nyuuzyou/wikis huggingface dataset:
Proper Integer Heronian Triangle whose Area is 24

== Theorem ==
There exists exactly one proper integer Heronian triangle whose area equals $24$.
That is, the obtuse triangle whose sides are of length $4$, $13$ and $15$.
== Proof ==
First we show that the $\tuple {4, 13, 15}$ triangle is actually Heronian.
Heron's Formula gives us that:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where:
:$\AA$ denotes the area of the triangle
:$a$, $b$ and $c$ denote the lengths of the sides of the triangle
:$s = \dfrac {a + b + c} 2$ denotes the semiperimeter of the triangle.
Hence:
It can be constructed by taking a $9-12-15$ Pythagorean triangle and removing a $5-12-13$ Pythagorean triangle from it:
:
Let $\tuple {a, b, c}$ be the sides of such a triangle.
Using Heron's Formula, we have:
Note that:
:$\paren {s - a} + \paren {s - b} + \paren {s - c} = 3 s - a - b - c = s$
Hence by substituting $x = s - a$, $y = s - b$, $z = s - c$:
:$x y z \paren {x + y + z} = 576$
By Semiperimeter of Integer Heronian Triangle is Composite, $s$ is an integer.
Hence $s, x, y, z \in \N_{>0}$.
 suppose $x \le y \le z$.
Then:
so we need to check $1 \le x \le 3$.
Note that for fixed $x$:
For $x = 1$:
:$1 \le y \le \sqrt [3] {\frac {576} 2} \approx 6.60$
For $x = 2$:
:$2 \le y \le \sqrt [3] {\frac {576} 4} \approx 5.24$
For $x = 3$:
:$3 \le y \le \sqrt [3] {\frac {576} 6} \approx 4.58$
Finally, for fixed $x$, $y$:
Since $z \in \Z$:
:$\sqrt {\paren {x + y}^2 + \dfrac {2304} {x y} } = 2 z + x + y \in \Z$
We check the value of $\sqrt {\paren {x + y}^2 + \dfrac {2304} {x y} }$ for the valid values of $x, y$.
We do not need to check $y = 5$ since $5 \nmid 576$.
We have:
:$\tuple {1, 1}: \sqrt {\paren {1 + 1}^2 + \dfrac {2304} {1 \times 1} } = \sqrt {2308} \notin \Z$
:$\tuple {1, 2}: \sqrt {\paren {1 + 2}^2 + \dfrac {2304} {1 \times 2} } = \sqrt {1161} \notin \Z$
:$\tuple {1, 3}: \sqrt {\paren {1 + 3}^2 + \dfrac {2304} {1 \times 3} } = \sqrt {784} = 28$
:$\tuple {1, 4}: \sqrt {\paren {1 + 4}^2 + \dfrac {2304} {1 \times 4} } = \sqrt {601} \notin \Z$
:$\tuple {1, 6}: \sqrt {\paren {1 + 6}^2 + \dfrac {2304} {1 \times 6} } = \sqrt {433} \notin \Z$
:$\tuple {2, 2}: \sqrt {\paren {2 + 2}^2 + \dfrac {2304} {2 \times 2} } = \sqrt {592} \notin \Z$
:$\tuple {2, 3}: \sqrt {\paren {2 + 3}^2 + \dfrac {2304} {2 \times 3} } = \sqrt {409} \notin \Z$
:$\tuple {2, 4}: \sqrt {\paren {2 + 4}^2 + \dfrac {2304} {2 \times 4} } = \sqrt {324} = 18$
:$\tuple {3, 3}: \sqrt {\paren {3 + 3}^2 + \dfrac {2304} {3 \times 3} } = \sqrt {292} \notin \Z$
:$\tuple {3, 4}: \sqrt {\paren {3 + 4}^2 + \dfrac {2304} {3 \times 4} } = \sqrt {241} \notin \Z$
Now we calculate $z$ for $\tuple {x, y} = \tuple {1, 3}, \tuple {2, 4}$.
:$\tuple {1, 3}: z = \dfrac {- 1 - 3 + 28} 2 = 12$
:$\tuple {2, 4}: z = \dfrac {- 2 - 4 + 18} 2 = 6$
Using:
:$a = s - x = x + y + z - x = y + z$
:$b = s - y = x + z$
:$c = s - z = x + y$
the possible sets of side lengths are:
:$\tuple {15, 13, 4}, \tuple {10, 8, 6}$
But note that the $\tuple {6, 8, 10}$ triangle is right-angled, and hence not a proper Heronian Triangle.
Hence the result.
== Sources ==
* : Bachet: $111$
Category:Heronian Triangles
Proper Integer Heronian Triangle whose Area is 24/Mistake

