nyuuzyou/wikis huggingface dataset:
Three Points in Ultrametric Space have Two Equal Distances/Corollary 4

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm{\,\cdot\,}$,
Let $x, y \in R$.
Then:
:* $\norm {x + y} \lt \norm y \implies \norm x = \norm y$
:* $\norm {x - y} \lt \norm y \implies \norm x = \norm y$
:* $\norm {y - x} \lt \norm y \implies \norm x = \norm y$
== Proof ==
The contrapositive statements are proved.
Let $\norm x \ne \norm y$
By Corollary 2 then:
:$\norm {x + y} = \norm {x - y} = \norm {y - x} = \max \set {\norm x, \norm y} \ge \norm y$
The result follows.
== Sources ==
* : $\S 1.2$ Normed Fields: Proposition $1.15$
Category:Normed Division Rings
Three Points in Ultrametric Space have Two Equal Distances/Corollary 5

== Theorem ==
Let $\norm {\, \cdot \,}$ be a non-trivial non-Archimedean norm on the rational numbers $\Q$.
Let $a, b \in \Z_{\ne 0}$ be coprime:
:$a \perp b$
Then:
:$\norm a = 1$ or $\norm b = 1$
== Proof ==
By Bézout's Identity then:
:$\exists n, m \in \Z : m a + n b = 1$ 
By Norm of Unity:
:$\norm {m a + n b} = 1$
By :
:$\norm a, \norm b, \norm n, \norm m \le 1$
Let $\norm a < 1$.
By :
:$\norm {m a} = \norm m \norm a < 1$
Hence:
:$\norm {m a} < \norm {m a + n b}$
By :
:$\norm {n b} = \norm {m a + n b} = 1$
By :
:$\norm n \norm b = 1$
Hence:
:$\norm b = 1$
The result follows.
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Theorem $3.1.3$
* : $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski’s Theorem, Theorem $1.50$
Category:Normed Division Rings
Three Points on Sphere in Same Hemisphere

== Theorem ==
Let $S$ be a sphere.
Let $A$, $B$ and $C$ be points on $S$ which do not all lie on the same great circle.
Then it is possible to divide $S$ into two hemispheres such that $A$, $B$ and $C$ all lie on the same hemisphere.
== Proof ==
Because $A$, $B$ and $C$ do not lie on the same great circle, no two of these points are the endpoints of the same diameter of $S$.
Otherwise it would be possible to construct a great circle passing through all $3$ points $A$, $B$ and $C$.
Let a great circle $E$ be constructed through $A$ and $B$.
Then as $C$ is not on $E$, it is on either one side or the other.
Hence there is a finite spherical angle $\phi$ between $E$ and $C$.
Let a diameter $PQ$ of $S$ be constructed whose endpoints are on $E$ such that:
:neither $P$ nor $Q$ coincides with $A$ or $B$
:both $A$ and $B$ are on the same semicircle of $E$ into which $PQ$ divides $E$.
Let two great circles $F$ and $G$ be constructed through $PQ$ which are at a spherical angle $\dfrac \phi 2$ from $E$, one in one direction and one in the other.
Then $F$ and $G$ both divide $S$ into two hemispheres:
:one such division has a hemisphere which contains $A$ and $B$ but does not contain $E$
:the other such division has a hemisphere which contains $A$, $B$ and $E$.
Hence the result.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $2$. The spherical triangle.
Category:Spherical Geometry
Three Regular Tessellations

== Theorem ==
There exist exactly $3$ regular tessellations of the plane.
=== Triangles ===
=== Squares ===
=== Hexagons ===
== Proof ==
Let $m$ be the number of sides of each of the regular polygons that form the regular tessellation.
Let $n$ be the number of those regular polygons which meet at each vertex.
From Internal Angles of Regular Polygon, the internal angles of each polygon measure $\dfrac {\paren {m - 2} 180 \degrees} m$.
The sum of the internal angles at a point is equal to $360 \degrees$ by Sum of Angles between Straight Lines at Point form Four Right Angles
So:
But $m$ and $n$ are both greater than $2$.
So:
:if $m = 3$, $n = 6$.
:if $m = 4$, $n = 4$.
:if $m = 5$, $n = \dfrac {10} 3$, which is not an integer.
:if $m = 6$, $n = 3$.
Now suppose $m > 6$.
We have:
But there are no integers between $2$ and $3$, so $m \not > 6$.
There are $3$ possibilities in all.
Therefore all regular tessellations have been accounted for.
== Also see ==
* Five Platonic Solids
== Sources ==
* : $3$
* : $3$
* 
* 
* 
Category:Three Regular Tessellations
Category:Regular Tessellations
Category:3
Three Regular Tessellations/Hexagons

