nyuuzyou/wikis huggingface dataset:
Epsilon-Function Differentiability Condition

== Theorem ==
Let $f: D \to \R$ be a continuous function, where $D \subseteq \R$ is an open set.
Let $x \in \R$.
Then $f$ is differentiable at $x$ there exist $\alpha \in \R$ and $r \in \R_{>0}$ such that for all $h \in \openint {-r} r \setminus \set 0$:
:$\map f {x + h} = \map f x + h \paren {\alpha + \map \epsilon h}$
where:
:$\openint {-r} r$ denotes an open interval of $0$
:$\epsilon: \openint {-r} r \setminus \set 0 \to \R$ is a real function with $\ds \lim_{h \mathop \to 0} \map \epsilon h = 0$.
If the conditions are true, then:
:$\alpha = \map {f'} x$
If $f$ is continuously differentiable at $x$, then $\epsilon$ is a continous function.
=== Complex Case ===
== Proof ==
=== Necessary Condition ===
Assume that $f$ is differentiable in $x$.
By definition of open set, there exists $r \in \R_{>0}$ such that $\openint {x - r} {x + r} \subseteq D$.
Define $\epsilon: \openint {-r} r \setminus \set 0 \to \R$ by:
:$\map \epsilon h = \dfrac {\map f {x + h} - \map f x} h - \map {f'} x$
If $h \in \openint {-r} r \setminus \set 0$, then:
:$x + h \in \openint {x - r} {x + r} \setminus \set x \subseteq D$
so $\epsilon$ is well-defined.
As $f$ is differentiable at $x$, it follows that:
:$\ds \lim_{h \mathop \to 0} \map \epsilon h = \lim_{h \mathop \to 0} \dfrac {\map f {x + h} - \map f x} h - \map {f'} x = \map {f'} x - \map {f'} x = 0$
If we put:
:$\alpha = \map {f'} x$
it follows that for any $h \in \openint {-r} r \setminus \set 0$:
:$\map f {x + h} = \map f x + h \paren {\alpha + \map \epsilon h}$
If $f$ is continuously differentiable, we can choose $r \in \R_{>0}$ such that $f'$ is continuous in $\openint {-r} r$.
From Continuity of Composite Mapping: Corollary, it follows that $\epsilon$ will be continuous in $\openint {-r} r \setminus \set 0$.
=== Sufficient condition ===
Rewrite the equation of the assumption to get:
:$\dfrac {\map f {x + h} - \map f x} h = \alpha + \map \epsilon h$
From Sum Rule for Limits of Real Functions, it follows that:
By definition of differentiability, $f$ is differentiable at $x$ with:
:$\map {f'} x = \alpha$
== Also see ==
* Characterization of Differentiability
== Sources ==
* : $\S 7.1$
Category:Differential Calculus
Category:Epsilon-Function Differentiability Condition
Epsilon-Function Differentiability Condition/Complex Case

