nyuuzyou/wikis huggingface dataset:
No Non-Trivial Norm on Rational Numbers is Complete

== Theorem ==
No non-trivial norm on the set of the rational numbers is complete.
== Proof ==
By P-adic Norm not Complete on Rational Numbers, no $p$-adic norm $\norm{\,\cdot\,}_p$ on the set of the rational numbers, for any prime $p$, is complete.
By Rational Number Space is not Complete Metric Space, the absolute value $\size{\,\cdot\,}$ on the set of the rational numbers is not complete.
By Norm is Complete Iff Equivalent Norm is Complete, no norm is complete if it is equivalent to either the absolute value $\size{\,\cdot\,}$ or the $p$-adic norm $\norm{\,\cdot\,}_p$ for some prime $p$.
By Ostrowski's Theorem, every non-trivial norm is equivalent to either the absolute value $\size{\,\cdot\,}$ or the $p$-adic norm $\norm{\,\cdot\,}_p$ for some prime $p$.
The result follows.
== Sources ==
* : $\S 3.2$ Completions, Lemma $3.2.3$
Category:Normed Division Rings
Category:Complete Metric Spaces
No Order Isomophism Between Distinct Initial Segments of Woset

== Theorem ==
Let $E$ be a well-ordered set.
Let $S_\alpha, S_\beta$ be initial segments of $E$ that are order isomorphic.
Then $S_\alpha = S_\beta$.
== Proof ==
 $S_\alpha \ne S_\beta$.
Then $\alpha \ne \beta$.
By the trichotomy law, $\alpha \prec \beta$ or $\beta \prec \alpha$.
 assume $\alpha \prec \beta$.
Then $S_\alpha \subsetneqq S_\beta$.
That is, $S_\alpha$ is an initial segment of $S_\beta$.
We have that $S_\alpha$ and $S_\beta$ are order isomorphic.
Thus there is an order isomorphism between $S_\beta$ and an initial segment of $S_\beta$.
This contradicts Well-Ordered Class is not Isomorphic to Initial Segment.
Thus $S_\alpha = S_\beta$.
== Sources ==
* : Supplementary Exercise $1.2$
Category:Well-Orderings
No Perfect Magic Cube of Order Less than 5 Exists

== Theorem ==
Apart from the trivial order $1$ case, no perfect magic cube exists whose order is $4$ or less.
== Proof ==
=== Order $2$ ===
Consider a layer of the order $2$ perfect magic cube:
:$\begin{array}{|c|c|}
\hline a & b \\
\hline c & d \\
\hline
\end{array}$
Then we must have $a + b = a + c$.
So $b = c$, so they are not distinct, so this array cannot be a layer of a perfect magic cube.
=== Order $3$ ===
Let $C$ be the magic constant of an order $3$ perfect magic cube.
Consider a layer of the perfect magic cube:
:$\begin{array}{|c|c|c|}
\hline a & b & c \\
\hline d & e & f \\
\hline g & h & i \\
\hline
\end{array}$
The sum of all elements in this layer is $3 C$.
We also have:
This gives $e = \dfrac C 3$.
However these equations applies for every layer of the perfect magic cube.
So the center of every layer is equal to $\dfrac C 3$.
They are not distinct, so there cannot exist a order $3$ perfect magic cube.
=== Order $4$ ===
Let $C$ be the magic constant of an order $4$ perfect magic cube.
Consider a layer of the perfect magic cube:
:$\begin{array}{|c|c|c|c|}
\hline a & b & c & d \\
\hline e & f & g & h \\
\hline i & j & k & l \\
\hline m & n & o & p \\
\hline
\end{array}$
The sum of all elements in this layer is $4 C$.
We also have:
So the sum of the corners is equal to $C$.
These equations apply to all layers, including the diagonal layers.
Now consider the full perfect magic cube, but only the corners:
:$\begin{array}{|c|c|c|c|}
\hline a & \phantom b & \phantom b & c \\
\hline & & & \\
\hline & & & \\
\hline b & & & d \\
\hline
\end{array}
\begin{array}{|c|c|c|c|}
\hline \phantom b & \phantom b & \phantom b & \phantom b \\
\hline & & & \\
\hline & & & \\
\hline & & & \\
\hline
\end{array}
\begin{array}{|c|c|c|c|}
\hline \phantom b & \phantom b & \phantom b & \phantom b \\
\hline & & & \\
\hline & & & \\
\hline & & & \\
\hline
\end{array}
\begin{array}{|c|c|c|c|}
\hline e & \phantom b & \phantom b & * \\
\hline & & & \\
\hline & & & \\
\hline f & & & * \\
\hline
\end{array}$
From our discussion above, we have:
:$\paren {a + b + c + d} = \paren {a + b + e + f} = \paren {c + d + e + f} = C$
Hence:
Therefore any two adjacent corners sum up to $\dfrac C 2$.
However this implies $a + c = \dfrac C 2 = a + b$.
So $b = c$, so they are not distinct, so there cannot exist a order $4$ perfect magic cube.
== Sources ==
* : $8$
* : $8$
Category:Perfect Magic Cubes
No Quadruple of Consecutive Sums of Squares Exists

