Interderivability}} == Proof of Logical Implication == {{:Method of Truth Tables/Proof of Logical Implication}} == Indirect Technique == {{:Method of Truth Tables/Indirect Technique}} == Notational Convenience == It is not actually necessary to include the truth values of the variables themselves (as we have done in the leftmost columns). One is equally justified to write this: $\begin{array}{ccccccc} ((p & \implies & q) & \implies & p) & \implies & p \\ \hline F & T & F & F & F & T & F \\ F & T & T & F & F & T & F \\ T & F & F & T & T & T & T \\ T & T & T & T & T & T & T \\ \end{array}$ and it serves just as well. However, it can help to clarify the derivation, as well as making the truth table easier to construct, if they ''are'' included. It's a matter of personal taste. == Also known as == The '''method of truth tables''' is also sometimes referred to as the '''method of matrices'''. However, it can be argued that the term '''truth table''' is more specific than '''matrix''' and hence preferable. == Comment == Note that solution by truth table is valid only for Aristotelian logic, as it takes for granted the Law of Excluded Middle and the Principle of Non-Contradiction. Within that context, it is a completely mechanical procedure and about as exciting as a strip-tease artist who starts the performance naked. == Sources == * {{BookReference|Introduction to Logic and to the Methodology of Deductive Sciences|1946|Alfred Tarski|ed = 2nd|edpage = Second Edition|prev = Hypothetical Syllogism|next = Definition:Logical Not/Notational Variants}}: $\S \text{II}.13$: Symbolism of sentential calculus * {{BookReference|Introduction to Symbolic Logic|1959|A.H. Basson|author2 = D.J. O'Connor|ed = 3rd|edpage = Third Edition|prev = Definition:Therefore|next = Method of Truth Tables/Proof of Tautology}}: $\S 3.3$: The Construction and Application of Truth-Tables * {{BookReference|Logic: Techniques of Formal Reasoning|1964|Donald Kalish|author2 = Richard Montague|prev = Definition:Truth Table|next = Definition:Proper Name}}: $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 6$ * {{BookReference|Beginning Logic|1965|E.J. Lemmon|prev = Definition:Biconditional/Truth Table/Matrix Form|next = Method of Truth Tables/Proof of Tautology}}: $\S 2.3$: Truth-Tables * {{BookReference|Symbolic Logic|1973|Irving M. Copi|ed = 4th|edpage = Fourth Edition|prev = Modus Tollendo Ponens/Sequent Form/Case 1/Proof 2|next = Modus Ponendo Ponens}}: $2.3$: Argument Forms and Truth Tables * {{BookReference|Elementary Logic|1980|D.J. O'Connor|author2 = Betty Powell|prev = Definition:Conditional/Truth Table|next = Method of Truth Tables/Proof of Tautology}}: $\S \text{I}: 4$: Using the Definitions * {{BookReference|Logic for Mathematicians|1988|Alan G. Hamilton|ed = 2nd|edpage = Second Edition|prev = Bottom-Up Specification of Propositional Logic/Examples/Example 1|next = Rule of Material Implication/Formulation 1/Proof}}: $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables * {{BookReference|Mathematical Logic and Computability|1996|H. Jerome Keisler|author2 = Joel Robbin|prev = Definition:Truth Table|next = Definition:Tautology/Formal Semantics}}: $\S 1.6$: Truth Tables and Tautologies * {{BookReference|Logic in Computer Science: Modelling and reasoning about systems|2000|Michael R.A. Huth|author2 = Mark D. Ryan|prev = Count of Rows of Truth Table|next = Definition:Boolean Interpretation}}: $\S 1.4.1$: The meaning of logical connectives Category:Proof Techniques Category:Truth Tables gsj9vhqlxuhxgg48vdazq5xccjnroqr


Finished Set Lemma

Finished Set Lemma 0 4617 360477 320261 2018-07-14T09:28:51Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $\Delta$ be a finished set of WFFs of propositional logic. Then any model of the set of basic WFFs in $\Delta$ is a model of $\Delta$. === Corollary === {{:Finished Set Lemma/Corollary}} == Proof == Let $\mathcal M$ be a model of the set of basic WFFs in $\Delta$. We need to show that $\mathcal M \models \Delta$. That is, that $\mathcal M \models \mathbf C$ for each $\mathbf C \in \Delta$. Now, let $R \left({n}\right)$ be a propositional function on the set of natural numbers $\N$ such that: :$R \left({n}\right) = T$ iff: for every WFF $\mathbf C$, if $\mathbf C \in \Delta$ and $\mathbf C$ has length at most $n$, then $\mathcal M \models \mathbf C$. It is clear that $R \left({0}\right), R \left({1}\right), R \left({2}\right)$ are true because every WFF length 2 or less is basic, and $\mathcal M$ models every basic WFF in $\Delta$. So, assume $R \left({k}\right)$ is true for some $k \in \N$. Suppose $\mathbf C$ has length at most $k+1$ and belongs to $\Delta$. By examining each of the cases in the definition of finished set, we see that since $\mathcal M$ models every WFF in $\Delta$ of length at most $k$, then $\mathcal M$ models $\mathbf C$. Thus $R \left({k+1}\right)$ is true. Thus by strong induction, $R \left({n}\right)$ is true for all $n \in \N$. Hence the result. {{qed}} == Sources == * {{BookReference|Mathematical Logic and Computability|1996|H. Jerome Keisler|author2 = Joel Robbin|prev = Finished Set Lemma/Corollary|next = Completeness Theorem for Propositional Tableaus and Boolean Interpretations}}: $\S 1.9$: Finished Sets: Lemma $1.9.1$ Category:Propositional Tableaus Category:Boolean Interpretations 7zy3lgsjgzkgyb9ht1nctbm3oyc0rf0