== Source Work ==
: 
:The Puzzles:
::Bachet
:::$111$
== Mistake ==
:What is the unique Heronian triangle with area $24$?
== Correction ==
The fact that the only Heronian triangles under discussion are integer ones is implicit.
While the solution gives the solution $\tuple{ 4, 13, 15}$, the $\tuple {6, 8, 10}$ solution is overlooked.
Note, however, that this is a right triangle.
But it is indeed an integer Heronian triangle, just not a proper one.
Hence the question should be worded:
:What is the unique proper Heronian triangle with area $24$?
== Sources ==
* : Bachet: $111$
Category:Mistakes/Curious and Interesting Puzzles
Proper Linear Subspace of Topological Vector Space has Empty Interior

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\GF$.
Let $L$ be a proper linear subspace of $X$. 
Then:
:$L^\circ = \O$
where $L^\circ$ denotes the interior of $L$. 
== Proof ==
 that $L^\circ \ne \O$.
Let $x \in L^\circ$. 
Then there exists an open neighborhood $U$ of $x$ such that $U \subseteq L$. 
From Translation of Open Set in Topological Vector Space is Open, $U - x$ is an open neighborhood of ${\mathbf 0}_X$ with $U - x \subseteq L - x$.
Since $L$ is a linear subspace of $X$, we have $L - x = L$.
Let $V = U - x$.
From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, we have:
:$\ds X = \bigcup_{n \mathop = 1}^\infty n V$
On the other hand, since $L$ is a linear subspace of $X$ and $V \subseteq L$, we also have:
:$\ds \bigcup_{n \mathop = 1}^\infty n V \subseteq L$
and so $L = X$.
This contradicts that $L$ is a proper linear subspace.
Hence $L^\circ = \O$.
Category:Topological Vector Spaces
Proper Lower Section under Well-Ordering is Initial Segment

== Theorem ==
Let $A$ be a class under a well-ordering $\preccurlyeq$.
Let $B$ be a proper lower section of $A$.
Let $b$ be the smallest element of $A \setminus B$.
Then $B$ is the initial segment of $b$ in $A$.
== Proof ==
By definition of proper lower section of $A$:
:$\forall x \in B: \forall a \in A \setminus B: x \preccurlyeq a$
while:
:$B \ne \O$
:$B \ne A$
Because $B \ne A$ we have that $A \setminus B \ne \O$.
Because $\preccurlyeq$ is a well-ordering, $A \setminus B$ does indeed have a smallest element, which we can indeed call $b$.
We have that:
:$\forall x \in B: \forall a \in A \setminus B: x \preccurlyeq a$
This applies in particular to $b \in A \setminus B$:
:$\forall x \in B: \exists b \in A \setminus B: x \preccurlyeq b$
That is, $B$ is the initial segment of $b$ in $A$ by definition.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries
Category:Initial Segments
Category:Lower Sections
Category:Well-Orderings
Proper Modular Ideal of Algebra is Contained in Maximal Ideal