== Theorem ==
It is not possible for a quadruple of consecutive positive integers each of which is the sum of two squares.
== Proof ==
$4$ consecutive positive integers will be in the forms:
in some order.
But from Sum of Two Squares not Congruent to 3 modulo 4, $n_3$ cannot be the sum of two squares.
The result follows.
== Sources ==
* : $232$
* : $232$
Category:Sums of Squares
No Simple Graph is Perfect

== Theorem ==
Let $G$ be a simple graph whose order is $2$ or greater.
Then $G$ is not perfect.
== Proof ==
Recall that a perfect graph is one where each vertex is of different degree.
We note in passing that the simple graph consisting of one vertex trivially fulfils the condition for perfection.
 $G$ is a simple graph of order $n$ where $n \ge 2$ such that $G$ is perfect.
First, suppose that $G$ has no isolated vertices.
By the Pigeonhole Principle, for all vertices to have different degrees, one of them must be of degree at least $n$.
That means it must connect to at least $n$ other vertices.
But there are only $n - 1$ other vertices to connect to.
Therefore $G$ cannot be perfect.
Now suppose $G$ has an isolated vertex.
There can be only one, otherwise there would be two vertices of degree zero, and so $G$ would not be perfect.
Again by the Pigeonhole Principle, for all vertices to have different degrees, one of them must be of degree at least $n - 1$.
But of the remaining $n - 1$ vertices, one of them is of degree zero.
So it cannot be adjacent to any vertex.
So there are only $n - 2$ other vertices to connect to.
Therefore $G$ cannot be perfect.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $12$
Category:Simple Graphs
Category:Perfect Graphs
No Valid Categorical Syllogism contains two Particular Premises

== Theorem ==
Let $Q$ be a valid categorical syllogism.
Then at least one of the premises of $Q$ is universal.
== Proof ==
Suppose both premises of $Q$ are particular.
Then the pattern of $Q$ is one of $\text{II}x$, $\text{IO}x$, $\text{OI}x$ or $\text{OO}x$, where $x$ is the conclusion.
$\text{I}$ is neither universal nor negative.
Thus the $\text{II}x$ pattern does not distribute the middle term of $Q$.
So $\text{II}x$ violates the rule Middle Term of Valid Categorical Syllogism is Distributed at least Once.
$\text{O}$ is negative.
Thus $\text{OO}x$ violates the rule No Valid Categorical Syllogism contains two Negative Premises.
It remains to investigate $\text{IO}x$ and $\text{OI}x$.
By Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative, the conclusion of $Q$ is negative, either $\text{E}$ or $\text{O}$.
By the definition of Distributed Predicate of Categorical Syllogism, the predicate of the conclusion of $Q$ is distributed.
Thus, by construction, the primary term $P$ of $Q$ is distributed.
By Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise, it follows that $P$ is distributed in the major premise of $Q$.
This eliminates $\text{IO}x$ as the particular affirmative $\text{I}$ distributes neither its subject nor its predicate.
The final pattern to be investigated is $\text{OI}x$.
By definition, $\text{O}$ distributes only its predicate.
As $\text{O}$ needs to distribute the primary term $P$, it cannot also distribute the middle term $M$ of $Q$.
But the particular affirmative $\text{I}$ (as has been seen above) also does not distribute $M$.
Thus $M$ remains undistributed, so violating the rule Middle Term of Valid Categorical Syllogism is Distributed at least Once.
This eliminates $\text{OI}x$.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism: Exercise $\text{(a)}$
Category:Categorical Syllogisms
No Valid Categorical Syllogism with Particular Premise has Universal Conclusion