Finite Main Lemma of Propositional Tableaus 0 4618 320147 319775 2017-10-08T18:35:01Z Lord Farin 560 wikitext text/x-wiki {{mergeto|Main Lemma of Propositional Tableaus}} == Lemma == Let $\mathbf H$ be a finite set of WFFs of propositional logic. Either $\mathbf H$ has a tableau confutation or $\mathbf H$ has a model. == Proof == Let $\mathbf H$ be a finite set of WFFs of propositional logic which does not have a tableau confutation. By the Tableau Extension Lemma, the tableau which consists only of a root node with hypothesis set $\mathbf H$ can be extended into a finite finished tableau $T$. The tableau $T$ still has root $\mathbf H$. Since $T$ is not a confutation, it has a finished branch $\Gamma$. By the Corollary to the Finished Branch Lemma, $\Gamma$ has a model, $\mathcal M$, say. In particular, $\mathcal M$ is a model of $\mathbf H$ as required. {{qed}} == Comment == From Tableau Confutation implies Unsatisfiable, we already know that $\mathbf H$ can not have both a tableau confutation ''and'' a model. This result gives us that $\mathbf H$ has a tableau confutation iff $\mathbf H$ does not have a model. == Sources == * {{BookReference|Mathematical Logic and Computability|1996|H. Jerome Keisler|author2=Joel Robbin|prev=Completeness Theorem for Propositional Tableaus and Boolean Interpretations|next=Definition:Finished Branch of Propositional Tableau}}: $\S 1.10$: Completeness: Lemma $1.10.1$ Category:Propositional Tableaus i42my72x1ywpf40t08rxv3d9x03p0kr


Tableau Extension Lemma

Tableau Extension Lemma 0 4626 360478 320137 2018-07-14T09:29:24Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $T$ be a finite propositional tableau. Let its hypothesis set $\mathbf H$ be finite. Then $T$ can be extended to a finished finite propositional tableau $T'$ which also has hypothesis set $\mathbf H$. === General Statement === {{:Tableau Extension Lemma/General Statement}} == Proof == Let $\mathbf A$ be a WFF of propositional logic at a node $t$ such that: * $\mathbf A$ is not a basic WFF; * There is a non-contradictory branch through $t$ on which $\mathbf A$ is not used. Such a WFF $\mathbf A$ will be called '''unused'''. Note that a tableau is finished iff there are no unused WFFs in the tableau. Let $u \left({T}\right)$ be the length of the longest unused WFF in $T$. If $T$ is finished, we set $u \left({T}\right) = 0$. Since there are only finitely many WFFs in $T$, the number $u \left({T}\right)$ must exist. Let $R \left({n}\right)$ be the proposition that every finite propositional tableau with hypothesis set $\mathbf H$ and with $u \left({T}\right) < n$ can be extended to a finite finished tableau. That is, $R \left({n}\right)$ is an assertion that this lemma is true whenever $u \left({T}\right) < n$. The statement $R \left({1}\right)$ is true, because a tableau $T$ such that $u \left({T}\right) < 1$ is already finished. So, let us assume the truth of $R \left({k}\right)$. Suppose that $u \left({T}\right) < k+1$. We extend $T$ to a new tableau $T'$ by using every unused WFF $\mathbf A$ in $T$ once on every non-contradictory branch through $\mathbf A$. Each of the unused WFFs in $T$ is used in the new tableau $T'$. Also, each new WFF which was added when forming $T'$ has a length less than $u \left({T}\right)$, because the added WFFs are always shorter than the used WFFs. So $u \left({T'}\right) < u \left({T}\right) < k + 1$, so $u \left({T'}\right) < k$. So, by the induction hypothesis $R \left({k}\right)$, there is a finite finished extension $T''$ of $T'$. $T''$ is also a finished extension of $T$. Hence $R \left({k+1}\right)$, and the result follows by strong induction. {{qed}} == Sources == * {{BookReference|Mathematical Logic and Computability|1996|H. Jerome Keisler|author2 = Joel Robbin|prev = Definition:Extension of Propositional Tableau/Definition 1|next = Extended Completeness Theorem for Propositional Tableaus and Boolean Interpretations}}: $\S 1.10$: Completeness: Lemma $1.10.2$ Category:Propositional Tableaus 0ojqs4qlskowxutkv14mp49d96ibfip