== Theorem ==
Let $R$ be a ring.
Let $\struct {A, \ast}$ be an $R$-algebra.
Let $\struct {J, \ast}$ be a proper modular ideal of $\struct {A, \ast}$. 
Then there exists a maximal ideal $M$ of $\struct {A, \ast}$ such that $J \subseteq M$. 
== Proof ==
Let $u \in A$ be such that:
:$a - u a \in J$ and $a - a u \in J$
for each $a \in A$.
Let $S$ be the set of ideals $I$ of $\struct {A, \ast}$ such that:
:$u \not \in I$ and $J \subseteq I$.
Consider the partially ordered set $\struct {S, \subseteq}$ where $\subseteq$ is the subset relation.
We use Zorn's Lemma on $S$.
Let $\CC$ be a chain in $\struct {S, \subseteq}$.
We show that $\bigcup \CC$ is an upper bound for $\CC$. 
From Union of Chain of Ideals is Ideal: Algebra over Ring, we have:
:$\bigcup \CC$ is an ideal. 
From Set Union Preserves Subsets, we have $J \subseteq \bigcup \CC$. 
Since $u \not \in I$ for each $I \in \CC$, we also have $u \not \in \bigcup \CC$.
So every chain in $\struct {S, \subseteq}$ has an upper bound. 
Hence from Zorn's Lemma, $\struct {S, \subseteq}$ has a maximal element $I$. 
We show that $I$ is a maximal ideal. 
We know that there is no proper ideal $K$ with $u \not \in K$ such that $I \subsetneq K$ since $I$ is a maximal element of $\struct {S, \subseteq}$.
 that there exists an ideal $K$ with $u \in K$ and $I \subseteq K$.
We then have $a u \in K$ for each $a \in A$ since $K$ is an ideal.
Since $J$ is modular and $J \subseteq K$, we also have $a - a u \in K$ for each $a \in A$.
Hence for each $a \in A$ we have $a = \paren {a - a u} + a u \in K$.
Hence $K = A$.
So there is no proper ideal $K$ such that $I \subsetneq K$.
Hence $I$ is a maximal ideal.
== Sources ==
* : $\text{I}.3$: Strictly Cyclic Modules. Modular Right Ideals.
* : $1.1$: Banach Algebras
Category:Modular Ideals of Algebras
Category:Maximal Ideals of Algebras
Proper Subset/Examples/Arbitrary Example 1

== Example of Proper Subset ==
Let $A = \set {1, 2, 3}$.
Let $B = \set {1, 2, 3, 4}$.
Let $C = \set {1, 2, 3}$.
We see by inspection that $A$ is a proper subset of $B$.
However, while $A$ is a subset of $C$, it is not the case that $A$ is a proper subset of $C$.
== Sources ==
* 
* 
Category:Examples of Proper Subsets
Proper Subset of Finite Ordinal is Equivalent to Smaller Ordinal

== Theorem ==
Let $n$ be a finite ordinal.
Let $x \subsetneq n$.
Then for some finite ordinal $m < n$:
:$m \sim x$
where $m \sim x$ denotes that $m$ is (set) equivalent to $x$.
That is, every proper subset of a finite ordinal $n$ is equivalent to some finite ordinal smaller than $n$.
== Proof ==
Proof by induction:
For all finite ordinals $n$, let $\map P n$ be the proposition:
:$x \subsetneq n \implies \exists m \in \N: m < n: m \sim x$
$\map P 0$ is vacuously true, as there are no proper subsets of $0 = \O$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.
So this is our induction hypothesis:
:$x \subsetneq k \implies \exists m \in \N: m < k: m \sim x$
Then we need to show:
:$x \subsetneq k^+ \implies \exists m \in \N: m < k^+: m \sim x$
=== Induction Step ===
This is our induction step:
Let $x \subsetneq k^+$.
Then either:
:$(1) \quad x \subsetneq k$, in which case the induction hypothesis applies
or:
:$(2) \quad x = k$, in which case the result is trivially true
or:
:$(3) \quad k \in x$.
In case $(3)$, we find a finite ordinal $j \in k$ such that $j \notin x$.
Then we define a mapping $f$ on $x$ as:
:$\forall i \in x: \map f i = \begin {cases}
i & : i \ne k \\
j & : i = k
\end {cases}$
Clearly $f$ is injective and $f$ maps $x$ into $k$.
So the image of $x$ under $f$ is either equal to $k$ or by the induction hypothesis equivalent to some element of $k$.
Consequently, $x$ is always equivalent to some element of $k$.
So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: x \subsetneq n \implies \exists m \in \N: m < n: m \sim x$
== Also see ==
* No Bijection between Finite Set and Proper Subset
== Sources ==
* : $\S 13$: Arithmetic
Category:Finite Ordinals
Category:Proofs by Induction
Proper Subset of Matroid Circuit is Independent

