nyuuzyou/wikis huggingface dataset:
Third Sylow Theorem/Proof 2

== Theorem ==
== Proof ==
Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$.
Let $H$ be a Sylow $p$-subgroup of $G$.
We have that:
:$\order H = p^n$
:$\index G H = m$
Let $S_1, S_2, \ldots, S_m$ denote the left cosets of $G \pmod H$.
We have that $G$ acts on $G / H$ by the rule:
:$g * S_i = g S_i$.
Let $H_i$ denote the stabilizer of $S_i$.
By the Orbit-Stabilizer Theorem:
:$\order {H_i} = p^n$
while:
:$S_i = g H \implies g H g^{-1} \subseteq H_i$
Because $\order {g H g^{-1} } = \order H = \order {H_i}$, we have:
:$g H g^{-1} \subseteq H_i$
Let $H'$ be a second Sylow $p$-subgroup of $G$.
Then $H'$ acts on $G / H$ by the same rule as $G$.
Since $p \nmid m$, there exists at least one orbit under $H'$ whose cardinality is not divisible by $p$.
Suppose that $S_1, S_2, \ldots, S_r$ are the elements of an orbit where $p \nmid r$.
Let $K = H' \cap H_1$.
Then $K$ is the stabilizer of $S_1$ under the action of $H'$.
Therefore:
:$\index {H'} K = r$
However:
:$\order {H'} = p^n$
and:
:$p \nmid r$
from which it follows that:
:$r = 1$
and:
:$K = H'$
Therefore:
:$\order K = \order {H'} = \order {H_1} = p^n$
and:
:$H' = K = H_1$
Thus $H'$ and $H$ are conjugates.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 57$: Second Sylow Theorem
Category:Third Sylow Theorem
Third of 12 by 7 over 5 is 11

== Puzzle ==
:One third of twelve if you divide
:By just one fifth of seven,
:The true result (it has been tried)
:Exactly is eleven.
== Solution ==
One third of $\mathtt {TWELVE}$ is $\mathtt {LV}$ which is $55$ in Roman numerals.
One fifth of $\mathtt {SEVEN}$ is $\mathtt {V}$ which is $5$ in Roman numerals.
Then we have that:
:$\dfrac {55} 5 = 11$
== Sources ==
* 
* : Rational Amusements for Winter Evenings: $149$
Category:Trick Puzzles
Category:Roman Numerals
Thirteen Archimedean Polyhedra

== Theorem ==
There exist exactly $13$ distinct Archimedean polyhedra:
:Truncated Tetrahedron
:Cuboctahedron
:Truncated Cube
:Truncated Octahedron
:Rhombicuboctahedron
:Truncated Cuboctahedron
:Snub Cube
:Icosidodecahedron
:Truncated Dodecahedron
:Truncated Icosahedron
:Rhombicosidodecahedron
:Truncated Icosidodecahedron
:Snub Dodecahedron
== Proof ==
== Sources ==
* : $13$
* 
* : $13$
* 
* 
* 
Category:Archimedean Polyhedra
Thirteen Catalan Polyhedra

== Theorem ==
There exist exactly $13$ distinct Catalan polyhedra:
:Triakis tetrahedron
:Triakis octahedron
:Disdyakis dodecahedron
:Tetrakis hexahedron
:Triakis icosahedron
:Disdyakis triacontahedron
:Pentakis dodecahedron
:Rhombic dodecahedron
:Rhombic triacontahedron
:Deltoidal icositetrahedron
:Deltoidal hexecontahedron
:Pentagonal icositetrahedron
:Pentagonal hexecontahedron
== Proof ==
By definition, the Catalan polyhedra are the dual polyhedra of the Archimedean polyhedra.
There are $13$ Archimedean polyhedra, and so there are $13$ Catalan polyhedra.
== Sources ==
* : $13$
* : $13$
Category:Catalan Polyhedra
Thomae's Transformation