Countable Hypothesis Set has Finished Tableau 0 4630 319779 227105 2017-10-04T21:10:26Z Lord Farin 560 wikitext text/x-wiki == Lemma == Let $\mathbf H$ be a countable set of WFFs of propositional logic. Then there exists a finished tableau whose root node is $\mathbf H$. == Proof == The Tableau Extension Lemma shows that each finite hypothesis set $\mathbf H$ is the root of some finished tableau. It remains, then, to show that the result still applies when $\mathbf H = \left\{{\mathbf A_1, \ldots, \mathbf A_n, \ldots}\right\}$ is countably infinite. Let $\mathbf H_n = \left\{{\mathbf A_1, \ldots, \mathbf A_n}\right\} \subset \mathbf H$. We will say that a finite tableau $T_n$ with root $\mathbf H$ is '''finished for $\mathbf H_n$''' if the tableau $T'_n$ is finished where: : $T'_n$ is the same as $T_n$ except: : $T'_n$ has the root $\mathbf H_n$ instead of $\mathbf H$. We can use the Tableau Extension Lemma countably many times, and get a sequence of finite tableaus $T_0, T_1, \ldots, T_n, \ldots$ such that: : $T_0$ has only a root node; : For each $n > 0$, $T_n$ is an extension of $T_{n-1}$ such that $T_n$ is '''finished for $\mathbf H_n$'''; : For each $n > 0$, $T_n$ has the property that no contradictory branch $\Gamma$ of $T_{n-1}$ gets extended when forming $T_n$. Now, let $T$ be the union $\displaystyle T = \bigcup_{k \mathop = 0}^\infty T_k$. Let $\Gamma$ be a branch of $T$. Suppose $\Gamma$ is contradictory, with complementary pair $\mathbf A, \neg \mathbf A$. Then $\exists n \in \N$ such that both $\mathbf A$ and $\neg \mathbf A$ are in $T_n$. Then $\Gamma \cap T_n$ is already a contradictory branch of $T_n$. So, by our method of construction of $T$, this branch $\Gamma \cap T_n$ is never extended past stage $n$, so $\Gamma = \Gamma \cap T_n$ and $\Gamma$ is finite. On the other hand, suppose $\Gamma$ is not contradictory. Then the construction ensures that $\Gamma$ is a finished branch. So $T$ is a finished tableau whose root node is $\mathbf H$. {{qed}} == Sources == * {{BookReference|Mathematical Logic and Computability|1996|H. Jerome Keisler|author2 = Joel Robbin|prev = Main Lemma of Propositional Tableaus|next = König's Tree Lemma/Proof 1}}: $\S 1.11$: Compactness: Lemma $1.11.2$ Category:Propositional Tableaus klz60bypzjrzi04gxv4n71qk3hi6os3


Main Lemma of Propositional Tableaus

Main Lemma of Propositional Tableaus 0 4639 320139 320121 2017-10-08T18:30:09Z Lord Farin 560 wikitext text/x-wiki == Lemma == Let $\mathbf H$ be a countable set of WFFs of propositional logic. Either $\mathbf H$ has a tableau confutation or $\mathbf H$ has a model. == Proof == If $\mathbf H$ is finite, then the Finite Main Lemma of Propositional Tableaus applies. So, assume that $\mathbf H$ is countably infinite. Suppose $\mathbf H$ does not have a tableau confutation. By Countable Hypothesis Set has Finished Tableau, there is a finished tableau $T$ with hypothesis set $\mathbf H$. By Finished Propositional Tableau has Finished Branch or is Confutation, as $T$ is (by hypothesis) not a confutation, it must have a finished branch; call it $\Gamma$. By the Finished Set Lemma, the set of WFFs of propositional logic on $\Gamma$ has a model $\mathcal M$. Finally, because all the WFFs in the hypothesis set occur on $\Gamma$, $\mathcal M \models \mathbf H$. {{qed}} == Sources == * {{BookReference|Mathematical Logic and Computability|1996|H. Jerome Keisler|author2=Joel Robbin|prev=Extended Completeness Theorem for Propositional Tableaus and Boolean Interpretations|next=Countable Hypothesis Set has Finished Tableau}}: $\S 1.11$: Compactness: Lemma $1.11.1$ Category:Propositional Tableaus 2mrm61u2wv571x6yzfbqp8a4sb3wg80


