nyuuzyou/wikis huggingface dataset:
Point in Convex Set is Extreme Point iff Singleton is Extreme Set

== Theorem ==
Let $X$ be a vector space over $\R$. 
Let $K$ be a convex subset of $X$. 
Let $a \in K$.
Then $a$ is an extreme point of $K$ :
:$\set a$ is an extreme set in $K$.
== Proof ==
We have: 
:$a$ is an extreme point of $K$.
: 
:whenever $a = t x + \paren {1 - t} y$ for some $x, y \in K$ and $t \in \openint 0 1$, we have $x = y = a$.
We can rewrite this: 
:whenever $t x + \paren {1 - t} y \in \set a$ for some $x, y \in K$ and $t \in \openint 0 1$, we have $x, y \in \set a$.
This is precisely the criteria for $\set a$ being an extreme set in $K$.
== Sources ==
* $21.6$: The Krein-Milman Theorem
Category:Extreme Sets
Category:Extreme Points of Convex Sets
Point in Discrete Space is Adherent Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $U \subseteq S$.
Then $x$ is an adherent point of $U$ $x \in U$.
== Proof ==
Let $U \subseteq S$.
From Set in Discrete Topology is Clopen it follows that $U$ is open in $T$.
Let $x \in U$.
Then $U$ is an open neighborhood of $x$ containing (trivially) an element of $U$, that is, $x$.
So, by definition, $x$ is an adherent point of $U$.
Now suppose $x \notin U$.
Then $\set x$ is an open neighborhood of $x$ such that $\set x \cap U = \O$.
So $x$ can not be an adherent point of $U$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $3$
Category:Discrete Topology
Category:Adherent Points
Point in Discrete Space is Neighborhood

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $x \in S$.
Then $\set x$ is a neighborhood of $x$ in $T$.
== Proof ==
By definition, a neighborhood $N_x$ of $x$ is any subset of $S$ containing an open set which itself contains $x$.
That is:
:$\exists U \in \tau: x \in U \subseteq N_x \subseteq S$
From Set in Discrete Topology is Clopen we have that $\set x$ is open set in $S$.
So by Set is Subset of Itself, $\set x$ is a subset of $S$ containing an open set $\set x$ which itself contains $x$.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $7$
Category:Discrete Topology
Category:Neighborhoods
Point in Finite Hausdorff Space is Isolated

== Theorem ==
Let $T = \struct {S, \tau}$ be a Hausdorff space.
Let $X \subseteq S$ such that $X$ is finite.
Let $x \in X$.
Then $x$ is isolated in $X$.
== Proof 1 ==
== Proof 2 ==
Category:Finite Topological Spaces
Category:Isolated Points
Category:Hausdorff Spaces
Category:Point in Finite Hausdorff Space is Isolated
Point in Finite Hausdorff Space is Isolated/Proof 1

== Theorem ==
== Proof ==
As $X$ is finite, its elements can be placed in one-to-one correspondence with the elements of $\set {1, 2, \ldots, n}$ for some $n \in \N$.
So let $X = \set {x_1, x_2, \ldots, x_n}$.
From the definition of Hausdorff space:
: $\forall x_i, x_j \in X: x_i \ne x_j: \exists U, V \in \tau: x_i \in U, x_j \in V: U \cap V = \O$
Let $x_k \in X$.
Then by definition:
:$\forall x_i \in X, x_i \ne x_k: \exists U_i, V \in \tau: x_k \in U_i, x_i \in V: U_i \cap V = \O$
Note that $U_i$ may be different for each $X_i$.
That is:
:$\forall x_i \in X: \exists U_i \in \tau: x_k \in U_i, x_i \notin U_i$
Now consider:
:$\ds U = \bigcap_{i \mathop \ne k} U_i$
We have by definition of a topology that $U \in \tau$, as the intersection $U$ is of a finite number of sets $U_i$.
We have that:
: $\forall i: x_k \in U_i$
: $\forall i: x_i \notin U_i$
So:
: $U \cap X = \set {x_k}$
and so $U$ is an open set of $T$ which contains no points of $X$ other than $x_k$.
So by definition, $x_k$ is an isolated point of $X$.
Hence the result.
Category:Point in Finite Hausdorff Space is Isolated
Point in Finite Hausdorff Space is Isolated/Proof 2

