hcju/proofwikips huggingface dataset:
Derivative of Complex Power Series/Proof 2/Lemma
== Lemma == Let $n \in N_{\ge 1}$. Then:  :$\displaystyle \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} = 1$  == 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: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} = 1$
Expansion Theorem for Determinants/Corollary
== 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 $D$.  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.  == 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.
Reductio ad Absurdum/Sequent Form
== Proof Rule ==  The Reductio ad Absurdum can be symbolised by the sequent: : $\left({\neg p \vdash \bot}\right) \vdash p$  == 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}}   {{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
Sum of Squares of Sine and Cosine/Corollary 1
== Corollary to Sum of Squares of Sine and Cosine == For all $x \in \C$:  :$\sec^2 x - \tan^2 x = 1 \quad \text {(when $\cos x \ne 0$)}$  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}}  == 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: $4$
Biconditional Introduction
== 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
Modus Ponendo Tollens/Variant
== Theorems ==  ==== Formulation 1 ==== {{:Modus Ponendo Tollens/Variant/Formulation 1}} ==== Formulation 2 ==== {{:Modus Ponendo Tollens/Variant/Formulation 2}}
Rule of Transposition/Variant 1/Formulation 2/Forward Implication/Proof
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 ==  {{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 ==  {{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}}
Rule of Transposition/Variant 1/Formulation 2/Reverse Implication/Proof
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 ==  {{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 ==  {{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}}
Rule of Transposition/Variant 1/Formulation 1/Forward Implication/Proof
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 ==  {{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 ==  {{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}}
Rule of Transposition/Variant 1/Formulation 1/Reverse Implication/Proof
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 ==  {{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 ==  {{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}}
Axiom:Axiom of Dependent Choice/Right-Total
== Axiom ==  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: :$\forall n \in \N: x_{n + 1} \mathrel {\mathcal R} x_n$  Dependent Choice 55o04uhf17seubw32jq7zib415n7yjq
Transitive Closure is Closure Operator
== 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.
Set Union Preserves Subsets/Corollary
== Corollary to Set Union Preserves Subsets ==  Let $A, B, S$ be sets. Then: :$A \subseteq B \implies A \cup S \subseteq B \cup S$ :$A \subseteq B \implies S \cup A \subseteq S \cup B$  == 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 $1 \ \text{(d)}$
Set Union Preserves Subsets/Corollary/Proof 2
== Corollary to Set Union Preserves Subsets == {{:Set Union Preserves Subsets/Corollary}} == Proof ==  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.
Polynomial Factor Theorem/Corollary
== Corollary to Polynomial Factor Theorem ==  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 $k \in \R$.  === 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.  == 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 $\S 3.11 \ (3)$
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Forward Implication
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 ==  : $\vdash \left({p \land \left({q \lor r}\right)}\right) \implies \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right)$ </onlyincl..." wikitext text/x-wiki == Theorem ==  : $\vdash \left({p \land \left({q \lor r}\right)}\right) \implies \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right)$  == 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}}
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Reverse Implication
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 ==  :$\vdash \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right) \implies \left({p \land \left({q \lor r}\right)}\right)$ </onlyinclu..." wikitext text/x-wiki == Theorem ==  :$\vdash \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right) \implies \left({p \land \left({q \lor r}\right)}\right)$  == 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}}
Degree of Product of Polynomials over Ring/Corollary 1
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 $f$.  Let $R$ have no proper zero divisors. Then: :$\forall f, g \in R \left[{X}\right]: \deg \left({f g}\right) = \deg \left({f}\right) + \deg \left({g}\right)$  == 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: :$\deg \left({f g}\right) = \deg \left({f}\right) + \deg \left({g}\right)$
Union of Bijections with Disjoint Domains and Codomains is Bijection/Corollary
== Corollary to Union of Bijections with Disjoint Domains and Codomains is Bijection ==  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.  == 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.
Set Difference and Intersection form Partition/Corollary 1
== Corollary to Set Difference and Intersection form Partition ==  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 $T$.  == 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.
Inclusion-Exclusion Principle/Corollary
== Corollary to Inclusion-Exclusion Principle ==  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: :$\displaystyle \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$  == 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.  == 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.
Independent Events are Independent of Complement/Corollary
== Corollary to Independent Events are Independent of Complement == Let $A$ and $B$ be events in a probability space $\struct {\Omega, \Sigma, \Pr}$.  $A$ and $B$ are independent  $\Omega \setminus A$ and $\Omega \setminus B$ are independent.  == 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.
Condition for Independence from Product of Expectations/Corollary
== Corollary to Condition for Independence from Product of Expectations == Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.  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.  === 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.  == 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 $3 \text{C}$
Condition for Independence from Product of Expectations/Corollary/General Result
== Corollary to Condition for Independence from Product of Expectations == Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.  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.  == 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: :$\displaystyle \forall n \in \Z_{> 0}: \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$
Sum of Infinite Geometric Sequence/Corollary 1
== 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:  :$\displaystyle \sum_{n \mathop = 1}^\infty z^n = \frac z {1 - z}$  == Proof 1 == {{:Sum of Infinite Geometric Sequence/Corollary 1/Proof 1}} == Proof 2 == {{:Sum of Infinite Geometric Sequence/Corollary 1/Proof 2}}
Axiom:Axiom of Countable Choice for Finite Sets
== 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.}}
Axiom:Axiom of Countable Choice/Form 2
== Axiom ==  Let $S$ be a countable set of non-empty sets. Then $S$ has a choice function.  == Also see == * Equivalence of Forms of Axiom of Countable Choice == Sources ==  Countable Choice 9gilr1ghka7rq545viyvdgkfh6xvnkl
Axiom:Axiom of Countable Choice/Form 1
== Axiom ==  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 $n \in \N$.  == Also see == * Equivalence of Forms of Axiom of Countable Choice Countable Choice 5571sm8cd5w30rp4kupm6da3uthox0z
Between two Rational Numbers exists Irrational Number/Lemma 1
== Lemma for Between two Rational Numbers exists Irrational Number ==  Let $\alpha \in \Q \setminus \set 0$ and $\beta \in \R \setminus \Q$.  Then:  :$\alpha \cdot \beta \in \R \setminus \Q$  == Proof ==  $\alpha \cdot \beta \in \Q$. 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  $\alpha \cdot \beta \in \R \setminus \Q$.