nyuuzyou/wikis huggingface dataset:
Accurate to n Significant Figures/Examples/e^-3 to 3 Figures

== Example of Accurate to $n$ Significant Figures ==
$e^{-3} expressed accurate to $3$ significant figures is:
:$e^{-3} \approx 0 \cdotp 0498$
== Proof ==
We note that:
:$e^{-3} \approx 0.049787 \ldots$
Then we see:
:$0.0498 - 0.049787 \ldots < 0.049787 \ldots - 0.0497$
and so $e^{-3}$ rounded to $3$ significant figures is:
:$0.0498$
== Sources ==
* 
* 
Category:Examples of Accuracy
Accurate to n Significant Figures/Examples/e to 3 Figures

== Example of Accurate to $n$ Significant Figures ==
The Euler's number $e$ expressed accurate to $3$ significant figures is:
:$e \approx 2 \cdotp 72$
== Proof ==
We start with the decimal expansion of $e$:
We note that:
:$2.72 - 2.718 \ldots < 2.718 \ldots - 2.71$
and so $e$ rounded to $3$ significant figures is:
:$2.72$
== Sources ==
* 
* 
Category:Examples of Accuracy
Accurate to n Significant Figures/Examples/e to 4 Figures

== Example of Accurate to $n$ Significant Figures ==
The Euler's number $e$ expressed accurate to $4$ significant figures is:
:$e \approx 2 \cdotp 718$
== Proof ==
We start with the decimal expansion of $e$:
We note that:
:$2.719 - 2.7182 \ldots > 2.7182 \ldots - 2.718$
and so $e$ rounded to $4$ significant figures is:
:$2.718$
== Sources ==
* 
* 
Category:Examples of Accuracy
Achilles Paradox

== Paradox ==
 and a are to have a .
Achilles (not surprisingly) is expected to run considerably faster than the tortoise. So, to make it marginally more fair, he gives the tortoise a head start.
But it is apparent that Achilles cannot actually catch up with the tortoise.
Suppose he gives the tortoise a headstart of $x_0$.
By the time he has got to $x_0$, the tortoise has moved on, to $x_1$, say.
But by the time Achilles has reached $x_1$, the tortoise has moved on, to $x_2$, say.
You can continue this indefinitely.
However, it is intuitively obvious that Achilles will eventually overtake the tortoise, as Achilles is faster.
== Resolution ==
It is clear that there is a problem with this reasoning, as it is tantamountly clear that someone running faster than another will overtake, sooner or later.
The solution depends on the concept of a limit.
The sum of the distances run by Achilles in catching up the tortoise is an infinite series whose sequence of partial sums is bounded above.
As such, once Achilles reaches that limit, any further distance he travels will bring him further than the tortoise.
The problem lies in the assumption that Achilles is bounded to points only previously set by the tortoise, however in all practicality this is not the case (Achilles' step size). 
The error lies in measuring the distance Achilles has traveled after a finite number of steps.
We must instead measure the distance after an infinite number of steps with the summation of a series.
Let $a$ and $b$ be the velocities at which Achilles and the tortoise respectively travel.
As Achilles is faster than the tortoise, let $a > b$.
Say Achilles is distance $d_n$ behind the tortoise.
In the time $t = \dfrac {d_n} a$ it will take him to travel that distance, the tortoise will have traveled $b t$ further.
Thus:
:$d_{n + 1} = d_n \paren {\dfrac b a}$
The sequence of distances that Achilles must travel is thus the recursive sequence:
:$d_n = \begin {cases} h & : n = 0 \\ d_{n - 1} \paren {\dfrac b a} & : \text {otherwise} \end {cases}$
where $h$ is the tortoise's head start.
This is the geometric sequence:
:$h, h \paren {\dfrac b a}, h \paren {\dfrac b a}^2, h \paren {\dfrac b a}^3, \ldots$
$a > b$ implies $\size {\dfrac b a} < 1$.
Thus, by Sum of Infinite Geometric Sequence: Corollary 2, the sum $S$ of the sequence is:
:$(1): \quad \ds \sum_{i \mathop = 0}^\infty h \paren {\dfrac b a}^i = \dfrac h {1 - \frac b a}$
Thus it takes Achilles:
time to catch up to the tortoise.
In this time, the tortoise should travel the same distance as Achilles minus the head start.
That is, the following should hold:
:$S - h = b t $
Indeed:
== Also known as ==
== Historical Note ==
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : $\S 2$: Continuum Property: $\S 2.1$: Achilles and the tortoise
* 
* 
* 
* 
* 
* 
* 
* 
Category:Achilles Paradox
Category:Zeno's Paradoxes
Category:Antinomies
Achilles Paradox/Historical Note