== Theorem ==
== Proof ==
Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Let $X \subseteq T$ be finite.
From Separation Properties Preserved in Subspace, it follows that $\struct {X, \tau_X}$ is also a $T_2$ (Hausdorff) space.
From $T_2$ Space is $T_1$ Space it follows that $\struct {X, \tau_X}$ is a $T_1$ (Fréchet) space.
From Finite $T_1$ Space is Discrete, it follows that $\struct {X, \tau_X}$ is a discrete space.
The result follows from Topological Space is Discrete iff All Points are Isolated.
Category:Point in Finite Hausdorff Space is Isolated
Point in Finite Metric Space is Isolated

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $X \subseteq A$ such that $X$ is finite.
Let $x \in X$.
Then $x$ is isolated in $X$.
== Proof 1 ==
== Proof 2 ==
Category:Isolated Points
Category:Finite Metric Spaces
Category:Point in Finite Metric Space is Isolated
Point in Finite Metric Space is Isolated/Proof 1

== Theorem ==
== Proof ==
As $X$ is finite, its elements can be placed in one-to-one correspondence with the elements of $\N^*_n$ for some $n \in \N$.
So let $X = \left\{{x_1, x_2, \ldots, x_n}\right\}$.
Now let $\epsilon := \min \left\{{\forall i, j \in \N^*_n: i \ne j: d \left({x_i, x_j}\right)}\right\}$.
That is, $\epsilon$ is the minimum distance between any two elements of $X$.
We have that $\epsilon > 0$ from the definition of metric.
Let $N_\epsilon \left({x_k}\right)$ be the open $\epsilon$-ball of any $x_k \in X$.
By the method of construction of $\epsilon$, it is clear that $N_\epsilon \left({x_k}\right) = \left\{{x_k}\right\}$.
Hence $x_k$ is an isolated point.
Category:Point in Finite Metric Space is Isolated
Point in Hilbert Sequence Space has no Compact Neighborhood

== Theorem ==
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} x_i^2$ is convergent.
Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$.
Then no point of $\ell^2$ has a compact neighborhood.
== Proof ==
From Hilbert Sequence Space is Metric Space, $\ell^2$ is a metric space.
Let $x = \sequence {x_i} \in A$ be a point of $\ell^2$.
Consider the closed $\epsilon$-ball of $x$ in $\ell^2$:
:$\map { {B_\epsilon}^-} x := \set {y \in A: \map {d_2} {x, y} \le \epsilon}$
for some $\epsilon \in \R_{>0}$.
Consider the point:
:$\sequence {y_n} = \tuple {x_1, x_2, \ldots, x_{n - 1}, x_n + \epsilon, x_{n + 1}, \ldots}$
We have that $\sequence {y_n} \in \map { {B_\epsilon}^-} x$.
But for $m \ne n$ we have that:
:$\map {d_2} {y_m, y_n} = \epsilon \sqrt 2$
and so $\sequence {y_n}$ has no convergent subsequence.
Thus $\map { {B_\epsilon}^-} x$ is not compact.
So $x$ has no compact neighborhood.
== Sources ==
* : Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $3$
Category:Hilbert Sequence Space
Point in Metric Space has Countable Neighborhood Basis

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Then there exists a basis for the neighborhood system of $a$ which is countable.
== Proof ==
Consider the countable set:
:$\BB_a := \set {\map {B_\epsilon} a: \exists n \in \Z_{>0}: \epsilon = \dfrac 1 n}$
That is, let $\BB_a$ be the set of all open $\epsilon$-balls of $a$ such that $\epsilon$ is of the form $\dfrac 1 n$ for (strictly) positive integral $n$.
Let $N$ be a neighborhood of $a$.
Then by definition of neighborhood:
:$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$
From Between two Real Numbers exists Rational Number:
:$\exists \epsilon \in \Q: 0 < \epsilon < \epsilon'$
Let $\epsilon$ be expressed in canonical form as:
:$\epsilon = \dfrac p q$
where $p$ and $q$ are coprime integers and $q > 0$.
Then:
:$\epsilon := \dfrac 1 q \le \dfrac p q$
Thus $0 < \epsilon < \epsilon'$.
So:
From Open Ball is Neighborhood of all Points Inside, $\map {B_\epsilon} a$ is a neighborhood of $a$.
But as $\epsilon = \dfrac 1 q$ where $q \in \Z_{>0}$ it follows that:
:$\map {B_\epsilon} a \in \set {\map {B_\epsilon} a: \exists n \in \Z_{>0}: \epsilon = \dfrac 1 n}$
Hence the result.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $5$
Category:Neighborhoods
Point in Metric Space has Neighborhood

