nyuuzyou/wikis huggingface dataset:
Tangent of Zero

== Theorem ==
:$\tan 0 = 0$
where $\tan$ denotes tangent.
== Proof ==
== Also see ==
* Sine of Zero is Zero
* Cosine of Zero is One
* Cotangent of Zero
* Secant of Zero
* Cosecant of Zero
== Sources ==
* : $\text V$. Trigonometry: Special angles
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
* : Appendix $12$: Trigonometric formulae: Trigonometric values for some special angles
* : Appendix $14$: Trigonometric formulae: Trigonometric values for some special angles
Category:Tangent Function
Tangent of i

== Theorem ==
:$\tan i = \paren {\dfrac {e^2 - 1} {e^2 + 1} } i$
where $\tan$ denotes the complex tangent function and $i$ is the imaginary unit.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Sine of $i$
* Cosine of $i$
* Cotangent of $i$
* Secant of $i$
* Cosecant of $i$
Category:Tangent of i
Category:Tangent Function
Category:Imaginary Unit
Tangent times Tangent plus Cotangent

== Theorem ==
:$\tan x \paren {\tan x + \cot x} = \sec^2 x$
where $\tan$, $\cot$ and $\sec$ denote tangent, cotangent and secant respectively.
== Proof 1 ==
== Proof 2 ==
Category:Trigonometric Identities
Category:Tangent times Tangent plus Cotangent
Tangent to Archimedean Spiral at Point/Historical Note

== Historical Note on Tangent to Archimedean Spiral at Point ==
The Tangent to Archimedean Spiral at Point was first determined by in his book .
His proof appears as Proposition $20$.
It is suggested that the method used by anticipated the methods of differential calculus.
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
Category:Historical Notes
Tangent to Cycloid

== Theorem ==
Let $C$ be a cycloid generated by the equations:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
Then the tangent to $C$ at a point $\tuple {x, y}$ on $C$ is given by the equation:
:$y - a \paren {1 - \cos \theta} = \dfrac {\sin \theta} {1 - \cos \theta} \paren {x - a \theta + a \sin \theta}$
== Proof ==
From Slope of Tangent to Cycloid, the slope of the tangent to $C$ at the point $\tuple {x, y}$ is given by:
:$\dfrac {\d y} {\d x} = \cot \dfrac \theta 2$
This tangent to $C$ also passes through the point $\tuple {a \paren {\theta - \sin \theta}, a \paren {1 - \cos \theta} }$.
We have:
By Equation of Straight Line in Plane we have:
:$y - y_0 = m \paren {x - x_0}$
where $m$ is the slope and $\tuple {x_0, y_0}$ a given point on the straight line.
Substituting for the slope $m$ we obtain:
:$y - y_0 = \dfrac {\sin \theta} {1 - \cos \theta} \paren {x - x_0}$
Substituting for the given point $\tuple {a \paren {\theta - \sin \theta}, a \paren {1 - \cos \theta} }$
:$y - a \paren {1 - \cos \theta} = \dfrac {\sin \theta} {1 - \cos \theta} \paren {x - a \paren {\theta - \sin \theta} }$
Hence the result.
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Example $3$
Category:Cycloids
Tangent to Cycloid is Vertical at Cusps

== Theorem ==
The tangent to the cycloid whose locus is given by:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
is vertical at the cusps.
== Proof ==
From Slope of Tangent to Cycloid, the slope of the tangent to $C$ at the point $\tuple {x, y}$ is given by:
:$\dfrac {\d y} {\d x} = \cot \dfrac \theta 2$
At the cusps, $\theta = 2 n \pi$ for $n \in \Z$.
Thus at the cusps, the slope of the tangent to $C$ is $\cot n \pi$.
From Shape of Cotangent Function:
:$\ds \lim_{\theta \mathop \to n \pi} \cot \theta \to \infty$
Hence the result by definition of vertical tangent line.
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid
Category:Cycloids
Tangent to Cycloid passes through Top of Generating Circle