== Theorem ==
Let $a, b, c, e, f, s \in \C$.
Let $s = e + f - a - b - c$
Then:
:$\ds \map { {}_3 \operatorname F_2} { { {a, b, c} \atop {e, f} } \, \middle \vert \, 1} = \dfrac {\map \Gamma e \map \Gamma f \map \Gamma s } {\map \Gamma a \map \Gamma {s + b} \map \Gamma {s + c} } \map { {}_3 \operatorname F_2} { { {e - a, f - a, s} \atop {s + b, s + c} } \, \middle \vert \, 1} $
where:
:$\ds \map { {}_3 \operatorname F_2} { { {a, b, c} \atop {e, f} } \, \middle \vert \, 1}$ is the generalized hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} c^{\overline k} } { e^{\overline k} f^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
== Proof ==
First, we observe that:
We now have:
== Also known as ==
* Thomae's Transformation is also known as Thomae's Theorem.
== Also see ==
* Euler's Transformation
* Kummer's Quadratic Transformation
* Pfaff's Transformation
== Sources ==
* Chapter $\text {3}$. The hypergeometric series
* 
Category:Thomae's Transformation
Category:Gamma Function
Category:Hypergeometric Functions
Thomae Function is Continuous at Irrational Numbers

== Theorem ==
Let $D_M: \R \to \R$ denote the Thomae function:
:$\forall x \in \R: \map {D_M} x = \begin {cases} 0 & : x = 0 \text { or } x \notin \Q \\ \dfrac 1 q & : x = \dfrac p q : p, q \in \Z, p \perp q, q > 0 \end {cases}$
where:
:$\Q$ denotes the set of rational numbers
:$\Z$ denotes the integers
:$p \perp q$ denotes that $p$ and $q$ are coprime (that is, $x$ is a rational number expressed in canonical form)
Then $\map {D_M} x$ is:
:continuous at all irrational $x$ and at $x = 0$
:discontinuous at all rational $x$ such that $x \ne 0$.
== Proof ==
=== Rational $x$ ===
Let $x = \dfrac p q \in \Q \setminus \set 0$ such that $\dfrac p q$ is the canonical form of $x$.
Then we have:
:$\map {D_M} x = \dfrac 1 q$ 
Let $\epsilon = \dfrac 1 {2 q}$.
Let $\delta \in \R_{>0}$.
Then from Between two Real Numbers exists Irrational Number:
:$\exists z \in \R \setminus \Q: x < z < x + \delta$
Hence there exists $z \in \R$ such that:
:$z: \size {x - z} < \delta: \map {D_M} z = 0$
that is, such that:
:$\size {\map {D_M} x - \map {D_M} z} = \dfrac 1 q > \epsilon$
That is, there exists an $\epsilon \in \R_{>0}$ such that for all $\delta \in \R_{>0}$ it is possible to find $z$ such that:
:$\size {x - z} < \delta$
but such that:
:$\size {\map {D_M} x - \map {D_M} z} > \epsilon$
Hence when $x$ is rational $\map {D_M} x$ is discontinuous.
=== Irrational $x$ ===
In the following it is to be understood that all rational numbers expressed in the form $\dfrac p q$ are in canonical form.
Let $x \in \R \setminus \Q$ or $x = 0$.
Let $\Q$ be ordered in the following way:
:$\dfrac {p_1} {q_1} \prec \dfrac {p_2} {q_2} \iff \begin {cases} q_1 < q_2 & : q_1 \ne q_2 \\ p_1 < p_2 & : q_1 = q_2 \end {cases}$
and so we can denote $\Q$ with this ordering as $\struct {\Q, \prec}$
Let $\epsilon$ be arbitrary.
Let $q$ be the smallest positive integer such that $\dfrac 1 q < \epsilon$.
=== Lemma ===
Thus:
:$\forall y \in C: \map {D_M} y \le \dfrac 1 q$
and so:
:$\forall y \in C: \size {\map {D_M} y - \map {D_M} x} \le \epsilon$
because $\map {D_M} x = 0$ by definition.
Letting $\delta = \min \set {\size {x - a}, \size {b - x} }$ gives us our $\delta$.
Thus we have shown that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in \R: \size {y - x} < \delta \implies \size {\map {D_M} y - \map {D_M} x} < \epsilon$
That is, $D_M$ is continuous at $x$.
Hence, when $x$ is irrational or $0$, $\map {D_M} x$ is continuous.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 21$
Category:Thomae Function
Thomae Function is Continuous at Irrational Numbers/Lemma