== Theorem ==
Let $Q$ be a valid categorical syllogism.
Let one of the premises of $Q$ be particular.
Then the conclusion of $Q$ is also particular.
== Proof ==
Let the major premise of $Q$ be denoted $\text{Maj}$.
Let the minor premise of $Q$ be denoted $\text{Min}$.
Let the conclusion of $Q$ be denoted $\text{C}$.
From No Valid Categorical Syllogism contains two Particular Premises, either $\text{Maj}$ or $\text{Min}$ has to be universal.
Let the other premise of $Q$ be particular.
Suppose $\text{C}$ is the universal affirmative $\map {\mathbf A} {S, P}$.
From Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative it follows that $\text{Maj}$ and $\text{Min}$ are also both affirmative.
The only patterns fitting the criteria are $\text{AI}x$ and $\text{IA}x$, both of which distribute only one term in $Q$.
We have that $\text{C}$ distributes $S$.
From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise, $S$ must also be distributed in $\text{Min}$.
From Middle Term of Valid Categorical Syllogism is Distributed at least Once, the middle term $M$ of $Q$ needs to be distributed.
Thus we require two distributed terms in the premises for $\text{C}$ to be the universal affirmative.
So if $\text{C}$ is affirmative, neither premise of $Q$ can be particular.
From this contradiction it follows that $\text{C}$ is not $\map {\mathbf A} {S, P}$.
Suppose $\text{C}$ is the universal negative $\map {\mathbf E} {S, P}$.
In this case, both $S$ and $P$ are distributed in $C$, as the universal negative distributes both $S$ and $P$.
	
By Middle Term of Valid Categorical Syllogism is Distributed at least Once, the middle term $M$ of $Q$ needs to be distributed.
	
Thus there are three terms needing to be distributed in two premises.
So at least one premises needs to be the universal negative.
The other premise is not $\mathbf O$ as this violates No Valid Categorical Syllogism contains two Negative Premises.
But it also is not $\mathbf I$, as $\mathbf I$ does not distribute any terms.
So if $\text{C}$ is the universal negative, neither premise can be particular.
From this contradiction it follows that $\text{C}$ is not $\map {\mathbf E} {S, P}$.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism: Exercise $\text{(b)}$
Category:Categorical Syllogisms
Node of Rooted Tree with Multiple Children is on Multiple Branches

== Theorem ==
Let $T$ be a rooted tree with root node $r_T$.
Let $t$ be a node of $T$.
Let $t$ have more than one child nodes.
Then $t$ is on more than one branch of $T$.
== Proof ==
== Sources ==
* : $\S 1.7$: Tableaus
Category:Rooted Trees
Noether's Theorem (Calculus of Variations)