Polynomial Long Division

Polynomial Long Division 0 4873 240053 141276 2015-12-06T10:08:51Z Prime.mover 59 wikitext text/x-wiki == Technique == Let $P_n \left({x}\right)$ be a polynomial in $x$ of degree $n$. Let $Q_m \left({x}\right)$ be a polynomial in $x$ of degree $m$ where $m \le n$. Then $P_n \left({x}\right)$ can be expressed in the form: :$P_n \left({x}\right) \equiv Q_m \left({x}\right) D_{n-m} \left({x}\right) + R_k \left({x}\right)$ {{explain|what is $\equiv$ being used for here?}} where: :$D_{n-m} \left({x}\right)$ is a polynomial in $x$ of degree $n-m$ :$R_k \left({x}\right)$ is a polynomial in $x$ of degree $k$, where $k < m$, or may be null. Hence we can define $\dfrac {P_n \left({x}\right)} {Q_m \left({x}\right)}$: :$\dfrac {P_n \left({x}\right)} {Q_m \left({x}\right)} = D_{n - m} \left({x}\right) + \dfrac {R_k \left({x}\right)} {Q_m \left({x}\right)}$ The polynomial $R_k \left({x}\right)$ is called the '''remainder'''. The procedure for working out what $D_{n - m} \left({x}\right)$ and $R_k \left({x}\right)$ are is called '''(polynomial) long division'''. {{explain|Establish the precise versions of polynomial that are invoked via the links on this page.}} == Proof == Let $\displaystyle P_n \left({x}\right) = \sum_{j \mathop = 0}^n p_j x^j$. Let $\displaystyle Q_m \left({x}\right) = \sum_{j \mathop = 0}^m q_j x^j$. First calculate $\displaystyle Q'_m \left({x}\right) = Q_m \left({x}\right) \times \frac {p_n} {q_m} x^{n-m}$. This gives: {{begin-eqn}} {{eqn | l = Q'_m \left({x}\right)       | r = \sum_{j \mathop = 0}^m \frac {p_n q_j} {q_m} x^{n - m + j}       | c =  }} {{eqn | r = \sum_{j \mathop = n-m}^n \frac {p_n q_{j - n + m} } {q_m} x^j       | c =  }} {{eqn | r = p_n x^n + \sum_{j \mathop = n - m}^{n - 1} \frac {p_n q_{j - n + m} } {q_m} x^j       | c =  }} {{end-eqn}} Then evaluate: :$P'_{n - 1} \left({x}\right) = P_n \left({x}\right) - Q'_m \left({x}\right)$ which (after some algebra) works out as: :$\displaystyle P_n \left({x}\right) - Q'_m \left({x}\right) = \sum_{j \mathop = n - m}^{n - 1} \frac {p_n q_{j - n + m} } {q_m} x^j + \sum_{j \mathop = 0}^{n - m - 1} p_j x^j$ So we see that $P_n \left({x}\right) - Q'_m \left({x}\right)$ is a polynomial in $x$ of degree $n-1$. Let $\dfrac {p_n} {q_m} = d_{n - m}$. Hence we have: :$P_n \left({x}\right) = d_{n - m} x^{n - m} Q_m \left({x}\right) + P'_{n-1} \left({x}\right)$ We can express $P'_{n-1} \left({x}\right)$ as: :$\displaystyle P'_{n-1} \left({x}\right) = \sum_{j \mathop = 0}^{n - 1} p'_j x^j$ Repeat the above by subtracting $\displaystyle \frac {p'_{n - 1} } {q_m} x^{n - m - 1} Q_m \left({x}\right)$ from $P'_{n - 1} \left({x}\right)$, and letting $\dfrac {p'_{n - 1} } {q_m} = d_{n - m - 1}$. Hence: :$P'_{n - 1} \left({x}\right) = d_{n - m - 1} x^{n - m - 1} Q_m \left({x}\right) + P''_{n - 2} \left({x}\right)$ The process can be repeated $n-m$ times. It can be seen that after the last stage, we have: :$P_n \left({x}\right) = D_{n-m} \left({x}\right) Q_m \left({x}\right) + R_k \left({x}\right)$ where: : $\displaystyle D_{n-m} \left({x}\right) = \sum_{j \mathop = 0}^{n-m} d_j x^j$ : $R_k \left({x}\right)$ is a polynomial of degree at most $m - 1$. {{qed}} Category:Proof Techniques 5wfxx82wynr2yb1wbb73q5ul4ef8e5a


