hcju/proofwikiqa huggingface dataset:
By Congruence of Powers: :$a \equiv b \pmod 5 \iff a^4 \equiv b^4 \pmod 5$ so it is sufficient to demonstrate the result for $n \in \set {0, 1, 2, 3, 4}$. Thus: {{begin-eqn}} {{eqn | l = 0^4       | m = 0       | mo= \equiv       | r = 0       | rr= \pmod 5       | c =  }} {{eqn | l = 1^4       | m = 1       | mo= \equiv       | r = 1       | rr= \pmod 5       | c =  }} {{eqn | l = 2^4       | m = 16       | mo= \equiv       | r = 1       | rr= \pmod 5       | c =  }} {{eqn | l = 3^4       | m = 81       | mo= \equiv       | r = 1       | rr= \pmod 5       | c =  }} {{eqn | l = 4^4       | m = 256       | mo= \equiv       | r = 1       | rr= \pmod 5       | c =  }} {{end-eqn}} The result follows. {{qed}} Category:Fourth Powers 2a82dfb5jyv3nyxyy58zi9gjf88h0dg
By Lagrange's Theorem, the elements of $G \setminus K$ can be of order $1$, $2$, $p$ or $2 p$. $1$ is not possible because Identity is Only Group Element of Order 1. Then we have that $G$ is non-abelian. Hence from Cyclic Group is Abelian, $G$ is not cyclic. Thus $\order b \ne 2 p$. It remains to investigate $2$ and $p$. Let $b \in G \setminus K$. By Subgroup of Index 2 contains all Squares of Group Elements: :$b^2 \in K$ {{AimForCont}} $b$ has order $p$. Then by Intersection of Subgroups of Prime Order: :$K \cap \gen b = e$ which contradicts $b^2 \in K$. Thus by Proof by Contradiction $\order b \ne p$ Hence it must follow that $\order b = 2$. {{qed|lemma}} We have that: :$b a \in G \setminus K$ Thus: :$\paren {b a}^2 = e$ {{begin-eqn}} {{eqn | l = b a       | o = \in       | r = G \setminus K       | c =  }} {{eqn | ll= \leadsto       | l = \paren {b a}^2       | r = e       | c =  }} {{eqn | ll= \leadsto       | l = b a b       | r = \paren {b a}^2 a^{-1}       | c =  }} {{eqn | r = a^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = b a b^{-1}       | r = a^{-1}       | c = as $b = b^{-1}$ }} {{end-eqn}} {{qed}}
The result Non-Abelian Order 2p Group has Order p Element demonstrates that such an element $a$ exists in $G$. By definition of generator of cyclic group, $K$ is the cyclic group $C_p$ of order $p$. By Lagrange's Theorem, the index of $K$ is: :$\index G K = \dfrac {\order G} {\order K} = \dfrac {2 p} p = 2$ The result follows from Subgroup of Index 2 is Normal. {{qed}}
Let $G$ be a group whose identity is $e$. Let $\kappa_x: G \to G$ be the inner automorphism defined as: :$\forall g \in G: \kappa_x \paren g = x g x^{-1}$ We see that: :$\Inn G \ne \O$ as $\kappa_x$ is defined for all $x \in G$. We show that: :$\kappa_x, \kappa_y \in \Inn G: \kappa_x \circ \paren {\kappa_y}^{-1} \in \Inn G$ So: {{begin-eqn}} {{eqn | ll= \forall g \in G:       | l = \map {\paren {\kappa_x \circ \paren {\kappa_y}^{-1} } } g       | r = \map {\kappa_x} {\map {\paren {\kappa_y}^{-1} } g}       | c =  }} {{eqn | r = \map {\kappa_x} {\map {\kappa_{y^{-1} } } g}       | c = Inverse of Inner Automorphism }} {{eqn | r = \map {\kappa_x} {y^{-1} g y}       | c =  }} {{eqn | r = x y^{-1} g y x^{-1}       | c =  }} {{eqn | r = \paren {x y^{-1} } g \paren {x y^{-1} }^{-1}       | c = Inverse of Product }} {{eqn | r = \map {\kappa_{x y^{-1} } } g       | c =  }} {{end-eqn}} As $x y^{-1} \in G$, it follows that: :$\kappa_{x y^{-1} } \in \Inn G$ By the One-Step Subgroup Test: :$\Inn G \le \Aut G$ {{qed}}
