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23073
|
\section{Union of Total Ordering with Lower Sets is Total Ordering}
Tags: Lower Sets, Order Theory, Total Orderings
\begin{theorem}
Let $\struct {Y, \preceq}$ be a totally ordered set.
Let $X$ be the disjoint union of $Y$ with the set of lower sets of $Y$.
Define a relation $\preceq'$ on $X$ extending $\preceq$ by letting:
:$y_1 \preceq' y_2 \iff y_1 \preceq y_2$
:$y \preceq' L \iff y \in L$
:$L_1 \preceq' L_2 \iff L_1 \subseteq L_2$
:$L \preceq' y \iff y \in Y \setminus L$
Then $\preceq'$ is a total ordering.
\end{theorem}
\begin{proof}
First note that by Lower Sets in Totally Ordered Set form Nest:
:$\subseteq$ is a total ordering on the set of lower sets.
Also note that by Complement of Lower Set is Upper Set, the complement of each $\preceq$-lower set is a $\preceq$-upper set.
\end{proof}
|
23074
|
\section{Union of Transitive Class is Subset}
Tags: Class Union, Class is Transitive iff Union is Subset, Transitive Classes
\begin{theorem}
Let $A$ be a transitive class.
Let $\ds \bigcup A$ denote the union of $A$.
Then:
:$\ds \bigcup A \subseteq A$
\end{theorem}
\begin{proof}
$A$ be transitive.
Let $x \in \ds \bigcup A$.
Then by definition:
:$\exists y \in A: x \in y$
By definition of transitive class:
:$x \in y \land y \in A \implies x \in A$
and so:
:$x \in A$
Hence the result by definition of subclass.
{{qed}}
\end{proof}
|
23075
|
\section{Union of Transitive Class is Transitive}
Tags: Class Union, Union of Transitive Class is Transitive, Transitive Classes
\begin{theorem}
Let $A$ be a class.
Let $\ds \bigcup A$ denote the union of $A$.
Let $A$ be transitive.
Then $\ds \bigcup A$ is also transitive.
\end{theorem}
\begin{proof}
Let $x \in \displaystyle \bigcup A$.
By Class is Transitive iff Union is Subset we have that
:$\displaystyle \bigcup A \subseteq A$
Thus by definition of subclass:
:$x \in A$
As $A$ is transitive:
:$x \subseteq A$
Let $z \in x$.
As $x \subseteq A$, it follows by definition of subclass that:
:$z \in A$
Thus we have that:
:$\exists x \in A: z \in x$
and so by definition of union of class:
:$z \in \displaystyle \bigcup A$
Thus we have that:
:$z \in x \implies z \in \displaystyle \bigcup A$
and so by definition of subclass:
:$x \subseteq \displaystyle \bigcup A$
Thus we have that:
:$x \in \displaystyle \bigcup A \implies x \subseteq \displaystyle \bigcup A$
Hence $\displaystyle \bigcup A$ is a transitive class by definition.
{{qed}}
\end{proof}
|
23076
|
\section{Union of Transitive Relations Not Always Transitive}
Tags: Set Union, Transitive Relations, Relations, Union
\begin{theorem}
The union of transitive relations is not necessarily itself transitive.
\end{theorem}
\begin{proof}
Proof by counterexample:
Let $S = \set {a, b, c, d}$.
Let $\RR_1$ be the transitive relation $\set {\tuple {a, b}, \tuple {b, c}, \tuple {a, c} }$.
Let $\RR_2$ be the transitive relation $\set {\tuple {b, c}, \tuple {c, d}, \tuple {b, d} }$.
Then we have that $\tuple {a, b} \in \RR_1 \cup \RR_2$ and $\tuple {b, d} \in \RR_1 \cup \RR_2$.
However, $\tuple {a, d} \notin \RR_1 \cup \RR_2$, and so $\RR_1 \cup \RR_2$ is not transitive.
{{qed}}
Category:Transitive Relations
Category:Set Union
\end{proof}
|
23077
|
\section{Union of Two Compact Sets is Compact}
Tags: Set Union, Compact Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological spaces.
Let $H$ and $K$ be compact subsets of $T$.
Then $H \cup K$ is compact in $T$.
\end{theorem}
\begin{proof}
Let $\CC$ be an open cover of $H \cup K$.
Then $\CC$ is an open cover of both $H$ and $K$.
As $H$ and $K$ are both compact in $T$:
:$H$ has a finite subcover $C_H$ of $\CC$
:$K$ has a finite subcover $C_K$ of $\CC$.
Their union $C_H \cup C_K$ is a finite subcover of $\CC$ for $H \cup K$.
From Union of Finite Sets is Finite it follows that $C_H \cup C_K$ is finite.
As $\CC$ is arbitrary, it follows by definition that $H \cup K$ is compact in $T$.
\end{proof}
|
23078
|
\section{Union of Union of Cartesian Product with Empty Factor}
Tags: Set Union, Cartesian Product
\begin{theorem}
Let $A$ and $B$ be sets such that either $A = \O$ or $B = \O$.
Let the ordered pair $\tuple {a, b}$ be defined using the Kuratowski formalization:
:$\tuple {a, b} := \set {\set a, \set {a, b} }$
Then:
:$\ds \bigcup \bigcup \paren {A \times B} = A \cup B \iff A = B = \O$
where:
:$\cup$ denotes union
:$\times$ denotes Cartesian product.
That is, if either $A$ or $B$ is empty:
:$\ds \bigcup \bigcup \paren {A \times B} = A \cup B$
holds {{iff}} they are ''both'' empty
\end{theorem}
\begin{proof}
Let $A = \O$ or $B = \O$.
From Cartesian Product is Empty iff Factor is Empty:
:$A \times B = \O$
Hence from Union of Empty Set:
:$\ds \bigcup \bigcup \paren {A \times B} = \O$
However, from Union is Empty iff Sets are Empty:
:$A \cup B = \O \iff A = \O \text { and } B = \O$
The result follows.
{{qed}}
\end{proof}
|
23079
|
\section{Union of Union of Relation is Union of Domain with Image}
Tags: Class Union, Relation Theory
\begin{theorem}
Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $\Dom \RR$ denote the domain of $\RR$.
Then:
:$\map \bigcup {\bigcup \RR} = \Dom \RR \cup \Img \RR$
where:
:$\bigcup \RR$ denotes the union of $\RR$
:$\Dom \RR$ denotes the domain of $\RR$
:$\Img \RR$ denotes the image of $\RR$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \bigcup \RR
| r = \set {z: \exists \tuple {x, y} \in \RR: z \in \tuple {x, y} }
| c = {{Defof|Union of Class}}
}}
{{eqn | r = \set {z: \exists \set {\set x, \set {x, y} } \in \RR: z \in \tuple {x, y} }
| c = {{Defof|Kuratowski Formalization of Ordered Pair}}
}}
{{eqn | r = \set {\set x, \set {x, y}: x \in \Dom \RR, \tuple {x, y} \in \RR}
| c =
}}
{{eqn | ll= \leadsto
| l = \map \bigcup {\bigcup \RR}
| r = \set {z: \exists X \in \bigcup \RR: z \in X}
| c = {{Defof|Union of Class}}
}}
{{eqn | r = \set {x, y: x \in \Dom \RR, \tuple {x, y} \in \RR}
| c = Definition of $\bigcup \RR$
}}
{{eqn | r = \set {x, y: x \in \Dom \RR, y \in \Img \RR}
| c = {{Defof|Image of Relation}}
}}
{{eqn | r = \set {x: x \in \Dom \RR} \cup \set {y: y \in \Img \RR}
| c = {{Defof|Class Union}}
}}
{{eqn | r = \Dom \RR \cup \Img \RR
| c =
}}
{{end-eqn}}
{{qed}}
Category:Class Union
Category:Relation Theory
\end{proof}
|
23080
|
\section{Union of Unordered Tuples}
Tags: Set Theory
\begin{theorem}
Let $x_1, \dots, x_n, x_{n+1}, \dots, x_m$ be arbitrary.
Then
:$\set {x_1, \dots, x_n} \cup \set {x_{n + 1}, \dots, x_m} = \set {x_1, \dots, x_n, x_{n + 1}, \dots, x_m}$
\end{theorem}
\begin{proof}
Let $a$ be arbitrary.
:$a \in \set {x_1, \dots, x_n} \cup \set {x_{n + 1}, \dots, x_m}$
{{iff}}
:$a \in \set {x_1, \dots, x_n}$ or $a \in \set {x_{n + 1}, \dots, x_m}$ by definition of union
{{iff}}
:$a = x_1 \lor \dots \lor a = x_n$ or $a = x_{n + 1} \lor \dots \lor a = x_m$ by definition of unordered tuple
{{iff}}
:$a = x_1 \lor \dots \lor a = x_n \lor a = x_{n + 1} \lor \dots \lor a = x_m$
{{iff}}
:$a \in \set {x_1, \dots, x_n, x_{n + 1}, \dots, x_m}$ by definition of unordered tuple
Thus result follows by definition of set equality.
{{qed}}
Category:Set Theory
\end{proof}
|
23081
|
\section{Union of Upper Sets is Upper}
Tags: Upper Sets
\begin{theorem}
Let $\left({S, \preceq}\right)$ be a preordered set.
Let $A$ be a set of subsets of $S$ such that
:$\forall X \in A: X$ is an upper set.
Then:
:$\bigcup A$ is also an upper set.
\end{theorem}
\begin{proof}
Let $x \in \bigcup A, y \in S$ such that
:$x \preceq y$
By definition of union:
:$\exists Y \in A: x \in Y$
By assumption:
:$Y$ is an upper set.
By definition of upper set:
:$y \in Y$
Thus by definition of union:
:$y \in \bigcup A$
Hence
:$\bigcup A$ is an upper set.
{{qed}}
\end{proof}
|
23082
|
\section{Union of Vertical Sections is Vertical Section of Union}
Tags: Set Union, Vertical Section of Sets
\begin{theorem}
Let $X$ and $Y$ be sets.
Let $\set {E_\alpha : \alpha \in A}$ be a set of subsets of $X \times Y$.
Let $x \in X$.
Then:
:$\ds \paren {\bigcup_{\alpha \in A} E_\alpha}_x = \bigcup_{\alpha \in A} \paren {E_\alpha}_x$
where:
:$\ds \paren {\bigcup_{\alpha \in A} E_\alpha}_x$ is the $x$-vertical section of $\ds \bigcup_{\alpha \in A} E_\alpha$
:$\paren {E_\alpha}_x$ is the $x$-vertical section of $E_\alpha$.
\end{theorem}
\begin{proof}
Note that:
:$\ds y \in \bigcup_{\alpha \in A} \paren {E_\alpha}_x$
{{iff}}:
:$y \in \paren {E_\alpha}_x$ for some $\alpha \in A$.
From the definition of the $x$-vertical section, this is equivalent to:
:$\tuple {x, y} \in E_\alpha$ for some $\alpha \in A$.
This in turn is equivalent to:
:$\ds \tuple {x, y} \in \bigcup_{\alpha \in A} E_\alpha$
Again applying the definition of the $x$-vertical section:
:$\ds y \in \paren {\bigcup_{\alpha \in A} E_\alpha}_x$
So:
:$\ds y \in \bigcup_{\alpha \in A} \paren {E_\alpha}_x$ {{iff}} $\ds y \in \paren {\bigcup_{\alpha \in A} E_\alpha}_x$
giving:
:$\ds \paren {\bigcup_{\alpha \in A} E_\alpha}_x = \bigcup_{\alpha \in A} \paren {E_\alpha}_x$
{{qed}}
Category:Set Union
Category:Vertical Section of Sets
\end{proof}
|
23083
|
\section{Union with Complement}
Tags: Set Union, Set Complement, Union, Universe
\begin{theorem}
The union of a set and its complement is the universe:
:$S \cup \map \complement S = \mathbb U$
\end{theorem}
\begin{proof}
Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $T \cup \relcomp S T = S$ from Union with Relative Complement.
{{qed}}
{{LEM|Union with Relative Complement}}
\end{proof}
|
23084
|
\section{Union with Disjoint Singleton is Dependent if Element Depends on Subset}
Tags: Matroid Theory
\begin{theorem}
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $A \subseteq S$.
Let $x \in S : x \notin A$.
If $x$ depends on $A$ then $A \cup \set x$ is dependent
\end{theorem}
\begin{proof}
We proceed by Proof by Contraposition.
Let $A \cup \set x$ be independent.
By matroid axiom $( \text I 2)$:
:$A$ is independent
We have:
{{begin-eqn}}
{{eqn | l = \map \rho {A \cup \set x}
| r = \size {A \cup \set x}
| c = Rank of Independent Subset Equals Cardinality
}}
{{eqn | r = \size A + \size{\set x}
| c = Corollary to Cardinality of Set Union
}}
{{eqn | r = \size A + 1
| c = Cardinality of Singleton
}}
{{eqn | o = >
| r = \size A
}}
{{eqn | r = \map \rho A
| c = Rank of Independent Subset Equals Cardinality
}}
{{end-eqn}}
Then $x$ does not depend on $A$ by definition.
The theorem holds by the Rule of Transposition.
{{qed}}
Category:Matroid Theory
\end{proof}
|
23085
|
\section{Union with Superset is Superset}
Tags: Set Union, Subsets, Subset, Union
\begin{theorem}
:$S \subseteq T \iff S \cup T = T$
where:
:$S \subseteq T$ denotes that $S$ is a subset of $T$
:$S \cup T$ denotes the union of $S$ and $T$.
\end{theorem}
\begin{proof}
Let $S \cup T = T$.
Then by definition of set equality:
:$S \cup T \subseteq T$
Thus:
{{begin-eqn}}
{{eqn | l = S
| o = \subseteq
| r = S \cup T
| c = Subset of Union
}}
{{eqn | ll= \leadsto
| l = S
| o = \subseteq
| r = T
| c = Subset Relation is Transitive
}}
{{end-eqn}}
Now let $S \subseteq T$.
From Subset of Union, we have $S \cup T \supseteq T$.
We also have:
{{begin-eqn}}
{{eqn | l = S
| o = \subseteq
| r = T
| c =
}}
{{eqn | ll= \leadsto
| l = S \cup T
| o = \subseteq
| r = T \cup T
| c = Set Union Preserves Subsets
}}
{{eqn | ll= \leadsto
| l = S \cup T
| o = \subseteq
| r = T
| c = Union is Idempotent
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = S \cup T
| o = \subseteq
| r = T
| c =
}}
{{eqn | l = S \cup T
| o = \supseteq
| r = T
| c =
}}
{{end-eqn}}
By definition of set equality:
:$S \cup T = T$
So:
{{begin-eqn}}
{{eqn | l = S \cup T = T
| o = \implies
| r = S \subseteq T
| c =
}}
{{eqn | l = S \subseteq T
| o = \implies
| r = S \cup T = T
| c =
}}
{{end-eqn}}
and so:
:$S \subseteq T \iff S \cup T = T$
from the definition of equivalence.
{{Qed}}
\end{proof}
|
23086
|
\section{Unique Code for State of URM Program}
Tags: URM Programs
\begin{theorem}
Every state of a URM program can be assigned a unique '''code number'''.
This code number is called the '''state code''' (or '''situation code''').
\end{theorem}
\begin{proof}
The state of a URM program at a particular point in time is defined as:
: the value of the instruction pointer
: the value, at that point, of each of the registers that are used by the program.
Let $P$ be a URM program.
Suppose that, at a given stage of computation:
: the value of the instruction pointer is $a$;
: the value of register $R_k$ is $r_k$.
Let $b = \rho \left({P}\right)$ be the number of registers used by $P$.
Then we can define the '''state code''' $s$ as:
:$s = p_1^a p_2^{r_1} p_3^{r_2} \cdots p_{b+1}^{r_b}$
where $p_j = p \left({j}\right)$ is defined as the $j$th prime number.
Hence it follows from the Fundamental Theorem of Arithmetic that $s$ is uniquely specified for any given state.
{{qed}}
\end{proof}
|
23087
|
\section{Unique Composition of Group Element whose Order is Product of Coprime Integers}
Tags: Order of Group Elements
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $g \in G$ be an element of $g$.
Let:
:$\order g = m n$
where:
:$\order g$ denotes the order of $g$ in $G$
:$m$ and $n$ are coprime integers.
Then $g$ can be expressed uniquely as the product of two commuting elements $a$ and $b$ of order $m$ and $n$ respectively.
\end{theorem}
\begin{proof}
Let $g_1 = g^n$ and $g_2 = g^m$.
By Powers of Group Element Commute, $g_1$ and $g_2$ commute.
We have:
:${g_1}^m = g^{n m} = e$
:${g_2}^n = g^{m n} = e$
It follows that:
:the order of $g_1$ is $m$
:the order of $g_2$ is $n$
otherwise if either $g_1$ or $g_2$ were of a smaller order then $g$ would also be of a smaller order.
By Bézout's Lemma:
:$\exists u, v \in \Z: u n + v m = 1$
as $m \perp n$.
