id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-15600 | Order of Boolean Group is Power of 2 | Let $G$ be a Boolean group.
Let $\order G$ denote the order of $G$.
Then:
:$\order G = 2^n$
where $n \in \Z_{\ge 0}$ is a positive integer. | The case where $n = 0$ is clear:
:$\order {\set e} = 1$
and $e^2 = e$.
{{AimForCont}} $\order G = m \times 2^k$ for some odd integer $m$.
Then $m$ itself has an odd prime $p$ as a integer (which may of course equal $m$ if $m$ is itself prime).
Then by Cauchy's Lemma (Group Theory) there exists $g \in G$ such that $\ord... | Let $G$ be a [[Definition:Boolean Group|Boolean group]].
Let $\order G$ denote the [[Definition:Order of Group|order]] of $G$.
Then:
:$\order G = 2^n$
where $n \in \Z_{\ge 0}$ is a [[Definition:Positive Integer|positive integer]]. | The case where $n = 0$ is clear:
:$\order {\set e} = 1$
and $e^2 = e$.
{{AimForCont}} $\order G = m \times 2^k$ for some [[Definition:Odd Integer|odd integer]] $m$.
Then $m$ itself has an [[Definition:Odd Prime|odd prime]] $p$ as a [[Definition:Divisor of Integer|integer]] (which may of course equal $m$ if $m$ is it... | Order of Boolean Group is Power of 2 | https://proofwiki.org/wiki/Order_of_Boolean_Group_is_Power_of_2 | https://proofwiki.org/wiki/Order_of_Boolean_Group_is_Power_of_2 | [
"Boolean Groups"
] | [
"Definition:Boolean Group",
"Definition:Order of Structure",
"Definition:Positive/Integer"
] | [
"Definition:Odd Integer",
"Definition:Odd Prime",
"Definition:Divisor (Algebra)/Integer",
"Definition:Odd Prime",
"Cauchy's Lemma (Group Theory)",
"Definition:Prime Factor",
"Definition:Odd Integer"
] |
proofwiki-15601 | Subgroup of Index 2 contains all Squares of Group Elements | Let $G$ be a group.
Let $H$ be a subgroup of $G$ whose index is $2$.
Then:
:$\forall x \in G: x^2 \in H$ | By Subgroup of Index 2 is Normal, $H$ is normal in $G$.
Hence the quotient group $G / H$ exists.
Then we have:
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = \paren {x^2} H
| r = \paren {x H}^2
| c =
}}
{{eqn | r = H
| c = as $G / H$ is of order $2$
}}
{{end-eqn}}
{{qed}} | Let $G$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ whose [[Definition:Index of Subgroup|index]] is $2$.
Then:
:$\forall x \in G: x^2 \in H$ | By [[Subgroup of Index 2 is Normal]], $H$ is [[Definition:Normal Subgroup|normal]] in $G$.
Hence the [[Definition:Quotient Group|quotient group]] $G / H$ exists.
Then we have:
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = \paren {x^2} H
| r = \paren {x H}^2
| c =
}}
{{eqn | r = H
| c = as $... | Subgroup of Index 2 contains all Squares of Group Elements | https://proofwiki.org/wiki/Subgroup_of_Index_2_contains_all_Squares_of_Group_Elements | https://proofwiki.org/wiki/Subgroup_of_Index_2_contains_all_Squares_of_Group_Elements | [
"Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Index of Subgroup"
] | [
"Subgroup of Index 2 is Normal",
"Definition:Normal Subgroup",
"Definition:Quotient Group",
"Definition:Order of Structure"
] |
proofwiki-15602 | Subgroup of Index 3 does not necessarily contain all Cubes of Group Elements | Let $G$ be a group.
Let $H$ be a subgroup of $G$ whose index is $3$.
Then it is not necessarily the case that:
:$\forall x \in G: x^3 \in H$ | Proof by Counterexample:
Consider $S_3$, the symmetric group on $3$ letters.
From Subgroups of Symmetric Group on 3 Letters, the subsets of $S_3$ which form subgroups of $S_3$ are:
{{begin-eqn}}
{{eqn | o =
| r = S_3
}}
{{eqn | o =
| r = \set e
}}
{{eqn | o =
| r = \set {e, \tuple {123}, \tuple {13... | Let $G$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ whose [[Definition:Index of Subgroup|index]] is $3$.
Then it is not necessarily the case that:
:$\forall x \in G: x^3 \in H$ | [[Proof by Counterexample]]:
Consider $S_3$, the [[Definition:Symmetric Group on 3 Letters|symmetric group on $3$ letters]].
From [[Subgroups of Symmetric Group on 3 Letters]], the [[Definition:Subset|subsets]] of $S_3$ which form [[Definition:Subgroup|subgroups]] of $S_3$ are:
{{begin-eqn}}
{{eqn | o =
| r =... | Subgroup of Index 3 does not necessarily contain all Cubes of Group Elements | https://proofwiki.org/wiki/Subgroup_of_Index_3_does_not_necessarily_contain_all_Cubes_of_Group_Elements | https://proofwiki.org/wiki/Subgroup_of_Index_3_does_not_necessarily_contain_all_Cubes_of_Group_Elements | [
"Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Index of Subgroup"
] | [
"Proof by Counterexample",
"Symmetric Group on 3 Letters",
"Symmetric Group on 3 Letters/Subgroups",
"Definition:Subset",
"Definition:Subgroup",
"Definition:Index of Subgroup"
] |
proofwiki-15603 | Center of Non-Abelian Group of Order pq is Trivial | Let $p$ and $q$ be distinct prime numbers.
Let $G$ be a non-abelian group of order $p q$ whose identity is $e$.
Then the center of $G$ is trivial:
:$\map Z G = \set e$ | From Center of Group is Normal Subgroup, $\map Z G$ is a normal subgroup of $G$.
By Lagrange's Theorem, the order of $\map Z G$ is either $1$, $p$, $q$ or $p q$.
Because $G$ is not abelian, $G \ne \map Z G$.
Hence $\order {\map Z G} \ne p q$.
From Quotient of Group by Center Cyclic implies Abelian:
:$G / \map Z G$ cann... | Let $p$ and $q$ be [[Definition:Distinct Elements|distinct]] [[Definition:Prime Number|prime numbers]].
Let $G$ be a non-[[Definition:Abelian Group|abelian]] [[Definition:Group|group]] of [[Definition:Order of Group|order]] $p q$ whose [[Definition:Identity Element|identity]] is $e$.
Then the [[Definition:Center of ... | From [[Center of Group is Normal Subgroup]], $\map Z G$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
By [[Definition:Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], the [[Definition:Order of Group|order]] of $\map Z G$ is either $1$, $p$, $q$ or $p q$.
Because $G$ is not [[Definition:Abelian Gr... | Center of Non-Abelian Group of Order pq is Trivial | https://proofwiki.org/wiki/Center_of_Non-Abelian_Group_of_Order_pq_is_Trivial | https://proofwiki.org/wiki/Center_of_Non-Abelian_Group_of_Order_pq_is_Trivial | [
"Centers of Groups",
"Groups of Order p q"
] | [
"Definition:Distinct/Plural",
"Definition:Prime Number",
"Definition:Abelian Group",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Trivial Subgroup"
] | [
"Center of Group is Normal Subgroup",
"Definition:Normal Subgroup",
"Lagrange's Theorem (Group Theory)",
"Definition:Order of Structure",
"Definition:Abelian Group",
"Quotient of Group by Center Cyclic implies Abelian",
"Definition:Cyclic Group",
"Definition:Trivial Subgroup",
"Definition:Centralize... |
proofwiki-15604 | Number of Elements of Order p in Group of Order pq is Multiple of q | Let $p$ and $q$ be distinct prime numbers.
Let $G$ be a non-abelian group of order $p q$.
Then the number of elements of $G$ of order $p$ is a multiple of $q$. | Let $x$ be an element of $G$ of order $p$.
From Center of Non-Abelian Group of Order pq is Trivial:
:$p \notin \map Z G$
where $\map Z G$ denotes the center of $G$.
As $x \notin \map Z G$:
:$\map C x \subsetneq G$
where $\map C x$ is the centralizer of $x$.
From Order of Element divides Order of Centralizer:
:$p \divid... | Let $p$ and $q$ be [[Definition:Distinct Elements|distinct]] [[Definition:Prime Number|prime numbers]].
Let $G$ be a non-[[Definition:Abelian Group|abelian]] [[Definition:Group|group]] of [[Definition:Order of Group|order]] $p q$.
Then the number of [[Definition:Element|elements]] of $G$ of order $p$ is a [[Definiti... | Let $x$ be an [[Definition:Element|element]] of $G$ of order $p$.
From [[Center of Non-Abelian Group of Order pq is Trivial]]:
:$p \notin \map Z G$
where $\map Z G$ denotes the [[Definition:Center of Group|center]] of $G$.
As $x \notin \map Z G$:
:$\map C x \subsetneq G$
where $\map C x$ is the [[Definition:Centraliz... | Number of Elements of Order p in Group of Order pq is Multiple of q | https://proofwiki.org/wiki/Number_of_Elements_of_Order_p_in_Group_of_Order_pq_is_Multiple_of_q | https://proofwiki.org/wiki/Number_of_Elements_of_Order_p_in_Group_of_Order_pq_is_Multiple_of_q | [
"Groups of Order p q"
] | [
"Definition:Distinct/Plural",
"Definition:Prime Number",
"Definition:Abelian Group",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Element",
"Definition:Integral Multiple/Real Numbers"
] | [
"Definition:Element",
"Center of Non-Abelian Group of Order pq is Trivial",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Centralizer/Group Element",
"Order of Element divides Order of Centralizer",
"Lagrange's Theorem (Group Theory)",
"Number of Conjugates is Number of Cosets of Centralizer... |
proofwiki-15605 | Intersection of Normal Subgroup with Center in p-Group | Let $p$ be a prime number
Let $G$ be a $p$-group.
Let $N$ be a non-trivial normal subgroup of $G$.
Let $\map Z G$ denote the center of $G$.
Then:
:$N \cap \map Z G$ is a non-trivial normal subgroup of $G$. | First we note that:
:Center of Group is Normal Subgroup
and from Intersection of Normal Subgroups is Normal:
:$N \cap \map Z G$ is normal in $G$.
Suppose $G$ is abelian.
By definition:
:$\map Z G = G$
Then:
:$N \cap \map Z G = N$
which is non-trivial.
From Prime Group is Cyclic and Cyclic Group is Abelian, this will al... | Let $p$ be a [[Definition:Prime Number|prime number]]
Let $G$ be a [[Definition:P-Group|$p$-group]].
Let $N$ be a [[Definition:Non-Trivial Subgroup|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Let $\map Z G$ denote the [[Definition:Center of Group|center]] of $G$.
Then:
:$N \cap \map Z G$ i... | First we note that:
:[[Center of Group is Normal Subgroup]]
and from [[Intersection of Normal Subgroups is Normal]]:
:$N \cap \map Z G$ is [[Definition:Normal Subgroup|normal]] in $G$.
Suppose $G$ is [[Definition:Abelian Group|abelian]].
By definition:
:$\map Z G = G$
Then:
:$N \cap \map Z G = N$
which is [[Defini... | Intersection of Normal Subgroup with Center in p-Group | https://proofwiki.org/wiki/Intersection_of_Normal_Subgroup_with_Center_in_p-Group | https://proofwiki.org/wiki/Intersection_of_Normal_Subgroup_with_Center_in_p-Group | [
"Normal Subgroups",
"P-Groups"
] | [
"Definition:Prime Number",
"Definition:P-Group",
"Definition:Non-Trivial Subgroup",
"Definition:Normal Subgroup",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Non-Trivial Subgroup",
"Definition:Normal Subgroup"
] | [
"Center of Group is Normal Subgroup",
"Intersection of Normal Subgroups is Normal",
"Definition:Normal Subgroup",
"Definition:Abelian Group",
"Definition:Non-Trivial Subgroup",
"Prime Group is Cyclic",
"Cyclic Group is Abelian",
"Definition:Abelian Group",
"Center of Group of Prime Power Order is No... |
proofwiki-15606 | Normal Subgroup of p-Group of Order p is Subset of Center | Let $p$ be a prime number.
Let $G$ be a $p$-group.
Let $N$ be a normal subgroup of $G$ of order $p$.
Then:
:$N \subseteq \map Z G$
where $\map Z G$ denotes the center of $G$. | From Intersection of Normal Subgroup with Center in p-Group:
:$\order {N \cap \map Z G} > 1$
From Intersection of Subgroups is Subgroup, $N \cap \map Z G$ is a subgroup of $N$.
It follows from Lagrange's Theorem that:
:$\order {N \cap \map Z G} = p$
and so:
:$N \cap \map Z G = N$
But from Intersection of Subgroups is S... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $G$ be a [[Definition:P-Group|$p$-group]].
Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$ of [[Definition:Order of Group|order]] $p$.
Then:
:$N \subseteq \map Z G$
where $\map Z G$ denotes the [[Definition:Center of Group|center]] of $G$... | From [[Intersection of Normal Subgroup with Center in p-Group]]:
:$\order {N \cap \map Z G} > 1$
From [[Intersection of Subgroups is Subgroup]], $N \cap \map Z G$ is a [[Definition:Subgroup|subgroup]] of $N$.
It follows from [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]] that:
:$\order {N \cap \map Z G} = p... | Normal Subgroup of p-Group of Order p is Subset of Center | https://proofwiki.org/wiki/Normal_Subgroup_of_p-Group_of_Order_p_is_Subset_of_Center | https://proofwiki.org/wiki/Normal_Subgroup_of_p-Group_of_Order_p_is_Subset_of_Center | [
"Normal Subgroups",
"Centers of Groups",
"P-Groups"
] | [
"Definition:Prime Number",
"Definition:P-Group",
"Definition:Normal Subgroup",
"Definition:Order of Structure",
"Definition:Center (Abstract Algebra)/Group"
] | [
"Intersection of Normal Subgroup with Center in p-Group",
"Intersection of Subgroups is Subgroup",
"Definition:Subgroup",
"Lagrange's Theorem (Group Theory)",
"Intersection of Subgroups is Subgroup",
"Definition:Subgroup",
"Definition:Subgroup"
] |
proofwiki-15607 | Non-Abelian Group of Order p Cubed has Exactly One Normal Subgroup of Order p | Let $p$ be a prime number.
Let $G$ be a non-abelian group of order $p^3$.
Then $G$ contains exactly one normal subgroup of order $p$. | From Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is not the trivial subgroup.
From Quotient of Group by Center Cyclic implies Abelian, $G / \map G Z$ cannot be cyclic and non-trivial.
Thus $\order {G / \map G Z}$ cannot be $p$ and so must be $p^2$.
Thus $\order {\map G Z} = p$.
Let $N$ be a normal s... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $G$ be a non-[[Definition:Abelian Group|abelian group]] of [[Definition:Order of Group|order]] $p^3$.
Then $G$ contains [[Definition:Unique|exactly one]] [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $p$. | From [[Center of Group of Prime Power Order is Non-Trivial]], $\map Z G$ is not the [[Definition:Trivial Subgroup|trivial subgroup]].
From [[Quotient of Group by Center Cyclic implies Abelian]], $G / \map G Z$ cannot be [[Definition:Cyclic Group|cyclic]] and [[Definition:Non-Trivial Subgroup|non-trivial]].
Thus $\ord... | Non-Abelian Group of Order p Cubed has Exactly One Normal Subgroup of Order p | https://proofwiki.org/wiki/Non-Abelian_Group_of_Order_p_Cubed_has_Exactly_One_Normal_Subgroup_of_Order_p | https://proofwiki.org/wiki/Non-Abelian_Group_of_Order_p_Cubed_has_Exactly_One_Normal_Subgroup_of_Order_p | [
"P-Groups"
] | [
"Definition:Prime Number",
"Definition:Abelian Group",
"Definition:Order of Structure",
"Definition:Unique",
"Definition:Normal Subgroup",
"Definition:Order of Structure"
] | [
"Center of Group of Prime Power Order is Non-Trivial",
"Definition:Trivial Subgroup",
"Quotient of Group by Center Cyclic implies Abelian",
"Definition:Cyclic Group",
"Definition:Non-Trivial Subgroup",
"Definition:Normal Subgroup",
"Definition:Order of Structure",
"Normal Subgroup of p-Group of Order ... |
proofwiki-15608 | Quaternion Group has Normal Subgroup without Complement | Let $Q$ denote the quaternion group.
There exists a normal subgroup of $Q$ which has no complement. | From Subgroups of Quaternion Group:
{{:Subgroups of Quaternion Group}}
From Quaternion Group is Hamiltonian, all these subgroups are normal.
For two subgroups to be complementary, they need to have an intersection which is trivial.
However, apart from $\set e$ itself, all these subgroups contain $a^2$.
Hence none of th... | Let $Q$ denote the [[Definition:Quaternion Group|quaternion group]].
There exists a [[Definition:Normal Subgroup|normal subgroup]] of $Q$ which has no [[Definition:Complement of Subgroup|complement]]. | From [[Subgroups of Quaternion Group]]:
{{:Subgroups of Quaternion Group}}
From [[Quaternion Group is Hamiltonian]], all these [[Definition:Subgroup|subgroups]] are [[Definition:Normal Subgroup|normal]].
For two [[Definition:Subgroup|subgroups]] to be [[Definition:Complement of Subgroup|complementary]], they need t... | Quaternion Group has Normal Subgroup without Complement | https://proofwiki.org/wiki/Quaternion_Group_has_Normal_Subgroup_without_Complement | https://proofwiki.org/wiki/Quaternion_Group_has_Normal_Subgroup_without_Complement | [
"Subgroup Complements",
"Quaternion Group"
] | [
"Definition:Dicyclic Group/Quaternion Group",
"Definition:Normal Subgroup",
"Definition:Complement of Subgroup"
] | [
"Quaternion Group/Subgroups",
"Quaternion Group is Hamiltonian",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Subgroup",
"Definition:Complement of Subgroup",
"Definition:Set Intersection",
"Definition:Trivial Subgroup",
"Definition:Subgroup",
"Definition:Subgroup",
"Definitio... |
proofwiki-15609 | Non-Abelian Order 2p Group has Order p Element | Let $p$ be an odd prime.
Let $G$ be a non-abelian group of order $2 p$.
Then $G$ has at least one element of order $p$. | By Lagrange's Theorem, all the elements of $G$ have orders $1$, $2$, $p$ or $2p$.
From Identity is Only Group Element of Order 1, $2 p - 1$ elements of $G$ have orders greater than $1$.
From Cyclic Group is Abelian, $G$ is not the cyclic group $2 p$.
If $g \in G$ was of order $2 p$ then $g$ would generate the cyclic gr... | Let $p$ be an [[Definition:Odd Prime|odd prime]].
Let $G$ be a non-[[Definition:Abelian Group|abelian group]] of [[Definition:Order of Structure|order $2 p$]].
Then $G$ has at least one [[Definition:Element|element]] of [[Definition:Order of Group Element|order $p$]]. | By [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], all the [[Definition:Element|elements]] of $G$ have [[Definition:Order of Group Element|orders]] $1$, $2$, $p$ or $2p$.
From [[Identity is Only Group Element of Order 1]], $2 p - 1$ [[Definition:Element|elements]] of $G$ have [[Definition:Order of Group Elem... | Non-Abelian Order 2p Group has Order p Element | https://proofwiki.org/wiki/Non-Abelian_Order_2p_Group_has_Order_p_Element | https://proofwiki.org/wiki/Non-Abelian_Order_2p_Group_has_Order_p_Element | [
"Order of Group Elements",
"Groups of Order 2 p"
] | [
"Definition:Odd Prime",
"Definition:Abelian Group",
"Definition:Order of Structure",
"Definition:Element",
"Definition:Order of Group Element"
] | [
"Lagrange's Theorem (Group Theory)",
"Definition:Element",
"Definition:Order of Group Element",
"Identity is Only Group Element of Order 1",
"Definition:Element",
"Definition:Order of Group Element",
"Cyclic Group is Abelian",
"Definition:Cyclic Group",
"Definition:Order of Group Element",
"Defini... |
proofwiki-15610 | Fourth Power Modulo 5 | Let $n \in \Z$ be an integer.
Then:
:$n^4 \equiv m \pmod 5$
where $m \in \set {0, 1}$. | 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= \equ... | Let $n \in \Z$ be an [[Definition:Integer|integer]].
Then:
:$n^4 \equiv m \pmod 5$
where $m \in \set {0, 1}$. | 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
| ... | Fourth Power Modulo 5 | https://proofwiki.org/wiki/Fourth_Power_Modulo_5 | https://proofwiki.org/wiki/Fourth_Power_Modulo_5 | [
"Fourth Powers"
] | [
"Definition:Integer"
] | [
"Congruence of Powers",
"Category:Fourth Powers"
] |
proofwiki-15611 | Subgroup of Order p in Group of Order 2p is Normal/Corollary | Let $G$ be non-abelian.
Every element of $G \setminus K$ is of order $2$, and:
:$\forall b \in G \setminus K: b a b^{-1} = a^{-1}$ | 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 $G$ be non-[[Definition:Abelian Group|abelian]].
