Pairwise Disjoint Subsets in Semiring Part of Partition 0 6093 170249 46671 2013-12-01T03:19:00Z Prime.mover 59 wikitext text/x-wiki == 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. {{qed}} Category:Semirings of Sets 4n6l469lda9nu2qd9ceqytlwls70idh


Axiom:Kolmogorov Axioms

Axiom:Kolmogorov Axioms 100 6103 496131 496130 2020-10-23T22:51:46Z Prime.mover 59 wikitext text/x-wiki == 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 == <onlyinclude> {{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}} </onlyinclude> == 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 Category:Axioms/Probability Theory 9jfz5uval44btpxxwqib2ffcx34f5gw


Proof by Counterexample

Proof by Counterexample 0 6686 410774 405754 2019-06-28T18:44:11Z Prime.mover 59 wikitext text/x-wiki == 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 Category:Proof Techniques Category:Logic tuc80mry1q1ge4pc5ehgiuxi5rkpajh


Talk:Surjection Induced by Powerset is Induced by Surjection 1 6966 403435 217992 2019-04-30T08:58:03Z Prime.mover 59 wikitext text/x-wiki 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 0 6991 244569 244568 2016-01-16T13:25:20Z Prime.mover 59 wikitext text/x-wiki == 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 Category:Conjunction Category:Negation Category:Modus Ponendo Tollens mjbrmbzhpqjtf2emif4b8qxlkrnc3qr


Modus Tollendo Ponens

Modus Tollendo Ponens 0 6992 466201 462043 2020-05-06T06:33:14Z Prime.mover 59 wikitext text/x-wiki == 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''}} Category:Disjunction Category:Negation Category:Modus Tollendo Ponens 017gfwoo0yla3186sm4z7ejx993e3ok


Axiom:Playfair's Axiom

Axiom:Playfair's Axiom 100 7470 425120 236994 2019-09-13T17:03:37Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> :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. </onlyinclude> == 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}} Category:Euclid's Fifth Postulate osaymmlkj8dywhyqbm780ttcwea7yuv


Axiom:Parallel Postulate

Axiom:Parallel Postulate 100 7472 430027 295117 2019-10-08T12:23:33Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> :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. </onlyinclude> == 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 Category:Euclid's Fifth Postulate 2kzf2g76z8xc4fmnjs4vzx50eanrhnt


Newton's Laws of Motion/Third Law

Newton's Laws of Motion/Third Law 0 8131 488527 448467 2020-09-17T05:11:17Z Prime.mover 59 wikitext text/x-wiki == Physical Law == '''Newton's third law of motion''' is one of three physical laws that forms the basis for classical mechanics. === Statement of Law === <onlyinclude> :To every force there is always an equal and opposite force. That is, the forces of two bodies on each other are always equal and are directed in opposite directions. </onlyinclude> As {{AuthorRef|Isaac Newton}} himself put it: :''Whenever one body exerts a force on a second body, the second body exerts a force on the first body. These forces are equal in magnitude and opposite in direction.'' == Also known as == This law is also referred to as the '''law of action and reaction'''. It is also often referred to as just '''Newton's third law'''. == Also see == * Newton's First Law of Motion * Newton's Second Law of Motion {{Namedfor|Isaac Newton}} == Sources == * {{BookReference|Men of Mathematics|1937|Eric Temple Bell|prev = Newton's Second Law of Motion|next = Definition:Rate of Change}}: Chapter $\text{VI}$: On the Seashore * {{BookReference|Classical Mechanics|1965|J.W. Leech|ed = 2nd|edpage = Second Edition|prev = Newton's Second Law of Motion|next = Definition:Force}}: Chapter $\text {I}$: Introduction: $(3)$ * {{BookReference|Understanding Physics|1966|Isaac Asimov|prev = Definition:Force/Unit|next = Newton's Law of Universal Gravitation/Historical Note}}: $\text {I}$: Motion, Sound and Heat: Chapter $3$: The Laws of Motion: Action and Reaction * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Newton's Second Law of Motion|next = Newton's Method|entry = Newton's laws of motion}} Category:Newton's Laws of Motion f8htgrjffg7efsxw47gb5iij11ydk9c


Newton's Laws of Motion/First Law

Newton's Laws of Motion/First Law 0 8136 493529 488525 2020-10-10T10:20:55Z Prime.mover 59 wikitext text/x-wiki == Physical Law == '''Newton's first law of motion''' is one of three physical laws that forms the basis for classical mechanics. === Statement of Law === <onlyinclude> :Every body remains in a state of constant velocity unless it is acted upon by an external unbalanced force.   This of course includes it being stationary, that is, with a constant velocity of zero. </onlyinclude> This tendency is sometimes called inertia. As {{AuthorRef|Isaac Newton}} himself put it: :''A body remains at rest or, if already in motion, remains in uniform motion with constant speed in a straight line, unless it is acted on by an unbalanced external force.'' == Also known as == This law is also sometimes called the '''principle of inertia'''. It is also often referred to as just '''Newton's first law'''. == Also see == * Newton's Second Law of Motion * Newton's Third Law of Motion It is worth pointing out that '''Newton's first law of motion''' is in fact a special case of Newton's Second Law of Motion, putting $\mathbf F = 0$. {{Namedfor|Isaac Newton}} == Sources == * {{BookReference|Men of Mathematics|1937|Eric Temple Bell|prev = Newton's Laws of Motion|next = Newton's Second Law of Motion}}: Chapter $\text{VI}$: On the Seashore * {{BookReference|Classical Mechanics|1965|J.W. Leech|ed = 2nd|edpage = Second Edition|prev = Definition:Mass|next = Newton's Second Law of Motion}}: Chapter $\text {I}$: Introduction: $(1)$ * {{BookReference|Understanding Physics|1966|Isaac Asimov|prev = Newton's Laws of Motion|next = Definition:Stationary}}: $\text {I}$: Motion, Sound and Heat: Chapter $3$: The Laws of Motion: Inertia * {{BookReference|Tensor Calculus and Relativity|1975|Derek F. Lawden|ed = 3rd|edpage = Third Edition|next = Definition:Frame of Reference}}: Chapter $1$ Special Principle of Relativity. Lorentz Transformations: $1$. Newton's laws of motion * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Newton's Laws of Motion|next = Newton's Second Law of Motion|entry = Newton's laws of motion}} Category:Newton's Laws of Motion cdmovc4vas7r9fenkwfb9xet9pau00e


Newton's Laws of Motion/Second Law

Newton's Laws of Motion/Second Law 0 8137 493976 488526 2020-10-11T11:25:21Z Prime.mover 59 wikitext text/x-wiki == Physical Law == '''Newton's second law of motion''' is one of three physical laws that forms the basis for classical mechanics. === Statement of Law === <onlyinclude> :The total force applied on a body is equal to the derivative with respect to time of the linear momentum of the body: :$\mathbf F = \map {\dfrac \d {\d t} } {m \bsv}$ </onlyinclude> As {{AuthorRef|Isaac Newton}} himself put it: :''The acceleration produced by a particular force acting on a body is directly proportional to the magnitude of the force and inversely proportional to the mass of the body.'' == Also known as == This law is also often referred to as just '''Newton's second law'''. == Also see == * At velocities near the speed of light, see Einstein's Law of Motion. * Newton's First Law of Motion * Newton's Third Law of Motion {{Namedfor|Isaac Newton}} == Sources == * {{BookReference|Men of Mathematics|1937|Eric Temple Bell|prev = Newton's First Law of Motion|next = Newton's Third Law of Motion}}: Chapter $\text{VI}$: On the Seashore * {{BookReference|Classical Mechanics|1965|J.W. Leech|ed = 2nd|edpage = Second Edition|prev = Newton's First Law of Motion|next = Newton's Third Law of Motion}}: Chapter $\text {I}$: Introduction: $(2)$ * {{BookReference|Understanding Physics|1966|Isaac Asimov|prev = Definition:Mass|next = Definition:Gram}}: $\text {I}$: Motion, Sound and Heat: Chapter $3$: The Laws of Motion: Mass * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Definition:Derivative|next = Definition:Free Fall}}: $1$: The Nature of Differential Equations: $\S 1$: Introduction: $(1)$ * {{BookReference|Engineering Mathematics|1977|A.J.M. Spencer|volume = I|prev = Definition:Differential Equation|next = Definition:Ordinary Differential Equation}}: Chapter $1$ Ordinary Differential Equations: $1.1$ Introduction * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Volume of Sphere/Proof by Archimedes|next = Definition:Linear Momentum}}: Chapter $\text {B}.7$: A Simple Approach to $E = M c^2$ * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Einstein's Mass-Energy Equation|next = Motion of Rocket in Outer Space/Proof 2}}: Chapter $\text {B}.8$: Rocket Propulsion in Outer Space * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Volume of Solid of Revolution/Historical Note|next = Definition:Velocity}}: Chapter $8$: The System of the World: Newton * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Speed of Light|next = Definition:Newton (Unit)|entry = constant|subentry = in physical laws}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Newton's First Law of Motion|next = Newton's Third Law of Motion|entry = Newton's laws of motion}} Category:Newton's Laws of Motion ft6ilao12vjwlk0aei8sr6bfcckfhik


Newton's Law of Cooling

Newton's Law of Cooling 0 8240 486172 486171 2020-09-07T21:29:24Z Prime.mover 59 wikitext text/x-wiki == Physical Law == The rate at which a hot body loses heat is proportional to the difference in temperature between it and its surroundings. {{Namedfor|Isaac Newton|cat = Newton}} == Historical Note == {{:Newton's Law of Cooling/Historical Note}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Law of Mass Action|next = Newton's Law of Cooling/Historical Note}}: $1$: The Nature of Differential Equations: $\S 4$: Growth, Decay and Chemical Reactions: Problem $5$ Category:Thermodynamics grksipow1gt13we3qj3atxf1kot5yc9


Acceleration Due to Gravity

Acceleration Due to Gravity 0 8242 448198 448180 2020-02-11T21:54:54Z Prime.mover 59 wikitext text/x-wiki == Physical Law == A body $B$ in a uniform gravitational field experiences a force which gives rise to a constant acceleration independent of the mass of the body. If the force due to the gravitational field is the '''only''' force on the body, it is said to be in free fall. == Derivation == This law can be derived from Newton's Law of Universal Gravitation. Let the mass of $B$ be $m$. Let the mass of the body $P$ which gives rise to the gravitational field be $M$. Then the force on $B$ is given by: :$F = G \dfrac {M m} {r^2}$ where: :$G$ is the gravitational constant :$r$ is the distance between the centers of gravity of $B$ and $P$. The assumption is that $M$ is orders of magnitude greater than $m$, and $r$ is also several orders of magnitude greater than the displacements observed on $B$ in the local frame. Then: :$F = m \dfrac {G M} {r^2}$ But from Newton's Second Law of Motion: :$F = m a$ where $a$ is the magnitude of the acceleration which would be imparted to the body if no other force were acting on it. Hence: :$a = \dfrac {G M} {r^2}$ which is, as we said, independent of the mass of the object. {{qed}} == Gravity on Earth == {{:Acceleration Due to Gravity/Earth's Surface|Gravity on Earth}} == Historical Note == {{:Acceleration Due to Gravity/Historical Note}} == Sources == * {{BookReference|Understanding Physics|1966|Isaac Asimov|prev = Definition:Constant of Proportion|next = Definition:Gravity}}: $\text {I}$: Motion, Sound and Heat: Chapter $2$: Falling Bodies: Free Fall Category:Mechanics Category:Gravity 5h2pl882vzxxfjnk5n71dggi0box3la


Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation 0 8245 462330 462318 2020-04-16T13:17:44Z Prime.mover 59 wikitext text/x-wiki == Physical Law == Let $a$ and $b$ be particles with mass $m_a$ and $m_b$ respectively. Then $a$ and $b$ exert a force upon each other whose magnitude and direction are given by '''Newton's law of universal gravitation''': :$\mathbf F_{a b} \propto \dfrac {m_a m_b {\mathbf r_{b a} } } {r^3}$ where: :$\mathbf F_{a b}$ is the force exerted on $b$ by the electric charge on $a$ :$\mathbf r_{b a}$ is the displacement vector from $b$ to $a$ :$r$ is the distance between $a$ and $b$. Thus it is seen that the direction of $\mathbf F_{a b}$ is specifically '''towards''' $a$. By exchanging $a$ and $b$ in the above, it is seen that $b$ exerts the same force on $a$ as $a$ does on $b$, but in the opposite direction, that is, towards $b$. === Gravitational Constant === {{:Definition:Gravitational Constant}} Thus the equation becomes: :$\mathbf F_{a b} = \dfrac {G m_a m_b \mathbf r_{b a} } {r^3}$ == Also presented as == :$\mathbf F_{a b} \propto \dfrac {m_a m_b \hat {\mathbf r}_{b a} } {r^2}$ where $\hat {\mathbf r}_{a b}$ is the unit vector in the direction from $b$ to $a$. == Also known as == '''Newton's Law of Universal Gravitation''' is also known as just '''Newton's Law of Gravitation'''. Some sources refer to it as the '''inverse square law of gravitation'''. {{Namedfor|Isaac Newton|cat = Newton}} == Historical Note == {{:Newton's Law of Universal Gravitation/Historical Note}} == Sources == * {{BookReference|Men of Mathematics|1937|Eric Temple Bell|prev = Kepler's Third Law of Planetary Motion|next = Newton's Laws of Motion}}: Chapter $\text{VI}$: On the Seashore * {{BookReference|Understanding Physics|1966|Isaac Asimov|prev = Mathematician:Frederick William Herschel|next = Definition:Gravitational Constant}}: $\text {I}$: Motion, Sound and Heat: Chapter $4$: Gravitation: The Gravitational Constant * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Kepler's Laws of Planetary Motion/Historical Note|next = Definition:Gravitational Constant}}: $\S 3.21$: Newton's Law of Gravitation: $(11)$ * {{BookReference|Electromagnetism|1990|I.S. Grant|author2 = W.R. Phillips|ed = 2nd|edpage = Second Edition|next = Coulomb's Law of Electrostatics}}: Chapter $1$: Force and energy in electrostatics * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Kepler's Laws of Planetary Motion|next = Book:Isaac Newton/Philosophiae Naturalis Principia Mathematica}}: Chapter $\text {B}.25$: Kepler's Laws and Newton's Law of Gravitation * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Acceleration is Second Derivative of Displacement with respect to Time|next = Newton's Law of Universal Gravitation/Historical Note}}: Chapter $8$: The System of the World: Newton * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Inverse for Complex Multiplication|next = Definition:Inverse Trigonometric Function|entry = inverse square law of gravitation}} Category:Gravity feriqxzt5bc0wgv9sdr31pc7kdlycd7


Fermat's Principle of Least Time

Fermat's Principle of Least Time 0 8267 486203 448998 2020-09-07T21:50:19Z Prime.mover 59 wikitext text/x-wiki == Physical Law == === Original formulation === The path taken by a ray of light from one point to another is the one that can be traversed in the least time. === Modern version === The optical path length must be stationary, that is, be either a minimum, maximum or a point of inflection. == Proof == This follows from the Huygens-Fresnel Principle. {{proof wanted}} == Also see == * Snell-Descartes Law {{Namedfor|Pierre de Fermat|cat = Fermat}} == Historical Note == {{:Fermat's Principle of Least Time/Historical Note}} == Sources == * {{BookReference|Men of Mathematics|1937|Eric Temple Bell|prev = Definition:Euler-Lagrange Equation/Historical Note|next = Fermat's Principle of Least Time/Historical Note}}: Chapter $\text{IV}$: The Prince of Amateurs * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Mathematician:Willebrord van Royen Snell|next = Brachistochrone is Cycloid/Proof 1}}: $1$: The Nature of Differential Equations: $\S 6$: The Brachistochrone. Fermat and the Bernoullis * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Heron's Principle of Reflection|next = Fermat's Principle of Least Time/Historical Note}}: Chapter $\text {A}.7$: Heron (first century A.D.) * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Snell-Descartes Law/Historical Note|next = Definition:Folium of Descartes/Historical Note}}: Chapter $\text {A}.13$: Fermat ({{DateRange|1601|1665}}) Category:Optics 23ym34cgy2myo527vebsvhnb1brob0o


Snell-Descartes Law

Snell-Descartes Law 0 8272 486201 409925 2020-09-07T21:49:27Z Prime.mover 59 wikitext text/x-wiki == Physical Law == Consider a ray of light crossing the threshold between two media. Let its speed: :in medium $1$ be $v_1$ :in medium $2$ be $v_2$. Let it meet the threshold at: :an angle $\alpha_1$ from the vertical in medium 1 :an angle $\alpha_2$ from the vertical in medium 2. Then the '''Snell-Descartes law''' states that: :$\dfrac {\sin \alpha_1} {v_1} = \dfrac {\sin \alpha_2} {v_2}$ == Proof == The '''Snell-Descartes law''' can be derived from Fermat's Principle of Least Time as follows: Let the ray of light travel from $A$ to $P$ in the medium $1$. Then let it travel from $P$ to $B$ in medium $2$. :350px The total time $T$ required for that journey is: :$T = \dfrac {\sqrt {a^2 + x^2} } {v_1} + \dfrac {\sqrt {b^2 + \paren {c - x}^2} } {v_2}$ from the geometry of the above diagram. From Fermat's Principle of Least Time, this time will be a minimum. From Derivative at Maximum or Minimum, we need: :$\dfrac {\d T} {\d x} = 0$. So: :$\dfrac x {v_1 \sqrt {a^2 + x^2} } = \dfrac {c - x} {v_2 \sqrt {b^2 + \paren {c - x}^2} }$ which leads directly to: :$\dfrac {\sin \alpha_1} {v_1} = \dfrac {\sin \alpha_2} {v_2}$ by definition of sine. {{qed}} == Also presented as == This law can also be seen expressed as: :$\dfrac {\sin \alpha_1} {\sin \alpha_2} = \dfrac {v_1} {v_2}$ {{Namedfor|Willebrord van Royen Snell|name2 = René Descartes|cat = Snell|cat2 = Descartes}} == Historical Note == {{:Snell-Descartes Law/Historical Note}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Definition:Brachistochrone|next = Mathematician:Willebrord van Royen Snell}}: $1$: The Nature of Differential Equations: $\S 6$: The Brachistochrone. Fermat and the Bernoullis * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Largest Rectangle with Given Perimeter is Square/Historical Note|next = Snell-Descartes Law/Historical Note}}: Chapter $\text {A}.13$: Fermat ({{DateRange|1601|1665}}) * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Length of Arc of Nephroid|next = Brachistochrone is Cycloid/Proof 1}}: Chapter $\text {B}.22$: Bernoulli's Solution of the Brachistochrone Problem Category:Optics ljffje6b8d5mp6djnxrc1st4h40ggeq


Einstein's Mass-Velocity Equation

Einstein's Mass-Velocity Equation 0 8348 256262 256255 2016-05-19T06:05:06Z Prime.mover 59 wikitext text/x-wiki == Physical Law == The mass $m$ of a body is not constant. It varies with the body's velocity, according to the equation: : $m = \dfrac {m_0} {\sqrt{1 - \dfrac {v^2} {c^2} } }$ where: : $v$ is the magnitude of the velocity of the body : $c$ is the speed of light in vacuum : $m_0$ is the rest mass of the body. The value $m$ is known as the '''relativistic mass''' of the body. The factor $\dfrac 1 {\sqrt{1 - \dfrac {v^2} {c^2} } }$ is known as the Lorentz Factor. == Proof == {{Proofread|grammar}} Imagine a comet that flies towards a planet on which you are resting. The comet's velocity $u$ towards the planet is much smaller than the speed of light. Now imagine the impact caused by the comet striking the planet as a deformation of the planet. That impact can be seen as proportional to the momentum of the comet which is: : $p = m u$ where: : $p$ is the magnitude of the comet's momentum : $m$ is the comet's rest mass : $u$ is the magnitude of the comet's velocity. If someone else watches the crash from a space ship passing by with a relativistic velocity (for example $v = 0.7c$) he will find that the comet appears to move more slowly than it does from your stationary perspective on the planet. This is due to the time dilation, which is given by: : $\Delta t' = \dfrac {\Delta t} {\sqrt{1 - \dfrac {v^2} {c^2}}}$ where: : $\Delta t'$ is the time interval measured from the space ship : $\Delta t$ is the time interval measured in the inertial system containing planet and comet : $v$ is the magnitude of the space ship's velocity : $c$ is the speed of light in vacuum. Because the time measured in the space ship is less, the comet will appear to need more time to cover a certain distance. Thus, its velocity seems smaller from the perspective of the space ship (Note: The space ship's trajectory be perpendicular to the comet's trajectory towards the planet, so there is no length contraction parallel to the trajectory of the comet). The comet's velocity measured from the planet is: : $u = \dfrac {\mathrm d s} {\mathrm d t}$ where the comet's velocity measured from the space ship is: : $u' = \dfrac {\mathrm d s} {\mathrm d t'}$ and as we know from the time dilation, the term for $u'$ is thus: : $u' = u \sqrt{1 - \frac {v^2}{c^2}}$ The observer in the space ship will nevertheless find out that the impact is equal to the one observed by the resting person. That means that the comet's momentum doesn't change, no matter from what inertial system you measure it. That can only be possible, if -- seen from the space ship -- the comet's mass increases, as its velocity decreases. The comet's momentum from the perspective of the space ship is: : $p' = m' u'$ where: : $p'$ is the magnitude of the comet's momentum measured from the inertial system of the space ship : $m'$ is the comet's relativistic mass measured from the inertial system of the space ship : $u'$ is the magnitude of the comet's velocity measured from the inertial system of the space ship. And because the measured momentums from both observers are the same, you can write: : $p = p'$  : $m u = m' u'$ : $m' = m \dfrac u {u'}$ : $m' = m \dfrac u {u \sqrt{1 - \frac {v^2} {c^2}}}$ : $m' = \dfrac m {\sqrt{1 - \frac {v^2} {c^2}}}$ {{qed}} {{Namedfor|Albert Einstein|cat = Einstein}} == Historical Note == {{:Einstein's Mass-Velocity Equation/Historical Note}} {{MissingLinks}}  == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Definition:Burnout Height|next = Einstein's Mass-Energy Equation}}: Miscellaneous Problems for Chapter $2$: Problem $32$ * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Definition:Linear Momentum|next = Einstein's Law of Motion}}: Chapter $\text {B}.7$: A Simple Approach to $E = M c^2$ Category:Physics 8rli0ucu43dqg2cj99uo08s38lsth5f


