Polynomial Forms over Field form Principal Ideal Domain/Corollary 1 0 18767 489213 488791 2020-09-19T13:30:54Z Prime.mover 59 wikitext text/x-wiki == Corollary to Polynomial Forms over Field form Principal Ideal Domain == Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$. Let $X$ be transcendental over $F$. Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$. <onlyinclude> Let $f$ be an irreducible element of $F \sqbrk X$. Then $F \sqbrk X / \ideal f$ is a field, where $\ideal f$ denotes the ideal generated by $f$. </onlyinclude> == Proof == It follows from Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal that $\ideal f$ is maximal for irreducible $f$. Therefore by Maximal Ideal iff Quotient Ring is Field, $F \sqbrk X / \ideal f$ is a field. {{qed}} == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Polynomial Forms over Field form Unique Factorization Domain|next = Field of Quotients of Ring of Polynomial Forms on Reals that yields Complex Numbers}}: $\S 65.3$ Some properties of $F \sqbrk X$, where $F$ is a field Category:Polynomial Forms over Field form Principal Ideal Domain klnder8eo8jiru8ocohfjwk67o1dh9c


No Arithmetic Sequence of 4 Primes with Common Difference 2 0 18783 456429 456424 2020-03-19T17:27:22Z Prime.mover 59 wikitext text/x-wiki == Definition == There exist no $n \in \Z_{>0}$ such that $n, n + 2, n + 4, n + 6$ are all prime. === Corollary === {{:No Arithmetic Sequence of 4 Primes with Common Difference 2/Corollary}} == Proof == {{AimForCont}} $S$ is a set of $4$ prime numbers of the form $n, n + 2, n + 4, n + 6$. $S$ must contain as a subset a prime triplet. From Prime Triplet is Unique, the only one of these is $\set {3, 5, 7}$. The only sets of the form $\set {n, n + 2, n + 4, n + 6}$ containing $\set {3, 5, 7}$ are: :$(1): \quad \set {1, 3, 5, 7}$: as $1$ is by convention not a prime, then this is not $S$. :$(2): \quad \set {3, 5, 7, 9}$: as $9 = 3 \times 3$ is not a prime, then this is not $S$. There are no more possible $\set {n, n + 2, n + 4, n + 6}$ all prime. Hence, by Proof by Contradiction, $S$ does not exist. {{qed}} Category:Prime Numbers Category:Arithmetic Sequences meyav5fl4qmca6v01w3blr8myf3737d


Cardinal Number Less than Ordinal/Corollary 0 18870 113041 105265 2012-10-28T08:27:31Z Prime.mover 59 wikitext text/x-wiki == Corollary to Cardinal Number Less than Ordinal == <onlyinclude> Let $x$ be an ordinal. Let $\left|{x}\right|$ denote the cardinal number of $x$. Then: :$\left|{x}\right| \le x$ </onlyinclude> == Proof == By Set Equivalence is Equivalence Relation: : $x \sim x$ By Cardinal Number Less than Ordinal: : $\left|{x}\right| \le x$ {{qed}} == Sources == * {{BookReference|Introduction to Axiomatic Set Theory|1971|Gaisi Takeuti|author2=Wilson M. Zaring}}: $\S 10.13$ Category:Cardinals Category:Ordinals lluo5cfjxrroelppyxemowssnp4ensp


Set of Words Generates Group/Corollary

Set of Words Generates Group/Corollary 0 19302 440598 107132 2019-12-23T11:30:02Z Prime.mover 59 wikitext text/x-wiki == Corollary to Set of Words Generates Group == Let $G$ be a group. Let $T \subseteq G$. Let $\map W T$ be the set of words of $T$. <onlyinclude> If $T$ is closed under taking inverses, then $\map W T$ is a subgroup of $G$. </onlyinclude> == Proof == This follows directly from Set of Words Generates Group and the fact that $T$ has the same properties as $\hat S$ in that result. {{qed}} Category:Group Theory ojhkc24jnwmrt2q5h30afhh2bmg1sw2


Subset Product within Semigroup is Associative/Corollary 0 19550 378263 371306 2018-11-24T16:28:47Z Prime.mover 59 wikitext text/x-wiki == Corollary to Subset Product within Semigroup is Associative == <onlyinclude> Let $\struct {S, \circ}$ be a semigroup. Then: {{begin-eqn}} {{eqn | l = x \paren {y S}       | r = \paren {x y} S }} {{eqn | l = x \paren {S y}       | r = \paren {x S} y }} {{eqn | l = \paren {S x} y       | r = S \paren {x y} }} {{end-eqn}} </onlyinclude> == Proof == From the definition of Subset Product with Singleton: {{begin-eqn}} {{eqn | l = x \paren {y S}       | r = \set x \paren {\set y S} }} {{eqn | l = x \paren {S y}       | r = \set x \paren {S \set y} }} {{eqn | l = \paren {S x} y       | r = \paren {S \set x} \set y }} {{end-eqn}} The result then follows directly from Subset Product within Semigroup is Associative. {{qed}} == Also see == * Subset Product within Commutative Structure is Commutative == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Definition:Subset Product with Singleton|next = Definition:Right Coset}}: $\S 6.1$. The quotient sets of a subgroup Category:Subset Products 428at3duvbkpwh0be1wa186nd0yyl9q


Partition Topology is T4

Partition Topology is T4 0 19748 470070 469187 2020-05-23T21:52:16Z Prime.mover 59 wikitext text/x-wiki == Corollary to Partition Topology is T5 == Let $S$ be a set and let $\PP$ be a partition on $S$ which is not the (trivial) partition of singletons. Let $T = \struct {S, \tau}$ be the partition space whose basis is $\PP$. Then: <onlyinclude> :$T$ is a $T_4$ space. </onlyinclude> == Proof == We have that the Partition Topology is $T_5$. We also have that a $T_5$ Space is $T_4$ Space. The result follows. {{qed}} == Sources == * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Partition Topology is T3 1/2|next = Partition Topology is T5}}: Part $\text {II}$: Counterexamples: $5$. Partition Topology: $2$ Category:T4 Spaces Category:Partition Topology krnrmpuxydyeriig2u8po6xswjdg64a


Pointwise Addition on Real-Valued Functions is Commutative 0 19758 473610 472906 2020-06-12T08:19:56Z Prime.mover 59 wikitext text/x-wiki == Definition == Let $S$ be a set. <onlyinclude> Let $f, g: S \to \R$ be real-valued functions. Let $f + g: S \to \R$ denote the pointwise sum of $f$ and $g$. Then: :$f + g = g + f$ </onlyinclude> That is, pointwise addition of real-valued functions is commutative. == Proof == {{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {f + g} } x       | r = \map f x + \map g x       | c = {{Defof|Pointwise Addition of Real-Valued Functions}} }} {{eqn | r = \map g x + \map f x       | c = Real Addition is Commutative }} {{eqn | r = \map {\paren {g + f} } x       | c = {{Defof|Pointwise Addition of Real-Valued Functions}} }} {{end-eqn}} {{qed}} Category:Pointwise Addition Category:Real Addition Category:Commutativity g2yuv1giptnoh4uoh5b2ierg1xnwcbi


Compact Subspace of Hausdorff Space is Closed/Corollary 0 19761 310455 179959 2017-08-12T06:50:28Z Prime.mover 59 wikitext text/x-wiki == Corollary to Compact Subspace of Hausdorff Space is Closed == <onlyinclude> A finite subspace of a Hausdorff space is closed. </onlyinclude> == Proof == Follows directly from: : Finite Topological Space is Compact and: : Compact Subspace of Hausdorff Space is Closed. {{qed}} Category:Finite Topological Spaces Category:Compact Subspace of Hausdorff Space is Closed qlgcadopcukk0um0hgwvfipvbwlswp4


Brahmagupta-Fibonacci Identity/Extension 0 19835 396426 396423 2019-03-20T22:11:06Z Prime.mover 59 wikitext text/x-wiki == Extension to Brahmagupta-Fibonacci Identity == <onlyinclude> Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be integers. Then: :$\displaystyle \prod_{j \mathop = 1}^n \paren { {a_j}^2 + {b_j}^2} = c^2 + d^2$ where $c, d \in \Z$. That is: the product of any number of sums of two squares is also a sum of two squares. </onlyinclude> === General Version === {{:Brahmagupta-Fibonacci Identity/Extension/General}} == Proof 1 == {{:Brahmagupta-Fibonacci Identity/Extension/Proof 1}} == Proof 2 == {{:Brahmagupta-Fibonacci Identity/Extension/Proof 2}} == Proof 3 == {{:Brahmagupta-Fibonacci Identity/Extension/Proof 3}} Category:Brahmagupta-Fibonacci Identity jw9pxxhqrn182b29d5c2czrn1rvtyhc


Sum of Squared Deviations from Mean/Corollary 1 0 19995 488868 348299 2020-09-18T12:17:29Z Prime.mover 59 wikitext text/x-wiki == Theorem  == Let $x_1, x_2, \ldots, x_n$ be real data about some quantitative variable. Let $\overline x$ be the arithmetic mean of the above data. Then:  <onlyinclude> :$\displaystyle \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2 = \sum_{i \mathop = 1}^n x_i^2 - n \overline x^2$ </onlyinclude>  == Proof == {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2       | r = \sum_{i \mathop = 1}^n \paren {x_i^2 - \overline x^2}       | c = Sum of Squared Deviations from Mean }} {{eqn | r = \sum_{i \mathop = 1}^n x_i^2 - \sum_{i \mathop = 1}^n \overline x^2       | c = Summation is Linear }} {{eqn | r = \sum_{i \mathop = 1}^n x_i^2 - n \overline x^2       | c = Sum of Identical Terms }} {{end-eqn}} {{qed}} Category:Sum of Squared Deviations from Mean d80mi4gqu49iwugeua4n833grnqgns6


Limit of Sequence to Zero Distance Point/Corollary 1 0 20153 408658 408654 2019-06-16T22:26:39Z Prime.mover 59 wikitext text/x-wiki == Corollary to Limit of Sequence to Zero Distance Point == Let $S$ be a non-empty subset of $\R$. Let the distance $\map d {\xi, S} = 0$ for some $\xi \in \R$. <onlyinclude> If $S$ is bounded above, then there exists a sequence $\sequence {x_n}$ in $S$ such that $\displaystyle \lim_{n \mathop \to \infty} x_n = \sup S$. </onlyinclude> == Proof == Let  $\xi = \sup S$. Then from Distance from Subset of Real Numbers: :$\map d {\xi, S} = 0$ The result then follows directly from Limit of Sequence to Zero Distance Point. Note that the terms of this sequence do not necessarily have to be distinct. {{Qed}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Limit of Sequence to Zero Distance Point|next = Limit of Sequence to Zero Distance Point/Corollary 2}}: $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (6)$ Category:Limit of Sequence to Zero Distance Point rjfax6iujwpafbe3bcdtct9a2b86yvn


Derivative of Function of Constant Multiple/Corollary 0 20203 447516 447515 2020-02-08T19:18:17Z Prime.mover 59 wikitext text/x-wiki == Corollary of Derivative of Function of Constant Multiple == Let $f$ be a real function which is differentiable on $\R$. <onlyinclude> Let $a, b \in \R$ be constants. Then: :$\map {D_x} {\map f {a x + b} } = a \, \map {D_{a x + b} } {\map f {a x + b} }$ </onlyinclude> == Proof == First it is shown that $\map {D_x} {a x + b} = a$: {{begin-eqn}} {{eqn | n = 1       | l = \map {D_x} {a x + b}       | r = \map {D_x} {a x} + \map {D_x} b       | c = Sum Rule for Derivatives }} {{eqn | r = a + 0       | c = Derivative of Function of Constant Multiple and Derivative of Constant }} {{eqn | r = a       | c =  }} {{end-eqn}} Next: {{begin-eqn}} {{eqn | l = \map {D_x} {\map f {a x + b} }       | r = \map {D_x} {a x + b} \, \map {D_{a x + b} } {\map f {a x + b} }       | c = Chain Rule for Derivatives }} {{eqn | r = a \, \map {D_{a x + b} } {\map f {a x + b} }       | c = from $(1)$ }} {{end-eqn}} {{qed}} Category:Differential Calculus aj264wkpsc0lsl0e6edhp2m9ze8ee0l


Open Real Interval is Open Set/Corollary 0 20317 467826 202377 2020-05-14T06:21:02Z Prime.mover 59 wikitext text/x-wiki == Corollary to Open Real Interval is Open Set == <onlyinclude> Let $\R$ be the real number line considered as an Euclidean space. Let $A := \openint a \infty \subset \R$ be an open interval of $\R$. Let $B := \openint {-\infty} b \subset \R$ be an open interval of $\R$. Then both $A$ and $B$ are open sets of $\R$. </onlyinclude> == Proof == From Open Real Interval is Open Set we have that for any $c \in \openint a b$ there exists an open $\epsilon$-ball of $c$ lying wholly within $\openint a b$. When either of $a \to -\infty$ or $b \to \infty$ the result still holds. The result follows by definition of open set. {{qed}} Category:Real Intervals Category:Open Sets 8gsilbpdhc1wyoz828748s8ia5w373g