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$ be a point in $M$.
Then there exists some neighborhood of $a$ in $M$.
== Proof ==
Let $a \in A$.
Then $A$ is a neighborhood of $a$ in $M$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.8: \ N \, 1$
Category:Neighborhoods
Point in Metric Space is Element of its Neighborhood

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$ be a point in $M$.
Let $N$ be a neighborhood of $a$ in $M$.
Then $a \in N$.
== Proof ==
Trivially follows by definition of neighborhood of $a$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.8: \ N \, 2$
Category:Neighborhoods
Point in Particular Point Space is not Omega-Accumulation Point

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $x \in S$ such that $x \ne p$.
Let $H \subseteq S$ such that $p \in H$.
Then $x$ is not an $\omega$-accumulation point of $H$.
== Proof ==
Let $x \in S, x \ne p$.
By Limit Points in Particular Point Space, $x$ is a ‎limit point of $H$.
Consider the set $U = \set {x, p} \subseteq S$.
By definition of the particular point topology, $U$ is open in $T$.
But as $U$ contains only $x$ and $p$, it is clear that $U$ does not contain an infinite number of points of $H$.
Hence, by definition, $x$ can not be an $\omega$-accumulation point of $H$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $3$
Category:Omega-Accumulation Points
Category:Particular Point Topology
Point in Plane is Constructible iff Coordinates in Extension of Degree Power of 2

== Theorem ==
Let $\CC$ be a Cartesian plane.
Let $S$ be a set of points in $\CC$.
Let $F$ be the smallest field containing all the coordinates of the points in $S$.
Let $P = \tuple {a, b}$ be a point in $\CC$.
Then:
:$P$ is constructible from $S$ using a compass and straightedge construction
:
:the coordinates of $P$ are contained in a finite extension $K$ of $F$ whose degree is given by:
::$\index K F = 2^m$
:for some $m \in \Z_{\ge 0}$.
== Proof ==
A point $P$ is constructed in a compass and straightedge construction from one of $3$ basic operations:
:$(1): \quad$ the intersection of $2$ straight lines
:$(2): \quad$ the intersection of a straight line and the circumference of a circle
:$(3): \quad$ the intersection of the circumferences of $2$ circle.
Let $A$, $B$, $C$ and $D$ be points in $S$.
In case $(1)$, the equations defining the straight lines $AB$ and $CD$ are polynomials of degree $1$.
Hence the coordinates of $P$ can be found by solving the linear simultaneous equations defining $AB$ and $CD$.
It follows that the solution is in $F$.
In case $(2)$:
:the equation defining the circle whose center is at $A$ and whose radius is $AB$ is of the form $x^2 + y^2 + 2 f x + 2 g y + c = 0$, where $c, f, g \in F$.
while:
:the equation defining the straight line $CD$ is a polynomial of degree $1$.
The $x$-coordinate of $P$ is then the root of a quadratic equation with coefficients in $F$.
The $y$-coordinate of $P$ is then found by solving a linear equation.
Thus the coordinates of $P$ are elements of either a field extension of $F$ of degree $2$ or of $F$ itself.
In case $(3)$:
:the equation defining the circle whose center is at $A$ and whose radius is $AB$ is of the form $x^2 + y^2 + 2 f x + 2 g y + c = 0$, where $c, f, g \in F$
:the equation defining the circle whose center is at $C$ and whose radius is $CD$ is of the form $x^2 + y^2 + 2 f' x + 2 g' y + c' = 0$, where $c', f', g' \in F$.
The solution of these equations is the same as the solution of either circle and the straight line $2 \paren {f - f'} x + 2 \paren {g - g'} y + \paren {c - c'} = 0$, which is the same as case $(2)$.
We have that the coordinates of the various points belong to the field $F = F_0$, say.
Suppose $P$ is constructed using $k$ of the basic constructions $(1)$, $(2)$ and $(3)$.
Let $F_i$ be the smallest field containing all the points so far obtained by $i$ such constructions, for $i = 1, 2, \ldots, k$.
Either $F_i = F_{i - 1}$ or $F_i$ is an finite extension of $F_i$ of degree $2$.
Thus for each $i$, either:
:$\index {F_i} {F_{i - 1} } = 1$
or:
:$\index {F_i} {F_{i - 1} } = 2$
Therefore:
:$\index {F_k} {F_0} = \index {F_k} {F_{k - 1} } \index {F_{k - 1} } {F_{k - 2} } \dotsm {F_1} {F_0} = 2^m$
where $0 \le m \le k$.
Hence the result.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 40$. Construction with Ruler and Compasses: Theorem $75$
Category:Analytic Geometry
Category:Compass and Straightedge Constructions
Point in Set Closure iff Limit of Net