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $C \subseteq S$ be a circuit of $M$.
Then every proper subset $A$ of $C$ is independent.
== Proof ==
By definition of a circuit of $M$:
:$C$ is a minimum dependent subset of $M$
By definition of the minimum dependent subset of $M$:
:every proper subset $A$ of $C$ is not a dependent subset.
By definition of a dependent subset:
:every proper subset $A$ of $C$ is an independent subset.
== Sources ==
* Chapter $1.$ $\S 9.$ Circuits
Category:Matroid Circuits
Proper Subtower is Initial Segment

== Theorem ==
Let $\struct {T_1, \preccurlyeq}$ be a proper subtower of $\struct {T_2, \preccurlyeq}$.
Then $\struct {T_1, \preccurlyeq}$ is an initial segment of $\struct {T_2, \preccurlyeq}$.
== Proof ==
Define the set:
:$Y = \set {y \in T_1: S_y \text { is an initial segment of } \struct {T_2, \preccurlyeq} }$.
Then:
By Induction on Well-Ordered Set, $Y = T_1$.
That is, $\struct {T_1, \preccurlyeq}$ is an initial segment in $\struct {T_2, \preccurlyeq}$.
== Sources ==
* : Supplementary Exercises $1.7$
Category:Well-Orderings
Proper Well-Ordering determines Smallest Elements

== Theorem ==
Let $S$ be a class.
Let $\preceq$ be a proper well-ordering on $S$.
Let $B$ be a nonempty subclass of $S$.
Then $B$ has a $\preceq$-smallest element.
== Proof ==
Because $B \ne \O$ it follows that $B$ has an element $x$.
If $x$ is the smallest element of $B$ then the theorem holds.
Otherwise, there is an element $y \in B$ such that $x \npreceq y$.
By Well-Ordering is Total Ordering, $y \prec x$.
Thus $S_x$, the $\preceq$-initial segment of $x$, is not empty.
By the hypothesis, $S_x$ is a set.
By Intersection is Subset, $B \cap S_x \subseteq S_x$
Thus by the subset axiom, $B \cap S_x$ is a set.
Thus by the definition of proper well-ordering, $B \cap S_x$ has a smallest element $a$.
Let $z$ be any element of $B$.
 $z \prec a$.
Then by Extended Transitivity, $z \prec x$, so $z \in B \cap S_x$.
But this contradicts the fact that $a$ is the smallest element of $B \cap S_x$.
Thus $z \nprec a$.
Since $\preceq$ is a total ordering, $a \preceq z$.
As this holds for all $z \in B$, $a$ is the smallest element of $B$.
== Sources ==
* : $\S 6.26$
Category:Class Theory
Category:Well-Orderings
Proper Zero Divisor/Examples/Multiplication Modulo 6

== Examples of Proper Zero Divisors ==
Consider the multiplicative monoid of integers modulo $6$, defined by its Cayley table:
Thus we have:
:$\eqclass 2 6 \times \eqclass 3 6 = \eqclass 0 6$
and:
:$\eqclass 4 6 \times \eqclass 3 6 = \eqclass 0 6$
Hence in the ring of integers modulo $6$, there are seen to be $3$ proper zero divisors: $\eqclass 2 6$, $\eqclass 3 6$ and $\eqclass 4 6$.
== Sources ==
* : $\S 55$. Special types of ring and ring elements: $(5)$
* 
Category:Examples of Proper Zero Divisors
Proper Zero Divisor/Examples/Ring of Order 2 Complex Matrices