Let $\psi: G \to G$ be the mapping: :$\forall x \in G: \map \psi x = x^{-1} \, \map \phi x$ From Property of Group Automorphism which Fixes Identity Only: :$\forall x, y \in G: x^{-1} \, \map \phi x = y^{-1} \, \map \phi y \implies x = y$ That is, $\psi$ is an injection. From Injection from Finite Set to Itself is Surjection, it follows that $\psi$ is a surjection. That is: :$\forall x \in G: \exists g \in G: x = \map \psi g$ which is a restatement of the hypothesis. {{qed}}
Let $G = \gen g$ Let $\phi, \psi \in \Aut G$. As $G$ is cyclic: {{begin-eqn}} {{eqn | lo= \exists a \in \Z:       | l = \map \phi g        | r = g^a }} {{eqn | lo= \exists b \in \Z:       | l = \map \psi g        | r = g^b }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \map {\phi \circ \psi} g       | r = \paren {g^a}^b       | c =  }} {{eqn | r = g^{a b}       | c =  }} {{eqn | r = g^{b a}       | c =  }} {{eqn | r = \paren {g^b}^a       | c =  }} {{eqn | r = \map {\psi \circ \phi} g       | c =  }} {{end-eqn}} Thus in particular, $\phi \circ \psi$ and $\psi \circ \phi$ are equal on the generator $g$. Since $g$ generates $G$, they must be equal as automorphisms. {{qed}}
From Prime Group is Cyclic we have that $G$ is a cyclic group. From Order of Automorphism Group of Cyclic Group: :$\order {\Aut G} = \map \phi p$ where $\map \phi n$ denotes the Euler $\phi$ function. The result follows from Euler Phi Function of Prime. {{qed}}
This is an instance of Ring of Square Matrices over Field is Ring with Unity. {{qed}}
In the following, let: :$0_R$ denote the zero of $R$ :$1_R$ denote the unity of $R$ :$R^*$ denote the set of elements of $R$ without $0_R$. === $(1)$ implies $(2)$ === Let $\struct {R, +, \circ}$ be a Division Ring by definition 1. Then by definition: :$\forall x \in R^*: \exists! x^{-1} \in R^*: x^{-1} \circ x = x \circ x^{-1} = 1_R$ That is, $x^{-1}$ is the (unique) product inverse of $x$. Thus, by definition, $x$ is a unit of $R$. Thus $\struct {R, +, \circ}$ is a Division Ring by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $\struct {R, +, \circ}$ be a Division Ring by definition 2. Then by definition: :$\forall x \in \R^*: x$ is a unit. Thus, by definition, $x$ has a product inverse $x^{-1}$. From Product Inverse in Ring is Unique it follows that: :$\forall x \in R^*: \exists! x^{-1} \in R^*: x^{-1} \circ x = x \circ x^{-1} = 1_R$ Thus $\struct {R, +, \circ}$ is a Division Ring by definition 1. {{qed|lemma}} === $(2)$ is equivalent to $(3)$ === By definition, a unit of $R$ is an element of $R$ which has a product inverse. Also by definition, a proper element of $R$ is a non-zero element of $R$ which does not have a product inverse. Hence: :a ring with unity whose non-zero elements are all units and: :a ring with unity whose non-zero elements are all not proper elements are the same thing. {{qed}} Category:Division Rings f8ydh0swavqutk5yqin3q01r478aqoo