Einstein's Law of Motion

Einstein's Law of Motion 0 8349 447608 258812 2020-02-09T00:54:06Z Prime.mover 59 wikitext text/x-wiki == Physical Law == The force and acceleration on a body of constant rest mass are related by the equation: : $\mathbf F = \dfrac {m_0 \mathbf a} {\left({1 - \dfrac{v^2}{c^2}}\right)^{\tfrac 3 2}}$ where: : $\mathbf F$ is the force on the body : $\mathbf a$ is the acceleration induced on the body : $v$ is the magnitude of the velocity of the body : $c$ is the speed of light : $m_0$ is the rest mass of the body. == Proof == Into Newton's Second Law of Motion: : $\mathbf F = \dfrac {\mathrm d}{\mathrm d t} \left({m \mathbf v}\right)$ we substitute Einstein's Mass-Velocity Equation: : $m = \dfrac {m_0} {\sqrt {1 - \dfrac {v^2} {c^2}}}$ to obtain: : $\mathbf F = \dfrac {\mathrm d} {\mathrm d t} \left({\dfrac {m_0 \mathbf v}{\sqrt{1 - \dfrac {v^2}{c^2}}}}\right)$ Then we perform the differentiation {{WRT|Differentiation}} time: {{begin-eqn}} {{eqn | l = \frac{\mathrm d}{\mathrm d t} \left({\frac {\mathbf v}{\sqrt{1 - \dfrac {v^2}{c^2} } } }\right)       | r = \frac{\mathrm d}{\mathrm d v} \left({\frac {\mathbf v}{\sqrt{1 - \dfrac {v^2}{c^2} } } }\right) \frac{\mathrm d v}{\mathrm d t}       | c = Chain Rule for Derivatives }} {{eqn | r = \mathbf a \left({\frac {\sqrt{1 - \dfrac {v^2}{c^2} } - \dfrac v 2 \dfrac 1 {\sqrt{1 - \dfrac {v^2}{c^2} } } \dfrac{-2 v}{c^2} } {1 - \dfrac {v^2}{c^2} } }\right)       | c = Chain Rule for Derivatives, Quotient Rule, etc. }} {{eqn | r = \mathbf a \left({\frac {c^2 \left({1 - \dfrac {v^2}{c^2} }\right) + v^2} {c^2 \left({1 - \dfrac {v^2}{c^2} }\right)^{3/2} } }\right)       | c =  }} {{eqn | r = \mathbf a \left({\frac 1 {\left({1 - \dfrac {v^2}{c^2} }\right)^{3/2} } }\right)       | c =  }} {{end-eqn}} Thus we arrive at the form: : $\mathbf F = \dfrac {m_0 \mathbf a} {\left({1 - \dfrac{v^2}{c^2}}\right)^{\tfrac 3 2}}$ {{qed}} == Comment == Thus we see that at low velocities (i.e. much less than that of light), the well-known equation $\mathbf F = m \mathbf a$ holds to a high degree of accuracy. {{namedfor|Albert Einstein|cat = Einstein}} == Sources == * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Einstein's Mass-Velocity Equation|next = Einstein's Mass-Energy Equation}}: Chapter $\text {B}.7$: A Simple Approach to $E = M c^2$ Category:Physics mxvf0tynbofgs7y6tmuxz6slcyrlg7i


Hooke's Law

Hooke's Law 0 8451 496104 451938 2020-10-23T20:49:30Z Prime.mover 59 wikitext text/x-wiki == Physical Law == '''Hooke's Law''' applies to an ideal spring: :$\mathbf F = -k \mathbf x$ where: :$\mathbf F$ is the force caused by a displacement $\mathbf x$ :$k$ is the spring force constant. The negative sign indicates that the force pulls in the opposite direction to that of the displacement imposed. :''The strain is proportional to the stress.'' == Also see == * Dimension of Spring Force Constant: the dimension of $k$ is $\mathsf {M T}^{-2}$. == Sources == * {{BookReference|Understanding Physics|1966|Isaac Asimov|prev = Definition:Elasticity|next = Definition:Elastic Limit}}: $\text {I}$: Motion, Sound and Heat: Chapter $4$: Gravitation: The Gravitational Constant * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Mathematician:Robert Hooke|next = Definition:Hopf Bifurcation|entry = Hooke's law}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Mathematician:Robert Hooke|next = Definition:Hopf Bifurcation|entry = Hooke's law}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Homotopy|next = Mathematician:Guillaume de l'Hôpital|entry = Hooke's law}} Category:Physics lfeol0b5fc6vyjku946eh2tcrrzgpon


Leibniz's Formula for Pi/Lemma

Leibniz's Formula for Pi/Lemma 0 8459 456449 258605 2020-03-19T17:45:58Z Prime.mover 59 wikitext text/x-wiki == Lemma == {{begin-eqn}} {{eqn | l = \frac 1 {1 + t^2}       | r = 1 - t^2 + t^4 - t^6 + \cdots + t^{4 n} - \frac {t^{4 n + 2} } {1 + t^2}       | c =  }} {{eqn | r = \paren {\sum_{k \mathop = 0}^{2 n} \paren {-1}^k t^{2 k} } - \frac {t^{4 n + 2} } {1 + t^2}       | c =  }} {{end-eqn}} This holds for all real $t \in \R$. == Proof == {{begin-eqn}} {{eqn | l = \frac {1 - \paren {-t^2}^{2 n + 1} } {1 - \paren {-t^2} }       | r = \sum_{k \mathop = 0}^{2 n} \paren {-t^2}^k       | c = Sum of Geometric Sequence }} {{eqn | ll= \leadsto       | l = \frac {1 + \paren {t^2}^{2 n + 1} } {1 + t^2}       | r = 1 - t^2 + t^4 - t^6 + \cdots + t^{4 n}       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 {1 + t^2} + \frac {t^{4 n + 2} } {1 + t^2}       | r = 1 - t^2 + t^4 - t^6 + \cdots + t^{4 n}       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 {1 + t^2}       | r = 1 - t^2 + t^4 - t^6 + \cdots + t^{4 n} - \frac {t^{4 n + 2} } {1 + t^2}       | c =  }} {{end-eqn}} From Square of Real Number is Non-Negative, we have that: :$t^2 \ge 0$ for all real $t$. So $- t^2 \le 0$ and so $- t^2 \ne 1$. So the conditions of Sum of Geometric Sequence are satisfied, and so the above argument holds for all real $t$. {{qed}} Category:Analysis p4j6tw6gbc96p47fxnnxya9iq3srmwp


Stabilizer is Subgroup/Corollary

Stabilizer is Subgroup/Corollary 0 8606 479172 378345 2020-07-22T06:48:35Z Prime.mover 59 Prime.mover moved page Stabilizer is Subgroup/Corollary 2 to Stabilizer is Subgroup/Corollary wikitext text/x-wiki == Corollary to Stabilizer is Subgroup == <onlyinclude> Let $G$ be a group whose identity is $e$. Let $G$ act on a set $X$. Let $x \in X$. Then: :$\forall g, h \in G: g * x = h * x \iff g^{-1} h \in \Stab x$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = h * x       | r = g * x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = g^{-1} * \paren {h * x}       | r = g^{-1} * \paren {g * x}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {g^{-1} h} * x       | r = \paren {g^{-1} g} * x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {g^{-1} h} * x       | r = e * x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {g^{-1} h} * x       | r = x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = g^{-1} h       | o = \in       | r = \Stab x       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Definition:Transitive Group Action|next = Orbit-Stabilizer Theorem/Proof 2}}: $\S 6.5$. Orbits: Lemma $\text {(ii)}$ Category:Stabilizer is Subgroup bdt6cxy88ib67897zz493mozynn8f9a


Upper Bound is Lower Bound for Inverse Ordering 0 8810 212639 212638 2015-04-07T12:51:49Z Prime.mover 59 wikitext text/x-wiki {{refactor|Not trivial. Result may need rephrasing using iff instead}} == Definition == Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$. Let $M$ be an upper bound for $\left({T, \preceq}\right)$. Let $\succeq$ be the dual ordering of $\preceq$. Then $M$ is a lower bound for $\left({T, \succeq}\right)$. == Proof == Let $M$ be an upper bound for $\left({T, \preceq}\right)$. That is: :$\forall a \in T: a \preceq M$ By definition of dual ordering, it follows that: :$\forall a \in T: M \succeq a$ That is, $M$ is a lower bound for $\left({T, \succeq}\right)$.  {{qed}} {{expand|Proof via duality}} == Also see == * Lower Bound is Upper Bound for Inverse Ordering Category:Order Theory d06yyjlwxyk51wcwiot068yvvtw6ecq