== Theorem ==
Regular hexagons form a regular tessellation:
:
== Proof ==
Let $s \in \R_{>0}$ be the side length of the regular hexagons.
For all $x, y \in \Z$, let the center $Q_{x,y}$ of each regular hexagon have Cartesian coordinates:
:$Q_{x,y} = s \tuple {x, \sqrt 3 y + \dfrac {\sqrt 3 \paren {3 - m} } 4}$
where:
:$m = \begin{cases} 1 & : \textrm {for $x$ even} \\ -1 & : \textrm{for $x$ odd} \end{cases}$
From Regular Hexagon is composed of Equilateral Triangles, it follows that each hexagon can be triangulated into six congruent equilateral triangles.
Also, each of the six equilateral triangles has side length $s$, and $Q_{x,y}$ as one of its vertices.
By Three Regular Tessellations:Triangles, there exists a regular tessellation of equilateral triangles with side length s.
Set:
:$i := 3 x$
:$j := 2 y + \dfrac {1-m} 2$
The hexagon center $Q_{x,y}$ corresponds to the six vertices:
:$A_{i+1, j}, A_{i+1, j+1}, B_{i-1,j}, B_{i-1,j+1}, C_{i,j}, C_{i,j+1} $
as they are labelled in the proof of Three Regular Tessellations:Triangles.
The six triangles in the triangulation of the hexagon correspond to the triangles of the regular tessellation:
:$\triangle A_{i+1, j} B_{i+1, j} C_{i+1, j}$
:$\triangle A_{i+1, j+1} B_{i+1, j+1} C_{i+1, j+1} $
:$\triangle A_{i-1, j} B_{i-1, j} C_{i-1, j} $
:$\triangle A_{i-1, j+1} B_{i-1, j+1} C_{i-1, j+1} $
:$\triangle A_{i, j} B_{i, j} C_{i, j}$
:$\triangle A_{i, j+1} B_{i, j+1} C_{i, j+1} $
as they are labelled in the proof of Three Regular Tessellations:Triangles.
For all $x' \in \Z$, there exist exactly one $x \in \Z$ and one $k \in \set {-1, 0, 1}$ such that:
:$x' = 3 x + k$
For all $y' \in \Z$, there exist exactly one $y \in \Z$ and one $k \in \set {0, 1}$ such that:
:$y' = 2 y + k$
which shows that each $\triangle A_{x', y'} B_{x', y'} C_{x', y'}$ is part of exactly one of the hexagons.
As the triangles form a regular tessellation of the plane, it follows that the hexagons also form a regular tessellation of the plane.
== Sources ==
* : $6$
* : $6$
Category:Three Regular Tessellations
Category:Hexagons
Three Regular Tessellations/Triangles