Thus:
:$g = g^{u n + v m} = \paren {g^n}^u \paren {g^m}^v = {g_1}^u {g_2}^v$
Also by Bézout's Lemma:
:$u \perp m$
and:
:$v \perp n$
Thus by Order of Group Element equals Order of Coprime Power:
:$\order { {g_1}^u} = m$
and:
:$\order { {g_2}^v} = n$
We have that $g_1$ and $g_2$ commute.
So by Commutativity of Powers in Group, ${g_1}^u$ and ${g_2}^v$ also commute.
Putting $a = g_1^u$ and $b = g_2^v$, we have $a$ and $b$ which satisfy the required conditions.
It remains to prove uniqueness.
Suppose that:
:$(1): \quad g = r_1 r_2 = s_1 s_2$
where:
:$r_1$ and $r_2$ commute
:$s_1$ and $s_2$ commute
:$\order {r_1} = \order {s_1} = m$
:$\order {r_2} = \order {s_2} = n$
Raising $(1)$ to the $n u$th power:
:$g^{n u} = {r_1}^{n u} {r_2}^{n u} = {s_1}^{n u} {s_2}^{n u}$
and so:
{{begin-eqn}}
{{eqn | l = {r_1}^{n u} {r_2}^{n u}
| r = {s_1}^{n u} {s_2}^{n u}
| c =
}}
{{eqn | l = {r_1}^{n u} \paren { {r_2}^n}^u
| r = {s_1}^{n u} \paren { {s_2}^n}^u
| c =
}}
{{eqn | l = {r_1}^{n u} e^u
| r = {s_1}^{n u} e^u
| c = as $\order {r_2} = \order {s_2} = n$
}}
{{eqn | ll= \leadsto
| l = {r_1}^{n u}
| r = {s_1}^{n u}
| c =
}}
{{eqn | ll= \leadsto
| l = {r_1}^{1 - m v}
| r = {s_1}^{1 - m v}
| c =
}}
{{eqn | ll= \leadsto
| l = r_1 \paren { {r_1}^m}^{-v}
| r = s_1 \paren { {s_1}^m}^{-v}
| c =
}}
{{eqn | ll= \leadsto
| l = r_1 e^{-v}
| r = s_1 e^{-v}
| c =
}}
{{eqn | ll= \leadsto
| l = r_1
| r = s_1
| c =
}}
{{end-eqn}}
It follows directly from $(1)$ that $r_2 = s_2$.
Hence the result.
{{qed}}
\end{proof}
|
23088
|
\section{Unique Constant in Category of Monoids}
Tags: Category of Monoids
\begin{theorem}
Let $\mathbf{Mon}$ be the category of monoids.
Then every object $M$ of $\mathbf{Mon}$ has precisely one constant.
\end{theorem}
\begin{proof}
From Trivial Monoid is Terminal Element, we obtain that a constant of $M$ is a morphism $f: \left\{{e}\right\} \to M$.
By Trivial Monoid is Initial Element, there is precisely one such morphism.
Hence $M$ has one constant.
{{qed}}
\end{proof}
|
23089
|
\section{Unique Epimorphism from Quotient Group to Quotient Group}
Tags: Quotient Groups, Group Epimorphisms
\begin{theorem}
Let $\struct {G_1, \circledcirc}$ and $\struct {G_2, \circledast}$ be groups.
Let $\phi: \struct {G_1, \circledcirc} \to \struct {G_2, \circledast}$ be a group epimorphism.
Let $K$ be the kernel of $\phi$.
Let $H_1$ be a normal subgroup of $\struct {G_1, \circledcirc}$.
Let $H_2 := \phi \sqbrk {H_1}$ be the image of $H_1$ under $\phi$.
Then there exists a unique epimorphism $\psi$ from $G_1 / H_1$ to $G_2 / H_2$ such that:
:$\psi \circ q_{H_1} = q_{H_2} \circ \phi$
where:
:$G_1 / H_1$ and $G_2 / H_2$ are quotient groups of $G_1$ by $H_1$ and of $G_2$ by $H_2$ respectively
:$q_{H_1}: G_1 \to G_1 / H_1$ and $q_{H_2}: G_2 \to G_2 / H_2$ are their respective quotient mappings.
The kernel of $\psi$ is given by:
:$\map \ker \psi = \dfrac {H_1 \circledcirc K} {H_1}$
\end{theorem}
\begin{proof}
First we note that by Group Epimorphism Preserves Normal Subgroups, $H_2$ is indeed a normal subgroup of $\struct {G_2, \circledast}$.
Hence $G_2 / H_2$ and $q_{H_2}: G_2 \to G_2 / H_2$ are appropriately defined.
Let $\RR$ be the congruence relation induced on $\struct {G_1, \circledcirc}$ by $H$.
Let $q_\RR$ be the quotient mapping induced by $\RR$.
Then by definition $q_\RR = q_{H_1}$.
We are given that $\phi$ is a group epimorphism.
We also have from Quotient Group Epimorphism is Epimorphism that $q_{H_2}$ is also a group epimorphism.
Hence from Composite of Group Epimorphisms is Epimorphism, $q_{H_2} \circ \phi$ is also a group epimorphism.
So from Quotient Theorem for Epimorphisms, there exists a unique epimorphism $\psi$ from $G_1 / H_1$ to $G_2 / H_2$ such that:
:$\psi \circ q_{H_1} = q_{H_2} \circ \phi$
It remains to be shown that:
:$\map \ker \psi = \dfrac {H_1 \circledcirc K} {H_1}$
{{ProofWanted}}
\end{proof}
|
23090
|
\section{Unique Factorization Domain is Integrally Closed}
Tags: Algebraic Number Theory, Definitions: Factorization, Unique Factorization Domains, Factorization
\begin{theorem}
Let $A$ be a unique factorization domain (UFD).
Then $A$ is integrally closed.
\end{theorem}
\begin{proof}
Let $K$ be the field of quotients of $A$.
Let $x \in K$ be integral over $A$.
Let:
$x = a / b$
for $a, b \in A$ with $\gcd \set {a, b} \in A^\times$.
This makes sense because a UFD is GCD Domain.
There is an equation:
:$\paren {\dfrac a b}^n + a_{n - 1} \paren {\dfrac a b}^{n - 1} + \dotsb + a_0$
{{explain|This is not actually an equation, it has no equals sign in it.}}
with $a_i \in A$, $i = 0, \dotsc, n - 1$.
Multiplying by $b^n$, we obtain:
:$a^n + b c = 0$
with $c \in A$.
Therefore:
:$b \divides a^n$
{{AimForCont}} $b$ is not a unit.
Then:
:$\gcd \set {a, b} \notin A^\times$
which is a contradiction.
So $b$ is a unit, and:
:$a b^{-1} \in A$
{{explain|Make it clear why this fulfils the conditions for the statement of the result}}
{{qed}}
Category:Algebraic Number Theory
Category:Factorization
Category:Unique Factorization Domains
\end{proof}
|
23091
|
\section{Unique Factorization Theorem}
Tags: Factorization, Unique Factorization Domains, Euclidean Domains
\begin{theorem}
Let $\struct {D, +, \times}$ be a Euclidean domain.
Then $\struct {D, +, \times}$ is a unique factorization domain.
\end{theorem}
\begin{proof}
By the definition of unique factorization domain, we need to show that:
For all $x \in D$ such that $x$ is non-zero and not a unit of $D$:
:$(1): \quad x$ has a complete factorization in $D$
:$(2): \quad$ Any two complete factorizations of $x$ in $D$ are equivalent.
\end{proof}
|
23092
|
\section{Unique Integer Close to Rational in Valuation Ring of P-adic Norm}
Tags: P-adic Number Theory
\begin{theorem}
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime number $p$.
Let $x \in \Q$ such that $\norm{x}_p \le 1$.
Then for all $i \in \N$ there exists a unique $\alpha \in \Z$ such that:
:$(1): \quad \norm {x - \alpha}_p \le p^{-i}$
:$(2): \quad 0 \le \alpha \le p^i - 1$
\end{theorem}
\begin{proof}
Let $i \in \N$.
From Integer Arbitrarily Close to Rational in Valuation Ring of P-adic Norm:
:$\exists \mathop {\alpha'} \in \Z: \norm{x - \alpha'}_p \le p^{-i}$
By Integer is Congruent to Integer less than Modulus, then there exists $\alpha \in \Z$:
:$\alpha \equiv \alpha' \pmod {p^i}$.
:$0 \le \alpha \le p^i - 1$
Then $\norm {\alpha' - \alpha}_p \le p^{-i}$
Hence:
{{begin-eqn}}
{{eqn | l = \norm {x - \alpha}_p
| r = \norm {\paren {x - \alpha'} + \paren {\alpha' - \alpha} }_p
}}
{{eqn | o = \le
| r = \max \set {\norm {x - \alpha'}_p, \norm {\alpha' - \alpha}_p }
| c = {{NormAxiomNonArch|4}}
}}
{{eqn | o = \le
| r = p^{-i}
}}
{{end-eqn}}
Now suppose $\beta \in \Z$ satisfies:
:$(\text a): \quad 0 \le \beta \le p^i - 1$
:$(\text b): \quad \norm {x -\beta}_p \le p^{-i}$
Then:
{{begin-eqn}}
{{eqn | l = \norm {\alpha - \beta}_p
| r = \norm {\paren{\alpha - x} + \paren {x - \beta} }_p
}}
{{eqn | o = \le
| r = \max \set {\norm{\alpha - x}_p, \: \norm {x - \beta}_p}
| c = {{NormAxiomNonArch|4}}
}}
{{eqn | o = \le
| r = \max \set {\norm {x - \alpha}_p, \: \norm {x - \beta}_p}
| c = Norm of Negative
}}
{{eqn | o = \le
| r = p^{-i}
}}
{{end-eqn}}
Hence $p^i \divides \alpha - \beta$, or equivalently, $\alpha \equiv \beta \pmod {p^i}$
By Initial Segment of Natural Numbers forms Complete Residue System then $\alpha = \beta$.
The result follows.
{{qed}}
\end{proof}
|
23093
|
\section{Unique Isomorphism between Equivalent Finite Totally Ordered Sets}
Tags: Set Theory, Order Isomorphisms, Orderings, Total Orderings, Order Morphisms
\begin{theorem}
Let $S$ and $T$ be finite sets such that:
:$\card S = \card T$
Let $\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be totally ordered sets.
Then there is exactly one order isomorphism from $\struct {S, \preceq}$ to $\struct {T, \preccurlyeq}$.
\end{theorem}
\begin{proof}
It is sufficient to consider the case where $\struct {T, \preccurlyeq}$ is $\struct {\N_n, \le}$ for some $n \in \N$.
Let $A$ be the set of all $n \in \N$ such that if:
:$(1): \quad S$ is any set such that $\card S = n$, and
:$(2): \quad \preceq$ is any total ordering on $S$
then there is exactly one isomorphism from $\struct {S, \preceq}$ to $\struct {\N_n, \le}$.
$\O \in A$ from Empty Mapping is Mapping and Equivalence of Mappings between Sets of Same Cardinality.
Now let $n \in A$.
Let $\struct {S, \preceq}$ be a totally ordered set with $n + 1$ elements.
By Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements, $S$ has a greatest element $b$.
Then $\card {S \setminus \set b} = n$ by Cardinality Less One.
The total ordering on $S \setminus \set b$ is the one induced from that on $S$.
So as $n \in A$, there exists a unique order isomorphism $f: S \setminus \set b \to \struct {\N_n, \le}$.
Let us define the mapping $g: S \to \N_{n + 1}$ as follows:
:$\forall x \in S: \map g x = \begin{cases}
\map f x: & x \in S \setminus \set b \\
n: & x = b
\end{cases}$
This is the desired order isomorphism from $\struct {S, \preceq}$ to $\struct {\N_{n + 1}, \le}$.
Now let $h: \struct {S, \preceq}$ to $\struct {\N_{n + 1}, \le}$ be an order isomorphism.
Then $\map h b = n$ (it has to be).
So the restriction of $h$ to $S \setminus \set b$ is an isomorphism from $S \setminus \set b$ to $\struct {\N_n, \le}$, and hence $h = f$ (as $n \in A$, any such order isomorphism is unique).
Thus:
:$\forall x \in S \setminus \set b: \map h x = \map f x = \map g x$
and:
:$\map h b = n = \map g b$
thus $h = g$.
Therefore $n + 1 \in A$.
The result follows from the Principle of Mathematical Induction.
{{qed}}
\end{proof}
|
23094
|
\section{Unique Isomorphism between Ordinal Subset and Unique Ordinal}
Tags: Order Isomorphisms, Ordinals
\begin{theorem}
Let $\On$ be the class of ordinals.
Let $S \subset \On$ where $S$ is a set.
Then there exists a unique mapping $\phi$ and a unique ordinal $x$ such that $\phi : x \to S$ is an order isomorphism.
\end{theorem}
\begin{proof}
Since $S \subset \On$, $\struct {S, \in}$ is a strict well-ordering.
{{explain|Link to result justifying this statement.}}
The result follows directly from Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping.
{{qed}}
\end{proof}
|
23095
|
\section{Unique Isomorphism from Quotient Mapping to Epimorphism Domain}
Tags: Isomorphisms, Quotient Mappings, Epimorphisms
\begin{theorem}
Let $\struct {S, \oplus}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an epimorphism.
Let $\RR_\phi$ be the equivalence induced by $\phi$.
Let $S / \RR_\phi$ be the quotient of $S$ by $\RR_\phi$.
Let $q_{\RR_\phi}: S \to S / \RR_\phi$ be the quotient mapping induced by $\RR_\phi$.
Let $\struct {S / \RR_\phi, \oplus_{\RR_\phi} }$ be the quotient structure defined by $\RR_\phi$.
Then there is one and only one isomorphism:
:$\psi: \struct {S / \RR_\phi, \oplus_{\RR_\phi} } \to \struct {T, *}$
which satisfies:
:$\psi \circ q_{\RR_\phi} = \phi$
where $\circ$ denotes composition of mappings.
\end{theorem}
\begin{proof}
From the Quotient Theorem for Surjections, there is a unique bijection from $S / \RR_\phi$ onto $T$ satisfying $\psi \circ q_{\RR_\phi} = \phi$.
Also:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
| l = \map \psi {\eqclass x {\RR_\phi} \oplus_{\RR_\phi} \eqclass y {\RR_\phi} }
| r = \map \psi {\eqclass {x \oplus y} {\RR_\phi} }
| c = {{Defof|Quotient Structure}}
}}
{{eqn | r = \map \phi {x \oplus y}
| c = {{Defof|Epimorphism (Abstract Algebra)}}
}}
{{eqn | r = \map \phi x * \map \phi y
| c = {{Defof|Epimorphism (Abstract Algebra)}}
}}
{{eqn | r = \map \psi {\eqclass x {\RR_\phi} } * \map \psi {\eqclass y {\RR_\phi} }
| c = {{Defof|Quotient Mapping}}
}}
{{end-eqn}}
Therefore $\psi$ is an isomorphism.
{{Qed}}
\end{proof}
|
23096
|
\section{Unique Linear Transformation Between Modules}
Tags: Linear Transformations, Modules
\begin{theorem}
Let $\struct {G, +_G, \times_G}_R$ and $\struct {H, +_H, \times_H}_R$ be unitary $R$-modules.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an ordered basis of $G$.
Let $\sequence {b_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of $n$ elements of $H$.
Then there exists a unique linear transformation $\phi: G \to H$ satisfying:
:$\forall k \in \closedint 1 n: \map \phi {a_k} = b_k$
\end{theorem}
\begin{proof}
By Unique Representation by Ordered Basis, the mapping $\phi: G \to H$ defined as:
:$(1): \quad \ds \map \phi {\sum_{j \mathop = 1}^n \lambda_j \times_G a_j} = \sum_{j \mathop = 1}^n \lambda_j \times_H b_j$
is well-defined.
{{begin-eqn}}
{{eqn | q = \forall k \in \closedint 1 n
| l = \map \phi {\sum_{j \mathop = 1}^n \lambda_{j k} \times_G a_j}
| r = \sum_{j \mathop = 1}^n \lambda_{j k} \times_H b_j
| c = from $(1)$
}}
{{eqn | ll= \leadsto
| l = \map \phi {\sum_{j \mathop = 1}^n \delta_{j k} \times_G a_j}
| r = \sum_{j \mathop = 1}^n \delta_{j k} \times_H b_j
| c = setting $\lambda_{j k} = \delta_{j k}$, where $\delta_{j k}$ is Kronecker Delta
}}
{{eqn | ll= \leadsto
| l = \map \phi {a_k}
| r = b_k
| c = {{Defof|Kronecker Delta}}
}}
{{end-eqn}}
To verify that $\phi$ is a linear transformation, we need to show the following:
:$(1): \quad \forall x, y \in G: \map \phi {x +_G y} = \map \phi x +_H \map \phi y$
:$(2): \quad \forall x \in G: \forall \lambda \in R: \map \phi {\lambda \times_G x} = \lambda \times_H \map \phi x$
Let $x, y \in G$ be arbitrary, such that:
:$\ds x = \sum_{k \mathop = 1}^n \lambda_k \times_G a_k$
:$\ds y = \sum_{k \mathop = 1}^n \mu_k \times_G a_k$
where:
:$a_k$ are elements of $\sequence {a_n}$
:$\lambda_k$ and $\mu_k$ are elements of $R$.