== Theorem ==
Let $f: D \to \C$ be a continuous function, where $D \subseteq \C$ is an open set.
Let $z \in \C$.
Then $f$ is differentiable at $z$ there exist $\alpha \in \C$ and $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + h \paren {\alpha + \map \epsilon h}$
where:
:$\map {B_r} 0$ denotes an open ball of $0$
:$\epsilon: \map {B_r} 0 \setminus \set 0 \to \C$ is a complex function with $\ds \lim_{h \mathop \to 0} \map \epsilon h = 0$.
If the conditions are true, then $\alpha = \map {f'} z$.
== Proof ==
=== Necessary Condition ===
Assume that $f$ is differentiable in $z$.
By definition of open set, there exists $r \in \R_{>0}$ such that the open ball $\map {B_r} z \subseteq D$.
Define $\epsilon: \map {B_r} 0 \setminus \set 0 \to \C$ by:
:$\map \epsilon h = \dfrac {\map f {z + h} - \map f z} h - \map {f'} z$
Let $h \in \map {B_r} 0 \setminus \set 0$.
Then:
:$z + h \in \map {B_r} z \setminus \set z \subseteq D$
so $\epsilon$ is well-defined.
As $f$ is differentiable in $z$, it follows that:
:$\ds \lim_{h \mathop \to 0} \map \epsilon h = \lim_{h \mathop \to 0} \dfrac {\map f {z + h} - \map f z} h - \map {f'} z = \map {f'} z - \map {f'} z = 0$
Let us put $\alpha = \map {f'} z$.
It follows that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + h \paren {\alpha + \map \epsilon h}$
=== Sufficient condition ===
Let us rewrite the equation of the assumption as:
:$\dfrac {\map f {z + h} - \map f z} h = \alpha + \map \epsilon h$
From Sum Rule for Limits of Complex Functions:
:$\ds \lim_{h \mathop \to 0} \dfrac {\map f {z + h} - \map f z} h = \lim_{h \mathop \to 0} \paren { \alpha + \map \epsilon h} = \alpha$
By definition of differentiability, $f$ is differentiable at $z$ with $\map {f'} z = \alpha$.
== Also see ==
* Characterization of Differentiability
== Sources ==
* : $\S 7.1$
Category:Complex Differential Calculus
Category:Epsilon-Function Differentiability Condition
Epsilon-Neighborhood (Real Number Line)/Examples/1-Neighborhood of 2

== Examples of Epsilon-Neighborhoods ==
The $1$-neighborhood of $2$ is the set:
:$\map {N_1} 2 = \openint 1 3 = \set {x \in \R: \size {x - 2} < 1}$
== Sources ==
* : Chapter $\text {II}$: Topological Spaces: $\S 11$. Continuity on the Euclidean line
Category:Examples of Neighborhoods
Epsilon Induction

== Theorem ==
=== Formulation 1 ===
Let $A$ be a class.
Let $\Bbb U$ denote the universe.
:$\paren {\forall x: \paren {x \subseteq A \implies x \in A} } \implies A = \Bbb U$
=== Formulation 2 ===
Let $\map P x$ be a well-formed formula containing $x$ as a free variable.
Suppose that, from the truth of $\map P x$ for every member of a set $a$, the truth of $\map P a$ follows.
That is:
:$\paren {\forall x \in a : \map P x} \implies \map P a$
Then $\map P x$ is true for every set.
== Proof of Formulation 1 ==
Suppose that
:$\forall x: \paren {x \subseteq A \implies x \in A}$
Consider $\Bbb U \setminus A$.
:
:$\Bbb U \setminus A \ne \O$
Then by Axiom of Foundation (Strong Form), we have that:
:$\exists x \notin A: x \cap \paren {\Bbb U \setminus A} = \O$
But:
Thus:
:$x \setminus A = \O$
So from Set Difference with Superset is Empty Set:
:$x \subseteq A$
Thus :
:$x \in A$
contradicting the fact that $x \notin A$.
Therefore we can conclude that
:$\Bbb U \setminus A = \O$
and so from Set Difference with Superset is Empty Set:
:$\Bbb U \subseteq A$
Furthermore, by definition of universe:
:$A \subseteq \Bbb U$
so by definition of set equality:
:$A = \Bbb U$
== Proof of Formulation 2 ==
 there is a set $b$ for which $\neg \map P b$.
Define:
:$\map f x = \set {y \in x : \neg \map P y}$
From the contrapositive of the assumption:
:$\neg \map P x \implies \exists y \in x : \neg \map P y$
it follows that $\map f x$ is non-empty whenever $\neg \map P x$.
Define:
:$\map g p = \bigcup \set{f(q) : q \in p}$
Whenever $\exists x \in p : \neg \map P x$:
:$\exists y \in \map g p : \neg \map P y$
By the Principle of Recursive Definition, define a sequence $c_n$ such that:
:$c_0 = \set{b}$
:$c_{n+1} = \map g {c_n}$
Let $c = \bigcup \set{c_n : n \in \omega}$.
$c$ is non-empty, as $b \in c_0 \subseteq c$.
By the Axiom of Foundation, $\exists x_0 \in c : \paren{\forall x \in d : x \notin c}$.
By construction of $c_n$:
:$\neg \map P {x_0}$
so by the contrapositive of the assumption:
:$\exists y_0 \in x_0 : \neg \map P {y_0}$
But:
:$x_0 \in c_{n_0}$
for some natural number $n_0$.
Then:
:$y_0 \in c_{n_0 + 1}$
Therefore:
:$y_0 \in c$
a contradiction.
Thus, by Reductio ad Absurdum, $\map P x$ is true for all $x$.
== Sources ==
* : $\S 5.24$
Category:Class Theory
Epsilon Relation is Proper