== Theorem ==
Equilateral triangles form a regular tessellation:
:
== Proof ==
Let $s \in \R_{>0}$ be the side length of the equilateral triangles.
For all $x, y \in \Z$, let the center $P_{x, y}$ of each triangle have Cartesian coordinates:
:$P_{x, y} = s \tuple {\dfrac 1 2 x, \dfrac {\sqrt 3} 2 y + \dfrac {3 - m} {4 \sqrt 3} }$
where $m = \begin{cases} 1 & \textrm {for $x + y$ even} \\ -1 & \textrm {for $x + y$ odd} \end{cases}$.
Let each triangle be defined as $\triangle A_{x, y} B_{x, y} C_{x, y}$, where the coordinates of its vertices are:
We show that $\triangle A_{x, y} B_{x, y} C_{x, y}$ is an equilateral triangle.
Translate the plane by the translation mapping $\tau: \R^2 \to \R^2$, defined by:
:$\map \tau {\mathbf p} = \mathbf p - P_{x, y}$
so the translation of the triangle center $\map \tau {P_{x, y} }$ is equal to the origin $\tuple {0, 0}$.
From Translation Mapping is Isometry, it follows that this does not change the length of the sides of $\triangle A_{x, y} B_{x, y} C_{x, y}$.
By the Distance Formula, the length of each side is:
From Triangle Side-Side-Side Congruence, it follows that equilateral triangles with the same length of their sides are congruent.
By direct calculations, it follows that each vertex of $\triangle A_{x, y} B_{x, y} C_{x, y}$ is identical to five other vertices:
:$A_{x, y} = A_{x, y - m} = B_{x - 2, y} = B_{x - 2, y - m} = C_{x - 1, y} = C_{x - 1, y - m}$
:$B_{x, y} = B_{x, y - m} = A_{x + 2, y} = A_{x - 2, y - m} = C_{x + 1, y} = C_{x + 1, y - m}$
:$C_{x, y} = C_{x, y + m} = A_{x + 1, y} = A_{x + 1, y + m} = B_{x - 1, y} = B_{x - 1, y + m}$
which shows that each side of $\triangle A_{x, y} B_{x, y} C_{x, y}$ is identical to one other side:
:$A_{x, y} B_{x, y} = A_{x, y - m} B_{x, y - m}$
:$B_{x, y} C_{x, y} = C_{x + 1, y} A_{x + 1, y}$
:$C_{x, y} A_{x, y} = B_{x - 1, y} C_{x - 1, y}$
Hence, the triangles $\sequence {\triangle A_{x, y} B_{x, y} C_{x, y} }_{x \mathop \in \Z, y \mathop \in \Z}$ form a regular tesselation of the plane.
Category:Three Regular Tessellations
Category:Equilateral Triangles
Three Regular n-Dimensional Polytopes over 4 Dimensions

== Theorem ==
Let $n \in \N$ be a natural number such that $n > 4$.
There exist exactly $3$ $n$-dimensional regular polytopes:
:the $n$-simplex
:the $n$-dimensional regular polytope corresponding to the cube
:the $n$-dimensional regular polytope corresponding to the octohedron.
== Proof ==
== Sources ==
* 
* 
Category:Polytopes
Category:3
Three Times Sum of Cubes of Three Indeterminates Plus 6 Times their Product

== Theorem ==
:$3 \paren {a^3 + b^3 + c^3 + 6 a b c} = \paren {a + b + c}^3 + \paren {a + b \omega + c \omega^2}^3 + \paren {a + b \omega^2 + c \omega}^3$
where:
: $\omega = -\dfrac 1 2 + \dfrac {\sqrt 3} 2$
== Proof ==
Multiplying out:
Replacing $b$ with $b \omega$ and $c$ with $c \omega^2$ in $(1)$:
Replacing $b$ with $b \omega^2$ and $c$ with $c \omega$ in $(1)$:
Adding together $(1)$, $(2)$ and $(3)$:
== Sources ==
* : $\S 3$. Roots of Unity: Exercise $5 \ \text {(ii)}$
Category:Cube Roots of Unity
Category:Algebra
Three Tri-Automorphic Numbers for each Number of Digits

== Theorem ==
Let $d \in \Z_{>0}$ be a (strictly) positive integer.
Then there exist exactly $3$ tri-automorphic numbers with exactly $d$ digits.
These tri-automorphic numbers all end in $2$, $5$ or $7$.
== Proof ==
Let $n$ be a tri-automorphic number with $d$ digits.
Let $n = 10 a + b$.
Then:
:$3 n^2 = 300a^2 + 60 a b + 3 b^2$
As $n$ is tri-automorphic, we have:
:$(1): \quad 300 a^2 + 60 a b + 3 b^2 = 1000 z + 100 y + 10 a + b$
and:
:$(2): \quad 3 b^2 - b = 10 x$
where $x$ is an integer.
This condition is only satisfied by $b = 2$, $b = 5$, or $b = 7$
Substituting $b = 2$ in equation $(1)$:
:$a = 9$
Substituting $b = 5$ in equation $(1)$:
:$a = 7$
Substituting $b = 7$ in equation $(1)$:
:$a = 6$
== Examples ==
== Sources ==
* : $6667$
* : $6667$
Category:Tri-Automorphic Numbers
Three Tri-Automorphic Numbers for each Number of Digits/Examples/10 Digits/Historical Note