So:
{{begin-eqn}}
{{eqn | l = \map \phi {x +_G y}
| r = \map \phi {\sum_{k \mathop = 1}^n \lambda_k \times_G a_k +_G \sum_{k \mathop = 1}^n \mu_k \times_G a_k}
| c = Definition of $x$ and $y$
}}
{{eqn | r = \map \phi {\sum_{k \mathop = 1}^n \paren {\lambda_k + \mu_k} \times_G a_k}
| c = {{Module-axiom|2}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {\lambda_k + \mu_k} \times_H b_k
| c = from $(1)$
}}
{{eqn | r = \sum_{k \mathop = 1}^n \lambda_k \times_H b_k +_H \sum_{k \mathop = 1}^n \mu_k \times_H b_k
| c = {{Module-axiom|2}}
}}
{{eqn | r = \map \phi x +_H \sum_{k \mathop = 1}^n \map \phi y
| c = Definition of $x$ and $y$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map \phi {\lambda \times_G x}
| r = \map \phi {\lambda \times_G \sum_{k \mathop = 1}^n \lambda_k \times_G a_k}
| c = Definition of $x$
}}
{{eqn | r = \map \phi {\sum_{k \mathop = 1}^n \lambda \times_G \paren {\lambda_k \times_G a_k} }
| c = {{Module-axiom|1}}
}}
{{eqn | r = \map \phi {\sum_{k \mathop = 1}^n \paren {\lambda \lambda_k} \times_G a_k}
| c = {{Module-axiom|3}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {\lambda \lambda_k} \times_H b_k
| c = from $(1)$
}}
{{eqn | r = \sum_{k \mathop = 1}^n \lambda \times_H \paren {\lambda_k \times_H b_k}
| c = {{Module-axiom|1}}
}}
{{eqn | r = \lambda \times_H \map \phi x
| c = Definition of $x$
}}
{{end-eqn}}
By Linear Transformation of Generated Module, $\phi$ is the only linear transformation whose value at $a_k$ is $b_k$ for all $k \in \closedint 1 n$.
{{Qed}}
\end{proof}
|
23097
|
\section{Unique Minimal Element may not be Smallest}
Tags: Minimal Elements, Smallest Elements
\begin{theorem}
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $\struct {S, \preccurlyeq}$ have a unique minimal element.
Then it is not necessarily the case that $\struct {S, \preccurlyeq}$ has a smallest element.
\end{theorem}
\begin{proof}
Let $S$ be the set defined as:
:$S = \Z \cup \set m$
where:
:$\Z$ denotes the set of integers
:$m$ is an arbitrary object such that $m \ne \Z$.
Let $\preccurlyeq$ be the relation on $\Z$ defined as:
:$\forall a, b \in S: a \preccurlyeq b \iff \begin {cases} a \le b & : a, b \in \Z \\ a = m = b & : a, b \notin \Z \end{cases}$
where $\le$ is the usual ordering on $\Z$.
Note that for all $x \in \Z$, we have that $x$ and $m$ are non-comparable.
Then $\struct {S, \preccurlyeq}$ is an ordered set such that:
:$m$ is the only minimal element of $\struct {S, \preccurlyeq}$
:$\struct {S, \preccurlyeq}$ has no smallest element.
This is shown as follows:
Let $x \in S$ such that $x \preccurlyeq m$.
Then by definition $x = m$.
Hence $m$ is a minimal element of $\struct {S, \preccurlyeq}$.
{{AimForCont}} $y \in \Z$ is another minimal element of $\struct {S, \preccurlyeq}$.
Let $z = y - 1$.
By definition of $\Z$, we have that $z \in \Z$.
But $z \le y$ and so $z \preccurlyeq y$.
Hence for all $y \in \Z$, it is not the case that $y$ is a minimal element of $\struct {S, \preccurlyeq}$.
Hence by Proof by Contradiction it follows that the minimal element $m$ of $\struct {S, \preccurlyeq}$ is unique.
{{AimForCont}} $x \in S$ is the smallest element of $\struct {S, \preccurlyeq}$.
Suppose $x \in \Z$.
Then because $x$ and $m$ are non-comparable, it is not the case that $x \preccurlyeq m$.
Hence $x$ cannot be the smallest element.
Suppose $x = m$.
Let $a \in S: a \ne m$.
Then $x$ and $a$ are again non-comparable.
Hence it is not the case that $x \preccurlyeq a$.
Hence $x$ again cannot be the smallest element of $\struct {S, \preccurlyeq}$.
So by Proof by Cases $x$ cannot be the smallest element of $\struct {S, \preccurlyeq}$.
It follows by Proof by Contradiction that there can be no smallest element of $\struct {S, \preccurlyeq}$.
{{qed}}
\end{proof}
|
23098
|
\section{Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space}
Tags: Hilbert Spaces, Convex Sets (Vector Spaces)
\begin{theorem}
Let $H$ be a Hilbert space, and let $h \in H$.
Let $K \subseteq H$ be a closed, convex, non-empty subset of $H$.
Then there is a unique point $k_0 \in K$ such that:
:$\norm {h - k_0} = \map d {h, K}$
where $d$ denotes distance to a set.
{{refactor|Own page, and prove it}}
Furthermore, if $K$ is a linear subspace, this point is characterised by:
:$\norm {h - k_0} = \map d {h, K} \iff \paren {h - k_0} \perp K$
where $\perp$ signifies orthogonality.
\end{theorem}
\begin{proof}
Let $\mathbf 0_H$ be the zero of $H$.
Since for every $k \in K$, we have:
:$\map d {h, k} = \norm {h - k} = \map d {\mathbf 0_H, k - h}$
it follows that:
:$\map d {h, K} = \map d {\mathbf 0_H, K - h}$
{{WLOG}}, we may therefore assume that $h = \mathbf 0_H$.
The problem has therefore reduced to finding $k_0 \in K$ such that:
:$\norm {k_0} = \map d {\mathbf 0_H, K} = \inf \set {\norm k : k \in K}$
Let $d = \map d {\mathbf 0_H, K}$.
By definition of infimum, there exists a sequence $\sequence {k_n}_{n \mathop \in \N}$ such that:
:$\ds \lim_{n \mathop \to \infty} \norm {k_n} = d$
By the Parallelogram Law, we have that for all $m, n \in \N$:
:$(1): \quad \norm {\dfrac {k_n - k_m} 2 } = \dfrac 1 2 \paren {\norm {k_n}^2 + \norm {k_m}^2} - \norm {\dfrac {k_n + k_m} 2 }^2$
Since $K$ is convex, $\dfrac {k_n + k_m} 2 \in K$.
Hence:
:$\norm {\dfrac {k_n + k_m} 2 }^2 \ge d^2$
Now given $\epsilon > 0$, choose $N$ such that for all $n \ge N$:
:$\norm {k_n}^2 < d^2 + \epsilon$
From $(1)$, it follows that:
:$\norm {\dfrac {k_n - k_m} 2} < d^2 + \epsilon - d^2 = \epsilon$
and hence that $\sequence {k_n}_{n \mathop \in \N}$ is a Cauchy sequence.
Since $H$ is a Hilbert space and $K$ is closed, it follows that there is a $k_0 \in K$ such that:
:$\ds \lim_{n \mathop \to \infty} k_n = k_0$
From Norm is Continuous, we infer that $\norm {k_0} = d$.
This demonstrates existence of $k_0$.
For uniqueness, suppose that $h_0 \in K$ has $\norm {h_0} = d$.
Since $K$ is convex, it follows that $\dfrac {h_0 + k_0} 2 \in K$.
This implies that $\norm {\dfrac {h_0 + k_0} 2} \ge d$.
Now from the Triangle Inequality:
:$\norm {\dfrac {h_0 + k_0} 2} \le \dfrac {\norm {h_0} + \norm {k_0} } 2 = d$
meaning that $\norm {\dfrac {h_0 + k_0} 2} = d$.
Thus, the Parallelogram Law implies that:
:$d^2 = \norm {\dfrac {h_0 + k_0} 2}^2 = d^2 - \norm {\dfrac {h_0 - k_0} 2}^2$
from which we conclude that $h_0 = k_0$.
{{qed}}
\end{proof}
|
23099
|
\section{Unique Readability for Polish Notation}
Tags: Formal Systems
\begin{theorem}
Let $\AA$ be an alphabet.
Then Polish notation for $\AA$ has the unique readability property.
\end{theorem}
\begin{proof}
Let $\phi$ be a WFF of Polish notation for $\AA$.
Apply the Principle of Mathematical Induction on the length of $\phi$ to prove:
:$(1): \quad$ No initial part of $\phi$ is a WFF, except $\phi$ itself;
:$(2): \quad$ If the first symbol of $\phi$ has arity $n$, then there exist unique WFFs $\phi_1, \ldots, \phi_n$ such that $\phi = s \phi_1 \cdots \phi_n$.
Let $\phi'$ be a WFF that is an initial part of $\phi$.
Because all WFFs have at least length $1$, it follows that:
:$\phi' = s \phi'_1 \cdots \phi'_n$
Now it must be that either $\phi_1$ is an initial part of $\phi'_1$ or vice versa.
By the induction hypothesis on $(1)$, it must be that $\phi_1 = \phi'_1$ because $\phi_1$ and $\phi'_1$ are both strictly shorter than $\phi$.
Now if, given $1 < j \le n$:
:$\phi_i = \phi'_i$ for all $i < j$
it follows that $\phi_j$ and $\phi'_j$ start at the same position in $\phi$.
Hence again, inductively, it follows that $\phi_j = \phi'_j$.
Thus $\phi_i = \phi'_i$ for all $1 \le i \le n$; that is:
:$\phi = \phi'$
The result follows by the Principle of Mathematical Induction.
{{qed}}
\end{proof}
|
23100
|
\section{Unique Representation in Polynomial Forms}
Tags: Polynomials, Polynomial_Theory, Polynomial Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Let $X \in R$ be transcendental over $D$.
Let $D \sqbrk X$ be the ring of polynomials in $X$ over $D$.
Then each non-zero member of $D \left[{X}\right]$ can be expressed in just one way in the form:
:$\ds f \in D \sqbrk X: f = \sum_{k \mathop = 0}^n {a_k \circ X^k}$
\end{theorem}
\begin{proof}
Suppose $f \in D \sqbrk X \setminus \set {0_R}$ has more than one way of being expressed in the above form.
Then you would be able to subtract one from the other and get a polynomial in $D \sqbrk X$ equal to zero.
As $f$ is transcendental, the result follows.
{{qed}}
\end{proof}
|
23101
|
\section{Unique Representation in Polynomial Forms/General Result}
Tags: Polynomial Theory
\begin{theorem}
Let $f$ be a polynomial form in the indeterminates $\set {X_j: j \in J}$ such that $f: \mathbf X^k \mapsto a_k$.
For $r \in \R$, $\mathbf X^k \in M$, let $r \mathbf X^k$ denote the polynomial form that takes the value $r$ on $\mathbf X^k$ and zero on all other monomials.
Let $Z$ denote the set of all multiindices indexed by $J$.
Then the sum representation:
:$\ds \hat f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
has only finitely many non-zero terms.
Moreover it is everywhere equal to $f$, and is the unique such sum.
\end{theorem}
\begin{proof}
Suppose that the sum has infinitely many non-zero terms.
Then infinitely many $a_k$ are non-zero, which contradicts the definition of a polynomial.
Therefore the sum consists of finitely many non-zero terms.
Let $\mathbf X^m \in M$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map {\hat f} {\mathbf X^m}
| r = \paren {\sum_{k \mathop \in Z} a_k \mathbf X^k} \paren {\mathbf X^m}
}}
{{eqn | r = \paren {a_m \mathbf X^m} \paren {\mathbf X^m} + \sum_{k \mathop \ne m \mathop \in Z} \paren {a_k \mathbf X^k} \paren {\mathbf X^m}
}}
{{eqn | r = a_m
}}
{{end-eqn}}
So $\hat f = f$.
Finally suppose that:
:$\ds \tilde f = \sum_{k \mathop \in Z} b_k \mathbf X^k$
is another such representation with $b_m \ne a_m$ for some $m \in Z$.
Then:
:$\map {\tilde f} {\mathbf X^m} = b_m \ne a_m = \map f {\mathbf X^m}$
Therefore $\hat f$ as defined above is the only such representation.
{{qed}}
Category:Polynomial Theory
\end{proof}
|
23102
|
\section{Unique Subgroup of a Given Order is Normal}
Tags: Normal Subgroups
\begin{theorem}
Let a group $G$ have only one subgroup of a given order.
Then that subgroup is normal.
\end{theorem}
\begin{proof}
Let $H \le G$, where $\le$ denotes that $H$ is a subgroup of $G$.
Let $H$ be the only subgroup of $G$ whose order is $\order H$.
Let $g \in G$.
From Conjugate of Subgroup is Subgroup:
:$g H g^{-1} \le G$
From Order of Conjugate of Subgroup:
: $\order {g H g^{-1} } = \order H$
But $H$ is the only subgroup of $G$ of order $\order H$.
Hence any subgroup whose order is $\order H$ must in fact ''be'' $H$.
That is, $g H g^{-1} = H$.
The result follows from Subgroup equals Conjugate iff Normal.
{{Qed}}
\end{proof}
|
23103
|
\section{Uniqueness Condition for Relation Value}
Tags: Relation Theory
\begin{theorem}
Let $\RR$ be a relation.
Let $\tuple {x, y} \in \RR$.
Let:
:$\exists ! y: \tuple {x, y} \in \RR$
Then:
:$\map \RR x = y$
where $\map \RR x$ denotes the image of $\RR$ at $x$.
If $y$ is not unique, then:
:$\map \RR x = \O$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | n = 1
| l = z \in \map \RR x
| o = \implies
| r = \exists y: \paren {z \in y \land \tuple {x, y} \in \RR} \land \exists ! y: \tuple {x, y} \in \RR
| c = {{Defof|Image (Set Theory)/Relation/Element/Singleton|Image of Element under Relation}}
}}
{{eqn | o = \implies
| r = \forall w: \paren {\tuple {x, w} \in \RR \iff w = y}
| c = {{Defof|Unique}} and {{hypothesis}} that $\tuple {x, y} \in \RR$
}}
{{eqn | o = \implies
| r = z \in w \land \tuple {x, w} \in \RR
| c = Existential Instantiation from $(1)$
}}
{{eqn | o = \implies
| r = z \in y
| c = Substitutivity of Equality from $z \in w \land w = y$
}}
{{end-eqn}}
Conversely:
{{begin-eqn}}
{{eqn | l = z \in y
| o = \implies
| r = \paren {z \in y \land \tuple {x, y} \in \RR}
| c = {{hypothesis}} that $\tuple {x, y} \in \RR$
}}
{{eqn | o =
| r = \exists ! y: \tuple {x, y} \in \RR
| c = {{hypothesis}}
}}
{{eqn | o = \implies
| r = z \in \map \RR x
| c = {{Defof|Image (Set Theory)/Relation/Element/Singleton|Image of Element under relation}}
}}
{{end-eqn}}
Generalizing:
:$\forall z: \paren {z \in y \iff z \in \map \RR x}$
Therefore:
:$y = \map \RR x$
by the definition of class equality.
{{qed|lemma}}
Suppose that $\neg \exists ! y: \tuple {x, y} \in \RR$.
Then:
{{begin-eqn}}
{{eqn | l = \neg \exists ! y: \tuple {x, y} \in \RR
| o = \implies
| r = z \notin \map \RR x
| c = {{Defof|Image (Set Theory)/Relation/Element/Singleton|Image of Element under relation}}
}}
{{end-eqn}}
Thus:
:$\forall z: z \notin \map \RR x$
Therefore:
:$\map \RR x = \O$
{{qed}}
\end{proof}
|
23104
|
\section{Uniqueness of Analytic Continuation}
Tags: Complex Analysis
\begin{theorem}
Let $U \subset V \subset \C$ be open subsets of the complex plane.
Let $V$ be connected.
Suppose:
:$(1): \quad F_1, F_2$ are functions defined on $V$
:$(2): \quad f$ is a function defined on $U$.
Let $F_1$ and $F_2$ be analytic continuations of $f$ to $V$.
Then $F_1 = F_2$.
\end{theorem}
\begin{proof}
Let $\map g z = \map {F_1} z - \map {F_2} z$.
Then:
:$\forall z \in U: \map g z = 0$
Because Zeroes of Analytic Function are Isolated, and the zeroes of $g$ are not isolated, $g$ must be constant everywhere in its domain.
Since $\map g z = 0$ for some $z$, it follows that $\map g z = 0$ for all $z$.
Hence:
:$\forall z \in V: \map {F_1} z - \map {F_2} z = 0$ and so $F_1 = F_2$.
{{qed}}
Category:Complex Analysis
\end{proof}
|
23105
|
\section{Uniqueness of Jordan Decomposition}
Tags: Uniqueness of Jordan Decomposition, Signed Measures
\begin{theorem}
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {P_1, N_1}$ and $\tuple {P_2, N_2}$ be Hahn decompositions of $\mu$.