== Theorem ==
Let $\mathbb U$ be the universal class.
Let $\Epsilon$ be the epsilon relation.
Then $\struct {\mathbb U, \Epsilon}$ is a proper relational structure.
== Proof ==
Let $x \in \mathbb U$.
Then by the Axiom of Extension:
:$x = \map {\Epsilon^{-1} } x$
where $\map {\Epsilon^{-1} } x$ denotes the preimage of $x$ under $\Epsilon$.
Since $x$ is a set, $\map {\prec^{-1} } x = x$ is a set.
As this holds for all $x \in \mathbb U$, $\struct {\mathbb U, \Epsilon}$ is a proper relational structure.
Category:Class Theory
Epsilon Relation is Strictly Well-Founded

== Theorem ==
Let $\Epsilon$ denote the epsilon relation.
Then $\Epsilon$ is a strictly well-founded relation on every class $A$.
== Proof ==
By the :
:$\forall S: \paren {\exists x: x \in S \implies \exists y \in S: \forall x \in S: \neg x \in y}$
That is, by Nonempty Class has Members:
:$\forall S: \paren {S \ne \O \implies \exists y \in S: \forall x \in S: \neg x \in y}$ 
This holds for all sets $S$ whose construction is based on the Zermelo-Fraenkel axioms.
We can weaken the antecedent of the above statement with this statement:
:$\forall S: \paren {\paren {S \ne \O \land S \subseteq A} \implies \exists y \in S: \forall x \in S: \neg x \in y}$
Note that this step does not require that $A$ be a set: it can be any class, even a proper class.
By definition, it follows that $\Epsilon$ is a strictly well-founded relation on every class $A$.
Category:Well-Founded Relations
Category:Axiom of Foundation
Category:Class Theory
Equal Alternate Angles implies Parallel Lines

== Theorem ==
Given two infinite straight lines which are cut by a transversal, if the alternate angles are equal, then the lines are parallel.
== Proof ==
:
Let $AB$ and $CD$ be two infinite straight lines.
Let $EF$ be a transversal that cuts them.
Let at least one pair of alternate angles be equal.
, let $\angle AHJ = \angle HJD$.
 that $AB$ and $CD$ are not parallel.
Then they meet at some point $G$.
, let $G$ be on the same side as $B$ and $D$.
Since $\angle AHJ$ is an exterior angle of $\triangle GJH$, from External Angle of Triangle is Greater than Internal Opposite, $\angle AHJ > \angle HJG$, which is a contradiction.
Similarly, they cannot meet on the side of $A$ and $C$.
Therefore, by definition, $AB$ and $CD$ are parallel.
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.4$
Category:Transversals (Geometry)
Category:Parallel Lines
Equal Angles in Equal Circles

== Theorem ==
In equal circles, equal angles stand on equal arcs, whether at the center or at the circumference of those circles.
== Proof ==
Let $ABC$ and $DEF$ be equal circles.
Let $\angle BGC = \angle EHF$ and $\angle BAC = \angle EDF$.
:
Let $BC$ and $EF$ be joined.
Since the circles $ABC$ and $DEF$ are equal, their radii are equal.
So $BG = EH$ and $CG = FH$.
We also have that $\angle BGC = \angle EHF$.
So from Triangle Side-Angle-Side Congruence it follows that $BC = EF$.
Since $\angle BAC = \angle EDF$ we have from that segment $BAC$ is similar to segment $EDF$.
Moreover, these segments have equal bases.
So from Similar Segments on Equal Bases are Equal, segment $BAC$ is equal to segment $EDF$. 
But as $ABC$ and $DEF$ are equal circles, it follows that arc $BKC$ equals arc $ELF$.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Equal Arcs of Circles Subtended by Equal Straight Lines

