nyuuzyou/wikis huggingface dataset:
Antilexicographic Product of Totally Ordered Sets is Totally Ordered

== Theorem ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be totally ordered sets.
Let $S_1 \otimes^a S_2 = \struct {S_1 \times S_2, \preccurlyeq_a}$ be the antilexicographic product of $S_1$ and $S_2$.
Then $\struct {S_1 \times S_2, \preccurlyeq_a}$ is itself a totally ordered set.
=== General Result ===
== Proof ==
From Antilexicographic Order is Ordering, we have that $\preccurlyeq_a$ is an ordering.
It remains to be shown that $\preccurlyeq_a$ is connected.
By definition of antilexicographic product, we have that:
::$\tuple {a, b} \preccurlyeq_a \tuple {c, d} \iff \tuple {c \prec_2 d} \lor \paren {c = d \land a \preccurlyeq_1 b}$
We note that as $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ are both totally ordered sets, then $\preccurlyeq_1$ and $\preccurlyeq_2$ are both connected.
Hence either:
:$c \prec_2 d$ or $d \prec_2 c$ or $c = d$
and also either 
:$a \preccurlyeq_2 b$ or $b \preccurlyeq_2 a$
If $c \prec_2 d$ or $d \prec_2 c$, it follows directly that:
:$\tuple {a, b} \preccurlyeq_a \tuple {c, d}$ or $\tuple {c, d} \preccurlyeq_a \tuple {a, b}$
Suppose $c = d$.
We have that either:
:$a \preccurlyeq_2 b$ or $b \preccurlyeq_2 a$
and so it follows that:
:$\tuple {a, b} \preccurlyeq_a \tuple {c, d}$ or $\tuple {c, d} \preccurlyeq_a \tuple {a, b}$
Hence, by definition, $\preccurlyeq_a$ is connected.
So we have shown that:
:$\preccurlyeq_a$ is connected
:$\preccurlyeq_a$ is reflexive, transitive and antisymmetric.
Thus by definition, $\preccurlyeq_a$ is a total ordering and so $\struct {S_1 \times S_2, \preccurlyeq_a}$ is a totally ordered set.
== Sources ==
* : $\S 3.4$: Ordered sums and products of ordered sets
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $31 \ \text{(a)}$
Category:Total Orderings
Category:Antilexicographic Order
Category:Antilexicographic Product of Totally Ordered Sets is Totally Ordered
Antilexicographic Product of Totally Ordered Sets is Totally Ordered/General Result

== Theorem ==
Let $S_1, S_2, \ldots, S_n$ all be totally ordered sets.
Let $T_n$ be the antilexicographic product of $S_1, S_2, \ldots, S_n$:
:$\forall n \in \N_{>0}: T_n = \begin {cases}
S_1 & : n = 1 \\
T_{n - 1} \otimes^a S_n & : n > 1
\end {cases}$
Then $T_n$ is a totally ordered set.
== Proof ==
From Antilexicographic Product of Totally Ordered Sets is Totally Ordered, $S_1 \otimes^a S_2$ is a totally ordered set.
Suppose $T_{n - 1}$ is a totally ordered set.
Given that $S_n$ is a totally ordered set, it follows from Antilexicographic Product of Totally Ordered Sets is Totally Ordered that $T_{n - 1} \otimes^a S_n$ is also a totally ordered set.
The result follows by induction.
Category:Antilexicographic Product of Totally Ordered Sets is Totally Ordered
Antilexicographic Product on Pair of Well-Ordered Sets is Well-Ordered