Let $\tuple {\mu^+_1, \mu^-_1}$ be the Jordan decomposition of $\mu$ corresponding to $\tuple {P_1, N_1}$.
Let $\tuple {\mu^+_2, \mu^-_2}$ be the Jordan decomposition of $\mu$ corresponding to $\tuple {P_2, N_2}$.
Then:
:$\mu^+_1 = \mu^+_2$
and:
:$\mu^-_1 = \mu^-_2$
So:
:the Jordan decomposition of a signed measure is unique.
\end{theorem}
\begin{proof}
We first prove two useful identities.
\end{proof}
|
23106
|
\section{Uniqueness of Jordan Decomposition/Lemma}
Tags: Uniqueness of Jordan Decomposition
\begin{theorem}
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {P, N}$ be a Hahn decomposition of $\mu$.
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$ corresponding to $\tuple {P, N}$.
Then, for each $A \in \Sigma$, we have:
:$\map {\mu^+} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
and:
:$\map {\mu^-} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
\end{theorem}
\begin{proof}
Since $\tuple {\mu^+, \mu^-}$ is a Jordan decomposition of $\mu$, we have:
:$\mu = \mu^+ - \mu^-$
with $\mu^+$ and $\mu^-$ measures.
Let $A \in \Sigma$.
We first show:
:$\map {\mu^+} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
Let $B \in \Sigma$ have $B \subseteq A$.
We have:
{{begin-eqn}}
{{eqn | l = \map \mu B
| r = \map {\mu^+} B - \map {\mu^-} B
}}
{{eqn | o = \le
| r = \map {\mu^+} B
| c = $\mu^+ \ge 0$ since $\mu^+$ is a measure
}}
{{eqn | o = \le
| r = \map {\mu^+} A
| c = Measure is Monotone
}}
{{end-eqn}}
So:
:$\map {\mu^+} A$ is an upper bound for $\set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
Note that from Intersection is Subset, we have:
:$A \cap P \subseteq A$
From Sigma-Algebra Closed under Countable Intersection, we also have $A \cap P \in \Sigma$ so:
:$\map \mu {A \cap P} \in \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
By definition of the Jordan decomposition:
:$\map {\mu^+} A = \map \mu {A \cap P}$
so:
:$\map {\mu^+} A \in \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
So:
:$\map {\mu^+} A$ is the greatest element of $\set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
From Greatest Element is Supremum, we therefore have:
:$\map {\mu^+} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
We now show that:
:$\map {\mu^-} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
Let $B \in \Sigma$ have $B \subseteq A$.
We have:
{{begin-eqn}}
{{eqn | l = -\map \mu B
| r = \map {\mu^-} B - \map {\mu^+} B
}}
{{eqn | o = \le
| r = \map {\mu^-} B
| c = $\mu^- \ge 0$ since $\mu^-$ is a measure
}}
{{eqn | o = \le
| r = \map {\mu^-} A
| c = Measure is Monotone
}}
{{end-eqn}}
So:
:$\map {\mu^-} A$ is an upper bound for $\set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
Note from Intersection is Subset, we have:
:$A \cap N \subseteq A$
From Sigma-Algebra Closed under Countable Intersection, we also have $A \cap N \in \Sigma$ so:
:$-\map \mu {A \cap N} \in \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
By definition of the Jordan decomposition:
:$\map {\mu^-} A = -\map \mu {A \cap N}$
so:
:$\map {\mu^-} A \in \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
So:
:$\map {\mu^-} A$ is the greatest element of $\set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
From Greatest Element is Supremum, we have:
:$\map {\mu^-} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
hence the result.
{{qed}}
Category:Uniqueness of Jordan Decomposition
\end{proof}
|
23107
|
\section{Uniqueness of Polynomial Ring in One Variable}
Tags: Polynomial Theory
\begin{theorem}
Let $R$ be a commutative ring with unity.
Let $\struct {R \sqbrk X, \iota, X}$ and $\struct {R \sqbrk Y, \kappa, Y}$ be polynomial rings in one variable over $R$.
Then there exists a unique ring homomorphism $f: R \sqbrk X \to R \sqbrk Y$ such that:
:$f \circ \iota = \kappa$
:$\map f X = Y$
and it is an isomorphism.
\end{theorem}
\begin{proof}
The existence and uniqueness of $f$ follows from the universal property.
Likewise, there is a unique ring homomorphism $g: R \sqbrk Y \to R \sqbrk X$ such that:
:$g \circ \kappa = \iota$
:$\map g Y = X$
and a unique ring homomorphism $h: R \sqbrk X \to R \sqbrk X$ such that:
:$h \circ \iota = \iota$
:$\map X = X$
By uniqueness and Identity Mapping is Ring Homomorphism, $h = \operatorname{id}$ is the identity mapping on $R \sqbrk X$.
Again by uniqueness and Composition of Ring Homomorphisms is Ring Homomorphism, $g\circ f = \operatorname{id}_{R \sqbrk X}$.
By symmetry, $f \circ g = \operatorname{id}_{R \sqbrk Y}$.
Thus $f$ is an isomorphism.
{{qed}}
Category:Polynomial Theory
\end{proof}
|
23108
|
\section{Uniqueness of Positive Root of Positive Real Number/Positive Exponent}
Tags: Uniqueness of Positive Root of Positive Real Number
\begin{theorem}
Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n > 0$.
Then there is at most one $y \in \R: y \ge 0$ such that $y^n = x$.
\end{theorem}
\begin{proof}
Let the real function $f: \hointr 0 \to \to \hointr 0 \to$ be defined as:
:$\map f y = y^n$
First let $n > 0$.
By Identity Mapping is Order Isomorphism, the identity function $I_\R$ on $\hointr 0 \to$ is strictly increasing.
We have that:
:$\map f y = \paren {\map {I_\R} y}^n$
By Product of Positive Strictly Increasing Mappings is Strictly Increasing, $f$ is strictly increasing on $\hointr 0 \to$.
From Strictly Monotone Mapping with Totally Ordered Domain is Injective:
:there is at most one $y \in \R: y \ge 0$ such that $y^n = x$.
{{Qed}}
Category:Uniqueness of Positive Root of Positive Real Number
442140
442137
2020-01-03T21:15:58Z
Prime.mover
59
442140
wikitext
text/x-wiki
\end{proof}
|
23109
|
\section{Uniqueness of Real z such that x Choose n+1 Equals y Choose n+1 Plus z Choose n}
Tags: Binomial Coefficients
\begin{theorem}
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $x, y \in \R$ be real numbers which satisfy:
:$n \le y \le x \le y + 1$
Then there exists a unique real number $z$ such that:
:$\dbinom x {n + 1} = \dbinom y {n + 1} + \dbinom z n$
where $n - 1 \le z \le y$.
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = \dbinom y {n + 1}
| o = \le
| r = \dbinom x {n + 1}
| c = Ordering of Binomial Coefficients
}}
{{eqn | o = \le
| r = \dbinom {y + 1} {n + 1}
| c =
}}
{{eqn | r = \dbinom y {n + 1} + \dbinom y n
| c = Pascal's Rule
}}
{{end-eqn}}
{{finish}}
\end{proof}
|
23110
|
\section{Uniqueness of Representing Objects}
Tags: Category Theory
\begin{theorem}
Let $C$ be a locally small category.
Let $\mathbf{Set}$ be the category of sets.
Let $F : \mathbf C \to \mathbf{Set}$ be a covariant functor.
Let $(A, \eta)$ and $(B, \xi)$ be representations of $F$.
Then there exists a unique isomorphism $f : A \to B$ such that $\eta \circ h^f = \xi$, where:
:$h^f$ is the precomposition natural transformation
:$\circ$ denotes vertical composition of natural transformations
\end{theorem}
\begin{proof}
{{proof wanted}}
Category:Category Theory
\end{proof}
|
23111
|
\section{Unit Cylinder as Surface of Revolution}
Tags: Induced Metric, Surfaces of Revolution, Solid Geometry
\begin{theorem}
Let $\struct {\R^3, d}$ be the Euclidean space.
Let $S_C \subseteq \R^3$ be the surface of revolution.
Let $C$ be a straight line in the open upper half-plane.
Let the smooth local parametrization of $C$ be:
:$\map \gamma t = \tuple {t, 1}$
Then the induced metric on $S_C$ is:
:$g = d t^2 + d \theta^2$
\end{theorem}
\begin{proof}
We have that:
:$\map {\gamma'} t = \tuple {1, 0}$
Hence, $\map \gamma t$ is a unit-speed curve.
By the corollary of the induced metric on the surface of revolution:
:$g = d t^2 + d \theta^2$
{{qed}}
\end{proof}
|
23112
|
\section{Unit Interval is Path-Connected in Real Numbers}
Tags: Path-Connectedness, Connected Spaces, Real Intervals
\begin{theorem}
Let $\R$ be the real number line with the usual (Euclidean} metric.
The closed unit interval $\mathbf I = \closedint 0 1$ is a path-connected metric subspace of $\R$.
\end{theorem}
\begin{proof}
Follows directly from Subset of Real Numbers is Path-Connected iff Interval.
{{qed}}
\end{proof}
|
23113
|
\section{Unit Matrix is Identity for Matrix Multiplication/Left}
Tags: Unit Matrix is Identity for Matrix Multiplication
\begin{theorem}
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $m, n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map {\MM_R} {m, n}$ denote the $m \times n$ metric space over $R$.
Let $I_m$ denote the unit matrix of order $m$.
Then:
:$\forall \mathbf A \in \map {\MM_R} {m, n}: \mathbf I_m \mathbf A = \mathbf A$
\end{theorem}
\begin{proof}
Let $\sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
Let $\sqbrk b_{m n} = \mathbf I_m \sqbrk a_{m n}$.
Then:
{{begin-eqn}}
{{eqn | q = \forall i \in \closedint 1 m, j \in \closedint 1 n
| l = b_{i j}
| r = \sum_{k \mathop = 1}^m \delta_{i k} a_{k j}
| c = where $\delta_{i k}$ is the Kronecker delta: $\delta_{i k} = 1_R$ when $i = k$ otherwise $0_R$
}}
{{eqn | r = a_{i j}
| c =
}}
{{end-eqn}}
Thus $\sqbrk b_{m n} = \sqbrk a_{m n}$ and $\mathbf I_m$ is shown to be a left identity.
{{qed}}
Category:Unit Matrix is Identity for Matrix Multiplication
\end{proof}
|
23114
|
\section{Unit Matrix is Identity for Matrix Multiplication/Right}
Tags: Unit Matrix is Identity for Matrix Multiplication
\begin{theorem}
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $m, n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map {\MM_R} {m, n}$ denote the $m \times n$ metric space over $R$.
Let $I_n$ denote the unit matrix of order $n$.
Then:
:$\forall \mathbf A \in \map {\MM_R} {m, n}: \mathbf A \mathbf I_n = \mathbf A$
\end{theorem}
\begin{proof}
Let $\sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
Let $\sqbrk b_{m n} = \sqbrk a_{m n} \mathbf I_n$.
Then:
{{begin-eqn}}
{{eqn | q = \forall i \in \closedint 1 m, j \in \closedint 1 n
| l = b_{i j}
| r = \sum_{k \mathop = 1}^n a_{i k} \delta_{k j}
| c = where $\delta_{k j}$ is the Kronecker delta: $\delta_{k j} = 1_R$ when $k = j$ otherwise $0_R$
}}
{{eqn | r = a_{i j}
| c =
}}
{{end-eqn}}
Thus $\sqbrk b_{m n} = \sqbrk a_{m n}$ and $\mathbf I_n$ is shown to be a right identity.
{{qed}}
Category:Unit Matrix is Identity for Matrix Multiplication
\end{proof}
|
23115
|
\section{Unit Matrix is Orthogonal}
Tags: Unit Matrices, Orthogonal Matrices
\begin{theorem}
The unit matrix $\mathbf I_n$ of order $n$ is orthogonal.
\end{theorem}
\begin{proof}
By Unit Matrix is its own Inverse the inverse $I_n^{-1}$ of $I_n$ is $I_n$.
By definition a unit matrix is a diagonal matrix.
Hence by Diagonal Matrix is Symmetric:
:$I_n = I_n^\intercal$
where $I_n^\intercal$ is the transpose of $I_n$.
Thus:
:$I_n^{-1} = I_n^\intercal$
and the result follows by definition of orthogonal.
{{qed}}
Category:Unit Matrices
Category:Orthogonal Matrices
\end{proof}
|
23116
|
\section{Unit Matrix is Proper Orthogonal}
Tags: Unit Matrices, Orthogonal Matrices
\begin{theorem}
The unit matrix $\mathbf I_n$ of order $n$ is proper orthogonal.
\end{theorem}
\begin{proof}
By Unit Matrix is Orthogonal, $\mathbf I_n$ is an orthogonal matrix.
By Determinant of Unit Matrix, the determinant of $\mathbf I_n$ is equal to $1$.
The result follows by definition of proper orthogonal.
{{qed}}
Category:Unit Matrices
Category:Orthogonal Matrices
\end{proof}
|
23117
|
\section{Unit Matrix is Unity of Ring of Square Matrices}
Tags: Matrix Product, Rings of Square Matrices, Unit Matrix is Unity of Ring of Square Matrices, Matrix Algebra, Unit Matrices, Conventional Matrix Multiplication
\begin{theorem}
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\struct {\map {\MM_R} n, +, \times}$ denote the ring of square matrices of order $n$ over $R$.
The unit matrix over $R$:
:$\mathbf I_n = \begin {pmatrix} 1_R & 0_R & 0_R & \cdots & 0_R \\ 0_R & 1_R & 0_R & \cdots & 0_R \\ 0_R & 0_R & 1_R & \cdots & 0_R \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0_R & 0_R & 0_R & \cdots & 1_R \end {pmatrix}$
is the identity element of $\struct {\map {\MM_R} n, +, \times}$.
\end{theorem}
\begin{proof}
In Unit Matrix is Identity for Matrix Multiplication, it is demonstrated that:
:$\forall \mathbf A \in \map {\MM_R} n: \mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$
Hence the result, by definition of identity element
{{qed}}
\end{proof}
|
23118
|
\section{Unit Matrix is its own Inverse}
Tags: Unit Matrices, Inverse Matrices, Matrix Inverses, Linear Algebra
\begin{theorem}
The inverse of the unit matrix $\mathbf I_n$ of order $n$ is $\mathbf I_n$.
That is, a unit matrix it its own inverse.
\end{theorem}
\begin{proof}
By definition, a unit matrix is a diagonal matrix.
From Inverse of Diagonal Matrix, the inverse of a diagonal matrix:
:$\mathbf D = \begin{bmatrix}
a_{11} & 0 & \cdots & 0 \\
0 & a_{22} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{nn} \\
\end{bmatrix}$
is the diagonal matrix:
: $\mathbf D^{-1} = \begin{bmatrix}
\dfrac 1 {a_{11}} & 0 & \cdots & 0 \\
0 & \dfrac 1 {a_{22}} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \dfrac 1 {a_{nn}} \\
\end{bmatrix}$
When $\mathbf D$ is the unit matrix $\mathbf I_n$, all elements $a_{kk}$ are equal to $1$:
:$\mathbf I_n = \begin{bmatrix}
1 & 0 & \cdots & 0 \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1 \\
\end{bmatrix}$
Hence:
:$\mathbf I_n^{-1} = \begin{bmatrix}
\dfrac 1 1 & 0 & \cdots & 0 \\
0 & \dfrac 1 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \dfrac 1 1 \\
\end{bmatrix}$
Hence the result.
{{qed}}
Category:Unit Matrices
Category:Inverse Matrices
\end{proof}
|
23119
|
\section{Unit Sphere as Surface of Revolution}
Tags: Induced Metric, Surfaces of Revolution, Solid Geometry
\begin{theorem}
{{WIP|Resolving consistency issues with previous results}}
Let $\struct {\R^3, d}$ be the Euclidean space.
Let $S_C \subseteq \R^3$ be the surface of revolution.
Let $C$ be a semi-circle defined by $x^2 + y^2 = 1$ in the open upper half-plane.
Let the smooth local parametrization of $C$ be:
:$\map \gamma \phi = \tuple {\cos \phi, \sin \phi}$
where $\phi \in \openint 0 \pi$.
Then the induced metric on $S_C$ is:
:$g = d \phi^2 + \sin^2 \phi \, d \theta^2$
\end{theorem}
\begin{proof}
By Smooth Local Parametrization of Surface of Revolution, the smooth local parametrization of $S_C$ can be written as:
:$\map X {\phi, \theta} = \tuple {\sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi}$
By Induced Metric on Surface of Revolution:
{{begin-eqn}}
{{eqn | l = g
| r = \paren {\paren {\cos' \phi}^2 + \paren {\sin' \phi}^2} d \phi^2 + \sin^2 \phi \, d \theta^2
}}
{{eqn | r = \paren {\sin^2 \phi + \cos^2 \phi} d \phi^2 + \sin^2 \phi \, d \theta^2
}}
{{eqn | r = d \phi^2 + \sin^2 \phi \, d \theta^2
}}
{{end-eqn}}
This is a metric of a unit sphere with the points on the $x$-axis removed.