User talk:Linus44

User talk:Linus44 3 9010 346671 346666 2018-03-11T09:36:20Z Prime.mover 59 wikitext text/x-wiki ==== Welcome ==== Welcome to ProofWiki! Since you're new, you may want to check out the general help page. It's the best first stop to see how things are done (next to reading proofs, of course!). Please feel free to contribute to whichever area of mathematics interests you, either by adding new proofs, or fixing up existing ones. If you have any questions please feel free to contact me, or post your question on the questions page. Here are some useful pages to help get you started: * Community Portal - To see what needs to be done, and keep up to date with the community. * Recent Changes - To keep up with what's new, and what's being added. * Main Page talk - This is where most of the main discussions regarding the direction of the site take place. If you have any ideas, please share them! Cheers, prime.mover (talk) == Properties of Degree == We're obviously on precisely the same page here (literally!) as this was the next proof (after tidying up after the infrastructure update) that I was going to post! You're not working your way through Hartley and Hawkes yourself, are you?  Keep up the damn fine work, bro - and bear with me when I reformat stuff into our (so far unwritten) house style. --prime mover 00:48, 8 February 2011 (CST) : Just read your comment (later deleted) on polynomials: if you have a strategy for axiomatic foundation of Galois theory, then feel free to go for it. You've noticed there's a lot of groundwork done already - but the author of most of this (ahem, me) has little background in it except what's been self-taught, so feel free to rewrite or amend if stuff is insufficiently precise or accurate. And if you have an idea of how to turn into crystal the amorphous sludge that currently constitutes the section on polynomials, then go for it. --prime mover 00:59, 8 February 2011 (CST) == The big picture == You're obviously focused here - I'll let you get on with it as you definitely seem to be in the zone. You just happened to turn up as we're in the middle of changing the infrastructure to use a new LaTeX package MathJax. Some of the old commands no longer work like <nowiki>\or</nowiki> and <nowiki>\and</nowiki> - I'm replacing them with <nowiki>\lor</nowiki> and <nowiki>\land</nowiki> as I go - and display text formatting insists on centre justifying, which doesn't work too well here. It'll shake itself down in time, no doubt, but in the meantime expect plenty of pages to not look as good as they should. IF you see any then feel free to try and fix them, and if you have problems let me know and I'll see what I can do. --prime mover 00:22, 9 February 2011 (CST) : ... Just pick it up as you go along, you're all right.--prime mover 00:46, 9 February 2011 (CST) == Comments, then ... == You said "comments welcome, okay then ... Nice job. In the groove. There's a few stylistic and "house standard" issues which you may want to take on board (not to worry if not, it'll get changed). * Good to separate each entity in LaTeX with a space - a rule of thumb is put a space between each variable and before each backslash starting a command. It's not important, but it makes it readable and allows for a saner line breaking. * More tediously, anything parenthesised ought to be between $\left({ left and right }\right)$ delimiters. Not sure if this is relevant any more actually, but with MediaWiki in particular this enforced the browser to space things neatly, even in cases where what's in the middle is a single letter. E.g. $f \left({x}\right)$ not $f(x)$. This needs to be waived, of course, if the brackets involved are in different latex strings. * You've noticed that the delimiter of choice is now dollar not < math >. The latter renders still, but interestingly not in transclusions (see Trigonometric Identities for an example of lots of transclusions). There is an ongoing exercise to convert all math delimiters to dollars. * Finally (this time round), the writing style here is (barring the occasional page or two in symbolic logic written by someone who didn't get it) "short and sweet." : Long sentences are likely to be split up into shorter ones. : Each sentence needs to go on a separate line.  : The result is that it is easy to follow. : The above fact continues to hold even when the proof is tortuously complex. * Other minor stylistic quibbles are relatively inconsequential: display equations aren't punctuated with a full stop (that's an Ian Stewartism), lines before a display equation end with a colon: :because this is a display equation and punctuation does not go inside a latex object (despite what they say on wikipedia). That is: :Hence the result for $f(x)$. not: :Hence the result for $f(x).$ And yes I know I've broken more than one of the above rules in the above. Keep on trucking.--prime mover 13:34, 11 February 2011 (CST) == Numbering equations and referring back == Like this: :$\label{einstein} e = m c^2$ Then you can refer back to \eqref{einstein} like this. This is new with the MathJax package. I'm not sure how limited we are on how such equations may be formatted. I'm not a fan of centre justification on a wiki page as it doesn't lend itself to the eye when the sentences are short. I haven't found a way to define the default for display equations to default to left justification. I believe you can do it in standard LaTeX documents (long time since I wrote one) so it ought to be doable here. What I tend to do in the meantime is do it manually: :$(1) \quad e = m c^2$ Then you can refer back to $(1)$ like this. --prime mover 02:13, 12 February 2011 (CST) == Help page == Wow! Nice one. That was something I was going to get round to doing one time. Saves me a pain! Thx. --prime mover 02:18, 12 February 2011 (CST) ... in fact it encouraged me to work on it. --prime mover 04:11, 12 February 2011 (CST) == Pointwise operations == We already have a page for this. I think it's in the Abstract Algebra category. Might take some hunting down - sorry, been called away for domestic reasons - I'll get back to you on it. --prime mover 06:21, 13 February 2011 (CST) : You wrote: "... the definition of induced structure is quite robust so it covers addition (although interestingly not [I think] multiplication) of polynomial forms as well ..." I don't see why not, if you treat the concept of $\oplus$ as being either multiplication or addition. Can we continue this discussion in the talk page of Definition:Induced Structure? I need to understand what you're thinking. --prime mover 08:35, 13 February 2011 (CST) ::... Goodness. It'll take me a while to digest all that lot ... bear with me, I'm only able to bear a subset of my braincells at the moment. I never dreamed it would have been so involved. --prime mover 09:56, 13 February 2011 (CST) == Defining operators == If you need to define an operator in $\LaTeX$, rather than use \text{op} for example, use \operatorname{op} instead. There's a reason for this: apparently it gets rendered differently so as to be more appropriate to being an operator (more space afterwards perhaps) in certain browsers. Or something. It's what we've been doing up till now. Probably no big deal, it's just a nicety. We may find that \operatorname isn't what's supposed to be used now anyway, it's probably \mathop now - but I'm not in a position to experiment. --prime mover 00:45, 17 February 2011 (CST) : Sure thing, it appears that \mathop is for turning a composition of symbols into a single operator, e.g. ::$\displaystyle \mathop{\bigoplus \bigotimes}_{1\leq i< j \leq n} (ij)$ : while \operatorname is for operators with textual names, hence makes a better standard of \id etc. At least I think those their intended purposes. Linus44 13:39, 17 February 2011 (CST) :::backslashed your LaTeX for yez --prime mover 14:53, 17 February 2011 (CST) == Macros == Replied on my page: --prime mover 15:07, 20 February 2011 (CST) == Internal links == I'm interested as to your techniques for including internal links in your articles because I see they have the underscores in them. Not a big problem (can make it more tedious to produce a form for reading because then it forces you to format it) - but I'm curous as to what your technique is that makes this easier than writing the link without the underscores. --prime mover 00:25, 23 February 2011 (CST) == ... unrelatedly == While the Principle of Mathematical Induction implies Well-Ordering Principle, as proved on Equivalence of Well-Ordering Principle and Induction, it is not the case that the Well-Ordering Principle implies the Axiom of Choice. The latter is implied by the Well-Ordering Theorem (not documented yet), which states that any set can have an ordering under which that set is a well-ordered set. But that latter is NOT the Well-Ordering Principle.  Good question. --prime mover 00:41, 23 February 2011 (CST) : Ah, my book has the WOP as "every set can be well ordered". I was wondering why no-one had told me about that incredible set of equivalences. Thanks! --Linus44 00:48, 23 February 2011 (CST) :: There is a word in the WOP defn on this site warning not to confuse the WOP with the WOT - perhaps it needs a further word about how the WOT is also sometimes called the WOP! Scary biscuits ... --prime mover 01:57, 23 February 2011 (CST) == Listen here wolfchild == ... I speak to you of the science of mythology ... :-) Good catch. --prime mover 14:43, 23 February 2011 (CST) == big thankx == for helping with this tedious migration. Your effort is appreciated. --prime mover 15:14, 3 March 2011 (CST) : S'alright, I just figured out external editing in Chrome; the equations load about 10 times quicker than they did in Firefox so it's quite relaxing to do now --Linus44 23:50, 3 March 2011 (CST) == Sorting definition pages == When you categorise a new definition page, plz remember to add <nowiki>|{{SUBPAGENAME}}</nowiki> to the definitions category - otherwise the category page will sort it on "Definition:..." and therefore put it under D where it ought to be sorted as the subpage name. (Note to self: check this is stated in the edit help pages.) --prime mover 17:35, 4 March 2011 (CST) : Ah yes sorry, I had picked up on that....then forgotten again --Linus44 01:46, 5 March 2011 (CST) :: ... no big deal, we can correct them as we find them. :: Underneath the edit pane there's a "MediaWiki Functions" box which you can use to put the template for a category, definition category, etc. in place without having to type it in. Just click on the appropriate blue thing and a copy appears where your cursor is. Might be worth getting the habit then it's just muscle memory. --prime mover 03:23, 5 March 2011 (CST) == Minor changes == Something to beware of (I'm always doing it myself!) is hitting the "minor edit" check box when you don't want to, and you especially don't want to do it with a new proof. In the latter case what happens is that the proof does not get logged as the "Latest proof" (on front page) or added to the list of "recently added proofs" (see Community portal), and you lose the kudos. --prime mover 17:02, 27 March 2011 (CDT) : Ah yes, had it marking as minor by default, now changed. --Linus44 17:08, 27 March 2011 (CDT) == Properties of Gamma Function == Just a heads up: what I did with Properties of Gamma Function was split it up into 4 separate pages linked by transclusion. It's the same technique as I used on Properties of Binomial Coefficients, q.v. for comparison. The individual results all merit a page of their own (particularly the ones with names on them) and besides they'll probably end up being too big for comfort. Big pages don't work too well on this wiki. --prime mover 15:59, 29 March 2011 (CDT) : Works for me --Linus44 09:36, 30 March 2011 (CDT) == Does not divide == Using nmid. See Definition:Divisor Notation where the notation is discussed. It's not a perfect solution but using slash to negate a backslash makes something ugly. --prime mover 01:34, 3 May 2011 (CDT) == Using newcommand == I see you're experimenting with newcommand (fair play to you) but just a heads-up: I've found it doesn't work too well when transcluding pages with it on. I also think it doesn't work when the newcommand isn't right at the top of the page (had to change a Spec of yours back to an operatorname a couple days ago) and I haven't got to finding out why this is. Be aware you may get rendering issues. If we can solve these problems (I confess they're not high on my list ATM), then I wonder whether it might be feasible to set up some "standard" abbrevs for some stuff, a) because it will streamline certain pages, and b) because it will then give us a basis for a "uniform" symbology. One thing I've been meaning to get round to (but there's always exciting results to document instead) is a list of standard symbols (as part of "house style") like using $\varnothing$ for the empty set rather than $\emptyset$, and $\backslash$ for "divides" and not "\mid" and so on. Using a standard header file with "newcommand" commands in may give us a headstart on that experiment.--prime mover 16:10, 6 May 2011 (CDT) : Yeah, the newcommand stuff was just because I prefered $\operatorname{Max}\:\operatorname{Spec}$ to $\operatorname{MaxSpec}$, but the former takes forever to type. : Uniform notation would have it's advantages, but then it'd be good to have a more obvious link on each page saying "The notation used is explained here", otherwise there's a risk of being difficult to read for anyone unfamiliar with these pages, especially with stuff like $\backslash$, where $\mid$ is more commonly used. --Linus44 07:22, 7 May 2011 (CDT) :: My approach to this is always, when using a piece of notation which is possibly non-standard or unusual, to add "... where $\backslash$ denotes divides" or whatever. How do I know whether it's non-standard enough to merit this? Someone complains they can't understand it. :-) --prime mover 08:14, 7 May 2011 (CDT) == Welcome back == I thought you'd gone forever! --prime mover (talk) 05:51, 28 September 2012 (UTC) :I've been making do with the internet on my phone for ages, since I never used it for much more than emails.  I'm not sure how often I'll be on here, the spare time I do have is usually a much needed break from mathematics, but we'll see. --Linus44 (talk) 16:41, 28 September 2012 (UTC) == Hold your horses, bro' == The recent edit you made to Definition:Differential has completely obscured the basic simplicity that it originally was and cast it into language which is inaccessible to undergraduates. We're trying not to lose the basic understandability here, and besides, the citation at the bottom of the page no longer has anything to do with what's on the page. You might want to take some time to refamiliarise yourself with the house style and the general philosophy of this site - the idea is for it to be more of a dictionary than an encyclopedia. In particular, I believe that what we might want to do with this page is to write the more complicated stuff in a separate page and transclude it. --prime mover (talk) 19:26, 28 September 2012 (UTC) : This is a point. It might be good to split the definition up into sections of successive generality, with the $\R \to \R$ setting at the top. I'll rearrange it after I've eaten some chocolate bread.  : BTW, is there a page for vector spaces of the same dimension are isomorphic? There seemed to be enough linear algebra that it ought to be there, but I couldn't find it. --Linus44 (talk) 20:26, 28 September 2012 (UTC) == SourceReview template == Please add an invocation of Template:SourceReview at the bottom of any pages which you have refactored, like for the Gauss's Lemma page. It allows someone coming after to note the fact that the process flow of the works that are invoked can be amended as appropriate. --prime mover (talk) 06:23, 19 March 2013 (UTC) : ok --Linus44 (talk) 11:11, 19 March 2013 (UTC) ==  Fundamental Theorem of Algebra == Thanks for polishing up my proof! :-) It was my first attempt to write a proof here, so I'm really grateful for your corrections and additions. --BFeuerbacher (talk) 20:36, 16 April 2013 (UTC) :No worries, and thanks for your contribution. My changes were just cosmetic stuff and adding links to the results used, see here for an explanation of the changes. --Linus44 (talk) 22:23, 16 April 2013 (UTC) == Question about Definition:Partition of Unity (Topology) == I have addressed some of the unclarities in this page, but there is still an outstanding question about what is specifically meant by "support" - there are two definitions on that page which appear to be mutually incompatible. Are you in a position to be able to go back to this page and see whether (a) it is still accurate, and (b) it needs improvement? I gather you may be more familiar with this area of maths than I am - and besides, you wrote the original (presumably in support of the Stokes' Theorem page) ... --prime mover (talk) 07:52, 15 September 2013 (UTC) m21g5cp0m2ex8l1na3mzsx5hyu2kgk0


User:Linus44

User:Linus44 2 9084 455719 221265 2020-03-18T21:30:25Z Prime.mover 59 wikitext text/x-wiki Things to do: == Definition:Differential == Let $(E, \| \cdot \|_E)$, $(F, \| \cdot \|_F)$ be normed vector spaces. Let $U \subseteq E$ be an open set. Let $f : U \to F$ be a mapping. Let $a \in U$ be an element of $U$. Then $f$ is '''differentiable''' at $a$ if there exists a continuous and linear map $df_a \in \mathcal L(E,F)$ such that :$\displaystyle \lim_{h \to 0} \| f(a+h) - f(a) - df_a \cdot h \|_F \| h \|_E^{-1} = 0$ Then $df_a$ is called the '''differental''' or the '''tangent map''' of $f$ at $a$. We say that $f$ is '''continuously differentiable''' if: :$\displaystyle df : (U, \| \cdot \|_E) \to \mathcal (L(E,F),\| \cdot \|_{L(E,F)})$ :$\displaystyle \ : a \mapsto df_a$ is continuous. If $E = \R^n$, this is true iff the first order partial derivatives of $f$ exist and are continuous. == Induced polynomial homomorphism == Even this needs serious thought if it's to be any good. == Permutations == Definition:Cyclic Permutation $k$ well defined. Add canonicality. Incorrect: Definition:Permutation on n Letters/Cycle Notation permutation/cycle confusion? Also $\rho$ should be $\pi$ for consistency. Then here: Equality of Cycles == Rings, properties, equivalent definitions == GCD domain: *Equivalent Definitions of a GCD Domain *Properties of GCD Domains Bézout domain *Equivalent Definitions of a Bézout domain *Properties of Bézout Domains Definition:Unique Factorization Domain: *Equivalent Definitions of a Unique Factorization Domain *Properties of Unique Factorization Domains etc...needs organizing into something more standardized == The rest == *Vinogradov's Theorem: Pick a method of proof and think of a good structure for it. *Figure out conditions for the "function-form epimorphism" to have trivial kernel (see Epimorphism from Polynomial Forms to Polynomial Functions and especially Equality of Polynomials) === Proof of prime number theorem and Siegel Walfiz === See also stuff on Dirichlet's Theorem Poisson Summation Formula Definition:Schwarz Function Harmonic Properties of Schwarz Functions Hadamard Factorisation Theorem Jensen's Formula Definition:Completed Riemann Zeta Function Estimation Lemma Uniform Limit of Analytic Functions is Analytic Poles of Gamma Function Properties of Gamma Function Stirling's Formula for Gamma Function Definition:Order of Entire Function Completed Riemann Zeta Function has Order One Product Equation for Riemann Zeta Function Zeros of Functions of Finite Order Poles of Riemann Zeta Function Riemann Zeta Has No Zeros With Real Part One Unsymmetric Functional Equation for Riemann Zeta Function === Dirichlet: Finished, but check for errors === Analytic Continuation of Dirichlet L-Function L-Function does not Vanish at One Logarithm of Dirichlet L-Functions Dirichlet's Theorem on Arithmetic Sequences Dirichlet L-Function from Trivial Character Convergence of Dirichlet Series with Bounded Coefficients Convergence of Dirichlet Series with Bounded Partial Sums Definition:Dirichlet L-function Definition:Dirichlet Character Definition:Completed Dirichlet L-Function Functional Equation for Dirichlet L-Functions Orthogonality Relations for Characters === Unfinished Pages === Completeness Criterion (Metric Spaces) Open Sets in Real Number Line Equivalence of Definitions of Riemann Zeta Function Definition:Category Hardy-Littlewood Circle Method Vinogradov Circle Method Characterisation of Totally Ordered Field Properties of Totally Ordered Field Gauss's Lemma (Ring Theory) 8s9j00q06ec1pixsutawbecgaerh4p4