== Theorem ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be well-ordered sets.
Let $S_1 \otimes^a S_2 = \struct {S_1 \times S_2, \preccurlyeq_a}$ be the antilexicographic product of $S_1$ and $S_2$.
Then $\struct {S_1 \times S_2, \preccurlyeq_a}$ is itself a well-ordered set.
== Proof ==
By definition, a well-ordered set is a totally ordered set which is well-founded.
From Antilexicographic Product of Totally Ordered Sets is Totally Ordered, we have that $\preccurlyeq_a$ is a totally ordered set.
It remains to be shown that $\preccurlyeq_a$ is a well-founded relation.
Let $T = S_1 \times S_2$.
Let $A$ be a non-empty subset of $T$.
Let $A_2$ be the set of all of the second cooordinates of the ordered pairs that comprise $A$:
:$A_2 = \set {b \in S_2: \tuple {a, b} \in A}$
We have that:
:$A_2$ is a non-empty subset of $S_2$
:$S_2$ is well-ordered by $\preccurlyeq_2$
Hence $A_2$ contains a minimal element $y$ by $\preccurlyeq_a$.
Let $A_y$ be the subset of $A$ in which the second cooordinate equals $y$.
:$A_y = \set {\tuple {a, b} \in A: b = y}$
Let $A_x$ be the set of first cooordinates of $A_y$.
:$A_x = \set {a \in S_1: \tuple {a, y} \in A}$
Then:
:$A_x$ is a non-empty subset of $S_1$
:$S_1$ is well-ordered by $\preccurlyeq_1$
So $A_x$ contains a minimal element $x$.
Thus we have constructed $\tuple {x, y}$, which is the minimal element of $A$ by $\preccurlyeq_a$.
As $A$ is arbitrary, it follows that every non-empty subset of $T$ has a minimal element by $\preccurlyeq_a$.
That is, $\preccurlyeq_a$ is a well-founded relation.
Hence, by definition, $T$ is well-ordered by $\preccurlyeq_a$.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $31 \ \text{(b)}$
Category:Well-Orderings
Category:Antilexicographic Order
Antiperiodic Element is Multiple of Antiperiod

== Theorem ==
Let $f: \R \to \R$ be a real anti-periodic function with anti-period $A$.
Let $L$ be an anti-periodic element of $f$.
Then $A \divides L$.
== Proof ==
 that $A \nmid L$.
By the Division Theorem we have:
:$\exists! q \in \Z, r \in \R: L = q A + r, 0 < r < A$
By Even and Odd Integers form Partition of Integers, it follows that $q$ must be either even or odd.
=== Case 1 ===
Suppose $q$ is even.
Then:
And so $r$ is an anti-periodic element of $f$ that is less than $A$.
But then $A$ cannot be the anti-period of $f$.
Therefore by contradiction $q$ cannot be even.
=== Case 2 ===
Suppose $q$ is odd.
Then:
So:
:$-\map f x = - \map f {x + r} \implies \map f x = \map f {x + r}$
It is seen that $r$ is a periodic element of $f$ such that $0 < r < A$.
But consider $0 < A - r < A$:
This contradicts the fact that $A$ is the anti-period of $f$.
Hence the result.
Category:Antiperiodic Functions
Antireflexive Relation/Examples/Distinctness

== Example of Antireflexive Relation ==
Let $S$ be a set.
Let $\RR$ be the relation on $S$ defined as:
:$\forall x, y \in S: x \mathrel \RR y$ $x$ is distinct from $y$
Then $\RR$ is antireflexive.
== Proof ==
No element of $S$ is distinct from itself.
Hence the result by definition of distinct.
== Sources ==
* 
Category:Examples of Antireflexive Relations
Antireflexive Relation/Examples/Non-Equality

== Example of Antireflexive Relation ==
The relation $\ne$ on the set of natural numbers $\N$ is antireflexive.
== Proof ==
We have:
:$\forall a \in \N: \lnot \paren {a \ne a}$
Hence the result by definition of antireflexive relation.
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations: Exercise $1 \ \text{(i)}$
Category:Examples of Antireflexive Relations
Antireflexive Relation/Examples/Strict Ordering

== Example of Antireflexive Relation ==
The relation $<$ on one of the standard number systems $\N$, $\Z$, $\Q$ and $\R$ is antireflexive.
== Proof ==
We have:
:$\forall a \in \N: \lnot \paren {a < a}$
Hence the result by definition of antireflexive relation.
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations: Exercise $1 \ \text{(ii)}$
* : $\S 6$. Indexed families; partitions; equivalence relations: Exercise $4 \ \text{(b)}$
* 
* 
Category:Examples of Antireflexive Relations
Antireflexive and Transitive Relation is Antisymmetric

== Theorem ==
Let $\RR \subseteq S \times S$ be a relation which is not null.
Let $\RR$ be antireflexive and transitive.
Then $\RR$ is also antisymmetric.
== Proof ==
Let $\RR \subseteq S \times S$ be antireflexive and transitive.
From Antireflexive and Transitive Relation is Asymmetric it follows that $\RR$ is asymmetric.
The result follows from Asymmetric Relation is Antisymmetric.
== Also see ==
If $\RR = \O$ then Null Relation is Antireflexive, Symmetric and Transitive applies instead.
== Sources ==
* : $\S 1.19$: Some Important Properties of Relations: Exercise $4$
Category:Antireflexive Relations
Category:Antisymmetric Relations
Category:Transitive Relations
Antireflexive and Transitive Relation is Asymmetric