== Historical Note on Achilles Paradox ==
The Achilles Paradox is one of Zeno's Paradoxes, as famously raised by .
Some attention was given to this problem by in around $1646$. However, his solution would not be accepted by today's standards of mathematical rigor.
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : Chapter $\text{III}$: Gentleman, Soldier and Mathematician
* : $\S 2$: Continuum Property: $\S 2.1$: Achilles and the tortoise
Category:Historical Notes
Ackermann-Péter Function/Examples

== Examples of Ackermann-Péter Function ==
The Ackermann-Péter function $A: \Z_{\ge 0} \times \Z_{\ge 0} \to \Z_{> 0}$ is defined as:
:$\map A {m, n} = \begin{cases} n + 1 & : m = 0 \\
\map A {m - 1, 1} & : m > 0, n = 0 \\
\map A {m - 1, \map A {m, n - 1} } & : \text{otherwise} \end{cases}$
:$\begin {array} {c|c|c|c}
\map A {m, n} & m = 0 & m = 1 & m = 2 & m = 3 \\
\hline
n = 0 & 1 & \map A {0, 1} & \map A {1, 1} & \map A {2, 1} \\
n = 1 & 2 & \map A {0, \map A {1, 0} } & \map A {1, \map A {2, 0} } & \map A {2, \map A {3, 0} } \\
n = 2 & 3 & \map A {0, \map A {1, 1} } & \map A {1, \map A {2, 1} } & \map A {2, \map A {3, 1} } \\
n = 3 & 4 & \map A {0, \map A {1, 2} } & \map A {1, \map A {2, 2} } & \map A {2, \map A {3, 2} } \\
n = 4 & 5 & \map A {0, \map A {1, 3} } & \map A {1, \map A {2, 3} } & \map A {2, \map A {3, 3} } \\
\end{array}$
which leads to:
:$\begin{array}{c|c|c|c}
\map A {m, n} & m = 0 & m = 1 & m = 2 & m = 3 \\
\hline
n = 0 & 1 & 2 & 3 & 5 \\
n = 1 & 2 & 3 & 5 & 13 \\
n = 2 & 3 & 4 & 7 & \map A {2, 13} \\
n = 3 & 4 & 5 & 9 & \map A {2, \map A {3, 2} } \\
n = 4 & 5 & 6 & 11 & \map A {2, \map A {3, 3} } \\
\end{array}$
Category:Ackermann-Péter Function
Ackermann-Péter Function at (1,y)

== Theorem ==
For every $y \in \N$:
:$\map A {1, y} = y + 2$
where $A$ is the Ackermann-Péter function.
== Proof ==
Proceed by induction on $y$.
=== Basis for the Induction ===
Suppose $y = 0$.
Then:
This is the basis for the induction
=== Induction Hypothesis ===
This is the induction hypothesis:
Suppose that:
:$\map A {1, y} = y + 2$
We need to show that:
:$\map A {1, y + 1} = y + 3$
=== Induction Step ===
This is the induction step:
Thus, by Principle of Mathematical Induction, the result holds for all $y$.
Category:Ackermann-Péter Function
Ackermann-Péter Function at (2,y)