== Lemma for Thomae Function is Continuous at Irrational Numbers ==
In the following it is to be understood that all rational numbers expressed in the form $\dfrac p q$ are in canonical form.
Let $x \in \R \setminus \Q$.
Let $\Q$ be ordered in the following way:
:$\dfrac {p_1} {q_1} \prec \dfrac {p_2} {q_2} \iff \begin {cases} q_1 < q_2 & : q_1 \ne q_2 \\ p_1 < p_2 & : q_1 = q_2 \end {cases}$
and so we can denote $\Q$ with this ordering as $\struct {\Q, \prec}$
Let $q \in \Z_{>0}$ be a positive integer.
Let $S \subseteq \struct {\Q, \prec}$ defined as:
:$S = \set {z \in \Q: z \prec \dfrac 1 q}$
That is, $S$ is the set of all rational numbers whose denominators are all less than or equal to $q$.
Let $a$ be the supremum of the set:
:$\set {z \in S: z < x}$
Let $b$ be the infimum of the set:
:$\set {z \in S: x < z}$
Then we have that the open interval:
:$C = \openint a b$
contains $x$ and no rational numbers whose denominators are less than $q$.
== Proof ==
Let $F := \floor x$ be the floor of $x$.
Let $C := \ceiling x$ be the ceiling of $x$.
Consider the interval:
:$I = \closedint F C$
As $x$ is irrational, we have that:
:$F < x < C$
and so:
:$x \in I$
Because $F$ and $C$ are integers by definition, they can be expressed as:
:$F = \dfrac F 1$
:$C = \dfrac C 1$
and so $F, C \in S$
Let $r \le q$.
Consider the set $S_r$:
:$S_r = \set {F, F + \dfrac 1 r, F + \dfrac 2 r, \ldots, F + \dfrac {r - 1} r, C}$
which is a finite set of rational numbers.
We have that:
:$I = S_1 \cup S_2 \cup S_3 \cup \cdots \cup S_{q - 1} \cup S_q$
and so it follows that $I$ is itself finite.
Then we have that:
But as $I$ is finite, then so are $\set {z \in I: z < x}$ and $\set {z \in I: x < z}$.
Because $F \in \set {z \in I: z < x}$ and $C \in \set {z \in I: x < z}$ we have that neither $\set {z \in I: z < x}$ nor $\set {z \in I: x < z}$ are empty.
Hence from Finite Subset of Ordered Set contains Supremum:
:$a \in \set {z \in I: z < x}$
and from Finite Subset of Ordered Set contains Infimum:
:$b \in \set {z \in I: x < z}$
Hence $\exists a, b, \in S: a < x < b$
It follows from the construction that there are no elements of $S$ in \openint a b$.
Category:Thomae Function
Thomsen Graph is not Planar

== Theorem ==
The Thomsen graph is not planar.
== Proof ==
Recall that the complete bipartite graph $K_{2, 3}$ is planar:
:
Let $A = \set {A_1, A_2, A_3}$ and $B = \set {B_1, B_2, B_3}$ be the two subsets which partition the vertex set of $K_{3, 3}$ which we are to demonstrate non-planarity.
Now we attempt to create $K_{3, 3}$ by building it from its subgraph $K_{2, 3}$, using $\set {A_1, A_2}$ (black) and $\set {B_1, B_2, B_3}$ (white) as the partition of the vertex set of $K_{2, 3}$.
Each of the faces of $K_{2, 3}$ is in the same form as each of the others.
That is, each face consists of $4$ vertices, which is not incident to the $5$th vertex of $K_{2, 3}$.
So exactly where in the plane we place the $6$th vertex does not matter.
 let us place vertex $A_3$ outside the face $A_1 B_1 A_2 B_3$ as shown below:
:
This face is not incident to $B_2$.
Edges $A_3 B_1$ and $A_3 B_3$ can be constructed without crossing an existing edge of $K_{2, 3}$, as indicated by dotted black lines.
However, it is not possible to construct edge $A_3 B_2$ without crossing one of $A_1 B_1$ or $A_1 B_3$, $A_2 B_1$ or $A_2 B_3$.
This is shown in dotted $\color{red} {\text {red} }$.
Hence the result.
== Sources ==
* 
* 
Category:Thomsen Graph
Category:Planar Graphs
Category:Complete Bipartite Graphs
Three-Point Form of Equation of Plane/Determinant Form

== Theorem ==
Let $P_1 := \tuple {x_1, y_1, z_1}$, $P_2 := \tuple {x_2, y_2, z_2}$ and $P_3 := \tuple {x_3, y_3, z_3}$ be points in a cartesian $3$-space.
Let $P$ be the plane passing through $P_1$, $P_2$ and $P_3$.
$P$ can be expressed in the form:
:$\begin {vmatrix} x & y & z & 1 \\ x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \end {vmatrix} = 0$
== Proof ==
== Sources ==
* 
* 
Category:Three-Point Form of Equation of Plane
Three-Point Form of Equation of Plane/Vector Form