== Theorem ==
== Proof ==
Let $ABC$ and $DEF$ be equal circles.
Let equal arcs $BGC$ and $EHF$ be cut off by the straight lines $BC$ and $EF$.
:
Let $K$ and $L$ be the centers of the circles $ABC$ and $DEF$ respectively.
Let $BK, KC, EL, LF$ be joined.
We have that the arcs $BGC$ and $EHF$ are equal.
So from Angles on Equal Arcs are Equal $\angle BKC = \angle ELF$.
Since circles $ABC$ and $DEF$ are equal, so are their radii.
So $BK = EL$ and $KC = LF$, and they contain equal angles.
So from Triangle Side-Angle-Side Congruence base $BC$ is equal to base $EF$.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Equal Chords in Circle

== Theorem ==
In a circle, equal chords are equally distant from the center, and chords that are equally distant from the center are equal in length.
== Proof ==
:
Let $ABDC$ be a circle, and let $AB$ and $CD$ be equal chords on it.
From Finding Center of Circle, let $E$ be the center of $ABDC$.
Construct $EF$, $EG$ perpendicular to $AB$ and $CD$, and join $AE$ and $EC$.
From Conditions for Diameter to be Perpendicular Bisector, $EF$ bisects $AD$.
So $AF = FB$ and so $AB = 2 AF$.
Similarly $CD = 2CG$.
As $AB = CD$ it follows that $AE = CG$.
Since $AE = EC$ the square on $AE$ equals the square on $EC$.
We have that $\angle AFE$ is a right angle.
So from Pythagoras's Theorem the square on $AE$ equals the sum of the squares on $AF$ and $FE$.
Similarly, the square on $EC$ equals the sum of the squares on $EG$ and $GC$.
Thus the square on $EF$ equals the square on $EG$.
So $EF = EG$.
So from Definition III: 4, chords are at equal distance from the center when the perpendiculars drawn to them from the center are equal.
So $AB$ and $CD$ are at equal distance from the center.
Now suppose that $AB$ and $CD$ are at equal distance from the center.
That is, let $EF = EG$.
Using the same construction as above, it is proved similarly that $AB = 2AF$ and $CD = 2CG$.
Since $AE = CE$, the square on $AE$ equals the square on $CE$.
From Pythagoras's Theorem:
: the square on $AE$ equals the sum of the squares on $AF$ and $FE$
: the square on $EC$ equals the sum of the squares on $EG$ and $GC$.
So the sum of the squares on $AF$ and $FE$ equals the sum of the squares on $EG$ and $GC$.
But the square on $FE$ equals the square on $EG$.
So the square on $AF$ equals the square on $CG$, and so $AF = CG$.
As $AB = 2AF$ and $CD = 2CG$ it follows that $AB = CD$.
Hence the result.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Equal Consecutive Prime Number Gaps are Multiples of Six

== Theorem ==
If you list the gaps between consecutive primes greater than $5$:
:$2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, \ldots$
you will notice that consecutive gaps that are equal are of the form $6 x$.
This is always the case.
== Proof ==
Suppose there were two consecutive gaps between $3$ consecutive prime numbers that were equal, but not divisible by $6$.
Then the difference is $2 k$ where $k$ is not divisible by $3$.
Therefore the (supposed) prime numbers will be:
:$p, p + 2 k, p + 4 k$
But then $p + 4 k$ is congruent modulo $3$ to $p + k$.
That makes the three numbers congruent modulo $3$ to $p, p + k, p + 2k$.
One of those is divisible by $3$ and so cannot be prime.
So two consecutive gaps must be divisible by $3$ and therefore (as they have to be even) by $6$.
Category:Prime Numbers
Equal Corresponding Angles implies Parallel Lines