== Historical Note on Squares Ending in 5 Occurrences of 2-Digit Pattern ==
According to in his $1997$ book , this result is attributed to , in volume $5$ of the .
No online corroboration of this can be found online, as there appears to be no consolidated archive of this magazine.
The author of this page would give his right arm to have a copy of the entire series resting on a bookshelf in his study.
== Sources ==
* : $6667$
* : $6667$
Category:Historical Notes
Three times Number whose Divisor Sum is Square

== Theorem ==
Let $n \in \Z_{>0}$ be a positive integer.
Let the divisor sum $\map {\sigma_1} n$ of $n$ be square.
Let $3$ not be a divisor of $n$.
Then the divisor sum of $3 n$ is square.
== Proof 1 ==
== Proof 2 ==
Category:Numbers whose Divisor Sum is Square
Category:Three times Number whose Divisor Sum is Square
Thurston's Geometrization Conjecture

== Theorem ==
When a topological manifold of dimension $3$ has been split into its connected sum and the Jaco-Shalen-Johannson torus decomposition, the remaining components each admit exactly one of the following geometries:
:$(1): \quad$ Euclidean geometry
:$(2): \quad$ Hyperbolic geometry
:$(3): \quad$ Spherical geometry
:$(4): \quad$ The geometry of $\mathbb S^2 \times \R$
:$(5): \quad$ The geometry of $\mathbb H^2 \times \R$
:$(6): \quad$ The geometry of the universal cover $S L_2 \R^~$ of the Lie group $S L_2 \R$
:$(7): \quad$ Nil geometry
:$(8): \quad$ Sol geometry
where:
:$\mathbb S^2$ is the 2-sphere
:$\mathbb H^2$ is the hyperbolic plane.
== Proof ==
== Sources ==
* 
Category:Manifolds
Tietze Extension Theorem

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is normal.
Let $A \subseteq S$ be a closed set in $T$.
Let $f: A \to \R$ be a continuous mapping from $A \subseteq S$ to the real number line under the usual (Euclidean) topology.
Then there exists a continuous extension $g: S \to \R$, that is, such that:
:$\forall s \in A: \map f s = \map g s$
== Proof ==
=== Lemma ===
First suppose that for any continuous mapping on a closed set there is a continuous extension.
Let $C$ and $D$ be disjoint sets which are closed in $S$.
Define $f: C \cup D \to \R$ by:
:$\map f x = \begin {cases}
0 & : x \in C \\
1 & : x \in D \end {cases}$
Now $f$ is continuous and we can extend it to a continuous mapping $g: X \to \R$.
By Urysohn's Lemma, $S$ is normal because $g$ is a continuous mapping such that $\map g x = 0$ for $x \in C$ and $\map g x = 1$ for $x \in D$.
Conversely, let $S$ be normal and $A$ be closed in $S$.
By the lemma, there exists a continuous mapping $g_0: S \to \R$ such that:
:$\forall x \in S: \size {\map {g_0} x} \le \dfrac 1 3$
:$\forall x \in A: \size {\map f x − \map {g_0} x} \le \dfrac 2 3$
Since $\paren {f − g_0}: A \to \R$ is continuous, the lemma tells us there is a continuous mapping $g_1: X \to \R$ such that:
:$\forall x \in S: \size {\map {g_1} x} \le \dfrac 1 3 \paren {\dfrac 2 3}$
:$\forall x \in A: \size {\map f x − \map {g_0} x - \map {g_1} x} \le \dfrac 2 3 \paren {\dfrac 2 3}$
By repeated application of the lemma we can construct a sequence of continuous mappings $g_0, g_1, g_2, \ldots$ such that:
:$\forall x \in S: \size {\map {g_n} x} \le \dfrac 1 3 \paren {\dfrac 2 3}^n$
:$\forall x \in A: \size {\map f x − \map {g_0} x − \map {g_1} x - \map {g_2} x - \cdots - \map {g_n} x} \le \dfrac 2 3 \paren {\dfrac 2 3}^n$
Define:
:$\ds \map g x = \sum_{n \mathop = 0}^\infty \map {g_n} x$
We have that:
:$\size {\map {g_n} x} \le \dfrac 1 3 \paren {\dfrac 2 3}^n$
and:
:$\ds \sum_{n \mathop = 0}^\infty \dfrac 1 3 \paren {\dfrac 2 3}^n$
converges as a geometric series.
Thus $\ds \sum_{n \mathop = 0}^\infty \map {g_n} x$ converges absolutely and uniformly.
So $g$ is a continuous mapping defined everywhere.
Also, $\ds \sum_{n \mathop = 0}^\infty \dfrac 1 3 \paren {\dfrac 2 3}^n = 1$ implies that $\size {\map g x} \le 1$.
Now for $x \in A$, we have that:
:$\ds \size {\map f x − \sum_{n \mathop = 0}^k \map {g_n} x} \le \paren {\dfrac 2 3}^{k + 1}$
As $k \to \infty$, the $\to 0$ and so the sum goes to $\map g x$.
Thus:
:$\size {\map f x − \map g x} = 0$
Therefore $g$ extends $f$. 
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
* 
Category:Normal Spaces
Category:Continuous Mappings
Tietze Extension Theorem/Lemma