=== Lemma === {{:Product Formula for Norms on Non-zero Rationals/Lemma}}{{qed|lemma}} Let $a = \dfrac b c$, where $b, c \in \Z_{\ne 0}$. From the Lemma, the following infinite products converge: :$\size b \displaystyle \times \prod_{p \mathop \in \Bbb P} \norm b_p = 1$ :$\size c \displaystyle \times \prod_{p \mathop \in \Bbb P} \norm c_p = 1$ From Quotient Rule for Real Sequences, the following infinite product converges: :$\displaystyle \dfrac {\size b} {\size c} \times \prod_{p \mathop \in \Bbb P } \dfrac {\norm b_p} {\norm c_p} = \dfrac 1 1 = 1 $ We have: {{begin-eqn}} {{eqn | l = \size a \times \mathop \prod_{p \mathop \in \Bbb P} \norm a_p       | r = \size {\dfrac b c} \times \prod_{p \mathop \in \Bbb P} \norm {\dfrac b c}_p }} {{eqn | r = \dfrac {\size b} {\size c} \times \prod_{p \mathop \in \Bbb P} \dfrac {\norm b_p} {\norm c_p}       | c = Norm of Quotient }} {{eqn | r = 1 }} {{end-eqn}} {{qed}}
$R$ is a ring and $R \subseteq R$. It follows by definition that $R$ is a subring of $R$. {{qed}}
From Null Ring is Ring, the null ring $\struct {\set {0_R}, +, \circ}$ is a ring. We also have that $\set {0_R}$ is a subset of $R$ Hence the result by definition of subring. {{qed}}
We have that the set of integers $\Z$ are a subset of the real numbers $\R$. The field of real numbers is, a fortiori, also a ring. Hence the result, by definition of subring. {{qed}}
Let $\struct {S, + \times}$ be the ring formed by the set of order $2$ square matrices over the real numbers $R$ under (conventional) matrix addition and (conventional) matrix multiplication. Let $T$ be the subset of $S$ consisting of the matrices of the form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ for $x \in \R$. We have that: :$\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} -y & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} x - y & 0 \\ 0 & 0 \end{bmatrix}$ and so $T$ is closed under subtraction. From Matrices of the Form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ we have that $\struct {T, \times}$ is a subsemigroup of $\struct {S, \times}$. It follows from the Subring Test that $\struct {T, + \circ}$ be a subbring of $\struct {S, + \circ}$. From (some result somewhere) $\struct {S, \times}$ has an identity $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ which is not in $\struct {T, \times}$. However, note that: {{begin-eqn}} {{eqn | l = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}       | r = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}       | c =  }} {{eqn | r = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}       | c =  }} {{end-eqn}} demonstrating that $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ is the identity of $\struct {T, \times}$. The result follows. {{qed}}
By definition, $\struct {F, +, \times}$ and $\struct {K, +, \times}$ are both rings. Thus $\struct {K, +, \times}$ is a subring of $\struct {F, +, \times}$ The result follows from Zero of Subring is Zero of Ring. {{qed}}
By definition, $\struct {K, +, \times}$ is a subset of $F$ which is a field. By definition of field, $\struct {K, +}$ and $\struct {F, +}$ are groups such that $K \subseteq F$. So, by definition, $\struct {K, +}$ is a subgroup of $\struct {F, +}$. By Identity of Subgroup, the identity of $\struct {F, +}$, which is $0$, is also the identity of $\struct {K, +}$. {{qed}}
By definition, $\struct {K, +, \times}$ is a subset of $F$ which is a field. By definition of field, $\struct {K^*, \times}$ and $\struct {F^*, \times}$ are groups such that $K \subseteq F$. So $\struct {K^*, \times}$ is a subgroup of $\struct {F^*, \times}$. By Identity of Subgroup, the identity of $\struct {F^*, \times}$, which is $1$, is also the identity of $\struct {K^*, \times}$. {{qed}}