Euclid's Lemma for Irreducible Elements

Euclid's Lemma for Irreducible Elements 0 9515 402085 352815 2019-04-22T07:55:42Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $\struct {D, +, \times}$ be a Euclidean domain whose unity is $1$. Let $p$ be an irreducible element of $D$. Let $a, b \in D$ such that: :$p \divides a \times b$ where $\divides$ means '''is a divisor of'''. Then $p \divides a$ or $p \divides b$. === General Result === {{:Euclid's Lemma for Irreducible Elements/General Result}} == Proof == Let $p \divides a \times b$. Suppose $p \nmid a$. Then from the definition of irreducible: : $p \perp a$ Thus from Euclid's Lemma for Euclidean Domains: :$p \divides b$ Similarly, if $p \nmid b$: :$p \divides a$ So: :$p \divides a b \implies p \divides a$ or: :$p \divides b$ as we needed to show. {{qed}} {{Namedfor|Euclid}} == Also see == * Euclid's Lemma for Prime Divisors, for the usual statement of this result, which is this lemma as applied specifically to the integers. == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Euclid's Lemma for Euclidean Domains|next = Euclid's Lemma for Irreducible Elements/General Result}}: Chapter $6$: Polynomials and Euclidean Rings: $\S 29$. Irreducible elements: Theorem $56 \ \text{(i)}$ Category:Euclidean Domains Category:Euclid's Lemma 4ejlquo28pgy35i9ww9c5y30e0814ov


Continuous Image of Separable Space is Separable 0 9604 469520 468928 2020-05-21T13:49:26Z Prime.mover 59 wikitext text/x-wiki == Definition == Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces. Let $f: T_1 \to T_2$ be a continuous mapping. If $T_1$ is separable, then so is the image $f \sqbrk {T_1}$. That is, separability is a continuous invariant. == Proof == By definition, $T_1 = \struct {S_1, \tau_1}$ is separable {{iff}} there exists a countable subset $D \subset S_1$ which is (everywhere) dense in $T_1$. We need to show that if there exists a continuous mapping $f: T_1 \to T_2$, then $f \sqbrk {T_1}$ is also separable. That is, that there exists a countable subset of $f \sqbrk {S_1}$ which is dense in $T_2$. Let $x_2$ be any point in the image $f \sqbrk {S_1}$ of $S_1$ under $f$. Let $U \in \tau_2$ be an arbitrary open set such that $x_2 \in U$. By definition of image set, there exists some $x_1 \in S_1$ with $\map f {x_1} = x_2$. Since $f$ is continuous, $f^{-1} \sqbrk U$ is open in $T_1$. By definition of preimage, $x_1$ is in this set. We are given that $D$ is a countable subset of $S_1$ which is dense in $T_1$. By definition of dense, $D \cap f^{-1} \sqbrk U \ne \O$ Thus there exists some $d \in D$ such that $d \in f^{-1} \sqbrk U$. Therefore $\map f d \in U$. Since $U$ was arbitrary, it follows that $f \sqbrk D$ is dense in $T_2$. By Image of Countable Set under Mapping is Countable, $f \sqbrk D$ is countable. Hence $T_2$ is separable. {{qed}} == Sources == * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Definition:Closed Invariant|next = Continuous Image of Compact Space is Compact}}: Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Lindelöf Property is Preserved under Continuous Surjection|next = Compactness Properties Preserved under Projection Mapping}}: Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties Category:Separable Spaces Category:Continuous Invariants o526ot6ldkjyrmhw1zn4p9o4efhl777


Urysohn's Lemma

Urysohn's Lemma 0 9642 469196 469183 2020-05-20T06:59:31Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $T = \struct {S, \tau}$ be a $T_4$ topological space. Let $A, B \subseteq S$ be closed sets of $T$ such that $A \cap B = \O$. Then there exists an Urysohn function for $A$ and $B$. == Proof == Let $T = \struct {S, \tau}$ be a $T_4$ space. Let $A, B \subseteq S$ be closed sets of $T$ such that $A \cap B = \O$. Let $P = \Q \cap \closedint 0 1$ where $\closedint 0 1$ is the closed unit interval. $\Q$ is countable, therefore so is $P$. === Creation of Domain === We are going to construct a set $\Bbb U \subseteq \tau$ of open sets with $P$ as an indexing set: :$\Bbb U = \set {U_i: i \in P}$ such that: :$\forall p, q \in P: p < q \implies {U_p}^- \subseteq U_q$ where ${U_p}^-$ denotes the set closure of $U_p$. We define $U_p$ by induction, as follows. List the elements of $P$ in the form of an infinite sequence $\sequence z$. Let $z_0 = 1, z_1 = 0$. In general, let $P_n$ denote the set consisting of the first $n$ elements of $\sequence z$. Let $\map \PP n$ be the proposition: :$U_p$ is defined for all $p \in P_n$, and: ::$(1): \quad \forall p, q \in P_n: p < q \implies {U_p}^- \subseteq U_q$ ==== Basis for the Induction ==== As $S$ is a $T_4$ space we have that: :$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U_1, V \in \tau: A \subseteq U_1, B \subseteq V$  We have that $A \subseteq U_1$ is a closed set of $S$. We define $U_1 = S \setminus B$ where $S \setminus B$ denotes the complement of $B$ in $S$. As $S$ is $T_4$, we can choose an open set $U_0 \in \tau$ such that $A \subseteq U_0$ and ${U_0}^- \subseteq U_1$. Thus $\map \PP 1$ is shown to hold. This is our basis for the induction. ==== Induction Hypothesis ==== This is our induction hypothesis: Let $\map \PP k$ be the proposition: :$U_p$ is defined for all $p \in P_k$, and: :: $(1): \quad \forall p, q \in P_k: p < q \implies {U_p}^- \subseteq U_q$ We want to show that if $\map \PP k$ holds, then: :$U_p$ is defined for all $p \in P_{k + 1}$, and: :: $(1): \quad \forall p, q \in P_{k + 1}: p < q \implies {U_p}^- \subseteq U_q$ ==== Induction Step ==== This is our induction step: Let $r = z_{k + 1}$ be the next element in $\sequence z$. Consider $P_{k + 1} = P_k \cup \set r$. It is a finite subset of the closed unit interval $\closedint 0 1$. We consider the usual $<$ ordering on $P_{k + 1}$, which is a subset of $\closedint 0 1$ which in turn is a subset of $\R$.  From Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements, $P_{k + 1}$ has both a minimal element $m$ and a maximal element $M$. From Predecessor and Successor of Finite Toset, every element other than $m$ and $M$ has an immediate predecessor and immediate successor. We already know that $z_1 = 0$ is the minimal element and $z_0 = 1$ is the maximal element of $P_{k + 1}$. So $r$ must be neither of these. Thus: :$r$ has an immediate predecessor $p$ :$r$ has an immediate successor $q$ in $P_{k + 1}$. The sets $U_p$ and $U_q$ are already defined by the inductive hypothesis. As $T$ is a $T_4$ space, there exists an open set $U_r \subseteq \tau$ such that: :${U_p}^- \subseteq U_r$ :${U_r}^- \subseteq U_q$ We now show that $(1)$ holds for every pair of elements in $P_{k + 1}$. If both elements are in $P_n$, then $(1)$ is true by the inductive hypothesis. If one is $r$ and the other is $s \in P_k$, then: :$s < p \implies {U_s}^- \subseteq {U_p}^- \subseteq U_r$ and: :$s \ge q \implies U_r \subseteq {U_q}^- \subseteq {U_s}^-$ Thus $(1)$ holds for every pair of elements in $P_{k + 1}$. Therefore by induction, $U_p$ is defined for all $p \in P$. We have defined $U_p$ for all rational numbers in $\closedint 0 1$. We now extend this definition to every rational $p$ by defining: :$U_p = \begin{cases} \O & : p < 0 \\ S & : p > 1 \end{cases}$ It is easily checked that $(1)$ still holds. === Definition of Function === Let $x \in S$. Define $\map \Q x = \set {p: x \in U_p}$. This set contains no rational number less than $0$ and contains every rational number greater than $1$ by the definition of $U_p$ for $p < 0$ and $p > 1$. Thus $\map \Q x$ is bounded below and its infimum is an element in $\closedint 0 1$. We define: :$\map f x = \map \inf {\map \Q x}$ Now we need to show that $f$ satisfies the conditions of Urysohn's Lemma. Let $x \in A$. Then $x \in U_p$ for all $p \ge 0$, so $\map \Q x$ equals the set $\Q_+$ of all non-negative rationals. Hence $\map f x = 0$. Let $x \in B$. Then $x \notin U_p$ for $p \le 1$ so $\map \Q x$ equals the set of all the rationals greater than $1$. Hence $\map f x = 1$. === Continuity of Function === To show that $f$ is continuous, we first prove two smaller results: :$(a): \quad x \in {U_r}^- \implies \map f x \le r$ We have: $x \in {U_r}^- \implies \forall s > r: x \in U_s$ so $\map \Q x$ contains all rationals greater than $r$. Thus $\map f x \le r$ by definition of $f$. {{qed|lemma}} :$(b): \quad x \notin U_r \implies \map f x \ge r$ We have: $x \notin U_r \implies \forall s < r: x \notin U_s$ so $\map \Q x$ contains no rational less than $r$. Thus $\map f x \ge r$. {{qed|lemma}} Let $x_0 \in S$ and let $\openint c d$ be an open real interval containing $\map f x$. We will find a neighborhood $U$ of $x_0$ such that $\map f U \subseteq \openint c d$. Choose $p, q \in \Q$ such that: :$c < p < \map f {x_0} < q < d$ Let $U = U_q \setminus {U_p}^-$. Then since $\map f {x_0} < q$, we have that $(b)$ implies vacuously that $x \in U_q$. Since $\map f {x_0} > p$, $(a)$ implies that $x_0 \notin U_p$. Hence $x_0 \in U$ and thus $U$ is a neighborhood of $x_0$ because $U$ is open. Finally, let $x \in U$. Then $x \in U_q \subseteq {U_q}^-$, so $\map f x \le q$ by $(a)$. Also, $x \notin {U_p}^-$ so $x \notin U_p$ and $\map f x \ge p$ by $(b)$. Thus: $\map f x \in \closedint p q \subseteq \openint c d$ Therefore $f$ is continuous. {{qed}} == Also see == * Urysohn's Lemma Converse {{Namedfor|Pavel Samuilovich Urysohn|cat = Urysohn}} == Sources == * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Definition:Urysohn Function|next = Urysohn's Lemma Converse}}: Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces * {{planetmath|url = proofofurysohnslemma|title = Urysohn's Lemma}}  Category:T4 Spaces Category:Closed Sets Category:Urysohn Functions lk4wrekvug3s7ehj3mfaidaplglyvka