{{qed}}
\end{proof}
|
23120
|
\section{Unit Sphere is Closed/Normed Vector Space}
Tags: Closed Sets (Normed Vector Spaces), Normed Vector Spaces, Closed Sets
\begin{theorem}
Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\Bbb S := \set {x \in X : \norm x = 1}$ be a unit sphere in $M$.
Then $\Bbb S$ is closed in $M$.
\end{theorem}
\begin{proof}
Let $\map {B_1} 0 = \set {x \in X : \norm x < 1}$ be an open ball.
Let $\map { {B_1}^-} 0 = \set {x \in X : \norm x \le 1}$ be a closed ball.
Then:
:$\ds X = \Bbb S \bigcup \relcomp X {\Bbb S}$
where
:$\ds \relcomp X {\Bbb S} = \map {B_1} 0 \bigcup \paren {X \setminus \map { {B_1}^-} 0}$
is the relative complement of $\Bbb S$ in $X$.
We have that Closed Ball is Closed in Normed Vector Space.
By definition, $X \setminus \map { {B_1}^-} 0$ is open.
Furthermore, Open Ball is Open Set in Normed Vector Space.
By Union of Open Sets of Normed Vector Space is Open, $\relcomp X {\Bbb S}$ is open.
By definition, $\Bbb S$ is closed.
{{qed}}
\end{proof}
|
23121
|
\section{Unit Vector in Direction of Vector}
Tags: Unit Vectors
\begin{theorem}
Let $\mathbf v$ be a vector quantity.
The '''unit vector''' $\mathbf {\hat v}$ in the direction of $\mathbf v$ is:
:$\mathbf {\hat v} = \dfrac {\mathbf v} {\norm {\mathbf v} }$
where $\norm {\mathbf v}$ is the magnitude of $\mathbf v$.
\end{theorem}
\begin{proof}
From Vector Quantity as Scalar Product of Unit Vector Quantity:
:$\mathbf v = \norm {\mathbf v} \mathbf {\hat v}$
whence the result.
{{qed}}
\end{proof}
|
23122
|
\section{Unit Vector in terms of Direction Cosines}
Tags: Direction Cosines, Unit Vectors
\begin{theorem}
Let $\mathbf r$ be a vector quantity embedded in a Cartesian $3$-space.
Let $\mathbf i$, $\mathbf j$ and $\mathbf k$ be the unit vectors in the positive directions of the $x$-axis, $y$-axis and $z$-axis respectively.
Let $\cos \alpha$, $\cos \beta$ and $\cos \gamma$ be the direction cosines of $\mathbf r$ with respect to the $x$-axis, $y$-axis and $z$-axis respectively.
Let $\mathbf {\hat r}$ denote the unit vector in the direction of $\mathbf r$.
Then:
:$\mathbf {\hat r} = \paren {\cos \alpha} \mathbf i + \paren {\cos \beta} \mathbf j + \paren {\cos \gamma} \mathbf k$
\end{theorem}
\begin{proof}
From Components of Vector in terms of Direction Cosines:
:$(1): \quad \mathbf r = r \cos \alpha \mathbf i + r \cos \beta \mathbf j + r \cos \gamma \mathbf k$
where $r$ denotes the magnitude of $\mathbf r$, that is:
:$r := \size {\mathbf r}$
By Unit Vector in Direction of Vector:
:$\mathbf {\hat r} = \dfrac {\mathbf r} r$
The result follows by multiplication of both sides of $(1)$ by $\dfrac 1 r$.
{{qed}}
\end{proof}
|
23123
|
\section{Unit Vectors in Complex Plane which are Vertices of Equilateral Triangle}
Tags: Equilateral Triangles, Geometry of Complex Numbers, Geometry of Complex Plane
\begin{theorem}
Let $\epsilon_1, \epsilon_2, \epsilon_3$ be complex numbers embedded in the complex plane such that:
:$\epsilon_1, \epsilon_2, \epsilon_3$ all have modulus $1$
:$\epsilon_1 + \epsilon_2 + \epsilon_3 = 0$
Then:
:$\paren {\dfrac {\epsilon_2} {\epsilon_1} }^3 = \paren {\dfrac {\epsilon_3} {\epsilon_2} }^2 = \paren {\dfrac {\epsilon_1} {\epsilon_3} }^2 = 1$
\end{theorem}
\begin{proof}
We have that:
{{begin-eqn}}
{{eqn | l = \epsilon_1 + \epsilon_2 + \epsilon_3
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = \epsilon_1 - \paren {-\epsilon_2}
| r = -\epsilon_3
| c =
}}
{{end-eqn}}
Thus by Geometrical Interpretation of Complex Subtraction, $\epsilon_1$, $\epsilon_2$ and $\epsilon_3$ form the sides of a triangle.
As the modulus of each of $\epsilon_1$, $\epsilon_2$ and $\epsilon_3$ equals $1$, $\triangle \epsilon_1 \epsilon_2 \epsilon_3$ is equilateral.
:420px
By Complex Multiplication as Geometrical Transformation:
{{begin-eqn}}
{{eqn | l = \arg \epsilon_1
| r = \arg \epsilon_2 \pm \dfrac {2 \pi} 3
| c = Complex Multiplication as Geometrical Transformation
}}
{{eqn | ll= \leadsto
| l = \arg \paren {\dfrac {\epsilon_2} {\epsilon_1} }
| r = \pm \dfrac {2 \pi} 3
| c =
}}
{{eqn | ll= \leadsto
| l = \dfrac {\epsilon_2} {\epsilon_1}
| r = \cos \dfrac {2 \pi} 3 \pm i \sin \dfrac {2 \pi} 3
| c =
}}
{{eqn | r = \omega \text { or } \overline \omega
| c = where $\omega$, $\overline \omega$ are the complex cube roots of unity
}}
{{eqn | ll= \leadsto
| l = \paren {\dfrac {\epsilon_2} {\epsilon_1} }^3
| r = 1
| c =
}}
{{end-eqn}}
The same analysis can be done to the other two pairs of sides of $\triangle \epsilon_1 \epsilon_2 \epsilon_3$.
Hence the result.
{{qed}}
\end{proof}
|
23124
|
\section{Unit n-Sphere under Euclidean Metric is Metric Subspace of Euclidean Real Vector Space}
Tags: Metric Subspaces
\begin{theorem}
Let $\Bbb S^n$ be the unit $n$-sphere.
Let $d_S: \Bbb S^n \times \Bbb S^n \to \R$ be the real-valued function defined as:
:$\ds \forall x, y \in \Bbb S^n: \map {d_S} {x, y} = \sqrt {\sum_{i \mathop = 1}^{n + 1} \paren {x_i - y_i}^2}$
where $x = \tuple {x_1, x_2, \ldots, x_{n + 1} }, y = \tuple {y_1, y_2, \ldots, y_{n + 1} }$.
Then $\struct {\Bbb S^n, d_S}$ is a metric subspace of $\struct {\R^{n + 1}, d}$, where $d$ is the Euclidean metric on the real vector space $\R^{n + 1}$.
\end{theorem}
\begin{proof}
The metric given is the Euclidean metric restricted to the subset $\Bbb S^n$ of the real vector space $\R^{n + 1}$.
The result follows from Subspace of Metric Space is Metric Space.
{{qed}}
\end{proof}
|
23125
|
\section{Unit of Ring is not Zero Divisor}
Tags: Units of Rings, Ring Theory, Rings, Zero Divisors, Rings with Unity
\begin{theorem}
Let $\struct {R, +, \circ}$ be a non-null ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $x$ be a unit of $\struct {R, +, \circ}$.
Then $x$ is neither a left zero divisor nor a right zero divisor of $\struct {R, +, \circ}$.
\end{theorem}
\begin{proof}
{{AimForCont}} $x$ is either a left zero divisor or a right zero divisor of $\struct {R, +, \circ}$.
{{WLOG}}, suppose $x$ is a left zero divisor of $\struct {R, +, \circ}$.
That is:
:$x \circ y = 0_R$
for some $y \in R \setminus \set {0_R}$.
Then:
{{begin-eqn}}
{{eqn | l = y
| r = 1_R \circ y
| c = {{Defof|Unity of Ring}}
}}
{{eqn | r = x^{-1} \circ x \circ y
| c = {{Defof|Unit of Ring}}
}}
{{eqn | r = x^{-1} \circ 0_R
| c = $x \circ y = 0_R$
}}
{{eqn | r = 0_R
| c = Ring Product with Zero
}}
{{end-eqn}}
This contradicts the deduction that $y \ne 0_R$.
Thus by Proof by Contradiction $x$ is neither a left zero divisor nor a right zero divisor of $\struct {R, +, \circ}$.
{{qed}}
Category:Rings with Unity
Category:Units of Rings
Category:Zero Divisors
\end{proof}
|
23126
|
\section{Unit of Ring of Mappings iff Image is Subset of Ring Units}
Tags: Abstract Algebra, Rings of Mappings, Unit of Ring of Mappings iff Image is Subset of Ring Units, Ring Theory, Mapping Theory, Rings with Unity
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring with unity $1_R$.
Let $U_R$ be the set of units in $R$.
Let $S$ be a set.
Let $\struct {R^S, +', \circ'}$ be the ring of mappings on the set of mappings $R^S$.
Then:
:$f \in R^S$ is a unit of $R^S$ {{iff}} $\Img f \subseteq U_R$
where $\Img f$ is the image of $f$.
In this case, the inverse of $f$ is the mapping $f^{-1} : S \to U_R$ defined by:
:$\forall x \in S : \map {f^{-1} } x = \map f x^{-1}$
\end{theorem}
\begin{proof}
From Structure Induced by Ring with Unity Operations is Ring with Unity, $\struct {R^S, +', \circ'}$ has a unity $f_{1_R}$ defined by:
:$\forall x \in S: \map {f_{1_R}} x = 1_R$
\end{proof}
|
23127
|
\section{Unit of Ring of Mappings iff Image is Subset of Ring Units/Image is Subset of Ring Units implies Unit of Ring of Mappings}
Tags: Unit of Ring of Mappings iff Image is Subset of Ring Units, Mapping Theory, Ring Theory, Abstract Algebra
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring with unity $1_R$.
Let $U_R$ be the set of units in $R$.
Let $S$ be a set.
Let $\struct {R^S, +', \circ'}$ be the ring of mappings on the set of mappings $R^S$.
Let $\Img f \subseteq U_R$ where $\Img f$ is the image of $f$.
Then:
:$f \in R^S$ is a unit of $R^S$
and the inverse of $f$ is the mapping $f^{-1} : S \to U_R$ defined by:
:$\forall x \in S : \map {f^{-1}} {x} = \map f x^{-1}$
\end{theorem}
\begin{proof}
From Structure Induced by Ring with Unity Operations is Ring with Unity, $\struct {R^S, +', \circ'}$ has a unity $f_{1_R}$ defined by:
:$\forall x \in S: \map {f_{1_R}} x = 1_R$
By assumption:
:$\forall x \in S: \exists \map f x^{-1} : \map f x \circ \map f x^{-1} = \map f x^{-1} \circ \map f x = 1_R$
Let $f^{-1} : S \to U_R$ be defined by:
:$\forall x \in S : \map {f^{-1}} {x} = \map f x^{-1}$
Consider the mapping $f \circ’ f^{-1}$.
For all $x \in S$:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \circ’ f^{-1} } } x
| r = \map f x \circ’ \map {f^{-1} } x
}}
{{eqn | r = \map f x \circ \map f x^{-1}
}}
{{eqn | r = 1_R
}}
{{end-eqn}}
Hence $f \circ’ f^{-1} = f_{1_R}$.
Similarly, $f^{-1} \circ’ f = f_{1_R}$.
Hence $f$ is a unit of $\struct {R^S, +', \circ'}$ and the inverse of $f$ is the mapping $f^{-1}$.
{{qed}}
Category:Unit of Ring of Mappings iff Image is Subset of Ring Units
\end{proof}
|
23128
|
\section{Unit of Ring of Mappings iff Image is Subset of Ring Units/Unit of Ring of Mappings implies Image is Subset of Ring Units}
Tags: Unit of Ring of Mappings iff Image is Subset of Ring Units, Mapping Theory, Ring Theory, Abstract Algebra
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring with unity $1_R$.
Let $U_R$ be the set of units in $R$.
Let $S$ be a set.
Let $\struct {R^S, +', \circ'}$ be the ring of mappings on the set of mappings $R^S$.
Let $f \in R^S$ be a unit of $R^S$.
Then:
:$\Img f \subseteq U_R$
where $\Img f$ is the image of $f$.
In which case, the inverse of $f$ is the mapping $f^{-1} : S \to U_R$ defined by:
:$\forall x \in S : \map {f^{-1} } x = \map f x^{-1}$
\end{theorem}
\begin{proof}
From Structure Induced by Ring with Unity Operations is Ring with Unity, $\struct {R^S, +', \circ'}$ has a unity $f_{1_R}$ defined by:
:$\forall x \in S: \map {f_{1_R} } x = 1_R$
Let $f^{-1}$ be the product inverse of $f$.
Let $x \in R$.
Then:
{{begin-eqn}}
{{eqn | l = 1_R
| r = \map {f_{1_R} } x
| c = Structure Induced by Ring with Unity Operations is Ring with Unity
}}
{{eqn | r = \map {\paren {f \circ' f^{-1} } } x
| c = {{Defof|Product Inverse|product inverse}}
}}
{{eqn | r = \map f x \circ \map {f^{-1} } x
| c = {{Defof|Pointwise Operation}} $\circ'$
}}
{{end-eqn}}
Similarly, $1_R = \map {f^{-1} } x \circ \map f x$
Hence $\map {f^{-1}} x$ is the product inverse of $\map f x$.
So $\map f x$ is a unit in $R$ and $\map f x^{-1} = \map {f^{-1} } x$.
The result follows.
{{qed}}
Category:Unit of Ring of Mappings iff Image is Subset of Ring Units
\end{proof}
|
23129
|
\section{Unit of System of Sets is Unique}
Tags: Set Systems
\begin{theorem}
The unit of a system of sets, if it exists, is unique.
If $U$ is the unit of a system of sets $\SS$, then $\forall A \in \SS: A \subseteq U$.
\end{theorem}
\begin{proof}
Let $\SS$ be a system of sets.
Suppose $U$ and $U'$ are both units of $\SS$.
Then, by definition:
:$\forall A \in \SS: A \cap U = A$
:$\forall A \in \SS: A \cap U' = A$
This applies to both $U$ and $U'$, of course.
So $U \cap U' = U$ and $U' \cap U = U'$.
From Intersection with Subset is Subset it follows that $U \subseteq U'$ and $U' \subseteq U$.
By definition of set equality:
:$U = U'$
We also see that from Intersection with Subset is Subset, $A \cap U = A \iff A \subseteq U$, which shows that:
:$\forall A \in \SS: A \subseteq U$.
{{qed}}
Category:Set Systems
\end{proof}
|
23130
|
\section{Unital Ring Homomorphism by Idempotent}
Tags: Ring Theory, Ring Homomorphisms, Commutative Rings
\begin{theorem}
Let $A$ be a commutative ring with unity.
Let $e \in A$ be an idempotent element.
Let $\ideal e$ be the ideal of $A$ generated by $e$.
Then the mapping:
:$f: A \to \ideal e: a \mapsto e a$
is a surjective unital ring homomorphism from $\struct {A, +, \circ}$ to $\struct {\ideal e, +, \circ}$ with kernel the ideal $\ideal {1 - e}$ generated by $1 - e$.
\end{theorem}
\begin{proof}
By Ring Homomorphism by Idempotent $f$ is a surjective ring homomorphism with kernel $\ideal {1 - e}$.
By Ring by Idempotent $A$ is a commutative ring with unity whose unity is $e$.
Then
{{begin-eqn}}
{{eqn | l = \map f 1
| r = e \cdot 1
| c = Definition of $f$
}}
{{eqn | r = e
| c = $1$ is the unity of $A$.
}}
{{end-eqn}}
Thus $f$ is a unital ring homomorphism.
{{qed}}
Category:Ring Theory
Category:Commutative Rings
Category:Ring Homomorphisms
\end{proof}
|
23131
|
\section{Unitary Module of All Mappings is Unitary Module}
Tags: Examples of Unitary Modules, Module Theory
\begin{theorem}
Let $\struct {R, +_R, \times_R}$ be a ring with unity whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Let $S$ be a set.
Let $\struct {G^S, +_G', \circ}_R$ be the module of all mappings from $S$ to $G$.
Then $\struct {G^S, +_G', \circ}_R$ is a unitary module.
\end{theorem}
\begin{proof}
From Module of All Mappings is Module, we have that $\struct {G^S, +_G', \circ}_R$ is an $R$-module.
To show that $\struct {G^S, +_G', \circ}_R$ is a unitary $R$-module, we verify the following:
:$\forall f \in G^S: 1_R \circ f = f$
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Then:
:$\forall x \in G: 1_R \circ x = x$
Thus:
{{begin-eqn}}
{{eqn | l = \map {\paren {1_R \circ f} } x
| r = 1_R \circ \paren {\map f x}
| c =
}}
{{eqn | r = \map f x
| c =
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
23132
|
\section{Unitary R-Modules with n-Element Bases are Isomorphic}
Tags: Unitary Modules, Modules
\begin{theorem}
Let $G$ and $H$ be unitary $R$-modules for some ring with unity $R$.