Every [[Definition:Element|element]] of $G \setminus K$ is of [[Definition:Order of Group Element|order]] $2$, and:
:$\forall b \in G \setminus K: b a b^{-1} = a^{-1}$ | By [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], the [[Definition:Element|elements]] of $G \setminus K$ can be of [[Definition:Order of Group Element|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-[[Definition:Abelia... | Subgroup of Order p in Group of Order 2p is Normal/Corollary | https://proofwiki.org/wiki/Subgroup_of_Order_p_in_Group_of_Order_2p_is_Normal/Corollary | https://proofwiki.org/wiki/Subgroup_of_Order_p_in_Group_of_Order_2p_is_Normal/Corollary | [
"Groups of Order 2 p"
] | [
"Definition:Abelian Group",
"Definition:Element",
"Definition:Order of Group Element"
] | [
"Lagrange's Theorem (Group Theory)",
"Definition:Element",
"Definition:Order of Group Element",
"Identity is Only Group Element of Order 1",
"Definition:Abelian Group",
"Cyclic Group is Abelian",
"Definition:Cyclic Group",
"Subgroup of Index 2 contains all Squares of Group Elements",
"Definition:Ord... |
proofwiki-15612 | Subgroup of Order p in Group of Order 2p is Normal | Let $p$ be an odd prime.
Let $G$ be a group of order $2 p$.
Let $a \in G$ be of order $p$.
Let $K = \gen a$ be the subgroup of $G$ generated by $a$.
Then $K$ is normal in $G$. | 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 ... | Let $p$ be an [[Definition:Odd Prime|odd prime]].
Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Structure|order $2 p$]].
Let $a \in G$ be of [[Definition:Order of Group Element|order]] $p$.
Let $K = \gen a$ be the [[Definition:Generated Subgroup|subgroup of $G$ generated]] by $a$.
Then $K$ is [[... | The result [[Non-Abelian Order 2p Group has Order p Element]] demonstrates that such an [[Definition:Element|element]] $a$ exists in $G$.
By definition of [[Definition:Generator of Cyclic Group|generator of cyclic group]], $K$ is the [[Definition:Cyclic Group|cyclic group]] $C_p$ of [[Definition:Order of Group Element... | Subgroup of Order p in Group of Order 2p is Normal | https://proofwiki.org/wiki/Subgroup_of_Order_p_in_Group_of_Order_2p_is_Normal | https://proofwiki.org/wiki/Subgroup_of_Order_p_in_Group_of_Order_2p_is_Normal | [
"Groups of Order 2 p"
] | [
"Definition:Odd Prime",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Order of Group Element",
"Definition:Generated Subgroup",
"Definition:Normal Subgroup"
] | [
"Non-Abelian Order 2p Group has Order p Element",
"Definition:Element",
"Definition:Cyclic Group/Generator",
"Definition:Cyclic Group",
"Definition:Order of Group Element",
"Lagrange's Theorem (Group Theory)",
"Definition:Index of Subgroup",
"Subgroup of Index 2 is Normal"
] |
proofwiki-15613 | Inner Automorphisms form Subgroup of Automorphism Group | Let $G$ be a group.
Then the set $\Inn G$ of all inner automorphisms of $G$ forms a normal subgroup of the automorphism group $\Aut G$ of $G$:
:$\Inn G \le \Aut G$ | Let $G$ be a group whose identity is $e$.
Let $\kappa_x: G \to G$ be the inner automorphism defined as:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
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 \I... | Let $G$ be a [[Definition:Group|group]].
Then the [[Definition:Set|set]] $\Inn G$ of all [[Definition:Inner Automorphism|inner automorphisms]] of $G$ forms a [[Definition:Subgroup|normal subgroup]] of the [[Definition:Automorphism Group of Group|automorphism group]] $\Aut G$ of $G$:
:$\Inn G \le \Aut G$ | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\kappa_x: G \to G$ be the [[Definition:Inner Automorphism|inner automorphism]] defined as:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
We see that:
:$\Inn G \ne \O$
as $\kappa_x$ is defined for all $x \in G$.
... | Inner Automorphisms form Subgroup of Automorphism Group | https://proofwiki.org/wiki/Inner_Automorphisms_form_Subgroup_of_Automorphism_Group | https://proofwiki.org/wiki/Inner_Automorphisms_form_Subgroup_of_Automorphism_Group | [
"Automorphism Groups",
"Inner Automorphisms"
] | [
"Definition:Group",
"Definition:Set",
"Definition:Inner Automorphism",
"Definition:Subgroup",
"Definition:Automorphism Group/Group"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inner Automorphism",
"Inverse of Inner Automorphism",
"Inverse of Product",
"One-Step Subgroup Test"
] |
proofwiki-15614 | Property of Group Automorphism which Fixes Identity Only | Let $G$ be a finite group whose identity is $e$.
Let $\phi: G \to G$ be a group automorphism.
Let $\phi$ have the property that:
:$\forall g \in G \setminus \set e: \map \phi t \ne t$
That is, the only fixed element of $\phi$ is $e$.
Then:
:$\forall x, y \in G: x^{-1} \, \map \phi x = y^{-1} \, \map \phi y \implies x =... | {{begin-eqn}}
{{eqn | l = x^{-1} \, \map \phi x
| r = y^{-1} \, \map \phi y
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi x
| r = x y^{-1} \, \map \phi y
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi x \paren {\map \phi y}^{-1}
| r = x y^{-1}
| c =
}}
{{eqn | ll= \lead... | Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\phi: G \to G$ be a [[Definition:Group Automorphism|group automorphism]].
Let $\phi$ have the property that:
:$\forall g \in G \setminus \set e: \map \phi t \ne t$
That is, the only [[Definition:Fixed El... | {{begin-eqn}}
{{eqn | l = x^{-1} \, \map \phi x
| r = y^{-1} \, \map \phi y
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi x
| r = x y^{-1} \, \map \phi y
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi x \paren {\map \phi y}^{-1}
| r = x y^{-1}
| c =
}}
{{eqn | ll= \lead... | Property of Group Automorphism which Fixes Identity Only | https://proofwiki.org/wiki/Property_of_Group_Automorphism_which_Fixes_Identity_Only | https://proofwiki.org/wiki/Property_of_Group_Automorphism_which_Fixes_Identity_Only | [
"Group Automorphisms",
"Property of Group Automorphism which Fixes Identity Only"
] | [
"Definition:Finite Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Automorphism",
"Definition:Fixed Element"
] | [] |
proofwiki-15615 | Property of Group Automorphism which Fixes Identity Only/Corollary 1 | :$\forall x \in G: \exists g \in G: x = g^{-1} \, \map \phi g$ | 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 Surj... | :$\forall x \in G: \exists g \in G: x = g^{-1} \, \map \phi g$ | Let $\psi: G \to G$ be the [[Definition:Mapping|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 [[Definition:Injection|injecti... | Property of Group Automorphism which Fixes Identity Only/Corollary 1 | https://proofwiki.org/wiki/Property_of_Group_Automorphism_which_Fixes_Identity_Only/Corollary_1 | https://proofwiki.org/wiki/Property_of_Group_Automorphism_which_Fixes_Identity_Only/Corollary_1 | [
"Property of Group Automorphism which Fixes Identity Only"
] | [] | [
"Definition:Mapping",
"Property of Group Automorphism which Fixes Identity Only",
"Definition:Injection",
"Injection from Finite Set to Itself is Surjection",
"Definition:Surjection"
] |
proofwiki-15616 | Automorphism Group of Cyclic Group is Abelian | Let $G$ be a cyclic group.
Let $\Aut G$ denote the automorphism group of $G$.
Then $\Aut G$ is abelian. | Let $G = \gen g$
Let $\phi, \psi \in \Aut G$.
As $G$ is cyclic:
{{begin-eqn}}
{{eqn | q = \exists a \in \Z
| l = \map \phi g
| r = g^a
}}
{{eqn | q = \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
... | Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
Let $\Aut G$ denote the [[Definition:Automorphism Group|automorphism group]] of $G$.
Then $\Aut G$ is [[Definition:Abelian Group|abelian]]. | Let $G = \gen g$
Let $\phi, \psi \in \Aut G$.
As $G$ is [[Definition:Cyclic Group|cyclic]]:
{{begin-eqn}}
{{eqn | q = \exists a \in \Z
| l = \map \phi g
| r = g^a
}}
{{eqn | q = \exists b \in \Z
| l = \map \psi g
| r = g^b
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \map {\phi \circ \p... | Automorphism Group of Cyclic Group is Abelian | https://proofwiki.org/wiki/Automorphism_Group_of_Cyclic_Group_is_Abelian | https://proofwiki.org/wiki/Automorphism_Group_of_Cyclic_Group_is_Abelian | [
"Cyclic Groups",
"Automorphism Groups"
] | [
"Definition:Cyclic Group",
"Definition:Automorphism Group",
"Definition:Abelian Group"
] | [
"Definition:Cyclic Group",
"Definition:Cyclic Group/Generator",
"Definition:Group Automorphism"
] |
proofwiki-15617 | Order of Automorphism Group of Prime Group | Let $p$ be a prime number.
Let $G$ be a group of order $p$.
Let $\Aut G$ denote the automorphism group of $G$.
Then:
:$\order {\Aut G} = p - 1$
where $\order {\, \cdot \,}$ denotes the order of a group. | 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}} | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $p$.
Let $\Aut G$ denote the [[Definition:Automorphism Group|automorphism group]] of $G$.
Then:
:$\order {\Aut G} = p - 1$
where $\order {\, \cdot \,}$ denotes the [[Definition:Order... | From [[Prime Group is Cyclic]] we have that $G$ is a [[Definition:Cyclic Group|cyclic group]].
From [[Order of Automorphism Group of Cyclic Group]]:
:$\order {\Aut G} = \map \phi p$
where $\map \phi n$ denotes the [[Definition:Euler Phi Function|Euler $\phi$ function]].
The result follows from [[Euler Phi Function of... | Order of Automorphism Group of Prime Group | https://proofwiki.org/wiki/Order_of_Automorphism_Group_of_Prime_Group | https://proofwiki.org/wiki/Order_of_Automorphism_Group_of_Prime_Group | [
"Prime Groups",
"Automorphism Groups"
] | [
"Definition:Prime Number",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Automorphism Group",
"Definition:Order of Structure",
"Definition:Group"
] | [
"Prime Group is Cyclic",
"Definition:Cyclic Group",
"Order of Automorphism Group of Cyclic Group",
"Definition:Euler Phi Function",
"Euler Phi Function of Prime"
] |
proofwiki-15618 | Square Order 2 Matrices over Real Numbers form Ring with Unity | Let $S$ denote the set of square matrices of order $2$ whose entries are the set of real numbers.
Then $S$ forms a non-commutative '''ring with unity''' whose unity is the matrix $\begin {pmatrix} 1 & 0 \\ 0 & 1 \end {pmatrix}$. | This is an instance of Ring of Square Matrices over Field is Ring with Unity.
{{qed}} | Let $S$ denote the [[Definition:Set|set]] of [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order $2$]] whose [[Definition:Matrix Entry|entries]] are the [[Definition:Real Number|set of real numbers]].
Then $S$ forms a [[Definition:Non-Commutative Ring|non-commutative]] '''[[Defini... | This is an instance of [[Ring of Square Matrices over Field is Ring with Unity]].
{{qed}} | Square Order 2 Matrices over Real Numbers form Ring with Unity | https://proofwiki.org/wiki/Square_Order_2_Matrices_over_Real_Numbers_form_Ring_with_Unity | https://proofwiki.org/wiki/Square_Order_2_Matrices_over_Real_Numbers_form_Ring_with_Unity | [
"Examples of Rings with Unity"
] | [
"Definition:Set",
"Definition:Matrix/Square Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Matrix/Element",
"Definition:Real Number",
"Definition:Non-Commutative Ring",
"Definition:Ring with Unity",
"Definition:Unity",
"Definition:Matrix/Square Matrix"
] | [
"Ring of Square Matrices over Field is Ring with Unity"
] |
proofwiki-15619 | Product Formula for Norms on Non-zero Rationals | Let $\Q_{\ne 0}$ be the set of non-zero rational numbers.
Let $\Bbb P$ denote the set of prime numbers.
Let $a \in \Q_{\ne 0}$.
Then the following infinite product converges:
:$\size a \times \ds \prod_{p \mathop \in \Bbb P}^{} \norm a_p = 1$
where:
:$\size {\,\cdot\,}$ is the absolute value on $\Q$
:$\norm {\,\cdot\,... | === 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:
:$\size b \ds \times \prod_{p \mathop \in \Bbb P} \norm b_p = 1$
:$\size c \ds \times \prod_{p \mathop \in \Bbb P} \norm c_p ... | Let $\Q_{\ne 0}$ be the [[Definition:Set|set]] of non-zero [[Definition:Rational Numbers|rational numbers]].
Let $\Bbb P$ denote the [[Definition:Set|set]] of [[Definition:Prime Number|prime numbers]].
Let $a \in \Q_{\ne 0}$.
Then the following [[Definition:Infinite Product|infinite product]] [[Definition:Converge... | === [[Product Formula for Norms on Non-zero Rationals/Lemma|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 [[Product Formula for Norms on Non-zero Rationals/Lemma|Lemma]], the following [[Definition:Infinite Product|infi... | Product Formula for Norms on Non-zero Rationals | https://proofwiki.org/wiki/Product_Formula_for_Norms_on_Non-zero_Rationals | https://proofwiki.org/wiki/Product_Formula_for_Norms_on_Non-zero_Rationals | [
"P-adic Number Theory"
] | [
"Definition:Set",
"Definition:Rational Number",
"Definition:Set",
"Definition:Prime Number",
"Definition:Continued Product/Infinite",
"Definition:Convergent Sequence/Real Numbers",
"Definition:Absolute Value",
"Definition:P-adic Norm",
"Definition:Prime Number"
] | [
"Product Formula for Norms on Non-zero Rationals/Lemma",
"Product Formula for Norms on Non-zero Rationals/Lemma",
"Definition:Continued Product/Infinite",
"Definition:Convergent Sequence/Real Numbers",
"Combination Theorem for Sequences/Real/Quotient Rule",
"Definition:Continued Product/Infinite",
"Defi... |
proofwiki-15620 | Ring is Subring of Itself | Let $R$ be a ring.
Then $R$ is a subring of itself. | $R$ is a ring and $R \subseteq R$.
It follows by definition that $R$ is a subring of $R$.
{{qed}} | Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Then $R$ is a [[Definition:Subring|subring]] of itself. | $R$ is a [[Definition:Ring (Abstract Algebra)|ring]] and $R \subseteq R$.
It follows by definition that $R$ is a [[Definition:Subring|subring]] of $R$.
{{qed}} | Ring is Subring of Itself | https://proofwiki.org/wiki/Ring_is_Subring_of_Itself | https://proofwiki.org/wiki/Ring_is_Subring_of_Itself | [
"Subrings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Subring"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Subring"
] |
proofwiki-15621 | Null Ring is Subring of Ring | Let $R$ be a ring.
Then the null ring is a subring of $R$. | 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}} | Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Then the [[Definition:Null Ring|null ring]] is a [[Definition:Subring|subring]] of $R$. | From [[Null Ring is Ring]], the [[Definition:Null Ring|null ring]] $\struct {\set {0_R}, +, \circ}$ is a [[Definition:Ring (Abstract Algebra)|ring]].
We also have that $\set {0_R}$ is a [[Definition:Subset|subset]] of $R$
Hence the result by definition of [[Definition:Subring|subring]].
{{qed}} | Null Ring is Subring of Ring | https://proofwiki.org/wiki/Null_Ring_is_Subring_of_Ring | https://proofwiki.org/wiki/Null_Ring_is_Subring_of_Ring | [
"Subrings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Null Ring",
"Definition:Subring"
] | [
"Null Ring is Ring",
"Definition:Null Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Subset",
"Definition:Subring"
] |
proofwiki-15622 | Integers form Subring of Reals | The ring of integers $\struct {\Z, +, \times}$ forms a subring of the field of real numbers. | 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}} | The [[Definition:Ring of Integers|ring of integers]] $\struct {\Z, +, \times}$ forms a [[Definition:Subring|subring]] of the [[Definition:Field of Real Numbers|field of real numbers]]. | We have that the [[Definition:Integer|set of integers]] $\Z$ are a [[Definition:Subset|subset]] of the [[Definition:Real Numbers|real numbers]] $\R$.
The [[Definition:Field of Real Numbers|field of real numbers]] is, [[Definition:A Fortiori|a fortiori]], also a [[Definition:Ring (Abstract Algebra)|ring]].
Hence the r... | Integers form Subring of Reals | https://proofwiki.org/wiki/Integers_form_Subring_of_Reals | https://proofwiki.org/wiki/Integers_form_Subring_of_Reals | [
"Integers",
"Real Numbers",
"Subrings"
] | [
"Definition:Ring of Integers",
"Definition:Subring",
"Definition:Field of Real Numbers"
] | [
"Definition:Integer",
"Definition:Subset",
"Definition:Real Number",
"Definition:Field of Real Numbers",
"Definition:A Fortiori",
"Definition:Ring (Abstract Algebra)",
"Definition:Subring"
] |
proofwiki-15623 | Unity of Subring is not necessarily Unity of Ring | Let $\struct {S, +, \circ}$ be a ring with unity whose unity is $1_S$.
Let $\struct {T, + \circ}$ be a subring of $\struct {S, + \circ}$ whose unity is $1_T$.
Then it is not necessarily the case that $1_T = 1_S$. | 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... | Let $\struct {S, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Unity of Ring|unity]] is $1_S$.
Let $\struct {T, + \circ}$ be a [[Definition:Subring|subring]] of $\struct {S, + \circ}$ whose [[Definition:Unity of Ring|unity]] is $1_T$.
Then it is not necessarily the case that $1_T ... | Let $\struct {S, + \times}$ be the [[Definition:Ring (Abstract Algebra)|ring]] formed by the [[Definition:Set|set]] of [[Definition:Order of Square Matrix|order $2$]] [[Definition:Square Matrix|square matrices]] over the [[Definition:Real Number|real numbers]] $R$ under [[Definition:Matrix Entrywise Addition|(conventio... | Unity of Subring is not necessarily Unity of Ring | https://proofwiki.org/wiki/Unity_of_Subring_is_not_necessarily_Unity_of_Ring | https://proofwiki.org/wiki/Unity_of_Subring_is_not_necessarily_Unity_of_Ring | [
"Subrings"
] | [
"Definition:Ring with Unity",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Subring",
"Definition:Unity (Abstract Algebra)/Ring"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Set",
"Definition:Matrix/Square Matrix/Order",
"Definition:Matrix/Square Matrix",
"Definition:Real Number",
"Definition:Matrix Entrywise Addition",
"Definition:Matrix Product (Conventional)",
"Definition:Subset",
"Definition:Matrix/Square Matrix",
... |
proofwiki-15624 | Zero of Subfield is Zero of Field | The zero of $\struct {K, +, \times}$ is also $0$. | 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}} | The [[Definition:Field Zero|zero]] of $\struct {K, +, \times}$ is also $0$. | By definition, $\struct {F, +, \times}$ and $\struct {K, +, \times}$ are both [[Definition:Ring (Abstract Algebra)|rings]].
Thus $\struct {K, +, \times}$ is a [[Definition:Subring|subring]] of $\struct {F, +, \times}$
The result follows from [[Zero of Subring is Zero of Ring]].
{{qed}} | Zero of Subfield is Zero of Field/Proof 1 | https://proofwiki.org/wiki/Zero_of_Subfield_is_Zero_of_Field | https://proofwiki.org/wiki/Zero_of_Subfield_is_Zero_of_Field/Proof_1 | [
"Subfields",
"Zero of Subfield is Zero of Field"
] | [
"Definition:Field Zero"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Subring",
"Zero of Subring is Zero of Ring"
] |
proofwiki-15625 | Zero of Subfield is Zero of Field | The zero of $\struct {K, +, \times}$ is also $0$. | 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 $0$,... | The [[Definition:Field Zero|zero]] of $\struct {K, +, \times}$ is also $0$. | By definition, $\struct {K, +, \times}$ is a [[Definition:Subset|subset]] of $F$ which is a [[Definition:Field (Abstract Algebra)|field]].
By definition of [[Definition:Field (Abstract Algebra)|field]], $\struct {K, +}$ and $\struct {F, +}$ are [[Definition:Group|groups]] such that $K \subseteq F$.
So, by definition,... | Zero of Subfield is Zero of Field/Proof 2 | https://proofwiki.org/wiki/Zero_of_Subfield_is_Zero_of_Field | https://proofwiki.org/wiki/Zero_of_Subfield_is_Zero_of_Field/Proof_2 | [
"Subfields",
"Zero of Subfield is Zero of Field"
] | [
"Definition:Field Zero"
] | [
"Definition:Subset",
"Definition:Field (Abstract Algebra)",
"Definition:Field (Abstract Algebra)",
"Definition:Group",
"Definition:Subgroup",
"Identity of Subgroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-15626 | Unity of Subfield is Unity of Field | The unity of $\struct {K, +, \times}$ is also $1$. | 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^*, \tim... | The [[Definition:Unity of Ring|unity]] of $\struct {K, +, \times}$ is also $1$. | By definition, $\struct {K, +, \times}$ is a [[Definition:Subset|subset]] of $F$ which is a [[Definition:Field (Abstract Algebra)|field]].
By definition of [[Definition:Field (Abstract Algebra)|field]], $\struct {K^*, \times}$ and $\struct {F^*, \times}$ are [[Definition:Group|groups]] such that $K \subseteq F$.