Negative of Absolute Value/Corollary 1

Negative of Absolute Value/Corollary 1 0 20525 407044 407015 2019-06-08T13:50:48Z Prime.mover 59 wikitext text/x-wiki == Corollary to Negative of Absolute Value == Let $x, y \in \R$ be a real numbers. Let $\size x$ be the absolute value of $x$. Then: <onlyinclude> :$\size x < y \iff -y < x < y$ </onlyinclude> == Proof == === Necessary Condition === Let $\size x < y$. Then from Negative of Absolute Value: :$x \le \size x$ and: :$\size x \ge -x$ So $x < y$ and $-x < y$, and so $x > -y$ from Ordering of Inverses in Ordered Monoid. It follows that $-y < x < y$. {{qed|lemma}} === Sufficient Condition === Let $-y < x < y$. Then $x < y$ and $-x < y$. For all $x$: :$\size x = x$ or: :$\size x = -x$ Thus it follows that $\size x < y$. {{qed}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Reverse Triangle Inequality/Real and Complex Fields/Examples/6 - (-1)|next = Reverse Triangle Inequality/Real and Complex Fields/Proof 2}}: $\S 1$: Real Numbers: Exercise $\S 1.20 \ (1)$ Category:Negative of Absolute Value 0hochhv3flq6ud3fsdquw2ik490b1n6


Continuous Image of Connected Space is Connected/Corollary 1 0 20580 337657 328301 2018-01-16T23:21:45Z Prime.mover 59 wikitext text/x-wiki == Corollary to Continuous Image of Connected Space is Connected == <onlyinclude> Connectedness is a topological property. </onlyinclude> == Proof == Follows directly from Continuous Image of Connected Space is Connected and the definition of topological property. {{qed}} == Sources == * {{BookReference|Introduction to Metric and Topological Spaces|1975|W.A. Sutherland|prev = Continuous Image of Connected Space is Connected/Proof 1|next = Continuous Image of Connected Space is Connected/Corollary 2}}: $6.2$: Connectedness: Corollary $6.2.11$ Category:Connected Spaces Category:Continuous Image of Connected Space is Connected o4lnwd6b381lp6ql3phnjk8i8c7rscv


Continuous Image of Connected Space is Connected/Corollary 2 0 20581 337658 328302 2018-01-16T23:22:29Z Prime.mover 59 wikitext text/x-wiki == Corollary to Continuous Image of Connected Space is Connected == <onlyinclude> Let $T$ be a connected topological space. Let $f: T \to \R$ be a continuous real-valued mapping. Then $f \left({T}\right)$ is a real interval. </onlyinclude> == Proof == From Continuous Image of Connected Space is Connected, the continuous image of the connected space $T$ is connected. The result follows from Subset of Real Numbers is Interval iff Connected. {{qed}} == Sources == * {{BookReference|Introduction to Metric and Topological Spaces|1975|W.A. Sutherland|prev = Continuous Image of Connected Space is Connected/Corollary 1|next = Continuous Image of Connected Space is Connected/Corollary 3}}: $6.2$: Connectedness: Corollary $6.2.13$ Category:Connected Spaces Category:Continuous Mappings Category:Continuous Functions Category:Analysis Category:Continuous Image of Connected Space is Connected lyx1ep9b71ma9m3njstfgccykewvbcz


Image of Countable Set under Mapping is Countable 0 20635 336022 189918 2018-01-12T08:25:39Z Barto 3079 stable link to image of mapping wikitext text/x-wiki == Definition == Let $S$ be a countable set. Let $T$ be a set. Let $f: S \to T$ be a mapping. Then the image of $f$ is countable. == Proof == Let $A$ be the image of $f$. Let $g: S \to A$ be the restriction of $f$ to the cartesian product $S \times A$. By Surjection by Restriction of Codomain, $g$ is a surjection. By Surjection from Natural Numbers iff Countable, there exists a surjection $\phi: \N \to S$. Since the composition of surjections is a surjection, $g \circ \phi: \N \to A$ is a surjection. Hence, by Surjection from Natural Numbers iff Countable, it follows that $A$ is countable. {{qed}} Category:Countable Sets Category:Mapping Theory qqh9cn9h9q8i9ky1p5fbcwqrkel1phv


Infinite Sequence in Countably Compact Space has Accumulation Point 0 20668 471376 352748 2020-05-29T17:31:18Z Prime.mover 59 wikitext text/x-wiki == Corollary to Countably Infinite Set in Countably Compact Space has Omega-Accumulation Point == Let $T = \struct {S, \tau}$ be a countably compact topological space. <onlyinclude> Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $S$. Then $\sequence {x_n}$ has an accumulation point in $T$. </onlyinclude> == Proof == Let $A \subseteq S$ be the range of $\sequence {x_n}$: :$A = \set {x_n: n \in \N}$ If $A$ is finite, then consider the equality: :$\displaystyle \N = \bigcup_{y \mathop \in A} \set {n \in \N: x_n = y}$ Therefore, there exists a $y \in A$ such that $\set {n \in \N: x_n = y}$ is an infinite set. Hence, $y$ is an accumulation point of $\sequence {x_n}$. Otherwise, $A$ is countably infinite. Then $A$ has an $\omega$-accumulation point in $T$. It follows that $\sequence {x_n}$ has an accumulation point in $T$. {{explain|Link to a result that states that if $T$ has an $\omega$-accumulation point, then $\sequence {x_n}$ has an accumulation point}} {{qed}} == Sources == * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Definition:Countable Finite Intersection Axiom|next = Countably Compact Space satisfies Countable Finite Intersection Axiom}}: Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties Category:Countably Compact Spaces Category:Sequences Category:Accumulation Points dzx7hnopemrgv97o7xm8894azbpua0x


L'Hôpital's Rule/Corollary 1

L'Hôpital's Rule/Corollary 1 0 20803 480849 480847 2020-08-04T05:21:44Z Prime.mover 59 wikitext text/x-wiki == Corollary to L'Hôpital's Rule == Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$. Suppose that $\forall x \in \openint a b: \map {g^{\prime} } x \ne 0$. <onlyinclude> Suppose that $\exists c \in \openint a b: \map f c = \map g c = 0$. Then: :$\displaystyle \lim_{x \mathop \to c} \frac {\map f x} {\map g x} = \lim_{x \mathop \to c} \frac {\map {f^{\prime} } x} {\map {g^{\prime} } x}$ provided that the second limit exists. </onlyinclude> == Proof == This follows directly from the definition of limit. If $\displaystyle \lim_{x \mathop \to c} \frac {\map {f^{\prime} } x} {\map {g^{\prime} } x}$ exists, it follows that: :$\displaystyle \lim_{x \mathop \to c} \frac {\map {f^{\prime} } x} {\map {g^{\prime} } x} = \lim_{x \to c^+} \frac {\map {f^{\prime} } x} {\map {g^{\prime} } x}$ That is, if there exists such a limit, it is also a limit from the right. {{qed}} {{Namedfor|Guillaume de l'Hôpital|cat = L'Hôpital}} == Historical Note == {{:L'Hôpital's Rule/Historical Note}} Category:Differential Calculus Category:Limits of Functions Category:L'Hôpital's Rule g3ve7t2qrgnf63pv1732omz1jws77lb


L'Hôpital's Rule/Corollary 2

L'Hôpital's Rule/Corollary 2 0 20804 480850 408110 2020-08-04T05:32:14Z Prime.mover 59 wikitext text/x-wiki == Corollary to L'Hôpital's Rule == Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$. Suppose that $\forall x \in \openint a b: \map {g'} x \ne 0$. <onlyinclude> Suppose that $\map f x \to \infty$ and $\map g x \to \infty$ as $x \to a^+$. Then: :$\displaystyle \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$ provided that the second limit exists. </onlyinclude> == Proof == Let $\displaystyle \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x} = L$. Let $\sequence {x_n}$ be a sequence such that: :$\quad x_n \in \openint a b$ for all $n \in \N$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = a$ From Intermediate Value Theorem for Derivatives and the definition of limit of real function, it follows that: :$\displaystyle \lim_{n \mathop \to \infty} \map g {x_n} = \infty$ and $\sequence {\map g {x_n} }$ is strictly increasing. Consider the range $\closedint {x_{n - 1} } {x_n} \subset \openint a b$ where $n \ge 2$. By Cauchy Mean Value Theorem, there exists $c_n \in \openint {x_{n - 1} } {x_n}$ such that: :$\displaystyle \frac {\map f {x_n} - \map f {x_{n - 1} } } {\map g {x_n} - \map g {x_{n - 1} } } = \frac {\map {f'} {c_n} } {\map {g'} {c_n} }$ From the above and the Squeeze Theorem for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} c_n = a$ and: :$\displaystyle \lim_{n \mathop \to \infty} \frac {\map f {x_n} - \map f {x_{n - 1} } } {\map g {x_n} - \map g {x_{n - 1} } } = \lim_{n \mathop \to \infty} \frac {\map {f'} {c_n} } {\map {g'} {c_n} } = L$ So, by Stolz-Cesàro Theorem: :$\displaystyle \lim_{n \mathop \to \infty} \frac {\map f {x_n} } {\map g {x_n} } = L$ From the definition of limit of real function, we deduce that: :$\displaystyle \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = L = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$ {{qed}} == Remarks == * The proof does not actually use the assumption $\displaystyle \lim_{x \mathop \to a^+} \map f x = \infty$. {{refactor|Add these as separate corollaries}} * Cases $x \to b^-$, $x \to \pm \infty$ and $\map g x \to -\infty$ can be proven similarly. {{Namedfor|Guillaume de l'Hôpital|cat = L'Hôpital}} == Historical Note == {{:L'Hôpital's Rule/Historical Note}} Category:Differential Calculus Category:Limits of Functions Category:L'Hôpital's Rule gbkfejy6y0xc3hs9j3bdnnnzzzp7mxh


Law of Inverses (Modulo Arithmetic)/Corollary 1 0 20969 392611 264171 2019-02-17T23:13:43Z Prime.mover 59 wikitext text/x-wiki == Corollary to Law of Inverses (Modulo Arithmetic) == <onlyinclude> Let $m, n \in \Z$ such that: :$m \perp n$ that is, such that $m$ and $n$ are coprime. Then: :$\exists n' \in \Z: n n' \equiv 1 \pmod m$ </onlyinclude> == Proof == By definition of coprime: :$m \perp n \iff \gcd \set {m, n} = 1$ The result follows directly from Law of Inverses (Modulo Arithmetic). {{qed}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Chinese Remainder Theorem|next = Common Factor Cancelling in Congruence/Corollary 1}}: $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $19$ Category:Modulo Arithmetic jmay40sjb275qvretmgl0auqlkzbo3u


Continuous Image of Compact Space is Compact/Corollary 2 0 21155 316453 179573 2017-09-09T06:59:47Z Barto 3079 wikitext text/x-wiki == Corollary to Continuous Image of Compact Space is Compact == <onlyinclude> A continuous mapping from a compact topological space to a metric space is bounded. </onlyinclude> == Proof == Follows from Continuous Image of Compact Space is Compact and Compact Subspace of Metric Space is Bounded. {{Qed}} == Sources == * {{BookReference|Principles of Mathematical Analysis|1953|Walter Rudin|prev = Continuous Image of Compact Space is Compact|next = Continuous Image of Compact Space is Compact/Corollary 3/Proof 1}}: $4.15$ * {{BookReference|Introduction to Metric and Topological Spaces|1975|W.A. Sutherland|prev=Continuous Image of Compact Space is Compact/Corollary 1|next=Continuous Image of Compact Space is Compact/Corollary 3/Proof 1}}: $5.5$: Continuous maps on compact spaces: Corollary $5.5.3$ Category:Compact Spaces Category:Continuous Invariants fmt1at4n9zvce3nub5xzkjzspun1yyd


Continuous Image of Compact Space is Compact/Corollary 3 0 21156 316616 316495 2017-09-09T16:43:02Z Barto 3079 wikitext text/x-wiki == Corollary to Continuous Image of Compact Space is Compact == <onlyinclude> Let $S$ be a compact topological space. Let $f: S \to \R$ be a continuous real-valued function. Then $f$ attains its bounds on $S$. </onlyinclude> == Proof 1 == {{:Continuous Image of Compact Space is Compact/Corollary 3/Proof 1}} == Proof 2 == {{:Continuous Image of Compact Space is Compact/Corollary 3/Proof 2}} Category:Compact Spaces Category:Continuous Invariants mpyx4tzrb8si8eo1v0otk1yu41sdea9