== Lemma ==
Let $T = \struct {S, \tau}$ be a topological space which is normal.
Let $A \subseteq S$ be closed in $T$.
Let $f: A \to \R$ be a continuous mapping such that $\size {\map f x} \le 1$.
Then there exists a continuous mapping $g: S \to \R$ such that:
:$\forall x \in S: \size {\map g x} \le \dfrac 1 3$
:$\forall x \in A: \size {\map f x − \map g x} \le \dfrac 2 3$
== Proof ==
The sets $f^{−1} \sqbrk {\hointl {−\infty} {−\dfrac 1 3} }$ and $f^{−1} \sqbrk {\hointr {\dfrac 1 3} \infty}$ are disjoint and closed in $A$.
Since $A$ is closed, they are also closed in $S$.
We have that $S$ is normal.
Then by:
:Urysohn's Lemma
and:
:the fact that $\closedint 0 1$ is homeomorphic to $\closedint {−\dfrac 1 3} {\dfrac 1 3}$
there exists a continuous mapping $g: S \to \closedint {−\dfrac 1 3} {\dfrac 1 3}$ such that:
:$g \sqbrk {f^{-1} \sqbrk {\hointl {-\infty} {-\dfrac 1 3} } } = -\dfrac 1 3$
and
:$g \sqbrk {f^{-1} \sqbrk {\hointr {\dfrac 1 3} \infty} } = \dfrac 1 3$
Thus:
:$\forall x \in S: \size {\map g x} \le \dfrac 1 3$
Now if $−1 \le \map f x \le −\dfrac 1 3$, then:
:$\map g x = −\dfrac 1 3$
and thus:
:$\size {\map f x − \map g x} \le \dfrac 2 3$
Similarly if $\dfrac 1 3 \le \map f x \le 1$, then:
:$\map g x = \dfrac 1 3$
and thus:
:$\size {\map f x − \map g x} \le \dfrac 2 3$
Finally, for $\size {\map f x} \le \dfrac 1 3$ we have that:
:$\size {\map g x} \le \dfrac 1 3$
and so:
:$\size {\map f x − \map g x} \le \dfrac 2 3$
Hence, for all $x \in S$:
:$\size {\map f x − \map g x} \le \dfrac 2 3$
== Sources ==
Category:Normal Spaces
Category:Continuous Mappings
Time Taken for Body to Fall at Earth's Surface

== Theorem ==
Let an object $m$ be released above ground from a point near the Earth's surface and allowed to fall freely.
Let $m$ fall a distance $s$ in time $t$.
Then:
:$s = \dfrac 1 2 g t^2$
or:
:$t = \sqrt {\dfrac {2 s} g}$
where $g$ is the Acceleration Due to Gravity at the height through which $m$ falls.
It is supposed that the distance $s$ is small enough that $g$ can be considered constant throughout.
== Proof ==
From Body under Constant Acceleration: Distance after Time:
:$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$
Here the body falls from rest, so:
:$\mathbf u = \mathbf 0$
Thus:
:$\mathbf s = \dfrac {\mathbf g t^2} 2$
and so taking magnitudes:
:$s = \dfrac {g t^2} 2$
It follows by multiplying by $\dfrac 2 g$ that:
:$t^2 = \dfrac {2 s} g$
whence:
:$t = \sqrt {\dfrac {2 s} g}$
== Sources ==
* : Chapter $4$: Lure of the Unknown: The logic of species
Category:Mechanics
Category:Gravity
Time of Travel down Brachistochrone