So $... | Unity of Subfield is Unity of Field | https://proofwiki.org/wiki/Unity_of_Subfield_is_Unity_of_Field | https://proofwiki.org/wiki/Unity_of_Subfield_is_Unity_of_Field | [
"Subfields"
] | [
"Definition:Unity (Abstract Algebra)/Ring"
] | [
"Definition:Subset",
"Definition:Field (Abstract Algebra)",
"Definition:Field (Abstract Algebra)",
"Definition:Group",
"Definition:Subgroup",
"Identity of Subgroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-15627 | Zero of Subring is Zero of Ring | The zero of $\struct {S, +, \times}$ is also $0$. | 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 $0$, i... | The [[Definition:Ring Zero|zero]] of $\struct {S, +, \times}$ is also $0$. | By definition, $\struct {S, +, \times}$ is a [[Definition:Subset|subset]] of $R$ which is a [[Definition:Ring (Abstract Algebra)|ring]].
By definition of [[Definition:Ring (Abstract Algebra)|ring]], $\struct {S, +}$ and $\struct {R, +}$ are [[Definition:Group|groups]] such that $S \subseteq R$.
So, by definition, $\s... | Zero of Subring is Zero of Ring | https://proofwiki.org/wiki/Zero_of_Subring_is_Zero_of_Ring | https://proofwiki.org/wiki/Zero_of_Subring_is_Zero_of_Ring | [
"Subrings"
] | [
"Definition:Ring Zero"
] | [
"Definition:Subset",
"Definition:Ring (Abstract Algebra)",
"Definition:Ring (Abstract Algebra)",
"Definition:Group",
"Definition:Subgroup",
"Identity of Subgroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Category:Subrings"
] |
proofwiki-15628 | Zero of Subfield is Zero of Field/Proof 1 | Let $\struct {F, +, \times}$ be a field whose zero is $0$.
Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \times}$.
{{:Zero of Subfield is Zero of Field}} | 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}} | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0$.
Let $\struct {K, +, \times}$ be a [[Definition:Subfield|subfield]] of $\struct {F, +, \times}$.
{{:Zero of Subfield is Zero of Field}} | By definition, $\struct {F, +, \times}$ and $\struct {K, +, \times}$ are both [[Definition:Ring (Abstract Algebra)|rings]].
Thus $\struct {K, +, \times}$ is a [[Definition:Subring|subring]] of $\struct {F, +, \times}$
The result follows from [[Zero of Subring is Zero of Ring]].
{{qed}} | Zero of Subfield is Zero of Field/Proof 1 | https://proofwiki.org/wiki/Zero_of_Subfield_is_Zero_of_Field/Proof_1 | https://proofwiki.org/wiki/Zero_of_Subfield_is_Zero_of_Field/Proof_1 | [
"Zero of Subfield is Zero of Field"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Subfield"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Subring",
"Zero of Subring is Zero of Ring"
] |
proofwiki-15629 | Zero of Subfield is Zero of Field/Proof 2 | Let $\struct {F, +, \times}$ be a field whose zero is $0$.
Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \times}$.
{{:Zero of Subfield is Zero of Field}} | 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 $0$,... | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0$.
Let $\struct {K, +, \times}$ be a [[Definition:Subfield|subfield]] of $\struct {F, +, \times}$.
{{:Zero of Subfield is Zero of Field}} | By definition, $\struct {K, +, \times}$ is a [[Definition:Subset|subset]] of $F$ which is a [[Definition:Field (Abstract Algebra)|field]].
By definition of [[Definition:Field (Abstract Algebra)|field]], $\struct {K, +}$ and $\struct {F, +}$ are [[Definition:Group|groups]] such that $K \subseteq F$.
So, by definition,... | Zero of Subfield is Zero of Field/Proof 2 | https://proofwiki.org/wiki/Zero_of_Subfield_is_Zero_of_Field/Proof_2 | https://proofwiki.org/wiki/Zero_of_Subfield_is_Zero_of_Field/Proof_2 | [
"Zero of Subfield is Zero of Field"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Subfield"
] | [
"Definition:Subset",
"Definition:Field (Abstract Algebra)",
"Definition:Field (Abstract Algebra)",
"Definition:Group",
"Definition:Subgroup",
"Identity of Subgroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-15630 | P-adic Norm not Complete on Rational Numbers/Proof 1/Case 1 | Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p > 3$.
Then:
:$\struct {\Q, \norm {\,\cdot\,}_p}$ is not a complete normed division ring.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$. | 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:
:$\norm {a^{p^{n + 1} } - a^{p^n} }_p = \norm {a^{p^n} (a^{p^n \paren {p - 1} } - 1) }_p$
From {{Corollary|Euler's Theorem (Number Th... | Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime]] $p > 3$.
Then:
:$\struct {\Q, \norm {\,\cdot\,}_p}$ is not a [[Definition:Complete Normed Division Ring|complete normed division ring]].
That i... | Let $p > 3$.
Then there exists $a \in \Z: 1 < a < p-1$.
Consider the [[Definition:Sequence|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:
:$\norm {a^{p^{n + 1} } - a^{p^n} }_p = \norm {a^{p^n} (a^{p^n \paren {p - 1} } - 1) }_p$
From {{Coroll... | P-adic Norm not Complete on Rational Numbers/Proof 1/Case 1 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_1/Case_1 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_1/Case_1 | [
"P-adic Norm not Complete on Rational Numbers"
] | [
"Definition:P-adic Norm",
"Definition:Rational Number",
"Definition:Prime Number",
"Definition:Complete Normed Division Ring",
"Definition:Cauchy Sequence/Normed Division Ring",
"Definition:Convergent Sequence/Normed Division Ring",
"Definition:Limit of Sequence/Normed Division Ring"
] | [
"Definition:Sequence",
"Characterisation of Cauchy Sequence in Non-Archimedean Norm",
"Definition:Cauchy Sequence",
"Definition:Convergent Sequence",
"Modulus of Limit/Normed Division Ring",
"Definition:Norm/Division Ring",
"Limit of Subsequence equals Limit of Sequence",
"Combination Theorem for Sequ... |
proofwiki-15631 | P-adic Norm not Complete on Rational Numbers/Proof 1/Case 2 | Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for $p = 2$ or $3$.
Then:
:$\struct {\Q, \norm {\,\cdot\,}_p}$ is not a complete normed division ring.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$. | {{WIP}} | Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for $p = 2$ or $3$.
Then:
:$\struct {\Q, \norm {\,\cdot\,}_p}$ is not a [[Definition:Complete Normed Division Ring|complete normed division ring]].
That is, there exists a [[Definition:C... | {{WIP}} | P-adic Norm not Complete on Rational Numbers/Proof 1/Case 2 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_1/Case_2 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_1/Case_2 | [
"P-adic Norm not Complete on Rational Numbers"
] | [
"Definition:P-adic Norm",
"Definition:Rational Number",
"Definition:Complete Normed Division Ring",
"Definition:Cauchy Sequence/Normed Division Ring",
"Definition:Convergent Sequence/Normed Division Ring",
"Definition:Limit of Sequence/Normed Division Ring"
] | [] |
proofwiki-15632 | Element in Integral Domain is Divisor iff Principal Ideal is Superset | :$x \divides y \iff \ideal y \subseteq \ideal x$ | 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... | :$x \divides y \iff \ideal y \subseteq \ideal x$ | Let that $x \divides y$.
Then by definition of [[Definition:Divisor of Ring Element|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|Principa... | Element in Integral Domain is Divisor iff Principal Ideal is Superset | https://proofwiki.org/wiki/Element_in_Integral_Domain_is_Divisor_iff_Principal_Ideal_is_Superset | https://proofwiki.org/wiki/Element_in_Integral_Domain_is_Divisor_iff_Principal_Ideal_is_Superset | [
"Integral Domains",
"Principal Ideals of Rings",
"Factorization"
] | [] | [
"Definition:Divisor (Algebra)/Ring with Unity"
] |
proofwiki-15633 | Element in Integral Domain is Unit iff Principal Ideal is Whole Domain | :$x \in U_D \iff \ideal x = D$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = U_D
| c =
}}
{{eqn | ll= \leadsto
| q = \exists u \in U_D
| l = u
| o = \in
| r = \ideal x
| c = Ideal of Unit is Whole Ring
}}
{{eqn | ll= \leadsto
| l = \ideal x
| r = D
| c = Ideal of Unit is Whole Ring
}}
... | :$x \in U_D \iff \ideal x = D$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = U_D
| c =
}}
{{eqn | ll= \leadsto
| q = \exists u \in U_D
| l = u
| o = \in
| r = \ideal x
| c = [[Ideal of Unit is Whole Ring]]
}}
{{eqn | ll= \leadsto
| l = \ideal x
| r = D
| c = [[Ideal of Unit is Whole Ri... | Element in Integral Domain is Unit iff Principal Ideal is Whole Domain | https://proofwiki.org/wiki/Element_in_Integral_Domain_is_Unit_iff_Principal_Ideal_is_Whole_Domain | https://proofwiki.org/wiki/Element_in_Integral_Domain_is_Unit_iff_Principal_Ideal_is_Whole_Domain | [
"Integral Domains",
"Principal Ideals of Rings"
] | [] | [
"Ideal of Unit is Whole Ring",
"Ideal of Unit is Whole Ring"
] |
proofwiki-15634 | Residue of Gamma Function | Let $\Gamma$ be the Definition:Gamma Function.
Let $n$ be a non-negative integer.
Then:
:$\Res \Gamma {-n} = \dfrac {\paren {-1}^n} {n!}$ | 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 - \paren {-n} } \map \Gamma z
| c = Residue at Simple Pole
}}
{{eqn | r = \lim_{z \mathop \to -n} \paren {... | Let $\Gamma$ be the [[Definition:Gamma Function]].
Let $n$ be a non-[[Definition:Negative Integer|negative integer]].
Then:
:$\Res \Gamma {-n} = \dfrac {\paren {-1}^n} {n!}$ | By [[Poles of Gamma Function]], $\Gamma$ has [[Definition:Simple Pole|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 - \paren {-n} } \map \Gamma z
| c = [[Residue at Simple Pole]]
}}
{{eqn |... | Residue of Gamma Function | https://proofwiki.org/wiki/Residue_of_Gamma_Function | https://proofwiki.org/wiki/Residue_of_Gamma_Function | [
"Gamma Function"
] | [
"Definition:Gamma Function",
"Definition:Negative/Integer"
] | [
"Poles of Gamma Function",
"Definition:Order of Pole/Simple Pole",
"Residue at Simple Pole",
"Gamma Difference Equation",
"Category:Gamma Function"
] |
proofwiki-15635 | Periodic Element is Multiple of Period | Let $f: \R \to \R$ be a real periodic function with period $P$.
Let $L$ be a periodic element of $f$.
Then $P \divides L$. | {{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 Pro... | Let $f: \R \to \R$ be a [[Definition:Real Periodic Function|real periodic function]] with [[Definition:Period of Periodic Real Function|period]] $P$.
Let $L$ be a [[Definition:Periodic Element|periodic element]] of $f$.
Then $P \divides L$. | {{AimForCont}} that $P \nmid L$.
Then by the [[Division Theorem/Real Number Index|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... | Periodic Element is Multiple of Period | https://proofwiki.org/wiki/Periodic_Element_is_Multiple_of_Period | https://proofwiki.org/wiki/Periodic_Element_is_Multiple_of_Period | [
"Periodic Functions"
] | [
"Definition:Periodic Function/Real",
"Definition:Periodic Real Function/Period",
"Definition:Periodic Function/Periodic Element"
] | [
"Division Theorem/Real Number Index",
"General Periodicity Property",
"Definition:Periodic Function/Periodic Element",
"Definition:Periodic Real Function/Period",
"Proof by Contradiction",
"Category:Periodic Functions"
] |
proofwiki-15636 | Equivalence of Definitions of Associate in Integral Domain | {{TFAE|def = Associate in Integral Domain|view = associate|context = Integral Domain|contextview = integral domains}}
Let $\struct {D, +, \circ}$ be an integral domain.
Let $x, y \in D$. | === $(1)$ is Equivalent to $(2)$ ===
{{:Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 2}} | {{TFAE|def = Associate in Integral Domain|view = associate|context = Integral Domain|contextview = integral domains}}
Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]].
Let $x, y \in D$. | === [[Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 2|$(1)$ is Equivalent to $(2)$]] ===
{{:Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 2}} | Equivalence of Definitions of Associate in Integral Domain | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Associate_in_Integral_Domain | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Associate_in_Integral_Domain | [
"Integral Domains",
"Associates",
"Equivalence of Definitions of Associate in Integral Domain"
] | [
"Definition:Integral Domain"
] | [
"Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 2"
] |
proofwiki-15637 | Finite Set of Elements in Principal Ideal Domain has GCD | Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $a_1, a_2, \dotsc, a_n$ be non-zero elements of $D$.
Then $a_1, a_2, \dotsc, a_n$ all have a greatest common divisor. | 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 ... | Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
Let $a_1, a_2, \dotsc, a_n$ be non-[[Definition:Ring Zero|zero]] [[Definition:Element|elements]] of $D$.
Then $a_1, a_2, \dotsc, a_n$ all have a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor... | Let $0_D$ and $1_D$ be the [[Definition:Ring Zero|zero]] and [[Definition:Unity of Ring|unity]] respectively of $D$.
Let $J$ be the [[Definition:Set|set]] of all [[Definition:Linear Combination|linear combinations]] in $D$ of $\set {a_1, a_2, \dotsc, a_n}$.
From [[Set of Linear Combinations of Finite Set of Elements ... | Finite Set of Elements in Principal Ideal Domain has GCD | https://proofwiki.org/wiki/Finite_Set_of_Elements_in_Principal_Ideal_Domain_has_GCD | https://proofwiki.org/wiki/Finite_Set_of_Elements_in_Principal_Ideal_Domain_has_GCD | [
"Principal Ideal Domains",
"GCD Domains",
"Greatest Common Divisor"
] | [
"Definition:Principal Ideal Domain",
"Definition:Ring Zero",
"Definition:Element",
"Definition:Greatest Common Divisor/Integral Domain"
] | [
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Set",
"Definition:Linear Combination",
"Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal",
"Definition:Principal Ideal of Ring",
"Definition:Generator of Ideal of Ring",
"De... |
proofwiki-15638 | Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal | Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $a_1, a_2, \dotsc, a_n$ be non-zero elements of $D$.
Let $J$ be the set of all linear combinations in $D$ of $\set {a_1, a_2, \dotsc, a_n}$
Then for some $x \in D$:
:$J = \ideal x$
where $\ideal x$ denotes the principal ideal generated by $x$. | Let the unity of $D$ be $1_D$.
By definition of principal ideal:
:$\ds \ideal a = \set {\sum_{i \mathop = 1}^n r_i \circ a \circ s_i: n \in \N, r_i, s_i \in D}$
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... | Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
Let $a_1, a_2, \dotsc, a_n$ be non-[[Definition:Ring Zero|zero]] [[Definition:Element|elements]] of $D$.
Let $J$ be the [[Definition:Set|set]] of all [[Definition:Linear Combination|linear combinations]] in $D$ of $\set {a_... | Let the [[Definition:Unity of Ring|unity]] of $D$ be $1_D$.
By definition of [[Definition:Principal Ideal of Ring|principal ideal]]:
:$\ds \ideal a = \set {\sum_{i \mathop = 1}^n r_i \circ a \circ s_i: n \in \N, r_i, s_i \in D}$
Let $x, y \in J$.
By definition of [[Definition:Linear Combination|linear combination... | Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal | https://proofwiki.org/wiki/Set_of_Linear_Combinations_of_Finite_Set_of_Elements_of_Principal_Ideal_Domain_is_Principal_Ideal | https://proofwiki.org/wiki/Set_of_Linear_Combinations_of_Finite_Set_of_Elements_of_Principal_Ideal_Domain_is_Principal_Ideal | [
"Principal Ideal Domains"
] | [
"Definition:Principal Ideal Domain",
"Definition:Ring Zero",
"Definition:Element",
"Definition:Set",
"Definition:Linear Combination",
"Definition:Principal Ideal of Ring",
"Definition:Generator of Ideal of Ring"
] | [
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Principal Ideal of Ring",
"Definition:Linear Combination",
"Product with Ring Negative/Corollary",
"Product with Ring Negative/Corollary",
"Product with Ring Negative",
"Definition:Commutative/Operation",
"Definition:Integral Domain",
"Test for... |
proofwiki-15639 | Greatest Common Divisors in Principal Ideal Domain are Associates | Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $S = \set {a_1, a_2, \dotsc, a_n}$ be a set of non-zero elements of $D$.
Let $y_1$ and $y_2$ be greatest common divisors of $S$.
Then $y_1$ and $y_2$ are associates. | 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... | Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
Let $S = \set {a_1, a_2, \dotsc, a_n}$ be a [[Definition:Set|set]] of non-[[Definition:Ring Zero|zero]] [[Definition:Element|elements]] of $D$.
Let $y_1$ and $y_2$ be [[Definition:Greatest Common Divisor of Ring Elements|g... | From [[Finite Set of Elements in Principal Ideal Domain has GCD]] we have that at least one such [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] exists.
So, let $y_1$ and $y_2$ be [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisors]] of $S$.
Then:
{{begin-e... | Greatest Common Divisors in Principal Ideal Domain are Associates | https://proofwiki.org/wiki/Greatest_Common_Divisors_in_Principal_Ideal_Domain_are_Associates | https://proofwiki.org/wiki/Greatest_Common_Divisors_in_Principal_Ideal_Domain_are_Associates | [
"Principal Ideal Domains",
"Greatest Common Divisor",
"Associates"
] | [
"Definition:Principal Ideal Domain",
"Definition:Set",
"Definition:Ring Zero",
"Definition:Element",
"Definition:Greatest Common Divisor/Integral Domain",
"Definition:Associate/Integral Domain"
] | [
"Finite Set of Elements in Principal Ideal Domain has GCD",
"Definition:Greatest Common Divisor/Integral Domain",
"Definition:Greatest Common Divisor/Integral Domain",
"Definition:Greatest Common Divisor/Integral Domain",
"Definition:Greatest Common Divisor/Integral Domain",
"Definition:Divisor (Algebra)/... |
proofwiki-15640 | Bézout's Identity/Principal Ideal Domain | Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $S = \set {a_1, a_2, \dotsc, a_n}$ be a set of non-zero elements of $D$.
Let $y$ be a greatest common divisor of $S$.
Then $y$ is expressible in the form:
:$y = d_1 a_1 + d_2 a_2 + \dotsb + d_n a_n$
where $d_1, d_2, \dotsc, d_n \in D$. | 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 ... | Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
Let $S = \set {a_1, a_2, \dotsc, a_n}$ be a [[Definition:Set|set]] of non-[[Definition:Ring Zero|zero]] [[Definition:Element|elements]] of $D$.
Let $y$ be a [[Definition:Greatest Common Divisor of Ring Elements|greatest co... | From [[Finite Set of Elements in Principal Ideal Domain has GCD]] we have that at least one such [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] exists.
So, let $y$ be a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] of $S$.
Let $J$ be the [[Definition... | Bézout's Identity/Principal Ideal Domain | https://proofwiki.org/wiki/Bézout's_Identity/Principal_Ideal_Domain | https://proofwiki.org/wiki/Bézout's_Identity/Principal_Ideal_Domain | [
"Principal Ideal Domains",
"Bézout's Identity"
] | [
"Definition:Principal Ideal Domain",
"Definition:Set",
"Definition:Ring Zero",
"Definition:Element",
"Definition:Greatest Common Divisor/Integral Domain"
] | [
"Finite Set of Elements in Principal Ideal Domain has GCD",
"Definition:Greatest Common Divisor/Integral Domain",
"Definition:Greatest Common Divisor/Integral Domain",
"Definition:Set",
"Definition:Linear Combination",
"Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Prin... |
proofwiki-15641 | Complete Factorizations of Proper Element in Principal Ideal Domain are Equivalent | Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $x \in D$ be a proper element of $D$.
Let there be two complete factorizations of $x$:
:$x = u_y \circ y_1 \circ y_2 \circ \cdots \circ y_m = F_1$
:$x = u_z \circ z_1 \circ z_2 \circ \cdots \circ z_n = F_2$
Then $F_1$ and $F_2$ are equivalent. | {{ProofWanted|Whitelaw leaves this unresolved at the end of $\S 62$ as an exercise for the student. I haven't read ahead that far, but it may be proved in the exercises. Will return to this later.}}# | Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
Let $x \in D$ be a [[Definition:Proper Element of Ring|proper element]] of $D$.
Let there be two [[Definition:Complete Factorization|complete factorizations]] of $x$:
:$x = u_y \circ y_1 \circ y_2 \circ \cdots \circ y_m = F... | {{ProofWanted|Whitelaw leaves this unresolved at the end of $\S 62$ as an exercise for the student. I haven't read ahead that far, but it may be proved in the exercises. Will return to this later.}}# | Complete Factorizations of Proper Element in Principal Ideal Domain are Equivalent | https://proofwiki.org/wiki/Complete_Factorizations_of_Proper_Element_in_Principal_Ideal_Domain_are_Equivalent | https://proofwiki.org/wiki/Complete_Factorizations_of_Proper_Element_in_Principal_Ideal_Domain_are_Equivalent | [
"Factorization",
"Principal Ideal Domains"
] | [
"Definition:Principal Ideal Domain",
"Definition:Proper Element of Ring",
"Definition:Complete Factorization",
"Definition:Equivalent Factorizations"
] | [] |
proofwiki-15642 | Moment Generating Function of Poisson Distribution | Let $X$ be a discrete random variable with a Poisson distribution with parameter $\lambda$ for some $\lambda \in \R_{> 0}$.