Area of Triangle in Terms of Two Sides and Angle 0 21377 451001 393586 2020-02-28T10:20:41Z Prime.mover 59 wikitext text/x-wiki == Corollary to Area of Triangle in Terms of Side and Altitude == <onlyinclude> The area of a triangle $ABC$ is given by: :$\dfrac 1 2 a b \sin C$ where: : $a, b$ are two of the sides : $C$ is the angle of the vertex opposite its other side $c$. </onlyinclude> == Proof 1 == {{:Area of Triangle in Terms of Two Sides and Angle/Proof 1}} == Proof 2 == {{:Area of Triangle in Terms of Two Sides and Angle/Proof 2}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Area of Triangle in Terms of Side and Altitude|next = Heron's Formula}}: $\S 4$: Geometric Formulas: $4.5$ * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Area of Triangle in Terms of Side and Altitude|next = Law of Sines|entry = triangle|subentry = i}} Category:Areas of Triangles Category:Area of Triangle in Terms of Two Sides and Angle p0pjtycjchc38qyryusgp6rc6wthb2i


Derivative of Cosine Function/Corollary

Derivative of Cosine Function/Corollary 0 21389 485861 310524 2020-09-06T19:34:05Z Prime.mover 59 wikitext text/x-wiki == Corollary to Derivative of Cosine Function == <onlyinclude> :$\map {\dfrac \d {\d x} } {\cos a x} = -a \sin a x$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\cos x}       | r = -\sin x       | c = Derivative of $\cos x$ }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\cos a x}       | r = -a \sin a x       | c = Derivative of Function of Constant Multiple }} {{end-eqn}} {{qed}} == Also see == * Derivative of $\sin a x$ * Derivative of $\tan a x$ * Derivative of $\cot a x$ * Derivative of $\sec a x$ * Derivative of $\csc a x$ Category:Derivative of Cosine Function ghumvt1h1tgyvdqixu8iuxia17moyc0


Derivative of Sine Function/Corollary

Derivative of Sine Function/Corollary 0 21393 485860 310492 2020-09-06T19:33:06Z Prime.mover 59 wikitext text/x-wiki == Corollary to Derivative of Sine Function == <onlyinclude> :$\map {\dfrac \d {\d x} } {\sin a x} = a \cos a x$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\sin x}       | r = \cos x       | c = Derivative of $\sin x$ }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\sin a x}       | r = a \cos a x       | c = Derivative of Function of Constant Multiple }} {{end-eqn}} {{qed}} == Also see == * Derivative of $\cos a x$ * Derivative of $\tan a x$ * Derivative of $\cot a x$ * Derivative of $\sec a x$ * Derivative of $\csc a x$ Category:Derivative of Sine Function spsr35wfgis0woq4ajfod9z23jfa51r


Cosine of Sum/Corollary

Cosine of Sum/Corollary 0 21397 451129 396172 2020-02-28T16:07:44Z Prime.mover 59 wikitext text/x-wiki == Corollary to Cosine of Sum == <onlyinclude> :$\map \cos {a - b} = \cos a \cos b + \sin a \sin b$ </onlyinclude> where $\sin$ denotes the sine and $\cos$ denotes the cosine. == Proof == {{begin-eqn}} {{eqn | l = \map \cos {a - b}       | r = \cos a \, \map \cos {-b} - \sin a \, \map \sin {-b}       | c = Cosine of Sum }} {{eqn | r = \cos a \cos b - \sin a \, \map \sin {-b}       | c = Cosine Function is Even }} {{eqn | r = \cos a \cos b + \sin a \sin b       | c = Sine Function is Odd }} {{end-eqn}} {{qed}} == Historical Note == {{:Cosine of Sum/Historical Note}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Sine of Difference|next = Sum of Squares of Sine and Cosine/Proof 1}}: $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$ * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Cosine of Sum|next = Tangent of Sum}}: $\S 5$: Trigonometric Functions: $5.35$ * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Cosine of Sum|next = Tangent of Sum}}: $2$: Functions, Limits and Continuity: The Elementary Functions: $4$ * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Cosine of Sum/Mistake|next = Tangent of Sum|entry = compound angle formulae|subentry = in trigonometry}} Category:Cosine Function Category:Cosine of Sum Category:Compound Angle Formulas 3oleic61i3pc7bknsxjwtqj59ky1jkt


Union of Connected Sets with Non-Empty Intersections is Connected/Corollary 0 21403 339341 339340 2018-01-21T23:04:22Z Prime.mover 59 wikitext text/x-wiki == Corollary to Union of Connected Sets with Non-Empty Intersections is Connected == <onlyinclude> Let $T = \left({S, \tau}\right)$ be a topological space. Let $I$ be an indexing set. Let $\mathcal A = \left \langle{A_\alpha}\right \rangle_{\alpha \mathop \in I}$ be an indexed family of subsets of $S$, all connected in $T$. Let $B$ be a connected set of $T$ such that: : $\forall C \in \mathcal A: B \cap C \ne \varnothing$ Then $\displaystyle B \cup \bigcup \mathcal A$ is connected. </onlyinclude> == Proof 1 == {{:Union of Connected Sets with Non-Empty Intersections is Connected/Corollary/Proof 1}} == Proof 2 == {{:Union of Connected Sets with Non-Empty Intersections is Connected/Corollary/Proof 2}} == Proof 3 == {{:Union of Connected Sets with Non-Empty Intersections is Connected/Corollary/Proof 3}} Category:Connected Spaces Category:Union of Connected Sets with Non-Empty Intersections is Connected nmd29npt3fbj0q9eijt52aifpflhnlp


Sine of Sum/Corollary

Sine of Sum/Corollary 0 21561 451122 396168 2020-02-28T15:55:07Z Prime.mover 59 wikitext text/x-wiki == Corollary to Sine of Sum == <onlyinclude> :$\map \sin {a - b} = \sin a \cos b - \cos a \sin b$ </onlyinclude> where $\sin$ denotes the sine and $\cos$ denotes the cosine. == Proof == {{begin-eqn}} {{eqn | l = \map \sin {a - b}       | r = \sin a \, \map \cos {-b} + \cos a \, \map \sin {-b}       | c = Sine of Sum }} {{eqn | r = \sin a \cos b + \cos a \, \map \sin {-b}       | c = Cosine Function is Even }} {{eqn | r = \sin a \cos b - \cos a \sin b       | c = Sine Function is Odd }} {{end-eqn}} {{qed}} == Historical Note == {{:Sine of Sum/Historical Note}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Cosine of Sum|next = Cosine of Difference}}: $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$ * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Sine of Sum|next = Cosine of Sum}}: $\S 5$: Trigonometric Functions: $5.34$ * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Sine of Sum|next = Cosine of Sum}}: $2$: Functions, Limits and Continuity: The Elementary Functions: $4$ * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Sine of Sum|next = Cosine of Sum|entry = compound angle formulae|subentry = in trigonometry}} Category:Sine Function Category:Sine of Sum Category:Compound Angle Formulas ooshkhgan2izsxahq8ukue2ch9xey7z


Ideal of Unit is Whole Ring/Corollary

Ideal of Unit is Whole Ring/Corollary 0 21944 401535 398664 2019-04-19T22:29:39Z Prime.mover 59 wikitext text/x-wiki == Corollary to Ideal of Unit is Whole Ring == <onlyinclude> Let $\struct {R, +, \circ}$ be a ring with unity. Let $J$ be an ideal of $R$. If $J$ contains the unity of $R$, then $J = R$. </onlyinclude> == Proof == Follows directly from Ideal of Unit is Whole Ring and Unity is Unit. {{qed}} == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Ideals of Ring of Integers Modulo m/Examples/Order 12|next = Field has 2 Ideals}}: Chapter $5$: Rings: $\S 21$. Ideals: Theorem $35$ * {{BookReference|Rings, Modules and Linear Algebra|1970|B. Hartley|author2 = T.O. Hawkes|prev = Subring of Integers is Ideal|next = Ideals of Field}}: $\S 2$: Exercise $1$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Ideal of Unit is Whole Ring|next = Field has 2 Ideals}}: $\S 58.2$ Ideals Category:Ideal Theory ci3cu5ftvwct6sxi4zy7je8l1f71n6g


Sum of Geometric Sequence/Corollary 1

Sum of Geometric Sequence/Corollary 1 0 22078 456670 456638 2020-03-20T13:05:38Z Prime.mover 59 wikitext text/x-wiki == Corollary to Sum of Geometric Sequence == <onlyinclude> Let $a, a r, a r^2, \ldots, a r^{n-1}$ be a geometric sequence. Then: :$\displaystyle \sum_{j \mathop = 0}^{n - 1} a r^j = \frac {a \paren {r^n - 1} } {r - 1}$ </onlyinclude> {{:Euclid:Proposition/IX/35}} == Proof 1 == {{:Sum of Geometric Sequence/Corollary 1/Proof 1}} == Proof 2 == {{:Sum of Geometric Sequence/Corollary 1/Proof 2}} == Also presented as == This result is also seen presented as: :$\displaystyle \sum_{j \mathop = 0}^{n - 1} a r^j = \frac {a \paren {1 - r^n} } {1 - r}$ which is usually more manageable when $r < 1$. {{Euclid Note|35|IX}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Number neither whose Half is Odd nor Power of Two is both Even-Times Even and Even-Times Odd|next = Theorem of Even Perfect Numbers/Sufficient Condition}}: Book $\text{IX}$. Propositions * {{BookReference|Handbook of Mathematical Functions|1964|Milton Abramowitz|author2 = Irene A. Stegun|prev = Sum of Arithmetic Sequence|next = Sum of Infinite Geometric Sequence/Corollary 2}}: $3.1.10$: Sum of Geometric Progression to $n$ Terms * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Definition:Geometric Series|next = Sum of Infinite Geometric Sequence/Corollary 2}}: $\S 19$: Geometric Series: $19.4$ * {{BookReference|Elementary Number Theory|1980|David M. Burton|ed = revised|edpage = Revised Printing|prev = Sum of Sequence of Cubes/Proof by Induction|next = Difference of Two Powers/Proof by Induction}}: Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $2$ * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Definition:Geometric Series|next = Sum of Infinite Geometric Sequence/Corollary 2|entry = geometric series}} * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Definition:Common Ratio of Geometric Sequence|next = Definition:Geometric Series|entry = geometric progression (geometric sequence)}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Definition:Common Ratio of Geometric Sequence|next = Definition:Geometric Series|entry = geometric progression (geometric sequence)}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Geometric Series|next = Sum of Infinite Geometric Sequence/Corollary 2|entry = geometric series}} Category:Geometric Sequences Category:Sum of Geometric Sequence 2zf6hscfcls8ov9wr6ci34j8piu17c5


Upper and Lower Bounds of Integral/Corollary 0 22213 495020 348011 2020-10-16T15:07:45Z Prime.mover 59 wikitext text/x-wiki == Corollary to Upper and Lower Bounds of Integral == <onlyinclude> Let $f$ be a real function which is continuous on the closed interval $\closedint a b$. Suppose that $\forall t \in \closedint a b: \size {\map f t} < \kappa$. Then: :$\ds \forall \xi, x \in \closedint a b: \size {\int_x^\xi \map f t \rd t} < \kappa \size {x - \xi}$ </onlyinclude> == Proof == Follows directly from Upper and Lower Bounds of Integral. {{Qed}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Integral of Constant/Definite|next = Definite Integral of Function plus Constant}}: $\S 13.6$ Category:Integral Calculus 2w7ztever04t4na0aazjqtgs2l5xw6y


Double Negation/Double Negation Introduction 0 22278 244793 244787 2016-01-17T09:07:26Z Prime.mover 59 wikitext text/x-wiki == Proof Rule == <onlyinclude> The rule of '''double negation introduction''' is a valid deduction sequent in propositional logic. === Proof Rule === {{:Double Negation/Double Negation Introduction/Proof Rule}} === Sequent Form === {{:Double Negation/Double Negation Introduction/Sequent Form}} </onlyinclude> Category:Proof Rules Category:Double Negation clgtfyvaav8oiviqjve6uup56dbwesn


User:Dfeuer/CRG2

User:Dfeuer/CRG2 2 22614 125790 2013-01-11T19:57:12Z Dfeuer 1672 Dfeuer moved page User:Dfeuer/CRG2 to Properties of Relation Compatible with Group Operation/CRG2 wikitext text/x-wiki #REDIRECT Properties of Relation Compatible with Group Operation/CRG2 brgxg8uewnap1es2qrv2q92h4j7pmhw