== Theorem ==
Let a wire $AB$ be curved into the shape of a brachistochrone.
Let $AB$ be embedded in a constant and uniform gravitational field where Acceleration Due to Gravity is $g$.
Let a bead $P$ be released at $A$ to slide down without friction to $B$.
Then the time taken for $P$ to slide from $A$ to $B$ is:
:$T = \pi \sqrt {\dfrac a g}$
where $a$ is the radius of the generating circle of the cycloid which forms $AB$.
=== Corollary ===
== Proof ==
That the curve $AB$ is indeed a cycloid is demonstrated in Brachistochrone is Cycloid.
Let $A$ be located at the origin of a cartesian plane.
We have the equations of the cycloid:
Differentiating with respect to theta:
Let $s$ be the distance along $AB$ from the origin.
Then:
Thus:
:$\dfrac {\d s} {\d \theta} = 2 a \map \sin {\dfrac \theta 2}$
Now:
We also have, from the Principle of Conservation of Energy, that:
Thus:
Let $T$ be the time taken to slide down the brachistochrone.
At the top:
:$\theta = 0$
and at the bottom:
:$\theta = \pi$
Then:
So the time to slide down this brachistochrone from the top to the bottom is:
:$T = \pi \sqrt {\dfrac a g}$
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 6$: The Brachistochrone. Fermat and the Bernoullis: Problem $1$
Category:Cycloids
Time of Travel down Brachistochrone/Corollary

== Theorem ==
Let a wire $AB$ be curved into the shape of a brachistochrone.
Let $AB$ be embedded in a constant and uniform gravitational field where Acceleration Due to Gravity is $g$.
Let a bead $P$ be released from anywhere on the wire between $A$ and $B$ to slide down without friction to $B$.
Then the time taken for $P$ to slide to $B$ is:
:$T = \pi \sqrt{\dfrac a g}$
where $a$ is the radius of the generating circle of the cycloid which forms $AB$.
== Proof ==
That the curve $AB$ is indeed a cycloid is demonstrated in Brachistochrone is Cycloid.
Let $A$ be located at the origin of a cartesian plane.
Let the bead slide from an intermediate point $\theta_0$.
We have:
:$v = \dfrac {\d s} {\d t} = \sqrt {2 g \paren {y - y_0} }$
which leads us, via the same route as for Time of Travel down Brachistochrone, to:
Using the Half Angle Formula for Cosine and Half Angle Formula for Sine, this gives:
:$\ds T = \sqrt {\frac a g} \int_{\theta_0}^\pi \frac {\map \sin {\theta / 2} } {\sqrt {\map \cos {\theta_0 / 2} - \map \cos {\theta / 2} } } \rd \theta$
Now we make the substitution:
Recalculating the limits:
:when $\theta = \theta_0$ we have $u = 1$
:when $\theta = \pi$ we have $u = 0$.
So:
Thus the time to slide down a brachistochrone from any arbitrary point $\theta_0$ is:
:$T = \pi \sqrt {\dfrac a g}$
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 6$: The Brachistochrone. Fermat and the Bernoullis: Problem $2$
Category:Cycloids
Time when Hour Hand and Minute Hand at Right Angle

== Theorem ==
Let the time of day be such that the hour hand and minute hand are at a right angle to each other.
Then the time happens $22$ times in every $12$ hour period:
:when the minute hand is $15$ minutes ahead of the hour hand
:when the minute hand is $15$ minutes behind of the hour hand.
In the first case, this happens at $09:00$ and every $1$ hour, $5$ minutes and $27 . \dot 2 \dot 7$ seconds after
In the second case, this happens at $03:00$ and every $1$ hour, $5$ minutes and $27 . \dot 2 \dot 7$ seconds after.
Thus the times are, to the nearest second:
$\begin {array}
09:00:00 & 03:00:00 \\ 
10:05:27 & 04:05:27 \\
11:10:55 & 05:10:55 \\
12:16:22 & 06:16:22 \\
13:21:49 & 07:21:49 \\
14:27:16 & 08:27:16 \\
15:32:44 & 09:32:44 \\
16:38:11 & 10:38:11 \\
17:43:38 & 11:43:38 \\
18:49:05 & 12:49:05 \\
19:54:33 & 13:54:33 \\
\end{array}$
and times $12$ hours different.
== Proof ==
Obviously the hands are at right angles at $3$ and $9$ o'clock.
Thus we only need to show that the angle between the hands will be the same after every $1$ hour, $5$ minutes and $27 . \dot 2 \dot 7$ seconds.
Note that:
In $\dfrac {12} {11}$ hours:
:The minute hand has rotated $\dfrac {12} {11} \times 360^\circ$
:The hour hand has rotated $\dfrac {12} {11} \times 30^\circ$
Thus the angle between the hands has changed by:
which is a full rotation.
Hence after $\dfrac {12} {11}$ hours the angle between the hands would remain unchanged.
Category:Clocks
Titanic Prime consisting of 111 Blocks of each Digit plus Zeroes