Prüfer Sequence from Labeled Tree

Prüfer Sequence from Labeled Tree 0 5179 477053 477052 2020-07-04T15:53:07Z Prime.mover 59 wikitext text/x-wiki == Algorithm == Given a finite labeled tree, it is possible to generate a Prüfer sequence corresponding to that tree. Let $T$ be a labeled tree of order $n$, where the labels are assigned the values $1$ to $n$. :'''Step 1''': If there are two (or less) nodes in $T$, then '''stop'''. Otherwise, continue on to '''step 2'''. :'''Step 2''': Find all the nodes of $T$ of degree $1$. There are bound to be some, from Finite Tree has Leaf Nodes. Choose the one $v$ with the lowest label. :'''Step 3''': Look at the node $v'$ adjacent to $v$, and assign the label of $v\,'$ to the first available element of the Prüfer sequence being generated. :'''Step 4''': Remove the node $v$ and its incident edge. This leaves a smaller tree $T'$. Go back to '''step 1'''. The above constitutes an algorithm, for the following reasons: === Finiteness === For each iteration through the algorithm, '''step 4''' is executed, which reduces the number of nodes by $1$. Therefore, after $n-2$ iterations, at '''step 1''' there will be $2$ nodes left, and the algorithm will stop. === Definiteness === :'''Step 1''': There are either more than $2$ nodes in a tree or there are $2$ or less. :'''Step 2''': There are bound to be some nodes of degree $1$, from Finite Tree has Leaf Nodes. As integers are totally ordered, it is always possible to find the lowest label. :'''Step 3''': As the node $v$ is of degree $1$, there is a unique node $v'$ to which it is adjacent. (Note that this node will not ''also'' have degree $1$, for then $v v'$ would be a tree of order $2$, and we have established from step $1$ that this is not the case.) :'''Step 4''': The node and edge to be removed are unique and specified precisely, as this is a tree we are talking about. === Inputs === The input to this algorithm is the tree $T$. === Outputs === The output to this algorithm is the Prüfer sequence $\tuple {\mathbf a_1, \mathbf a_2, \ldots, \mathbf a_{n - 2} }$. === Effective === Each step of the algorithm is basic enough to be done exactly and in a finite length of time. == Example == Let $T$ be the following labeled tree: :300px This tree has $8$ nodes, so the corresponding Prüfer sequence will have $6$ elements. === Iteration 1 === : '''Step 1''': There are $8$ nodes, so continue to '''step 2'''. : '''Step 2''': The nodes of degree $1$ are $8, 2, 6, 4, 3$. Of these, $2$ is the lowest. : '''Step 3''': $2$ is adjacent to $1$, so add $\mathbf 1$ to the Prüfer sequence. : '''Step 4''': Removing node $2$ leaves the following tree: :300px At this stage, the Prüfer sequence is $\tuple {\mathbf 1}$. === Iteration 2 === : '''Step 1''': There are $7$ nodes, so continue to '''step 2'''. : '''Step 2''': The nodes of degree $1$ are $8, 6, 4, 3$. Of these, $3$ is the lowest. : '''Step 3''': $3$ is adjacent to $7$, so add $\mathbf 7$ to the Prüfer sequence. : '''Step 4''': Removing node $3$ leaves the following tree: :300px At this stage, the Prüfer sequence is $\tuple {\mathbf 1, \mathbf 7}$. === Iteration 3 === : '''Step 1''': There are $6$ nodes, so continue to '''step 2'''. : '''Step 2''': The nodes of degree $1$ are $8, 6, 4$. Of these, $4$ is the lowest. : '''Step 3''': $4$ is adjacent to $5$, so add $\mathbf 5$ to the Prüfer sequence. : '''Step 4''': Removing node $4$ leaves the following tree: :250px At this stage, the Prüfer sequence is $\tuple {\mathbf 1, \mathbf 7, \mathbf 5}$. === Iteration 4 === : '''Step 1''': There are $5$ nodes, so continue to '''step 2'''. : '''Step 2''': The nodes of degree $1$ are $8, 6, 5$. Of these, $5$ is the lowest. : '''Step 3''': $5$ is adjacent to $7$, so add $\mathbf 7$ to the Prüfer sequence. : '''Step 4''': Removing node $5$ leaves the following tree: :200px At this stage, the Prüfer sequence is $\tuple {\mathbf 1, \mathbf 7, \mathbf 5, \mathbf 7}$. === Iteration 5 === : '''Step 1''': There are $4$ nodes, so continue to '''step 2'''. : '''Step 2''': The nodes of degree $1$ are $8, 6$. Of these, $6$ is the lowest. : '''Step 3''': $6$ is adjacent to $7$, so add $\mathbf 7$ to the Prüfer sequence. : '''Step 4''': Removing node $6$ leaves the following tree: :200px At this stage, the Prüfer sequence is $\tuple {\mathbf 1, \mathbf 7, \mathbf 5, \mathbf 7, \mathbf 7}$. === Iteration 6 === : '''Step 1''': There are $3$ nodes, so continue to '''step 2'''. : '''Step 2''': The nodes of degree $1$ are $8, 7$. Of these, $7$ is the lowest. : '''Step 3''': $7$ is adjacent to $1$, so add $\mathbf 1$ to the Prüfer sequence. : '''Step 4''': Removing node $7$ leaves the following tree: :150px At this stage, the Prüfer sequence is $\tuple {\mathbf 1, \mathbf 7, \mathbf 5, \mathbf 7, \mathbf 7, \mathbf 1}$. === Iteration 7 === : '''Step 1''': There are $2$ nodes, so '''stop'''. The Prüfer sequence is $\tuple {\mathbf 1, \mathbf 7, \mathbf 5, \mathbf 7, \mathbf 7, \mathbf 1}$. == Also see == Compare with the example given in Labeled Tree from Prüfer Sequence. These two results are pulled together in Bijection between Prüfer Sequences and Labeled Trees. Category:Tree Theory Category:Combinatorics ksok75784rgp5zboypq2ednespi5ojl