== Theorem ==
Let $P_1$, $P_2$ and $P_3$ be points in a cartesian $3$-space whose position vectors are $\mathbf a$, $\mathbf b$ and $\mathbf c$.
Let $P$ be the plane passing through $P_1$, $P_2$ and $P_3$.
$P$ can be expressed in the form:
:$\mathbf r = \lambda \paren {\mathbf b - \mathbf a} + \mu \paren {\mathbf c - \mathbf a}$
where:
:$\mathbf r$ is the position vector of an arbitrary point on $P$
:$\lambda$ and $\mu$ are real parameters.
== Proof ==
== Sources ==
* 
* 
Category:Three-Point Form of Equation of Plane
Three-Way Exclusive Or and Equivalence

== Theorem ==
Let $p \iff q$ be the biconditional operator, and $p \oplus q$ be the exclusive or operator.
Then:
: $p \iff q \iff r \dashv \vdash p \oplus q \oplus r$
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||ccccc|} \hline
(p & \iff & q) & \iff & r & (p & \oplus & q) & \oplus & r \\
\hline
F & T & F & F & F & F & F & F & F & F \\
F & T & F & T & T & F & F & F & T & T \\
F & F & T & T & F & F & T & T & T & F \\
F & F & T & F & T & F & T & T & F & T \\
T & F & F & T & F & T & T & F & T & F \\
T & F & F & F & T & T & T & F & F & T \\
T & T & T & F & F & T & F & T & F & F \\
T & T & T & T & T & T & F & T & T & T \\
\hline
\end{array}$
From Biconditional is Associative and Exclusive Or is Associative, we have that both $\iff$ and $\oplus$ are associative, which justifies the rendition of this result without parentheses.
== Comment ==
A bizarrely non-intuitive result which is rarely documented.
Category:Biconditional
Category:Exclusive Or
Three Consecutive Sabbath Days on Same Day

== Problem ==
:Christians the week's first day for Sabbath hold.
:The Jews the seventh, as they did of old,
:The Turks the sixth, as we have oft been told.
:How can these three, in the same place, and day,
:Have each his own true Sabbath, tell, I pray.
It perhaps needs to be elaborated that:
:The word Sabbath is taken to mean religious holy day, whatever that day might be, rather than the more contemporary meaning of just Saturday
:The first day of the week is taken to be Sunday, rather than the more contemporary understanding of this being Monday
:Hence the sixth and seventh days are Friday and Saturday, considered as the holy day of the week by Muslims and Jews respectively
:In the milieu in which this question was asked, Turk was understood as being synonymous with Muslim, which is no longer an accurate position to hold.
== Solution ==
The three travel (or not) as follows.
:The Jew stays where he is.
:The Christian travels around the world towards the east.
:The Turk travels around the world towards the west.
When they meet again, it is (according to the Jew) Saturday.
But:
:the Christian has lost a day because those days are shorter
:the Turk has gained a day because those days are longer.
So by their own reckoning:
:according to the Christian, it is Sunday
:according to the Turk, it is Friday.
Note of course that in the modern world, they will have taken into account the , which has the effect of resetting the date so that it is globally consistent.
== Also see ==
* Horses Born and Died at Same Time but with Different Ages, a variant of this
== Sources ==
* : Geographical Paradoxes
* : Rational Amusements for Winter Evenings: $163$
Category:Geography Puzzles
Three Daughters

== Classic Riddles ==
One of my neighbours has $3$ daughters.
"How old are they?" I asked.
"The product of their ages is $36$," he replied.
I said, "That's not a great deal of help. Throw me a bone, here."
"Okay," he said, "the sum of their ages equals our house number."
I went outside to look at the front door to remind myself what that was, and said, "Nope, still can't work it out."
"My oldest daughter is the only one in the family who has green eyes."
"Aha," I replied, "got it. Their ages are ..."
and I told him.
=== Solution ===
Category:Three Daughters
Category:Classic Riddles
Three Daughters/Solution