== Theorem ==
Let $C$ be a cycloid generated by the equations:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
Then the tangent to $C$ at a point $P$ on $C$ passes through the top of the generating circle of $C$.
== Proof ==
From Tangent to Cycloid, the equation for the tangent to $C$ at a point $P = \tuple {x, y}$ is given by:
:$(1): \quad y - a \paren {1 - \cos \theta} = \dfrac {\sin \theta} {1 - \cos \theta} \paren {x - a \theta + a \sin \theta}$
From Equation of Cycloid, the point at the top of the generating circle of $C$ has coordinates $\tuple {2 a, a \theta}$.
Substituting $x = 2 a$ in $(1)$:
That is, the tangent to $C$ passes through $\tuple {a \theta, 2 a}$ as was required.
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Example $3$
Category:Cycloids
Tangential Component of Acceleration of Uniform Circular Motion is Zero

== Theorem ==
Let $P$ be a particle moving in space in uniform circular motion.
Let $\mathbf a$ denote the acceleration of $P$.
Then the tangential component of $\mathbf a$ is zero.
== Proof ==
From Acceleration of Point in Plane in Intrinsic Coordinates, the motion of $P$ can be expressed in intrinsic coordinates as:
:$\mathbf a = \dfrac {\d \map v t} {\d t} \mathbf s + \dfrac {\paren {\map v t}^2} \rho \bspsi$
where:
:$\mathbf s$ denotes the unit vector along the tangential direction of $P$
:$\bspsi$ denotes the unit vector toward the center of curvature of the motion of $P$
:$\map v t$ is the speed of $P$ at the time $t$
:$\rho$ is the radius of curvature of the motion of $P$ at time $t$.
By definition, the tangential component of $\mathbf a$ is defined as $\dfrac {\d v} {\d t}$.
Let $\map v t$ denote the speed of $P$ at time $t$.
By definition of uniform circular motion, $\map v t$ is constant.
Hence by Derivative of Constant it follows that:
:$\dfrac {\d \map v t} {\d t} = 0$
Hence the result.
== Also see ==
* Definition:Tangential Component of Acceleration
== Sources ==
* 
* 
Category:Acceleration
Tangents to Circle from Point are of Equal Length

== Theorem ==
Let $\CC$ be a circle.
Let $P$ be a point in the exterior of $\CC$.
Let $PA$ and $PB$ be tangents to $\CC$ from $P$ touching $\CC$ at $A$ and $B$ respectively.
:
Then:
:$PA = PB$
== Proof ==
Let $O$ be the center of $\CC$.
Construct $OA$ and $OB$.
From Radius at Right Angle to Tangent:
:$PA \perp OA$ and $PB \perp OB$
and so $\angle OAP = \angle OBP$ which equals a right angle.
:
Consider the right triangles $\triangle OAP$ and $\triangle OBP$.
We have:
:$OA = OB$ by definition of radius of circle
:$\angle OAP = \angle OBP$
:$OP$ is their common hypotenuse.
Hence by Triangle Right-Angle-Hypotenuse-Side Congruence:
:$\triangle OAP \cong \triangle OBP$
and so $PA = PB$.
== Sources ==
* $(4) \ \text {(a)}$
* $(4) \ \text {(a)}$
Category:Circles
Category:Tangents to Circles
Tangents to Circle from Point subtend Equal Angles at Center