== Theorem ==
Let $y_i$, $F$, $\Psi_i$, $\Phi$ be real functions.
Let $x, \epsilon \in \R$.
Let $\mathbf y = \sequence {y_i}_{1 \mathop \le i \mathop \le n}$ and $\mathbf \Psi = \sequence{\Psi_i}_{1 \mathop \le i \mathop \le n}$ be vectors.
Let 
:$\Phi = \map \Phi {x, \mathbf y, \mathbf y'; \epsilon}, \quad \Psi_i = \map {\Psi_i} {x, \mathbf y, \mathbf y'; \epsilon}$ 
such that:
:$\map \Phi {x, \mathbf y, \mathbf y'; 0} = x, \quad \map {\Psi_i} {x, \mathbf y, \mathbf y'; 0} = y_i$
where $x, \mathbf y, \mathbf y', \epsilon$ are variables.
Let:
:$\ds J \sqbrk {\mathbf y} = \int_{x_0}^{x_1} \map F {x, \mathbf y, \mathbf y'} \rd x$
be a functional.
Let:
:$X = \map \Phi {x, \mathbf y, \mathbf y'; \epsilon}, \quad \mathbf Y = \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon}$
Suppose, $J \sqbrk {\mathbf y}$ is invariant under transformations $x \rightarrow X$ and $\mathbf y \rightarrow \mathbf Y$ for arbitrary $x_0$ and $x_1$.
Then:
:$\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F} \phi = C$
where $C$ is a constant and:
:$\map {\boldsymbol \psi} {x, \mathbf y, \mathbf y'} = \valueat {\dfrac {\partial \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} } {\epsilon \mathop = 0}$
:$\map \phi {x, \mathbf y, \mathbf y'} = \valueat {\dfrac {\partial \map \Phi {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} } {\epsilon \mathop = 0}$
== Proof ==
Apply Taylor's Theorem to the transformations $X$, $\mathbf Y$ at the point $\epsilon = 0$:
Use the general variation formula, and suppose that the curve $\mathbf y = \map {\mathbf y} x$ is an extremal of $J \sqbrk {\mathbf y}$:
:$\delta x = \epsilon \phi$
:$\delta y = \epsilon \psi$
:$\delta J = \epsilon \bigintlimits {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F }\phi} {x \mathop = x_0} {x \mathop = x_1}$
Since $J \sqbrk {\mathbf y}$ is invariant under the transformation, the variation vanishes.
Then for any $\epsilon \ne 0$ we have:
:$\bigvalueat {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y'\cdot \nabla_{\mathbf y'} F} \phi} {x \mathop = x_0} = \bigvalueat {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F}\phi} {x \mathop = x_1}$
This has to hold for arbitrary $x_0$ and $x_1$.
Since only a constant mapping produces the same result for any input, the term in brackets has to be a constant.
== Sources ==
* : $\S 4.20$: Noether's Theorem
Category:Partial Differential Equations
Category:Calculus of Variations
Noether's Theorem (Hamiltonian Mechanics)

== Theorem ==
Let there be an infinitesimal transformation of generalised coordinates such that:
:$q_i \to \tilde q_i = q_i + q_i^\alpha \tuple {q, \dot q, t} \varepsilon_\alpha + \hbox {terms vanishing on shell}$
where $\varepsilon$ is not time-dependent.
Under this transformation, let the variation of the Lagrangian be:
:$L \tuple {q + \delta q, \dot q + \delta \dot q, t} - L \tuple{q, \dot q, t} = \dfrac {\d} {\d t} \LL^\alpha \tuple {q, \dot q, t} \varepsilon_\alpha$
Let $s$ be the number of degrees of freedom of the system.
Then the quantity:
:$\ds \JJ^\alpha = \sum_{i \mathop = 1}^s \frac {\partial L} {\partial \dot q_i} q_i^\alpha - \LL^\alpha$
is conserved.
== Proof ==
Category:Partial Differential Equations
Category:Physics
Category:Hamiltonian Mechanics
Category:Calculus of Variations
Noether Normalization Lemma