== Theorem ==
The integer defined as:
:$\paren {123456789}_{111} \paren 0_{2284} 1$
where $\paren a_b$ means $b$ instances of $a$ in a string, is a titanic prime.
== Proof ==
It is noted that it has $9 \times 111 + 2284 + 1 = 3284$ digits, making it titanic.
It can be expressed arithmetically as:
:$123456789 \times \dfrac {10^{999} - 1} {10^9 - 1} \times 10^{2285} + 1$
== Historical Note ==
== Sources ==
* 
* : $\paren 1_{111} \paren 2_{111} \paren 3_{111} \ldots \paren 8_{111} \paren 9_{111} \paren 0_{2284} 1$
Category:Titanic Primes
Titanic Prime consisting of 111 Blocks of each Digit plus Zeroes/Historical Note

== Historical Note on Titanic Prime consisting of 111 Blocks of each Digit plus Zeroes ==
This titanic prime was discovered by in December $1985$.
When reported on this number in his of $1997$, he mistakenly expressed the number as:
:$\paren 1_{111} \paren 2_{111} \paren 3_{111} \paren 4_{111} \paren 5_{111} \paren 6_{111} \paren 7_{111} \paren 8_{111} \paren 9_{111} \paren 0_{2284} 1$
which it is not.
== Sources ==
* : $\paren 1_{111} \paren 2_{111} \paren 3_{111} \ldots \paren 8_{111} \paren 9_{111} \paren 0_{2284} 1$
Category:Historical Notes
Titanic Prime consisting of 111 Blocks of each Digit plus Zeroes/Mistake

== Source Work ==
:
:The Dictionary
::$\paren 1_{111} \paren 2_{111} \paren 3_{111} \ldots \paren 8_{111} \paren 9_{111} \paren 0_{2284} 1$
== Mistake ==
:$\paren 1_{111} \paren 2_{111} \paren 3_{111} \ldots \paren 8_{111} \paren 9_{111} \paren 0_{2284} 1 \qquad \qquad$ [$3284$ digits]
:The notation indicates that the digits $1$ to $9$ are each repeated $111$ times, followed by $2284$ zeros and a $1$.
:The number is prime.
== Correction ==
Sorry, but this is badly wrong.
This prime is reported in Prime Pages as:
:$123456789 \times \dfrac {\map {\mathrm R} {999} } {\map {\mathrm R} 9} \times 10^{2285} + 1$
where $\map {\mathrm R} n$ denotes the $n$-digit repunit.
This has been misinterpreted.
The actual number is:
:$\paren {123456789}_{111} \paren 0_{2284} 1$
which is not the same thing at all.
We can establish that $\paren 1_{111} \paren 2_{111} \paren 3_{111} \ldots \paren 8_{111} \paren 9_{111} \paren 0_{2284} 1$ is composite, with a factor $397$.
There are primes of the form $\paren 1_{111} \paren 2_{111} \paren 3_{111} \ldots \paren 8_{111} \paren 9_{111} \paren 0_x 1$, where $0 \le x \le 5000$: as follows
:$x = 399$
:$x = 1667$
:$x = 1918$
but this is far off the mark of what discovered.
== Sources ==
* : $\paren 1_{111} \paren 2_{111} \paren 3_{111} \ldots \paren 8_{111} \paren 9_{111} \paren 0_{2284} 1$
Category:Mistakes/Curious and Interesting Numbers