Then the moment generating function $M_X$ of $X$ is given by:
:$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$ | 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:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{n \mathop = 0}^\infty \map \Pr {X = n} e^{t n}$
So:
{{begin-eqn}}
{{eqn | ... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]] for some $\lambda \in \R_{> 0}$.
Then the [[Definition:Moment Generating Function|moment generating function]] $M_X$ of $X$ is given by:
:$\map {M_X} ... | From the definition of the [[Definition:Poisson Distribution|Poisson distribution]], $X$ has [[Definition:Probability Mass Function|probability mass function]]:
:$\map \Pr {X = n} = \dfrac {\lambda^n e^{-\lambda} } {n!}$
From the definition of a [[Definition:Moment Generating Function|moment generating function]]:
:... | Moment Generating Function of Poisson Distribution | https://proofwiki.org/wiki/Moment_Generating_Function_of_Poisson_Distribution | https://proofwiki.org/wiki/Moment_Generating_Function_of_Poisson_Distribution | [
"Moment Generating Functions",
"Poisson Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Poisson Distribution",
"Definition:Moment Generating Function"
] | [
"Definition:Poisson Distribution",
"Definition:Probability Mass Function",
"Definition:Moment Generating Function",
"Power Series Expansion for Exponential Function",
"Exponential of Sum"
] |
proofwiki-15643 | Moment Generating Function of Binomial Distribution | Let $X$ be a discrete random variable with a binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$:
:$X \sim \Binomial n p$
Then the moment generating function $M_X$ of $X$ is given by:
:$\map {M_X} t = \paren {1 - p + p e^t}^n$ | From the definition of the Binomial distribution, $X$ has probability mass function:
:$\map \Pr {X = k} = \dbinom n k p^k \paren {1 - p}^{n - k}$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{k \mathop = 0}^n \map \Pr {X = k} e^{t k}$
So:
{{begin-eqn}}
{{eqn | l =... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Binomial Distribution|binomial distribution with parameters $n$ and $p$]] for some $n \in \N$ and $0 \le p \le 1$:
:$X \sim \Binomial n p$
Then the [[Definition:Moment Generating Function|moment generating function]] $M_... | From the definition of the [[Definition: Binomial Distribution|Binomial distribution]], $X$ has [[Definition:Probability Mass Function|probability mass function]]:
:$\map \Pr {X = k} = \dbinom n k p^k \paren {1 - p}^{n - k}$
From the definition of a [[Definition:Moment Generating Function|moment generating function]]... | Moment Generating Function of Binomial Distribution | https://proofwiki.org/wiki/Moment_Generating_Function_of_Binomial_Distribution | https://proofwiki.org/wiki/Moment_Generating_Function_of_Binomial_Distribution | [
"Moment Generating Function of Binomial Distribution",
"Moment Generating Functions",
"Binomial Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Binomial Distribution",
"Definition:Moment Generating Function"
] | [
"Definition: Binomial Distribution",
"Definition:Probability Mass Function",
"Definition:Moment Generating Function",
"Binomial Theorem"
] |
proofwiki-15644 | Moment Generating Function of Exponential Distribution | Let $X$ be a continuous random variable with an exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$.
Then the moment generating function $M_X$ of $X$ is given by:
:$\map {M_X} t = \dfrac 1 {1 - \beta t}$
for $t < \dfrac 1 \beta$, and is undefined otherwise. | From the definition of the exponential distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \int_0^\infty e^{t x} \map {f_X} x \rd x$
Then:
{{begin-eqn}}
{{eqn | l = \map {M... | Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] with an [[Definition:Exponential Distribution|exponential distribution]] with parameter $\beta$ for some $\beta \in \R_{> 0}$.
Then the [[Definition:Moment Generating Function|moment generating function]] $M_X$ of $X$ is given by:
:$\m... | From the definition of the [[Definition:Exponential Distribution|exponential distribution]], $X$ has [[Definition:Probability Density Function|probability density function]]:
:$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$
From the definition of a [[Definition:Moment Generating Function|moment generating function... | Moment Generating Function of Exponential Distribution | https://proofwiki.org/wiki/Moment_Generating_Function_of_Exponential_Distribution | https://proofwiki.org/wiki/Moment_Generating_Function_of_Exponential_Distribution | [
"Moment Generating Function of Exponential Distribution",
"Moment Generating Functions",
"Exponential Distribution"
] | [
"Definition:Random Variable/Continuous",
"Definition:Exponential Distribution",
"Definition:Moment Generating Function"
] | [
"Definition:Exponential Distribution",
"Definition:Probability Density Function",
"Definition:Moment Generating Function",
"Exponential of Sum",
"Primitive of Exponential Function",
"Exponential Tends to Zero and Infinity",
"Exponential Tends to Zero and Infinity",
"Exponential Tends to Zero and Infin... |
proofwiki-15645 | Moment Generating Function of Discrete Uniform Distribution | Let $X$ be a discrete random variable with a discrete uniform distribution with parameter $n$ for some $n \in \N$.
Then the moment generating function $M_X$ of $X$ is given by:
:$\map {M_X} t = \dfrac {e^t \paren {1 - e^{n t} } } {n \paren {1 - e^t} }$ | From the definition of the discrete uniform distribution, $X$ has probability mass function:
:$\map \Pr {X = N} = \dfrac 1 n$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{N \mathop = 1}^n \map \Pr {X = N} e^{N t}$
So:
{{begin-eqn}}
{{eqn | l = \map {M_X} t
| r =... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Discrete Uniform Distribution|discrete uniform distribution with parameter $n$]] for some $n \in \N$.
Then the [[Definition:Moment Generating Function|moment generating function]] $M_X$ of $X$ is given by:
:$\map {M_X} ... | From the definition of the [[Definition:Discrete Uniform Distribution|discrete uniform distribution]], $X$ has [[Definition:Probability Mass Function|probability mass function]]:
:$\map \Pr {X = N} = \dfrac 1 n$
From the definition of a [[Definition:Moment Generating Function|moment generating function]]:
:$\ds \map... | Moment Generating Function of Discrete Uniform Distribution | https://proofwiki.org/wiki/Moment_Generating_Function_of_Discrete_Uniform_Distribution | https://proofwiki.org/wiki/Moment_Generating_Function_of_Discrete_Uniform_Distribution | [
"Moment Generating Functions",
"Discrete Uniform Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Uniform Distribution/Discrete",
"Definition:Moment Generating Function"
] | [
"Definition:Uniform Distribution/Discrete",
"Definition:Probability Mass Function",
"Definition:Moment Generating Function",
"Sum of Geometric Sequence",
"Category:Moment Generating Functions",
"Category:Discrete Uniform Distribution"
] |
proofwiki-15646 | Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field | Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $J$ be a maximal ideal.
Then the quotient ring $R / J$ is a field. | Let $J$ be a maximal ideal.
Because $J \subset R$, it follows from Quotient Ring of Commutative Ring is Commutative and Quotient Ring of Ring with Unity is Ring with Unity that $R / J$ is a commutative ring with unity.
We now need to prove that every non-zero element of $\struct {R / J, +, \circ}$ has a product invers... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
Let $J$ be a [[Definition:Maximal Ideal of Ring|maximal... | Let $J$ be a [[Definition:Maximal Ideal of Ring|maximal ideal]].
Because $J \subset R$, it follows from [[Quotient Ring of Commutative Ring is Commutative]] and [[Quotient Ring of Ring with Unity is Ring with Unity]] that $R / J$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
We now ne... | Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field | https://proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field/Proof_1/Maximal_Ideal_implies_Quotient_Ring_is_Field | https://proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field/Proof_1/Maximal_Ideal_implies_Quotient_Ring_is_Field | [
"Maximal Ideal iff Quotient Ring is Field"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ideal of Ring",
"Definition:Maximal Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Field (Abstract Algebra)"
] | [
"Definition:Maximal Ideal of Ring",
"Quotient Ring of Commutative Ring is Commutative",
"Quotient Ring of Ring with Unity is Ring with Unity",
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero",
"Definition:Element",
"Definition:Product Inverse",
"Definition:Ring Zero",
"Definition:Su... |
proofwiki-15647 | Maximal Ideal iff Quotient Ring is Field/Proof 1/Quotient Ring is Field implies Ideal is Maximal | Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let the quotient ring $R / J$ be a field.
Then $J$ is a maximal ideal. | Let $R / J$ be a field.
Let $K$ be a left ideal of $R$ such that $J \subsetneq K \subseteq R$.
We have that $J$ is the zero of $R / J$.
Let $x \in K \setminus J$.
Because $x \notin J$ then $x + J \ne J$.
Because $R / J$ is a field then $x + J \in R / J$ has a product inverse, say $s + J$.
Hence:
:$1_R + J = \paren {s +... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
Let the [[Definition:Quotient Ring|quotient ring]] $R /... | Let $R / J$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $K$ be a [[Definition:Left Ideal of Ring|left ideal]] of $R$ such that $J \subsetneq K \subseteq R$.
We have that $J$ is the [[Definition:Ring Zero|zero]] of $R / J$.
Let $x \in K \setminus J$.
Because $x \notin J$ then $x + J \ne J$.
Because $R /... | Maximal Ideal iff Quotient Ring is Field/Proof 1/Quotient Ring is Field implies Ideal is Maximal | https://proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field/Proof_1/Quotient_Ring_is_Field_implies_Ideal_is_Maximal | https://proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field/Proof_1/Quotient_Ring_is_Field_implies_Ideal_is_Maximal | [
"Maximal Ideal iff Quotient Ring is Field"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Field (Abstract Algebra)",
"Definition:Maximal Ideal of Ring"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Ideal of Ring/Left Ideal",
"Definition:Ring Zero",
"Definition:Field (Abstract Algebra)",
"Definition:Product Inverse",
"Left Cosets are Equal iff Product with Inverse in Subgroup",
"Definition:Ideal of Ring"
] |
proofwiki-15648 | Ring of Polynomial Forms is not necessarily Isomorphic to Ring of Polynomial Functions | Let $D$ be an integral domain.
Let $D \sqbrk X$ be the ring of polynomial forms in $X$ over $D$.
Let $\map P D$ be the ring of polynomial functions over $D$.
Then it is not necessarily the case that $D \sqbrk X$ is isomorphic with $\map P D$. | Proof by Counterexample:
Consider the integral domain $\struct {\Z_2, +, \times}$.
From Ring of Integers Modulo Prime is Integral Domain, it is seen that $\struct {\Z_2, +, \times}$ is indeed an integral domain.
Consider the ring of polynomial forms $\Z_2 \sqbrk X$.
This is an infinite ring, as it can be seen that $S \... | Let $D$ be an [[Definition:Integral Domain|integral domain]].
Let $D \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] in $X$ over $D$.
Let $\map P D$ be the [[Definition:Ring of Polynomial Functions|ring of polynomial functions]] over $D$.
Then it is not necessarily the case that $D... | [[Proof by Counterexample]]:
Consider the [[Definition:Integral Domain|integral domain]] $\struct {\Z_2, +, \times}$.
From [[Ring of Integers Modulo Prime is Integral Domain]], it is seen that $\struct {\Z_2, +, \times}$ is indeed an [[Definition:Integral Domain|integral domain]].
Consider the [[Definition:Ring of P... | Ring of Polynomial Forms is not necessarily Isomorphic to Ring of Polynomial Functions | https://proofwiki.org/wiki/Ring_of_Polynomial_Forms_is_not_necessarily_Isomorphic_to_Ring_of_Polynomial_Functions | https://proofwiki.org/wiki/Ring_of_Polynomial_Forms_is_not_necessarily_Isomorphic_to_Ring_of_Polynomial_Functions | [
"Polynomial Theory"
] | [
"Definition:Integral Domain",
"Definition:Ring of Polynomial Forms",
"Definition:Ring of Polynomial Functions",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism"
] | [
"Proof by Counterexample",
"Definition:Integral Domain",
"Ring of Integers Modulo Prime is Integral Domain",
"Definition:Integral Domain",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Infinite Set",
"Definition:Ring (Abstract Algebra)",
"Definition:Ring of Polynomial Functions",
"De... |
proofwiki-15649 | Moment Generating Function of Bernoulli Distribution | Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$ for some $0 \le p \le 1$.
Then the moment generating function $M_X$ of $X$ is given by:
:$\map {M_X} t = q + p e^t$
where $q = 1 - p$. | From the definition of the Bernoulli distribution, $X$ has probability mass function:
:<nowiki>$\map \Pr {X = n} = \begin{cases}
q & : n = 0 \\
p & : n = 1 \\
0 & : n \notin \set {0, 1} \\
\end{cases}$</nowiki>
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{n \matho... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]] for some $0 \le p \le 1$.
Then the [[Definition:Moment Generating Function|moment generating function]] $M_X$ of $X$ is given by:
:$\map {M_X} t = q + ... | From the definition of the [[Definition:Bernoulli Distribution|Bernoulli distribution]], $X$ has [[Definition:Probability Mass Function|probability mass function]]:
:<nowiki>$\map \Pr {X = n} = \begin{cases}
q & : n = 0 \\
p & : n = 1 \\
0 & : n \notin \set {0, 1} \\
\end{cases}$</nowiki>
From the definition of a [[D... | Moment Generating Function of Bernoulli Distribution | https://proofwiki.org/wiki/Moment_Generating_Function_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Moment_Generating_Function_of_Bernoulli_Distribution | [
"Moment Generating Functions",
"Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Moment Generating Function"
] | [
"Definition:Bernoulli Distribution",
"Definition:Probability Mass Function",
"Definition:Moment Generating Function",
"Exponential of Zero",
"Category:Moment Generating Functions",
"Category:Bernoulli Distribution"
] |
proofwiki-15650 | Field of Quotients of Ring of Polynomial Forms on Reals that yields Complex Numbers | Let $\struct {\R, +, \times}$ denote the field of real numbers.
Let $X$ be transcendental over $\R$.
Let $\R \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Consider the field of quotients:
:$\R \sqbrk X / \ideal p$
where:
:$p = X^2 + 1$
:$\ideal p$ denotes the ideal generated by $p$.
Then $\R \sqbrk X / \ideal p... | It is taken as read that $X^2 + 1$ is irreducible in $\R \sqbrk X$.
Hence by Polynomial Forms over Field form Principal Ideal Domain: Corollary 1, $\R \sqbrk X / \ideal p$ is indeed a field.
Let $\nu$ be the quotient epimorphism from $\R \sqbrk X$ onto $\R \sqbrk X / \ideal p$.
From Quotient Ring Epimorphism is Epimorp... | Let $\struct {\R, +, \times}$ denote the [[Definition:Field of Real Numbers|field of real numbers]].
Let $X$ be [[Definition:Transcendental over Field|transcendental over $\R$]].
Let $\R \sqbrk X$ be the [[Definition:Ring of Polynomials|ring of polynomials]] in $X$ over $F$.
Consider the [[Definition:Field of Quotie... | It is taken as read that $X^2 + 1$ is [[Definition:Irreducible Element of Ring|irreducible]] in $\R \sqbrk X$.
Hence by [[Polynomial Forms over Field form Principal Ideal Domain/Corollary 1|Polynomial Forms over Field form Principal Ideal Domain: Corollary 1]], $\R \sqbrk X / \ideal p$ is indeed a [[Definition:Field (... | Field of Quotients of Ring of Polynomial Forms on Reals that yields Complex Numbers | https://proofwiki.org/wiki/Field_of_Quotients_of_Ring_of_Polynomial_Forms_on_Reals_that_yields_Complex_Numbers | https://proofwiki.org/wiki/Field_of_Quotients_of_Ring_of_Polynomial_Forms_on_Reals_that_yields_Complex_Numbers | [
"Polynomial Theory",
"Fields of Quotients",
"Real Numbers",
"Complex Numbers"
] | [
"Definition:Field of Real Numbers",
"Definition:Transcendental (Abstract Algebra)/Field Extension/Element",
"Definition:Polynomial Ring",
"Definition:Field of Quotients",
"Definition:Ideal of Ring",
"Definition:Generator of Ideal of Ring",
"Definition:Field of Complex Numbers"
] | [
"Definition:Irreducible Element of Ring",
"Polynomial Forms over Field form Principal Ideal Domain/Corollary 1",
"Definition:Field (Abstract Algebra)",
"Definition:Quotient Epimorphism/Ring",
"Quotient Epimorphism is Epimorphism/Ring",
"Definition:Ring Monomorphism",
"Definition:Isomorphism (Abstract Al... |
proofwiki-15651 | Double of Antiperiodic Element is Periodic | Let $f: \R \to \R$ be a real function.
Let $L \in \R_{>0}$ be an anti-periodic element of $f$.
Then $2 L$ is a periodic element of $f$.
In other words, every anti-periodic function is also periodic. | By Non-Zero Real Numbers Closed under Multiplication we have that $2 L \in \R_{>0}$.
Then:
{{begin-eqn}}
{{eqn | l = \map f {x + 2 L}
| r = \map f {x + \paren {L + L} }
}}
{{eqn | r = \map f {\paren {x + L} + L}
| c = Real Addition is Associative
}}
{{eqn | r = -\map f {x + L}
}}
{{eqn | r = \map f x
... | Let $f: \R \to \R$ be a [[Definition:Real Function|real function]].
Let $L \in \R_{>0}$ be an [[Definition:Antiperiodic Element|anti-periodic element]] of $f$.
Then $2 L$ is a [[Definition:Periodic Element|periodic element]] of $f$.
In other words, every [[Definition:Antiperiodic Function|anti-periodic function]] i... | By [[Non-Zero Real Numbers Closed under Multiplication]] we have that $2 L \in \R_{>0}$.
Then:
{{begin-eqn}}
{{eqn | l = \map f {x + 2 L}
| r = \map f {x + \paren {L + L} }
}}
{{eqn | r = \map f {\paren {x + L} + L}
| c = [[Real Addition is Associative]]
}}
{{eqn | r = -\map f {x + L}
}}
{{eqn | r = \map ... | Double of Antiperiodic Element is Periodic | https://proofwiki.org/wiki/Double_of_Antiperiodic_Element_is_Periodic | https://proofwiki.org/wiki/Double_of_Antiperiodic_Element_is_Periodic | [
"Antiperiodic Functions"
] | [
"Definition:Real Function",
"Definition:Antiperiodic Function/Antiperiodic Element",
"Definition:Periodic Function/Periodic Element",
"Definition:Antiperiodic Function",
"Definition:Periodic Function/Real"
] | [
"Non-Zero Real Numbers Closed under Multiplication",
"Real Addition is Associative",
"Negative of Negative Real Number",
"Category:Antiperiodic Functions"
] |
proofwiki-15652 | Idempotent Ring has Characteristic Two/Corollary | :$\forall x \in R: -x = x$ | Let $0_R$ denote the zero of $R$.
Let $x \in R$.
Then:
{{begin-eqn}}
{{eqn | l = x + x
| r = 0_R
| c = Idempotent Ring has Characteristic Two
}}
{{eqn | ll= \leadsto
| l = -x + x + x
| r = -x + 0_R
| c =
}}
{{eqn | ll= \leadsto
| l = x
| r = -x
| c =
}}
{{end-eqn}}
Henc... | :$\forall x \in R: -x = x$ | Let $0_R$ denote the [[Definition:Ring Zero|zero]] of $R$.
Let $x \in R$.
Then:
{{begin-eqn}}
{{eqn | l = x + x
| r = 0_R
| c = [[Idempotent Ring has Characteristic Two]]
}}
{{eqn | ll= \leadsto
| l = -x + x + x
| r = -x + 0_R
| c =
}}
{{eqn | ll= \leadsto
| l = x
| r = -x... | Idempotent Ring has Characteristic Two/Corollary | https://proofwiki.org/wiki/Idempotent_Ring_has_Characteristic_Two/Corollary | https://proofwiki.org/wiki/Idempotent_Ring_has_Characteristic_Two/Corollary | [
"Idempotent Rings"
] | [] | [
"Definition:Ring Zero",
"Idempotent Ring has Characteristic Two"
] |
proofwiki-15653 | Skewness of Normal Distribution | Let $X$ be a continuous random variable with a normal distribution with parameters $\mu$ and $\sigma^2$ for some $\mu \in \R$ and $\sigma \in \R_{> 0}$.
Then the skewness $\gamma_1$ of $X$ is equal to $0$. | From the definition of skewness:
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
From the definition of the normal distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {\sigma \sqrt{2 \pi} } \, \map \exp {-\dfrac { \paren {x - \mu}^2} {2 \sigma^2} }$
So, from Expectation of Function... | Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] with a [[Definition:Normal Distribution|normal distribution with parameters $\mu$ and $\sigma^2$]] for some $\mu \in \R$ and $\sigma \in \R_{> 0}$.
Then the [[Definition:Skewness|skewness]] $\gamma_1$ of $X$ is equal to $0$. | From the definition of [[Definition:Skewness|skewness]]:
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
From the definition of the [[Definition:Normal Distribution|normal distribution]], $X$ has [[Definition:Probability Density Function|probability density function]]:
:$\map {f_X} x = \dfrac 1 {\sigma \... | Skewness of Normal Distribution/Proof 1 | https://proofwiki.org/wiki/Skewness_of_Normal_Distribution | https://proofwiki.org/wiki/Skewness_of_Normal_Distribution/Proof_1 | [
"Skewness of Normal Distribution",
"Normal Distribution",
"Skewness"
] | [
"Definition:Random Variable/Continuous",
"Definition:Normal Distribution",
"Definition:Skewness"
] | [
"Definition:Skewness",
"Definition:Normal Distribution",
"Definition:Probability Density Function",
"Expectation of Function of Continuous Random Variable",
"Integration by Substitution",
"Definition:Odd Function",
"Definite Integral of Odd Function"
] |
proofwiki-15654 | Skewness of Normal Distribution | Let $X$ be a continuous random variable with a normal distribution with parameters $\mu$ and $\sigma^2$ for some $\mu \in \R$ and $\sigma \in \R_{> 0}$.