Labeled Tree from Prüfer Sequence

Labeled Tree from Prüfer Sequence 0 5191 477013 159217 2020-07-04T12:20:13Z Prime.mover 59 wikitext text/x-wiki == Algorithm == Given a Prüfer sequence, it is possible to generate a finite labeled tree corresponding to that sequence. Let $P = \tuple {\mathbf a_1, \mathbf a_2, \ldots, \mathbf a_{n - 2} }$ be a Prüfer sequence. This will be called '''the sequence'''. It is assumed the sequence is not empty. :'''Step 1''': Draw the $n$ nodes of the tree we are to generate, and label them from $1$ to $n$. This will be called '''the tree'''. :'''Step 2''': Make a list of all the integers $\left({1, 2, \ldots, n}\right)$. This will be called '''the list'''. :'''Step 3''': If there are two numbers left in '''the list''', connect them with an edge and then '''stop'''. Otherwise, continue on to '''step 4'''. :'''Step 4''': Find the smallest number in '''the list''' which is not in '''the sequence''', and also the first number in the '''the sequence'''. Add an edge to '''the tree''' connecting the nodes whose labels correspond to those numbers. :'''Step 5''': Delete the first of those numbers from '''the list''' and the second from '''the sequence'''. This leaves a smaller '''list''' and a shorter '''sequence'''. Then return to '''step 3'''. The above constitutes an algorithm, for the following reasons: === Finiteness === For each iteration through the algorithm, '''step 5''' is executed, which reduces the size of '''the list''' by $1$. Therefore, after $n-2$ iterations, at '''step 1''' there will be $2$ numbers left in '''the list''', and the algorithm will stop. === Definiteness === :'''Steps 1 and 2''': Trivially definite. :'''Step 3''': We are starting with a non-empty Prüfer sequence of length $n-2$, so '''the list''' must originally contain at least $3$ elements. As the number of elements in '''the list''' decreases by $1$ each iteration (see '''step 5'''), eventually there is bound to be just two elements in '''the list'''. :'''Step 4''': As there are more elements in '''the list''' than there are in '''the sequence''', by the Pigeonhole Principle there has to be at least one number in '''the list''' that is not in '''the sequence'''. :'''Step 5''': Trivially definite. === Inputs === The input to this algorithm is the Prüfer sequence $\tuple {\mathbf a_1, \mathbf a_2, \ldots, \mathbf a_{n - 2} }$. === Outputs === The output to this algorithm is the tree $T$. The fact that $T$ is in fact a tree follows from the fact that: * $T$ has $n$ nodes and (from the method of construction) $n - 1$ edges; * Each new edge connects two as yet unconnected parts of $T$, so every edge is a bridge. Therefore there are no cycles in $T$, from Condition for Edge to be Bridge. So $T$ is a tree from Equivalent Definitions for Finite Tree. === Effective === Each step of the algorithm is basic enough to be done exactly and in a finite length of time. == Example == Let the starting Prüfer sequence be $\tuple {\mathbf 1, \mathbf 7, \mathbf 5, \mathbf 7, \mathbf 7, \mathbf 1}$. :'''Step 1''': '''The sequence''' is length $6$, so the tree will have $8$ nodes: :250px :'''Step 2''': We generate '''the list''': $\tuple {1, 2, 3, 4, 5, 6, 7, 8}$. === Iteration 1 === :'''Step 3''': There are $8$ elements in '''the list''', so we move on to '''step 4'''. :'''Step 4''': The smallest number in '''the list''' which is not in '''the sequence''' is $2$, and the first number in '''the sequence''' is $1$. We join $1$ and $2$, to obtain this graph: :250px :'''Step 5''': We delete $2$ from '''the list''' to obtain $\tuple {1, 3, 4, 5, 6, 7, 8}$, and $1$ from the start of '''the sequence''' to obtain $\tuple {\mathbf 7, \mathbf 5, \mathbf 7, \mathbf 7, \mathbf 1}$. === Iteration 2 === :'''Step 3''': There are $7$ elements in '''the list''', so we move on to '''step 4'''. :'''Step 4''': The smallest number in '''the list''' which is not in '''the sequence''' is $3$, and the first number in '''the sequence''' is $7$. We join $3$ and $7$, to obtain this graph: :250px :'''Step 5''': We delete $3$ from '''the list''' to obtain $\tuple {1, 4, 5, 6, 7, 8}$, and $7$ from the start of '''the sequence''' to obtain $\tuple {\mathbf 5, \mathbf 7, \mathbf 7, \mathbf 1}$. === Iteration 3 === :'''Step 3''': There are $6$ elements in '''the list''', so we move on to '''step 4'''. :'''Step 4''': The smallest number in '''the list''' which is not in '''the sequence''' is $4$, and the first number in '''the sequence''' is $5$. We join $4$ and $5$, to obtain this graph: :250px :'''Step 5''': We delete $4$ from '''the list''' to obtain $\tuple {1, 5, 6, 7, 8}$, and $5$ from the start of '''the sequence''' to obtain $\tuple {\mathbf 7, \mathbf 7, \mathbf 1}$. === Iteration 4 === :'''Step 3''': There are $5$ elements in '''the list''', so we move on to '''step 4'''. :'''Step 4''': The smallest number in '''the list''' which is not in '''the sequence''' is $5$, and the first number in '''the sequence''' is $7$. We join $5$ and $7$, to obtain this graph: :250px :'''Step 5''': We delete $5$ from '''the list''' to obtain $\left({1, 6, 7, 8}\right)$, and $7$ from the start of '''the sequence''' to obtain $\tuple {\mathbf 7, \mathbf 1}$. === Iteration 5 === :'''Step 3''': There are $4$ elements in '''the list''', so we move on to '''step 4'''. :'''Step 4''': The smallest number in '''the list''' which is not in '''the sequence''' is $6$, and the first number in '''the sequence''' is $7$. We join $6$ and $7$, to obtain this graph: :250px :'''Step 5''': We delete $6$ from '''the list''' to obtain $\tuple {1, 7, 8}$, and $7$ from the start of '''the sequence''' to obtain $\tuple {\mathbf 1}$. === Iteration 6 === :'''Step 3''': There are $3$ elements in '''the list''', so we move on to '''step 4'''. :'''Step 4''': The smallest number in '''the list''' which is not in '''the sequence''' is $7$, and the first number in '''the sequence''' is $1$. We join $7$ and $1$, to obtain this graph: :250px :'''Step 5''': We delete $7$ from '''the list''' to obtain $\tuple {1, 8}$, and $1$ from the start of '''the sequence''', which is at this point empty. === Iteration 7 === :'''Step 3''': There are $2$ elements in '''the list''': $\tuple {1, 8}$, so we join them to obtain this graph: :250px Then we '''stop'''. The algorithm has terminated, and the tree is complete. Rearranging the positions of the nodes, we can draw it like this: :300px == Also see == Compare with the example given in Prüfer Sequence from Labeled Tree. These two results are pulled together in Bijection between Prüfer Sequences and Labeled Trees. Category:Tree Theory Category:Combinatorics 8pwuhc94aw63lbxf1frlf230hfxfxx5