== Theorem ==
Given two infinite straight lines which are cut by a transversal, if the corresponding angles are equal, then the lines are parallel.
== Proof ==
:
Let $AB$ and $CD$ be infinite straight lines.
Let $EF$ be a transversal that cuts them.
Let at least one pair of corresponding angles be equal.
, let $\angle EGB = \angle GHD$.
By the Vertical Angle Theorem:
:$\angle GHD = \angle EGB = \angle AGH$
Thus by Equal Alternate Angles implies Parallel Lines:
:$AB \parallel CD$
== Sources ==
* : Book $\text{I}$. Propositions
Category:Angles
Category:Parallel Lines
Equal Corresponding Angles or Supplementary Interior Angles implies Parallel Lines

== Theorem ==
=== Equal Corresponding Angles implies Parallel Lines ===
=== Supplementary Interior Angles implies Parallel Lines ===
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.5$
Category:Transversals (Geometry)
Category:Parallel Lines
Category:Supplementary Angles
Equal Elements of Field of Quotients

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
Let $\struct {K, +, \circ}$ be the field of quotients of $\struct {D, +, \circ}$.
Let $x = \dfrac p q \in K$.
Then:
:$\forall k \in D^*: x = \dfrac {p \circ k} {q \circ k}$
where:
:$D^* := D \setminus \set {0_D}$
that is, $D$ with its zero removed.
== Proof ==
We have that the field of quotients $\struct {K, +, \circ}$ of an integral domain is its inverse completion.
Thus we have:
:$\forall x_1, x_2 \in D, y_1, y_2 \in D^*: \dfrac {x_1} {y_1} = \dfrac {x_2} {y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
So:
Hence the result.
Note that in order for $\dfrac {p \circ k} {q \circ k}$ to be defined, $q \circ k \ne 0_D$, that is, $k \ne 0_D$.
Category:Integral Domains
Category:Inverse Completions
Category:Fields of Quotients
Equal Images of Mappings to Hausdorff Space form Closed Set

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $T_B$ be a Hausdorff space.
Let $f, g: T_A \to T_B$ be continuous mappings.
Let $W$ be the set defined as:
:$W = \set {x \in T_A: \map f x = \map g x}$
Then $W$ is closed in $T_A$.
== Proof ==
Consider the set $V = S_A \setminus W$.
Hence:
:$V = \set {x \in T_A: \map f x \ne \map g x}$
Let $x \in V$.
Then:
:$\map f x \ne \map g x$
and as $T_B$ is Hausdorff:
:$\exists U_1, U_2 \in \tau_B: \map f x \in U_1, \map g x \in U_2, U_1 \cap U_2 = \O$
As $f$ and $g$ are continuous mappings:
:$f^{-1} \sqbrk {U_1}$ and $g^{-1} \sqbrk {U_2}$ are open in $T_B$.
Because $f^{-1} \sqbrk {U_1}$ and $g^{-1} \sqbrk {U_2}$ are open in $T_A$, $f^{-1} \sqbrk U_1 \cap g^{-1} \sqbrk U_2$ is also open in $T_A$.
Then we have that:
:$x \in f^{-1} \sqbrk {U_1} \cap g^{-1} \sqbrk {U_2}$
such that:
:$f^{-1} \sqbrk {U_1} \cap g^{-1} \sqbrk {U_2} \subseteq V$
Hence $V$ is open in $T_A$.
Hence $W = T_A \setminus V$ is closed in $T_A$.
== Sources ==
* : $4$: The Hausdorff condition: Exercise $4.3: 6$
Category:Hausdorff Spaces
Equal Numbers are Congruent