Biconditional is Commutative/Formulation 1 0 22685 209536 177784 2015-03-20T20:06:29Z Prime.mover 59 wikitext text/x-wiki == Theorems == The biconditional operator is commutative: <onlyinclude> :$p \iff q \dashv \vdash q \iff p$ </onlyinclude> == Proof 1 == {{:Biconditional is Commutative/Formulation 1/Proof 1}} == Proof 2 == {{:Biconditional is Commutative/Formulation 1/Proof 2}} == Proof 3 == {{:Biconditional is Commutative/Formulation 1/Proof 3}} == Sources == * {{BookReference|Mathematical Logic for Computer Science|2012|M. Ben-Ari|ed=3rd|edpage=Third Edition|prev=Rule of Commutation/Conjunction/Formulation 1|next=Exclusive Or is Commutative}}: $\S 2.3.3$ Category:Biconditional is Commutative 1025jdfsxybvnmq4dfp3kj2vo92qpsg


De Morgan's Laws (Logic)/Disjunction of Negations 0 22816 445727 445716 2020-02-02T11:35:06Z Prime.mover 59 wikitext text/x-wiki == Theorems == ==== Formulation 1 ==== <onlyinclude>{{:De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1}}</onlyinclude> ==== Formulation 2 ==== {{:De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 2}} == Also see == * De Morgan's Laws (Set Theory) for a set theoretic application. * De Morgan's Laws (Predicate Logic) for a predicate logic application. == Sources == * {{BookReference|Undergraduate Topology|1971|Robert H. Kasriel|prev = Conjunction implies Disjunction|next = Definition:Subset}}: $\S 1.2$: Some Remarks on the Use of the Connectives ''and'', ''or'', ''implies'': Exercise $4$ * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = De Morgan's Laws (Set Theory)/Set Complement/Complement of Union|next = De Morgan's Laws (Logic)/Conjunction of Negations|entry = De Morgan's Laws}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = De Morgan's Laws (Set Theory)/Set Complement/Complement of Union|next = De Morgan's Laws (Logic)/Conjunction of Negations|entry = De Morgan's Laws}} Category:De Morgan's Laws (Logic) e3ja0teled439hfeurhls927e8duis0


Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Forward Implication 0 22866 244691 242036 2016-01-16T19:37:03Z Prime.mover 59 wikitext text/x-wiki == Definition == <onlyinclude> :$p \land \left({q \lor r}\right) \vdash \left({p \land q}\right) \lor \left({p \land r}\right)$ </onlyinclude> == Proof == {{BeginTableau|p \land \left({q \lor r}\right) \vdash \left({p \land q}\right) \lor \left({p \land r}\right)}} {{Premise|1|p \land \left({q \lor r}\right)}} {{Simplification|2|1|p|1|1}} {{Simplification|3|1|q \lor r|1|2}} {{Assumption|4|q}} {{Conjunction|5|1, 4|p \land q|2|4}} {{Addition|6|1, 4|\left({p \land q}\right) \lor \left({p \land r}\right)|5|1}} {{Assumption|7|r}} {{Conjunction|8|1, 7|p \land r|2|7}} {{Addition|9|1, 7|\left({p \land q}\right) \lor \left({p \land r}\right)|8|2}} {{ProofByCases|10|1|\left({p \land q}\right) \lor \left({p \land r}\right)|3|4|6|7|9}} {{EndTableau}} {{qed}} == Sources == * {{BookReference|Logic in Computer Science: Modelling and reasoning about systems|2000|Michael R.A. Huth|author2 = Mark D. Ryan|prev = Rule of Association/Disjunction/Formulation 1/Proof 1|next = Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1}}: $\S 1.2.1$: Rules for natural deduction: Example $1.18$ Category:Rule of Distribution i1s2qhget7rr6g4ove1z8elkj2itrrs


Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Reverse Implication 0 22867 244692 143371 2016-01-16T19:37:35Z Prime.mover 59 wikitext text/x-wiki == Definition == <onlyinclude> :$\left({p \land q}\right) \lor \left({p \land r}\right) \vdash p \land \left({q \lor r}\right)$ </onlyinclude> == Proof == {{BeginTableau|\left({p \land q}\right) \lor \left({p \land r}\right) \vdash p \land \left({q \lor r}\right)}} {{Premise|1|\left({p \land q}\right) \lor \left({p \land r}\right)}} {{Assumption|2|p \land q}} {{Simplification|3|2|p|2|1}} {{Assumption|4|p \land r}} {{Simplification|5|4|p|2|1}} {{ProofByCases|6|1|p|1|2|3|4|5}} {{Assumption|7|p \land q}} {{Simplification|8|7|q|7|2}} {{Addition|9|7|q \lor r|7|1}} {{Assumption|10|p \land r}} {{Simplification|11|10|r|10|2}} {{Addition|12|10|q \lor r|11|2}} {{ProofByCases|13|1|q \lor r|1|7|9|10|12}} {{Conjunction|14|1|p \land \left({q \lor r}\right)|6|13}} {{EndTableau}} {{qed}} == Sources == * {{BookReference|Logic in Computer Science: Modelling and reasoning about systems|2000|Michael R.A. Huth|author2 = Mark D. Ryan|prev = Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Forward Implication|next = Definition:Contradiction}}: $\S 1.2.1$: Rules for natural deduction: Exercises $1.4: \ 2 \ \text{(q)}$ Category:Rule of Distribution 33gp2h9yg0azjmaudagosx4p0xyigwf


Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1 0 22868 427188 181061 2019-09-22T22:23:15Z Prime.mover 59 wikitext text/x-wiki == Definition == The conjunction operator is left distributive over the disjunction operator: <onlyinclude> :$p \land \paren {q \lor r} \dashv \vdash \paren {p \land q} \lor \paren {p \land r}$ </onlyinclude> This can be expressed as two separate theorems: === Forward Implication === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Forward Implication}} === Reverse Implication === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Reverse Implication}} == Proof == {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Proof}} == Sources == * {{BookReference|Beginning Logic|1965|E.J. Lemmon|prev = Rule of Idempotence/Conjunction/Formulation 1|next = Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}}: $\S 1.5$: Further Proofs: Résumé of Rules: Exercise $1 \ \text{(c)}$ * {{BookReference|Mathematical Logic for Computer Science|2012|M. Ben-Ari|ed = 3rd|edpage = Third Edition|prev = Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1|next = Rule of Material Equivalence/Formulation 1}}: $\S 2.3.3$ * {{BookReference|Mathematical Logic for Computer Science|2012|M. Ben-Ari|ed = 3rd|edpage = Third Edition|prev = Completeness Theorem for Semantic Tableaus|next = De Morgan's Laws (Logic)/Disjunction/Formulation 1}}: $\S 2.10$: Exercise $2.3$ Category:Rule of Distribution 8qqk1syg44309nyzyukv8z0hqsksq5t


Conjunction Equivalent to Negation of Implication of Negative 0 22964 133377 127713 2013-02-08T09:57:52Z Prime.mover 59 wikitext text/x-wiki == Theorems == <onlyinclude> ==== Formulation 1 ==== {{:Conjunction Equivalent to Negation of Implication of Negative/Formulation 1}} ==== Formulation 2 ==== {{:Conjunction Equivalent to Negation of Implication of Negative/Formulation 2}}</onlyinclude> Category:Implication Category:Conjunction heid7i5x21oznmqxorxkb728hkr9c3g


Operations of Boolean Algebra are Idempotent 0 23151 445333 422670 2020-01-31T07:44:36Z Prime.mover 59 wikitext text/x-wiki == Definition == Let $\left({S, \vee, \wedge}\right)$ be a Boolean algebra. Then: :$\forall x \in S: x \wedge x = x = x \vee x$ That is, both $\vee$ and $\wedge$ are idempotent operations. == Proof == Let $x \in S$. Then: {{begin-eqn}} {{eqn | l = x       | r = x \vee \bot       | c = as $\bot$ is the identity of $\vee$ }} {{eqn | r = x \vee \left({x \wedge \neg x}\right)       | c = as $x \wedge \neg x = \bot$ }} {{eqn | r = \left({x \vee x}\right) \wedge \left({x \vee \neg x}\right)       | c = both $\vee$ and $*$ distribute over the other }} {{eqn | r = \left({x \vee x}\right) \wedge \top       | c = as $x \vee \neg x = \top$ }} {{eqn | r = x \vee x       | c =  }} {{end-eqn}} So $x = x \vee x$. {{qed|lemma}} The result $x = x \wedge x$ follows from Duality Principle (Boolean Algebras). {{qed}} == Sources == * {{BookReference|Abstract Algebra|1964|W.E. Deskins|prev = Idempotent Elements for Integer Multiplication|next = Duality Principle (Boolean Algebras)}}: $\S 1.5$: Theorem $1.13$ * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Duality Principle (Boolean Algebras)|next = Operations of Boolean Algebra are Associative|entry = Boolean algebra}} * {{BookReference|Introduction to Boolean Algebras|2008|Paul Halmos|author2 = Steven Givant|prev = Complement of Complement (Boolean Algebras)|next = De Morgan's Laws (Boolean Algebras)}}: $\S 2$: Exercise $2$ * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Duality Principle (Boolean Algebras)|next = Operations of Boolean Algebra are Associative|entry = Boolean algebra}} Category:Boolean Algebras bkdgh8ozqpsvrh2sqakisbtqe2b6c0p


Method of Truth Tables/Proof of Interderivability 0 23237 170634 170350 2013-12-05T14:20:28Z Lord Farin 560 wikitext text/x-wiki == Proof Technique == <onlyinclude> Suppose we have two propositional formulas $P$ and $Q$ and we wish to see whether $P \dashv \vdash Q$ or not. :Example: Let $P$ be $p \uparrow q$ and let $Q$ be $\neg \left({p \land q}\right)$. There are two things we can do. === Express two statements as a single WFF === We express $P \dashv \vdash Q$ as a single WFF $\left({P \iff Q}\right)$ and perform the method of truth tables on that. Demonstrating this with the example given: $\begin{array}{cc||cccccccc} p & q & (p & \uparrow & q) & \iff & \neg & (p & \land & q) \\ \hline F & F & F & T & F & T & T & F & F & F \\ F & T & F & T & T & T & T & F & F & T \\ T & F & T & T & F & T & T & T & F & F \\ T & T & T & F & T & T & F & T & T & T \\ \end{array}$ As can be seen, the column under the main connective $\iff$ of $\left({P \iff Q}\right)$ is all $T$, so $\left({\left({p \uparrow q}\right) \iff \neg \left({p \land q}\right)}\right)$ is a tautology. Hence from Equivalences are Interderivable, $\left({\left({p \uparrow q}\right) \dashv \vdash \neg \left({p \land q}\right)}\right)$ and the two formulas are interderivable. {{Qed}} === Compare two WFFs in the same table === Alternatively, we can place the two WFFs side by side in the same truth table: $\begin{array}{cc||ccc||cccc} p & q & (p & \uparrow & q) & \neg & (p & \land & q) \\ \hline F & F & F & T & F & T & F & F & F \\ F & T & F & T & T & T & F & F & T \\ T & F & T & T & F & T & T & F & F \\ T & T & T & F & T & F & T & T & T \\ \end{array}$ This time, we need to make sure that the truth values in the columns under the main connectives of both formulae are the same. Note that this is exactly the same as putting a $\iff$ between the two and making a WFF out of the pair of them. {{Qed}} </onlyinclude> Category:Proof Techniques Category:Truth Tables 4bt34xg0u08wexqqnryeeedvw2pws0i


Derivative of Complex Power Series/Proof 2/Lemma 0 23330 364935 364934 2018-09-08T15:39:47Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $n \in N_{\ge 1}$. Then: <onlyinclude> :$\displaystyle \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} = 1$ </onlyinclude> == Proof == Choose any $\alpha > 1$.  It follows from the ratio test that: :$\displaystyle \lim_{n \mathop \to \infty} \dfrac 1 {\alpha^n} \frac {n \paren {n - 1} } 2 = 0$ Therefore, for all sufficiently large $n$: :$\dfrac {n \paren {n - 1} } 2 \le \alpha^n$ and so: :$\paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \le \alpha$ It follows that: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \le \alpha$ Since $\alpha > 1$ was arbitrary, we can conclude that: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \le 1$ It is clear that the following holds for sufficiently large $n$: :$\displaystyle \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \ge 1^{1/n} = 1$ Therefore: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} = 1$ {{qed}} Category:Derivative of Complex Power Series ju1d8vthg9el62cjkrqfy99uuv857dv


