hcju/proofwikips huggingface dataset:
Countable Hypothesis Set has Finished Tableau
== 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 $\mathbf H$.  == 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 $1.11.2$
Main Lemma of Propositional Tableaus
== 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 $\Gamma$, $\mathcal M \models \mathbf H$.  == 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 $1.11.1$
Polynomial Long Division
== 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 $m - 1$.
Kruskal's Algorithm
== 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}}
Prim's Algorithm
== 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}}
Continuum Hypothesis
== Hypothesis ==  There is no set whose cardinality is strictly between that of the integers and the real numbers.  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
Division by Zero
== 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}}
Abel's Lemma
== 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}}
Pairwise Disjoint Subsets in Semiring Part of Partition
== Lemma == Let $\mathbb S$ be a semiring of sets. Let $A, A_1, A_2, \ldots, A_n$ all belong to $\mathbb S$. Let $A_1, A_2, \ldots, A_n$ all be pairwise disjoint subsets of $A$. Then there exists a finite expansion of $A$: :$\displaystyle \exists s \ge n: A = \bigcup_{k \mathop = 1}^s A_k$ with $A_1, \ldots, A_n$ as its first $n$ elements, such that: : $(1): \quad \forall k, 1 \le k \le s: A_k \in \mathbb S$ : $(2): \quad \forall k, l, 1 \le k \le s, 1 \le l \le s: k \ne l \implies A_k \cap A_l = \varnothing$. That is, the nature of a semiring is such that every collection of pairwise disjoint subsets of a given set $A$ of that semiring is part of a larger collection of pairwise disjoint subsets of $A$ which forms a complete partition of $A$. == Proof == By the definition of a semiring of sets, the lemma holds for $n = 1$. Now we suppose that the lemma holds for $n = m$, and we attempt to show it consequently holds for $n = m+1$. So, let $A_1, A_2, \ldots, A_m, A_{m+1}$ all be pairwise disjoint subsets of $A$. By hypothesis: :$A = A_1 \cup A_2 \cup \cdots \cup A_m \cup B_1 \cup \cdots \cup B_p$ where $A_1, A_2, \ldots, A_m, B_1, \ldots, B_p$ are pairwise disjoint subsets of $A$, all belonging to $\mathbb S$. Let $B_{q1} = A_{m+1} \cap B_q$. By the definition of a semiring of sets: :$B_q = B_{q1} \cup \cdots \cup B_{q r_q}$ where all the $B_{qj}$ are pairwise disjoint subsets of $B_q$, all belonging to $\mathbb S$. But then we see that: :$\displaystyle A = A_1 \cup A_2 \cup \cdots \cup A_m \cup A_{m+1} \cup \bigcup_{q \mathop = 1}^p \left({\bigcup_{j \mathop = 2}^{r_q} B_{qj}}\right)$ and so the lemma is true for $m+1$. The result follows by induction.
Axiom:Kolmogorov Axioms
== Definition == Let $\EE$ be an experiment. Let $\struct {\Omega, \Sigma, \Pr}$ be a probability measure on $\EE$. Then $\EE$ can be defined as being a measure space $\struct {\Omega, \Sigma, \Pr}$, such that $\map \Pr \Omega = 1$. Thus $\Pr$ satisfies the '''Kolmogorov axioms''': == Axioms ==  {{begin-axiom}} {{axiom | n = 1         | q = \forall A \in \Sigma         | ml= 0         | mo= \le         | mr= \map \Pr A \le 1         | rc= The probability of an event occurring is a real number between $0$ and $1$ }} {{axiom | n = 2         | ml= \map \Pr \Omega         | mo= =         | mr= 1         | rc= The probability of some elementary event occurring in the sample space is $1$ }} {{axiom | n = 3         | ml= \map \Pr {\bigcup_{i \mathop \ge 1} A_i}         | mo= =         | mr= \sum_{i \mathop \ge 1} \map \Pr {A_i}         | rc= where $\set {A_1, A_2, \ldots}$ is a countable (possibly countably infinite) set of pairwise disjoint events }} {{axiom | rc= That is, the probability of any one of countably many pairwise disjoint events occurring }} {{axiom | rc= is the sum of the probabilities of the occurrence of each of the individual events }} {{end-axiom}}  == Also defined as == Some sources include: :$\map \Pr \O = 0$ but this is strictly speaking not axiomatic as it can be deduced from the other axioms. == Also see == * Elementary Properties of Probability Measure * Probability of Union of Disjoint Events is Sum of Individual Probabilities * Definition:Measure Space: the '''Kolmogorov axioms''' follow directly from the fact that $\struct {\Omega, \Sigma, \Pr}$ is an example of such. {{NamedforAxiom|Andrey Nikolaevich Kolmogorov|cat = Kolmogorov}} == Sources == * {{BookReference|Probability: An Introduction|1986|Geoffrey Grimmett|author2 = Dominic Welsh|prev = Definition:Probability Measure/Definition 2|next = Definition:Countable Set/Definition 3}}: $1$: Events and probabilities: $1.3$: Probabilities
Proof by Counterexample
== Proof Technique == Consider the definition of a counterexample: {{:Definition:Counterexample}} Proving, or disproving, a statement in the form of $X$ by establishing the truth or falsehood of a statement in the form of $Y$ is known as the technique of '''proof by counterexample'''. {{Languages|Counterexample}} {{Language|German|Gegenbeispiel}} {{End-languages}} == Sources == * {{BookReference|Undergraduate Topology|1971|Robert H. Kasriel|prev = Definition:Counterexample|next = Definition:Union of Set of Sets}}: $\S 1.7$: Counterexamples * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Definition:Counterexample|next = Definition:Conditional}}: $\S 3$: Statements and conditions; quantifiers * {{BookReference|Algebra Volume 1|1982|P.M. Cohn|edpage = Second Edition|ed = 2nd|prev = Square Root of 2 is Irrational/Classic Proof|next = Goldbach's Conjecture}}: Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic
Talk:Surjection Induced by Powerset is Induced by Surjection
Is there a trivial reason that I'm just missing to explain "For the second, it can be seen that neither $\left\{{y_1}\right\}$ nor $\left\{{y_2}\right\}$ can be in $\mathrm{Rng} \left({f_{\mathcal{R}} \left({\mathcal{P} \left({S}\right)}\right)}\right)$"?  --Cynic  (talk) 03:28, 17 July 2010 (UTC) :My thinking was: ::"Because $\left({x, y_1}\right) \in f$ and $\left({x, y_2}\right) \in f$, it follows that $\left\{{y_1, y_2}\right\} \in \mathrm{Rng} \left({f_{\mathcal{R}} \left({\mathcal{P} \left({S}\right)}\right)}\right)$. :: "But as $\left\{{y_1}\right\}$ is not the image of $x$ ..." : Except the more I look at this, the more rubbish it looks. Why can't $\exists x_1 \in S: \left({x_1, y_1}\right) \in f$. : Either the result doesn't hold or I need another way to justify it. I've studied a lot since I wrote this result up, I may be able to take a step back and rethink it. --Prime.mover 05:39, 17 July 2010 (UTC) 18lntpn6bx3a4e7dsf17w5bh983mzer
Modus Ponendo Tollens
== Modus Ponendo Tollens == The '''modus ponendo tollens''' is a valid deduction sequent in propositional logic: === Proof Rule === {{:Modus Ponendo Tollens/Proof Rule}} == Variants == The following forms can be used as variants of this theorem: === Variant === {{:Modus Ponendo Tollens/Variant}} == Explanation == {{:Modus Ponendo Tollens/Explanation}} == Linguistic Note == {{:Modus Ponendo Tollens/Linguistic Note}} == Also see == The following are related argument forms: * Modus Ponendo Ponens * Modus Tollendo Ponens * Modus Tollendo Tollens
Modus Tollendo Ponens
== Sequent == The '''modus tollendo ponens''' is a valid deduction sequent in propositional logic. === Proof Rule === {{:Modus Tollendo Ponens/Proof Rule}} === Sequent Form === {{:Modus Tollendo Ponens/Sequent Form}} == Variants == The following forms can be used as variants of this theorem: === Variant === {{:Modus Tollendo Ponens/Variant}} Note that the form: {{:Modus Tollendo Ponens/Variant/Formulation 1/Reverse Implication}} requires Law of Excluded Middle. Therefore it is not valid in intuitionistic logic. == Explanation == {{:Modus Tollendo Ponens/Explanation}} == Also known as == {{:Modus Tollendo Ponens/Also known as}} == Also see == The following are related argument forms: * Modus Ponendo Ponens * Modus Ponendo Tollens * Modus Tollendo Tollens == Linguistic Note == {{:Modus Tollendo Ponens/Linguistic Note}} == Sources == * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Definition:Disjunctive Normal Form|next = Definition:Ball|entry = disjunctive syllogism}} * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Modus Ponendo Ponens|next = Modus Tollendo Tollens|entry = ''modus tollendo ponens''}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Modus Ponendo Ponens|next = Modus Tollendo Tollens|entry = ''modus tollendo ponens''}}
Axiom:Playfair's Axiom
== Axiom ==  :Exactly one straight line can be drawn through any point not on a given line parallel to the given straight line in a plane. Or:  :Given any straight line and a point not on it, there exists one and only one line which passes through this point and does not intersect the first line no matter how far they are extended. :This unique line is defined as being parallel to the original line in question. Or: :Two straight lines which intersect one another cannot both be parallel to one and the same straight line.  == Comment == This is a frequently seen alternative presentation of Euclid's Fifth Postulate. It can easily seen to be equivalent to that given by {{AuthorRef|Euclid}}, but it can be argued that it is easier to understand. {{NamedforAxiom|John Playfair|cat = Playfair}} However, he did not originate it, merely published it. When he did so, he credited others, specifically {{AuthorRef|William Ludlam}}, for having used it earlier. == 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 Common Notions|next = Axiom:Euclid's Common Notions/Historical Note}}: Book $\text{I}$. Notes on Postulate $5$ * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Axiom:Euclid's Fifth Postulate|next = Axiom:Eudoxus' Axiom|entry = Euclid's axioms}}
Axiom:Parallel Postulate
== Axiom ==  :If two straight lines are drawn which intersect a third in such a way that the sum of the measures of the two interior angles on one side is less than the sum of the measures of two right angles, then the two lines must intersect each other on that side if extended far enough.  == Also see == * Axiom:Euclid's Fifth Postulate: This is the name and formulation by which Euclid's Fifth Postulate is usually known. As can be seen, its wording in its modern format gives it an intent very similar to {{AuthorRef|Euclid}}'s. == Sources == * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Definition:Axiom|next = Isosceles Triangle has Two Equal Angles}}: Chapter $2$: The Logic of Shape: Euclid