Urysohn's Lemma Converse

Urysohn's Lemma Converse 0 9643 469195 469190 2020-05-20T06:58:26Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $T = \struct {S, \tau}$ be a topological space. Let there exist an Urysohn function for any two $A, B \subseteq S$ which are closed sets in $T$ such that $A \cap B = \varnothing$. Then $T = \struct {S, \tau}$ is a $T_4$ space. == Proof == Let $A$ and $B$ be arbitrary closed sets of $T$ and let $f$ be an Urysohn function for $A$ and $B$. Let $C = \hointr 0 {\dfrac 1 4}$ and $D = \hointl {\dfrac 3 4} 1$. Then $C$ and $D$ are open in $\closedint 0 1$. {{explain|There's a link available somewhere for the above fact.}} Hence, $\map {f^{-1} } C$ and $\map {f^{-1} } D$ are open in $T$. Furthermore, by definition of Urysohn function, $A \subset \map {f^{-1} } C$ and $B \subset \map {f^{-1} } D$. Also, from Preimage of Intersection under Mapping: :$C \cap D = \O \implies \map {f^{-1} } C \cap \map {f^{-1} } D = \O$ Therefore, $T$ is a $T_4$ space. {{qed}} == Also see == * Urysohn's Lemma {{Namedfor|Pavel Samuilovich Urysohn|cat = Urysohn}} == Sources == * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Urysohn's Lemma|next = Existence of Urysohn Function does not guarantee Normal Space}}: Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces Category:Closed Sets Category:T4 Spaces Category:Urysohn Functions l7l3u68ugbucfzt8ihtrr5vbjl3ux0o


Localization of Ring Exists/Lemma 1

Localization of Ring Exists/Lemma 1 0 9941 324545 306587 2017-11-04T13:26:20Z Barto 3079 wikitext text/x-wiki == Lemma for Localization of Ring Exists == Let $A$ be a commutative ring with unity. Let $S \subseteq A$ be a multiplicatively closed subset with $0 \notin S$. Let $\sim$ be the relation defined on the Cartesian product $A \times S$ by: :$\left({a, s}\right) \sim \left({b, t}\right) \iff \exists u \in S: a t u = b s u$ <onlyinclude> The relation $\sim$ is an equivalence relation. </onlyinclude> == Proof ==  Since by definition $1 \in S$ and $a s = a s$ for all $a \in A, s \in S$, it follows that $\sim$ is reflexive. Also $\sim$ is clearly symmetric because if $a t u = b s u$ then $b s u = a t u$ by symmetry of $=$. Lastly suppose that: : $\left({a, s}\right) \sim \left({b, t}\right)$ and: : $\left({b, t}\right) \sim \left({c, u}\right)$ Then there are $v, w \in S$ such that: :$a t v = b s v$ :$b u w = c t w$ Therefore: :$a u t v w = b u w s v = c t w s v$ and so: : $a u \left({t v w}\right) = c s \left({t v w}\right)$ Since $S$ is multiplicatively closed: : $\left({a, s}\right) \sim \left({c, u}\right)$ Thus $\sim$ is transitive. {{qed}} Category:Commutative Algebra ize69zt3moayarm2v8m8ebbc6c41yg0


Localization of Ring Exists/Lemma 2

Localization of Ring Exists/Lemma 2 0 9943 306586 306585 2017-07-26T15:07:22Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> The operations $+$ and $\cdot$ are well defined on $A_S$. </onlyinclude> {{explain|Put this into the context in which it is relevant so as to make the page standalone.}} == Proof == Let: :$a / s = c / u$ :$b / t = d / v$ be two sets of representatives for two distinct equivalence classes in $A_S$. We have $w, z \in S$ such that: :$\left({a u - c s}\right) w = 0$ and :$\left({b v - d t}\right) z = 0$ Therefore: :$z w \left[{\left({a u - c s}\right) w - \left({b v - d t}\right) z}\right] = 0$ and: :$z w \left[{\left({a t + b s}\right) u v - \left({c v + d u}\right)s t}\right]$ So: :$\left({a t + b s, s t}\right) \sim \left({c v + d u, u v}\right)$ That is: :$\dfrac {a t + b s} {s t} = \dfrac {c v + d u} {u v}$ For multiplication, with $z, w$ as above we have: :$\left({a b u v - c d s t}\right)z w = 0$ So: :$\left({a b, s t}\right) \sim \left({d c, u v}\right)$ and: :$\dfrac {a b} {s t} = \dfrac {d c} {u v}$ {{qed}} Category:Commutative Algebra 6789zb5nsajxmkfe19f7fngtth7kwyv


Localization of Ring Exists/Lemma 3

Localization of Ring Exists/Lemma 3 0 9944 305246 305232 2017-07-16T13:25:21Z Prime.mover 59 wikitext text/x-wiki {{refactor|The definitions of the entities need to be included here as they are on the main page.}} == Lemma for Localization of Ring Exists == <onlyinclude> The map $h \left({a / s}\right) = g \left({a}\right) g \left({s}\right)^{-1}$ is a well defined ring homomorphism $A_S \to B$. </onlyinclude> == Proof == Let $a / s = b / t \in A_S$. Then there is $u \in S$ such that $a t u = b s u$. Therefore, by the homomorphism property, and the fact that $g \left[{S}\right] \subseteq B^\times$: :$g \left({a}\right) g \left({t}\right) = g \left({b}\right) g \left({s}\right) \to g \left({a}\right) g \left({s}\right)^{-1} = g \left({b}\right) g \left({t}\right)^{-1}$ We have that $g$ is a ring homomorphism. As $B$ is a commutative ring it follows that $h$ is also a ring homomorphism. {{qed}} Category:Commutative Algebra hjse67aa67h9mlbociitmtd58rzeita


Equivalence Relation on Cauchy Sequences 0 10209 441888 394769 2020-01-01T14:38:38Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $\struct {X, d}$ be a metric space. Let $\CC \sqbrk X$ denote the set of all Cauchy sequences of elements from $X$. Let a relation $\sim$ be defined on $\CC \sqbrk X$ by: :$\displaystyle \sequence {x_n} \sim \sequence {y_n}  \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$ Then $\sim$ is an equivalence relation on $\CC \sqbrk X$. == Proof == We must show that $\sim$ is :reflexive, :symmetric and :transitive on $\CC \sqbrk X$. To this end, let $\sequence {x_n}, \sequence {y_n}, \sequence {z_n} \in \CC \sqbrk X$ be arbitrary. For each $n \in \N$ we have that $\map d {x_n, x_n} = 0$ by metric space axiom '''M1'''. Therefore $\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, x_n} = 0$. This shows that $\sequence {x_n} \sim \sequence {x_n}$. Thus $\sim$ is reflexive. {{qed|lemma}} By metric space axiom '''M3''', $\map d {x_n, y_n} = \map d {y_n, x_n}$ for each $n \in \N$. Therefore: :$\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \lim_{n \mathop \to \infty} \map d {y_n, x_n}$ So $\sequence {x_n} \sim \sequence {y_n}$ implies that $\sequence {y_n} \sim \sequence {x_n}$. Thus $\sim$ is symmetric. {{qed|lemma}} Finally, by metric space axiom '''M2''', $\map d {x_n, z_n} \le \map d {x_n, y_n} + \map d {y_n, z_n}$ for each $n \in \N$. Therefore, by the sum rule for limits of sequences: :$\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, z_n} \le \lim_{n \mathop \to \infty} \map d {x_n, y_n} + \lim_{n \mathop \to \infty} \map d {y_n, z_n}$ Thus $\sequence {x_n} \sim \sequence {y_n}$ and $\sequence {y_n} \sim \sequence {z_n}$ together imply that $\sequence {x_n} \sim \sequence {z_n}$. Thus $\sim$ is transitive. {{qed|lemma}} So $\sim$ is shown to be reflexive, symmetric and transitive, and therefore an equivalence relation on $\CC \sqbrk X$. {{qed}} Category:Cauchy Sequences oyhdg80u6w2apmebp8s0go65tceb2dc


Field Adjoined Algebraic Elements is Algebraic 0 10490 315920 109939 2017-09-08T08:04:46Z Barto 3079 wikitext text/x-wiki == Definition == Let $L / K$ be a field extension and $S \subseteq L$ a subset. If each $x \in S$ is algebraic over $K$ then $K \left({S}\right)$ is algebraic over $K$. == Proof == Let $S \subseteq L$ be arbitrary, and $x \in K \left({S}\right)$. By Field Adjoined Set $x \in K \left({S}\right)$ {{iff}} $x \in K \left({\alpha_1, \ldots, \alpha_n}\right)$ for some $\alpha_1, \ldots, \alpha_n \in S$. We have that $K \left({\alpha_1, \ldots, \alpha_n}\right) / K$ is finite by Finitely Generated Algebraic Extension is Finite. Moreover a Finite Field Extension is Algebraic. Therefore $x$ is algebraic over $K$. {{qed}} Category:Field Extensions mwdjy617ad88plf9n1bhuk3irb02o9g


Vinogradov's Theorem/Minor Arcs/Lemma 1

Vinogradov's Theorem/Minor Arcs/Lemma 1 0 10754 456325 391779 2020-03-19T11:11:42Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> For $\beta \in \R$, define: :$\norm \beta := \min \set {\size {n - \beta}: n \in \Z}$ Then: :$\displaystyle \forall \alpha \in \R: \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} } \le \min \set {N_2 - N_1, \frac 1 {2 \norm \alpha} }$ </onlyinclude> == Proof == The bound $N_2 - N_1$ is trivial: Since $\size {\map e {\alpha k} } = 1$ for all $k$, by the Triangle Inequality: :$\displaystyle \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} } \le \sum_{k \mathop = N_1}^{N_2} 1 = N_2 - N_1$ To show the second bound we evaluate the sum as a geometric series. We have: :$\map e {\alpha k} = \map e \alpha^k$ So by Sum of Geometric Sequence: {{begin-eqn}} {{eqn | l = \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} }       | r = \size {\frac {\map e {\alpha \paren {N_2 + 1} } - \map e {\alpha N_1} } {\map e \alpha - 1} } }} {{eqn | r = \frac 2 {\size {\map e \alpha - 1} }       | o = \le }} {{end-eqn}} By the polar form of a complex number: :$\map e \alpha - 1 = \map e {\alpha / 2} \paren {\map e {\alpha / 2} - \map e {-\alpha / 2} }$ and: :$\map e {\alpha / 2} - \map e {-\alpha / 2} = \map \exp {\pi i \alpha} - \map \exp {-\pi i \alpha} = 2 i \map \sin {\pi \alpha}$ Therefore: :$\displaystyle \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} } \le \frac 1 {\size {\map \sin {\pi \alpha} } }$ We know that the sine function is concave on $\closedint 0 {\pi / 2}$. So: :$\map \sin {\pi \alpha} \ge 2 \alpha$ for $\alpha \in \closedint 0 {1/2}$. By definition there exists $n \in \Z$ such that: :$\alpha = n + \norm \alpha$ and: :$\norm \alpha \in \closedint 0 {\pi/2}$. So: {{begin-eqn}} {{eqn | l = \size {\map \sin {\pi \alpha} }       | r = \size {\sin {\pi \norm \alpha} }       | c = Sine and Cosine are Periodic on Reals }} {{eqn | r = \map \sin {\pi \norm \alpha}       | c = as $\sin x > 0$ for $x \in \closedint 0 {\pi/2}$ }} {{eqn | o = \ge       | r = 2 \norm \alpha       | c =  }} {{end-eqn}} This completes the proof. {{qed}} Category:Vinogradov's Theorem 3zgphhzfwhljp1fbsyanz1v0j0gr0um