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $E \subseteq X$ be a subset. 
Let $\map \cl E$ be the closure of $E$ in $\struct {X, \tau}$.
Then $x \in \map \cl E$ there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ in $E$ converging to $x$. 
== Proof ==
=== Necessary Condition ===
Let $x \in \map \cl E$.
Let $\Lambda$ be the set of open neighborhoods of $x$ in $\struct {X, \tau}$. 
From Open Neighborhoods of Point form Directed Ordering, $\struct {\Lambda, \supseteq}$ is a directed set.
For each $U \in \Lambda$, we have $U \cap E \ne \O$ from the definition of closure.
So for each $U \in \Lambda$ we can pick $x_U \in U \cap E$, forming a net $\family {x_U}_{U \in \Lambda}$ valued in $E$.
We show that $\family {x_U}_{U \in \Lambda}$ converges to $x$.
Let $V$ be an open neighborhood of $x$ in $\struct {X, \tau}$.
Then if $U \supseteq V$, we have $x_V \in V \cap E \subseteq U \cap E \subseteq U$.
Hence $\family {x_U}_{U \in \Lambda}$ converges to $x$.
=== Sufficient Condition ===
Let $x \in X$.
Suppose that there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ in $E$ converging to $x$.
Let $U$ be an open neighborhood of $x$ in $\struct {X, \tau}$.
Then since $\family {x_\lambda}_{\lambda \in \Lambda}$ converges, there exists $\lambda_0 \in \Lambda$ such that: 
:$x_\lambda \in U$ for all $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$.
In particular, $x_{\lambda_0} \in U$, hence $x_{\lambda_0} \in U \cap E$.
So $U \cap E \ne \O$ for each open neighborhood $U$ of $x$.
So $x \in \map \cl E$ from the definition of closure.
Category:Nets (Set Theory)
Category:Set Closures
Point in Standard Discrete Metric Space is Isolated

== Theorem ==
Let $M = \struct {S, d}$ be the standard discrete metric space on a set $A$.
Let $H \subseteq S$ be a subset of $S$.
Let $\alpha \in H$.
The $\alpha$ is an isolated point of $H$.
== Proof ==
By definition of the standard discrete metric:
:$\map d {x, y} = \begin {cases} 0 & : x = y \\ 1 & : x \ne y \end {cases}$
Let $\alpha \in H$.
Let $\map {B_1} \alpha$ be the open $1$-ball of $\alpha$ in $M$.
Thus:
Hence the result by definition of isolated point.
Category:Standard Discrete Metric
Category:Isolated Points
Point in Topological Space has Neighborhood

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Then there exists in $T$ at least one neighborhood of $x$.
That is:
:$\forall x \in S: \NN_x \ne \O$
where $\NN_x$ is the neighborhood filter of $x$.
== Proof ==
Let $x \in S$.
Then $S$ itself is a neighborhood of $x$.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Theorem $3.1: \ N 1$
Category:Neighborhoods
Point in Topological Space is Element of its Neighborhood

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $N$ be a neighborhood of $x$ in $T$.
Then $a \in N$.
That is:
:$\forall x \in S: \forall N \in \NN_x: x \in N$
where $\NN_x$ is the neighborhood filter of $x$.
== Proof ==
Trivially follows by definition of neighborhood of $a$.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Theorem $3.1: \ N2$
Category:Neighborhoods
Point in Topological Space is Open iff Isolated