== Theorem ==
Let $k$ be a field.
Let $A$ be a non-trivial finitely generated $k$-algebra.
Then there exists $n \in \N$ and a finite injective morphism of $k$-algebra:
:$k \sqbrk {x_1, \dotsc, x_n} \to A$
== Proof ==
Since $A$ is finitely generated, we prove this by induction on the number $m$ of generators as a $k$-algebra.
=== Basis for the induction ===
If $m = 0$, then $A = k$ and thus there is nothing to prove.
=== Induction step ===
Let $A$ be a $k$-algebra generated by the elements $y_1, \dotsc, y_m, y_{m + 1}$. 
If $y_1, \dotsc, y_m, y_{m + 1}$ are algebraically independent, then $A$ is isomorphic to $k \sqbrk {x_1, \dotsc, x_m, x_{m + 1} }$.
In this case the theorem is obvious, so assume that $y_{m + 1}$ depends algebraically on $y_1, \ldots, y_m$.
So there exists a polynomial $P \in k \sqbrk {x_1, \dotsc, x_m, X}$ such that $\map P {y_1, \dotsc, y_n, y_{n + 1} } = 0$ in $A$. 
Let $\mu \in \N^n$ and define:
:$f_\mu: k \sqbrk {x_1, \dotsc, x_n, X} \to k \sqbrk {x_1, \dotsc, x_n, X}: x_i \mapsto x_i + X^{\mu_i}, X \mapsto X$
This is easily seen to be an isomorphism.
Next, we want to show that there exists $\mu \in \N^n$ such that $\map {f_\mu} P$ is a polynomial in $X$ with leading coefficient in $k$.
Let $e$ be the biggest natural number such that $P$ involves $x_e$ non-trivially. 
If there is no such number, define $e$ to be $0$. 
We proceed by induction on $e$.
The case $e = 0$ is easy since we then have a polynomial $f \in k \sqbrk X$.
Thus $\mu = \tuple {0, \dotsc, 0}$ will suffice.
Suppose it holds for $e$ and suppose $P$ depends only on $x_1, \dotsc, x_e, x_{e + 1}$.
Then we can write $P$ as:
:$\ds P = \sum_{i \mathop = 0}^l a_i x_{e + 1}^i$
where $a_i \in k \sqbrk {x_1, \dotsc, x_e, X}$ and $a_l \ne 0$.
Hence by the induction hypothesis, there exists a $\mu' \in \N^n $ such that $\map {f_{\mu'} } {a_l}$ has a leading coefficient in $k$.
Note that we can choose that $\mu'_l$ for $l > e$ since $\map {f_\mu} {a_i}$ is independent of these components. 
Hence we define $\mu$ as $\mu_i = \mu'_i$ for every $i \ne e + 1$.
Now observe that:
:$\ds \map {f_\mu} P = \sum_{i \mathop = 0}^l \map {f_\mu} {a_i} \, \map {f_\mu} {x_{e + 1} }^i = \sum_{i \mathop = 0}^l \map {f_{\mu'} } {a_i} \paren {x_{e + 1} + X^{\mu_{e + 1} } }^i$
We have that $\map {f_{\mu'} } {a_l}$ has a leading coefficient in $k$.
Thus by choosing $\mu_{e + 1}$ big enough, $\map {f_\mu} P$ has a leading coefficient in $k$.
Consider now thus the elements $z_i = \map {f_\mu^{-1} } {y_i} \in A$. 
These elements still generate $A$ since $f_\mu$ is an isomorphism. 
On the other hand, we find that:
By the induction hypothesis, we have the existence of:
:$\alpha: k \sqbrk {x_1, \dotsc, x_n} \to A'$
where $A'$ is generated by $z_1, \dotsc, z_m$.
By extending this $\alpha$ to $A$ by the natural inclusion, we find that $\alpha$ is finite, since $z_i$ is integral by construction and $z_{m + 1}$ is integral over the others since it satisfies $\map {f_\mu} P$. 
Hence it is the wanted finite injective morphism.
== Also see ==
* Zariski's Lemma
* Preimage of Maximal Ideal of Finitely Generated Algebra is Maximal
Category:Commutative Algebra
Noetherian Domain is Factorization Domain