== Theorem ==
For every $y \in \N$:
:$\map A {2, y} = 2 y + 3$
where $A$ is the Ackermann-Péter function.
== Proof ==
Proceed by induction on $y$.
=== Basis for the Induction ===
Suppose $y = 0$.
Then:
This is the basis for the induction.
=== Induction Hypothesis ===
This is the induction hypothesis:
Suppose:
:$\map A {2, y} = 2 y + 3$
We want to show that:
:$\map A {2, y + 1} = 2 \paren {y + 1} + 3 = 2 y + 5$
=== Induction Step ===
This is the induction step:
By the Principle of Mathematical Induction, the result holds for all $y \in \N$.
Category:Ackermann-Péter Function
Ackermann-Péter Function is Greater than Second Argument

== Theorem ==
For all $x, y \in \N$:
:$\map A {x, y} > y$
where $A$ is the Ackermann-Péter function.
== Proof ==
First, perform induction on $x$.
=== Basis for the Induction ===
Suppose $x = 0$.
Then:
This is the basis for the induction.
=== Induction Hypothesis $\paren 1$ ===
This is the induction hypothesis
Suppose that, for every $y \in \N$:
:$\map A {x, y} > y$
We want to show that, for every $y \in \N$:
:$\map A {x + 1, y} > y$
=== Induction Step ===
This is the induction step:
We will now perform another induction on $y$.
==== Basis for the Induction ====
Suppose $y = 0$.
Then:
This is the basis for the induction.
==== Induction Hypothesis $\paren 2$ ====
This is the induction hypothesis:
Suppose that:
:$\map A {x + 1, y} > y$
We want to show that:
:$\map A {x + 1, y + 1} > y + 1$
==== Induction Step ====
This is the induction step:
By the Principle of Mathematical Induction, the Induction Step $\paren 1$ holds.
By the Principle of Mathematical Induction, the result holds.
Ackermann-Péter Function is Strictly Increasing on First Argument

== Theorem ==
For all $x, y \in \N$:
:$\map A {x + 1, y} > \map A {x, y}$
where $A$ is the Ackermann-Péter function.
=== General Result ===
== Proof ==
Category:Ackermann-Péter Function is Strictly Increasing on First Argument
Category:Ackermann-Péter Function
Ackermann-Péter Function is Strictly Increasing on First Argument/General Result

== Theorem ==
Forall $x, y, z \in \N$ such that:
:$x < y$
we have:
:$\map A {x, z} < \map A {y, z}$
where $A$ is the Ackermann-Péter function.
== Proof ==
Let $y$ be expressed as:
:$y = x + k$
for some $k \in \N_{>0}$.
Proceed by induction on $k$.
=== Basis for the Induction ===
Suppose $k = 1$.
Then:
:$\map A {x, z} < \map A {x + 1, z}$
by Ackermann-Péter Function is Strictly Increasing on First Argument.
This is the basis for the induction.
=== Induction Hypothesis ===
The induction hypothesis is:
:$\map A {x, z} < \map A {x + k, z}$
We want to show:
:$\map A {x, z} < \map A {x + k + 1, z}$
=== Induction Step ===
Thus, the induction step is satisfied.
Thus, by Principle of Mathematical Induction, the result holds forall $x, y, z \in \N$.
Category:Ackermann-Péter Function is Strictly Increasing on First Argument
Ackermann-Péter Function is Strictly Increasing on Second Argument

== Theorem ==
For all $x, y \in \N$:
:$\map A {x, y + 1} > \map A {x, y}$
where $A$ is the Ackermann-Péter function.
=== General Result ===
== Proof ==
Proceed by induction on $x$.
=== Basis for the Induction ===
Suppose $x = 0$.
Then:
This is the basis for the induction.
=== Induction Hypothesis ===
This is the induction hypothesis:
Suppose that:
:$\map A {x, y + 1} > \map A {x, y}$
We want to show that:
:$\map A {x + 1, y + 1} > \map A {x + 1, y}$
=== Induction Step ===
This is the induction step:
By the Principle of Mathematical Induction, the result holds.
Category:Ackermann-Péter Function is Strictly Increasing on Second Argument
Category:Ackermann-Péter Function
Ackermann-Péter Function is Strictly Increasing on Second Argument/General Result