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $x \in S$.
Then $\left\{{x}\right\}$ is open in $T$ $x$ is an isolated point of $T$.
== Proof ==
Let $\left\{{x}\right\}$ be open in $T$.
Then we have that:
:$\exists \left\{{x}\right\} \in \tau: x \in \left\{{x}\right\}\subseteq S$
This is precisely the condition which ensures that $x$ is an isolated point of $T$.
Now suppose that $x$ is an isolated point of $T$.
Then by definition there exists a open set of $T$ containing no points other than $x$:
:$\exists U \in \tau: U = \left\{{x}\right\}$
That is, if $x$ is an isolated point of $T$ then $\left\{{x}\right\}$ is open in $T$.
Category:Open Sets
Category:Isolated Points
Point is Contained in Irreducible Component

== Theorem ==
Let $X$ be a topological space.
Let $x\in X$ be a point.
Then $x$ is contained in some irreducible component of $X$.
== Proof ==
Because Trivial Topological Space is Irreducible, this is a special case of Irreducible Subspace is Contained in Irreducible Component.
Category:Irreducible Spaces
Point is Isolated iff belongs to Set less Derivative

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $H \subseteq S$.
Let $x \in S$.
Then:
:$x$ is an isolated point in $H$
:
:$x \in H \setminus H'$
where
:$H'$ denotes the derivative of $H$.
== Proof ==
$x$ is an isolated point in $H$
$\iff$ $x \in H$ and $x$ is not an accumulation point of $H$ by Point is Isolated iff not Accumulation Point
$\iff$ $x \in H$ and $x \notin H'$ by definition of derivative
$\iff$ $x \in H \setminus H'$ by definition of set difference.
== Sources ==
* 
* 
Category:Isolated Points
Category:Set Derivatives
Point is Isolated iff not Accumulation Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $x \in H$.
Then:
:$x$ is an isolated point in $H$
:
:$x$ is not an accumulation point of $H$
== Proof ==
=== Sufficient Condition ===
Let $x \in H$ be an isolated point in $H$.
Then by definition of isolated point:
:$\exists U \in \tau: H \cap U = \set x$
That is, by definition of uniqueness:
:$\lnot \forall U \in \tau: \paren {x \in U \implies \exists y \in S: \paren {y \in H \cap U \land x \ne y} }$
Hence by Characterization of Derivative by Open Sets:
:$x \notin A'$
where $A'$ denotes the derivative of $A$.
Thus by definition of derivative:
:$x$ is not an accumulation point of $H$.
=== Necessary Condition ===
Let $x \in H$ not be an accumulation point of $H$.
Thus by definition of derivative:
:$x \notin A'$
Hence:
Thus by definition of isolated point:
:$x$ is an isolated point in $H$.
Category:Isolated Points
Category:Accumulation Points
Point is Path-Connected to Itself

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $a \in S$.
Then $a$ is path-connected to itself.
== Proof ==
Consider the constant mapping on the closed unit interval $\mathbb I = \left[{0 \,.\,.\, 1}\right]$:
:$\forall x \in \mathbb I: f_a \left({x}\right) = a$
Thus, in particular:
:$f_a \left({0}\right) = a$
:$f_a \left({1}\right) = a$
As a Constant Mapping is Continuous, it follows that $f_a$ is a path in $X$.
Thus $a$ is path-connected to itself.
Category:Path-Connected Spaces
Point not in Subset of Metric Space iff Distance from Elements is Greater than Zero

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be an arbitrary subset of $A$.
Let $x \in A$ be arbitrary.
Then:
:$x \notin H$
:
:$\forall y \in H: \map d {x, y} > 0$
== Proof ==
Category:Distance Function
Point of Intersection with Line at Infinity

== Definition ==
Let $\LL$ be a straight line embedded in a cartesian plane $\CC$ given in homogeneous Cartesian coordinates by the equation:
:$l X + m Y + n_1 Z = 0$
Then $\LL$ intersects the line at infinity at the point:
:$\tuple {-m, l, 0}$
== Proof ==
By definition, the line at infinity is the line whose equation in homogeneous Cartesian coordinates is:
:$n_2 Z = 0$
for some $n_2$.
Let $\tuple {X, Y, Z}$ be the intersection of $\LL$ with the line at infinity.
From Intersection of Straight Lines in Homogeneous Cartesian Coordinate Form:
:$\tuple {X, Y, Z} = \tuple {m n_2 - 0 n_1, n_1 0 - l n_2, l 0 - 0 m}$
from which the result follows by dividing the homogeneous Cartesian coordinates by $-n2$.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $9$. Parallel lines. Points at infinity
Category:Point at Infinity
Point of Perpendicular Intersection on Real Line from Points in Complex Plane