Then the skewness $\gamma_1$ of $X$ is equal to $0$. | From the definition of skewness, we have:
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
where:
:$\mu$ is the expectation of $X$.
:$\sigma$ is the standard deviation of $X$.
By Expectation of Normal Distribution, we have:
:$\mu = \mu$
By Variance of Normal Distribution, we have:
:$\sigma = \sigma$
So:... | Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] with a [[Definition:Normal Distribution|normal distribution with parameters $\mu$ and $\sigma^2$]] for some $\mu \in \R$ and $\sigma \in \R_{> 0}$.
Then the [[Definition:Skewness|skewness]] $\gamma_1$ of $X$ is equal to $0$. | From the definition of [[Definition:Skewness|skewness]], we have:
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
where:
:$\mu$ is the [[Definition:Expectation|expectation]] of $X$.
:$\sigma$ is the [[Definition:Standard Deviation|standard deviation]] of $X$.
By [[Expectation of Normal Distribution]], ... | Skewness of Normal Distribution/Proof 2 | https://proofwiki.org/wiki/Skewness_of_Normal_Distribution | https://proofwiki.org/wiki/Skewness_of_Normal_Distribution/Proof_2 | [
"Skewness of Normal Distribution",
"Normal Distribution",
"Skewness"
] | [
"Definition:Random Variable/Continuous",
"Definition:Normal Distribution",
"Definition:Skewness"
] | [
"Definition:Skewness",
"Definition:Expectation",
"Definition:Standard Deviation",
"Expectation of Normal Distribution",
"Variance of Normal Distribution",
"Binomial Theorem/Examples/Cube of Difference",
"Expectation is Linear",
"Variance of Normal Distribution",
"Moment in terms of Moment Generating... |
proofwiki-15655 | Inverse of Unit in Centralizer of Ring is in Centralizer | Let $\struct {R, +, \circ}$ be a ring.
Let $S$ be a subset of $R$.
Let $\map {C_R} S$ denote the centralizer of $S$ in $R$
Let $u \in R$ be a unit of $R$.
Then:
:$u \in \map {C_R} S \implies u^{-1} \in \map {C_R} S$ | Let $u \in R$ be a unit of $R$.
Let $u \in \map {C_R} S$.
Then from Commutation with Inverse in Monoid:
:$u^{-1} \in \map {C_R} S$
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $S$ be a [[Definition:Subset|subset]] of $R$.
Let $\map {C_R} S$ denote the [[Definition:Centralizer of Ring Subset|centralizer]] of $S$ in $R$
Let $u \in R$ be a [[Definition:Unit of Ring|unit]] of $R$.
Then:
:$u \in \map {C_R} S \... | Let $u \in R$ be a [[Definition:Unit of Ring|unit]] of $R$.
Let $u \in \map {C_R} S$.
Then from [[Commutation with Inverse in Monoid]]:
:$u^{-1} \in \map {C_R} S$
{{qed}} | Inverse of Unit in Centralizer of Ring is in Centralizer | https://proofwiki.org/wiki/Inverse_of_Unit_in_Centralizer_of_Ring_is_in_Centralizer | https://proofwiki.org/wiki/Inverse_of_Unit_in_Centralizer_of_Ring_is_in_Centralizer | [
"Ring Theory",
"Subrings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Subset",
"Definition:Centralizer/Ring Subset",
"Definition:Unit of Ring"
] | [
"Definition:Unit of Ring",
"Commutation with Inverse in Monoid"
] |
proofwiki-15656 | Inverse of Central Unit of Ring is in Center | Let $R$ be a ring.
Let $\map Z R$ denote the center of $R$.
Let $u \in R$ be a unit of $R$.
Then:
:$u \in \map Z R \implies u^{-1} \in \map Z R$ | Follows directly from the definition of center and Inverse of Unit in Centralizer of Ring is in Centralizer.
{{qed}} | Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\map Z R$ denote the [[Definition:Center of Ring|center]] of $R$.
Let $u \in R$ be a [[Definition:Unit of Ring|unit]] of $R$.
Then:
:$u \in \map Z R \implies u^{-1} \in \map Z R$ | Follows directly from the definition of [[Definition:Center of Ring|center]] and [[Inverse of Unit in Centralizer of Ring is in Centralizer]].
{{qed}} | Inverse of Central Unit of Ring is in Center | https://proofwiki.org/wiki/Inverse_of_Central_Unit_of_Ring_is_in_Center | https://proofwiki.org/wiki/Inverse_of_Central_Unit_of_Ring_is_in_Center | [
"Ring Theory",
"Subrings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Center (Abstract Algebra)/Ring",
"Definition:Unit of Ring"
] | [
"Definition:Center (Abstract Algebra)/Ring",
"Inverse of Unit in Centralizer of Ring is in Centralizer"
] |
proofwiki-15657 | Skewness in terms of Non-Central Moments | Let $X$ be a random variable with mean $\mu$ and standard deviation $\sigma$.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$ | {{begin-eqn}}
{{eqn | l = \gamma_1
| r = \expect {\paren {\dfrac {X - \mu} \sigma}^3}
| c = {{Defof|Skewness}}
}}
{{eqn | r = \frac {\expect {X^3 - 3 \mu X^2 + 3 \mu^2 X - \mu^3} } {\sigma^3}
| c = Expectation is Linear, Cube of Difference
}}
{{eqn | r = \frac {\expect {X^3} - 3 \mu \expect {X^2} + 3... | Let $X$ be a [[Definition:Random Variable|random variable]] with [[Definition:Expectation|mean]] $\mu$ and [[Definition:Standard Deviation|standard deviation]] $\sigma$.
Then the [[Definition:Skewness|skewness]] $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$ | {{begin-eqn}}
{{eqn | l = \gamma_1
| r = \expect {\paren {\dfrac {X - \mu} \sigma}^3}
| c = {{Defof|Skewness}}
}}
{{eqn | r = \frac {\expect {X^3 - 3 \mu X^2 + 3 \mu^2 X - \mu^3} } {\sigma^3}
| c = [[Expectation is Linear]], [[Cube of Difference]]
}}
{{eqn | r = \frac {\expect {X^3} - 3 \mu \expect {... | Skewness in terms of Non-Central Moments | https://proofwiki.org/wiki/Skewness_in_terms_of_Non-Central_Moments | https://proofwiki.org/wiki/Skewness_in_terms_of_Non-Central_Moments | [
"Skewness"
] | [
"Definition:Random Variable",
"Definition:Expectation",
"Definition:Standard Deviation",
"Definition:Skewness"
] | [
"Expectation is Linear",
"Binomial Theorem/Examples/Cube of Difference",
"Expectation is Linear",
"Variance as Expectation of Square minus Square of Expectation",
"Category:Skewness"
] |
proofwiki-15658 | Intersection of Ring Ideals is Ideal | Let $\struct {R, +, \circ}$ be a ring
Let $\mathbb L$ be a non-empty set of ideals of $R$.
Then the intersection $\bigcap \mathbb L$ of the members of $\mathbb L$ is itself an ideal of $R$. | Let $L = \bigcap \mathbb L$.
From Intersection of Subrings is Subring, we have that $L$ is a subring of $R$.
Let $x \in L$ and $y \in R$.
Then:
:$\forall T \in \bigcap \mathbb L: x \circ y \in T, y \circ x \in T$
as every element of $\bigcap \mathbb L$, including $T$, is an ideal of $R$.
If $y \in R$, then $x \circ y$ ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]]
Let $\mathbb L$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Ideal of Ring|ideals]] of $R$.
Then the [[Definition:Set Intersection|intersection]] $\bigcap \mathbb L$ of the members of $\mathbb L$ is itself an [[Definition... | Let $L = \bigcap \mathbb L$.
From [[Intersection of Subrings is Subring]], we have that $L$ is a [[Definition:Subring|subring]] of $R$.
Let $x \in L$ and $y \in R$.
Then:
:$\forall T \in \bigcap \mathbb L: x \circ y \in T, y \circ x \in T$
as every [[Definition:Element|element]] of $\bigcap \mathbb L$, including $T... | Intersection of Ring Ideals is Ideal | https://proofwiki.org/wiki/Intersection_of_Ring_Ideals_is_Ideal | https://proofwiki.org/wiki/Intersection_of_Ring_Ideals_is_Ideal | [
"Set Intersection",
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Non-Empty Set",
"Definition:Ideal of Ring",
"Definition:Set Intersection",
"Definition:Ideal of Ring"
] | [
"Intersection of Subrings is Subring",
"Definition:Subring",
"Definition:Element",
"Definition:Ideal of Ring",
"Definition:Element",
"Definition:Ideal of Ring"
] |
proofwiki-15659 | Set of Ring Elements forming Zero Product with given Element is Ideal | Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $a \in R$ be an arbitrary element of $R$.
Let $A$ be the subset of $R$ defined as:
:$A = \set {x \in R: x \circ a = 0_R}$
Then $A$ is an ideal of $A$. | By definition of ring zero:
:$\forall x \in R: x \circ 0_R = 0_R$
Hence $0_R \in A$ and so $A \ne \O$.
Let $a, b \in A$.
{{begin-eqn}}
{{eqn | q = \forall x \in R
| l = x \circ b
| r = 0_R
| c =
}}
{{eqn | ll= \leadsto
| l = -\paren {x \circ b}
| r = 0_R
| c =
}}
{{eqn | ll= \leads... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $a \in R$ be an arbitrary [[Definition:Element|element]] of $R$.
Let $A$ be the [[Definition:Subset|subset]] of $R$ defined as:
:$A = \set {x \in R: x \circ a = 0_R}$
Then $A$ is an [... | By definition of [[Definition:Ring Zero|ring zero]]:
:$\forall x \in R: x \circ 0_R = 0_R$
Hence $0_R \in A$ and so $A \ne \O$.
Let $a, b \in A$.
{{begin-eqn}}
{{eqn | q = \forall x \in R
| l = x \circ b
| r = 0_R
| c =
}}
{{eqn | ll= \leadsto
| l = -\paren {x \circ b}
| r = 0_R
... | Set of Ring Elements forming Zero Product with given Element is Ideal | https://proofwiki.org/wiki/Set_of_Ring_Elements_forming_Zero_Product_with_given_Element_is_Ideal | https://proofwiki.org/wiki/Set_of_Ring_Elements_forming_Zero_Product_with_given_Element_is_Ideal | [
"Ideal Theory"
] | [
"Definition:Commutative Ring",
"Definition:Ring Zero",
"Definition:Element",
"Definition:Subset",
"Definition:Ideal of Ring"
] | [
"Definition:Ring Zero",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Test for Ideal"
] |
proofwiki-15660 | Skewness of Bernoulli Distribution | Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac {1 - 2 p} {\sqrt {p q} }$
where $q = 1 - p$. | From Skewness in terms of Non-Central Moments:
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where $\mu$ is the mean of $X$, and $\sigma$ the standard deviation.
We have, by Expectation of Bernoulli Distribution:
:$\mu = p$
By Variance of Bernoulli Distribution, we also have:
:$\var X = \s... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]].
Then the [[Definition:Skewness|skewness]] $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac {1 - 2 p} {\sqrt {p q} }$
where $q = 1 - p$. | From [[Skewness in terms of Non-Central Moments]]:
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where $\mu$ is the [[Definition:Expectation|mean]] of $X$, and $\sigma$ the [[Definition:Standard Deviation|standard deviation]].
We have, by [[Expectation of Bernoulli Distribution]]:
:$\mu... | Skewness of Bernoulli Distribution | https://proofwiki.org/wiki/Skewness_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Skewness_of_Bernoulli_Distribution | [
"Bernoulli Distribution",
"Skewness"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Skewness"
] | [
"Skewness in terms of Non-Central Moments",
"Definition:Expectation",
"Definition:Standard Deviation",
"Expectation of Bernoulli Distribution",
"Variance of Bernoulli Distribution",
"Raw Moment of Bernoulli Distribution",
"Difference of Two Squares",
"Category:Bernoulli Distribution",
"Category:Skew... |
proofwiki-15661 | Intersection of Subrings is Subring | Let $\struct {R, +, \circ}$ be a ring.
Let $\mathbb L$ be a non-empty set of subrings of $R$.
Then the intersection $\bigcap \mathbb L$ of the members of $\mathbb L$ is itself a subring of $R$. | Let $L = \bigcap \mathbb L$.
By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is a subgroup of $\struct {R, +}$.
By the One-Step Subgroup Test:
:$\forall x, y \in \struct {L, +}: x + \paren {-y} \in L$
By Intersection of Subsemigroups, $\struct {L, \circ}$ a subsemigroup of $\struct {R, \circ}$.
So by definit... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\mathbb L$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Subring|subrings]] of $R$.
Then the [[Definition:Set Intersection|intersection]] $\bigcap \mathbb L$ of the members of $\mathbb L$ is itself a [[Definition:Sub... | Let $L = \bigcap \mathbb L$.
By [[Intersection of Subgroups is Subgroup]], $\struct {L, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {R, +}$.
By the [[One-Step Subgroup Test]]:
:$\forall x, y \in \struct {L, +}: x + \paren {-y} \in L$
By [[Intersection of Subsemigroups]], $\struct {L, \circ}$ a [[Definitio... | Intersection of Subrings is Subring | https://proofwiki.org/wiki/Intersection_of_Subrings_is_Subring | https://proofwiki.org/wiki/Intersection_of_Subrings_is_Subring | [
"Set Intersection",
"Subrings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Non-Empty Set",
"Definition:Subring",
"Definition:Set Intersection",
"Definition:Subring"
] | [
"Intersection of Subgroups is Subgroup",
"Definition:Subgroup",
"One-Step Subgroup Test",
"Intersection of Subsemigroups",
"Definition:Subsemigroup",
"Definition:Subsemigroup",
"Subring Test",
"Definition:Subring"
] |
proofwiki-15662 | Median of Exponential Distribution | Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$.
Then the median of $X$ is equal to $\beta \ln 2$. | Let $M$ be the median of $X$.
From the definition of the exponential distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$
Note that $f_X$ is non-zero, so the median is unique.
{{Explain|Why the above follows}}
We have by the definition of a median:
:$\ds \map \Pr {X ... | Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] of the [[Definition:Exponential Distribution|exponential distribution]] with parameter $\beta$ for some $\beta \in \R_{> 0}$.
Then the [[Definition:Median of Continuous Random Variable|median]] of $X$ is equal to $\beta \ln 2$. | Let $M$ be the [[Definition:Median of Continuous Random Variable|median]] of $X$.
From the definition of the [[Definition:Exponential Distribution|exponential distribution]], $X$ has [[Definition:Probability Density Function|probability density function]]:
:$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$
Note th... | Median of Exponential Distribution | https://proofwiki.org/wiki/Median_of_Exponential_Distribution | https://proofwiki.org/wiki/Median_of_Exponential_Distribution | [
"Exponential Distribution",
"Medians"
] | [
"Definition:Random Variable/Continuous",
"Definition:Exponential Distribution",
"Definition:Median of Continuous Random Variable"
] | [
"Definition:Median of Continuous Random Variable",
"Definition:Exponential Distribution",
"Definition:Probability Density Function",
"Definition:Median of Continuous Random Variable",
"Definition:Median of Continuous Random Variable",
"Definition:Definite Integral",
"Primitive of Exponential Function",
... |
proofwiki-15663 | Intersection of Ring Ideals is Largest Ideal Contained in all Ideals | Let $\struct {R, +, \circ}$ be a ring
Let $\mathbb L$ be a non-empty set of ideals of $R$.
Then the intersection $\bigcap \mathbb L$ of the members of $\mathbb L$ is the largest ideal of $R$ contained in each member of $\mathbb L$. | Let $L = \bigcap \mathbb L$.
From Intersection of Ring Ideals is Ideal $L$ is indeed an ideal of $R$.
Let $L = \bigcap \mathbb L$.
From Intersection of Subrings is Largest Subring Contained in all Subrings, we have that $L$ is the largest subring of $R$ contained in each member of $\mathbb L$.
As $L$ is the largest su... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]]
Let $\mathbb L$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Ideal of Ring|ideals]] of $R$.
Then the [[Definition:Set Intersection|intersection]] $\bigcap \mathbb L$ of the members of $\mathbb L$ is the largest [[Definiti... | Let $L = \bigcap \mathbb L$.
From [[Intersection of Ring Ideals is Ideal]] $L$ is indeed an [[Definition:Ideal of Ring|ideal]] of $R$.
Let $L = \bigcap \mathbb L$.
From [[Intersection of Subrings is Largest Subring Contained in all Subrings]], we have that $L$ is the largest [[Definition:Subring|subring]] of $R$ c... | Intersection of Ring Ideals is Largest Ideal Contained in all Ideals | https://proofwiki.org/wiki/Intersection_of_Ring_Ideals_is_Largest_Ideal_Contained_in_all_Ideals | https://proofwiki.org/wiki/Intersection_of_Ring_Ideals_is_Largest_Ideal_Contained_in_all_Ideals | [
"Set Intersection",
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Non-Empty Set",
"Definition:Ideal of Ring",
"Definition:Set Intersection",
"Definition:Ideal of Ring"
] | [
"Intersection of Ring Ideals is Ideal",
"Definition:Ideal of Ring",
"Intersection of Subrings is Largest Subring Contained in all Subrings",
"Definition:Subring",
"Definition:Subring",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring",
"Definition:Subring",
"Definition:Ideal of Ring"
] |
proofwiki-15664 | Intersection of Subrings is Largest Subring Contained in all Subrings | Let $\struct {R, +, \circ}$ be a ring.
Let $\mathbb L$ be a non-empty set of subrings of $R$.
Then the intersection $\ds \bigcap \mathbb L$ of the members of $\mathbb L$ is the largest subring of $R$ contained in each member of $\mathbb L$. | Let $\ds L = \bigcap \mathbb L$.
From Intersection of Subrings is Subring, $L$ is indeed a subring of $R$.
By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is the largest subgroup of $\struct {R, +}$ contained in each member of $\mathbb L$.
By Intersection of Subsemigroups, $\struct {L, \circ}$ is the largest... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\mathbb L$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Subring|subrings]] of $R$.
Then the [[Definition:Set Intersection|intersection]] $\ds \bigcap \mathbb L$ of the members of $\mathbb L$ is the largest [[Definit... | Let $\ds L = \bigcap \mathbb L$.
From [[Intersection of Subrings is Subring]], $L$ is indeed a [[Definition:Subring|subring]] of $R$.
By [[Intersection of Subgroups is Subgroup]], $\struct {L, +}$ is the largest [[Definition:Subgroup|subgroup]] of $\struct {R, +}$ contained in each member of $\mathbb L$.
By [[Inter... | Intersection of Subrings is Largest Subring Contained in all Subrings | https://proofwiki.org/wiki/Intersection_of_Subrings_is_Largest_Subring_Contained_in_all_Subrings | https://proofwiki.org/wiki/Intersection_of_Subrings_is_Largest_Subring_Contained_in_all_Subrings | [
"Set Intersection",
"Subrings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Non-Empty Set",
"Definition:Subring",
"Definition:Set Intersection",
"Definition:Subring"
] | [
"Intersection of Subrings is Subring",
"Definition:Subring",
"Intersection of Subgroups is Subgroup",
"Definition:Subgroup",
"Intersection of Subsemigroups",
"Definition:Subsemigroup",
"Definition:Subring",
"Intersection is Largest Subset",
"Definition:Subring"
] |
proofwiki-15665 | Median of Continuous Uniform Distribution | Let $X$ be a continuous random variable which is uniformly distributed on a closed real interval $\closedint a b$.
Then the median $M$ of $X$ is given by:
:$M = \dfrac {a + b} 2$ | From the definition of the continuous uniform distribution, $X$ has probability density function:
:$\map {f_X} x = \begin{cases} \dfrac 1 {b - a} & : a \leq x \leq b \\ 0 & : \text{otherwise} \end{cases}$
Note that $f_X$ is non-zero, so the median is unique.
We have by the definition of a median:
{{begin-eqn}}
{{eqn ... | Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] which is [[Definition:Continuous Uniform Distribution|uniformly distributed]] on a [[Definition:Closed Real Interval|closed real interval]] $\closedint a b$.
Then the [[Definition:Median of Continuous Random Variable|median]] $M$ of $X$... | From the definition of the [[Definition:Continuous Uniform Distribution|continuous uniform distribution]], $X$ has [[Definition:Probability Density Function|probability density function]]:
:$\map {f_X} x = \begin{cases} \dfrac 1 {b - a} & : a \leq x \leq b \\ 0 & : \text{otherwise} \end{cases}$
Note that $f_X$ is no... | Median of Continuous Uniform Distribution | https://proofwiki.org/wiki/Median_of_Continuous_Uniform_Distribution | https://proofwiki.org/wiki/Median_of_Continuous_Uniform_Distribution | [
"Continuous Uniform Distribution",
"Medians"
] | [
"Definition:Random Variable/Continuous",
"Definition:Uniform Distribution/Continuous",
"Definition:Real Interval/Closed",
"Definition:Median of Continuous Random Variable"
] | [
"Definition:Uniform Distribution/Continuous",
"Definition:Probability Density Function",
"Definition:Median of Continuous Random Variable",
"Definition:Unique",
"Definition:Median of Continuous Random Variable",
"Primitive of Constant",
"Category:Continuous Uniform Distribution",
"Category:Medians"
] |
proofwiki-15666 | Intersection of All Ring Ideals Containing Subset is Smallest | Let $\struct {R, +, \circ}$ be a ring
Let $S \subseteq R$ be a subset of $R$.