== Theorem ==
Let $\CC$ be a circle.
Let $P$ be a point in the exterior of $\CC$.
Let $PA$ and $PB$ be tangents to $\CC$ from $P$ touching $\CC$ at $A$ and $B$ respectively.
:
Then:
:$\angle OPA = \angle OPB$
=== Corollary ===
== Proof ==
Let $O$ be the center of $\CC$.
Construct $OA$ and $OB$.
From Radius at Right Angle to Tangent:
:$PA \perp OA$ and $PB \perp OB$
and so $\angle OAP = \angle OBP$ which equals a right angle.
:
Consider the right triangles $\triangle OAP$ and $\triangle OBP$.
We have:
:$OA = OB$ by definition of radius of circle
:$\angle OAP = \angle OBP$
:$OP$ is their common hypotenuse.
Hence by Triangle Right-Angle-Hypotenuse-Side Congruence:
:$\triangle OAP \cong \triangle OBP$
and so $OPA = \angle OPB$.
== Sources ==
* $(4) \ \text {(b)}$
* $(4) \ \text {(b)}$
Category:Circles
Category:Tangents to Circles
Tangents to Circle from Point subtend Equal Angles at Center/Corollary

== Theorem ==
Let $\CC$ be a circle.
Let $P$ be a point in the exterior of $\CC$.
Let $PA$ and $PB$ be tangents to $\CC$ from $P$ touching $\CC$ at $A$ and $B$ respectively.
:
Then $OP$ is a bisector of $\angle APB$.
== Proof ==
From Tangents to Circle from Point subtend Equal Angles at Center:
:$\angle OPA = \angle OPB$
from which follows the result.
== Sources ==
* $(4) \ \text {(c)}$
* $(4) \ \text {(c)}$
Category:Circles
Category:Tangents to Circles
Tannery's Theorem

== Theorem ==
Let $\sequence {p_n}$ be an increasing, unbounded above sequence of natural numbers.
For every $r, n \in \N$, let $v_r$ be a mapping:
:$\map {v_r} n \in \C$
Let $\sequence {w_r}$ be a sequence of complex numbers such that, for every $r \in \N$:
:$\ds w_r = \lim_{n \mathop \to \infty} \map {v_r} n$
Let $\sequence {M_r}$ be a sequence of non-negative real numbers such that:
:$\forall n \in \N: \size {\map {v_r} n} \le M_r$
and:
:$\ds \sum_{r \mathop = 0}^\infty M_r$ is convergent
Then:
:$\ds \lim_{n \mathop \to \infty} \sum_{r \mathop = 0}^{p_n} \map {v_r} n = \sum_{r \mathop = 0}^\infty w_r$
== Proof ==
First, observe that for all $r \in \N$:
Let $\epsilon > 0$ be arbitrary.
By Tail of Convergent Series tends to Zero, select some $q$ such that:
:$\ds \sum_{r \mathop = q}^\infty M_r < \frac \epsilon 3$
As $\sequence {p_n}$ is unbounded above, there is some $M \in \N$ such that:
:$p_M \ge q$
Since $\sequence {p_n}$ is increasing:
:$\forall n \ge M: p_n \ge p_M \ge q$
Additionally:
Therefore, we can find some $N \in \N$ such that:
:$\ds \size {\sum_{r \mathop = 0}^{q - 1} \map {v_r} n - \sum_{r \mathop = 0}^{q - 1} w_r} < \frac \epsilon 3$
Let $n \ge \max \set {M, N}$ be arbitrary.
Then:
Thus, for any $\epsilon > 0$, we have:
:$\ds \forall n \ge \max \set {M, N}: \size {\sum_{r \mathop = 0}^{p_n} \map {v_r} n - \sum_{r \mathop = 0}^\infty w_r} < \epsilon$
Thus, by definition of a limit:
:$\ds \lim_{n \to \infty} \sum_{r \mathop = 0}^{p_n} \map {v_r} n = \sum_{r \mathop = 0}^\infty w_r$
== Sources ==
* : Chapter $\text {VII}$: Article $49$
Category:Analysis
Category:Series
Tarski's Undefinability Theorem