== Theorem ==
For all $x, y, z \in \N$ such that:
:$y < z$
we have:
:$\map A {x, y} < \map A {x, z}$
where $A$ is the Ackermann-Péter function.
== Proof ==
Let $z$ be expressed as:
:$z = y + k$
for some $k \in \N_{>0}$.
Proceed by induction on $k$.
=== Basis for the Induction ===
Suppose $k = 1$.
Then:
This is the basis for the induction.
=== Induction Hypothesis ===
This is the induction hypothesis:
Suppose that:
:$\map A {x, y} < \map A {x, y + k}$
We want to show that:
:$\map A {x, y} < \map A {x, y + k + 1}$
=== Induction Step ===
This is the induction step:
By Principle of Mathematical Induction, the result follows.
Category:Ackermann-Péter Function is Strictly Increasing on Second Argument
Ackermann Function/Examples

== Examples of Ackermann Function ==
The Ackermann function $A: \Z_{\ge 0} \times \Z_{\ge 0} \to \Z_{> 0}$ is defined as:
:$\map A {m, n} = \begin{cases} 2 n & : m = 1 \\
m & : m > 1, n = 1 \\
\map A {m - 1, \map A {m, n - 1} } & : \text{otherwise} \end{cases}$
:$\begin{array}{c|c|c|c}
\map A {m, n} & m = 1 & m = 2 & m = 3 & m = 4 & \cdots & m = k \\
\hline
n = 1 & 2 & 2 & 3 & 4 & & k \\
n = 2 & 4 & \map A {1, \map A {2, 1} } & \map A {2, \map A {3, 1} } & \map A {3, \map A {4, 1} } & & \map A {k - 1, \map A {k, 1} } \\
n = 3 & 6 & \map A {1, \map A {2, 2} } & \map A {2, \map A {3, 2} } & \map A {3, \map A {4, 2} } & & \map A {k - 1, \map A {k, 2} } \\
n = 4 & 8 & \map A {1, \map A {2, 3} } & \map A {2, \map A {3, 3} } & \map A {3, \map A {4, 3} } & & \map A {k - 1, \map A {k, 3} } \\
n = 5 & 10 & \map A {1, \map A {2, 4} } & \map A {2, \map A {3, 4} }\ & \map A {3, \map A {4, 4} } & & \map A {k - 1, \map A {k, 4} } \\
\vdots & & & & & & \\
n = j & 2 j & \map A {1, \map A {2, j - 1} } & \map A {2, \map A {3, j - 1} } & \map A {3, \map A {4, j - 1} } & & \map A {k - 1, \map A {k, j - 1} } \\
\end{array}$
which leads to:
:$\begin{array}{c|c|c|c}
\map A {m, n} & m = 1 & m = 2 & m = 3 & m = 4 & \cdots & m = k \\
\hline
n = 1 & 2 & 2 & 3 & 4 & & k \\
n = 2 & 4 & 4 & 8 & \map A {3, 4} & & \map A {k - 1, k} \\
n = 3 & 6 & 8 & 2^8 & \map A {3, \map A {4, 2} } & & \map A {k - 1, \map A {k, 2} } \\
n = 4 & 8 & 16 & 2^{2^8} & \map A {3, \map A {4, 3} } & & \map A {k - 1, \map A {k, 3} } \\
n = 5 & 10 & 32 & \map A {2, \map A {3, 4} } & \map A {3, \map A {4, 4} } & & \map A {k - 1, \map A {k, 4} } \\
\vdots & & & & \\
n = j & 2 j & 2^j & \map A {2, \map A {3, j - 1} } & \map A {3, \map A {4, j - 1} } & & \map A {k - 1, \map A {k, j - 1} } \\
\end{array}$
== Sources ==
* : $2^{65536}$
* : $2^{65,536}$
Category:Ackermann Functions
Acnode/Examples/Policeman on Point Duty

== Examples of Acnodes ==
Consider the locus of the equation:
:$x^2 y^2 = a^2 \paren {x^2 + y^2}$
This is the classic quartic curve "policeman on point duty":
:
which has an acnode at $\tuple {0, 0}$.
Category:Examples of Acnodes
Action in terms of Kinetic Energy