Expansion Theorem for Determinants/Corollary 0 23349 476037 350836 2020-06-26T07:39:26Z Prime.mover 59 wikitext text/x-wiki == Corollary to Expansion Theorem for Determinants == Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$. Let $D = \map \det {\mathbf A}$ be the determinant of $\mathbf A$. Let $a_{p q}$ be an element of $\mathbf A$. Let $A_{p q}$ be the cofactor of $a_{p q}$ in $D$. <onlyinclude> Let $\delta_{rs}$ be the Kronecker delta. Then: :$(1): \quad \displaystyle \forall r \in \closedint 1 n: \sum_{k \mathop = 1}^n a_{r k} A_{s k} = \delta_{r s} D$ :$(2): \quad \displaystyle \forall r \in \closedint 1 n: \sum_{k \mathop = 1}^n a_{k r} A_{k s} = \delta_{r s} D$ That is, if you multiply each element of a row or column by the cofactor of ''another'' row or column, the sum of those products is zero. </onlyinclude> == Proof == Let $D'$ be the determinant obtained by replacing row $s$ with row $r$. Let $a'_{ij}$ be an element of $D'$. Then: :$a'_{ij} = \begin {cases} a_{ij} & : i \ne s \\ a_{rj} & : i = s \end {cases}$ Let the cofactor of $a'_{ij}$ in $D'$ be denoted by $A'_{i j}$. Then: :$\forall k \in \closedint 1 n: A'_{s k} = A_{s k}$ So by the Expansion Theorem for Determinants: :$\displaystyle D' = \sum_{k \mathop = 1}^n a'_{s k} A'_{s k}$ But the $r$th and $s$th rows are identical. So by Square Matrix with Duplicate Rows has Zero Determinant: :$D' = 0$ Hence the result. The result for columns follows from Determinant of Transpose.  {{Qed}} Category:Determinants ov756fsll8it09frfmi3vb9s4wfgvvi


Reductio ad Absurdum/Sequent Form

Reductio ad Absurdum/Sequent Form 0 23556 245156 245131 2016-01-18T06:13:29Z Prime.mover 59 wikitext text/x-wiki == Proof Rule == <onlyinclude> The Reductio ad Absurdum can be symbolised by the sequent: : $\left({\neg p \vdash \bot}\right) \vdash p$ </onlyinclude> == Proof == {{BeginTableau|\left({\neg p \vdash \bot}\right) \vdash p}} {{Premise|1|\neg p \vdash \bot}} {{Assumption|2|\neg p}} {{SequentIntro|3|1|\bot|1|the premise itself}} {{Contradiction|4|1|\neg \neg p|2|3}} {{DoubleNegElimination|5|1|p|4}} {{EndTableau}} {{Qed}} {{LEM|Double Negation Elimination}} == Sources == * {{BookReference|Logic in Computer Science: Modelling and reasoning about systems|2000|Michael R.A. Huth|author2 = Mark D. Ryan|prev = Reductio ad Absurdum/Proof Rule|next = Law of Excluded Middle/Proof Rule}}: $\S 1.2.2$: Derived rules Category:Reductio ad Absurdum 4tjhlvqthkkgl3dr0ipn67izelvddny


Sum of Squares of Sine and Cosine/Corollary 1 0 23659 455235 451537 2020-03-17T15:29:42Z Prime.mover 59 wikitext text/x-wiki == Corollary to Sum of Squares of Sine and Cosine == For all $x \in \C$: <onlyinclude> :$\sec^2 x - \tan^2 x = 1 \quad \text {(when $\cos x \ne 0$)}$ </onlyinclude> where $\sec$, $\tan$ and $\cos$ are secant, tangent and cosine respectively. == Proof  == When $\cos x \ne 0$: {{begin-eqn}} {{eqn | l = \cos^2 x + \sin^2 x       | r = 1       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = 1 + \frac {\sin^2 x} {\cos^2 x}       | r = \frac 1 {\cos^2 x}       | c = dividing both sides by $\cos^2 x$, as $\cos x \ne 0$ }} {{eqn | ll= \leadsto       | l = 1 + \tan^2 x       | r = \sec^2 x       | c = {{Defof|Tangent Function}} and {{Defof|Secant Function}} }} {{eqn | ll= \leadsto       | l = \sec^2 x - \tan^2 x       | r = 1       | c = rearranging }} {{end-eqn}} {{qed}} == Also defined as == This result can also be reported as: :$\sec^2 x = 1 + \tan^2 x \quad \text {(when $\cos x \ne 0$)}$ or: :$\tan^2 x = \sec^2 x - 1 \quad \text {(when $\cos x \ne 0$)}$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Sum of Squares of Sine and Cosine|next = Difference of Squares of Cosecant and Cotangent}}: $\S 5$: Trigonometric Functions: $5.20$ * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Sum of Squares of Sine and Cosine/Proof 5|next = Difference of Squares of Cosecant and Cotangent}}: $2$: Functions, Limits and Continuity: The Elementary Functions: $4$ Category:Tangent Function Category:Secant Function Category:Sum of Squares of Sine and Cosine 80bwpr4yeift0whx7blgrh8b0n16tpo


Biconditional Introduction

Biconditional Introduction 0 23917 244863 244862 2016-01-17T12:07:31Z Prime.mover 59 wikitext text/x-wiki == Proof Rule == The rule of '''biconditional introduction''' is a valid deduction sequent in propositional logic. === Proof Rule === {{:Biconditional Introduction/Proof Rule}} === Sequent Form === {{:Biconditional Introduction/Sequent Form}} == Also known as == {{:Biconditional Introduction/Also known as}} == Also see == * Biconditional Elimination Category:Biconditional Category:Biconditional Introduction Category:Implication akjh8pd2536ep8297vske6irfv9su7t


Modus Ponendo Tollens/Variant

Modus Ponendo Tollens/Variant 0 23962 134035 133706 2013-02-10T19:17:44Z Prime.mover 59 wikitext text/x-wiki == Theorems == <onlyinclude> ==== Formulation 1 ==== {{:Modus Ponendo Tollens/Variant/Formulation 1}} ==== Formulation 2 ==== {{:Modus Ponendo Tollens/Variant/Formulation 2}}</onlyinclude> Category:Modus Ponendo Tollens othpbszkidaufqqjci9i6wkrwnle9bb


Jordan Polygon Parity Lemma

Jordan Polygon Parity Lemma 0 24068 443062 443061 2020-01-09T07:54:42Z Prime.mover 59 wikitext text/x-wiki == Definition == Let $P$ be a polygon embedded in $\R^2$. Let $U \subseteq \R^2 \setminus \partial P$ be a path-connected subset of $\R^2 \setminus \partial P$. Let $q \in U$, and let $\mathbf v \in \R^2 \setminus \set {\mathbf 0}$ be a non-zero vector. Let $\LL = \set {q + s \mathbf v: s \in \R_{\ge 0} }$ be a ray with start point $q$. Then: :$(1): \quad$ The parity $\map {\operatorname{par} } q$ is independent of the choice of $v$. :$(2): \quad$ All points in $U$ have the same parity. == Proof == Define $g: \R \to \R^2$ by $\map g \theta = \tuple {\cos \theta, \sin \theta}$. Define $\LL_\theta = \set {q + s \map g \theta: s \in \R_{\ge 0} }$ as a ray with start point $q$. Note that the image of $g$ is equal to the unit circle. Then: :$\dfrac v {\norm {\mathbf v} } = \map g {\theta'}$ for some $\theta' \in \R$, where $\norm {\mathbf v}$ denotes the Euclidean norm of $v$. Then $\LL = \LL_{\theta'}$, as $q + s \mathbf v = q + \paren {s \norm {\mathbf v} } \, \map g {\theta'}$. $\LL_\theta \cap \partial P$ consists of a finite number of line segments, some of which are crossings. Let $\map {N_\theta} q \in \N$ be the number of crossings of $\LL_\theta$. As the value of $\theta$ increases, the value of $\map {N_\theta} q \in \N$ only changes when $\LL_\theta$ crosses a vertex of $P$. If $\theta_0 \in \R$ is a value for which $\LL_{\theta_0}$ crosses exactly one vertex $A$ or side $S$, there are three possibilities: :$\displaystyle \map {N_{\theta_0} } q = \lim_{\theta \mathop \to \theta_0} \map {N_\theta} q$, if $A$ or $S$ is not part of a crossing. :$\displaystyle \map {N_{\theta_0} } q = \lim_{\theta \mathop \to \theta_0^-} \map {N_\theta} q = \lim_{\theta \mathop \to \theta_0^+} \map {N_\theta} q - 2$, if $A$ or $S$ is part of a crossing, and $\LL_{\theta_0}$ intersects both lines adjacent to $A$ or $S$ for some $\theta > \theta_0$. :$\displaystyle \map {N_{\theta_0} } q = \lim_{\theta \mathop \to \theta_0^-} \map {N_\theta} q - 2 = \lim_{\theta \mathop \to \theta_0^+} \map {N_\theta} q$, if $A$ or $S$ is part of a crossing, and $\LL_{\theta_0}$ intersects both lines adjacent to $A$ or $S$ for some $\theta < \theta_0$. Here, $\displaystyle \lim_{\theta \mathop \to \theta_0^-} $ denotes a limit from the left, and $\displaystyle \lim_{\theta \mathop \to \theta_0^+} $ denotes a limit from the right. :File:JordanPolygon2.png The figure shows the change of $\map {N_\theta} q$ each time $\LL_\theta$ intersects a vertex.  The change of $\map {N_\theta} q$ is for increasing values of $\theta$, which corresponds to a counterclockwise rotation of $\LL_\theta$ around $q$. {{explain|Prove the above statement.}} If $\LL_{\theta_0}$ crosses more than one vertex, $\displaystyle N_{\theta_0} \left({q}\right)$ may change be a larger number, but always by a multiple of $2$. Hence, $\map {N_\theta} q \bmod 2$ is independent of $\theta$. Thus we define the parity of $q$ as $\map {\operatorname{par} } q := \map {N_\theta} q \bmod 2$. Since $\LL = \LL_{\theta'}$ for some $\theta' \in \R$, it follows that $\map {\operatorname{par} } q$ can also be defined as the number of crossings modulo $2$ between $\LL$ and $\partial P$. Let $\sigma: \closedint 0 1 \to \R^2 \setminus \partial P$ be any path in $U$ with initial point $q$. For fixed $\theta \in \R$ and $t \in \closedint 0 1$, define $\LL_{\theta, t} = \set {\map \sigma t + s \map g \theta: s \in \R_{\ge 0} }$ as a ray with start point $\map \sigma t$. A similar argument to the one above shows that for all $t_0 \in \closedint 0 1$: :$\displaystyle \lim_{t \mathop \to t_0} \map {N_\theta} {\map \sigma t + s \map g \theta} \equiv \map {N_\theta} {\map \sigma {t_0} + s \, \map g \theta} \pmod 2$ So all points in $\Img \sigma$ have the same parity. This shows that all points in $U$ have the same parity. {{qed}} {{Namedfor|Marie Ennemond Camille Jordan|cat = Jordan}} == Sources == {{wtd|Replace this with a link to a hard-copy; links to web pages have been known to be ephemeral. See house style: Help:Sources.}} : [http://compgeom.cs.uiuc.edu/~jeffe/teaching/comptop/2009/notes/jordan-polygon-theorem.pdf Jeff Ericsson: Computational Topology] Category:Topology 3unndunkt1gy2dx7jajj737x1hyk4k6