== Classic Riddles ==
== Solution ==
The ages of the neighbour's three daughters are $9$, $2$ and $2$.
== Proof ==
There are $8$ ways of selecting $3$ positive integers whose product is $36$:
Indeed, it is not possible to know what the ages are from just knowing the product of their ages.
Now we learn that the sum of their ages equals the house number at which they live.
Let us investigate what those house numbers could be, by examining the $8$ possibilities above:
Having looked at the house number, I still could not tell the ages of his daughters.
For all the sums except for $13$, those sums are the only way you can make that number from $3$ positive integers whose product is $36$
So the house number must be $13$.
Finally, I was told some information about the oldest daughter.
If the ages were $1$, $6$ and $6$, technically speaking there is no "oldest" daughter, as the $6$-year-olds are twins.
So for there to be an "oldest daughter", ages must be $2$, $2$ and $9$.
== Quibbles ==
The assumption is that as they are both $6$ years old, the daughters of the $1$, $6$, $6$ family are twins.
However, by the convention that describing a child's age as $6$ means they can be anywhere between $6$ and $7$, it is possible that their ages are just-$6$ and nearly-$7$, and that the mother (poor woman) fell pregnant less than $3$ months after giving birth, or that the second child was severely premature.
Hence one of the $6$-year-olds could easily be referred to as the "oldest daughter".
Besides, it is of course quite possible for the neighbour to refer to the "older twin" as the "oldest daughter".
Category:Three Daughters
Three Intersecting Lines Perpendicular to Another Line are in One Plane

== Theorem ==
== Proof ==
:
Let $AB$ be a straight line set up at right angles to the three straight lines $BC$, $BD$ and $BE$ at their point of meeting $B$.
It is to be demonstrated that $BC$, $BD$ and $BE$ are all in the same plane.
Suppose they are not all in the same plane, but that while $BD$ and $BE$ are in the plane of reference, $BC$ is in a plane more elevated.
Thus the plane through $AB$ and $BC$ is different from the plane of reference.
From :
:the common section of the plane through $AB$ and $BC$ with the plane of reference is a straight line $BF$, say.
Therefore the three straight lines $AB$, $BC$ and $BF$ are in the same plane, that is, the one through $AB$ and $BC$.
We have that $AB$ is at right angles to each of the straight lines $BD$ and $BE$.
Therefore from :
:$AB$ is at right angles to the plane through $BD$ and $BE$.
But the plane through $BD$ and $BE$ is the plane of reference.
Therefore $AB$ is at right angles to the plane of reference.
Thus from :
:$AB$ will also make right angles with all the straight lines which meet it and are in the plane of reference.
But $BF$, which is in the plane of reference, meets $AB$.
Therefore $\angle ABF$ is a right angle.
But by hypothesis $\angle ABC$ is also a right angle.
Therefore $\angle ABF = \angle ABC$.
But $\angle ABF$ and $\angle ABC$ are in one plane, which is impossible.
Therefore $BC$ is not in a plane more elevated.
Therefore $BC$, $BD$ and $BE$ are all in the same plane.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Planes
Category:Perpendiculars
Three Non-Collinear Planes have One Point in Common

== Theorem ==
Three planes which are not collinear have exactly one point in all three planes.
== Proof ==
Let $A$, $B$ and $C$ be the three planes in question.
From Two Planes have Line in Common, $A$ and $B$ share a line, $p$ say.
From Propositions of Incidence: Plane and Line, $p$ meets $C$ in one point.
== Sources ==
* : Chapter $1$: Introduction: The Propositions of Incidence: $1.2$: The projective method: The propositions of incidence
Category:Projective Geometry
Three Pheasants

== Classic Problem ==
A man has $3$ pheasants that he wants to give to $2$ fathers and $2$ sons, giving each of them $1$ pheasant.
How can this be done?
== Solution ==
This is an instance of the riddle Two Fathers and Two Sons who are Three People.
To make this question tighter in structure, it should be stated that each of the $2$ fathers is actually the father of one of the $2$ sons.
In order to give $3$ pheasants to $4$ people so that each gets $1$ pheasant, it must be the case that $2$ of the $4$ people are actually the same person.
The only way to do that is to arrange that one of the fathers is also one of the sons.
So
:One of the fathers gets one pheasant.
:His son gets one pheasant.
:This son is also the other father.
:The son of this other father is the other son.
:He gets the third pheasant.
So the $3$ people involved are grandfather, father and son.
== Historical Note ==
== Sources ==
* : Exchanging the Knights: $102$
Category:Two Fathers and Two Sons who are Three People
Three Points are Coplanar

== Theorem ==
Let $P_1$, $P_2$ and $P_3$ be points in Euclidean $3$-space.
Then there exists a plane $\PP$ such that $P_1$, $P_2$ and $P_3$ all lie in $\PP$.
That is, $P_1$, $P_2$ and $P_3$ are coplanar.
== Proof ==
Let the straight line $P_1 P_2$ be constructed according to Euclid's first postulate.
Let the straight line $P_2 P_3$ be constructed according to Euclid's first postulate.
Thus $P_1 P_2$ and $P_2 P_3$ intersect at $P_2$.
From Two Intersecting Straight Lines are in One Plane, $P_1 P_2$ and $P_2 P_3$ are coplanar.
Hence as $P_1$, $P_2$ and $P_3$ all lie on either $P_1 P_2$ or $P_2 P_3$, it follows that $P_1$, $P_2$ and $P_3$ are likewise coplanar.
== Sources ==
* 
* 
Category:Coplanar Points
Three Points in Ultrametric Space have Two Equal Distances