Kruskal's Algorithm

Kruskal's Algorithm 0 5215 498169 476926 2020-11-08T22:51:12Z Prime.mover 59 wikitext text/x-wiki == Algorithm == The purpose of this algorithm is to produce a minimum spanning tree for any given weighted graph $G$. :'''Step 1''': Start with the edgeless graph $T$ whose vertices correspond with those of $G$. :'''Step 2''': Choose an edge $e$ of $G$ such that: ::'''a)''': Adding $e$ to $T$ would not make a cycle in $T$; ::'''b)''': $e$ has the minimum weight of all the edges remaining in $G$ that fulfil the condition in '''a)'''. :'''Step 3''': Add $e$ to $T$. :'''Step 4''': If $T$ spans $G$, '''stop'''. Otherwise, go to '''Step 2'''. The above constitutes an algorithm, for the following reasons: === Finiteness === For each iteration through the algorithm, '''step 3''' is executed, which increases the number of edges in $T$ by 1. As a tree with $n$ nodes has $n-1$ edges, the algorithm will terminate after $n-1$ iterations. === Definiteness === :'''Step 1''': Trivially definite. :'''Step 2''': As the edges of a graph can be arranged in order of weight, there is a definite edge (or set of edges) with minimal weight. It is straightforward to select an edge $e$ which does not make a cycle in $T$, by ensuring that at least one end of $e$ is incident to a vertex which has not so far been connected into $T$. :'''Step 3''': Trivially definite. :'''Step 4''': It is straightforward to determine whether all the vertices are connected. === Inputs === The input to this algorithm is the weighted graph $G$. === Outputs === The output to this algorithm is the minimum spanning tree $T$. === Effective === Each step of the algorithm is basic enough to be done exactly and in a finite length of time. == Note == It is clear that this is a greedy algorithm: at each stage the minimum possible weight is chosen, without any analysis as to whether there may be a combination of larger weights which may produce a smaller-weight spanning tree. For this reason, it is sometimes called '''Kruskal's greedy algorithm'''. In this case, the greedy algorithm ''does'' produce the minimum spanning tree. {{Namedfor|Joseph Bernard Kruskal|cat = Kruskal}} == Also see == * Kruskal's Algorithm produces Minimum Spanning Tree * Prim's Algorithm == Sources == * {{BookReference|Introductory Graph Theory|1977|Gary Chartrand|prev = Definition:Minimum Spanning Tree|next = Kruskal's Algorithm produces Minimum Spanning Tree}}: $\S 4.1$: The Minimal Connector Problem: An Introduction to Trees * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Definition:Kronecker Delta/Also denoted as|next = Kruskal-Wallis Test|entry = Kruskal's algorithm}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Definition:Kronecker Delta/Also denoted as|next = Kruskal-Wallis Test|entry = Kruskal's algorithm}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Kronecker's Lemma|next = Kruskal-Wallis Test|entry = Kruskal's algorithm}} Category:Graph Theory Category:Greedy Algorithms itasf7blibtyywvutl5n7lo5b7377po