== Example of Proper Zero Divisor ==
Consider the ring of square matrices $\struct {\map {\MM_\C} 2, +, \times}$ of order $2$ over the complex numbers $\C$.
We have:
:$\begin {pmatrix} 1 & i \\ i & -1 \end {pmatrix} \begin {pmatrix} 1 & i \\ i & -1 \end {pmatrix} = \begin {pmatrix} 0 & 0 \\ 0 & 0 \end {pmatrix}$
demonstrating that $\begin {pmatrix} 1 & i \\ i & -1 \end {pmatrix}$ is a proper zero divisor of $\struct {\map {\MM_\C} 2, +, \times}$.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
Category:Examples of Proper Zero Divisors
Proper and Prime iff Ultrafilter in Boolean Lattice

== Theorem ==
Let $B = \struct {S, \vee, \wedge, \neg, \preceq}$ be a Boolean lattice.
Let $F$ be a filter in $B$.
Then
:$F$ is a proper subset of $S$ and $F$ is a prime filter in $B$
:
:$F$ is ultrafilter on $B$
== Proof ==
=== Sufficient Condition ===
Let us assume
:$F$ is a proper subset of $S$ and $F$ is a prime filter in $B$.
Thus
:$F$ is a proper subset of $S$.
Let $G$ be a filter in $B$ such that
:$F \subseteq G$ and $F \ne G$.
By definitions of subset and set equality:
:$\exists x: x \in G \land x \notin F$
By definition of Boolean algebra:
:$x \vee \neg x = \top$
where $\top$ denotes the top of $B$.
By Top in Filter:
:$\top \in F$
By definition of prime filter:
:$\neg x \in F$
By definition of subset:
:$\neg x \in G$
By Filtered in Meet Semilattice:
:$x \wedge \neg x \in G$
By definition of Boolean algebra:
:$\bot \in G$
Thus by definition of subset:
:$G \subseteq S$
By definition of set equality it remains to prove that
:$S \subseteq G$
Let $y \in S$.
By definition of smallest element:
:$\bot \preceq y$
Thus by definition of upper section:
:$y \in G$
=== Necessary Condition ===
Let $F$ be ultrafilter on $B$.
Thus by definition of ultrafilter:
:$F$ is proper subset of $S$.
Let $x \in S$.
:
:$x \notin F$ and $\neg x \notin F$
By Finite Infima Set and Upper Closure is Smallest Filter:
:$F \cup \set x \subseteq \map {\operatorname{fininfs} } {F \cup \set x}^\succeq$
By Set is Subset of Union:
:$\set x \subseteq F \cup \set x$ and $F \subseteq F \cup \set x$
By definition of singleton:
:$x \in \set x$
By definition of subset:
:$x \in \map {\operatorname{fininfs} } {F \cup \set x}^\succeq$
By Finite Infima Set and Upper Closure is Filter:
:$\map {\operatorname{fininfs} } {F \cup \set x}^\succeq$ is filter in $L$.
By Subset Relation is Transitive:
:$F \subseteq \map {\operatorname{fininfs} } {F \cup \set x}^\succeq$
By definition of ultrafilter:
:$\map {\operatorname{fininfs} } {F \cup \set x}^\succeq = S$
By Finite Subset Bounds Element of Finite Infima Set and Upper Closure:
:$\exists a \in S: a \in F \land \neg x \succeq a \wedge \inf \set x$
By definition of greatest element:
:$a \preceq \top$
By Infimum of Singleton:
:$\inf \set x = x$
By Meet Precedes Operands:
:$a \wedge \neg x \preceq \neg x$
By definition of infimum:
:$a \preceq \neg x$
By definition of upper section:
:$\neg x \in F$
This contradicts $\neg x \notin F$
Thus by Proof by Contradiction:
:$F$ is a prime filter by Filter is Prime iff For Every Element Element either Negation Belongs to Filter in Boolean Lattice.
== Sources ==
* 
* 
Category:Prime Ideals (Order Theory)
Category:Boolean Lattices
Properties of 12,345,679