== Theorem ==
Let $\ZZ$ be the standard structure $\struct {\Z, +, \cdot, s, <, 0}$ for the language of arithmetic.
Let $\operatorname {Th}_\ZZ$ be the sentences which are true in $\ZZ$.
Let $\Theta$ be the set of Gödel numbers of those sentences in $\operatorname {Th}_\ZZ$.
$\Theta$ is not definable in $\operatorname {Th}_\ZZ$.
== Note ==
Informally, the theorem says that the set of true statements about arithmetic can't be defined arithmetically.
== Proof ==
$\operatorname {Th}_\ZZ$ is easily seen to be a consistent extension of minimal arithmetic. (In fact, the axioms in minimal arithmetic were selected based on the behavior of standard arithmetic.)
Thus, the theorem is a special case of Set of Gödel Numbers of Arithmetic Theorems Not Definable in Arithmetic (and can be seen to follow immediately).
== Sources ==
* : $\S 15$: Theorem $4$
Category:Mathematical Logic
Tartaglia's Formula

== Theorem ==
Let $T$ be a tetrahedron with vertices $\mathbf d_1, \mathbf d_2, \mathbf d_3$ and $\mathbf d_4$.
For all $i$ and $j$, let the distance between $\mathbf d_i$ and $\mathbf d_j$ be denoted $d_{ij}$.
Then the volume $V_T$ of $T$ satisfies:
$\quad V_T^2 = \dfrac 1 {288} \det \ \begin{vmatrix} 
0 & 1 & 1 & 1 & 1 \\
1 & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2 \\
1 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2 \\
1 & d_{13}^2 & d_{23}^2 & 0 & d_{34}^2 \\
1 & d_{14}^2 & d_{24}^2 & d_{34}^2 & 0
\end{vmatrix}$
== Proof ==
A proof of Tartaglia's Formula will be found in a proof of the Value of Cayley-Menger Determinant as a tetrahedron is a $3$-simplex.
== Also known as ==
== Also see ==
* Definition:Cayley-Menger Determinant
== Sources ==
* : Theorem $3.1$
Category:Tartaglia's Formula
Category:Tetrahedra
Tartaglia's Poem

== Mnemonic ==
Tartaglia's poem is a mnemonic for the solution of the cubic equation:
:When the cube and its things near
:Add to a new number, discrete
:Determine two new numbers different
:By that one; this feat
:Will be kept as a rule
:Their product always equal, the same
:To the cube of a third
:Of the number of things named.
:Then, generally speaking
:The remaining amount
:Of the cube roots subtracted
:Will be our desired count.
Category:Cardano's Formula
Category:Cubic Equations
Category:Mnemonics
Tautochrone Problem

== Problem ==
Let a point $A$ be joined by a wire to a lower point $B$.
Let the wire be allowed to be bent into whatever shape is required.
Let a bead be released at $A$ to slide down without friction to $B$.
What is the shape of the wire so that it takes same time to descend to $B$ from wherever you release the bead between $A$ and $B$?
== Also see ==
* Cycloid has Tautochrone Property
* Definition:Brachistochrone
* Definition:Tautochrone
* Definition:Cycloid
== Historical Note ==
Category:Tautochrone Problem
Category:Cycloids
Category:Classic Problems
Tautochrone Problem/Historical Note

== Historical Note on Tautochrone Problem ==
The fact that Cycloid has Tautochrone Property was discovered by in $1658$, during his work on developing a reliable and accurate pendulum clock.
The Tautochrone Problem was also solved independently by in $1823$, using the technique now known as Abel's integral equation.
== Also see ==
* Brachistochrone Problem/Historical Note
== Sources ==
* : Chapter $\text{V}$: "Greatness and Misery of Man"
* : Chapter $\text {B}.21$: The Cycloid
* : Chapter $\text {A}.27$: Abel ()
* 
* 
Category:Historical Notes
Tautological Antecedent

== Theorem ==
A conditional with a tautology as antecedent:
:$\top \implies p \dashv \vdash p$
== Proof by Natural Deduction ==
== Proof by Truth Table ==
== Also see ==
* Tautological Consequent
== Sources ==
* : $\S 2.3.3$
Category:Tautological Antecedent
Category:Tautology
Category:Conditional
Tautological Antecedent/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|ccc||c|ccc|} \hline
p & \top & \implies & p \\
\hline
F & T & F & F \\
T & T & T & T \\
\hline
\end{array}$
Category:Tautological Antecedent
Tautological Consequent