Jordan Polygon Theorem/Lemma 1

Jordan Polygon Theorem/Lemma 1 0 24070 472015 451015 2020-06-02T19:15:55Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $P$ be a polygon embedded in $\R^2$. Denote the boundary of $P$ as $\partial P$. <onlyinclude> $\R^2 \setminus \partial P$ is the union of at most two disjoint path-connected sets. </onlyinclude> == Proof == The polygon $P$ has $n$ sides, where $n \in \N$. Denote the vertices of $P$ as $A_1, \ldots, A_n$. Denote the sides of $P$ as $S_1, \ldots, S_n$, such that each vertex $A_i$ has adjacent sides $S_{i - 1}$ and $S_i$. We use the conventions that $S_0 = S_n$, and $A_{n + 1} = A_1$. Let $\displaystyle \delta_i = \map d {S_i, \bigcup_{j \mathop = 1, \ldots, n: \, \size {i - j} > 1 } S_j}$ be the Euclidean distance between a side $S_i$ and all sides not adjacent to $S_i$. From Distance between Closed Sets in Euclidean Space: :$\delta_i > 0$ Put $\displaystyle \delta = \min_{i \mathop = 1, \ldots, n} \delta_i$. From Boundary of Polygon is Jordan Curve, it follows that $\partial P$ is equal to the image of a Jordan curve: :$\gamma: \closedint 0 1 \to \R^2$ that is a concatenation of $n$ paths: :$\gamma_1, \ldots, \gamma_n$ Each $\gamma_i$ is a line segment that joins its initial point $A_i$ and its final point $A_{i + 1}$. Therefore the image of $\gamma_i$ is equal to the side $S_i$. Let $\mathbf v_i = \dfrac {A_{i + 1} - A_i} {\map d {A_{i + 1}, A_i} }$ be the direction vector of $\gamma_i$ with norm $\norm {\mathbf v_i} = 1$. Let $\mathbf w_i$ be the vector $v_i$ rotated $\dfrac \pi 2$ radians counterclockwise, so $\norm {\mathbf w_i} = 1$. {{explain|Invoke the fact that rotation is an isometry (still to be covered on {{ProofWiki}}).}} For any $\epsilon \in \openint 0 {\dfrac \delta 2}$, we intend to construct two Jordan curves $\sigma, \overline \sigma$ such that: :$\Img \sigma \cup \Img {\overline \sigma} = \set {q \in \R^2: \map d {q, \partial P} = \epsilon}$ rightthumbIllustration of the [[Definition:Jordan CurveJordan curves $\sigma, \overline \sigma$ which both have distance $\epsilon$ to $\partial P$.<br/>Note the rounded corners of the curves that occur in place of non-convex angles.]] For $i \in \set {1, \ldots, n}$, initially define the $\sigma_i$ as the line segment that joins its initial point $A_i + \epsilon \mathbf w_i$ with its final point $A_{i + 1} + \epsilon \mathbf w_i$. Suppose $\sigma_i$ and $\sigma_{i + 1}$ intersect at some point $p_{i + 1} \in R^2$. Then re-define the two line segments so that: :the final point of $\sigma_i$ becomes $p_{i + 1}$ and: :the initial point of $\sigma_{i + 1}$ becomes $p_{i + 1}$. Then define a path $\rho_i$ as the constant function $\map {\rho_i} t = p_{i + 1}$. Otherwise, define a path $\rho_i$ with initial point $A_{i + 1} + \epsilon \mathbf w_i$ and final point $A_{i + 1} + \epsilon \mathbf w_{i + 1}$, such that: :the image of $\rho_i$ is part of the circumference of the circle with center $A_{i + 1}$ and radius $\epsilon$. Define the path $\sigma: \closedint 0 1 \to \R^2$ as the concatenation: :$\sigma = \sigma_1 * \rho_1 * \sigma_2 * \rho_2 * \ldots \sigma_n * \rho_n$ Then $\sigma$ is a closed path, as $\sigma_1$ has initial point $A_1$ equal to the final point of $\rho_n$. Each of the paths $\sigma_i$ and $\rho_i$ is injective. For all $i, j \in \set {1, \ldots, n}$, $\sigma_i$ only intersects $\rho_{i - 1}$ and $\rho_i$ in their endpoints. Also, $\sigma_i$ can only possibly intersect $\sigma_{i - 1}$ in its endpoint. In that case the path $\rho_{i - 1}$ is constant. Similarly, $\sigma_i$ can only possibly intersect $\sigma_{i + 1}$ in its endpoint. In that case the path $\rho_{i + 1}$ is constant. For $\size {i - j} > 1$: :let $x_i \in \Img {\sigma_i} \cup \Img {\rho_i}$ :let $x_j \in \Img {\sigma_j} \cup \Img {\rho_j}$. Let $p_i, p_j \in \partial P$ be two points such that: :$\map d {x_i, p_i} = \map d {x_j, p_j} = \epsilon$ Suppose $x_i \in \Img {\sigma_i}$. Then we can put: :$p_i = x_i - \epsilon \mathbf w_i$ Suppose $x_i \in \Img {\rho_i}$. Then we can put $p_i = A_{i + 1}$. Then $x_i \ne x_j$, as: {{begin-eqn}} {{eqn | l = \map d {x_i, x_j}       | r = \map d {x_i, x_j} + \map d {x_i, p_i} + \map d {x_j, p_j} - 2 \epsilon }} {{eqn | o = \ge       | r = \map d {p_i, p_j} - 2 \epsilon       | c = Triangle Inequality for Vectors in Euclidean Space }} {{eqn | o = >       | r = \delta - 2 \epsilon       | c = as $p_i \in S_i, p_j \in S_j$, and $\map d {S_i, S_j} \ge \delta$ }} {{eqn | r = 0 }} {{end-eqn}} It follows that $\sigma_i$ and $\rho_i$ do not intersect $\sigma_j$ and $\rho_j$. It follows that $\sigma$ is a Jordan curve. Now for $i \in \set {1, \ldots, n}$, initially define the $\overline \sigma_i$ as the line segment that joins its initial point $A_i - \epsilon \mathbf w_i$ with its final point $A_{i + 1} - \epsilon \mathbf w_i$. Proceed to define $\overline \rho_i$ and re-define $\overline \sigma_i$ similarly to what was done with $\sigma_i$ and $\rho_i$. Finally, define $\overline \sigma$ as the concatenation: :$\overline \sigma = \overline \sigma_1 * \overline \rho_1 * \overline \sigma_2 * \overline \rho_2 * \ldots \overline \sigma_n * \overline \rho_n$ It follows that $\overline \sigma$ is a Jordan curve as above. We now show that $\sigma$ and $\overline \sigma$ do not intersect. First, $\sigma_i$ and $\overline \sigma_i$ do not intersect, as they are both line segments parallel to $S_i$ with a distance of $2 \epsilon$. Second, $\rho_i$ and $\overline \rho_i$ do not intersect, as one path, say $\rho_i$, is constant with image equal to the crossing point of $\sigma_{i - 1}$ and $\sigma_i$.  The other path, say $\overline \rho_i$, is part of the circumference of the circle that joins the final point of $\overline \sigma_{i - 1}$ with the initial point of $\overline \sigma_i$. Third, $\sigma_i$ and $\overline \sigma_{i + 1}$ do not intersect, since neither line segment crosses $\partial P$, which would be necessary if the two line segment should intersect. Similarly, $\sigma_i$ and $\overline \sigma_{i - 1}$ do not intersect. For all other combinations of $i, j \in \set {1, \ldots, n}$, $\sigma_i$ and $\rho_i$ do not intersect with $\overline \sigma_j$ and $\overline \rho_j$.  This follows as: :$\map d {\sigma_i, S_i} = \map d {\rho_i, S_i} = \map d {\rho_i, S_{i - 1} } = \epsilon$ and as $S_i$, or $S_{i - 1}$, are not adjacent sides to $S_j$, we have $\map d {S_i, S_j} > 2 \epsilon$. Now we can use the Triangle Inequality for Vectors in Euclidean Space as above, to prove that the paths do not intersect. {{qed|lemma}} Finally, let $q_1, q_2, q_3 \in R^2 \setminus \partial P$.  Let: :$\displaystyle \epsilon = \min \set {\map d {q_1, \partial P} / 2, \map d {q_2, \partial P} / 2, \map d {q_3, \partial P} / 2, \delta}$ For all $i \in \set {1, 2, 3}$, draw a line segment $\LL_i$ joining $q_i$ with any point on the boundary $\partial P$. As $\map d {q_i, \partial P} > \epsilon$, it follows from the Intermediate Value Theorem that there is a point $x_i$ on $\LL_i$ such that $\map d {x_i, \partial P} = \epsilon$. Then we have either: :$x_i \in \Img \sigma$ or: :$x_i \in \Img {\overline \sigma}$ when the Jordan curves $\sigma$ and $\overline \sigma$ gets defined from the new value of $\epsilon$. It follows that at least two out of three of the points $q_1, q_2, q_3$ is path-connected to the same Jordan curve. As a Jordan curve is a path, it follows that at least two of the points can be connected by a path. Hence, $R^2 \setminus \partial P$ is the union of at most two disjoint path-connected sets. {{qed}} == Sources == {{SourceReview|Revisit this link}} [http://compgeom.cs.uiuc.edu/~jeffe/teaching/comptop/2009/notes/jordan-polygon-theorem.pdf Jeff Ericsson: Computational Topology]   Category:Topology n10ulxyauewjglujhx6xsj8jzklkux3


Rule of Transposition/Variant 1/Formulation 2/Forward Implication/Proof 0 24245 135343 2013-02-15T21:32:04Z Prime.mover 59 Created page with "== Theorem ==  : $\vdash \left({p \implies \neg q}\right) \implies \left({q \implies \neg p}\right)$   == Proof == <onlyinclude> {{BeginTableau|\vdash \left({p \implies \neg q..." wikitext text/x-wiki == Theorem == : $\vdash \left({p \implies \neg q}\right) \implies \left({q \implies \neg p}\right)$ == Proof == <onlyinclude> {{BeginTableau|\vdash \left({p \implies \neg q}\right) \implies \left({q \implies \neg p}\right)}} {{Assumption|1|p \implies \neg q}} {{Assumption|2|q}} {{DoubleNegIntro|3|2|\neg \neg q|2}} {{ModusTollens|4|1, 2|\neg p|1|3}} {{Implication|5|1|q \implies \neg p|2|4}} {{Implication|6||\left({p \implies \neg q}\right) \implies \left({q \implies \neg p}\right)|1|5}} {{EndTableau}} {{Qed}} </onlyinclude> Category:Rule of Transposition 9o85lre25t88tpxflgvogpdc7zv4we3


Rule of Transposition/Variant 1/Formulation 2/Reverse Implication/Proof 0 24246 135345 2013-02-15T21:33:30Z Prime.mover 59 Created page with "== Theorem ==  : $\vdash \left({q \implies \neg p}\right) \implies \left({p \implies \neg q}\right)$   == Proof == <onlyinclude> {{BeginTableau|\vdash \left({q \implies \neg p..." wikitext text/x-wiki == Theorem == : $\vdash \left({q \implies \neg p}\right) \implies \left({p \implies \neg q}\right)$ == Proof == <onlyinclude> {{BeginTableau|\vdash \left({q \implies \neg p}\right) \implies \left({p \implies \neg q}\right)}} {{Assumption|1|q \implies \neg p}} {{Assumption|2|p}} {{DoubleNegIntro|3|2|\neg \neg p|2}} {{ModusTollens|4|1, 2|\neg q|1|3}} {{Implication|5|1|p \implies \neg q|2|4}} {{Implication|6||\left({q \implies \neg p}\right) \implies \left({p \implies \neg q}\right)|1|5}} {{EndTableau}} {{Qed}} </onlyinclude> Category:Rule of Transposition oqcax6kqn2a2wky8z2vjzwr4rafw2ok


Rule of Transposition/Variant 1/Formulation 1/Forward Implication/Proof 0 24254 135394 2013-02-15T22:18:32Z Prime.mover 59 Created page with "== Theorem ==  :$p \implies \neg q \vdash q \implies \neg p$   == Proof == <onlyinclude> {{BeginTableau|p \implies \neg q \vdash q \implies \neg p}} {{Premise|1|p \implies \ne..." wikitext text/x-wiki == Theorem == :$p \implies \neg q \vdash q \implies \neg p$ == Proof == <onlyinclude> {{BeginTableau|p \implies \neg q \vdash q \implies \neg p}} {{Premise|1|p \implies \neg q}} {{Assumption|2|q}} {{DoubleNegIntro|3|2|\neg \neg q|2}} {{ModusTollens|4|1, 2|\neg p|1|3}} {{Implication|5|1|q \implies \neg p|2|4}} {{EndTableau}} {{Qed}} </onlyinclude> Category:Rule of Transposition an10xkw6ifgwgb8m81kh2nyi2phlsik


Rule of Transposition/Variant 1/Formulation 1/Reverse Implication/Proof 0 24256 135396 2013-02-15T22:21:31Z Prime.mover 59 Created page with "== Theorem ==  : $q \implies \neg p \vdash p \implies \neg q$   == Proof == <onlyinclude> {{BeginTableau|q \implies \neg p \vdash p \implies \neg q}} {{Premise|1|q \implies \n..." wikitext text/x-wiki == Theorem == : $q \implies \neg p \vdash p \implies \neg q$ == Proof == <onlyinclude> {{BeginTableau|q \implies \neg p \vdash p \implies \neg q}} {{Premise|1|q \implies \neg p}} {{Assumption|2|p}} {{DoubleNegIntro|3|2|\neg \neg p|2}} {{ModusTollens|4|1, 2|\neg q|1|3}} {{Implication|5|1|p \implies \neg q|2|4}} {{EndTableau}} {{Qed}} </onlyinclude> Category:Rule of Transposition sybrt3850ifoizy9sqtgr37btaieihr


Axiom:Axiom of Dependent Choice/Right-Total 100 24480 444068 415825 2020-01-18T02:07:40Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> Let $\mathcal R$ be a binary relation on a non-empty set $S$. Suppose that: :$\forall a \in S: \exists b \in S: b \mathrel {\mathcal R} a$ that is, that $\mathcal R$ is a right-total relation. The '''axiom of dependent choice''' states that there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$ such that: :$\forall n \in \N: x_{n + 1} \mathrel {\mathcal R} x_n$ </onlyinclude> Dependent Choice 55o04uhf17seubw32jq7zib415n7yjq