== Theorem ==
Let $R$ be a noetherian integral domain.
Then $R$ is a factorization domain.
== Proof ==
Let $\FF$ be the set of ideals of $R$ of the form $x R$, with $x$ not a unit and such that $x$ cannot be decomposed in the form:
:$x = u p_1 \dotsm p_r$
where $u$ is a unit and $p_1, \dotsc, p_r$ irreducible.
We show by contradiction that $\FF = \O$.
 $\FF \ne \O$.
Since $R$ is noetherian, we can choose a maximal element $a R \in \FF$.
By construction, $a$ is not irreducible, so we can write:
:$a = b c$
with $b, c$ non-units and not associates.
By the definition of associates in a commutative ring this means that $b R \subsetneq a R$ and $a R \subsetneq b R$.
Since $a R$ is assumed maximal, this means that $b R$ and $c R$ do not belong to $\FF$.
Therefore there exist units $u, v$ and irreducible elements $p_1, \dotsc, p_r, q_1, \dotsc, q_s$ such that:
:$b = u p_1 \dotsm p_r$
and:
:$c = v q_1 \dotsm q_s$
But this implies that:
:$a = b c = \paren {u v} p_1 \cdots p_r \cdot q_1 \dotsm q_s$
This is a contradiction, since $a$ cannot be written in this form.
Category:Ring Theory
Category:Factorization
Noetherian Hausdorff Space is Finite Discrete Space

== Theorem ==
Let $\struct {S, \tau}$ be a Noetherian Hausdorff space.
Let $S \ne \O$.
Then $\struct {S, \tau}$ is a finite discrete space.
== Proof ==
Category:Hausdorff Spaces
Category:Noetherian Topological Spaces
Category:Discrete Topology
Category:Finite Topological Spaces
Noetherian Topological Space is Compact

== Theorem ==
Let $T = \struct {X, \tau}$ be a Noetherian topological space.
Then $T$ is compact.
== Proof 1 ==
== Proof 2 ==
Category:Noetherian Topological Space is Compact
Category:Noetherian Topological Spaces
Category:Examples of Compact Topological Spaces
Noetherian Topological Space is Compact/Proof 1

== Theorem ==
== Proof ==
Let $\family {U_i}_{i \mathop \in I}$ be a cover of $X$.
That is:
:$\ds \bigcup_{i \mathop \in I} U_i = X$
Let $V$ be the collection of finite cover of $\family {U_i}_{i \mathop \in I}$.
Let $W = \set {\bigcup Y: Y \in V}$.
Then $W$ is a collection of open sets of $T$.
By , $W$ has a maximal element the subset relation.
Let $\ds U' = \bigcup_{j \mathop = 1}^n U_{i_j}$ be the maximal element.
 $U' \subsetneq X$.
Let $x \in X \setminus U'$.
Let $U_{i_{n + 1} }$ be a neighborhood of $x$, where $i_{n + 1} \in I$.
Then $U' \cup U_{i_{n + 1} }$ is larger than $U'$.
This contradicts our hypothesis that $U'$ is maximal.
Hence $U'$ is a finite subcover.
This shows that $\struct {X, \tau}$ is compact.
Category:Noetherian Topological Space is Compact
Noetherian Topological Space is Compact/Proof 2