User:Yohji kinomoto

User:Yohji kinomoto 2 10774 56649 2011-06-08T20:58:56Z Yohji kinomoto 404 Created page with "'''<big><big>日本への栄光</big>></big>'''" wikitext text/x-wiki '''<big><big>日本への栄光</big>></big>''' jekumnbz28ww4r0l5dep1423smbmjhk


Vinogradov's Theorem/Major Arcs/Lemma 1

Vinogradov's Theorem/Major Arcs/Lemma 1 0 10842 458354 296435 2020-03-29T20:58:04Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> Let $\phi$ be the Euler $\phi$ function. Let $\mu$ be the Möbius function. Let $c_q$ be the Ramanujan sum modulo $q$. Let $P, N \ge 1$. {{Explain|what are $P$ and $N$: real, integer, ...?}} Let: :$\displaystyle \map {\SS_P} N := \sum_{q \mathop \le P} \frac {\map \mu q \map {c_q} N} {\map \phi q^3}$ :$\displaystyle \map \SS N := \lim_{P \mathop \to \infty} \map {\SS_P} N$ Then: :$\map \SS N = \map {\SS_P} N + \map \OO {P^{\epsilon -1} }$ and $\SS$ has the Euler product: :$\displaystyle \map \SS N = \prod_{p \mathop \nmid N} \paren {1 + \frac 1 {\paren {p - 1}^3} } \prod_{p \mathop \divides N} \paren {1 - \frac 1 {\paren {p - 1}^2} }$ where: :$p \nmid N$ denotes that $p$ is not a divisor of $N$ :$p \divides N$ denotes that $p$ is a divisor of $N$. </onlyinclude> == Proof == We have: {{begin-eqn}} {{eqn | l = \size {\map {c_q} N}       | o = \le        | r = \sum_{\substack {1 \mathop \le a \mathop \le q \\ \map \gcd {a, q} \mathop = 1} } 1       | c = Triangle Inequality }} {{eqn | r = \map \phi q       | c = Definition of Euler $\phi$ Function }} {{end-eqn}} Trivially we have $\size {\map \mu q} \le 1$. Therefore: {{begin-eqn}} {{eqn | l = \size {\map \SS N - \map {\SS_P} N}       | r = \sum_{q \mathop > P} \frac {\size {\map \mu q \map {c_q} N} } {\map \phi q^3}       | c =  }} {{eqn | o = \le       | r = \sum_{q \mathop > P} \frac 1 {\map \phi q^2}       | c =  }} {{eqn | o = \le       | r = \sum_{q \mathop > P} q^{\epsilon - 2}       | c = for large $P$ and $\epsilon \in \R_{>0}$, by Asymptotic Growth of Euler Phi Function }} {{eqn | o = \le       | r = P^{\delta - 1} \sum_{q \mathop > P} q^{\epsilon - \delta - 1}       | c = for any $\delta > \epsilon$ }} {{end-eqn}} Therefore by Convergence of Powers of Reciprocals: :$\map \SS N = \map {\SS_P} N + \map \OO {P^{\epsilon - 1} }$ as claimed. By Euler Product, and because $\map \mu {p^k} = 0$ for $k > 1$: :$\displaystyle \map \SS N = \prod_p \paren {1 + \frac {\map \mu q \map {c_q} N} {\map \phi q^3} }$ Now for a prime $p$ we have: :$\map \mu p = -1$ :$\map \phi p = p - 1$ We also have Kluyver's Formula for Ramanujan's Sum: :$\displaystyle \map {c_p} n = \sum_{d \mathop \divides \map \gcd {p, n} } \rd \map \mu {\frac p d}$ Let $p \divides N$. Then: $\map \gcd {p, N} = p$ which gives: :$\map {c_p} N = p \map \mu 1 + \map \mu p = p - 1$ Let $p \nmid N$. Then: :$\map \gcd {p, N} = 1$ and so: :$\map {c_p} N = -1$ Therefore: :$\displaystyle \map \SS N = \prod_{p \mathop \divides N} \paren {1 - \frac 1 {\paren {p - 1}^2} } \prod_{p \mathop \nmid N} \paren {1 + \frac 1 {\paren {p - 1}^3} }$ This completes the proof. {{qed}} Category:Vinogradov's Theorem 01jbgz0ddezyfq6pa4c347tpvo0r7x9


Axiom:Axiom of Limitation of Size

Axiom:Axiom of Limitation of Size 100 10934 444291 443991 2020-01-20T15:30:46Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> For any class $\CC$, a set $x$ such that $x = \CC$ exists {{iff}} there is no bijection between $\CC$ and the universe. </onlyinclude> {{explain|Check whether "universal class" is meant here}} Limitation of Size 7zgunwpct6381rvcoas04h8zhnakky0


Axiom:Class Comprehension Schema

Axiom:Class Comprehension Schema 100 10935 443994 443988 2020-01-18T01:05:48Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> For any formula $\phi$ containing no quantifiers over classes, there is a class $A$ such that: :$\forall x: \paren {x \in A \iff \map \phi x}$ </onlyinclude> Class Comprehension Schema bkn58cjsj43c7fdu63cibngfneffbfv


Bijection between Specific Elements

Bijection between Specific Elements 0 11017 491812 57625 2020-09-30T21:52:08Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $A$ and $B$ be sets. Let $f: A \leftrightarrow B$ be a bijection. Then given $a \in A$ and $b \in B$, there exists another bijection $g: A \leftrightarrow B$ such that $\map g a = b$. == Proof == Let $\map f a = b'$. Since $f$ is surjective, there is $a' \in A$ such that $\map f {a'} = b$. We define the mapping $g: A \to B$ as: :$\map g x = \begin {cases} b & : x = a \\ b' & : x = a' \\ \map f x & : x \in A \setminus \set {a, a'} \end {cases}$ === $g$ is Surjective === Let $y \in B \setminus \set {b, b'}$. Then: :$\exists x \in A: \map f x = y$ and so by definition of $g$: :$\exists x \in A: \map g x = y$ Otherwise: :$y = b \implies b = \map g a$ :$y = b' \implies b = \map g {a'}$ and so $y$ is seen to be surjective. === $g$ is Injective === Suppose $\map g x = \map g {x'}$. If $\map g x, \map g {x'} \in B \setminus \set {b, b'}$ then $\map f x = \map f {x'}$. As $f$ is bijective and so injective it follows that $x = x'$. If $\map g x, \map g {x'} \in \set {b, b'}$ then either: :$\map g x = \map g {x'} = b$ and so $x = x' = a$ or :$\map g x = \map g {x'} = b'$ and so $x = x' = a'$ For the other two cases: :$\map g x \in \set {b, b'}, \map g {x'} \notin \set {b, b'}$ :$\map g x \notin \set {b, b'}, \map g {x'} \in \set {b, b'}$ it is clear that $\map g x \ne \map g {x'}$. So $g$ has been shown to be injective. === $g$ is Bijective === So $g$ is both injective and surjective, and so by definition a bijection. {{qed}} == Comment == What this means is that given any two equivalent sets, it is always possible to find a bijection that connects any given pair of elements, one in one set and one in the other. Category:Bijections 4yekzc58r9rc2mkv5ztzgdnsq73zy6d


Tietze Extension Theorem/Lemma

Tietze Extension Theorem/Lemma 0 11212 338392 335952 2018-01-18T20:28:48Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $T = \left({S, \tau}\right)$ be a topological space which is normal. Let $A \subseteq S$ be closed in $T$. <onlyinclude> Let $f: A \to \R$ be a continuous mapping such that $\left|{f \left({x}\right)}\right| \le 1$. Then there exists a continuous mapping $g: S \to \R$ such that: : $\forall x \in S: \left|{g \left({x}\right)}\right| \le \dfrac 1 3$ : $\forall x \in A: \left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$ </onlyinclude> == Proof == The sets $f^{−1} \left({\left({−\infty \,.\,.\, −\dfrac 1 3}\right]}\right)$ and $f^{−1} \left({\left[{\dfrac 1 3 \,.\,.\, \infty}\right)}\right)$ are disjoint and closed in $A$. Since $A$ is closed, they are also closed in $S$. We have that $S$ is normal. Then by: : Urysohn's Lemma and: : the fact that $\left[{0 \,.\,.\, 1}\right]$ is homeomorphic to $\left[{−\dfrac 1 3 \,.\,.\, \dfrac 1 3}\right]$ there exists a continuous mapping $g: S \to \left[{−\dfrac 1 3 \,.\,.\, \dfrac 1 3}\right]$ such that: :$g \left({f^{-1} \left({ \left({-\infty \,.\,.\, -\dfrac 1 3}\right]}\right)}\right) = -\dfrac 1 3$ and :$g \left({f^{-1} \left({ \left[{\dfrac 1 3 \,.\,.\, \infty}\right)}\right)}\right) = \dfrac 1 3$ Thus: : $\forall x \in S: \left|{g \left({x}\right)}\right| \le \dfrac 1 3$ Now if $−1 \le f \left({x}\right) \le −\dfrac 1 3$, then: :$g \left({x}\right) = −\dfrac 1 3$ and thus: :$\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$ Similarly if $\dfrac 1 3 \le f \left({x}\right) \le 1$, then: :$g \left({x}\right) = \dfrac 1 3$ and thus: :$\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$ Finally, for $\left|{f \left({x}\right)}\right| \le \dfrac 1 3$ we have that: :$\left|{g \left({x}\right)}\right| \le \dfrac 1 3$ and so: : $\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$ Hence, for all $x \in S$: : $\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$ {{qed}} {{MissingLinks|Some of the statements above need to be justified.}} {{Namedfor|Heinrich Franz Friedrich Tietze}} == Sources == {{Planetmath|id = 5566|title = Tietze extension theorem}} Category:Normal Spaces Category:Continuous Mappings 3fm1tg8pgstmtgqce1rqqi08u7oq6bi


Finitely Many Reduced Associated Quadratic Irrationals 0 11491 273258 178390 2016-10-16T10:23:46Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $n \in \Z: n > 0$. There are finitely many reduced quadratic irrationals which are associated to $n$. == Proof == <!-- This was taken from Continued Fraction Expansion of Irrational Square Root and may require some tinkering. --> By definition, we can write an arbitrary reduced irrational as $\alpha = \dfrac{\sqrt n + P} Q$. By definition, $\alpha > 1$ and its conjugate $\tilde{\alpha} > -1$. So we know $\alpha + \tilde{\alpha} = \dfrac {2P} Q > 0$. Hence with the assumption $Q > 0$ we have $P > 0$. Since $\tilde{\alpha} < 0$ we also have $P < \sqrt n$. Also, since $\alpha > 1$ by assumption, we have $Q < P + \sqrt n < 2 \sqrt n$. Thus there are finitely many choices for both $P$ and $Q$, forcing finitely many reduced quadratic irrationals associated to a fixed $n$. This amount is strictly bounded above by $2n$. {{explain|Why the word "strictly" is needed here - a quantity is bounded above or it's not. Also, clarify "amount".}} {{qed}} Category:Quadratic Irrationals cpzhwfzhio3upcvlwexgn8wi5mko3hk