Let $G$ and $G$ both have bases with $n$ elements.
Then $G$ and $H$ are isomorphic.
\end{theorem}
\begin{proof}
From Isomorphism from $R^n$ via $n$-Term Sequence, both $G$ and $H$ are isomorphic to the $R$-module $R^n$.
{{qed}}
\end{proof}
|
23133
|
\section{Units of 5th Cyclotomic Ring}
Tags: Cyclotomic Rings
\begin{theorem}
Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
The units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are $1$ and $-1$.
\end{theorem}
\begin{proof}
Let $\map N z$ denote the field norm of $z \in \Z \sqbrk {i \sqrt 5}$.
Let $z_1 \in \Z \sqbrk {i \sqrt 5}$ be a unit of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
Thus by definition:
:$\exists z_2 \in \Z \sqbrk {i \sqrt 5}: z_1 \times z_2 = 1$
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.
Then:
{{begin-eqn}}
{{eqn | l = 1
| r = \map N 1
| c = Field Norm on 5th Cyclotomic Ring
}}
{{eqn | r = \map N {z_1 \times z_2}
| c =
}}
{{eqn | r = \map N {z_1} \times \map N {z_2}
| c =
}}
{{eqn | r = \paren { {x_1}^2 + 5 {y_1}^2} \paren { {x_2}^2 + 5 {y_2}^2}
| c = Field Norm on 5th Cyclotomic Ring
}}
{{eqn | ll= \leadsto
| l = {x_1}^2
| r = 1
| c =
}}
{{eqn | lo= \land
| l = {x_2}^2
| r = 1
| c =
}}
{{eqn | lo= \land
| l = y_1
| r = y_2 = 0
| c =
}}
{{end-eqn}}
The result follows.
{{qed}}
\end{proof}
|
23134
|
\section{Units of Field of Complex Numbers form Group}
Tags: Groups: Examples, Group Theory, Examples of Groups, Group Examples, Complex Numbers, Groups
\begin{theorem}
:$\C^\times = \C \setminus \set 0$
where $\C^\times$ denotes the group of units of $\C$.
\end{theorem}
\begin{proof}
By Complex Numbers form Field, $\C$ is a field.
From Group of Units of Field it follows that:
: $\C^\times = \C \setminus \set 0$
{{qed}}
Category:Examples of Groups
Category:Complex Numbers
\end{proof}
|
23135
|
\section{Units of Gaussian Integers}
Tags: Integral Domains, Complex Numbers, Units of Gaussian Integers, Gaussian Integers
\begin{theorem}
Let $\struct {\Z \sqbrk i, +, \times}$ be the ring of Gaussian integers.
The set of units of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$.
\end{theorem}
\begin{proof}
Let $a + bi$ be a unit of $\left({\Z \left[{i}\right], +, \times}\right)$.
Then $a$ and $b$ are not both $0$ as then $a + bi$ would be the zero of $\left({\Z \left[{i}\right], +, \times}\right)$.
Then:
:$\exists c, d \in \Z: \left({a + bi}\right) \left({c + di}\right) = 1 + 0i$
This leads (after algebra) to:
:$c = \dfrac a {a^2 + b^2}, d = \dfrac {-b} {a^2 + b^2}$
Let $a^2 + b^2 = n$.
We have that $n \in \Z, n > 0$.
If $c$ and $d$ are integers, then $a$ and $b$ must both be divisible by $n$.
Let $a = nx, b = n y$.
Then:
:$n^2 x^2 + n^2 y^2 = n$
and so:
:$n \left({x^2 + y^2}\right) = 1$
Thus $n = a^2 + b^2 = 1$ and so as $a, b \in \Z$ we have:
:$a^2 = 1, b^2 = 0$
or:
:$a^2 = 0, b^2 = 1$
from which the result follows.
{{qed}}
\end{proof}
|
23136
|
\section{Units of Gaussian Integers form Group}
Tags: Complex Roots of Unity, Roots of Unity, Gaussian Integers, Group of Gaussian Integer Units, Group of Gaussian Integers, Group Examples, Units of Gaussian Integers form Group
\begin{theorem}
Let $U_\C$ be the set of units of the Gaussian integers:
:$U_\C = \set {1, i, -1, -i}$
where $i$ is the imaginary unit: $i = \sqrt {-1}$.
Let $\struct {U_\C, \times}$ be the algebraic structure formed by $U_\C$ under the operation of complex multiplication.
Then $\struct {U_\C, \times}$ forms a cyclic group under complex multiplication.
\end{theorem}
\begin{proof}
From Units of Gaussian Integers, $U_\C$ is the set of units of the ring of Gaussian integers.
From Group of Units is Group, $\left({U_\C, \times}\right)$ forms a group.
It remains to note that:
{{begin-eqn}}
{{eqn | l = i^2
| r = -1
}}
{{eqn | l = i^3
| r = -i
}}
{{eqn | l = i^4
| r = 1
}}
{{end-eqn}}
thus demonstrating that $U_\C$ is generated by $i$.
Hence the result, by definition of cyclic group.
{{qed}}
\end{proof}
|
23137
|
\section{Units of Quadratic Integers over 2}
Tags: Quadratic Integers, Examples of Integral Domains, Units of Rings, Numbers of Type Integer a plus b root n, Integral Domains
\begin{theorem}
Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Let $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$ be the integral domain where $+$ and $\times$ are conventional addition and multiplication on real numbers.
Then numbers of the form $a + b \sqrt 2$ such that $a^2 - 2 b^2 = \pm 1$ are all units of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$.
\end{theorem}
\begin{proof}
For $a + b \sqrt 2$ to be a unit of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$, we require that:
:$\exists c, d \in \Z: \paren {a + b \sqrt 2} \paren {c + d \sqrt 2} = 1$
In Quadratic Integers over 2 are Not a Field it is shown that the product inverse of $\paren {a + b \sqrt 2}$ is $\dfrac a {a^2 - 2 b^2} + \dfrac {b \sqrt 2} {a^2 - 2 b^2}$.
So if $a^2 - 2 b^2 = \pm 1$ it follows that $c$ and $d$ are integers.
Hence the result.
{{qed}}
\end{proof}
|
23138
|
\section{Units of Ring of Arithmetic Functions}
Tags: Arithmetic Functions, Analytic Number Theory
\begin{theorem}
Let $f$ be an arithmetic function.
Then $f$ is a unit in the ring of arithmetic functions {{iff}}:
:$\map f 1 \ne 0$
\end{theorem}
\begin{proof}
Follows immediately from Invertibility of Arithmetic Functions and the definition of ring of arithmetic functions.
{{qed}}
Category:Arithmetic Functions
\end{proof}
|
23139
|
\section{Units of Ring of Polynomial Forms over Field}
Tags: Units of Rings, Polynomial Theory
\begin{theorem}
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $F \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $F$.
Then the units of $F \sqbrk X$ are all the elements of $F \sqbrk X$ whose degree is $0$.
\end{theorem}
\begin{proof}
An element of $F \sqbrk X$ whose degree is $0$ is merely an element of $F$.
But note that $0_F$, considered as an element of $F \sqbrk X$, has a degree which is not defined, so the null polynomial is seen to be excluded.
Any element $a$ of $F$ has an inverse $1_F / a$.
So all the elements of $F \sqbrk X$ whose degree is $0$ are units of $F$ and hence of $F \sqbrk X$.
Now suppose $\map a X \in F \sqbrk X$ is a unit of $F \sqbrk X$.
Then $\map a X \, \map q X = 1_F$ for some $\map q X \in F \sqbrk X$.
Neither $\map a X$ nor $\map q X$ can be null.
So by Properties of Degree $\map \deg {\map a X} + \map \deg {\map q X} = 0$.
So $\map \deg {\map a X} = 0$.
{{qed}}
\end{proof}
|
23140
|
\section{Units of Ring of Polynomial Forms over Integral Domain}
Tags: Polynomial Theory
\begin{theorem}
Let $\struct {D, +, \circ}$ be an integral domain.
Let $D \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $D$.
Then the group of units of $D \sqbrk X$ is precisely the group of elements of $D \sqbrk X$ of degree zero that are units of $D$.
\end{theorem}
\begin{proof}
It is immediate that a unit of $D$ is also a unit of $D \sqbrk X$.
Let $P$ be a unit of $D \sqbrk X$.
Then there exists $Q \in D \sqbrk X$ such that $P Q = 1$.
By Corollary 2 to Degree of Product of Polynomials over Ring we have:
:$0 = \map \deg 1 = \map \deg P + \map \deg Q$
Therefore:
:$\map \deg P = \map \deg Q = 0$
That is, $P \in R$ and $Q \in R$.
Moreover $P Q = 1$ in $R$, so it follows that $P$ is a unit of $R$.
{{Qed}}
Category:Polynomial Theory
\end{proof}
|
23141
|
\section{Unity Function is Completely Multiplicative}
Tags: Multiplicative Functions, Number Theory, Completely Multiplicative Functions
\begin{theorem}
Let $f_1: \Z_{> 0} \to \Z_{> 0}$ be the constant function:
:$\forall n \in \Z_{> 0}: f_1 \left({n}\right) = 1$
Then $f_1$ is completely multiplicative.
\end{theorem}
\begin{proof}
:$\forall m, n \in \Z_{> 0}: f_1 \left({m n}\right) = 1 = f_1 \left({m}\right) f_1 \left({n}\right)$
{{qed}}
Category:Completely Multiplicative Functions
\end{proof}
|
23142
|
\section{Unity and Negative form Subgroup of Units}
Tags: Rings, Subgroups, Rings with Unity
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring with unity.
Then:
:$\struct {\set {1_R, -1_R}, \circ} \le U_R$
That is, the set consisting of the unity and its negative forms a subgroup of the group of units.
\end{theorem}
\begin{proof}
From Unity is Unit:
:$1_R \in U_R$
It remains to be shown that $-1_R \in U_R$.
From {{Ring-axiom|A}}, $\struct {R, +}$ is an abelian group.
Therefore:
:$1_R \in R \implies -1_R \in R$.
From Product of Ring Negatives:
:$-1_R \circ -1_R = 1_R \circ 1_R = 1_R$
Thus $-1_R$ has a ring product inverse (itself) and therefore $-1_R \in U_R$.
We have:
{{begin-eqn}}
{{eqn | l = 1_R \circ 1_R
| r = 1_R
| c =
}}
{{eqn | l = 1_R \circ -1_R
| r = -1_R
| c =
}}
{{eqn | l = -1_R \circ 1_R
| r = -1_R
| c =
}}
{{eqn | l = -1_R \circ -1_R
| r = 1_R
| c =
}}
{{end-eqn}}
This exhausts all the ways we can form a ring product between $1_R$ and $-1_R$.
That is:
:$\forall x, y \in \set {1_R, -1_R}: x \circ y \in \set {1_R, -1_R}$
Hence the result from the Two-Step Subgroup Test.
{{qed}}
Category:Rings with Unity
Category:Subgroups
\end{proof}
|
23143
|
\section{Unity is Unit}
Tags: Rings, Rings with Unity
\begin{theorem}
The unity in a ring is a unit.
\end{theorem}
\begin{proof}
Let $\left({R, +, \circ}\right)$ be a ring with unity $1_R$.
From Identity is Self-Inverse:
:$1_R^{-1} = 1_R \in R \implies 1_R \in U_R$
{{qed}}
Category:Rings with Unity
\end{proof}
|
23144
|
\section{Unity is Unity in Ring of Idempotents}
Tags: Ring Theory
\begin{theorem}
Let $\left({R, +, \circ}\right)$ be a commutative and unitary ring whose unity is $1_R$.
Let $\left({A, \oplus, \circ}\right)$ be the ring of idempotents of $R$.
Then $1_R$ is also a unity for $\left({A, \oplus, \circ}\right)$.
\end{theorem}
\begin{proof}
From Unity of Ring is Idempotent, $1_R$ is an idempotent element of $R$.
Hence $1_R \in A$.
Recall that the ring product of $A$ is a restriction from that of $R$.
Hence, for each $x \in A$:
:$x \circ 1_R = x = 1_R \circ x$
so that $1_R$ is a unity for $A$, as desired.
{{qed}}
Category:Ring Theory
\end{proof}
|
23145
|
\section{Unity of Integral Domain is Unique}
Tags: Integral Domains
\begin{theorem}
Let $\struct {D, +, \times}$ be an integral domain.
Then the unity of $\struct {D, +, \times}$ is unique.
\end{theorem}
\begin{proof}
From the definition of an integral domain, $\struct {D, +, \times}$ is a commutative ring such that $\struct {D^*, \circ}$ is a monoid.
The result follows from Identity of Monoid is Unique.
{{qed}}
\end{proof}
|
23146
|
\section{Unity of Ordered Integral Domain is Strictly Positive}
Tags: Integral Domains, Ordered Integral Domains, Orderings
\begin{theorem}
Let $\struct {D, +, \times \le}$ be an ordered integral domain whose unity is $1_D$.
Then:
:$\map P {1_D}$
where $P$ is the (strict) positivity property.
\end{theorem}
\begin{proof}
We have by definition of the unity that:
:$\forall a \in D: 1_D \times a = a = a \times 1_D$
This particularly applies to $1_D$ itself:
:$1_D = 1_D \times 1_D$
But then by Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive:
:$\map P {1_D \times 1_D} \implies \map P {1_D}$
{{qed}}
\end{proof}
|
23147
|
\section{Unity of Ring is Idempotent}
Tags: Ring Theory
\begin{theorem}
Let $\left({R, +, \circ}\right)$ be a ring with unity whose unity is $1_R$.
Then $1_R$ is an idempotent element of $R$ under the ring product $\circ$:
:$1_R \circ 1_R = 1_R$
\end{theorem}
\begin{proof}
By definition of ring with unity, $\left({R, \circ}\right)$ is a monoid whose identity element is $1_R$.
From Identity Element is Idempotent (applied to $1_R$):
:$1_R \circ 1_R = 1_R$
which was to be proven.
{{qed}}
Category:Ring Theory
\end{proof}
|
23148
|
\section{Unity of Ring is Unique}
Tags: Rings, Rings with Unity
\begin{theorem}
A ring can have no more than one unity.
\end{theorem}
\begin{proof}
Let $\struct {R, +, \circ}$ be a ring.
If $\struct {R, \circ}$ has an identity, then it is a monoid.
From Identity of Monoid is Unique, it follows that such an identity is unique.
{{qed}}
\end{proof}
|
23149
|
\section{Unity of Subfield is Unity of Field}
Tags: Subfields
\begin{theorem}
Let $\struct {F, +, \times}$ be a field whose unity is $1$.
Let $\struct {K, +, \times}$ be a subfield of $F$.
The unity of $\struct {K, +, \times}$ is also $1$.
\end{theorem}
\begin{proof}
By definition, $\struct {K, +, \times}$ is a subset of $F$ which is a field.
By definition of field, $\struct {K^*, \times}$ and $\struct {F^*, \times}$ are groups such that $K \subseteq F$.
So $\struct {K^*, \times}$ is a subgroup of $\struct {F^*, \times}$.
By Identity of Subgroup, the identity of $\struct {F^*, \times}$, which is $1$, is also the identity of $\struct {K^*, \times}$.
{{qed}}
\end{proof}
|
23150
|
\section{Unity of Subring is not necessarily Unity of Ring}
Tags: Subrings
\begin{theorem}
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 $1_T = 1_S$.
\end{theorem}
\begin{proof}
Let $\struct {S, + \times}$ be the ring formed by the set of order $2$ square matrices over the real numbers $R$ under (conventional) matrix addition and (conventional) matrix multiplication.
Let $T$ be the subset of $S$ consisting of the matrices of the form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ for $x \in \R$.
We have that:
:$\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} -y & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} x - y & 0 \\ 0 & 0 \end{bmatrix}$
and so $T$ is closed under subtraction.
From Matrices of the Form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ we have that $\struct {T, \times}$ is a subsemigroup of $\struct {S, \times}$.
It follows from the Subring Test that $\struct {T, + \circ}$ be a subbring of $\struct {S, + \circ}$.
From (some result somewhere) $\struct {S, \times}$ has an identity $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ which is not in $\struct {T, \times}$.
However, note that:
{{begin-eqn}}
{{eqn | l = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}
| r = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}
| c =
}}
{{eqn | r = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}
| c =
}}
{{end-eqn}}
demonstrating that $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ is the identity of $\struct {T, \times}$.
The result follows.
{{qed}}
\end{proof}
|
23151
|
\section{Unity plus Negative of Nilpotent Ring Element is Unit}
Tags: Nilpotent Ring Elements
\begin{theorem}
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 $R$.
\end{theorem}
\begin{proof}
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^{n - 3} \circ b^2 + \dotsb + a \circ b^{n - 2} + b^{n - 1} }
| c =
}}
{{end-eqn}}
for $a, b \in R$.