Transitive Closure is Closure Operator

Transitive Closure is Closure Operator 0 24660 138579 138437 2013-03-03T23:47:46Z Dfeuer 1672 wikitext text/x-wiki == Definition == Let $S$ be a set. Let $\mathscr A$ be the set of all endorelations on $S$. Then the transitive closure operator is a closure operator on $\mathscr A$. == Proof == Let $\mathcal Q$ be the set of transitive relations on $S$. By Intersection of Transitive Relations is Transitive, the intersection of any subset of $\mathcal Q$ is in $\mathcal Q$. Recall the definition of transitive closure as the intersection of transitive supersets: :The transitive closure of a relation $\mathcal R$ on $S$ is the intersection of elements of $\mathcal Q$ that contain $S$. From Closure Operator from Closed Sets we conclude that transitive closure is a closure operator. {{qed}} Category:Transitive Closures Category:Closure Operators r8q0evdsyz24wk5t8or2growohwwxyq


Set Union Preserves Subsets/Corollary

Set Union Preserves Subsets/Corollary 0 24716 413984 413982 2019-07-20T22:28:07Z Prime.mover 59 wikitext text/x-wiki == Corollary to Set Union Preserves Subsets == <onlyinclude> Let $A, B, S$ be sets. Then: :$A \subseteq B \implies A \cup S \subseteq B \cup S$ :$A \subseteq B \implies S \cup A \subseteq S \cup B$ </onlyinclude> == Proof 1 == {{:Set Union Preserves Subsets/Corollary/Proof 1}} == Proof 2 == {{:Set Union Preserves Subsets/Corollary/Proof 2}} == Sources == * {{BookReference|Undergraduate Topology|1971|Robert H. Kasriel|prev = Intersection with Subset is Subset|next = Union with Empty Set}}: $\S 1.6$: Set Identities and Other Set Relations: Exercise $1 \ \text{(d)}$ Category:Set Union Preserves Subsets 7cieht9076f142282ozekvgadqvxu95


Set Union Preserves Subsets/Corollary/Proof 2 0 24718 413987 352645 2019-07-20T22:30:10Z Prime.mover 59 wikitext text/x-wiki == Corollary to Set Union Preserves Subsets == {{:Set Union Preserves Subsets/Corollary}} == Proof == <onlyinclude> Let $A$, $B$, and $S$ be sets. Let $A \subseteq B$. Let $x \in A \cup S$. By the definition of union: :$x \in A$ or $x \in S$ Suppose $x \in A$. Then by the definition of subset: :$x \in B$ Thus by the definition of union: :$x \in B \cup S$ Suppose instead that $x \in S$. Then by the definition of union: :$x \in B \cup S$ Thus for all $x \in A \cup S$: :$x \in B \cup S$ The second result follows from Union is Commutative. {{qed}} </onlyinclude> Category:Set Union Preserves Subsets ckr3kx3z8dmfi2hlkiqc2nw13p5kamh


Zero Simple Staircase Integral Condition for Primitive/Lemma 0 25235 442794 351912 2020-01-07T13:20:36Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain. Let $C$ be a closed staircase contour in $D$. Then there exists a contour $C'$ such that: :$\displaystyle \oint_C \map f z \rd z = \oint_{C'} \map f z \rd z$ This contour $C'$ has the property that for all $k \in \set {1, \ldots, n - 1}$, the intersection of the images of $C_k$ and $C_{k + 1}$ is equal to their common end point $\map {\gamma_k} {b_k}$. </onlyinclude> == Proof == The proof is by induction over $n \in \N$, the number of directed smooth curves that $C$ is a concatenation of. === Basis for the Induction === As induction basis, if $C$ consists of just one directed smooth curve, we can simply put $C' = C$. Then, we have to check a statement for all $k \in \O$, so it is trivially true. === Induction Hypothesis === As induction hypothesis, suppose that if $C$ is a concatenation of $n$ directed smooth curves, we can find a contour $C'$ with the properties above. === Induction Step === In the induction step, let $C$ be a concatenation of $n + 1$ directed smooth curves. Suppose that there exists $k \in \set {1, \ldots, n}$ such that the intersection of the images of $C_k$ and $C_{k + 1}$ contain more elements than their common end point $\map {\gamma_k} {b_k}$, otherwise we can put $C' = C$. Then, one of the two line segments must contain the other line segment, so: :$\Img {C_k} \cap \Img {C_{k + 1} } = \gamma_k \sqbrk {\closedint {a'} {b_k} } \cup \gamma_{k + 1} \sqbrk {\closedint {a_{k + 1} } {b'} }$ where either $a' = a_k$, or $b' = b_{k + 1}$. Define $C_k'$ as the contour that is parameterized as a line segment that goes from $\map {\gamma_k} {a_k}$ to $\map {\gamma_k} {a'}$, and then to $\map {\gamma_{k + 1} } {b_{k + 1} }$. Then, the image of $C_k'$ is equal to the part of the line segments that does not overlap. Define $\tilde C_k$ as the contour with the parameterization $\gamma_k \restriction_{\closedint {a'} {b_k} }$, and $\tilde C_{k + 1}$ as the contour with the parameterization $\gamma_{k + 1} \restriction_{\closedint {a_{k + 1} } {b'} }$. Here, $\gamma_k \restriction_{\closedint {a'} {b_k} }$ denotes the restriction of $\gamma_k$ to $\closedint {a'} {b_k}$. Now $\tilde C_k$ is the reversed contour of $\tilde C_{k + 1}$, so: {{begin-eqn}} {{eqn | l = \oint_C \map f z \rd z       | r = \oint_{C_1 \cup \ldots \cup C_{k - 1} } \map f z \rd z + \oint_{C_k} \map f z \rd z + \oint_{C_{k + 1} } \map f z \rd z + \oint_{C_{k + 2} \cup \ldots \cup C_{n + 1} } \map f z \rd z       | c = Contour Integral of Concatenation of Contours }} {{eqn | r = \oint_{C_1 \cup \ldots \cup C_{k - 1} } \map f z \rd z + \oint_{C_k'} \map f z \rd z + \oint_{\tilde C_k} \map f z \rd z + \oint_{\tilde C_{k + 1} } \map f z \rd z + \oint_{C_{k + 2} \cup \ldots \cup C_{n + 1} } \map f z \rd z       | c = Contour Integral of Concatenation of Contours }} {{eqn | r = \oint_{C_1 \cup \ldots \cup C_{k - 1} } \map f z \rd z + \oint_{C_k'} \map f z \rd z + \oint_{\tilde C_k} \map f z \rd z - \oint_{\tilde C_k} \map f z \rd z + \oint_{C_{k + 2} \cup \ldots \cup C_{n + 1} } \map f z \rd z       | c = Contour Integral along Reversed Contour }} {{eqn | r = \oint_{C_1 \cup \ldots \cup C_{k - 1} \cup \tilde C_k \cup C_{k + 2} \ldots \cup C_{n + 1} } \map f z \rd z       | c = Contour Integral of Concatenation of Contours }} {{end-eqn}} which shows that the contour integral of $f$ along $C$ is equal to the contour integral of $f$ along $C'' = C_1 \cup \ldots \cup C_{k - 1} \cup \tilde C_k \cup C_{k + 2} \ldots \cup C_{n + 1}$. This new contour $C''$ is a staircase contour which is a concatenation of $n$ directed smooth curves, so by the induction hypothesis, there exists a contour $C'$ such that: :$\displaystyle \oint_{C'} \map f z \rd z = \oint_{C''} \map f z \rd z$ where $C'$ has the desired property. {{qed}} Category:Complex Analysis j4it516u0i2u375g2cjnypku1rnupg1


Polynomial Factor Theorem/Corollary

Polynomial Factor Theorem/Corollary 0 25240 442610 407962 2020-01-06T20:43:45Z Prime.mover 59 wikitext text/x-wiki == Corollary to Polynomial Factor Theorem == <onlyinclude> Let $\map P x$ be a polynomial in $x$ over the real numbers $\R$ of degree $n$. Suppose there exists $\xi \in \R: \map P \xi = 0$. Then $\map P x = \paren {x - \xi} \map Q x$, where $\map Q x$ is a polynomial of degree $n - 1$. Hence, if $\xi_1, \xi_2, \dotsc, \xi_n \in \R$ such that all are different, and $\map P {\xi_1} = \map P {\xi_2} = \dotsb = \map P {\xi_n} = 0$, then: :$\displaystyle \map P x = k \prod_{j \mathop = 1}^n \paren {x - \xi_j}$ where $k \in \R$. </onlyinclude> === Complex Number form === {{:Polynomial Factor Theorem/Corollary/Complex Numbers}} == Proof == Recall that Real Numbers form Field. The result then follows from the Polynomial Factor Theorem. {{qed}} == Also known as == Some sources refer to this (and its more general form) as the Factor Theorem, but there are multiple theorems with this name. == Sources == * {{BookReference|A Course in Pure Mathematics|1960|Margaret M. Gow|prev = Polynomial Remainder Theorem|next = }}: Chapter $1$: Polynomials; The Remainder and Factor Theorems; Undetermined Coefficients; Partial Fractions: $1.2$. The remainder and factor theorems * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Definition:Polynomial Function/Real/Definition 1|next = Definition:Factorial/Definition 2}}: $\S 3$: Natural Numbers: Exercise $\S 3.11 \ (3)$ Category:Polynomial Theory kpusbhnxnnvoa2h5gxzlnh0uyxc72py


Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Forward Implication 0 25526 141832 2013-03-24T17:45:43Z Dfeuer 1672 Created page with "== Theorem == <onlyinclude> : $\vdash \left({p \land \left({q \lor r}\right)}\right) \implies \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right)$ </onlyincl..." wikitext text/x-wiki == Theorem == <onlyinclude> : $\vdash \left({p \land \left({q \lor r}\right)}\right) \implies \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right)$ </onlyinclude> == Proof == {{BeginTableau|\vdash \left({p \land \left({q \lor r}\right)}\right) \implies \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right)}} {{Assumption|1|p \land \left({q \lor r}\right)}} {{SequentIntro|2|1|\left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right)|1|Conjunction is Left Distributive over Disjunction: Formulation 1}} {{Implication|3||\left({p \land \left({q \lor r}\right)}\right) \implies \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right)|1|2}} {{EndTableau}} {{qed}} Category:Rule of Distribution 44o08ry9xd4hrjz9tdnxcsck49f143v


Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Reverse Implication 0 25536 141890 2013-03-25T04:36:09Z Dfeuer 1672 Created page with "== Theorem == <onlyinclude> :$\vdash \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right) \implies \left({p \land \left({q \lor r}\right)}\right)$ </onlyinclu..." wikitext text/x-wiki == Theorem == <onlyinclude> :$\vdash \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right) \implies \left({p \land \left({q \lor r}\right)}\right)$ </onlyinclude> == Proof == {{BeginTableau|\vdash \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right) \implies \left({p \land \left({q \lor r}\right)}\right)}} {{Assumption|1|\left({p \land q}\right) \lor \left({p \land r}\right)}} {{SequentIntro|2|1|p \land \left({q \lor r}\right)|1|Conjunction is Left Distributive over Disjunction: Formulation 1}} {{Implication|3||\left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right) \implies \left({p \land \left({q \lor r}\right)}\right)|1|2}} {{EndTableau}} {{qed}} Category:Rule of Distribution g9i1e7hay4zpyt39iurxetwxljvt9lm


Degree of Product of Polynomials over Ring/Corollary 1 0 25647 348160 348158 2018-03-21T18:38:32Z Barto 3079 terminology: ring of polynomials wikitext text/x-wiki {{mergeto|Degree of Product of Polynomials over Integral Domain}} == Corollary to Degree of Product of Polynomials over Ring == Let $\left({R, +, \circ}\right)$ be a ring with unity whose zero is $0_R$. Let $R \left[{X}\right]$ be the ring of polynomials over $R$ in the indeterminate $X$. For $f \in R \left[{X}\right]$ let $\deg \left({f}\right)$ be the degree of $f$. <onlyinclude> Let $R$ have no proper zero divisors. Then: :$\forall f, g \in R \left[{X}\right]: \deg \left({f g}\right) = \deg \left({f}\right) + \deg \left({g}\right)$ </onlyinclude> == Proof == Let the leading coefficient of: : $f \left({X}\right)$ be $a_n$ : $g \left({X}\right)$ be $b_m$. From Degree of Product of Polynomials over Ring: :$\deg \left({f g}\right) \le \deg \left({f}\right) + \deg \left({g}\right)$ From the definition of polynomial multiplication: : $f \left({X}\right) g \left({X}\right) = a_n b_m X^{n+m} + \cdots + a_0 b_0$ As $\left({R, +, \circ}\right)$ has no proper zero divisors the $X^{n+m}$ term can not equal $0_R$ and so: :$\deg \left({f g}\right) = \deg \left({f}\right) + \deg \left({g}\right)$ {{qed}} Category:Degree of Product of Polynomials over Ring 2t0ozz3b84kx5izs33ctam5bmqtt3xj