Let $L$ be the intersection of the set of all ideals of $R$ containing $S$.
Then $L$ is the smallest ideal of $R$ containing $S$. | From Intersection of All Subrings Containing Subset is Smallest, $L$ is the smallest subring of $R$ containing $S$.
From Intersection of Ring Ideals is Ideal, $L$ is an ideal of $R$.
As $L$ is the smallest subring of $R$ containing $S$, and it is an ideal of $R$, there can be no smaller ideal of $R$ containing $S$ as i... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]]
Let $S \subseteq R$ be a [[Definition:Subset|subset]] of $R$.
Let $L$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Ideal of Ring|ideals]] of $R$ containing $S$.
Then $L$ is the small... | From [[Intersection of All Subrings Containing Subset is Smallest]], $L$ is the smallest [[Definition:Subring|subring]] of $R$ containing $S$.
From [[Intersection of Ring Ideals is Ideal]], $L$ is an [[Definition:Ideal of Ring|ideal]] of $R$.
As $L$ is the smallest [[Definition:Subring|subring]] of $R$ containing $S... | Intersection of All Ring Ideals Containing Subset is Smallest | https://proofwiki.org/wiki/Intersection_of_All_Ring_Ideals_Containing_Subset_is_Smallest | https://proofwiki.org/wiki/Intersection_of_All_Ring_Ideals_Containing_Subset_is_Smallest | [
"Set Intersection",
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Set",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring"
] | [
"Intersection of All Subrings Containing Subset is Smallest",
"Definition:Subring",
"Intersection of Ring Ideals is Ideal",
"Definition:Ideal of Ring",
"Definition:Subring",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring",
"Definition:Subring",
"Definition:Set Intersection",
"Definition:Set... |
proofwiki-15667 | Intersection of All Subrings Containing Subset is Smallest | Let $\struct {R, +, \circ}$ be a ring.
Let $S \subseteq R$ be a subset of $R$.
Let $L$ be the intersection of the set of all subrings of $R$ containing $S$.
Then $L$ is the smallest subring of $R$ containing $S$. | From Intersection of Subrings is Subring, $L$ is indeed a subring of $R$.
Let $T$ be a subring of $R$ containing $S$.
Let $x, y \in L$.
By the Subring Test, we have that:
{{begin-eqn}}
{{eqn | l = x - y
| o = \in
| r = L
}}
{{eqn | l = x \circ y
| o = \in
| r = L
}}
{{end-eqn}}
By Intersection i... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $S \subseteq R$ be a [[Definition:Subset|subset]] of $R$.
Let $L$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Subring|subrings]] of $R$ containing $S$.
Then $L$ is the smallest... | From [[Intersection of Subrings is Subring]], $L$ is indeed a [[Definition:Subring|subring]] of $R$.
Let $T$ be a [[Definition:Subring|subring]] of $R$ containing $S$.
Let $x, y \in L$.
By the [[Subring Test]], we have that:
{{begin-eqn}}
{{eqn | l = x - y
| o = \in
| r = L
}}
{{eqn | l = x \circ y
... | Intersection of All Subrings Containing Subset is Smallest | https://proofwiki.org/wiki/Intersection_of_All_Subrings_Containing_Subset_is_Smallest | https://proofwiki.org/wiki/Intersection_of_All_Subrings_Containing_Subset_is_Smallest | [
"Set Intersection",
"Subrings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Set",
"Definition:Subring",
"Definition:Subring"
] | [
"Intersection of Subrings is Subring",
"Definition:Subring",
"Definition:Subring",
"Subring Test",
"Intersection is Largest Subset",
"Definition:Subring",
"Subring Test",
"Definition:Subset"
] |
proofwiki-15668 | Intersection of Division Subrings is Division Subring | Let $\struct {D, +, \circ}$ be a division ring.
Let $\mathbb K$ be a non-empty set of division subrings of $D$.
Then the intersection $\ds \bigcap \mathbb K$ of the members of $\mathbb K$ is itself a division subring of $D$. | Let $\ds L = \bigcap \mathbb K$.
Let $0$ be the zero of $\struct {D, +, \circ}$.
By Intersection of Subgroups is Subgroup: General Result, $\struct {L, +}$ is a subgroup of $\struct {D, +}$.
By the One-Step Subgroup Test:
:$\forall x, y \in L: x + \paren {-y} \in L$
By Non-Zero Elements of Division Ring form Group:
:$\... | Let $\struct {D, +, \circ}$ be a [[Definition:Division Ring|division ring]].
Let $\mathbb K$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Division Subring|division subrings]] of $D$.
Then the [[Definition:Set Intersection|intersection]] $\ds \bigcap \mathbb K$ of the members of $\mathbb K$ is itse... | Let $\ds L = \bigcap \mathbb K$.
Let $0$ be the [[Definition:Ring Zero|zero]] of $\struct {D, +, \circ}$.
By [[Intersection of Subgroups is Subgroup/General Result|Intersection of Subgroups is Subgroup: General Result]], $\struct {L, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {D, +}$.
By the [[One-Step S... | Intersection of Division Subrings is Division Subring | https://proofwiki.org/wiki/Intersection_of_Division_Subrings_is_Division_Subring | https://proofwiki.org/wiki/Intersection_of_Division_Subrings_is_Division_Subring | [
"Division Subrings",
"Set Intersection"
] | [
"Definition:Division Ring",
"Definition:Non-Empty Set",
"Definition:Division Subring",
"Definition:Set Intersection",
"Definition:Division Subring"
] | [
"Definition:Ring Zero",
"Intersection of Subgroups is Subgroup/General Result",
"Definition:Subgroup",
"One-Step Subgroup Test",
"Non-Zero Elements of Division Ring form Group",
"Definition:Group",
"Definition:Group",
"Set Difference over Subset",
"Definition:Subgroup",
"Set Difference is Right Di... |
proofwiki-15669 | Intersection of Subfields is Subfield | Let $\struct {F, +, \circ}$ be a field.
Let $\mathbb K$ be a non-empty set of subfields of $F$.
Then the intersection $\ds \bigcap \mathbb K$ of the members of $\mathbb K$ is itself a subfield of $F$. | Let $\ds L = \bigcap \mathbb K$.
A field is by definition also a division ring.
From Intersection of Division Subrings is Division Subring, $L$ is itself a division subring of $F$.
As $\struct {F, +, \circ}$ is a field, $\circ$ is commutative on $F$.
By Restriction of Commutative Operation is Commutative, it follows th... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $\mathbb K$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Subfield|subfields]] of $F$.
Then the [[Definition:Set Intersection|intersection]] $\ds \bigcap \mathbb K$ of the members of $\mathbb K$ is itself a [[Defini... | Let $\ds L = \bigcap \mathbb K$.
A [[Definition:Field (Abstract Algebra)|field]] is by definition also a [[Definition:Division Ring|division ring]].
From [[Intersection of Division Subrings is Division Subring]], $L$ is itself a [[Definition:Division Subring|division subring]] of $F$.
As $\struct {F, +, \circ}$ is ... | Intersection of Subfields is Subfield | https://proofwiki.org/wiki/Intersection_of_Subfields_is_Subfield | https://proofwiki.org/wiki/Intersection_of_Subfields_is_Subfield | [
"Set Intersection",
"Subfields"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Non-Empty Set",
"Definition:Subfield",
"Definition:Set Intersection",
"Definition:Subfield"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Division Ring",
"Intersection of Division Subrings is Division Subring",
"Definition:Division Subring",
"Definition:Field (Abstract Algebra)",
"Definition:Commutative/Operation",
"Restriction of Commutative Operation is Commutative",
"Definition:Commu... |
proofwiki-15670 | Intersection of Division Subrings is Largest Division Subring Contained in all Division Subrings | Let $\struct {D, +, \circ}$ be a division ring.
Let $\mathbb K$ be a non-empty set of division subrings of $D$.
Let $\ds \bigcap \mathbb K$ be the intersection of the elements of $\mathbb K$.
Then $\ds \bigcap \mathbb K$ is the largest division subring of $D$ contained in each element of $\mathbb K$. | Let $\ds L = \bigcap \mathbb K$.
Let $0$ be the zero of $\struct {D, +, \circ}$.
From Intersection of Division Subrings is Division Subring, $\struct {L, +, \circ}$ is a division subring of $\struct {D, +, \circ}$.
By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is the largest subgroup of $\struct {D, +}$ co... | Let $\struct {D, +, \circ}$ be a [[Definition:Division Ring|division ring]].
Let $\mathbb K$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Division Subring|division subrings]] of $D$.
Let $\ds \bigcap \mathbb K$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Element|elements... | Let $\ds L = \bigcap \mathbb K$.
Let $0$ be the [[Definition:Ring Zero|zero]] of $\struct {D, +, \circ}$.
From [[Intersection of Division Subrings is Division Subring]], $\struct {L, +, \circ}$ is a [[Definition:Division Subring|division subring]] of $\struct {D, +, \circ}$.
By [[Intersection of Subgroups is Subgro... | Intersection of Division Subrings is Largest Division Subring Contained in all Division Subrings | https://proofwiki.org/wiki/Intersection_of_Division_Subrings_is_Largest_Division_Subring_Contained_in_all_Division_Subrings | https://proofwiki.org/wiki/Intersection_of_Division_Subrings_is_Largest_Division_Subring_Contained_in_all_Division_Subrings | [
"Division Subrings"
] | [
"Definition:Division Ring",
"Definition:Non-Empty Set",
"Definition:Division Subring",
"Definition:Set Intersection",
"Definition:Element",
"Definition:Division Subring",
"Definition:Element"
] | [
"Definition:Ring Zero",
"Intersection of Division Subrings is Division Subring",
"Definition:Division Subring",
"Intersection of Subgroups is Subgroup",
"Definition:Subgroup",
"Definition:Element",
"Intersection of Subgroups is Subgroup",
"Definition:Subgroup",
"Definition:Element",
"Definition:Su... |
proofwiki-15671 | Intersection of Division Subrings Containing Subset is Smallest | Let $\struct {D, +, \circ}$ be a division ring.
Let $S \subseteq D$ be a subset of $D$.
Let $L$ be the intersection of the set of all division subrings of $D$ containing $S$.
Then $L$ is the smallest division subring of $D$ containing $S$. | From Intersection of Division Subrings is Division Subring, $L$ is indeed a division subring of $D$.
Let $T$ be a division subring of $D$ containing $S$.
Let $x, y \in L$.
By the Division Subring Test, we have that:
{{begin-eqn}}
{{eqn | l = x - y
| o = \in
| r = L
}}
{{eqn | l = x \circ y
| o = \in
... | Let $\struct {D, +, \circ}$ be a [[Definition:Division Ring|division ring]].
Let $S \subseteq D$ be a [[Definition:Subset|subset]] of $D$.
Let $L$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Division Subring|division subrings]] of $D$ containing $S$.
Then $L... | From [[Intersection of Division Subrings is Division Subring]], $L$ is indeed a [[Definition:Division Subring|division subring]] of $D$.
Let $T$ be a [[Definition:Division Subring|division subring]] of $D$ containing $S$.
Let $x, y \in L$.
By the [[Division Subring Test]], we have that:
{{begin-eqn}}
{{eqn | l = x ... | Intersection of Division Subrings Containing Subset is Smallest | https://proofwiki.org/wiki/Intersection_of_Division_Subrings_Containing_Subset_is_Smallest | https://proofwiki.org/wiki/Intersection_of_Division_Subrings_Containing_Subset_is_Smallest | [
"Division Subrings"
] | [
"Definition:Division Ring",
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Set",
"Definition:Division Subring",
"Definition:Division Subring"
] | [
"Intersection of Division Subrings is Division Subring",
"Definition:Division Subring",
"Definition:Division Subring",
"Division Subring Test",
"Intersection is Largest Subset",
"Definition:Division Subring",
"Division Subring Test",
"Definition:Subset"
] |
proofwiki-15672 | Intersection of Subfields Containing Subset is Smallest | Let $\struct {F, +, \circ}$ be a field.
Let $S \subseteq F$ be a subset of $F$.
Let $L$ be the intersection of the set of all subfields of $F$ containing $S$.
Then $L$ is the smallest subfield of $F$ containing $S$. | A field is by definition also a division subring.
Thus $L$ is the intersection of the set of all division subrings of $F$ containing $S$.
From Intersection of Division Subrings Containing Subset is Smallest, $L$ is the smallest division subring of $F$ containing $S$.
From Intersection of Subfields is Subfield, $L$ is a... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $S \subseteq F$ be a [[Definition:Subset|subset]] of $F$.
Let $L$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Subfield|subfields]] of $F$ containing $S$.
Then $L$ is the smal... | A [[Definition:Field (Abstract Algebra)|field]] is by definition also a [[Definition:Division Subring|division subring]].
Thus $L$ is the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Division Subring|division subrings]] of $F$ containing $S$.
From [[Intersection of Di... | Intersection of Subfields Containing Subset is Smallest | https://proofwiki.org/wiki/Intersection_of_Subfields_Containing_Subset_is_Smallest | https://proofwiki.org/wiki/Intersection_of_Subfields_Containing_Subset_is_Smallest | [
"Set Intersection",
"Subfields"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Set",
"Definition:Subfield",
"Definition:Subfield"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Division Subring",
"Definition:Set Intersection",
"Definition:Set",
"Definition:Division Subring",
"Intersection of Division Subrings Containing Subset is Smallest",
"Definition:Division Subring",
"Intersection of Subfields is Subfield",
"Definition... |
proofwiki-15673 | Intersection of Subfields is Largest Subfield Contained in all Subfields | Let $\struct {F, +, \circ}$ be a field.
Let $\mathbb K$ be a non-empty set of subfields of $F$.
Let $\bigcap \mathbb K$ be the intersection of the elements of $\mathbb K$.
Then $\bigcap \mathbb K$ is the largest subfield of $F$ contained in each element of $\mathbb K$. | Let $L = \bigcap \mathbb K$.
From Intersection of Subfields is Subfield, $\struct {L, +, \circ}$ is itself a subfield of $\struct {F, +, \circ}$.
A field is by definition also a division subring.
Thus $\struct {L, +, \circ}$ is the largest division subring of $F$ contained in each element of $\mathbb K$.
But as $\struc... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $\mathbb K$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Subfield|subfields]] of $F$.
Let $\bigcap \mathbb K$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Element|elements]] of $\mathbb K$... | Let $L = \bigcap \mathbb K$.
From [[Intersection of Subfields is Subfield]], $\struct {L, +, \circ}$ is itself a [[Definition:Subfield|subfield]] of $\struct {F, +, \circ}$.
A [[Definition:Field (Abstract Algebra)|field]] is by definition also a [[Definition:Division Subring|division subring]].
Thus $\struct {L, +, ... | Intersection of Subfields is Largest Subfield Contained in all Subfields | https://proofwiki.org/wiki/Intersection_of_Subfields_is_Largest_Subfield_Contained_in_all_Subfields | https://proofwiki.org/wiki/Intersection_of_Subfields_is_Largest_Subfield_Contained_in_all_Subfields | [
"Set Intersection",
"Subfields"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Non-Empty Set",
"Definition:Subfield",
"Definition:Set Intersection",
"Definition:Element",
"Definition:Subfield",
"Definition:Element"
] | [
"Intersection of Subfields is Subfield",
"Definition:Subfield",
"Definition:Field (Abstract Algebra)",
"Definition:Division Subring",
"Definition:Division Subring",
"Definition:Element",
"Definition:Field (Abstract Algebra)",
"Definition:Commutative/Operation",
"Restriction of Commutative Operation ... |
proofwiki-15674 | Equivalent Norms on Rational Numbers/Necessary Condition | Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be norms on the rational numbers $\Q$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be equivalent norms.
Then:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$ | Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be equivalent.
By Norm is Power of Other Norm then:
:$\exists \alpha \in \R_{\gt 0}: \forall q \in \Q: \norm q_1 = \norm q_2^\alpha$
In particular:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$ | Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be [[Definition:Norm on Division Ring|norms]] on the [[Definition:Rational Numbers|rational numbers]] $\Q$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be [[Definition:Equivalent Division Ring Norms|equivalent norms]].
Then:
:$\exists \alpha \in \R_... | Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be [[Definition:Equivalent Division Ring Norms|equivalent]].
By [[Definition:Equivalent Division Ring Norms/Norm is Power of Other Norm|Norm is Power of Other Norm]] then:
:$\exists \alpha \in \R_{\gt 0}: \forall q \in \Q: \norm q_1 = \norm q_2^\alpha$
In partic... | Equivalent Norms on Rational Numbers/Necessary Condition | https://proofwiki.org/wiki/Equivalent_Norms_on_Rational_Numbers/Necessary_Condition | https://proofwiki.org/wiki/Equivalent_Norms_on_Rational_Numbers/Necessary_Condition | [
"Normed Division Rings"
] | [
"Definition:Norm/Division Ring",
"Definition:Rational Number",
"Definition:Equivalent Division Ring Norms"
] | [
"Definition:Equivalent Division Ring Norms",
"Definition:Equivalent Division Ring Norms/Norm is Power of Other Norm"
] |
proofwiki-15675 | Equivalent Norms on Rational Numbers/Sufficient Condition | Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be norms on the rational numbers $\Q$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
Then $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent | Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
By Norm of Negative then:
:$\forall n \in \N: \norm {-n}_1 = \norm n_1 = \norm n_2^\alpha = \norm {-n}_2^\alpha$
Hence:
:$\forall k \in \Z: \norm k_1 = \norm k_2^\alpha$
By N... | Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be [[Definition:Norm on Division Ring|norms]] on the [[Definition:Rational Numbers|rational numbers]] $\Q$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
Then $\... | Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
By [[Properties of Norm on Division Ring/Norm of Negative|Norm of Negative]] then:
:$\forall n \in \N: \norm {-n}_1 = \norm n_1 = \norm n_2^\alpha = \norm {-n}_2^\alpha$
H... | Equivalent Norms on Rational Numbers/Sufficient Condition | https://proofwiki.org/wiki/Equivalent_Norms_on_Rational_Numbers/Sufficient_Condition | https://proofwiki.org/wiki/Equivalent_Norms_on_Rational_Numbers/Sufficient_Condition | [
"Normed Division Rings"
] | [
"Definition:Norm/Division Ring",
"Definition:Rational Number",
"Definition:Equivalent Division Ring Norms"
] | [
"Properties of Norm on Division Ring/Norm of Negative",
"Properties of Norm on Division Ring/Norm of Quotient",
"Definition:Equivalent Division Ring Norms/Norm is Power of Other Norm",
"Definition:Equivalent Division Ring Norms"
] |
proofwiki-15676 | Commutative and Unitary Ring with 2 Ideals is Field | Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$.
Let $\struct {R, +, \circ}$ be such that the only ideals of $\struct {R, +, \circ}$ are:
:$\set {0_R}$
and:
$\struct {R, +, \circ}$ itself.
That is, such that $\struct {R, +, \circ}$ has no non-null proper ideals.
Then $\struct {R, +, \ci... | From Null Ring is Ideal and Ring is Ideal of Itself, it is always the case that $\set {0_R}$ and $\struct {R, +, \circ}$ are ideals of $\struct {R, +, \circ}$.
Let $a \in R^*$, where $R^* := R \setminus \set {0_R}$.
Let $\ideal a$ be the principal ideal of $R$ generated by $a$.
We have that $\ideal a$ is a non-null ide... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $\struct {R, +, \circ}$ be such that the only [[Definition:Ideal of Ring|ideals]] of $\struct {R, +, \circ}$ are:
:$\set {0_R}$
and:
$\struct {R, +, \circ}$ itself.... | From [[Null Ring is Ideal]] and [[Ring is Ideal of Itself]], it is always the case that $\set {0_R}$ and $\struct {R, +, \circ}$ are [[Definition:Ideal of Ring|ideals]] of $\struct {R, +, \circ}$.
Let $a \in R^*$, where $R^* := R \setminus \set {0_R}$.
Let $\ideal a$ be the [[Definition:Principal Ideal of Ring|princ... | Commutative and Unitary Ring with 2 Ideals is Field | https://proofwiki.org/wiki/Commutative_and_Unitary_Ring_with_2_Ideals_is_Field | https://proofwiki.org/wiki/Commutative_and_Unitary_Ring_with_2_Ideals_is_Field | [
"Commutative Algebra",
"Field Theory",
"Ideal Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero",
"Definition:Ideal of Ring",
"Definition:Null Ideal",
"Definition:Ideal of Ring/Proper Ideal",
"Definition:Field (Abstract Algebra)"
] | [
"Null Ring is Ideal",
"Ring is Ideal of Itself",
"Definition:Ideal of Ring",
"Definition:Principal Ideal of Ring",
"Definition:Non-Null Ideal",
"Definition:Principal Ideal of Ring",
"Definition:Invertible Element",
"Definition:Invertible Element",
"Definition:Division Ring",
"Definition:Commutativ... |
proofwiki-15677 | Field has 2 Ideals | Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Then the only ideals of $\struct {F, +, \circ}$ are $\struct {F, +}$ and $\set {0_F}$.