By definition, $\struct {S, +, \times}$ is a subset of $R$ which is a ring. By definition of ring, $\struct {S, +}$ and $\struct {R, +}$ are groups such that $S \subseteq R$. So, by definition, $\struct {S, +}$ is a subgroup of $\struct {R, +}$. By Identity of Subgroup, the identity of $\struct {S, +}$, which is $0$, is also the identity of $\struct {R, +}$. {{qed}} Category:Subrings eu7v5l65ix0xuk1vhigy84hewdr458h
Let $p > 3$. Then there exists $a \in \Z: 1 < a < p-1$. Consider the sequence $\sequence {x_n} \subseteq \Q$ where $x_n = a^{p^n}$ for some $a \in \Z: 1 < a < p-1$. Let $n \in \N$. Then: :$\norm {a^{p^{n + 1} } - a^{p^n} }_p = \norm {a^{p^n} (a^{p^n \left({p - 1}\right)} - 1) }_p$ From the corollary to Euler's Theorem: :$a^{p^n \left({p - 1}\right)} - 1 \equiv 0 \pmod {p^n}$ so: :$\norm {a^{p^n} \paren {a^{p^n \paren {p - 1} } - 1} }_p \le p^{-n} \xrightarrow {n \to \infty} 0$ That is: :$\displaystyle \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$ By Characterisation of Cauchy Sequence in Non-Archimedean Norm :$\sequence {x_n }$ is a cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p }$. {{AimForCont}} $\sequence {x_n}$ converges to some $x \in \Q$. That is: :$x = \displaystyle \lim_{n \mathop \to \infty} x_n$ By Modulus of Limit on a Normed Division Ring: :$\displaystyle \lim_{n \mathop \to \infty} \norm {x_n }_p = \norm {x }_p$ Since $\forall n, p \nmid a^{p^n} = x_n$, then: :$ \norm {x_n }_p = 1$ So: :$\norm x_p =  \displaystyle \lim_{n \mathop \to \infty} \norm {x_n}_p = 1$  By Axiom (N1) of a norm on a division ring then: :$x \ne 0$. Since: {{begin-eqn}} {{eqn | l = x       | r = \lim_{n \mathop \to \infty} x_n       | c = Definition of $x$ }} {{eqn | r = \lim_{n \mathop \to \infty} x_{n + 1}       | c = Limit of Subsequence equals Limit of Sequence }} {{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n}^p       | c = Definition of $x_n$ }} {{eqn | r = \paren {\lim_{n \mathop \to \infty} x_n}^p       | c = Product rule for Normed Division Rings }} {{eqn | r = x^p       | c = Definition of $x$ }} {{end-eqn}} and $x \ne 0$ then: :$x^{p-1} = 1$ So: :$x = 1$ or $x = -1$ and so $a-x$ is an integer: :$0 < a - x < p$ It follows that: :$p \nmid \paren {a - x}$ and so: :$\norm {x - a}_p = 1$ Since $x_n \to x$ as $n \to \infty$ then: :$\exists N: \forall n > N: \norm {x_n - x}_p < \norm {x - a}_p$ That is: :$\exists N: \forall n > N: \norm {a^{p^n} - x}_p < \norm {x - a}_p$ Let $n > N$: {{begin-eqn}} {{eqn | l = \norm {x - a}_p       | r = \norm {x - a^{p^n} + a^{p^n} - a}_p }} {{eqn | o = \le       | r = \max \set {\norm {x - a^{p^n} }_p, \norm {a^{p^n} - a}_p}       | c = P-adic Norm is Non-Archimedean Norm }} {{end-eqn}} As $\norm {x - a^{p^n} }_p < \norm {x - a}_p$: {{begin-eqn}} {{eqn | l = \norm {x - a}_p       | r = \norm {a^{p^n} - a}_p       | c = Three Points in Ultrametric Space have Two Equal Distances  }} {{eqn | r = \norm a_p \norm {a^{p^n - 1} - 1}_p       | c = Axiom (N2) of a norm on a division ring }} {{eqn | r = \norm {a^{p^n - 1} - 1}_p       | c = as $\norm a_p = 1$ }} {{eqn | o = <       | r = 1       | c = corollary 4 to Fermat's Little Theorem }} {{end-eqn}} This contradicts the earlier assertion that $\norm {x - a}_p = 1$. In conclusion: :$\sequence {x_n}$ is a Cauchy sequence that does not converge in $\struct {\Q, \norm {\,\cdot\,}_p }$.