== Definition ==
Let $S$ be a dynamical system
Let $A$ be the action for a given segment from point $A$ to point $B$ on the trajectory of $S$ is defined by the line integral:
:$\ds \sum_i \int_A^B p_i \rd q_i$
Let $A$ act over a time interval of length $t$.
Then:
:$A = 2 K t$
where $K$ denotes the mean kinetic energy of $S$ over $t$.
== Proof ==
== Sources ==
* 
* 
Category:Action (Dynamics)
Category:Kinetic Energy
Action of Group on Coset Space is Group Action

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Let $*: G \times G / H \to G / H$ be the action on the (left) coset space:
:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
Then $G$ is a group action.
== Proof ==
demonstrating that $*$ satisfies .
Then:
demonstrating that $*$ satisfies .
== Sources ==
* : Chapter $6$: Cosets: Exercise $9$
* : Chapter $2$: The Sylow Theorems: $\S 53$
Category:Cosets
Category:Group Action on Coset Space
Action of Inverse of Group Element

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $S$ be a sets.
Let $*: G \times S \to S$ be a group action.
Then:
:$g * a = b \iff g^{-1} * b = a$
== Proof ==
and:
== Sources ==
* : $\S 5.6$. Stabilizers: Lemma
Category:Group Actions
Adapted Stochastic Process at Stopping Time is Measurable with respect to Stopped Sigma-Algebra