== Theorem ==
$12 \, 345 \, 679$ has the following properties:
In each product, the sequence $1$ to $9$, with the one given digit missing, can be read in order by cycling round it, skipping a fixed number of digits (counting an extra one when going from start to end), for example:
:$2 \ (4691) \ 3 \ (58?2) \ 4 \ (6913) \ 5 \ (8?24) \ 6 \ (9135) \ 8 (?246) \ 9$
== Also see ==
* Penholodigital Properties of 123,456,789
== Sources ==
* : $12,345,679$
* : $12,345,679$
Category:12,345,679
Category:Recreational Mathematics
Properties of 142,857

== Theorem ==
This page gathers together some properties of $142 \, 857$ which arise through its being the digits of the recurring part of the reciprocal of $7$.
Multiplication of $142 \, 857$ by numbers higher than $7$ produces a similar pattern to when you multiply it by a single digit, but with added complications.
For example:
This becomes $714 \, 285$ when you take the $1$ off the front and add it to the back.
The exception is when you multiply it by $7$ or a multiple of $7$:
From Number times Recurring Part of Reciprocal gives 9-Repdigit, it is seen that this property is shared of all numbers formed from the digits of the recurring part of a recurring fraction.
If you divide $142 \, 857$ into two equal parts and add them, you get $999$:
:$142 + 857 = 999$
Thus by Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999, $142 \, 857$ is divisible by $999$:
:$142 \, 857 = 143 \times 999$
Also, we have:
:$999 \, 999 = 1001 \times 999$
and so $999 \, 999$ is divisible by $999$.
But as $999 \, 999 = 7 \times 142 \, 857$ we have that $999 \, 999$ is divisible by $7$.
Thus it follows from Euclid's Lemma that $142 \, 857$ is divisible by $999$.
== Sources ==
* : $142,857$
* : $142,857$
Category:142,857
Properties of 47,619

== Theorem ==
If you split $047 \, 619$ into two halves, they add up to $666$:
:$047 + 619 = 666$
which is a multiple of $333$.
and:
: $047 \, 619 = 143 \times 333$
Similarly, you can split $047 \, 619$ into three thirds, and these add up to $99$:
:$04 + 76 + 19 = 99$
and:
: $047 \, 619 = 481 \times 99$
The square of $47 \, 619$:
:$47 \, 619^2 = 2 \, 267 \, 569 \, 161$
can itself be split into two $6$-digit halves which together add to the recurring part of $\dfrac 4 7$:
:$2267 + 569 \, 161 = 571 \, 428$
This is caused by the fact that $047 \, 619$ is the recurring part of the Reciprocal of 21:
:$\dfrac 1 {21} = 0 \cdotp \dot 04761 \, \dot 9$
where $21$ is the product of (the smallest) $2$ distinct primes which do not divide $10$.
== Sources ==
* : $47,619$
* : $47,619$
Category:47,619
Properties of 5,559,060,566,555,523

== Theorem ==
$3^{33} = 5 \, 559 \, 060 \, 566 \, 555 \, 523$ has the following properties:
: It has a remarkably large number of $5$s (half of its digits).
: Multiply it by $2$, $4$ or $6$, and the result has $10$ of a particular digit.
: Multiply it by $8$, and $9$ of the digits of the result are $4$.
== Proof ==
== Historical Note ==
== Sources ==
* : $5,559,060,566,555,523$
Category:5,559,060,566,555,523
Category:Recreational Mathematics
Properties of 5,559,060,566,555,523/Historical Note

== Historical Note on Properties of $5 \, 559 \, 060 \, 566 \, 555 \, 523$ ==
 reports in his of $1997$ that this result is attributed to , but no details are given.
== Sources ==
* : $5,559,060,566,555,523$
Category:Historical Notes
Properties of Affine Spaces

