nyuuzyou/wikis huggingface dataset:
Trivial Module is Not Unitary

== Theorem ==
Let $\struct {G, +_G}$ be an abelian group whose identity is $e_G$.
Let $\struct {R, +_R, \circ_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be the trivial $R$-module, such that:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$
Then unless $R$ is a ring with unity and $G$ contains only one element, this is not a unitary module.
== Proof ==
By definition, for a trivial module to be unitary, $R$ needs to be a ring with unity.
For to apply, we require that:
:$\forall x \in G: 1_R \circ x = x$
But for the trivial module:
:$\forall x \in G: 1_R \circ x = e_G$
So can apply only when:
:$\forall x \in G: x = e_G$
Thus for the trivial module to be unitary, it is necessary that $G$ be the trivial group, and thus to contain one element.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.6$
Category:Module Theory
Trivial Norm on Division Ring is Non-Archimedean

== Theorem ==
Let $\struct {R, +, \circ}$ be a division ring whose ring zero is $0_R$.
Then the trivial norm $\norm {\, \cdot \,}: R \to \R_{\ge 0}$, which is given by:
:$\norm x = \begin{cases}
 0 & : x = 0_R \\
 1 & : \text{ otherwise}
\end{cases}$
is non-archimedean:
:$\norm {x + y} \le \max \set {\norm x, \norm y}$
== Proof ==
Let $x, y = 0_R$.
Then:
:$\norm x, \norm y = 0$
Therefore:
:$\max \set {\norm x, \norm y} = 0$.
Hence:
Let $x \ne 0_R$ or $y \ne 0_R$.
Then:
:$\norm x = 1$ or $\norm y = 1$
Therefore:
:$\max \set {\norm x, \norm y} = 1$.
Hence:
Category:Trivial Norms
Trivial Norm on Division Ring is Norm

== Theorem ==
Let $\struct {R, +, \circ}$ be a division ring, and denote its ring zero by $0_R$.
Then the trivial norm $\norm {\, \cdot \,}: R \to \R_{\ge 0}$, which is given by:
:$ \norm x = \begin{cases} 0 & \text { if } x = 0_R\\ 1 & \text { otherwise} \end{cases}$
defines a norm on $R$.
== Proof ==
Proving each of the norm axioms one by one:
=== Proving : $\forall x \in R: \norm x = 0 \iff x = 0_R$ ===
This follows directly from the definition of the trivial norm.
=== Proving : $\forall x, y \in R: \norm {x \circ y} = \norm x \times \norm y$ ===
If $x = 0_R$:
The reasoning is similar if $y = 0_R$.
If $x,y \ne 0_R$, then $x \circ y \ne 0_R$ by alternative definition $(3)$ of division rings. We get:
=== Proving : $\forall x, y \in R: \norm {x + y} \le \norm x + \norm y$ ===
If $x = 0_R$:
The reasoning is similar if $y = 0_R$.
If $x, y \ne 0_R$:
Category:Division Rings
Category:Trivial Norms
Trivial Ordering Compatibility in Boolean Ring

== Theorem ==
Let $\struct {S, +, \circ}$ be a Boolean ring.
Then the trivial ordering is the only ordering on $S$ compatible with both its operations.
== Proof ==
That the trivial ordering is compatible with $\circ$ and $*$ follows from Trivial Ordering is Universally Compatible.
Conversely, suppose that $\preceq$ is an ordering compatible with $\circ$ and $*$.
We recall the definition of the trivial ordering:
Let $a, b \in S$ such that $a \preceq b$.
Since $\preceq$ is compatible with $\circ$ and $*$, we have:
We have that a Boolean ring is an idempotent ring.
Hence we have:
We have shown that:
:$0 \preceq b - a$
and:
:$b - a \preceq 0$
By definition of ordering, $\preceq$ is antisymmetric.
This means:
:$b - a = 0$
and so:
:$a = b$
Hence $\preceq$ is the trivial ordering.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Exercise $23.34$
Category:Boolean Rings
Category:Order Theory
Trivial Ordering is Universally Compatible