Putting $a = 1_R$ and $b = x$, we have:
{{begin-eqn}}
{{eqn | l = \paren {1_R}^n - x^n
| r = \paren {1_R - x} \circ \paren {1_R^{n - 1} + 1_R^{1_R - 2} \circ x + 1_R^{n - 3} \circ x^2 + \dotsb + 1_R \circ x^{n - 2} + x^{n - 1} }
| c =
}}
{{eqn | ll= \leadsto
| l = 1_R - 0_R
| r = \paren {1_R - x} \circ \paren {1_R + 1_R \circ x + 1_R \circ x^2 + \dotsb + 1_R \circ x^{n - 2} + x^{n - 1} }
| c = as $x^n = 0_R$, $1_R^k = 1_R$
}}
{{eqn | ll= \leadsto
| l = 1_R
| r = \paren {1_R - x} \circ \paren {1_R + x + x^2 + \dotsb + x^{n - 2} + x^{n - 1} }
| c =
}}
{{end-eqn}}
Thus by definition $1_R - x$ has a product inverse $1_R + x + x^2 + \dotsb + x^{n - 2} + x^{n - 1}$.
Hence by definition $1_R - x$ is a unit of $R$.
{{qed}}
\end{proof}
|
23152
|
\section{Universal Affirmative and Particular Negative are Contradictory}
Tags: Predicate Logic, Categorical Statements
\begin{theorem}
Consider the categorical statements:
:$\mathbf A: \quad$ The universal affirmative: $\forall x: \map S x \implies \map P x$
:$\mathbf O: \quad$ The particular negative: $\exists x: \map S x \land \neg \map P x$
Then $\mathbf A$ and $\mathbf O$ are contradictory.
Using the symbology of predicate logic:
:$\neg \paren {\paren {\forall x: \map S x \implies \map P x} \iff \paren {\exists x: \map S x \land \neg \map P x} }$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | o =
| r = \mathbf A
| c =
}}
{{eqn | ll= \leadsto
| q = \forall x
| o =
| r = \map S x \implies \map P x
| c = Definition of $\mathbf A$
}}
{{eqn | ll= \leadsto
| q = \forall x
| o =
| r = \neg \paren {\map S x \land \neg \map P x}
| c = Implication Equivalent to Negation of Conjunction with Negative
}}
{{eqn | ll= \leadsto
| q = \neg \exists x
| o =
| r = \map S x \land \neg \map P x
| c = De Morgan's Laws: Denial of Existence
}}
{{eqn | ll= \leadsto
| o =
| r = \neg \mathbf O
| c = Definition of $\mathbf O$
}}
{{end-eqn}}
The argument reverses:
{{begin-eqn}}
{{eqn | o =
| r = \mathbf O
| c =
}}
{{eqn | ll= \leadsto
| q = \exists x
| o =
| r = \map S x \land \neg \map P x
| c = Definition of $\mathbf O$
}}
{{eqn | ll= \leadsto
| q = \exists x
| o =
| r = \neg \paren {\map S x \implies \map P x}
| c = Conjunction with Negative Equivalent to Negation of Implication
}}
{{eqn | ll= \leadsto
| q = \neg \forall x
| o =
| r = \map S x \implies \map P x
| c = De Morgan's Laws: Denial of Universality
}}
{{eqn | ll= \leadsto
| o =
| r = \neg \mathbf A
| c = Definition of $\mathbf A$
}}
{{end-eqn}}
The result follows by definition of contradictory.
{{qed}}
\end{proof}
|
23153
|
\section{Universal Class is Proper}
Tags: Universal Class is Proper, Class Theory, Universal Class
\begin{theorem}
Let $V$ denote the universal class.
Then $V$ is a proper class.
\end{theorem}
\begin{proof}
Assume that $U$ is small.
Note that $\operatorname{Ru} \subseteq U$ where $\operatorname{Ru}$ denotes the Russell class.
By Axiom of Subsets Equivalents, $\operatorname{Ru}$ is also small.
This contradicts Russell's Paradox.
{{qed}}
\end{proof}
|
23154
|
\section{Universal Class less Set is not Transitive}
Tags: Examples of Transitive Classes
\begin{theorem}
Let $V$ be a basic universe.
Let $a \in V$ be a set.
Then:
:$V \setminus \set a$ is not a transitive class
where $\setminus$ denotes class difference.
\end{theorem}
\begin{proof}
By definition, $V$ is the class of all sets.
As $a \in V$, by definition of $V$ it follows that $a$ is a set.
Consider the power set $\powerset a$ of $a$.
From the axiom of powers:
:$\powerset a$ is also a set
and:
:$\powerset {\powerset a}$ is also a set.
By definition:
:$a \in \powerset a$
and so:
:$\set a \in \powerset {\powerset a}$
as $\powerset {\powerset a}$ is a set, it follows that:
:$\powerset {\powerset a} \in V$
But we have by definition of class difference that $\set a \notin V$.
Hence we have an element of an element of $V$ which is not itself in $V$.
Hence, by definition, $V \setminus \set a$ is not a transitive class.
{{qed}}
\end{proof}
|
23155
|
\section{Universal Closures are Semantically Equivalent}
Tags: Predicate Logic
\begin{theorem}
Let $\mathbf A$ be a WFF of predicate logic.
Let $\mathbf B, \mathbf B'$ be universal closures of $\mathbf A$.
Then $\mathbf B$ and $\mathbf B'$ are semantically equivalent.
\end{theorem}
\begin{proof}
Let $\AA$ be a structure for predicate logic.
Let $\mathbf B$ be any universal closure of $\mathbf A$.
Then $\mathbf B$ is a sentence of the form:
:$\forall x_1: \cdots \forall x_n: \mathbf A$
By definition of the models relation:
:$\AA \models_{\mathrm{PL}} \mathbf B$ {{iff}} $\map {\operatorname{val}_\AA} {\mathbf B} = \T$
Hence, recursively applying the definition of $\map {\operatorname{val}_\AA} \cdot \sqbrk \sigma$, we see:
:$\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \O = \T$ {{iff}} $\forall a_1, \ldots, a_n \in A: \mathop{\map {\operatorname{val}_\AA} {\mathbf A} } \sqbrk {\dfrac{x_1} {a_1} + \ldots + \dfrac{x_n} {a_n} } = \T$
where $\dfrac{x_1} {a_1} + \ldots + \dfrac{x_n} {a_n}$ denotes the iterated extension of an assignment.
By Value of Formula under Assignment Determined by Free Variables:
$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\dfrac{x_1} {a_1} + \ldots + \dfrac{x_n} {a_n} }$ only depends on the $a_i$ for the free variables $x_i$ in $\mathbf A$.
Because we check all possible $a_i \in A$ and all free variables $x_i$ in $\mathbf A$ are quantified over in $\mathbf B$, it follows that:
:$\forall a_1, \ldots, a_n \in A: \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\dfrac{x_1} {a_1} + \ldots + \dfrac{x_n} {a_n} } = \T$
{{iff}}:
:$\forall a_1, \ldots, a_k \in A: \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\dfrac{x_1} {a_1} + \ldots + \dfrac{x_k} {a_k} } = \T$
where $x_1, \ldots, x_k$ are the free variables of $\mathbf A$.
But this last condition does not depend on $\mathbf B$ beyond that it be a universal closure.
Hence, for any two universal closures $\mathbf B, \mathbf B'$ of $\mathbf A$:
:$\AA \models_{\mathrm{PL}} \mathbf B$ {{iff}} $\AA \models_{\mathrm{PL}} \mathbf B$
The result follows by definition of semantic equivalence.
{{qed}}
\end{proof}
|
23156
|
\section{Universal Generalisation}
Tags: Predicate Logic, Logic
\begin{theorem}
Let $\mathbf a$ be any arbitrarily selected object in the universe of discourse.
Then:
{{begin-eqn}}
{{eqn | l = \map P {\mathbf a}
| o =
}}
{{eqn | ll= \vdash
| q = \forall x
| l = \map P x
| o =
}}
{{end-eqn}}
In natural language:
:''Suppose $P$ is true of any arbitrarily selected $\mathbf a$'' in the universe of discourse.''
:''Then $P$ is true of everything in the universe of discourse.''
\end{theorem}
\begin{proof}
We can express $\forall x$ using its propositional expansion:
:$\map P {\mathbf X_1} \land \map P {\mathbf X_2} \land \map P {\mathbf X_3} \land \ldots$
where $\mathbf X_1, \mathbf X_2, \mathbf X_3 \ldots{}$ is the complete set of the objects in the universe of discourse.
The fact that any object we care to choose has the property in question means that they ''all'' must have this property.
The result then follows by generalising the Rule of Conjunction.
{{qed}}
\end{proof}
|
23157
|
\section{Universal Instantiation/Informal Statement}
Tags: Predicate Logic
\begin{theorem}
Suppose we have a universal statement:
:$\forall x: \map P x$
where $\forall$ is the universal quantifier and $\map P x$ is a propositional function.
Then we can deduce:
:$\map P {\mathbf a}$
where $\mathbf a$ is any arbitrary object we care to choose in the universe of discourse.
In natural language:
:''Suppose $P$ is true of everything in the universe of discourse.''
:''Let $\mathbf a$ be an element of the universe of discourse."
:''Then $P$ is true of $\mathbf a$.''
\end{theorem}
\begin{proof}
In the language of symbolic logic:
{{begin-eqn}}
{{eqn | q = \forall x
| l = \map P x
| o =
}}
{{eqn | ll= \therefore
| l = \map P {\mathbf a}
| o =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23158
|
\section{Universal Instantiation/Model}
Tags: Predicate Logic
\begin{theorem}
Let $\map {\mathbf A} x$ be a WFF of predicate logic.
Let $\tau$ be a term which is freely substitutable for $x$ in $\mathbf A$.
Then $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$ is a tautology.
\end{theorem}
\begin{proof}
Let $\AA$ be a structure on a set $A$, and let $\sigma$ be an assignment for $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$.
Define:
:$a_\tau := \mathop {\map {\operatorname{val}_\AA} \tau} \sqbrk \sigma$
the value of $\tau$ under $\sigma$.
From the definition of value under $\sigma$:
:$\map {\mathrm {val}_\AA} {\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau} \sqbrk \sigma = \map {f^\to} {\map {\mathrm {val}_\AA} {\forall x: \map {\mathbf A} x} \sqbrk \sigma, \map {\mathrm {val}_\AA} {\map {\mathbf A} \tau} \sqbrk \sigma}$
where $f^\to$ is the truth function of $\implies$.
We thus need to ascertain that if:
:$\map {\mathrm {val}_\AA} {\forall x: \map {\mathbf A} x} \sqbrk \sigma = T$
then also:
:$\map {\mathrm {val}_\AA} {\map {\mathbf A} \tau} \sqbrk \sigma = T$
By definition of value under $\sigma$, the former amounts to:
:$\map {\mathrm {val}_\AA} {\map {\mathbf A} x} \sqbrk {\sigma + \paren {x / a} } = T$
for all $a \in A$.
By the Substitution Theorem for Well-Formed Formulas:
:$\map {\mathrm {val}_\AA} {\map {\mathbf A} \tau} \sqbrk \sigma = \map {\mathrm {val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a_\tau} }$
Since $a_\tau \in A$, the conclusion follows, and $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$ is a tautology.
{{qed}}
\end{proof}
|
23159
|
\section{Universal Property for Field of Quotients}
Tags: Field Theory, Universal Properties, Integral Domains, Fields of Quotients, Quotient Fields
\begin{theorem}
Let $\struct {D, +, \circ}$ be an integral domain.
Let $\struct {F, \oplus, \cdot}$ be a field of quotients of $D$.
Then $F$ satisfies the following universal property:
There exists a (ring) homomorphism $\iota : D \to F$ such that:
::for every field $\tilde F$ and
:and:
::for every (ring) homomorphism $\phi: D \to \tilde F$
:there exists a unique field homomorphism $\psi: F \to \tilde F$ satisfying:
:::$\psi \iota = \phi$
That is, the following diagram commutes:
:Universal Property
Namely we may take:
:$\psi: a / b \mapsto \map \phi a \map \phi b^{-1}$
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Fields of Quotients
Category:Integral Domains
Category:Universal Properties
\end{proof}
|
23160
|
\section{Universal Property for Simple Field Extensions}
Tags: Universal Properties, Field Extensions
\begin{theorem}
Let $F / K$ be a field extension.
Let $\alpha \in F$ be an algebraic over $K$.
Let $\mu_\alpha$ be the minimal polynomial over $\alpha$ over $K$.
Let $\psi : K \left({\alpha}\right) \to L$ be a homomorphism.
Let $\phi = \psi \restriction_K$.
Let $\overline \phi: K \left[{X}\right] \to L \left[{X}\right]$ be the Induced Homomorphism of Polynomial Forms.
Then $\psi\left({\alpha}\right)$ is a root of $\overline \phi \left({\mu_\alpha}\right)$ in $L$.
Conversely let $L$ be a field.
Let $\phi: K \to L$ be a homomorphism.
Let $\beta$ be a root of $\overline \phi \left({\mu_\alpha}\right)$ in $L$.
Then there exists a unique field homomorphism $\psi : K \left({\alpha}\right) \to L$ extending $\phi$ that sends $\alpha$ to $\beta$.
\end{theorem}
\begin{proof}
Let $\psi: K \left({\alpha}\right) \to L$ be a homomorphism.
Let $\phi = \psi \big|_K$.
Let $\overline \phi: K \left[{X}\right] \to L \left[{X}\right]$ be the Induced Homomorphism of Polynomial Forms.
For any $f = a_0 + \dotsb + a_n X^n \in K \left[{X}\right]$ we have:
{{begin-eqn}}
{{eqn | l = \psi \left({f \left({\alpha}\right)}\right)
| r = \psi \left({a_0 + \cdots + a_n \alpha^n}\right)
| c =
}}
{{eqn | r = \phi \left({a_0}\right) + \cdots + \phi \left({a_n}\right) \psi \left({\alpha}\right)^n
| c = by the homomorphism property, and the fact that $\phi = \psi \big\vert_K$
}}
{{eqn | r = \overline \phi \left({f}\right) (\psi\left({\alpha}\right))
| c =
}}
{{end-eqn}}
Since $\mu_\alpha \left({\alpha}\right) = 0$ and a Ring Homomorphism Preserves Zero, the above yields:
:$\overline \phi \left({f}\right) \left({\psi \left({\alpha}\right)}\right) = 0$
as required.
{{qed|lemma}}
Conversely let $L$ be a field.
Let $\phi: K \to L$ be a homomorphism.
Let $\beta$ be a root of $\overline \phi \left({\mu_\alpha}\right)$ in $L$.
By Structure of Simple Algebraic Field Extension, there exists a unique isomorphism:
:$\Delta: K \left[{\alpha}\right] \to K \left[{X}\right] / \langle \mu_\alpha \rangle$
such that:
:$\Delta \left({\alpha}\right) = X + \left\langle{\mu_\alpha}\right\rangle$
and:
:$\Delta \big|_K$ is the identity mapping.
By Universal Property of Polynomial Ring there exists a unique homomorphism:
:$\chi: K \left[{X}\right] \to L$
such that:
:$\chi \left({X}\right) = \beta$
and:
:$\chi \big|_K = \phi$
By Universal Property of Quotient Ring there exists a unique homomorphism:
:$\psi: K \left[{X}\right] / \left\langle{\mu_\alpha}\right\rangle \to L$
such that:
:$\chi = \psi \circ \pi$
So we have the following diagram:
::$\begin{xy}\xymatrix@L+2mu@+1em {
K \left[{X}\right] \ar[r]^*{\pi}
\ar@{-->}[rd]_*{\exists ! \chi}
&
\dfrac {K \left[{X}\right]} {\left\langle{\mu_\alpha}\right\rangle} \ar@{-->}[d]^*{\exists ! \psi}
\ar[r]^*{\Delta}
&
K \left[{\alpha}\right]
\\
&
L
&
}\end{xy}$
Now we have:
:$\chi \left({X}\right) = \beta$
and:
:$\pi \left({X}\right) = X + \left\langle{\mu_\alpha}\right\rangle$
Therefore, because:
:$\chi = \psi \circ \pi$
we have:
:$\psi \left({X + \left\langle{\mu_\alpha}\right\rangle}\right) = \beta$
Also:
:$\Delta^{-1} \left({\alpha}\right) = X + \left\langle{\mu_\alpha}\right\rangle$
so by the above:
: $\psi \circ \Delta^{-1}\left({ \alpha }\right) = \beta$
: $\psi \circ \Delta^{-1} \big|_K = \psi\big|_K = \phi$
: the choice of $\Delta, \psi$ is unique.
{{qed}}
Category:Field Extensions
Category:Universal Properties
\end{proof}
|
23161
|
\section{Universal Property of Abelianization of Group}
Tags: Universal Properties, Group Theory
\begin{theorem}
Let $G$ be a group.
Let $G^{\operatorname {ab} }$ be its abelianization.
Let $\pi : G \to G^{\operatorname {ab} }$ be the quotient group epimorphism.
Let $H$ be an abelian group.
Let $f: G \to H$ be a group homomorphism.