== Theorem ==
:$\forall x, y, z \in \R: x = y \implies x \equiv y \pmod z$
where $x \equiv y \pmod z$ denotes congruence modulo $z$.
== Proof ==
== Also see ==
* Congruent Numbers are not necessarily Equal
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Example $\text {4-3}$
Category:Modulo Arithmetic
Equal Order Elements may not be Conjugate

== Theorem ==
Let $G$ be a group
Let $x, y \in G$ be elements of $G$ such that:
:$\order x = \order y$
where $\order x$ denotes the order of $x$.
Then it is not necessarily the case that $x$ and $y$ are conjugates.
== Proof ==
Consider the dihedral group $D_4$, whose group presentation is:
We have that:
:$\order {a^2} = 2$
and:
:$\order b = 2$
but $a^2$ and $b$ are not conjugate to each other.
== Also see ==
* Order of Conjugate Element equals Order of Element: the converse of this.
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.18$: Remark $2$
Category:Order of Group Elements
Category:Conjugacy
Equal Powers of Finite Order Element

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $g \in G$ be of finite order.
Let $\order g = k$.
Then:
:$g^r = g^s \iff k \divides \paren {r - s}$
== Proof ==
=== Necessary Condition ===
Suppose that $k \divides \paren {r - s}$.
From the definition of divisor:
:$k \divides \left({r - s}\right) \implies \exists t \in \Z: r - s = k t$
So:
:$g^{r - s} = g^{k t}$
Thus:
=== Sufficient Condition ===
Let $g^r = g^s$.
Then:
: $g^{r - s} = g^r g^{-s} = g^s g^{-s} = e$
 
By the Division Theorem:
: $r - s = q k + t$
for some $q \in \Z, 0 \le t < k$.
Thus:
:$e = g^{r - s} = g^{k q + t} = \paren {g^k}^q g^t = e^q g^t = g^t$
So by the definition of $k$:
:$\paren {t < k} \land \paren {e = g^t} \implies t = 0$
So:
:$r - s = q k + 0 = q k \implies k \divides \paren {r - s}$
== Sources ==
* : $\S 5.4$. Cyclic groups
* : Chapter $3$: Elementary consequences of the definitions: Proposition $3.10$
Category:Order of Group Elements
Equal Powers of Group Element implies Finite Order

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $g \in G$ such that $g^r = g^s$ where $r, s \in \Z: r \ne s$.
Then there exists $m \in \Z_{>0}$ such that:
:$(1): \quad g^m = e$
:$(2): \quad 0 \le i < j < m \implies g^i \ne g^j$
== Proof ==
, suppose that $r > s$.
From $g^r = g^s$ it follows from Powers of Group Elements that:
: $g^{r - s} = g^r g^{-s} = g^s g^{-s} = e$
Thus there exists $t \in \Z_{>0}$ such that $g^t = e$.
Let $m \in \Z_{>0}$ be the smallest such that $g^m = e$.
Suppose $0 \le i < j < m$ such that $g^i = g^j$.
Then $g^{j - i} = e$ and $0 < j - i < m$.
This contradicts the choice of $m$ as the smallest $m \in \Z_{>0}$ such that $g^m = e$.
Hence the result.
== Also see ==
* Definition:Order of Group Element
== Sources ==
* : $\S 1.4$: Theorem $3 \ (1)$
* : $0$: Some Conventions and some Basic Facts
Category:Order of Group Elements
Equal Relative Complements iff Equal Subsets

== Theorem ==
Let $S$ be a set.
Let $A, B \subseteq S$ be subsets of $S$.
Then:
:$\relcomp S B = \relcomp S A \iff A = B$
where $\complement_S$ denotes the complement relative to $S$.
== Proof ==
We have:
Category:Subsets
Category:Relative Complement
Equal Set Differences iff Equal Intersections

== Theorem ==
:$R \setminus S = R \setminus T \iff R \cap S = R \cap T$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \ \text{(c)}$
Category:Set Difference
Category:Set Intersection
Category:Equal Set Differences iff Equal Intersections
Equal Set Differences iff Equal Intersections/Proof 2