Union of Bijections with Disjoint Domains and Codomains is Bijection/Corollary 0 25665 200333 142500 2014-11-11T22:31:48Z Prime.mover 59 wikitext text/x-wiki == Corollary to Union of Bijections with Disjoint Domains and Codomains is Bijection == <onlyinclude> Let $A$, $B$, $C$, and $D$ be sets or classes. Let $A \cap B = C \cap D = \varnothing$. Let $f: A \to C$ and $g: D \to B$ be bijections. Then $f \cup g^{-1}: A \cup B \to C \cup D$ is also a bijection. </onlyinclude> == Proof == By definition of bijection, $g^{-1}: B \to D$ is a bijection. Hence the result by Union of Bijections with Disjoint Domains and Codomains is Bijection. {{qed}} Category:Bijections re730kt8jf0uk7g2td02l1aasv0v3z6


Set Difference and Intersection form Partition/Corollary 1 0 26175 496211 418193 2020-10-24T16:16:55Z Prime.mover 59 wikitext text/x-wiki == Corollary to Set Difference and Intersection form Partition == <onlyinclude> Let $S$ and $T$ be sets such that: :$S \setminus T \ne \O$ :$T \setminus S \ne \O$ :$S \cap T \ne \O$ Then $S \setminus T$, $T \setminus S$ and $S \cap T$ form a partition of $S \cup T$, the union of $S$ and $T$. </onlyinclude> == Proof == From Set Difference and Intersection form Partition: :$S \setminus T$ and $S \cap T$ form a partition of $S$ :$T \setminus S$ and $S \cap T$ form a partition of $T$ From Set Difference is Disjoint with Reverse: :$\paren {S \setminus T} \cap \paren {T \setminus S} = \O$ So: :$S \cup T = \paren {S \setminus T} \cup \paren {S \cap T} \cup \paren {T \setminus S} \cup \paren {S \cap T}$ and the result follows. {{qed}} Category:Set Difference Category:Set Intersection Category:Set Union 05cyy0o0qirvwojqxolxdkmyfrur5lo


Inclusion-Exclusion Principle/Corollary

Inclusion-Exclusion Principle/Corollary 0 26197 390368 352624 2019-01-28T13:36:09Z Prime.mover 59 wikitext text/x-wiki == Corollary to Inclusion-Exclusion Principle == <onlyinclude> Let $\mathcal S$ be an algebra of sets. Let $A_1, A_2, \ldots, A_n$ be finite sets which are pairwise disjoint. Let $f: \mathcal S \to \R$ be an additive function. Then: :$\displaystyle \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$ </onlyinclude> == Proof == As $A_1, A_2, \ldots, A_n$ are pairwise disjoint, their intersections are all empty. Then Inclusion-Exclusion Principle holds. However, from Cardinality of Empty Set, all the terms apart from the first vanish. {{qed}} == Comment == This result is usually quoted in the context of combinatorics, where $f$ is the cardinality function. It is also seen in the context of probability theory, in which $f$ is taken to be a probability measure. Category:Set Systems Category:Set Union Category:Probability Theory Category:Combinatorics srw2dqr9ylgcyhz7g4qkyn3o0o2ya7m


Independent Events are Independent of Complement/Corollary 0 26223 486487 189570 2020-09-08T14:58:16Z Prime.mover 59 wikitext text/x-wiki == Corollary to Independent Events are Independent of Complement == Let $A$ and $B$ be events in a probability space $\struct {\Omega, \Sigma, \Pr}$. <onlyinclude> $A$ and $B$ are independent {{iff}} $\Omega \setminus A$ and $\Omega \setminus B$ are independent. </onlyinclude> == Proof == Let $A$ and $B$ be independent. Then from Independent Events are Independent of Complement, $A$ and $\Omega \setminus B$ are independent. Setting $A' = \Omega \setminus B$ and $B' = A$, we see clearly that $A'$ and $B'$ are independent. So from the main result, $A'$ and $\Omega \setminus B'$ are independent. That is, $\Omega \setminus B$ and $\Omega \setminus A$ are independent. The "only if" part of the result follows directly from Relative Complement of Relative Complement and another application of this result. {{qed}} Category:Independent Events 5uhcjyyox39poiunckkjtnlx3vu6haf


Condition for Independence from Product of Expectations/Corollary 0 26304 475737 475727 2020-06-24T05:16:15Z Prime.mover 59 wikitext text/x-wiki == Corollary to Condition for Independence from Product of Expectations == Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. <onlyinclude> Let $X$ and $Y$ be independent discrete random variables on $\struct {\Omega, \Sigma, \Pr}$. Then: :$\expect {X Y} = \expect X \expect Y$ assuming the latter expectations exist. </onlyinclude> === General Result === {{:Condition for Independence from Product of Expectations/Corollary/General Result}} == Proof == From Condition for Independence from Product of Expectations, setting both $g$ and $h$ to the identity functions: :$\forall x \in \R: \map g x = x$ :$\forall y \in \R: \map h y = y$ It follows directly that if $X$ and $Y$ are independent, then: :$\expect {X Y} = \expect X \expect Y$ assuming the latter expectations exist. {{qed}} == Note on Converse == Note that the converse of the corollary does not necessarily hold. {{:Condition for Independence from Product of Expectations/Corollary/Converse}} == Sources == * {{BookReference|Probability: An Introduction|1986|Geoffrey Grimmett|author2 = Dominic Welsh|prev = Condition for Independence of Discrete Random Variables|next = Condition for Independence from Product of Expectations}}: $\S 3.3$: Independence of discrete random variables: Theorem $3 \text{C}$ Category:Expectation 9235x95dydyt0aurp8hvfi3pfu4dzhk


Condition for Independence from Product of Expectations/Corollary/General Result 0 26306 481558 475740 2020-08-13T10:30:02Z RandomUndergrad 3904 wikitext text/x-wiki == Corollary to Condition for Independence from Product of Expectations == Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. <onlyinclude> Let $X_1, X_2, \ldots, X_n$ be independent discrete random variables. Then: :$\displaystyle \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$ assuming the latter expectations exist. </onlyinclude> == Proof == Proof by induction: For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$ === Basis for the Induction === $\map P 1$ is the case: :$\displaystyle \expect {\prod_{k \mathop = 1}^1 {X_k} } = \expect {X_1} = \prod_{k \mathop = 1}^1 \expect {X_k}$ Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \expect {\prod_{k \mathop = 1}^r {X_k} } = \prod_{k \mathop = 1}^r \expect {X_k}$ from which it is to be shown that: :$\displaystyle \expect {\prod_{k \mathop = 1}^{r + 1} {X_k} } = \prod_{k \mathop = 1}^{r + 1} \expect {X_k}$ === Induction Step === This is the induction step: We have: {{begin-eqn}} {{eqn | l = \expect {\prod_{k \mathop = 1}^{r + 1} {X_k} }       | r = \expect {X_{r + 1} \prod_{k \mathop = 1}^r {X_k} } }} {{eqn | r = \expect {X_{r + 1} } \expect {\prod_{k \mathop = 1}^r {X_k} }       | c = Corollary to Condition for Independence from Product of Expectations }} {{eqn | r = \expect {X_{r + 1} } \prod_{k \mathop = 1}^r \expect {X_k}       | c = Induction Hypothesis }} {{eqn | r = \prod_{k \mathop = 1}^{r + 1} \expect {X_k} }} {{end-eqn}} So $\map P r \implies \map P {r + 1}$ and thus it follows by the Principle of Mathematical Induction that: :$\displaystyle \forall n \in \Z_{> 0}: \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$ {{qed}} Category:Expectation hl637f6rigaldm19zwxt3j8q8ql2s68


Sum of Infinite Geometric Sequence/Corollary 1 0 26315 456204 456125 2020-03-19T09:16:53Z Prime.mover 59 wikitext text/x-wiki == Corollary to Sum of Infinite Geometric Sequence == Let $S$ be a standard number field, i.e. $\Q$, $\R$ or $\C$. Let $z \in S$. Let $\size z < 1$, where $\size z$ denotes: :the absolute value of $z$, for real and rational $z$ :the complex modulus of $z$ for complex $z$. Then: <onlyinclude> :$\displaystyle \sum_{n \mathop = 1}^\infty z^n = \frac z {1 - z}$ </onlyinclude> == Proof 1 == {{:Sum of Infinite Geometric Sequence/Corollary 1/Proof 1}} == Proof 2 == {{:Sum of Infinite Geometric Sequence/Corollary 1/Proof 2}} Category:Sum of Infinite Geometric Sequence 18ydlbwfe2ktt3eos2413uldw1w9zgn


Axiom:Axiom of Countable Choice for Finite Sets 100 26471 453393 453350 2020-03-08T12:52:18Z Prime.mover 59 wikitext text/x-wiki == Axiom == Let $S$ be a countable set of non-empty finite sets. Then $S$ has a choice function. == Also see == This axiom is a weakening of the Axiom of Countable Choice and the Axiom of Choice for Finite Sets. == Sources == {{wtd|Have source; will add.}} Category:Axioms/Axiom of Choice 4zeewvmxrtth0ae6bvrkg2eqzxjginl


Axiom:Axiom of Countable Choice/Form 2

Axiom:Axiom of Countable Choice/Form 2 100 26473 444057 444053 2020-01-18T02:01:49Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> Let $S$ be a countable set of non-empty sets. Then $S$ has a choice function. </onlyinclude> == Also see == * Equivalence of Forms of Axiom of Countable Choice == Sources == {{NoSources}} Countable Choice 9gilr1ghka7rq545viyvdgkfh6xvnkl


Axiom:Axiom of Countable Choice/Form 1

Axiom:Axiom of Countable Choice/Form 1 100 26474 444056 444051 2020-01-18T02:01:38Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> Let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of non-empty sets. The '''axiom of countable choice''' states that there exists a sequence: :$\sequence {x_n}_{n \mathop \in \N}$ such that $x_n \in S_n$ for all $n \in \N$. </onlyinclude> == Also see == * Equivalence of Forms of Axiom of Countable Choice Countable Choice 5571sm8cd5w30rp4kupm6da3uthox0z


Between two Rational Numbers exists Irrational Number/Lemma 1 0 26902 369152 369112 2018-10-05T17:33:27Z Prime.mover 59 wikitext text/x-wiki == Lemma for Between two Rational Numbers exists Irrational Number == <onlyinclude> Let $\alpha \in \Q \setminus \set 0$ and $\beta \in \R \setminus \Q$.  Then:  :$\alpha \cdot \beta \in \R \setminus \Q$ </onlyinclude> == Proof == {{AimForCont}} $\alpha \cdot \beta \in \Q$. By the definition of rational numbers: :$\exists n, m, p, q \in \Z: \alpha = \dfrac n m$ :$\alpha \cdot \beta = \dfrac p q$ Thus: :$\beta = \dfrac p q \cdot \dfrac 1 \alpha = \dfrac p q \cdot \dfrac m n$ By Rational Multiplication is Closed, we have $\beta \in \Q$, which contradicts the statement that $\beta \in \R \setminus \Q$. Therefore  $\alpha \cdot \beta \in \R \setminus \Q$. {{qed}} Category:Between two Rational Numbers exists Irrational Number 35l02mdb33hx8y6ktvgn22juytppbwp


Between two Rational Numbers exists Irrational Number/Lemma 2 0 26903 483038 369153 2020-08-26T21:58:49Z Prime.mover 59 wikitext text/x-wiki == Lemma for Between two Rational Numbers exists Irrational Number == <onlyinclude> Let $\alpha \in \Q$ and $\beta \in \R \setminus \Q$.  Then:  :$\alpha + \beta \in \R \setminus \Q$ </onlyinclude> == Proof == {{AimForCont}} $\alpha + \beta \in \Q$. By the definition of rational numbers: :$\exists p, q \in \Z: \alpha + \beta = \dfrac p q$ Thus: :$\beta = \dfrac p q - \alpha$ By Rational Subtraction is Closed, we have $\beta \in \Q$, which contradicts the statement that $\beta \in \R \setminus \Q$. Therefore: :$\alpha + \beta \in \R \setminus \Q$ {{qed}} Category:Between two Rational Numbers exists Irrational Number 7pc2qb6kr0efk8crs9joxmhmq28bb48