== Theorem ==
Let $\RR \subseteq S \times S$ be a relation which is not null.
Let $\RR$ be antireflexive and transitive.
Then $\RR$ is also asymmetric.
== Proof ==
Let $\RR \subseteq S \times S$ be antireflexive and transitive.
That is:
We have that $\RR$ is not null.
 $\RR$ is not asymmetric.
So, by definition, $\exists \tuple {x, y} \in \RR: \tuple {y, x} \in \RR$.
Then from the transitivity of $\RR$ that would mean $\tuple {x, x} \in \RR$.
But that would contradict the antireflexivity of $\RR$.
Therefore:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \notin \RR$
and $\RR$ has been shown to be asymmetric.
== Also see ==
* Null Relation is Antireflexive, Symmetric and Transitive for the case where $\RR = \O$.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: $164$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $1 \ \text {(a)}$
Category:Antireflexive Relations
Category:Asymmetric Relations
Category:Transitive Relations
Antisymmetric Preordering is Ordering

== Theorem ==
Let $\RR \subseteq S \times S$ be a preordering on a set $S$.
Let $\RR$ also be antisymmetric.
Then $\RR$ is an ordering on $S$.
== Proof ==
By definition, a preordering on $S$ is a relation on $S$ which is:
:$(1): \quad$ reflexive
and:
:$(2): \quad$ transitive.
Thus $\RR$ is a relation on $S$ which is reflexive, transitive and antisymmetric.
Thus by definition $\RR$ is an ordering on $S$.
== Sources ==
* : $\S 7$: Exercise $7$
Category:Preorderings
Category:Orderings
Category:Asymmetric Relations
Antisymmetric Quotient of Preordered Set is Ordered Set

== Theorem ==
Let $\struct {S, \precsim}$ be a preordered set.
Let $\sim$ be the equivalence relation $S$ induced by $\precsim$.
Let $\struct {S / {\sim}, \preceq}$ be the antisymmetric quotient of $\struct {S, \precsim}$.
Then:
:$\struct {S / {\sim}, \preceq}$ is an ordered set.
:$\forall P, Q \in S / {\sim}: \paren {P \preceq Q} \land \paren {p \in P} \land \paren {q \in Q} \implies p \precsim q$
This second statement means that we could just as well have defined $\preceq$ by letting $P \preceq Q$ :
:$\forall p \in P: \forall q \in Q: p \precsim q$
== Proof ==
By the definition of equivalence relation, $\sim$ is transitive, reflexive, and symmetric.
By the definition of preordering, $\precsim$ is transitive and reflexive.
From Preordering induces Ordering, $\preceq$ is an ordering.
Let $P, Q \in S / {\sim}$ with $P \preceq Q$.
Then by the definition of $\preceq$, there are $p \in P$ and $q \in Q$ such that $p \precsim q$.
Let $p' \in P$ and $q' \in Q$.
By the definition of quotient set:
:$p' \sim p$
:$q \sim q'$
Thus by the definition of $\sim$:
:$p' \precsim p$
:$q \precsim q'$
Since $p \precsim q$ and $\precsim$ is transitive:
:$p' \precsim q'$
We have shown that:
:$\forall P, Q \in S / {\sim}: \paren {P \preceq Q} \land \paren {p \in P} \land \paren {q \in Q} \implies p \precsim q$
Category:Order Theory
Category:Preorder Theory
Category:Quotient Sets
Antitransitive Relation/Examples/Greater by One

== Example of Antitransitive Relation ==
Let $\RR$ denote the relation on the set of real numbers $\R$ defined as:
:$x \mathrel \RR y \iff \text {$x$ is greater than $y$ by $1$}$
that is:
:$x - y = 1$
Then $\RR$ is an antitransitive relation..
== Proof ==
Let $x \mathrel \RR y$ and $y \mathrel \RR z$.
Then:
Hence by definition:
:$\mathop \neg x \mathrel \RR z$
== Sources ==
* 
* 
Category:Examples of Antitransitive Relations
Antitransitive Relation is Antireflexive