== Theorem ==
Let $\EE$ be an affine space with difference space $V$.
Let $0$ denote the zero element of $V$.
Then the following hold for all $p,q,r \in \EE$ and all $u, v \in V$:
:$(1): \quad p - p = 0$
:$(2): \quad p + 0 = p$
:$(3): \quad p + u = p + v \iff u = v$
:$(4): \quad q - p = r - p \iff q = r$
== Proof ==
=== $(1): \quad p - p = 0$ ===
We have:
From Zero Element is Unique:
:$p - p = 0$
=== $(2): \quad p + 0 = p$ ===
Using $(1)$ we see that:
=== $(3): \quad p + u = p + v \iff u = v$ ===
Let $u = v$.
By definition a mapping has a unique image point on a given element.
It follows that:
:$p + u = p + v$
Let $p + u = p + v$.
We have:
=== $(4): \quad q - p = r - p \iff q = r$ ===
Let $q = r$.
By definition a mapping has a unique image point on a given element.
It follows that:
:$q - p = r - p$
Let $q - p = r - p \in V$.
Then:
Category:Affine Geometry
Properties of Algebras of Sets

== Theorem ==
Let $S$ be a set.
Let $\RR$ be an algebra of sets on $S$.
Then the following hold:
=== Algebra of Sets is Closed under Intersection‎ ===
=== Algebra of Sets is Closed under Set Difference ===
=== Algebra of Sets contains Underlying Set ===
=== Algebra of Sets contains Empty Set ===
Category:Algebras of Sets
Properties of Beta Function

== Theorem ==
Let $\Beta \left({x, y}\right)$ denote the Beta function:
$\Beta \left({x, y}\right)$ has the following properties:
=== Commutativity of Parameters of Beta Function ===
=== Beta Function of Real Number with 1 ===
=== Beta Function of $x$ with $y+1$ by $\dfrac {x+y} y$ ===
=== Beta Function of $x+1$ with $y$ plus Beta Function of $x$ with $y+1$ ===
Category:Beta Function
Properties of Binomial Coefficients

== Theorem ==
This page gathers together some of the simpler and more common identities concerning binomial coefficients.
== Symmetry Rule for Binomial Coefficients ==
== Negated Upper Index of Binomial Coefficient ==
== Moving Top Index to Bottom in Binomial Coefficient ==
== Factors of Binomial Coefficient ==
== Pascal's Rule ==
== Sum of Binomial Coefficients over Lower Index ==
== Alternating Sum and Difference of $r \choose k$ up to $n$ ==
== Alternating Sum and Difference of Binomial Coefficients for Given n ==
== Sum of Even Index Binomial Coefficients ==
== Sum of Odd Index Binomial Coefficients ==
== Sum of $r+k \choose k$ up to $n$ ==
== Rising Sum of Binomial Coefficients ==
== Sum of Binomial Coefficients over Upper Index ==
== Increasing Sum of Binomial Coefficients ==
== Increasing Alternating Sum of Binomial Coefficients ==
== Chu-Vandermonde Identity ==
== Sum of Squares of Binomial Coefficients ==
== Binomial Coefficient: $\left({-1}\right)^n \dbinom {-n} {k - 1} = \left({-1}\right)^k \dbinom {-k} {n - 1}$ ==
== Binomial Coefficient as Infinite Product ==
== Particular Values ==
== Also see ==
* Summations of Products of Binomial Coefficients
Category:Binomial Coefficients
Properties of Ceiling Function

== Theorem ==
This page gathers together some basic propeties of the ceiling function.
=== Number is between Ceiling and One Less ===
=== Ceiling is between Number and One More ===
=== Ceiling of Number plus Integer ===
=== Ceiling minus Real Number ===
=== Real Number is between Ceiling Functions ===
=== Real Number is Ceiling minus Difference ===
=== Ceiling Function is Idempotent ===
== Range of Values of Ceiling Function ==
== Also see ==
* Properties of Floor Function
Category:Ceiling Function
Properties of Class of All Ordinals

== Theorem ==
Let $\On$ denote the class of all ordinals.
$\On$ has the following properties:
=== Zero is an Ordinal ===
=== Successor of Ordinal is an Ordinal ===
=== Union of a Chain of Ordinals is an Ordinal ===
=== Superinduction Principle ===
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.2$
Category:Properties of Class of All Ordinals
Category:Class of All Ordinals