That is, $\struct {F, +, \circ}$ has no non-null proper ideals. | By definition, a field is a division ring.
From Null Ring is Ideal and Ring is Ideal of Itself, it is always the case that $\set {0_F}$ and $\struct {F, +}$ are ideals of $\struct {F, +, \circ}$.
From Ideals of Division Ring, it follows that the only ideals of $\struct {F, +, \circ}$ are $\struct {F, +}$ and $\set {0_F... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Then the only [[Definition:Ideal of Ring|ideals]] of $\struct {F, +, \circ}$ are $\struct {F, +}$ and $\set {0_F}$.
That is, $\struct {F... | By definition, a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Division Ring|division ring]].
From [[Null Ring is Ideal]] and [[Ring is Ideal of Itself]], it is always the case that $\set {0_F}$ and $\struct {F, +}$ are [[Definition:Ideal of Ring|ideals]] of $\struct {F, +, \circ}$.
From [[Ideals of... | Field has 2 Ideals | https://proofwiki.org/wiki/Field_has_2_Ideals | https://proofwiki.org/wiki/Field_has_2_Ideals | [
"Commutative Algebra",
"Field Theory",
"Ideal Theory"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Ideal of Ring",
"Definition:Null Ideal",
"Definition:Ideal of Ring/Proper Ideal"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Division Ring",
"Null Ring is Ideal",
"Ring is Ideal of Itself",
"Definition:Ideal of Ring",
"Ideals of Division Ring",
"Definition:Ideal of Ring"
] |
proofwiki-15678 | Non-Commutative Ring with Unity and 2 Ideals not necessarily Division Ring | Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {R, +, \circ}$ specifically not be commutative.
Let $\struct {R, +, \circ}$ be such that the only ideals of $\struct {R, +, \circ}$ are $\set {0_R}$ and $R$ itself.
Then it is not necessarily the case that $\stru... | Let $S$ be the set of square matrices of order $2$ over the real numbers $\R$.
$S$ is not a division ring, as for example:
:$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$
and so both $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_F$ and whose [[Definition:Unity of Ring|unity]] is $1_F$.
Let $\struct {R, +, \circ}$ specifically not be [[Definition:Commutative Ring|commutative]].
Let $\struct {R, +, \circ}$ be such that the ... | Let $S$ be the [[Definition:Set|set]] of [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order $2$]] over the [[Definition:Real Number|real numbers]] $\R$.
$S$ is not a [[Definition:Division Ring|division ring]], as for example:
:$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{... | Non-Commutative Ring with Unity and 2 Ideals not necessarily Division Ring | https://proofwiki.org/wiki/Non-Commutative_Ring_with_Unity_and_2_Ideals_not_necessarily_Division_Ring | https://proofwiki.org/wiki/Non-Commutative_Ring_with_Unity_and_2_Ideals_not_necessarily_Division_Ring | [
"Division Rings",
"Rings with Unity",
"Ideal Theory"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Commutative Ring",
"Definition:Ideal of Ring",
"Definition:Division Ring"
] | [
"Definition:Set",
"Definition:Matrix/Square Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Real Number",
"Definition:Division Ring",
"Definition:Proper Zero Divisor",
"Definition:Ideal of Ring",
"Definition:Zero Matrix",
"Definition:Matrix/Square Matrix",
"Definition:Matrix/Element"... |
proofwiki-15679 | General Antiperiodicity Property | Let $f: X \to X$ be an antiperiodic function, where $X$ is either $\R$ or $\C$.
Let $L$ be an antiperiodic element of $f$.
Let $n \in \Z$ be an integer.
:If $n$ is even, then $n L$ is a periodic element of $f$.
:If $n$ is odd, then $n L$ is an antiperiodic element of $f$. | Suppose that $X = \C$. | Let $f: X \to X$ be an [[Definition:Antiperiodic Function|antiperiodic function]], where $X$ is either $\R$ or $\C$.
Let $L$ be an [[Definition:Antiperiodic Element|antiperiodic element]] of $f$.
Let $n \in \Z$ be an [[Definition:Integer|integer]].
:If $n$ is [[Definition:Even Integer|even]], then $n L$ is a [[Defi... | Suppose that $X = \C$. | General Antiperiodicity Property | https://proofwiki.org/wiki/General_Antiperiodicity_Property | https://proofwiki.org/wiki/General_Antiperiodicity_Property | [
"Antiperiodic Functions"
] | [
"Definition:Antiperiodic Function",
"Definition:Antiperiodic Function/Antiperiodic Element",
"Definition:Integer",
"Definition:Even Integer",
"Definition:Periodic Function/Periodic Element",
"Definition:Odd Integer",
"Definition:Antiperiodic Function/Antiperiodic Element"
] | [] |
proofwiki-15680 | Antiperiodic Element is Multiple of Antiperiod | Let $f: \R \to \R$ be a real anti-periodic function with anti-period $A$.
Let $L$ be an anti-periodic element of $f$.
Then $A \divides L$. | {{AimForCont}} that $A \nmid L$.
By the Division Theorem we have:
:$\exists! q \in \Z, r \in \R: L = q A + r, 0 < r < A$
By Even and Odd Integers form Partition of Integers, it follows that $q$ must be either even or odd. | Let $f: \R \to \R$ be a [[Definition:Real Antiperiodic Function|real anti-periodic function]] with [[Definition:Antiperiod|anti-period]] $A$.
Let $L$ be an [[Definition:Antiperiodic Element|anti-periodic element]] of $f$.
Then $A \divides L$. | {{AimForCont}} that $A \nmid L$.
By the [[Division Theorem/Real Number Index|Division Theorem]] we have:
:$\exists! q \in \Z, r \in \R: L = q A + r, 0 < r < A$
By [[Even and Odd Integers form Partition of Integers]], it follows that $q$ must be either [[Definition:Even Integer|even]] or [[Definition:Odd Integer|odd]]... | Antiperiodic Element is Multiple of Antiperiod | https://proofwiki.org/wiki/Antiperiodic_Element_is_Multiple_of_Antiperiod | https://proofwiki.org/wiki/Antiperiodic_Element_is_Multiple_of_Antiperiod | [
"Antiperiodic Functions"
] | [
"Definition:Antiperiodic Function/Real",
"Definition:Antiperiodic Function/Antiperiod",
"Definition:Antiperiodic Function/Antiperiodic Element"
] | [
"Division Theorem/Real Number Index",
"Even and Odd Integers form Partition of Integers",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Odd Integer"
] |
proofwiki-15681 | Differential Entropy of Exponential Distribution | Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$.
Then the differential entropy of $X$, $\map h X$, is given by:
:$\map h X = 1 + \map \ln \beta$ | From the definition of the exponential distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$
From the definition of differential entropy:
:$\ds \map h X = -\int_0^\infty \map {f_X} x \map \ln {\map {f_X} x} \rd x$
So:
{{begin-eqn}}
{{eqn | l = \map h X
| r = -\fr... | Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] of the [[Definition:Exponential Distribution|exponential distribution]] with parameter $\beta$ for some $\beta \in \R_{> 0}$.
Then the [[Definition:Differential Entropy|differential entropy]] of $X$, $\map h X$, is given by:
:$\map h ... | From the definition of the [[Definition:Exponential Distribution|exponential distribution]], $X$ has [[Definition:Probability Density Function|probability density function]]:
:$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$
From the definition of [[Definition:Differential Entropy|differential entropy]]:
:$\ds \m... | Differential Entropy of Exponential Distribution | https://proofwiki.org/wiki/Differential_Entropy_of_Exponential_Distribution | https://proofwiki.org/wiki/Differential_Entropy_of_Exponential_Distribution | [
"Differential Entropy",
"Exponential Distribution"
] | [
"Definition:Random Variable/Continuous",
"Definition:Exponential Distribution",
"Definition:Differential Entropy"
] | [
"Definition:Exponential Distribution",
"Definition:Probability Density Function",
"Definition:Differential Entropy",
"Reciprocal of Logarithm",
"Sum of Logarithms",
"Primitive of Exponential Function",
"Integration by Parts",
"Exponential Tends to Zero and Infinity",
"Limit at Infinity of Polynomial... |
proofwiki-15682 | Skewness of Poisson Distribution | Let $X$ be a discrete random variable with a Poisson distribution with parameter $\lambda$.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac 1 {\sqrt \lambda}$ | From Skewness in terms of Non-Central Moments:
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where $\mu$ is the mean of $X$, and $\sigma$ the standard deviation.
We have, by Expectation of Poisson Distribution:
:$\expect X = \lambda$
By Variance of Poisson Distribution:
:$\var X = \sigma^2 ... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]].
Then the [[Definition:Skewness|skewness]] $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac 1 {\sqrt \lambda}$ | From [[Skewness in terms of Non-Central Moments]]:
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where $\mu$ is the [[Definition:Expectation|mean]] of $X$, and $\sigma$ the [[Definition:Standard Deviation|standard deviation]].
We have, by [[Expectation of Poisson Distribution]]:
:$\expe... | Skewness of Poisson Distribution | https://proofwiki.org/wiki/Skewness_of_Poisson_Distribution | https://proofwiki.org/wiki/Skewness_of_Poisson_Distribution | [
"Skewness",
"Poisson Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Poisson Distribution",
"Definition:Skewness"
] | [
"Skewness in terms of Non-Central Moments",
"Definition:Expectation",
"Definition:Standard Deviation",
"Expectation of Poisson Distribution",
"Variance of Poisson Distribution",
"Definition:Moment Generating Function",
"Moment Generating Function of Poisson Distribution",
"Moment in terms of Moment Ge... |
proofwiki-15683 | Periodic Element is Multiple of Antiperiod | Let $f: \R \to \R$ be a real anti-periodic function with anti-period $A$.
Let $L$ be a periodic element of $f$.
Then $A \divides L$. | Consider $A + L$:
{{begin-eqn}}
{{eqn | l = \map f {x + \paren {A + L} }
| r = \map f {\paren {x + A} + L}
}}
{{eqn | r = \map f {x + A}
}}
{{eqn | r = -\map f x
}}
{{end-eqn}}
Hence $A + L$ is an anti-periodic element of $f$.
Combining Antiperiodic Element is Multiple of Antiperiod, Divides is Reflexive, and Com... | Let $f: \R \to \R$ be a [[Definition:Real Antiperiodic Function|real anti-periodic function]] with [[Definition:Antiperiod|anti-period]] $A$.
Let $L$ be a [[Definition:Periodic Element|periodic element]] of $f$.
Then $A \divides L$. | Consider $A + L$:
{{begin-eqn}}
{{eqn | l = \map f {x + \paren {A + L} }
| r = \map f {\paren {x + A} + L}
}}
{{eqn | r = \map f {x + A}
}}
{{eqn | r = -\map f x
}}
{{end-eqn}}
Hence $A + L$ is an [[Definition:Antiperiodic Element|anti-periodic element]] of $f$.
Combining [[Antiperiodic Element is Multiple of ... | Periodic Element is Multiple of Antiperiod | https://proofwiki.org/wiki/Periodic_Element_is_Multiple_of_Antiperiod | https://proofwiki.org/wiki/Periodic_Element_is_Multiple_of_Antiperiod | [
"Antiperiodic Functions"
] | [
"Definition:Antiperiodic Function/Real",
"Definition:Antiperiodic Function/Antiperiod",
"Definition:Periodic Function/Periodic Element"
] | [
"Definition:Antiperiodic Function/Antiperiodic Element",
"Antiperiodic Element is Multiple of Antiperiod",
"Integer Divisor Results/Integer Divides Itself",
"Common Divisor Divides Difference",
"Category:Antiperiodic Functions"
] |
proofwiki-15684 | Skewness of Continuous Uniform Distribution | Let $X$ be a continuous random variable which is uniformly distributed on a closed real interval $\closedint a b$.
Then the skewness $\gamma_1$ of $X$ is equal to $0$. | From the definition of skewness:
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
where:
:$\mu$ is the mean of $X$
:$\sigma$ is the standard deviation of $X$.
From the definition of the continuous uniform distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {b - a}$
So, from Expect... | Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] which is [[Definition:Continuous Uniform Distribution|uniformly distributed]] on a [[Definition:Closed Real Interval|closed real interval]] $\closedint a b$.
Then the [[Definition:Skewness|skewness]] $\gamma_1$ of $X$ is equal to $0$. | From the definition of [[Definition:Skewness|skewness]]:
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
where:
:$\mu$ is the [[Definition:Expectation|mean]] of $X$
:$\sigma$ is the [[Definition:Standard Deviation|standard deviation]] of $X$.
From the definition of the [[Definition:Continuous Uniform ... | Skewness of Continuous Uniform Distribution | https://proofwiki.org/wiki/Skewness_of_Continuous_Uniform_Distribution | https://proofwiki.org/wiki/Skewness_of_Continuous_Uniform_Distribution | [
"Skewness",
"Continuous Uniform Distribution"
] | [
"Definition:Random Variable/Continuous",
"Definition:Uniform Distribution/Continuous",
"Definition:Real Interval/Closed",
"Definition:Skewness"
] | [
"Definition:Skewness",
"Definition:Expectation",
"Definition:Standard Deviation",
"Definition:Uniform Distribution/Continuous",
"Definition:Probability Density Function",
"Expectation of Function of Continuous Random Variable",
"Expectation of Continuous Uniform Distribution",
"Integration by Substitu... |
proofwiki-15685 | Double of Antiperiod is Period | Let $f: \R \to \R$ be a real antiperiodic function with an anti-period of $A$.
Then $f$ is also periodic with a period of $2A$. | Let $L_f$ be the set of all periodic elements of $f$.
By Periodic Element is Multiple of Antiperiod and Absolute Value of Real Number is not less than Divisors:
:$\forall p \in L_f: A \divides p \land A \le p$
Suppose there is a $p \in L_f$ such that $p = A$.
Then:
{{begin-eqn}}
{{eqn | l = \map f x
| r = \map ... | Let $f: \R \to \R$ be a [[Definition:Real Antiperiodic Function|real antiperiodic function]] with an [[Definition:Antiperiod|anti-period]] of $A$.
Then $f$ is also [[Definition:Real Periodic Function|periodic]] with a [[Definition:Period of Periodic Real Function|period]] of $2A$. | Let $L_f$ be the [[Definition:Set|set]] of all [[Definition:Periodic Element|periodic elements]] of $f$.
By [[Periodic Element is Multiple of Antiperiod]] and [[Absolute Value of Real Number is not less than Divisors]]:
:$\forall p \in L_f: A \divides p \land A \le p$
Suppose there is a $p \in L_f$ such that $p = A$... | Double of Antiperiod is Period | https://proofwiki.org/wiki/Double_of_Antiperiod_is_Period | https://proofwiki.org/wiki/Double_of_Antiperiod_is_Period | [
"Antiperiodic Functions"
] | [
"Definition:Antiperiodic Function/Real",
"Definition:Antiperiodic Function/Antiperiod",
"Definition:Periodic Function/Real",
"Definition:Periodic Real Function/Period"
] | [
"Definition:Set",
"Definition:Periodic Function/Periodic Element",
"Periodic Element is Multiple of Antiperiod",
"Absolute Value of Real Number is not less than Divisors",
"Constant Function has no Period",
"Definition:Smallest Element/Subset",
"Double of Antiperiodic Element is Periodic",
"Definition... |
proofwiki-15686 | Principal Ideal in Integral Domain generated by Power Plus One is Subset of Principal Ideal generated by Power | Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Let $a \in D$ be a proper element of $D$.
Then:
:$\forall n \in \Z_{\ge 0}: \ideal {a^{n + 1} } \subsetneq \ideal {a_n}$
where $\ideal x$ denotes the principal ideal of $D$ generated by $x$. | We have:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \ideal {a^{n + 1} }
| c =
}}
{{eqn | ll= \leadsto
| q = \exists r \in D
| l = x
| r = r \circ a^{n + 1}
| c =
}}
{{eqn | ll= \leadsto
| l = x
| r = \paren {r \circ a} \circ a^n
| c =
}}
{{eqn | ll= \leads... | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$.
Let $a \in D$ be a [[Definition:Proper Element of Ring|proper element]] of $D$.
Then:
:$\forall n \in \Z_{\ge 0}: \ideal {a^{n + 1} } \subsetneq \ideal {a_n}$
where $\ideal x$ denotes ... | We have:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \ideal {a^{n + 1} }
| c =
}}
{{eqn | ll= \leadsto
| q = \exists r \in D
| l = x
| r = r \circ a^{n + 1}
| c =
}}
{{eqn | ll= \leadsto
| l = x
| r = \paren {r \circ a} \circ a^n
| c =
}}
{{eqn | ll= \lead... | Principal Ideal in Integral Domain generated by Power Plus One is Subset of Principal Ideal generated by Power | https://proofwiki.org/wiki/Principal_Ideal_in_Integral_Domain_generated_by_Power_Plus_One_is_Subset_of_Principal_Ideal_generated_by_Power | https://proofwiki.org/wiki/Principal_Ideal_in_Integral_Domain_generated_by_Power_Plus_One_is_Subset_of_Principal_Ideal_generated_by_Power | [
"Integral Domains",
"Principal Ideals of Rings"
] | [
"Definition:Integral Domain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Proper Element of Ring",
"Definition:Principal Ideal of Ring"
] | [
"Cancellation Law for Ring Product of Integral Domain",
"Definition:Unit of Ring",
"Definition:Contradiction",
"Definition:Proper Element of Ring",
"Proof by Contradiction"
] |
proofwiki-15687 | Non-Field Integral Domain has Infinite Number of Ideals | Let $\struct {D, +, \circ}$ be an integral domain which is not a field.
Then $\struct {D, +, \circ}$ has an infinite number of distinct ideals. | Let $a \in D$ be a proper element of $D$.
Because $\struct {D, +, \circ}$ is not a field, such an element is known to exist.
From Principal Ideal in Integral Domain generated by Power Plus One is Subset of Principal Ideal generated by Power:
:$\forall n \in \Z_{\ge 0}: \ideal {a^{n + 1} } \subsetneq \ideal {a_n}$
where... | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] which is not a [[Definition:Field (Abstract Algebra)|field]].
Then $\struct {D, +, \circ}$ has an [[Definition:Infinite Set|infinite number]] of [[Definition:Distinct Elements|distinct]] [[Definition:Ideal of Ring|ideals]]. | Let $a \in D$ be a [[Definition:Proper Element of Ring|proper element]] of $D$.
Because $\struct {D, +, \circ}$ is not a [[Definition:Field (Abstract Algebra)|field]], such an [[Definition:Element|element]] is known to exist.
From [[Principal Ideal in Integral Domain generated by Power Plus One is Subset of Principa... | Non-Field Integral Domain has Infinite Number of Ideals | https://proofwiki.org/wiki/Non-Field_Integral_Domain_has_Infinite_Number_of_Ideals | https://proofwiki.org/wiki/Non-Field_Integral_Domain_has_Infinite_Number_of_Ideals | [
"Integral Domains",
"Ideals of Rings"
] | [
"Definition:Integral Domain",
"Definition:Field (Abstract Algebra)",
"Definition:Infinite Set",
"Definition:Distinct/Plural",
"Definition:Ideal of Ring"
] | [
"Definition:Proper Element of Ring",
"Definition:Field (Abstract Algebra)",
"Definition:Element",
"Principal Ideal in Integral Domain generated by Power Plus One is Subset of Principal Ideal generated by Power",
"Definition:Principal Ideal of Ring",
"Definition:Set",
"Definition:Infinite Set"
] |
proofwiki-15688 | Ring Homomorphism from Ring with Unity to Integral Domain Preserves Unity | Let $\struct {R, +_R, \circ_R}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {D, +_D, \circ_D}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $\phi: R \to D$ be a ring homomorphism such that:
:$\map \ker \phi \ne R$
where $\map \ker \phi$ denotes the kernel o... | {{AimForCont}} $\map \phi {1_R} = 0_D$.
Let $x \in R$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi x
| r = \map \phi {x \circ_R 1_R}
| c = {{Defof|Unity of Ring}}
}}
{{eqn | r = \map \phi x \circ_D \map \phi {1_R}
| c = {{Defof|Ring Homomorphism}}
}}
{{eqn | r = \map \phi x \circ_D 0_D
... | Let $\struct {R, +_R, \circ_R}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $\struct {D, +_D, \circ_D}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and w... | {{AimForCont}} $\map \phi {1_R} = 0_D$.
Let $x \in R$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi x
| r = \map \phi {x \circ_R 1_R}
| c = {{Defof|Unity of Ring}}
}}
{{eqn | r = \map \phi x \circ_D \map \phi {1_R}
| c = {{Defof|Ring Homomorphism}}
}}
{{eqn | r = \map \phi x \circ_D 0_D
... | Ring Homomorphism from Ring with Unity to Integral Domain Preserves Unity | https://proofwiki.org/wiki/Ring_Homomorphism_from_Ring_with_Unity_to_Integral_Domain_Preserves_Unity | https://proofwiki.org/wiki/Ring_Homomorphism_from_Ring_with_Unity_to_Integral_Domain_Preserves_Unity | [
"Ring Homomorphisms",
"Rings with Unity",
"Integral Domains"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Integral Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring Homomorphism",
"Definition:Kernel of Ring Homomorphism"
] | [
"Definition:By Hypothesis",
"Definition:Contradiction",
"Cancellation Law for Ring Product of Integral Domain"
] |
proofwiki-15689 | Unity plus Negative of Nilpotent Ring Element is Unit | Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $x \in R$ be nilpotent.