== Theorem ==
== Proof ==
Recall the definition for compact space:
We may assume $X \ne \O$, since the claim is otherwise trivial.
Let $\CC \subseteq \tau$ be an arbitrary cover for $X$.
We shall show that $\CC$ has a finite subcover.
Consider:
:$A := \leftset {\bigcup \eta: \eta}$ is a finite subset of $\rightset \CC$
$A$ has the following properties:
:$(1): \quad A \ne \O$
:$(2): \quad A \subseteq \tau$
Let $\alpha = \bigcup \eta$.
We have:
From $(1)$ and $(2)$, by , $A$ has a maximal element $\alpha$.
We now show that:
:$\alpha = X$
:
:$\exists x \in X \setminus \alpha$
Since $\CC$ is a cover for $X$:
:$\exists U \in \CC : x \in U$
so that:
:$\alpha \subsetneqq \alpha \cup U$
This contradicts $(3)$.
Hence by Proof by Contradiction:
:$\not \exists x \in X \setminus \alpha$
and so:
:$\alpha = X$
as required.
Because $\alpha \in A$, we can write it as:
:$\alpha = U_1 \cup \cdots \cup U_n$
using $U_1, \ldots, U_n \in \CC$ for some $n \in \N_{>0}$.
This means:
:$X = U_1 \cup \cdots \cup U_n$
Therefore:
:$\set {U_1, \ldots, U_n}$ is a finite subcover of $\CC$ for $X$.
Category:Noetherian Topological Space is Compact
Nominal Rate of Interest/Examples/Arbitrary Example 1

== Example of Nominal Rate of Interest ==
Let $\pounds 1000$ be invested for $2$ years at $8 \%$ per annum.
Let interest be compounded half-yearly.
The interest rate $8 \%$ per annum is a nominal rate of interest.
== Explanation ==
This the interest it would be if you applied it each conversion period onto the uncompounded principal.
Hence $2 \times 0 \cdotp 04 = 0 \cdotp 08$ and all is well.
== Sources ==
* 
* 
Category:Examples of Nominal Rates of Interest
Non-Abelian Group has Order Greater than 4

== Theorem ==
Let $\struct {G, \circ}$ be a non-abelian group.
Then the order of $\struct {G, \circ}$ is greater than $4$.
== Proof 1 ==
== Proof 2 ==
Category:Abelian Groups
Category:Non-Abelian Group has Order Greater than 4
Non-Abelian Group has Order Greater than 4/Proof 1

== Theorem ==
== Proof ==
Let $\left({G, \circ}\right)$ be a non-abelian group whose identity is $e$.
For a group $\left({G, \circ}\right)$ to be non-abelian, we require:
: $\exists x, y \in G: x \circ y \ne y \circ x$
Suppose $x \circ y \in \left\{ {x, y, e}\right\}$.
: $x \circ y = e \implies y \circ x = e$
and $\left({G, \circ}\right)$ is abelian.
, suppose $x \circ y = x$.
and again, $\left({G, \circ}\right)$ is abelian.
Similarly for $x \circ y = y$.
Again, the same applies if $y \circ x \in \left\{ {x, y, e}\right\}$.
So, if $x \circ y \ne y \circ x$, then;
: $x \circ y$ and $y \circ x$ must be different elements
: $x \circ y$ and $y \circ x$ must both different from $e, x$ and $y$.
Thus, in a non-abelian group, there needs to be at least $5$ elements:
:$e, x, y, x \circ y, y \circ x$
Category:Non-Abelian Group has Order Greater than 4
Non-Abelian Group has Order Greater than 4/Proof 2

== Theorem ==
== Proof ==
It follows from Trivial Group is Cyclic Group and Prime Group is Cyclic that groups of order less than $4$ are cyclic.
Therefore, by Cyclic Group is Abelian, all groups of order less than $4$ are abelian.
Let $G$ have order $4$.
From Order of Element Divides Order of Finite Group, every element of $G$ has order that divides $4$.
Therefore any element of $G$ has order $1$, $2$ or $4$.
Let $g \in G$ have order $4$.
Then $G$ is generated by $g$.
Therefore $G$ is cyclic, and therefore abelian by Cyclic Group is Abelian.
Let $G$ be such that it has no element of order $4$.
From Identity is Only Group Element of Order 1, all the elements apart from its identity must have order $2$.
That is, $G$ is a Boolean group.
By Boolean Group is Abelian, $G$ is abelian.
The result follows.
== Sources ==
* : $\S 44.3$ Some consequences of Lagrange's Theorem
Category:Non-Abelian Group has Order Greater than 4