Let that $x \divides y$. Then by definition of divisor:  {{begin-eqn}} {{eqn | l = x \divides y       | o = \leadsto       | r = \exists t \in D: y = t x       | c = {{Defof|Divisor of Ring Element}} }} {{eqn | o = \leadsto       | r = y \in \ideal x       | c = {{Defof|Principal Ideal of Ring}} }} {{eqn | o = \leadsto       | r = \ideal y \subseteq \ideal x       | c = {{Defof|Principal Ideal of Ring}}: $\ideal y$ is the smallest ideal containing $y$ }} {{end-eqn}} Conversely: {{begin-eqn}} {{eqn | l = \ideal y \subseteq \ideal x       | o = \leadsto       | r = y \in \ideal x       | c = as $y \in \ideal y$ }} {{eqn | o = \leadsto       | r = \exists t \in D: y = t x       | c = {{Defof|Principal Ideal of Ring}} }} {{eqn | o = \leadsto       | r = x \divides y       | c = {{Defof|Divisor of Ring Element}} }} {{end-eqn}} So: :$x \divides y \iff \ideal y \subseteq \ideal x$ {{qed}}
{{begin-eqn}} {{eqn | l = x \in U_D       | o = \leadsto       | r = \exists u \in U_D: u \in \ideal x       | c =  }} {{eqn | o = \leadsto       | r = \ideal x = D       | c = Ideal of Unit is Whole Ring }} {{end-eqn}} Conversely: {{begin-eqn}} {{eqn | l = \ideal x = D       | o = \leadsto       | r = 1_D \in \ideal x       | c =  }} {{eqn | o = \leadsto       | r = \exists t \in D: 1 = t x       | c =  }} {{eqn | o = \leadsto       | r = u \in U_D       | c =  }} {{end-eqn}} {{qed}}
By Poles of Gamma Function, $\Gamma$ has simple poles at the non-positive integers, so $-n$ is a simple pole of $\Gamma$.  Then: {{begin-eqn}} {{eqn	| l = \Res \Gamma {-n} 	| r = \lim_{z \mathop \to -n} \paren {z + n} \map \Gamma z 	| c = Residue at Simple Pole }} {{eqn	| r = \lim_{z \mathop \to -n} \paren {z + n} \frac {\map \Gamma {z + n + 1} } {z \paren {z + 1} \ldots \paren {z + n} } }} {{eqn	| r = \lim_{z \mathop \to -n} \frac {\map \Gamma {z + n + 1} } {z \paren {z + 1} \ldots \paren {z + n - 1} } }} {{eqn	| r = \frac {\map \Gamma 1} {-n \paren {-n + 1} \ldots \paren {-1} } }} {{eqn	| r = \frac 1 {\paren {-1}^n \paren {n \paren {n - 1} \ldots \paren 1} } 	| c = $\map \Gamma 1 = 1$ }} {{eqn	| r = \frac 1 {\paren {-1}^{n} n!} 	| c = {{Defof|Factorial}} }} {{eqn	| r = \frac {\paren {-1}^n} {n!} }} {{end-eqn}} {{qed}} Category:Gamma Function 33l04x5xfrse8mv2w8ojn26725r27l2
{{AimForCont}} that $P \nmid L$. Then by the Division Theorem we have $L = q P + r$ where $q \in \Z$ and $0 < r < P$. And so: {{begin-eqn}} {{eqn | l = \map f {x + L}       | r = \map f {x + \paren {q P + r} } }} {{eqn | r = \map f {\paren {x + r} + q P} }} {{eqn | r = \map f {x + r}       | c = General Periodicity Property }} {{eqn | r = \map f x       | c = {{Defof|Periodic Element}} }} {{end-eqn}} But then $r$ is a periodic element of $f$ that is less than $P$. Therefore $P$ cannot be the period of $f$. The result follows from Proof by Contradiction. {{qed}} Category:Periodic Functions 8ozj1819qzmmkun1rf2fdo4ry2m0wqc
=== $(1)$ is Equivalent to $(2)$ === {{:Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 2}} === $(1)$ is Equivalent to $(3)$ === {{:Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 3}} Category:Integral Domains Category:Associates Category:Equivalence of Definitions of Associate in Integral Domain lbb4s0ifmkn6vtxm5u7br1j16jo7mv3