Then $1_R - x$ is a unit of $R$. | By definition of nilpotent element:
:$x^n = 0_R$
for some $n \in \Z_{>0}$.
From Difference of Two Powers:
{{begin-eqn}}
{{eqn | l = a^n - b^n
| r = \paren {a - b} \circ \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j
| c =
}}
{{eqn | r = \paren {a - b} \circ \paren {a^{n - 1} + a^{n - 2} \circ b + a^{... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $x \in R$ be [[Definition:Nilpotent Ring Element|nilpotent]].
Then $1_R - x$ is a [[Definition:Unit of Ring|unit]] of $R$. | By definition of [[Definition:Nilpotent Ring Element|nilpotent element]]:
:$x^n = 0_R$
for some $n \in \Z_{>0}$.
From [[Difference of Two Powers]]:
{{begin-eqn}}
{{eqn | l = a^n - b^n
| r = \paren {a - b} \circ \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j
| c =
}}
{{eqn | r = \paren {a - b} \c... | Unity plus Negative of Nilpotent Ring Element is Unit | https://proofwiki.org/wiki/Unity_plus_Negative_of_Nilpotent_Ring_Element_is_Unit | https://proofwiki.org/wiki/Unity_plus_Negative_of_Nilpotent_Ring_Element_is_Unit | [
"Nilpotent Ring Elements"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Nilpotent Ring Element",
"Definition:Unit of Ring"
] | [
"Definition:Nilpotent Ring Element",
"Difference of Two Powers",
"Definition:Product Inverse",
"Definition:Unit of Ring"
] |
proofwiki-15690 | Quotient of Commutative Ring by Nilradical is Reduced | Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {N, +, \circ}$ denote the nilradical of $R$.
The quotient ring $R / N$ is a reduced ring. | From Nilpotent Elements of Commutative Ring form Ideal, $\struct {N, +, \circ}$ is an ideal of $\struct {R, +, \circ}$.
Hence the quotient ring $R / N$ is defined.
By definition of the ideal of $\struct {R, +, \circ}$, $N$ is the zero of $R / N$.
Let $\paren {x + N}^n \in N$.
Then:
:$x^n \in N$
and so:
:$\paren {x^n}^m... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $\struct {N, +, \circ}$ denote the [[Definition:Nilradical of Ring|nilradical]] of $R$.
The [[Definition:Quotient Ring|quotient rin... | From [[Nilpotent Elements of Commutative Ring form Ideal]], $\struct {N, +, \circ}$ is an [[Definition:Ideal of Ring|ideal]] of $\struct {R, +, \circ}$.
Hence the [[Definition:Quotient Ring|quotient ring]] $R / N$ is defined.
By definition of the [[Definition:Ideal of Ring|ideal]] of $\struct {R, +, \circ}$, $N$ is ... | Quotient of Commutative Ring by Nilradical is Reduced | https://proofwiki.org/wiki/Quotient_of_Commutative_Ring_by_Nilradical_is_Reduced | https://proofwiki.org/wiki/Quotient_of_Commutative_Ring_by_Nilradical_is_Reduced | [
"Nilpotent Ring Elements",
"Commutative Rings",
"Ideal Theory"
] | [
"Definition:Commutative Ring",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Nilradical of Ring",
"Definition:Quotient Ring",
"Definition:Reduced Ring"
] | [
"Nilpotent Elements of Commutative Ring form Ideal",
"Definition:Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Ideal of Ring",
"Definition:Ring Zero",
"Definition:Coset/Left Coset",
"Definition:Nilpotent Ring Element",
"Definition:Coset/Left Coset"
] |
proofwiki-15691 | Self-Inverse Element of Integral Domain is Unity or its Negative | Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $x \in D$ such that $x^2 = 1_D$.
Then either $x = 1_D$ or $x = -1_D$. | {{begin-eqn}}
{{eqn | l = x^2
| r = 1_D
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {x + 1_D} \paren {x + \paren {-1_D} }
| r = 0_D
| c =
}}
{{eqn | ll= \leadsto
| l = x + 1_D
| r = 0_D
| c =
}}
{{eqn | lo= \lor
| l = x + \paren {-1_D}
| r = 0_D
| c = ... | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
Let $x \in D$ such that $x^2 = 1_D$.
Then either $x = 1_D$ or $x = -1_D$. | {{begin-eqn}}
{{eqn | l = x^2
| r = 1_D
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {x + 1_D} \paren {x + \paren {-1_D} }
| r = 0_D
| c =
}}
{{eqn | ll= \leadsto
| l = x + 1_D
| r = 0_D
| c =
}}
{{eqn | lo= \lor
| l = x + \paren {-1_D}
| r = 0_D
| c = ... | Self-Inverse Element of Integral Domain is Unity or its Negative | https://proofwiki.org/wiki/Self-Inverse_Element_of_Integral_Domain_is_Unity_or_its_Negative | https://proofwiki.org/wiki/Self-Inverse_Element_of_Integral_Domain_is_Unity_or_its_Negative | [
"Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring"
] | [] |
proofwiki-15692 | Product of Units of Integral Domain with Finite Number of Units | Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $D$ have a finite number of units.
Let $U_D$ be the set of units of $\struct {D, +, \circ}$.
Then:
:$\ds \prod_{x \mathop \in U_D} x = -1_D$ | Consider the set $S$ defined as:
:$S = U_R \setminus \set {1_D, -1_D}$
If $S$ has even cardinality, it can be partitioned into doubletons of the form $\set {u, u^{-1} }$.
Each of these doubletons has a product of $1_D$.
The product of all these with $1_D$ and $-1_D$ is $-1_D$.
It remains to be shown that $S$ cannot be ... | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
Let $D$ have a [[Definition:Finite Set|finite number]] of [[Definition:Unit of Ring|units]].
Let $U_D$ be the [[Definition:Set|set]] of [... | Consider the [[Definition:Set|set]] $S$ defined as:
:$S = U_R \setminus \set {1_D, -1_D}$
If $S$ has [[Definition:Even Integer|even]] [[Definition:Cardinality|cardinality]], it can be [[Definition:Set Partition|partitioned]] into [[Definition:Doubleton|doubletons]] of the form $\set {u, u^{-1} }$.
Each of these [[Def... | Product of Units of Integral Domain with Finite Number of Units | https://proofwiki.org/wiki/Product_of_Units_of_Integral_Domain_with_Finite_Number_of_Units | https://proofwiki.org/wiki/Product_of_Units_of_Integral_Domain_with_Finite_Number_of_Units | [
"Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Finite Set",
"Definition:Unit of Ring",
"Definition:Set",
"Definition:Unit of Ring"
] | [
"Definition:Set",
"Definition:Even Integer",
"Definition:Cardinality",
"Definition:Set Partition",
"Definition:Doubleton",
"Definition:Doubleton",
"Definition:Ring (Abstract Algebra)/Product",
"Definition:Ring (Abstract Algebra)/Product",
"Definition:Odd Integer",
"Definition:Cardinality",
"Defi... |
proofwiki-15693 | Prime Ideals of Ring of Integers | Let $\struct {\Z, +, \times}$ denote the ring of integers.
Let $J$ be a prime ideal of $\Z$.
Then either:
:$J = \set 0$
or:
:$J = \ideal p$
where:
:$p$ is a prime number
:$\ideal p$ denotes the principal ideal of $\Z$ generated by $p$. | From Prime Ideal iff Quotient Ring is Integral Domain:
:$J$ is a prime ideal of $\Z$ {{iff}} $\Z / J$ is an integral domain.
From Quotient Ring of Integers and Zero:
:$\Z / \set 0 \cong \Z$
As $\Z$ is an integral domain, it follows that $\set 0$ is a prime ideal of $\Z$.
From Quotient Ring of Integers with Principal Id... | Let $\struct {\Z, +, \times}$ denote the [[Definition:Ring of Integers|ring of integers]].
Let $J$ be a [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]] of $\Z$.
Then either:
:$J = \set 0$
or:
:$J = \ideal p$
where:
:$p$ is a [[Definition:Prime Number|prime number]]
:$\ideal p$ denotes the [[De... | From [[Prime Ideal iff Quotient Ring is Integral Domain]]:
:$J$ is a [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]] of $\Z$ {{iff}} $\Z / J$ is an [[Definition:Integral Domain|integral domain]].
From [[Quotient Ring of Integers and Zero]]:
:$\Z / \set 0 \cong \Z$
As $\Z$ is an [[Definition:Int... | Prime Ideals of Ring of Integers | https://proofwiki.org/wiki/Prime_Ideals_of_Ring_of_Integers | https://proofwiki.org/wiki/Prime_Ideals_of_Ring_of_Integers | [
"Integers",
"Prime Ideals of Rings"
] | [
"Definition:Ring of Integers",
"Definition:Prime Ideal of Ring/Commutative and Unitary Ring",
"Definition:Prime Number",
"Definition:Principal Ideal of Ring"
] | [
"Prime Ideal iff Quotient Ring is Integral Domain",
"Definition:Prime Ideal of Ring/Commutative and Unitary Ring",
"Definition:Integral Domain",
"Quotient Ring of Integers and Zero",
"Definition:Integral Domain",
"Definition:Prime Ideal of Ring/Commutative and Unitary Ring",
"Quotient Ring of Integers w... |
proofwiki-15694 | Prime Power Mapping on Galois Field is Automorphism | Let $\GF$ be a Galois field whose zero is $0_\GF$ and whose characteristic is $p$.
Let $\sigma: \GF \to \GF$ be defined as:
:$\forall x \in \GF: \map \sigma x = x^p$
Then $\sigma$ is an automorphism of $\GF$. | Let $x, y \in \GF$.
Then:
{{begin-eqn}}
{{eqn | l = \map \sigma {x y}
| r = \paren {x y}^p
| c = Definition of $\sigma$
}}
{{eqn | r = x^p y^p
| c = Power of Product of Commutative Elements in Group
}}
{{eqn | r = \map \sigma x \map \sigma y
| c = Definition of $\sigma$
}}
{{end-eqn}}
and:
{{beg... | Let $\GF$ be a [[Definition:Galois Field|Galois field]] whose [[Definition:Field Zero|zero]] is $0_\GF$ and whose [[Definition:Characteristic of Field|characteristic]] is $p$.
Let $\sigma: \GF \to \GF$ be defined as:
:$\forall x \in \GF: \map \sigma x = x^p$
Then $\sigma$ is an [[Definition:Field Automorphism|automo... | Let $x, y \in \GF$.
Then:
{{begin-eqn}}
{{eqn | l = \map \sigma {x y}
| r = \paren {x y}^p
| c = Definition of $\sigma$
}}
{{eqn | r = x^p y^p
| c = [[Power of Product of Commutative Elements in Group]]
}}
{{eqn | r = \map \sigma x \map \sigma y
| c = Definition of $\sigma$
}}
{{end-eqn}}
an... | Prime Power Mapping on Galois Field is Automorphism | https://proofwiki.org/wiki/Prime_Power_Mapping_on_Galois_Field_is_Automorphism | https://proofwiki.org/wiki/Prime_Power_Mapping_on_Galois_Field_is_Automorphism | [
"Galois Fields",
"Field Isomorphisms"
] | [
"Definition:Galois Field",
"Definition:Field Zero",
"Definition:Characteristic of Field",
"Definition:Field Automorphism"
] | [
"Power of Product of Commutative Elements in Group",
"Binomial Theorem",
"Power of Sum Modulo Prime",
"Definition:Field Homomorphism",
"Congruence of Powers",
"Congruence of Powers",
"Kernel is Trivial iff Monomorphism",
"Definition:Ring Monomorphism",
"Definition:Injection",
"Injection from Finit... |
proofwiki-15695 | Additive Group and Multiplicative Group of Field are not Isomorphic | Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {F, +}$ denote the additive group of $F$.
Let $\struct {F_{\ne 0_F}, \times}$ denote the multiplicative group of $F$.
Then $\struct {F, +}$ and $\struct {F_{\ne 0_F}, \times}$ are not isomorphic to each other. | {{AimForCont}} $\phi: \struct {F_{\ne 0_F}, \times} \to \struct {F, +}$ is an isomorphism.
By definition:
:$0_F$ is the identity of $\struct {F, +}$
and
:$1_F$ is the identity of $\struct {F_{\ne 0_F}, \times}$.
We have that:
{{begin-eqn}}
{{eqn | l = 0_F
| r = \map \phi {1_F}
| c = Epimorphism Preserves Id... | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $\struct {F, +}$ denote the [[Definition:Additive Group of Ring|additive group]] of $F$.
Let $\struct {F_{\ne 0_F}, \times}$ denote ... | {{AimForCont}} $\phi: \struct {F_{\ne 0_F}, \times} \to \struct {F, +}$ is an [[Definition:Group Isomorphism|isomorphism]].
By definition:
:$0_F$ is the [[Definition:Identity Element|identity]] of $\struct {F, +}$
and
:$1_F$ is the [[Definition:Identity Element|identity]] of $\struct {F_{\ne 0_F}, \times}$.
We have ... | Additive Group and Multiplicative Group of Field are not Isomorphic | https://proofwiki.org/wiki/Additive_Group_and_Multiplicative_Group_of_Field_are_not_Isomorphic | https://proofwiki.org/wiki/Additive_Group_and_Multiplicative_Group_of_Field_are_not_Isomorphic | [
"Field Theory",
"Group Isomorphisms"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Additive Group of Ring",
"Definition:Multiplicative Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Epimorphism Preserves Identity",
"Definition:Characteristic of Field",
"Definition:Characteristic of Field",
"Definiti... |
proofwiki-15696 | First Central Moment is Zero | Let $X$ be a random variable on some probability space with mean $\mu$.
Then the first central moment $\mu_1$ of $X$ is equal to $0$. | {{begin-eqn}}
{{eqn | l = \mu_1
| r = \expect {X - \mu}
| c = {{Defof|Central Moment}}
}}
{{eqn | r = \expect X - \mu
| c = Expectation is Linear
}}
{{eqn | r = \mu - \mu
| c = $\expect X = \mu$
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{qed}}
Category:Expectation
9gwdh5c494qd88bkn754udl1c7f0djq | Let $X$ be a [[Definition:Random Variable|random variable]] on some [[Definition:Probability Space|probability space]] with [[Definition:Expectation|mean]] $\mu$.
Then the first [[Definition:Central Moment|central moment]] $\mu_1$ of $X$ is equal to $0$. | {{begin-eqn}}
{{eqn | l = \mu_1
| r = \expect {X - \mu}
| c = {{Defof|Central Moment}}
}}
{{eqn | r = \expect X - \mu
| c = [[Expectation is Linear]]
}}
{{eqn | r = \mu - \mu
| c = $\expect X = \mu$
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{qed}}
[[Category:Expectation]]
9gwdh5c494qd88bkn754udl1c7f0djq | First Central Moment is Zero | https://proofwiki.org/wiki/First_Central_Moment_is_Zero | https://proofwiki.org/wiki/First_Central_Moment_is_Zero | [
"Expectation"
] | [
"Definition:Random Variable",
"Definition:Probability Space",
"Definition:Expectation",
"Definition:Central Moment"
] | [
"Expectation is Linear",
"Category:Expectation"
] |
proofwiki-15697 | Principal Ideal Domain cannot have Infinite Strictly Increasing Sequence of Ideals | Let $\struct {D, +, \circ}$ be a principal ideal domain.
Then $D$ cannot have an infinite sequence of ideals $\sequence {j_n}_{n \mathop \in \N}$ such that:
:$\forall n \in \N: J_n \subsetneq j_{n + 1}$ | Let $K = \ds \bigcup_{n \mathop \in \N} J_n$.
Then from Increasing Union of Sequence of Ideals is Ideal, $K$ is an ideal of $D$.
We have that $D$ is a principal ideal domain.
Hence there exists $a \in D$ such that:
:$K = \ideal a$
where $\ideal a$ is the principal ideal of $D$ generated by $a$.
But $a \in J_m$ for some... | Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
Then $D$ cannot have an [[Definition:Infinite Sequence|infinite sequence]] of [[Definition:Ideal of Ring|ideals]] $\sequence {j_n}_{n \mathop \in \N}$ such that:
:$\forall n \in \N: J_n \subsetneq j_{n + 1}$ | Let $K = \ds \bigcup_{n \mathop \in \N} J_n$.
Then from [[Increasing Union of Sequence of Ideals is Ideal]], $K$ is an [[Definition:Ideal of Ring|ideal]] of $D$.
We have that $D$ is a [[Definition:Principal Ideal Domain|principal ideal domain]].
Hence there exists $a \in D$ such that:
:$K = \ideal a$
where $\ideal a... | Principal Ideal Domain cannot have Infinite Strictly Increasing Sequence of Ideals | https://proofwiki.org/wiki/Principal_Ideal_Domain_cannot_have_Infinite_Strictly_Increasing_Sequence_of_Ideals | https://proofwiki.org/wiki/Principal_Ideal_Domain_cannot_have_Infinite_Strictly_Increasing_Sequence_of_Ideals | [
"Principal Ideal Domains"
] | [
"Definition:Principal Ideal Domain",
"Definition:Sequence/Infinite Sequence",
"Definition:Ideal of Ring"
] | [
"Increasing Union of Ideals is Ideal/Sequence",
"Definition:Ideal of Ring",
"Definition:Principal Ideal Domain",
"Definition:Principal Ideal of Ring",
"Definition:Contradiction"
] |
proofwiki-15698 | Field Norm of Complex Number is Positive Definite | Let $\C$ denote the set of complex numbers.
Let $N: \C \to \R_{\ge 0}$ denote the field norm on complex numbers:
:$\forall z \in \C: \map N z = \cmod z^2$
where $\cmod z$ denotes the complex modulus of $z$.
Then $N$ is positive definite on $\C$. | First it is shown that $\map N z = 0 \iff z = 0$.
{{begin-eqn}}
{{eqn | l = z
| r = 0
| c =
}}
{{eqn | r = 0 + 0 i
| c =
}}
{{eqn | ll= \leadsto
| l = \map N z
| r = 0^2 + 0^2
| c = Definition of $N$
}}
{{eqn | r = 0
| c =
}}
{{end-eqn}}
Let $z = x + i y$.
{{begin-eqn}}
{{eq... | Let $\C$ denote the [[Definition:Complex Number|set of complex numbers]].
Let $N: \C \to \R_{\ge 0}$ denote the [[Definition:Field Norm of Complex Number|field norm on complex numbers]]:
:$\forall z \in \C: \map N z = \cmod z^2$
where $\cmod z$ denotes the [[Definition:Complex Modulus|complex modulus]] of $z$.
Then ... | First it is shown that $\map N z = 0 \iff z = 0$.
{{begin-eqn}}
{{eqn | l = z
| r = 0
| c =
}}
{{eqn | r = 0 + 0 i
| c =
}}
{{eqn | ll= \leadsto
| l = \map N z
| r = 0^2 + 0^2
| c = Definition of $N$
}}
{{eqn | r = 0
| c =
}}
{{end-eqn}}
Let $z = x + i y$.
{{begin-eqn}}
... | Field Norm of Complex Number is Positive Definite | https://proofwiki.org/wiki/Field_Norm_of_Complex_Number_is_Positive_Definite | https://proofwiki.org/wiki/Field_Norm_of_Complex_Number_is_Positive_Definite | [
"Field Norm of Complex Number"
] | [
"Definition:Complex Number",
"Definition:Field Norm of Complex Number",
"Definition:Complex Modulus",
"Definition:Positive Definite (Ring)"
] | [
"Square of Real Number is Non-Negative",
"Square of Real Number is Non-Negative",
"Definition:Positive Definite (Ring)",
"Category:Field Norm of Complex Number"
] |
proofwiki-15699 | Second Standardized Moment is One | Let $X$ be a random variable on some probability space with standard deviation $\sigma$.
Then the second standardized moment $\alpha_2$ of $X$ is equal to $1$. | {{begin-eqn}}
{{eqn | l = \alpha_2
| r = \frac {\mu_2} {\sigma^2}
| c = {{Defof|Standardized Moment}}
}}
{{eqn | r = \frac {\sigma^2} {\sigma^2}
| c = {{Defof|Central Moment}}
}}
{{eqn | r = 1
}}
{{end-eqn}}
{{qed}}
Category:Expectation
dlu8n91xg9zxqqrmbrwvdx0oozdkyaq | Let $X$ be a [[Definition:Random Variable|random variable]] on some [[Definition:Probability Space|probability space]] with [[Definition:Standard Deviation|standard deviation]] $\sigma$.
Then the second [[Definition:Standardized Moment|standardized moment]] $\alpha_2$ of $X$ is equal to $1$. | {{begin-eqn}}
{{eqn | l = \alpha_2
| r = \frac {\mu_2} {\sigma^2}
| c = {{Defof|Standardized Moment}}
}}
{{eqn | r = \frac {\sigma^2} {\sigma^2}
| c = {{Defof|Central Moment}}
}}
{{eqn | r = 1
}}
{{end-eqn}}
{{qed}}
[[Category:Expectation]]
dlu8n91xg9zxqqrmbrwvdx0oozdkyaq | Second Standardized Moment is One | https://proofwiki.org/wiki/Second_Standardized_Moment_is_One | https://proofwiki.org/wiki/Second_Standardized_Moment_is_One | [
"Expectation"
] | [
"Definition:Random Variable",
"Definition:Probability Space",
"Definition:Standard Deviation",
"Definition:Standardized Moment"
] | [
"Category:Expectation"
] |
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