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-adapted stochastic process.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let $X_T$ be the adapted stochastic process $\sequence {X_n}_{n \ge 0}$ stopped at $T$.
Let $\FF_T$ be the stopped $\sigma$-algebra associated with $T$.
Then $X_T$ is $\FF_T$-measurable. 
== Proof ==
We have that if $\map T \omega = t$ for $\omega \in \Omega$ and $t \in \Z_{\ge 0}$ then: 
:$\map {X_T} \omega = \map {X_t} \omega$
We aim to show that for each Borel set $A \subseteq \R$:
:$\ds \set {\omega \in \Omega : \map {X_T} \omega \in A} \in \FF_T$
That is, we want to show that:
:$\set {\omega \in \Omega : \map {X_T} \omega \in A} \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for $t \in \Z_{\ge 0}$.
Let $t \in \Z_{\ge 0}$. 
We have that:
:$\ds \set {\omega \in \Omega : \map {X_T} \omega \in A} \cap \set {\omega \in \Omega : \map T \omega \le t} = \bigcup_{s \in \Z_{\ge 0}, \, 0 \le s \le t} \set {\omega \in \Omega : \map T \omega = s} \cap \set {\omega \in \Omega : \map {X_s} \omega \in A}$
Since $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$, we have:
:$\set {\omega \in \Omega : \map T \omega = s} \in \FF_s$
Since $\sequence {\FF_n}_{n \ge 0}$ is a filtration of $\sigma$-algebra and $s \le t$:
:$\set {\omega \in \Omega : \map T \omega = s} \in \FF_t$
Since $\sequence {X_n}_{n \ge 0}$ is $\sequence {\FF_n}_{n \ge 0}$-adapted, we have that $X_s$ is $\FF_s$-measurable.
So, we have:
:$\set {\omega \in \Omega : \map {X_s} \omega \in A} \in \FF_s$
and so:
:$\set {\omega \in \Omega : \map {X_s} \omega \in A} \in \FF_t$ 
Since $\FF_t$ is closed under finite intersection, we have:
:$\set {\omega \in \Omega : \map T \omega = s} \cap \set {\omega \in \Omega : \map {X_s} \omega \in A} \in \FF_t$
Since $\FF_t$ is closed under finite union, we have:
:$\ds \bigcup_{s \in \Z_{\ge 0}, \, 0 \le s \le t} \set {\omega \in \Omega : \map T \omega = s} \cap \set {\omega \in \Omega : \map {X_s} \omega \in A} \in \FF_t$
and so:
:$\set {\omega \in \Omega : \map {X_T} \omega \in A} \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for all $t \in \Z_{\ge 0}$. 
We therefore have: 
:$\set {\omega \in \Omega : \map {X_T} \omega \in A} \in \FF_T$
So $X_T$ is $\FF_T$-measurable.
== Sources ==
* : Lemma $9.23$
Category:Stopped Sigma-Algebras
Category:Stopping Times
Category:Adapted Stochastic Processes
Adapted Stochastic Process is Martingale iff Supermartingale and Submartingale/Continuous Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $\sequence {X_t}_{t \ge 0}$ be a continuous-time $\sequence {\FF_t}_{t \ge 0}$-adapted stochastic process.
Then $\sequence {X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-martingale it is a $\sequence {\FF_t}_{t \ge 0}$-supermartingale and a $\sequence {\FF_t}_{t \ge 0}$-submartingale. 
== Proof ==
For each $t, s \in \hointr 0 \infty$ with $0 \le s < t$, we have: 
:$\expect {X_t \mid \FF_s} = X_s$ almost surely
:
:$\expect {X_t \mid \FF_s} \le X_s$ almost surely
and:
:$\expect {X_t \mid \FF_s} \ge X_s$ almost surely.
That is: 
:$\sequence {X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-martingale it is a $\sequence {\FF_t}_{t \ge 0}$-supermartingale and a $\sequence {\FF_t}_{t \ge 0}$-submartingale.
Category:Adapted Stochastic Process is Martingale iff Supermartingale and Submartingale
Adapted Stochastic Process is Martingale iff Supermartingale and Submartingale/Discrete Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \mathop \ge 0}, \Pr}$ be a discrete-time filtered probability space.
Let $\sequence {X_n}_{n \mathop \ge 0}$ be a discrete-time $\sequence {\FF_n}_{n \mathop \ge 0}$-adapted stochastic process.
Then $\sequence {X_n}_{n \mathop \ge 0}$ is a $\sequence {\FF_n}_{n \mathop \ge 0}$-martingale it is a $\sequence {\FF_n}_{n \mathop \ge 0}$-supermartingale and a $\sequence {\FF_n}_{n \mathop \ge 0}$-submartingale. 
== Proof ==
For each $n \in \Z_{\ge 0}$, we have: 
:$\expect {X_{n + 1} \mid \FF_n} = X_n$ almost surely
:
:$\expect {X_{n + 1} \mid \FF_n} \le X_n$ almost surely
and:
:$\expect {X_{n + 1} \mid \FF_n} \ge X_n$ almost surely.
That is: 
:$\sequence {X_n}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \mathop \ge 0}$-martingale it is a $\sequence {\FF_n}_{n \mathop \ge 0}$-supermartingale and a $\sequence {\FF_n}_{n \mathop \ge 0}$-submartingale.
== Sources ==
* : $10.3$: Martingale, supermartingale, submartingale
Category:Adapted Stochastic Process is Martingale iff Supermartingale and Submartingale
Adapted Stochastic Process is Supermartingale iff Negative is Submartingale/Continuous Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $\sequence {X_t}_{t \ge 0}$ be a $\sequence {\FF_t}_{t \ge 0}$-adapted stochastic process.
Then $\sequence {X_t}_{t \ge 0}$ is a supermartingale $\sequence {-X_t}_{t \ge 0}$ is a submartingale.
== Proof ==
Since $\sequence {X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-adapted stochastic process:
:$X_t$ is $\FF_t$-measurable for each $t \in \hointr 0 \infty$.
From Pointwise Scalar Multiple of Measurable Function is Measurable:
:$-X_t$ is $\FF_t$-measurable for each $t \in \hointr 0 \infty$.
So $\sequence {-X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-adapted stochastic process.
We then just need to check conditional expectations.
Let $s, t \in \hointr 0 \infty$ with $0 \le s < t$. 
From Conditional Expectation is Linear, we have: 
:$\expect {X_t \mid \FF_s} \le X_s$ almost surely $\expect {-X_t \mid \FF_s} \ge -X_s$ almost surely.
That is: 
:$\sequence {X_t}_{t \ge 0}$ is a supermartingale $\sequence {-X_t}_{t \ge 0}$ is a submartingale.
== Sources ==
* : $3.3$: Continuous Time Martingales and Supermartingales
Category:Adapted Stochastic Process is Supermartingale iff Negative is Submartingale
Adapted Stochastic Process is Supermartingale iff Negative is Submartingale/Discrete Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a discrete-time filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be a discrete-time $\sequence {\FF_n}_{n \ge 0}$-adapted stochastic process.
Then $\sequence {X_n}_{n \ge 0}$ is a supermartingale $\sequence {-X_n}_{n \ge 0}$ is a submartingale.
== Proof ==
Since $\sequence {X_n}_{n \ge 0}$ is a discrete-time $\sequence {\FF_n}_{n \ge 0}$-adapted stochastic process:
:$X_n$ is $\FF_n$-measurable for each $n \in \N$.
From Pointwise Scalar Multiple of Measurable Function is Measurable:
:$-X_n$ is $\FF_n$-measurable for each $n \in \N$.
So $\sequence {-X_n}_{n \ge 0}$ is a discrete-time $\sequence {\FF_n}_{n \ge 0}$-adapted stochastic process.
We then just need to check conditional expectations.
From Conditional Expectation is Linear, we have: 
:$\expect {X_{n + 1} \mid \FF_n} \le X_n$ almost surely $\expect {-X_{n + 1} \mid \FF_n} \ge -X_n$ almost surely.
That is: 
:$\sequence {X_n}_{n \ge 0}$ is a supermartingale $\sequence {-X_n}_{n \ge 0}$ is a submartingale.
== Sources ==
* : $10.3$: Martingale, supermartingale, submartingale
Category:Adapted Stochastic Process is Supermartingale iff Negative is Submartingale
Adding Edge to Tree Creates One Cycle