== Theorem ==
Let $\struct {X, d}$ be an ultrametric space.
Let $x, y, z \in X$ with $\map d {x, z} \ne \map d {y, z}$.
Then:
:$\map d {x, y} = \max \set {\map d {x, z}, \map d {y, z} }$
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
=== Corollary 4 ===
=== Corollary 5 ===
== Proof ==
, let $\map d {x, z} > \map d {y, z}$.
Then:
On the other hand:
Putting the two inequalities together then:
:$\map d {x, y} = \map d {x, z} = \max \set {\map d {x, z}, \map d {y, z} }$
== Sources ==
* : $\S 2.3$: Topology: Corollary $2.3.4$
Category:Metric Spaces
Three Points in Ultrametric Space have Two Equal Distances/Corollary

== Theorem ==
Let $\struct {X, d}$ be an ultrametric space.
Let $x, y, z \in X$.
Then:
:at least two of the distances $\map d {x, y}$, $\map d {x, z}$ and $\map d {y, z}$ are equal.
== Proof ==
Either:
:$\map d {x, z} = \map d {y, z}$
or:
:$\map d {x, z} \ne \map d {y, z}$
By Three Points in Ultrametric Space have Two Equal Distances:
:$\map d {x, z} = \map d {y, z}$ or $\map d {x, y} = \max \set {\map d {x, z}, \map d {y, z} }$
In either case two of the distances are equal.
== Sources ==
* : $\S 2.3$: Topology
Category:Metric Spaces
Three Points in Ultrametric Space have Two Equal Distances/Corollary 2

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm{\,\cdot\,}$,
Let $x, y \in R$ and $\norm x \ne \norm y$.
Then:
:$\norm {x + y} = \norm {x - y} = \norm {y - x} = \max \set {\norm x, \norm y}$
== Proof ==
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
By Non-Archimedean Norm iff Non-Archimedean Metric then $d$ is a non-Archimedean metric and $\struct {R, d}$ is an ultrametric space.
Let $x, y \in R$ and $\norm x \ne \norm y$.
By the definition of the non-Archimedean metric $d$ then:
:$\norm x = \norm {x - 0} = \map d {x, 0}$
and similarly:
:$\norm y = \map d {y, 0}$
By assumption then:
:$\map d {x, 0} \ne \map d {y, 0}$
By Three Points in Ultrametric Space have Two Equal Distances then:
:$\norm {x - y} = \map d {x, y} = \max \set {\map d {x, 0}, \map d {y, 0} } = \max \set {\norm x, \norm y}$
By Norm of Negative then:
:$\norm {y - x} = \norm {x - y} = \max \set {\norm x, \norm y}$
By the definition of the non-Archimedean metric $d$ then:
:$\norm y = \norm {0 - \paren {-y} } = \map d {0, -y} = \map d {-y, 0}$
By assumption then:
:$\map d {x, 0} \ne \map d {-y, 0}$
By Three Points in Ultrametric Space have Two Equal Distances then:
:$\map d {x, -y} = \max \set {\map d {x, 0}, \map d {-y, 0} } = \max \set {\norm x, \norm y}$
By the definition of the non-Archimedean metric $d$ then:
:$\map d {x, -y} = \norm {x - \paren {-y} } = \norm {x + y}$
So $\norm {x + y} = \max \set {\norm x, \norm y}$.
The result follows.
== Sources ==
* : $\S 2.3$: Topology: Proposition $2.3.3$
Category:Normed Division Rings
Three Points in Ultrametric Space have Two Equal Distances/Corollary 3

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm{\,\cdot\,}$,
Let $x, y \in R$ and $\norm x \lt \norm y$.
Then:
:$\norm {x + y} = \norm {x - y} = \norm {y - x} = \norm y$
== Proof ==
By Corollary 2 then:
:$\norm {x + y} = \norm {x - y} = \norm {y - x} = \max \set {\norm x, \norm y} = \norm y$
== Sources ==
* : $\S 1.2$ Normed Fields: Remark $1.16$
Category:Normed Division Rings