== Theorem ==
== Proof ==
From Set Difference and Intersection form Partition:
: $\paren {R \setminus S} \cup \paren {R \cap S} = R = \paren {R \setminus T} \cup \paren {R \cap T}$
: $\paren {R \cap S} \cap \paren {R \setminus S} = \O = \paren {R \cap T} \cap \paren {R \setminus T}$
whatever $R, S, T$ might be.
Let $R \setminus S = R \setminus T$.
Then:
Now, we have from Set Difference with Disjoint Set:
:$S \cap T = \O \iff S \setminus T = S$
and so:
:$\paren {R \cap S} \setminus \paren {R \setminus S} = R \cap S$
and:
:$\paren {R \cap T} \setminus \paren {R \setminus T} = R \cap T$
So:
:$R \cap S = R \cap T$
We can use exactly the same reasoning if we assume $R \cap S = R \cap T$:
and then because of Set Difference with Disjoint Set as above:
:$\paren {R \setminus S} \setminus \paren {R \cap S} = R \setminus S$
and:
:$\paren {R \setminus T} \setminus \paren {R \cap T} = R \setminus T$
So:
:$R \setminus S = R \setminus T$
Category:Equal Set Differences iff Equal Intersections
Equal Sized Triangles on Equal Base have Same Height

== Theorem ==
Triangles of equal area which are on equal bases, and on the same side of it, are also in the same parallels.
== Proof ==
:
Let $ABC$ and $CDE$ be equal-area triangles which are on equal bases $BC$ and $CD$, and on the same side.
Let $AE$ be joined.
Suppose $AE$ were not parallel to $BC$.
Then, by Construction of Parallel Line we draw $AF$ parallel to $BD$.
So by Triangles with Equal Base and Same Height have Equal Area, $\triangle ABC = \triangle FCD$.
But $\triangle ABC = \triangle DCE$, which means $\triangle FCD = \triangle DCE$.
But $\triangle DCE$ is bigger than $\triangle FCD$.
From this contradiction we deduce that $AF$ can not be parallel to $BD$.
In a similar way, we prove that no other line except $AE$ can be parallel to $BD$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Triangles
Equal Sized Triangles on Same Base have Same Height

== Theorem ==
Triangles of equal area which are on the same base, and on the same side of it, are also in the same parallels.
== Proof ==
:
Let $ABC$ and $DBC$ be equal-area triangles which are on the same base $BC$ and on the same side as it.
Let $AD$ be joined.
Suppose $AD$ were not parallel to $BC$.
Then, by Construction of Parallel Line we draw $AE$ parallel to $BC$.
So by Triangles with Same Base and Same Height have Equal Area:
: $\triangle ABC = \triangle EBC$
But $\triangle ABC = \triangle DBC$, which means:
: $\triangle DBC = \triangle EBC$
But $\triangle DBC$ is bigger than $\triangle EBC$.
From this contradiction we deduce that $AE$ can not be parallel to $BC$.
In a similar way, we prove that no other line except $AD$ can be parallel to $BC$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Triangles
Equal Surfaces do not Intersect

== Theorem ==
Let $R$ be a region of space, which may be the interior of a body.
Let there exist a point-function $F$ on $R$ giving rise to a scalar field.
Let $S_1$ and $S_2$ be equal surfaces in $R$ upon which the value of $F$ on $S_1$ is different from the value of $F$ on $S_2$.
Then $S_1$ and $S_2$ do not intersect.
== Proof ==
Let:
:$\forall p \in S_1: \map F p = C_1$
:$\forall p \in S_2: \map F p = C_2$
By hypothesis, $C_1 \ne C_2$.
 there exists a point $P$ in $R$ such that both $P \in S_1$ and $P \in S_2$.
Then $\map F p = C_1$ and also $\map F p = C_2$.
This contradicts the fact that $F$ is a function.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $\text I$: Definitions. Elements of Vector Algebra: $5$. Scalar and Vector Fields
Category:Equal Surfaces