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation on $S$.
Let $\RR$ be antitransitive.
Then $\RR$ is also antireflexive.
== Proof ==
Suppose $\RR \subseteq S \times S$ is not antireflexive.
Then $\exists x \in S: \tuple {x, x} \in \RR$. (In the case of $\RR$ being reflexive, the property holds for all $x \in S$.)
Thus $\RR$ is not antitransitive:
... which means $\exists x \in S$ such that the condition for antitransitivity is broken.
So $\RR \subseteq S \times S$ has to be antireflexive.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: $165$
Category:Antireflexive Relations
Category:Antitransitive Relations
Apothem of Regular Polygon equals Radius of Incircle

== Theorem ==
Let $P$ be a regular polygon.
Let $C$ be the incircle of $P$.
The apothem of $P$ is equal to the radius of $C$.
== Proof ==
By definition of incircle, $C$ is the circle such that all sides of $P$ are tangent to $C$.
From Regular Polygon can be Circumscribed around Circle, it is established that the center of $P$ and the center of $C$ are the same point.
Hence:
:the radius of $C$
is the same thing as:
:the perpendicular distance from $C$ of $P$.
Hence the result.
== Sources ==
* 
* 
* 
* 
Category:Regular Polygons
Apotome is Irrational

== Theorem ==
Every apotome is irrational, i.e.:
:$\ds \forall a, b \in \set {x \in \R_{>0} : x^2 \in \Q}: \paren {\frac a b \notin \Q \land \paren {\frac a b}^2 \in \Q} \implies \paren {\paren {a - b} \notin \Q \land \paren {a - b}^2 \notin \Q}$
== Proof ==
:
Let $AB$ be a rational straight line.
Let a rational straight line $BC$ which is commensurable in square only with $AB$ be cut off from $AB$.
We have that $AB$ is incommensurable in length with $BC$.
We also have:
:$AB : BC = AB^2 : AB \cdot AC$
Therefore from :
:$AB^2$ is incommensurable with $AB \cdot AC$.
But from :
:$AB^2 + BC^2$ is commensurable with $AB^2$.
and from :
:$2 \cdot AB \cdot AC$ is commensurable with $AB \cdot AC$.
From :
:$AB^2 + BC^2 = 2 \cdot AB \cdot AC + CA^2$
and so from:
:
and:
:
it follows that:
:$AB^2 + BC^2$ is incommensurable with $AC^2$.
But $AB$ and $BC$ are rational.
Therefore $AC$ is irrational.
Such a straight line is known as an apotome.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Apotome
Apotome not same with Binomial Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ be an apotome.
Suppose $AB$ were the same with a binomial.
Let $DC$ be a rational straight line.
Let the rectangle $CE$ be applied to $CD$ equal to the square on $AB$ and producing $DE$ as breadth.
We have that $AB$ is an apotome.
Therefore from :
:$DE$ is a first apotome.
Let $EF$ be the annex of $DE$.
Therefore by :
:$DF$ and $EF$ are rational straight lines which are commensurable in square only
:$DF^2 = FE^2 + \lambda^2$ where $\lambda$ is a straight line which is commensurable in length with $DF$
:$DF$ is commensurable in length with $DC$.
We have that $AB$ is a binomial.
Therefore by :
:$DE$ is a first binomial.
Let $DE$ be divided into its terms at $G$.
Let $DG$ be the greater term.
Then by :
:$DG$ and $GE$ are rational straight lines which are commensurable in square only
:$DG^2 = GE^2 + \mu^2$ where $\mu$ is a straight line which is commensurable in length with $DF$
:$DG$ is commensurable in length with $DC$.
Therefore by :
:$DG$ is commensurable in length with $DF$.
Therefore by :
:$GF$ is commensurable in length with $DG$.
But $DF$ and $EF$ are rational straight lines which are commensurable in square only.
So $DF$ is incommensurable in length with $EF$.
Therefore by :
:$FG$ is incommensurable in length with $EF$.
Therefore $GF$ and $EF$ are rational straight lines which are commensurable in square only.
Therefore, by definition, $EG$ is an apotome.
But $EG$ is also rational, which is impossible.
Hence the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Applesellers' Problem