== Theorem ==
A conditional with a tautology as consequent:
:$p \implies \top \dashv \vdash \top$
== Proof by Natural Deduction ==
== Proof by Truth Table ==
== Also see ==
* True Statement is implied by Every Statement
* Tautological Antecedent
== Sources ==
* : $\S 2.3.3$
Category:Tautological Consequent
Category:Tautology
Category:Conditional
Tautological Consequent/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|ccc||c|ccc|} \hline
p & p & \implies & \top & \top \\
\hline
F & F & T & T & T \\
T & T & T & T & T \\
\hline
\end{array}$
Category:Tautological Consequent
Tautology/Examples/Arbitrary Example 1

== Examples of Tautologies ==
The WFF of propositional logic:
:$\paren {\paren {\paren {\lnot p} \implies q} \implies \paren {\paren {\paren {\lnot p} \implies \paren {\lnot q} } \implies p} }$
is a tautology.
== Proof ==
Proof by truth table:
:$\begin{array}{|cccc|c|ccccccc|}
\hline
(((\lnot & p) & \implies & q) & \implies & (((\lnot & p) & \implies & (\lnot & q)) & \implies & p)) \\
\hline
 \T & \F & \F & \F & \T & \T & \F & \T & \T & \F & \F & \F \\
 \T & \F & \T & \T & \T & \T & \F & \F & \F & \T & \T & \F \\
 \F & \T & \T & \F & \T & \F & \T & \T & \T & \F & \T & \T \\
 \F & \T & \T & \T & \T & \F & \T & \T & \F & \T & \T & \T \\
\hline
\end{array}$
As can be seen by inspection, the truth value under the main connective is true for all rows.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Example $1.6 \ \text{(d)}$
Category:Examples of Tautologies
Tautology/Examples/Arbitrary Example 2

== Example of Tautology ==
The WFF of propositional logic:
:$\paren {\paren {\lnot p} \implies \paren {q \lor r} } \iff \paren {\paren {\lnot q} \implies \paren {\paren {\lnot r} \implies p} }$
is a tautology.
== Proof ==
Proof by truth table:
:$\begin{array}{|cccccc|c|ccccccc|}
\hline
((\lnot & p) & \implies & (q & \lor & r)) & \iff & ((\lnot & q) & \implies & ((\lnot & r) & \implies & p)) \\
\hline
 \T & \F & \F & \F & \F & \F & \T & \T & \F & \F & \T & \F & \F & \F \\
 \T & \F & \T & \F & \T & \T & \T & \T & \F & \T & \F & \T & \T & \F \\
 \T & \F & \T & \T & \T & \F & \T & \F & \T & \T & \T & \F & \F & \F \\
 \T & \F & \T & \T & \T & \T & \T & \F & \T & \T & \F & \T & \T & \F \\
 \F & \T & \T & \F & \F & \F & \T & \T & \F & \T & \T & \F & \T & \T \\
 \F & \T & \T & \F & \T & \T & \T & \T & \F & \T & \F & \T & \T & \T \\
 \F & \T & \T & \T & \T & \F & \T & \F & \T & \F & \T & \F & \T & \T \\
 \F & \T & \T & \T & \T & \T & \T & \F & \T & \T & \F & \T & \T & \T \\
\hline
\end{array}$
As can be seen by inspection, the truth value under the main connective is true for all rows.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Exercise $4$
Category:Examples of Tautologies
Tautology and Contradiction