== Theorem ==
Adding a new edge to a tree can create no more than one cycle.
== Proof ==
From Equivalent Definitions for Finite Tree, adding an edge creates at least one cycle.
Suppose that adding an edge $\tuple {u, v}$ to a tree $T$ creates two or more cycles.
Let two such cycles be $\tuple {u, v, \ldots, u_1, u_2, \ldots, u}$ and $\tuple {u, v, \ldots, v_1, v_2, \ldots, u}$.
By removing the edge $\tuple {u, v}$ from this cycle, we have two paths from $v$ to $u$:
:$\tuple {v, \ldots, u_1, u_2, \ldots, u}$
:$\tuple {v, \ldots, v_1, v_2, \ldots, u}$.
But that means $T$ has more than one path between two nodes.
From Path in Tree is Unique, that means $T$ can not be a tree.
Hence the result.
Category:Tree Theory
Addition Formulas for Hyperbolic Functions

== Theorem ==
=== Hyperbolic Sine of Sum ===
=== Hyperbolic Cosine of Sum ===
=== Hyperbolic Tangent of Sum ===
=== Hyperbolic Cotangent of Sum ===
== Also known as ==
== Also see ==
* Subtraction Formulas for Hyperbolic Functions
* Addition Formulas
* Subtraction Formulas
Category:Addition Formulas for Hyperbolic Functions
Category:Hyperbolic Identities
Addition Formulas for Hyperbolic Functions/Also known as

== Addition Formulas for Hyperbolic Functions: Also known as ==
The Addition Formulas for Hyperbolic Functions are also known as the compound angle formulas (for hyperbolic functions).
However, it is the view of that the arguments of the hyperbolic functions are in general not actually angles as they frequently are for the Compound Angle Formulas, and hence is a misnomer.
== Sources ==
* 
Category:Addition Formulas for Hyperbolic Functions
Addition Law of Probability

== Theorem ==
Let $\Pr$ be a probability measure on an event space $\Sigma$.
Let $A, B \in \Sigma$.
Then:
:$\map \Pr {A \cup B} = \map \Pr A + \map \Pr B - \map \Pr {A \cap B}$
That is, the probability of either event occurring equals the sum of their individual probabilities less the probability of them both occurring.
This is known as the addition law of probability.
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Also known as ==
== Sources ==
* 
* 
Category:Addition Law of Probability
Category:Probability Theory
Category:Named Theorems