== Theorem ==
Let $a, b \in \C$ be complex numbers represented by the points $A$ and $B$ respectively in the complex plane.
Let $x \in \R$ be a real number represented by the point $X$ on the real axis such that $AXB$ is a right triangle with $X$ as the right angle.
Then:
:$x = \dfrac {a_x - b_x \pm \sqrt {a_x^2 + b_x^2 + 2 a_x b_x - 4 a_y b_y} } 2$
where:
: $a = a_x + a_y i, b = b_x + b_y i$
== Proof ==
From Geometrical Interpretation of Complex Subtraction, the lines $XA$ and $XB$ can be represented by the complex numbers $a - x$ and $b - x$.
:
From Multiplication by Imaginary Unit is Equivalent to Rotation through Right Angle $a - x$ and $b - x$ are perpendicular either:
:$a - x = r i \paren {b - x}$
for some real numbers $r \in \R$.
That is, $\dfrac {a - x} {b - x}$ are purely imaginary.
Thus:
== Examples ==
Category:Geometry of Complex Plane
Category:Point of Perpendicular Intersection on Real Line from Points in Complex Plane
Point of Perpendicular Intersection on Real Line from Points in Complex Plane/Examples/a = 1-3i, b = -3+4i

== Examples of Point of Perpendicular Intersection on Real Line from Points in Complex Plane ==
Let $a, b \in \C$ be complex numbers represented by the points $A$ and $B$ respectively in the complex plane.
Let $x \in \R$ be a real number represented by the point $X$ on the real axis such that $AXB$ is a right triangle with $X$ as the right angle.
Let $a = 1 - 3 i, b = -3 + 4 i$.
The point $X$ on the positive half of the real axis is at:
:$x = 3$
== Proof ==
From Point of Perpendicular Intersection on Real Line from Points in Complex Plane:
:$x = \dfrac {a_x + b_x \pm \sqrt {a_x^2 + b_x^2 - 2 a_x b_x - 4 a_y b_y} } 2$
Setting $a = 1 - 3 i, b = -3 + 4 i$:
It is the positive solution that is needed, so:
:$x = 3$
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $1$.
Category:Point of Perpendicular Intersection on Real Line from Points in Complex Plane
Point with Zero Second Derivative is not necessarily Point of Inflection

== Theorem ==
Let $f$ be a real function which is twice differentiable on the open interval $\openint a b$.
Let
:$\map {f' '} \xi = 0$
where $\map {f' '} \xi$ denotes the second derivative of $f$ at $\xi \in \openint a b$.
Then it is not necessarily the case that $f$ has a point of inflection at $\xi$.
== Proof ==
Consider the function:
:$f: \R \to \R: \forall x \in \R: \map f x = x^4$
This has a local minimum at $x = 0$ at which $\map {f' '} x = 0$.
But $x = 0$ is not a point of inflection of $f$.
== Also see ==
* Second Derivative at Point of Inflection
== Sources ==
* 
* 
* 
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Category:Points of Inflection
Category:Differential Calculus
Points Defined by Adjacent Pairs of Digits of Reciprocal of 13 lie on Hyperbola

== Theorem ==
Consider the digits that form the recurring part of the reciprocal of $13$:
:$\dfrac 1 {13} = 0 \cdotp \dot 07692 \dot 3$
Take the digits in ordered pairs, and treat them as coordinates of a Cartesian plane.
It will be found that they all lie on a hyperbola:
:
== Proof ==
:
Let the points be labelled to simplify:
:$A := \left({0, 7}\right)$
:$B := \left({7, 6}\right)$
:$C := \left({6, 9}\right)$
:$D := \left({9, 2}\right)$
:$E := \left({2, 3}\right)$
:$F := \left({3, 0}\right)$
== Sources ==
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* : $142,857$
Category:Hyperbolas
Category:13