== Classic Problem ==
Two market-women were selling apples, one at $2$ for $1$ cent and the other at $3$ for $2$ cents.
They had $30$ apples apiece.
In order to end their competition they formed a trust, pooling their stocks and selling the apples at $5$ for $3$ cents.
This was to their advantage because under the new arrangement they took in total $36$ cents.
Under the old system they would have taken a total of only $35$ cents.
Their example was contagious.
Two other women, who also had $30$ apples apiece and who were selling them at $2$ for $1$ cent and $3$ for $1$ cent, formed a trust to sell their apples at $5$ for $2$ cents.
But instead of the $25$ cents which they would have taken in operating separate enterprises, their trust grossed only $24$ cents.
Why?
=== Variant ===
== Solution ==
Consider a general trust such as these.
Let one participant sell $a$ apples for $b$ cents, and the other sell $c$ apples for $d$ cents.
Then the first sells apples at $\dfrac b a$ cents each, and the other at $\dfrac d c$ cents each.
, let $\dfrac b a > \dfrac d c$.
The average price per apple is $\dfrac 1 2 \paren {\dfrac b a + \dfrac d c} = \dfrac {b c + a d} {2 a c}$.
The trust sets the price at $\dfrac {b + d} {a + c}$.
This trust price will be advantageous :
:$\dfrac {b + d} {a + c} > \dfrac {b c + a d} {2 a c}$
After algebra, this works out at:
:$\paren {b c - a d} \paren {a - c} > 0$
For this to hold, it is important for both factors to be positive.
We have that:
:$b c - a d > 0 \implies \dfrac b a > \dfrac d c$
This, combined with $a - c > 0$, means that the trust will be advantageous only if:
:the original prices are unequal
and:
:the denominator of the higher price is greater than that of the lower price.
For the first trust, we configure our variables such that:
:$a = 3$, $b = 2$
:$c = 2$, $d = 1$
in order to make $a - c > 0$.
Then we see:
:$b c - a d = 2 \times 2 - 3 \times 1 = 1$
and so the trust is favourable.
For the second trust, we configure our variables such that:
:$a = 3$, $b = 1$
:$c = 2$, $d = 1$
in order to make $a - c > 0$.
Then we see:
:$b c - a d = 2 \times 1 - 3 \times 1 = -1$
revealing that the trust is not favourable.
It is interesting to note that the trust price is arbitrarily dependent upon the form in which the original prices are presented.
Suppose the first participant in the second trust sold apples at $4$ for $2$ cents each.
Then the trust price of $\dfrac {2 + 1} {4 + 3} = \dfrac 3 7$ per apple is advantageous, as we see:
:$a = 4$, $b = 2$
:$c = 3$, $d = 1$
and:
:$b c - a d = 3 \times 2 - 4 \times 1 = 2$
As puts it:
:... the formula which by chance or accident fits both the presently considered trust prices is a wholly unsound one.
== Also known as ==
== Historical Note ==
== Sources ==
* : Chapter $2$. Ancient and Curious Problems: Hindu Problems: $41$. Imitation Punished
* : Indian Puzzles: $49$
Category:Classic Problems
Category:Maurice Kraitchik/Mathematical Recreations
Category:Applesellers' Problem
Applesellers' Problem/Historical Note

== Historical Note on the Applesellers' Problem ==
According to , this type of problem first appeared in by , in $800$ CE.
It hinges on the fact that a common mean is erroneously being calculated of $2$ for a monetary unit plus $3$ for a monetary unit equals $5$ for $2$ monetary units.
 goes on to remark that, while this problem appears everywhere in European puzzle-books since its appearance here, he has never seen a non-European version.
, on the other hand, includes it in his of $1960$ under the heading Hindu Problems.
, who includes 's version in his of $1992$, notes that this was taken by from a theme in 's , which dates from c. $850$.
However, this has not been corroborated, and it remains to be identified where in that source it was obtained.
== Sources ==
* 
* : Indian Puzzles: $49$
Category:Historical Notes
Applesellers' Problem/Variant