== Theorems ==
=== Contradiction is Negation of Tautology ===
=== Tautology is Negation of Contradiction ===
=== Conjunction with Tautology ===
=== Disjunction with Tautology ===
=== Conjunction with Contradiction ===
=== Disjunction with Contradiction ===
== Comment ==
Note that the proofs of:
* $\neg \bot \vdash \top$
* $\neg \top \vdash \bot$
* $p \vdash p \land \top$
* $p \lor \top \vdash \top$
rely (directly or indirectly) upon Law of Excluded Middle - and it can be seen that they are just another way of stating that truth.
The propositions:
:If it's not false, it must be true
and
:If it's not true, it must be false
are indeed valid only in the context where there are only two truth values.
From the intuitionistic perspective, these results do not hold.
== Sources ==
* : $\S \text{I}: 3$: Logical Constants $(2)$
Category:Tautology
Category:Contradiction
Tautology iff Negation is Unsatisfiable

== Theorem ==
Let $\mathbf A$ be a WFF of propositional logic.
Then $\mathbf A$ is a tautology its negation $\neg \mathbf A$ is unsatisfiable.
== Proof ==
=== Necessary Condition ===
Let $\mathbf A$ be a tautology.
Let $v$ be a boolean interpretation of $\mathbf A$.
Then $\map v {\mathbf A} = \T$.
Hence, by definition of the boolean interpretation of negation:
:$\map v {\neg \mathbf A} = \F$
Since $v$ was arbitrary, it follows that $\neg \mathbf A$ is unsatisfiable.
=== Sufficient Condition ===
Let $\neg \mathbf A$ be unsatisfiable.
Let $v$ be a boolean interpretation of $\neg \mathbf A$.
Then $\map v {\neg \mathbf A} = \F$.
Hence, by definition of the boolean interpretation of negation:
:$\map v {\mathbf A} = \T$
Since $v$ was arbitrary, it follows that $\mathbf A$ is a tautology.
== Sources ==
* : $\S 2.5$: Theorem $2.39$
Category:Boolean Interpretations
Category:Formal Semantics
Tautology is Negation of Contradiction

== Theorem ==
A tautology implies and is implied by the negation of a contradiction:
:$\top \dashv \vdash \neg \bot$
That is, a truth can not be false, and a non-falsehood must be a truth.
== Proof by Natural Deduction ==
== Proof by Truth Table ==
== Proof by Boolean Interpretation ==
That is, the proposition:
:If it's not false, it must be true
is valid only in the context where there are only two truth values.
== Also see ==
* Contradiction is Negation of Tautology
== Sources ==
* : $\S 3.5$: The Classification of Propositions
Category:Tautology
Category:Logical Negation
Category:Contradiction
Category:Tautology is Negation of Contradiction
Tautology is Negation of Contradiction/Proof 3

== Theorem ==
== Proof ==
Let $p$ be a propositional formula.
Let $v$ be an arbitrary boolean interpretation of $p$.
Then:
:$\map v p = T \iff \map v {\neg p} = F$ 
by the definition of the logical not.
Since $v$ is arbitrary, $p$ is true in all interpretations $\neg p$ is false in all interpretations.
Hence:
:$\top \dashv \vdash \neg \bot$
Category:Tautology is Negation of Contradiction
Tautology is Negation of Contradiction/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values in the appropriate columns match.
$\begin{array}{|c||cc|} \hline
\bot & \neg & \top \\
\hline
\F & \F & \T \\
\hline
\end{array}$
Category:Tautology is Negation of Contradiction
Taxicab Metric is Metric/Proof 1

== Theorem ==
== Proof ==
From the definition, the taxicab metric is as follows:
Let $M_{1'} = \struct {A_{1'}, d_{1'} }, M_{2'} = \struct {A_{2'}, d_{2'} }, \ldots, M_{n'} = \struct {A_{n'}, d_{n'} }$ be a finite number of metric spaces.
Let $\AA$ be the Cartesian product $\ds \prod_{i \mathop = 1}^n A_{i'}$.
The taxicab metric on $\AA$ is:
:$\ds \map {d_1} {x, y} = \sum_{i \mathop = 1}^n \map {d_{i'} } {x_{i'}, y_{i'} }$
for $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
=== Proof of ===
So holds for $d_1$.
=== Proof of ===
So holds for $d_1$.
=== Proof of ===
So holds for $d_1$.
=== Proof of ===
So holds for $d_1$.
Category:Taxicab Metric is Metric
Taxicab Metric is Topologically Equivalent to Chebyshev Distance on Real Vector Space