Prim's Algorithm

Prim's Algorithm 0 5221 476927 476864 2020-07-04T08:51:51Z Prime.mover 59 wikitext text/x-wiki == Algorithm == The purpose of this algorithm is to produce a minimum spanning tree for any given weighted graph $G$. :'''Step 1''': Choose any vertex of $G$, and add it to $T$. :'''Step 2''': Add an edge of minimum weight $e$ to join a vertex in $T$ to one not in $T$. :'''Step 3''': If $T$ spans $G$, '''stop'''. Otherwise, go to '''Step 2'''. == Proof == The above constitutes an algorithm, for the following reasons: === Finiteness === For each iteration through the algorithm, '''step 2''' is executed, which increases the number of edges in $T$ by 1. As a tree with $n$ nodes has $n-1$ edges, the algorithm will terminate after $n-1$ iterations. === Definiteness === :'''Step 1''': Trivially definite. :'''Step 2''': As the edges connecting $T$ to the remaining vertices can be arranged in order of weight, there is a definite edge (or set of edges) with minimal weight. :'''Step 3''': It is straightforward to determine whether all the vertices are connected. === Inputs === The input to this algorithm is the weighted graph $G$. === Outputs === The output to this algorithm is the minimum spanning tree $T$. === Effective === Each step of the algorithm is basic enough to be done exactly and in a finite length of time. == Also known as == It is clear that this is a greedy algorithm: at each stage the minimum possible weight is chosen, without any analysis as to whether there may be a combination of larger weights which may produce a smaller-weight spanning tree. For this reason, it is sometimes called '''Prim's greedy algorithm'''. In this case, the greedy algorithm ''does'' produce the minimum spanning tree. {{Namedfor|Robert Clay Prim|cat = Prim}} == Historical Note == {{:Prim's Algorithm/Historical Note}} == Also see == * Prim's Algorithm produces Minimum Spanning Tree * Kruskal's Algorithm == Sources == * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Primitive Polynomial (Ring Theory)|next = Definition:Principal Axes|entry = Prim's algorithm}} Category:Graph Theory Category:Greedy Algorithms tmbdpk4ake83t3myj8o2i55xaz3gy9j


Continuum Hypothesis

Continuum Hypothesis 0 5242 450153 443656 2020-02-20T13:39:06Z Prime.mover 59 wikitext text/x-wiki == Hypothesis == <onlyinclude> There is no set whose cardinality is strictly between that of the integers and the real numbers. </onlyinclude> Symbolically, the '''Continuum Hypothesis''' asserts that $\aleph_1 = \mathfrak c$. {{MissingLinks|$\aleph_1$}} === Generalized Continuum Hypothesis === {{:Generalized Continuum Hypothesis}} {{Hilbert23|1}} == Historical Note == {{:Continuum Hypothesis/Historical Note}} == Sources == * {{BookReference|Mathematical Logic and Computability|1996|H. Jerome Keisler|author2 = Joel Robbin|prev = Rational Numbers are Countably Infinite|next = Definition:Left Inverse Mapping}}: Appendix $\text{A}.6$: Cardinality * {{BookReference|A Handbook of Terms used in Algebra and Analysis|1972|A.G. Howson|prev = Definition:Countable Set/Countably Infinite/Definition 1|next = Definition:Integer}}: $\S 4$: Number systems $\text{I}$: A set-theoretic approach * {{BookReference|Introduction to Boolean Algebras|2008|Paul Halmos|author2 = Steven Givant|prev = Definition:Cardinal Number}}: Appendix $\text{A}$: Set Theory: Cardinal Numbers * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Power Set of Natural Numbers is Cardinality of Continuum|next = Generalized Continuum Hypothesis}}: Chapter $1$: General Background: $\S 5$ The continuum problem Category:Set Theory Category:Named Theorems Category:Open Questions gb2vdnxv9ysggtkor55vcgrx47i5ryl


Division by Zero

Division by Zero 0 5340 488685 484214 2020-09-17T20:30:42Z Prime.mover 59 wikitext text/x-wiki == Mistake == Let $x$ and $y$ be numbers such that $y = 0$. The quantity $\dfrac x y$ is '''undefined'''. It is a common mistake to forget this fact when evaluating formulas. == Also see == * L'Hôpital's Rule: while $\dfrac 0 0$ is undefined, the limit of a rational function whose denominator approaches $0$ is not necessarily undefined. == Historical Note == {{:Division by Zero/Historical Note}} == Sources == * {{BookReference|Calculus|1967|Michael Spivak|prev = Commutative Law of Multiplication|next = Multiplication by Non-Zero Numbers is Cancellable}}: Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers: $(\text P 8)$ * {{BookReference|Theory and Problems of Statistics|1972|Murray R. Spiegel|author2 = R.W. Boxer|ed = SI|edpage = SI Edition|prev = Definition:Right Hand Side|next = Definition:Simultaneous Equations}}: Chapter $1$: Equations * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Definition:Real Number Axioms|next = Definition:Infinity}}: $\S 1$: Real Numbers: $\S 1.3$: Arithmetic * {{KhanAcademySecure|algebra/x2f8bb11595b61c86:foundation-algebra/x2f8bb11595b61c86:division-zero/v/why-dividing-by-zero-is-undefined}} Category:Fallacies and Mistakes Category:Real Division 9o5mmpxqxqvjt6iuoqqofmq4xoqwlck


Abel's Lemma

Abel's Lemma 0 5969 454399 412780 2020-03-14T13:19:15Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $\sequence a$ and $\sequence b$ be sequences in an arbitrary ring $R$. === Formulation 1 === {{:Abel's Lemma/Formulation 1}} === Formulation 2 === {{:Abel's Lemma/Formulation 2}} == Also known as == {{:Abel's Lemma/Also known as}} {{Namedfor|Niels Henrik Abel|cat = Abel}} Category:Algebra Category:Finite Calculus Category:Proof Techniques Category:Abel's Lemma gxlhdof52p9s2dlo33wuk8ggllq2zd8