Then there exists a unique group homomorphism $g : G^{\operatorname {ab}} \to H$ such that $g \circ \pi = f$:
:$\xymatrix {
G \ar[d]_\pi \ar[r]^{\forall f} & H\\
G^{\operatorname {ab} } \ar[ru]_{\exists ! g} }$
\end{theorem}
\begin{proof}
By Universal Property of Quotient Group, it suffices to show that the kernel $\ker f$ contains the commutator subgroup $\sqbrk {G, G}$.
By definition of generated subgroup, this is equivalent to showing that $\ker f$ contains all commutators of $G$.
Let $x, y \in G$.
Let $e_H$ be the identity element of $H$.
Then:
{{begin-eqn}}
{{eqn | l = \map f {x^{-1} y^{-1} x y}
| r = \map f {x^{-1} } \paren {\map f {y^{-1} } \map f x} \map f y
| c = {{Defof|Group Homomorphism}}
}}
{{eqn | r = \map f {x^{-1} } \paren {\map f x \map f {y^{-1} } } \map f y
| c = {{Defof|Commutative Operation}}
}}
{{eqn | r = \paren {\map f x}^{-1} \paren {\map f x \paren {\map f y}^{-1} } \map f y
| c = Group Homomorphism Preserves Inverses
}}
{{eqn | r = \paren {\paren {\map f x}^{-1} \map f x} \paren {\paren {\map f y}^{-1} \map f y}
| c = Associativity on Four Elements
}}
{{eqn | r = e_H
| c =
}}
{{end-eqn}}
Thus the commutator $\sqbrk {x, y} = x^{-1} y^{-1} x y$ is in $\ker f$.
{{qed}}
Category:Group Theory
Category:Universal Properties
565569
565566
2022-03-28T12:20:37Z
Fake Proof
4193
565569
wikitext
text/x-wiki
\end{proof}
|
23162
|
\section{Universal Property of Field of Rational Fractions}
Tags: Universal Properties, Polynomial Theory
\begin{theorem}
Let $R$ be an integral domain.
Let $\struct {\map R x, \iota, x}$ be the field of rational fractions over $R$.
Let $\struct {K, f, a}$ be an ordered triple, where:
:$K$ is a field
:$f : R \to K$ is a unital ring homomorphism
:$a$ is a transcendental element of $K$.
Then there exists a unique unital ring homomorphism $\bar f : \map R x \to K$ such that $\bar f \circ \iota = f$ and $\map {\bar f} x = a$:
:$\xymatrix{
R \ar[d]^\iota \ar[r]^{\forall f} & K\\
\map R x \ar[ru]_{\exists ! \bar f} }$
\end{theorem}
\begin{proof}
Use Universal Property of Polynomial Ring and Universal Poperty of Field of Fractions.
{{ProofWanted}}
Category:Polynomial Theory
Category:Universal Properties
\end{proof}
|
23163
|
\section{Universal Property of Free Abelian Group on Set}
Tags: Universal Properties, Abelian Groups
\begin{theorem}
Let $S$ be a set.
Let $\struct {\Z^{\paren S}, \iota}$ be the free abelian group on $S$.
Let $\struct {G, +}$ be an abelian group.
Let $f: S \to G$ be a mapping.
Then there exists a unique group homomorphism $g : \Z^{\paren S} \to G$ with $g \circ \iota = f$:
:$\xymatrix{
S \ar[d]_\iota \ar[r]^{\forall f} & G\\
\Z^{\paren S} \ar[ru]_{\exists ! g} }$
\end{theorem}
\begin{proof}
For $x \in S$, the characteristic map $\phi_x$ of $x$ is defined as:
:$\map {\phi_x} s = \begin{cases} 1 & \quad \text{ if } s = x \\ 0 & \quad \text{ if } s \ne x \end{cases}$
The maps $\set {\phi_x \ : \ x \in S}$ form a basis for $\Z^{\paren S}$.
If $\phi \in \Z^{\paren S}$, then it can be expressed uniquely as a finite sum
:$\ds \phi = \sum_i n_i \phi_{x_i}$
where $\phi$ maps each $x_i$ to $n_i \in \Z$, and maps each element of $S$ not present in this sum to $0$.
$\iota$ maps $x \in S$ to $\phi_x \in \Z^{\paren S}$.
So, in order to satisfy $g \circ \iota = f$, we must have $\map g {\phi_x} = \map f x$ for every $x \in S$.
Since $g$ has to be a homomorphism, it must satisfy $\map g {\phi + \phi'} = \map g \phi + \map g {\phi'}$.
So, for $\phi \in \Z^{\paren S}$:
:$\ds \map g \phi = \map g {\sum_i n_i \phi_{x_i} } = \sum_i \map g {n_i \phi_{x_i} } = \sum_i n_i \map g {\phi_{x_i} } = \sum_i n_i \map f {x_i}$
Therefore, the value of $\map g \phi$ is unique, for every $\phi \in \Z^{\paren S}$.
So, the homomorphism $g$ that satisfies $g \circ \iota = f$ is unique.
{{Proofread}}
\end{proof}
|
23164
|
\section{Universal Property of Free Modules}
Tags: Universal Properties, Abstract Algebra, Module Theory, Free Modules
\begin{theorem}
Let $R$ be a ring.
Let $M$ be a free $R$-module with basis $\{e_i\mid i\in I\}$.
Let $N$ be an $R$-module.
Let $\{n_i\mid i\in I\}$ be a family of elements of $N$.
Then there exists a unique $R$-module homomorphism that maps $e_i$ to $n_i$ for all $i\in I$.
\end{theorem}
\begin{proof}
Combine Free Module is Isomorphic to Free Module Indexed by Set and Universal Property of Free Module on Set.
{{handwaving}}
Category:Free Modules
Category:Module Theory
Category:Universal Properties
\end{proof}
|
23165
|
\section{Universal Property of Group Ring}
Tags: Universal Properties, Group Rings
\begin{theorem}
Let $R$ be a commutative ring with unity.
Let $G$ be a group.
Let $R \sqbrk G$ be the corresponding group ring.
Let $S$ be a commutative ring with unity.
Let $\phi: R \to S$ be a ring homomorphism.
Let $\beta : G \to R^\times$ be a group homomorphism, where $R^\times$ is the multiplicative group of $R$.
Then there exists a unique ring homomorphism from $R \sqbrk G$ to $S$ which extends $\phi$ and $\beta$.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Universal Properties
Category:Group Rings
\end{proof}
|
23166
|
\section{Universal Property of Polynomial Ring/Free Monoid on Set}
Tags: Universal Properties, Polynomial Theory
\begin{theorem}
Let $R, S$ be commutative and unitary rings.
Let $\family {s_j}_{j \mathop \in J}$ be an indexed family of elements of $S$.
Let $\psi: R \to S$ be a ring homomorphism.
Let $R \sqbrk {\set {X_j: j \in J} }$ be a polynomial ring.
Then there exists a unique evaluation homomorphism $\phi: R \sqbrk {\set {X_j: j \in J} } \to S$ at $\family {s_j}_{j \mathop \in J}$ extending $\psi$.
\end{theorem}
\begin{proof}
Let $Z$ be the set of all multiindices indexed by $J$.
Let $k_j$ be the $j$th component of a multiindex $k$.
Let $\ds f = \sum_{k \mathop \in Z} a_k \prod_{j \mathop \in J} X_j^{k_j}$ be a polynomial over $R$.
Define:
:$\ds \map \phi f = \sum_{k \mathop \in Z} \map \psi {a_k} \prod_{j \mathop \in J} s_j^{k_j}$
It is clear that $\phi$ extends $\psi$.
If $\ds g = \sum_{k \mathop \in Z} b_k \prod_{j \mathop \in J} X_j^{k_j}$, then:
{{begin-eqn}}
{{eqn | l = \map \phi {f + g}
| r = \map \phi {\sum_{k \mathop \in Z} \paren {a_k + b_k} \prod_{j \mathop \in J} X_j^{k_j} }
| c = {{Defof|Addition of Polynomial Forms}}
}}
{{eqn | r = \sum_{k \mathop \in Z} \paren {\map \psi {a_k + b_k} } \prod_{j \mathop \in J} s_j^{k_j}
| c = Definition of $\phi$
}}
{{eqn | r = \sum_{k \mathop \in Z} \paren {\map \psi {a_k} + \map \psi {b_k} } \prod_{j \mathop \in J} s_j^{k_j}
| c = {{Defof|Ring Homomorphism}}
}}
{{eqn | r = \sum_{k \mathop \in Z} \map \psi {a_k} \prod_{j \mathop \in J} s_j^{k_j} + \sum_{k \mathop \in Z} \map \psi {b_k} \prod_{j \mathop \in J} s_j^{k_j}
| c = Ring Axioms of $S$
}}
{{eqn | r = \map \phi f + \map \phi g
| c = Definition of $\phi$
}}
{{end-eqn}}
Therefore $\phi$ preserves addition.
Also:
{{begin-eqn}}
{{eqn | l = \map \phi {f g}
| r = \map \phi {\sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_p b_q} \prod_{j \mathop \in J} X_j^{k_j} }
| c = {{Defof|Multiplication of Polynomial Forms}}
}}
{{eqn | r = \sum_{k \mathop \in Z} \map \psi {\sum_{p \mathop + q \mathop = k} a_p b_q} \prod_{j \mathop \in J} s_j^{k_j}
| c = Definition of $\phi$
}}
{{eqn | r = \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} \map \psi {a_p} \map \psi {b_q} } \prod_{j \mathop \in J} s_j^{k_j}
| c = {{Defof|Ring Homomorphism}}
}}
{{eqn | r = \paren {\sum_{k \mathop \in Z} \paren {\map \psi {a_k} } \prod_{j \mathop \in J} s_j^{k_j} } \paren {\sum_{k \mathop \in Z} \paren {\map \psi {b_k} } \prod_{j \mathop \in J} s_j^{k_j} }
| c = Ring Axioms of $S$
}}
{{eqn | r = \map \phi f \map \phi g
| c = Definition of $\phi$
}}
{{end-eqn}}
This shows that $\phi$ is a homomorphism.
Now suppose that $\phi'$ is another such homomorphism.
For each $j \in J$, $\phi'$ must satisfy $\map {\phi'} {X_j} = s_j$ and $\map {\phi'} r = \map \psi r$ for all $r \in R$.
In addition $\phi'$ must be a homomorphism, so we compute:
{{begin-eqn}}
{{eqn | l = \map {\phi'} {\sum_{k \mathop \in Z} a_k \prod_{j \mathop \in J} X_j^{k_j} }
| r = \sum_{k \mathop \in Z} \map {\phi'} {a_k \prod_{j \mathop \in J} X_j^{k_j} }
| c = $\phi'$ preserves Ring Addition
}}
{{eqn | r = \sum_{k \mathop \in Z} \map {\phi'} {a_k} \prod_{j \mathop \in J} \map {\phi'} {X_j}^{k_j}
| c = $\phi'$ preserves Ring Product
}}
{{eqn | r = \sum_{k \mathop \in Z} \map \psi {a_k} \prod_{j \mathop \in J} s_j^{k_j}
| c = as $\map {\phi'} {X_j} = s_j$ and $\map {\phi'} r = \map \psi r$ for all $r \in R$
}}
{{end-eqn}}
and therefore $\phi' = \phi$.
This concludes the proof.
{{qed}}
\end{proof}
|
23167
|
\section{Universal Property of Quotient of Topological Group}
Tags: Universal Properties, Topological Groups
\begin{theorem}
Let $G$ and $H$ be topological groups.
Let $N$ be a normal subgroup of $G$.
Let $\pi : G \to G/N$ be the quotient mapping.
Let $f : G \to H$ be a continuous group homomorphism whose kernel contains $N$.
Then there exists a unique continuous group homomorphism $\overline f: G / N \to H$ such that $f = \overline f \circ \pi$.
\end{theorem}
\begin{proof}
Because $N \subset \ker f$, $f$ is constant on the cosets of $N$.
Thus $f$ is invariant under congruence modulo $N$.
By Universal Property of Quotient Set, there exists a unique mapping $\overline f: G / N \to H$ such that $f = \overline f \circ \pi$.
It suffices to verify that it is a continuous group homomorphism.
By Universal Property of Quotient Space, there exists a continuous mapping $\overline g: G / N \to H$ such that $f = \overline g \circ \pi$.
By uniqueness of $f$, $g = f$.
By Universal Property of Quotient Group, there exists a group homomorphism $\overline h: G / N \to H$ such that $f = \overline h \circ \pi$.
By uniqueness of $f$, $h = f$.
{{qed}}
Category:Topological Groups
Category:Universal Properties
\end{proof}
|
23168
|
\section{Universal URM Computable Functions}
Tags: Recursive Functions, URM Programs
\begin{theorem}
For each integer $k \ge 1$, there exists a URM computable function:
:$\Phi_k: \N^{k+1} \to \N$
such that for each URM computable function $f: \N^k \to \N$ there exists a natural number $e$ such that:
:$\forall \left({n_1, n_2, \ldots, n_k}\right) \in \N^k: f \left({n_1, n_2, \ldots, n_k}\right) \approx \Phi_k \left({e, n_1, n_2, \ldots, n_k}\right)$.
This function $\Phi_k$ is '''universal for URM computable functions of $k$ variables'''.
\end{theorem}
\begin{proof}
Let $\Phi_k: \N^{k+1} \to \N$ be given by:
:$\Phi_k \left({e, n_1, n_2, \ldots, n_k}\right) = U \left({\mu z \ T_k \left({e, n_1, n_2, \ldots, n_k, z}\right)}\right)$
where $T_k$ and $U$ are as in Kleene's Normal Form Theorem.
Thus we have reinterpreted Kleene's Normal Form Theorem as being about URM computable functions.
This is legitimate, as a URM Computable Function is Recursive and vice versa.
{{qed}}
\end{proof}
|
23169
|
\section{Universal URM Programs}
Tags: URM Programs
\begin{theorem}
For each integer $k \ge 1$, there exists a URM program $P_k$ such that:
For each URM program $P$ there exists a natural number $e$ such that:
For all $\left({n_1, n_2, \ldots, n_k}\right) \in \N^k$, the computation using the program $P_k$ with input $\left({e, n_1, n_2, \ldots, n_k}\right)$
has the same output as the computation using the program $P$ with input $\left({n_1, n_2, \ldots, n_k}\right)$.
This function $P_k$ is a '''universal program for URM computations with $k$ inputs'''.
\end{theorem}
\begin{proof}
This follows directly from:
* Kleene's Normal Form Theorem;
* Universal URM Computable Functions.
{{qed}}
Category:URM Programs
\end{proof}
|
23170
|
\section{Unlike Electric Charges Attract}
Tags: Electric Charge
\begin{theorem}
Let $a$ and $b$ be stationary particles, each carrying an electric charge of $q_a$ and $q_b$ respectively.
Let $q_a$ and $q_b$ be of the opposite sign.
That is, let $q_a$ and $q_b$ be unlike charges.
Then the forces exerted by $a$ on $b$, and by $b$ on $a$, are such as to cause $a$ and $b$ to attract each other.
\end{theorem}
\begin{proof}
By Coulomb's Law of Electrostatics:
:$\mathbf F_{a b} \propto \dfrac {q_a q_b {\mathbf r_{a b} } } {r^3}$
where:
:$\mathbf F_{a b}$ is the force exerted on $b$ by the electric charge on $a$
:$\mathbf r_{a b}$ is the displacement vector from $a$ to $b$
:$r$ is the distance between $a$ and $b$.
{{WLOG}}, let $q_a$ be positive and $q_b$ be negative.
Then $q_a q_b$ is a positive number multiplied by a negative number.
Thus $q_a q_b$ is a negative number.
Hence $\mathbf F_{a b}$ is in the opposite direction to the displacement vector from $a$ to $b$.
That is, the force exerted on $b$ by the electric charge on $a$ is in the direction towards $a$.
The same applies to the force exerted on $a$ by the electric charge on $b$.
That is, the force exerted on $b$ by the electric charge on $a$ is in the direction towards $b$.
The effect of these forces is to cause $a$ and $b$ to pull together, that is, to attract each other.
\end{proof}
|
23171
|
\section{Unordered Pair is Finite}
Tags: Set Theory
\begin{theorem}
Let $x, y$ be arbitrary.
Then $\set {x, y}$ is finite.
\end{theorem}
\begin{proof}
By Pair is Union of Singletons:
:$\set {x, y} = \set x \cup \set y$
By Singleton is Finite:
:$\set x$ and $\set y$ are finite.
Thus by Union of Finite Sets is Finite:
:$\set {x, y}$ is finite.
{{qed}}
\end{proof}
|
23172
|
\section{Unsatisfiable Set Union Formula is Unsatisfiable}
Tags: Formal Semantics
\begin{theorem}
Let $\LL$ be a logical language.
Let $\mathscr M$ be a formal semantics for $\LL$.
Let $\FF$ be an $\mathscr M$-unsatisfiable set of formulas from $\LL$.
Let $\phi$ be a logical formula.
Then $\FF \cup \set \phi$ is also $\mathscr M$-unsatisfiable.
\end{theorem}
\begin{proof}
This is an immediate consequence of Superset of Unsatisfiable Set is Unsatisfiable.
{{qed}}
\end{proof}
|
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