== Theorem ==
For $n \in \N$, let $\R^n$ be a real vector space.
Let $d_1$ be the taxicab metric on $\R^n$.
Let $d_\infty$ be the Chebyshev distance on $\R^n$.
Then
:$\forall x, y \in \R^n: \map {d_\infty} {x, y} \le \map {d_1} {x, y} \le n \cdot \map {d_\infty} {x, y}$
It follows that $d_1$ and $d_\infty$ are Lipschitz equivalent.
== Proof ==
By definition of the Chebyshev distance on $\R^n$, we have:
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n {\size {x_i - y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$.
Let $j$ be chosen so that:
:$\ds \size {x_j - y_j} = \max_{i \mathop = 1}^n {\size {x_i - y_i} }$
Then:
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 2$: Metric Spaces: Exercise $3$
Category:Chebyshev Distance
Category:Taxicab Metric
Taxicab Metric on Metric Space Product is Continuous

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\AA$ be the Cartesian product $A \times A$.
Let $d_1$ be the taxicab metric on $\AA$:
:$\ds \map {d_1} {x, y} = \sum_{i \mathop = 1}^n \map {d_{i'} } {x_{i'}, y_{i'} }$
for $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
Then $d_1: \AA \to \AA$ is a continuous function.
== Proof ==
Recall the definition of continuous mapping in this context.
Given metric spaces $M_X = \struct {X, d_X}$ and $M_Y = \struct {Y, d_Y}$, and a mapping $f : X \to Y$, we say that $f$ is $\struct {X, d_X} \to \struct {Y, d_Y}$-continuous :
:$\forall x_0 \in X: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in X: \map {d_X} {x, x_0} < \delta \implies \map d {\map f x, \map f {x_0} } < \epsilon$
Hence it is necessary to prove that $d: \struct {A \times A, d_1} \to \struct {\R, \size {\,\cdot \,} }$ is continuous.
That is, given $\tuple {x_0, y_0} \in A \times A$ and $\epsilon>0$, it is necessary to find $\delta \in \R_{>0}$ such that:
:$\forall x, y \in A: \map {d_1} {\tuple {x, y}, \tuple {x_0, y_0} } < \delta \implies \size {\map d {x, y} - \map d {x_0, y_0} } < \epsilon$
Hence, let $\tuple {x_0, y_0} \in A \times A$.
If $\map {d_1} {\tuple {x, y}, \tuple {x_0, y_0} } = \map d {x, x_0} + \map d {y, y_0} < \epsilon$, then:
:$\size {\map d {x, y} - \map d {x_0, y_0} } \le \size {\map d {x, y} - \map d {x, y_0} } + \size {\map d {x, y_0} - \map d {x_0, y_0} } \le 2 \epsilon$
using .
The result follows.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 17$
Category:Taxicab Metric
Category:Continuous Mappings on Metric Spaces
Taxicab Metric on Real Number Plane is Translation Invariant

== Theorem ==
Let $\tau_{\mathbf t}: \R^2 \to \R^2$ denote the translation of the Euclidean plane by the vector $\mathbf t = \begin {pmatrix} a \\ b \end {pmatrix}$.
Let $d_1$ denote the taxicab metric on $\R^2$.
Then $d_1$ is unchanged by application of $\tau$:
:$\forall x, y \in \R^2: \map {d_1} {\map \tau x, \map \tau y} = \map {d_1} {x, y}$
== Proof ==
Let $x = \tuple {x_1, x_2}$ and $y = \tuple {y_1, y_2}$ be arbitrary points in $\R^2$.
Then:
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 22$
Category:Translation Mappings
Category:Taxicab Metric