We are to show that: :$x \divides y \text{ and } y \divides x \iff \ideal x = \ideal y$ Thus: {{begin-eqn}} {{eqn | o =        | r = x \divides y \text{ and } y \divides x       | c = {{Defof|Associate|subdef = Integral Domain|index = 1|Associate in Integral Domain}} }} {{eqn | o = \leadstoandfrom       | r = \ideal y \subseteq \ideal x \text{ and } \ideal x \subseteq \ideal y       | c = Element in Integral Domain is Divisor iff Principal Ideal is Superset }} {{eqn | o = \leadstoandfrom       | r = \ideal x = \ideal y       | c = {{Defof|Set Equality|index = 2}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = u \circ x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | r = u^{-1} \circ y       | c = {{Defof|Unit of Ring}} }} {{end-eqn}} By the definition of divisor: :$x \divides y$ and $y \divides x$ {{qed|lemma}} Let $x \divides y$ and $y \divides x$. Then $\exists s, t \in D$ such that: :$(1): \quad y = t \circ x$ and: :$(2): \quad x = s \circ y$ If either $x = 0_D$ or $y = 0_D$, then so must be the other (as an integral domain has no zero divisors by definition). So $x = 1_D \circ y$ and $y = 1_D \circ x$, and the result holds. Otherwise: {{begin-eqn}} {{eqn | l = 1_D \circ x       | r = x       | c = {{Defof|Unity of Ring}} }} {{eqn | r = s \circ y       | c = from $(2)$ }} {{eqn | r = s \circ \paren {t \circ x}       | c = from $(1)$ }} {{eqn | r = \paren {s \circ t} \circ x       | c = {{Defof|Associative Operation}} }} {{end-eqn}} So: :$s \circ t = 1_D$ and both $s \in U_D$ and $t \in U_D$. The result follows. {{qed}}
Let $0_D$ and $1_D$ be the zero and unity respectively of $D$. Let $J$ be the set of all linear combinations in $D$ of $\set {a_1, a_2, \dotsc, a_n}$. From Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal: :$J = \ideal x$ for some $x \in D$, where $\ideal x$ denotes the principal ideal generated by $x$. We have that each $a_i$ can be expressed as a linear combination of $\set {a_1, a_2, \dotsc, a_n}$: :$a_i = 0_D a_1 + 0_d a_2 + \dotsb + 1_D a_i + \dotsb + 0_D a_n$ Thus: :$\forall i \in \set {0, 1, \dotsc, n}: a_i \in J$ and so by definition of $J$: :$\forall i \in \set {0, 1, \dotsc, n}: a_i = t_i x$ for some $t_i \in D$. Thus $x$ is a common divisor of $a_1, a_2, \dotsc, a_n$. As $x \in \ideal x = J$, we have: :$x = c_1 a_1 + c_2 a_2 + \dotsb + c_n a_n$ for some $c_1, c_2, \dotsc, c_n \in D$. Thus every common divisor of $a_1, a_2, \dotsc, a_n$ also is a divisor of $x$. Thus $x$ is a greatest common divisor of $a_1, a_2, \dotsc, a_n$. {{qed}}
Let the unity of $D$ be $1_D$. By definition of principal ideal: :$\ideal a = \displaystyle \set {\sum_{i \mathop = 1}^n r_i \circ a \circ s_i: n \in \N, r_i, s_i \in D}$ Let $x, y \in J$. By definition of linear combination: {{begin-eqn}} {{eqn | l = x       | r = \sum_{i \mathop = 1}^n r_i \circ a_i       | c = for some $n \in \N$ and for some $r_i \in D$ where $i \in \set {1, 2, \dotsc, n}$ }} {{eqn | r = r_1 \circ a_1 + r_2 \circ a_2 + \dotsb + r_n \circ a_n       | c = for some $r_1, r_2, \dotsc, r_n \in D$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = y       | r = \sum_{i \mathop = 1}^n s_i \circ a_i       | c = for some $n \in \N$ and for some $s_i \in D$ where $i \in \set {1, 2, \dotsc, n}$ }} {{eqn | r = s_1 \circ a_1 + s_2 \circ a_2 + \dotsb + s_n \circ a_n       | c = for some $s_1, s_2, \dotsc, s_n \in R$ }} {{eqn | ll= \leadsto       | l = -y       | r = -\sum_{i \mathop = 1}^n s_i \circ a_i       | c =  }} {{eqn | r = -1_D \times \sum_{i \mathop = 1}^n s_i \circ a_i       | c = Product with Ring Negative: Corollary }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {-1_D} \times \paren {s_i \circ a_i}       | c = {{Ring-axiom|D}} }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {-\paren {s_i \circ a_i} }       | c = Product with Ring Negative: Corollary }} {{eqn | r = \sum_{i \mathop = 1}^n s_i \circ \paren {-a_i}       | c = Product with Ring Negative }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = x + \paren {-y}       | r = \sum_{i \mathop = 1}^n r_i \circ a_i + \sum_{i \mathop = 1}^n s_i \circ \paren {-a_i}       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i \circ a_i + s_i \circ \paren {-a_i} }       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i \circ a_i + \paren {-\paren {s_i \circ a_i} } }       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i \circ a_i + \paren {-s_i} \circ a_i}       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i + \paren {-s_i} } \circ a_i       | c = {{Ring-axiom|D}} }} {{eqn | r = \sum_{i \mathop = 1}^n t_i \circ a_i       | c = where $t_i - r_1 + \paren {-s_i}$ }} {{eqn | o = \in       | r = J       | c = as $t_i \in D$ }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | l = x \circ y       | r = \paren {\sum_{i \mathop = 1}^n r_i \circ a_i} \circ \paren {\sum_{i \mathop = 1}^n s_i \circ \a_i }       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {t_i \circ a_i}       | c = where $t_i \in D$ for $i \in \set {1, 2, \dotsc, n}$ }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {a_i \circ t_i}       | c = as $\circ$ is commutative in an integral domain }} {{end-eqn}} {{explain|There exists (or ought to) some convolution result which proves the above -- I just haven't found it yet.}} Thus by the Test for Ideal, $J$ is an ideal of $D$. As $D$ is a principal ideal domain, it follows that $J$ is a principal ideal. Thus by definition of principal ideal: :$J = \ideal x$ for some $x \in D$. {{qed}}
From Finite Set of Elements in Principal Ideal Domain has GCD we have that at least one such greatest common divisor exists. So, let $y_1$ and $y_2$ be greatest common divisors of $S$. Then: {{begin-eqn}} {{eqn | l = y_1       | o = \divides       | r = y_2       | c = as $y_2$ is a greatest common divisor }} {{eqn | l = y_2       | o = \divides       | r = y_1       | c = as $y_1$ is a greatest common divisor }} {{end-eqn}} Thus we have: :$y_1 \divides y_2$ and $y_2 \divides y_1$ where $\divides$ denotes divisibility. Hence the result, by definition of associates. {{qed}}
From Finite Set of Elements in Principal Ideal Domain has GCD we have that at least one such greatest common divisor exists. So, let $y$ be a greatest common divisor of $S$. Let $J$ be the set of all linear combinations in $D$ of $\set {a_1, a_2, \dotsc, a_n}$. From Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal: :$J = \ideal x$ for some $x \in D$, where $\ideal x$ denotes the principal ideal generated by $x$. From Finite Set of Elements in Principal Ideal Domain has GCD, $x$ is a greatest common divisor of $S$. From Greatest Common Divisors in Principal Ideal Domain are Associates, $y$ is an associate of $x$. By definition of associate: :$\ideal y = \ideal x$ Therefore: :$y \in J$ and so by definition, $y$ is expressible in the form: :$y = d_1 a_1 + d_2 a_2 + \dotsb + d_n a_n$ where $d_1, d_2, \dotsc, d_n \in D$. {{qed}}
From the definition of the Poisson distribution, $X$ has probability mass function: :$\map \Pr {X = n} = \dfrac {\lambda^n e^{-\lambda} } {n!}$ From the definition of a moment generating function: :$\displaystyle \map {M_X} t = \expect {e^{t X} } = \sum_{n \mathop = 0}^\infty \map \Pr {X = n} e^{t n}$ So:  {{begin-eqn}} {{eqn	| l = \map {M_X} t 	| r = \sum_{n \mathop = 0}^\infty \frac {\lambda^n e^{-\lambda} } {n!} e^{t n} }} {{eqn	| r = e^{-\lambda} \sum_{n \mathop = 0}^\infty \frac {\paren {\lambda e^t}^n} {n!} }} {{eqn	| r = e^{-\lambda} e^{\lambda e^t} 	| c = Power Series Expansion for Exponential Function }} {{eqn	| r = e^{\lambda \paren {e^t - 1} } 	| c = Exponential of Sum }} {{end-eqn}} {{qed}} Category:Moment Generating Functions Category:Poisson Distribution cqmgdzhn5lv8husq2qyzlg4f4qvlsvw