Expansion of Associated Reduced Quadratic Irrational 0 11492 273250 59859 2016-10-16T09:59:47Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $\alpha$ be a reduced quadratic irrational which is associated to $n$. Let $\alpha$ be transformed into its integer part and fractional part via: :$\alpha = \left \lfloor {\alpha} \right \rfloor + \dfrac 1{\alpha'}$ Then the resulting quadratic irrational $\alpha'$ is also reduced and associated to $n$. == Proof == Let $\alpha = \dfrac{\sqrt n + P} Q$ and $X = \left \lfloor {\alpha} \right \rfloor$ Then we have: :$\dfrac 1 {\alpha'} = \dfrac {\sqrt n - \left({Q X - P}\right)} Q$ Because $\sqrt n$ is irrational, we must have: :$\dfrac 1 {\alpha'} > 0$ Because $\dfrac 1 {\alpha'}$ is the fractional part: :$0 < \dfrac 1 {\alpha'} < 1 \implies \alpha' > 1$ Transforming: {{begin-eqn}} {{eqn | l = \alpha'       | r = \frac Q {\sqrt n - \left({Q X - P}\right)} \cdot \frac {\sqrt n + \left({Q X - P}\right)} {\sqrt n + \left({Q X - P}\right)}       | c =  }} {{eqn | r = \frac {\sqrt n + \left({Q X - P}\right)} {\frac 1 Q \left({n - \left({Q X - P}\right)^2}\right)}       | c =  }} {{end-eqn}} we have: :$P' = Q X - P$ and: :$Q' = \dfrac 1 Q \left({n - \left({Q X - P}\right)^2}\right)$ To show $Q' \in \Z$: We have that: :$n - \left({Q X - P}\right)^2 \equiv n - P^2 \pmod Q$ Because $\alpha$ is associated to $n$, $Q$ must divide this quantity. Hence $Q'$ is an integer as defined. Because $X = \left \lfloor{\dfrac{\sqrt n + P} Q}\right \rfloor$ is an integer and $\alpha$ is irrational: :$X < \dfrac {\sqrt n + P} Q$ Hence: :$P' = QX - P < \sqrt n $ forcing: :$\tilde{\alpha}' < 0$ Because $\alpha > 1$: :$X \ge 1 \iff 0 \le X - 1$ Thus: {{begin-eqn}} {{eqn | l = \tilde \alpha = \frac {P - \sqrt n} Q       | o = <       | r = 0 \le X - 1       | c =  }} {{eqn | ll= \leadsto       | l = Q       | o = <       | r = \sqrt n + \left({Q X - P}\right)       | c =  }} {{eqn | ll= \leadsto       | l = Q \left({\sqrt n - \left({Q X - P}\right)}\right)       | o = <       | r = n - \left({Q X - P}\right)^2       | c =  }} {{eqn | ll= \leadsto       | l = -\tilde \alpha' = \frac {\sqrt n - \left({Q X - P}\right)} {\frac 1 Q \left({n - \left({Q X - P}\right)^2}\right)}       | o = <       | r = 1       | c =  }} {{end-eqn}} Hence $\tilde \alpha' > -1$ and $\alpha'$ is reduced. Because $Q' = \dfrac 1 Q \left({n - \left({P'}\right)^2}\right)$: :$n - \left({P'}\right)^2 \equiv Q Q' \equiv 0 \pmod {Q'}$ Hence $\alpha'$ is associated to $n$. Thus $\alpha'$ is both reduced and associated to $n$. {{qed}} Category:Continued Fractions Category:Quadratic Irrationals h0c2qxftxwlwpmwvydczgxr5s5adwr5


Axiom:Area Axioms

Axiom:Area Axioms 100 11793 196396 195134 2014-10-12T07:51:29Z Prime.mover 59 wikitext text/x-wiki == Axioms of Area == <onlyinclude> When discussing areas of surfaces, we usually take the following statements to be axiomatic: :$(1) \quad$ The area of a square with a side of length one unit is defined to be one square unit. :$(2) \quad$ Let a surface be divided into a finite number of smaller non-overlapping surfaces. ::Let the smaller surfaces cover the entire larger surface. ::Then the sum of the areas of the smaller surfaces equals the area of the larger surface. :$(3) \quad$ Equal surfaces have equal areas. </onlyinclude> Category:Axioms/Geometry thsl5ryu5w2se2a3funl2fpd44tmf2y


Gelfond-Schneider Theorem/Lemma 1

Gelfond-Schneider Theorem/Lemma 1 0 11991 326204 175307 2017-11-11T10:19:20Z Barto 3079 "work to do" template wikitext text/x-wiki {{wtd|Links to Polynomial related results need to be resolved.}} == Lemma == <onlyinclude> Let $a_1 \left({t}\right), \ldots, a_n \left({t}\right)$ be non-zero polynomials in $\R \left[{t}\right]$ of degrees $d_1, \ldots, d_n$ respectively. Let $w_1, \ldots, w_n$ be pairwise distinct real numbers. Then: :$\displaystyle F \left({t}\right) = \sum_{j \mathop = 1}^n a_j \left({t}\right) e^{w_j t}$ has at most $d_1 + \cdots + d_n + n − 1$ real roots (counting multiplicities). </onlyinclude> == Proof == By multiplying through by $e^{−w_n t}$ if necessary, we may suppose that $w_n = 0$ and that otherwise $w_j \ne 0$. Let $k = d_1 + \cdots + d_n + n$. We use strong induction on $k$. If $k = 1$, then $n = 1$ and $d_1 = 0$, and the lemma easily follows. Let $l \ge 2$ be such that the lemma holds whenever $k < l$. Suppose $k = l$. Let $N$ be the number of real roots of $F \left({t}\right)$. By Rolle's Theorem, the number of real roots of $F' \left({t}\right)$ is at least $N − 1$. On the other hand: :$\displaystyle F' \left({t}\right) = \sum_{j=1}^n b_j \left({t}\right) e^{w_j t}$ where: :$b_j \left({t}\right) = a'_j \left({t}\right) + w_j a_j \left({t}\right)$ Note that for $1 \le j \le n − 1$, we have that $b_j \left({t}\right)$ has degree exactly $d_j$. Also, since $w_n = 0$, either: : there are only $n−1$ non-zero polynomials $b_j \left({t}\right)$ in the expression for $F' \left({t}\right)$ above or: : there are $n$ such polynomials and the degree of $b_n \left({t}\right)$ is one less than the degree of $a_n \left({t}\right)$. We get from the induction hypothesis that $F' \left({t}\right)$ has at most $d_1 + \cdots + d_n + n − 2$ real roots. Hence, $N − 1 \le d_1 + \cdots + d_n + n − 2$, and the result follows. {{qed|lemma}} Category:Transcendental Number Theory jxhani6jbvr3c03ovy9x4lyig5il3x4


Gelfond-Schneider Theorem/Lemma 2

Gelfond-Schneider Theorem/Lemma 2 0 11999 371662 371651 2018-10-17T20:15:31Z Prime.mover 59 wikitext text/x-wiki {{MissingLinks}} == Lemma == <onlyinclude> Let $\map f z$ be an analytic function in the disk $D \subseteq \C: D = \set {z : \size z < R}$ for some real number $R > 0$. Let $f$ also be continuous on the closure of $D$, that is, on $D^- = \set {z : \size z \le R}$. Then: :$\forall z \in D^-: \size {\map f z} \le \size f_R$, where we set $\size f_R = \max_{\size z = R} \size {\map f z}$. </onlyinclude> This lemma is essentially a restatement of the maximum modulus principle for analytic functions. == Proof == Since $D^-$ is compact, the continuous function $\size f$ has a maximum point $z_0 \in D^-$, meaning that $\size {\map f {z_0} } = \max_{D^-} \size f$. If $z_0$ belongs to the open disk $D$, then the maximum modulus principle implies that $f$ is constant, in which case the statement is trivial. If instead $z_0$ belongs to the boundary $D^- \setminus D$, then $\size {z_0} = R$ and so $\size {\map f {z_0} } \le \size f_R$. Since $\size {\map f z} \le \size {\map f {z_0} }$ for any $z \in D^-$, the statement follows. {{qed|lemma}} Category:Transcendental Number Theory 6201s4jahub0y0hw17zwn71uddz54op


Gelfond-Schneider Theorem/Lemma 3

Gelfond-Schneider Theorem/Lemma 3 0 12001 371650 350834 2018-10-17T15:24:47Z Kappaenne 3501 wikitext text/x-wiki == Lemma == <onlyinclude> Let $r$ and $R$ be two real numbers such that $0 < r \le R$. Let $f_1 \left({z}\right), f_2 \left({z}\right), \ldots, f_L \left({z}\right)$ be: : analytic in $D \subseteq \C: D = \left\{{z : \left|{z}\right| < R}\right\}$ : continuous on the closure $D$, that is, $D^- = \left\{{z : \left \vert{z}\right \vert \le R}\right\}$. Let $\zeta_1, \ldots, \zeta_L$ be complex numbers such that: :$\forall j \in \left\{{1, 2, \ldots, L}\right\}: \left|{\zeta_j}\right| \le r$ Then the determinant: :$\Delta = \det \begin{bmatrix} f_1 \left({\zeta_1}\right) & \cdots & f_L \left({\zeta_1}\right) \\ \vdots & \ddots & \vdots \\ f_1 \left({\zeta_L}\right) & \cdots & f_L \left({\zeta_L}\right) \end{bmatrix}$ satisﬁes: :$\displaystyle \left|{\Delta}\right| \le \left({\frac R r}\right)^{−L \left({L−1}\right) / 2} L! \prod_{\lambda \mathop = 1}^L \left|{f_λ}\right|_R$ </onlyinclude> == Proof == Let $h \left({z}\right)$ be the determinant of the $L \times L$ matrix $\left[{f_j \left({\zeta_i z}\right)}\right]$. Then $h \left({z}\right)$ is: : analytic in $D' = \left\{{z : \left \vert {z}\right \vert < R/r}\right\}$ : continuous on $D'^- = \left\{{z : \left \vert{z}\right \vert \le R/r}\right\}$. Let $M = L \left({L − 1}\right) / 2$, and write: :$\displaystyle f_j \left({\zeta_i z}\right) = \sum_{k \mathop = 0}^{M−1} b_k \left({j}\right) \zeta_i^k z^k + z^M g_{i,j} \left({z}\right)$ where: :$b_k \left({j}\right) \in \C$ for each $k$ :$g_{i,j} \left({z}\right)$ is analytic in $D'$ :$g_{i,j} \left({z}\right)$ is  continuous on $D'^-$ By evaluating along the columns it is seen that the determinant is linear in its columns. So we can view $h \left({z}\right)$ as $z^M$ times an analytic function plus terms involving the factor: :$z^{n_1 + \cdots + n_L} \det \left[{\zeta_i^{n_j} }\right]$ where the $n_j$ denote non-negative integers. Observe that the determinant in this last expression is zero if the $n_j$ are not distinct. Therefore, the non-zero terms of this form satisfy: :$n_1 + n_2 + \cdots + n_L \ge 0 + 1 + \cdots + \left({L − 1}\right) = \dfrac {L \left({L − 1}\right)} 2$ Hence, we deduce that $h \left({z}\right)$ is divisible by $z^M$. More precisely, $\dfrac {h \left({z}\right)} {z^M}$ is analytic in $D'$ and continuous on $D'^-$. Therefore, by Gelfond-Schneider Theorem: Lemma 2, for any $w \in D'$: :$\displaystyle \left|{\frac {h \left({w}\right)} {w^M}}\right| \le \left|{\frac {h \left({z}\right)} {z^M}}\right|_{R/r} = \left({\frac r R}\right)^M \left|{h \left({z}\right)}\right|_{R/r}$ For $\left|{z}\right| = R/r$, we get that $\left|{\zeta_i z}\right| \le R$. We create a bound for $\left|{h \left({z}\right)}\right|_{R/r}$ by multiplying the number of terms in $\det \left[{f_j \left({\zeta_i z}\right)}\right]$ by an obvious upper bound on the maximum such term. Thus: :$\displaystyle \left|{h \left({z}\right)}\right|_{R/r} \le L! \prod_{\lambda \mathop = 1}^L \left|{f_λ}\right|_R$ Observe that $\left|{\Delta}\right| = \left|{h \left({1}\right)}\right|$ and $1 \le R/r$. We deduce that :$\displaystyle \left|{\Delta}\right| \le \left({\frac r R}\right)^M \left|{h \left({z}\right)}\right|_{R/r} \le \left({\frac r R}\right)^M L! \prod_{\lambda \mathop = 1}^L \left|{f_λ}\right|_R$ giving the desired conclusion. {{qed|lemma}} Category:Transcendental Number Theory 173dtdvf0myyodnpyklcyk1gss1eemh