== Problem ==
:Three women, $A$, $B$ and $C$, carried apples to a market to sell.
::$A$ had sold $20$,
::$B$ had sold $30$,
::and $C$ had sold $40$.
:They sold at the same price, the one as the other,
:and, each having sold all their apples,
:brought home as much money as each other.
:How could this be?
== Solution ==
During the first part of the day, they sold their apples at $1$ penny each.
During the second part of the day, they sold them at $3$ pence each.
$A$ sold $2$ at $1$ penny and $18$ at $3$ pence each.
$B$ sold $17$ at $1$ penny and $13$ at $3$ pence each.
$C$ sold $32$ at $1$ penny and $8$ at $3$ pence each.
Hence each one made $56$ pence.
== Proof ==
In order for this to make sense, the two separate prices for the different times of day needs to be assumed.
Hence we need to determine:
:the two price tiers
:the total sum made
:the number each sold at each tier.
This is a system of Diophantine equations which has a number of solutions.
A general solution is given by:
$A$ sold $m$ at $k \paren {n - 10}$ pence and $\paren {20 - m}$ at $k n$ pence each.
$B$ sold $\paren {m + n}$ at $k \paren {n - 10}$ pence and $\paren {30 - m - n}$ at $k n$ pence each.
$C$ sold $\paren {m + 2 n}$ at $k \paren {n - 10}$ pence and $\paren {40 - m - 2 n}$ at $k n$ pence each.
Each one made $10 k \paren {2 n - m}$ pence.
where $m, n \in \N$, $0 \le m \le 20$, $n > 10$, $m + 2 n \le 40$, $k \in \R$.
Under these constraints there are $100$ choices for $\tuple {m, n}$ and (uncountably) infinite choices for $k$.
The solution given above is the particular solution $\tuple {m, n, k} = \tuple {2, 15, 0.2}$.
== Historical Note ==
== Sources ==
* 
* : Henry van Etten: $123$
Category:Applesellers' Problem
Approximate Formula for Number of Partitions

== Theorem ==
The number of partitions $\map p n$ of a (strictly) positive integer $n$ is given by the approximation:
:$\map p n \approx \dfrac {e^{\pi \sqrt {2 n / 3} } } {4 n \sqrt 3}$
== Proof ==
== Sources ==
* : Chapter $\text {A}.21$: Euler (): Footnote $7$
Category:Partition Theory
Approximate Identity for Everywhere Dense Subset is Approximate Identity of C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $\DD \subseteq A$ be everywhere dense.
Let $\tuple {\Lambda, \preceq}$ be a directed set. 
Let $\family {e_\lambda}_{\lambda \in \Lambda}$ be a net such that: 
:$e_\lambda$ is Hermitian for each $\lambda \in \Lambda$ with $\map {\sigma_A} {e_\lambda} \subseteq \closedint 0 1$
:for each $x \in \DD$, $\family {x e_\lambda}_{\lambda \mathop \in \Lambda}$ converges to $x$.
Then $\family {e_\lambda}_{\lambda \mathop \in \Lambda}$ is an approximate identity of $A$.
== Proof ==
We only need to show that for each $x \in A$, the net $\family {x e_\lambda}_{\lambda \mathop \in \Lambda}$ converges to $x$. 
First, for each $\lambda \in \Lambda$ we have:
:$\map {\sigma_A} {e_\lambda} \subseteq \closedint 0 1$
Hence, from the definition of spectral radius:
:$\map {r_A} {e_\lambda} \le 1$ 
From Spectral Radius of Normal Element of C*-Algebra Equal to Norm, we have $\norm {e_\lambda} \le 1$ for each $\lambda \in \Lambda$.
Let $x \in A$. 
Let $\epsilon > 0$.
Since $\DD$ is everywhere dense in $A$, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\DD$ with $x_n \to x$. 
Now, we have:
Take $n \in \N$ large enough to have:
:$\norm {x - x_n} < \dfrac \epsilon 3$ 
By hypothesis, since $x_n \in \DD$, we can take $a \in \Lambda$ such that for $\lambda \succeq a$ we have:
:$\norm {x_n - x_n e_\lambda} < \dfrac \epsilon 3$
Then for $\lambda \succeq a$ we have:
:$\norm {x - x e_\lambda} < \epsilon$
Hence the net $\family {x e_\lambda}_{\lambda \in \Lambda}$ converges to $x$. 
So $\family {e_\lambda}_{\lambda \mathop \in \Lambda}$ is an approximate identity of $A$.
Category:Approximate Identities of C*-Algebras
Approximate Motion of Compound Pendulum

== Theorem ==
Let $\PP$ be a compound pendulum free to swing about a pivot $A$.
Let $\PP$ be pulled to one side by a small angle $\alpha$ (less than about $10 \degrees$ or $15 \degrees$) from the vertical and then released.
Let $T$ be the period of $\PP$, that is, the time through which $\PP$ takes to travel from one end of its path to the other, and back again.
Then:
:$T = 2 \pi \sqrt {\dfrac a g}$
where:
:$g$ is the Acceleration Due to Gravity
:$h$ is the distance from the center of mass to $A$
:$k$ is the radius of gyration of $\PP$
:$I$ is the moment of inertia of $\PP$ about $A$
:$M$ is the mass of $\PP$.
== Proof ==
== Sources ==
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Category:Compound Pendulums
Category:Mechanics