== Theorem ==
Let $S$ be a set.
Let $\RR$ be the trivial ordering on $S$.
Then $\RR$ is universally compatible.
== Proof ==
To prove that the trivial ordering is in fact an ordering, we need to checking each of the criteria for an ordering:
=== Reflexivity ===
:$\forall a \in S: a \mathrel \RR a$:
From its definition, we have $\forall a, b \in S: a = b \implies a \mathrel \RR b$.
Thus, as $a = a$, we have $\forall a \in S: a \mathrel \RR a$.
So reflexivity is proved.
=== Transitivity ===
:$\forall a, b, c \in S: a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c$:
From the definition:
:$a \mathrel \RR b \iff a = b$
:$b \mathrel \RR c \iff b = c$
So as $a = b \land b = c \implies a = c$ from transitivity of equals, we have that $a \mathrel \RR c$ and thus transitivity is proved.
=== Antisymmetry ===
:$\forall a, b \in S: a \mathrel \RR b \land b \mathrel \RR a \implies a = b$:
From the definition:
:$a \mathrel \RR b \iff a = b$
:$b \mathrel \RR a \iff b = a$
Antisymmetry follows from symmetry of equals.
The trivial ordering is by definition the same as the diagonal relation, and is therefore universally compatible.
Category:Order Theory
Trivial Quotient Group is Quotient Group

== Theorem ==
Let $G$ be a group.
Then the trivial quotient group:
:$G / \set {e_G} \cong G$
where:
:$\cong$ denotes group isomorphism
:$e_G$ denotes the identity element of $G$
is a quotient group.
== Proof ==
From Trivial Subgroup is Normal:
:$\set {e_G} \lhd G$
Let $x \in G$.
Then:
:$x \set {e_G} = \set {x e_G} = \set x$
So each (left) coset of $G$ modulo $\set {e_G}$ has one element.
Now we set up the quotient epimorphism $\psi: G \to G / \set {e_G}$:
:$\forall x \in G: \map \phi x = x \set {e_G}$
which is of course a surjection.
We now need to establish that it is an injection.
Let $p, q \in G$.
So $\psi$ is a group isomorphism and therefore:
:$G / \set {e_G} \cong G$
== Sources ==
* : $\S 7.4$. Kernel and image: Example $140$
* : $1$: Introduction to Finite Group Theory: $1.7$
Category:Examples of Quotient Groups
Category:Group Isomorphisms
Trivial Quotient is a Bijection

== Theorem ==
Let $\Delta_S$ be the diagonal relation on a set $S$.
Let $q_{\Delta_S}: S \to S / \Delta_S$ be the trivial quotient of $S$.
Then $q_{\Delta_S}: S \to S / \Delta_S$ is a bijection.
== Proof ==
The diagonal relation is defined as:
:$\Delta_S = \set {\tuple {x, x}: x \in S} \subseteq S \times S$
From the fact that $q_{\Delta_S}$ is a quotient mapping, we know that it is a surjection.
It relates each $x \in S$ to the singleton $\set x$.
Thus:
:$\set x = \eqclass x {\Delta_S} \subseteq S$
So $\map {q_{\Delta_S} } x = \set x$, and it follows that:
:$\eqclass x {\Delta_S} = \eqclass y {\Delta_S} \implies x = y$
Thus $q_{\Delta_S}$ is injective, and therefore by definition a bijection.
== Comment ==
Some sources abuse notation and write $\map {q_{\Delta_S} } x = x$, which serves to emphasise its triviality, but can cause it to be conflated with the identity mapping.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Quotient Functions
Category:Quotient Mappings
Category:Bijections
Trivial Relation is Equivalence

== Theorem ==
The trivial relation on $S$:
:$\RR = S \times S$
is always an equivalence in $S$.
== Proof ==
Let us verify the conditions for an equivalence in turn.
=== Reflexivity ===
For $\RR$ to be reflexive means:
:$\forall x \in S: \tuple {x, x} \in S \times S$
which is trivial by definition of the Cartesian product $S \times S$.
=== Symmetry ===
For $\RR$ to be symmetric means:
:$\forall x, y \in S: \tuple {x, y} \in S \times S \land \tuple {y, x} \in S \times S$
Since we have by definition of Cartesian product that:
:$\forall x, y \in S: \tuple {y, x} \in S \times S$
this follows by True Statement is implied by Every Statement.
=== Transitivity ===
For $\RR$ to be transitive means:
:$\tuple {x, y} \in S \times S \land \tuple {y, z} \in S \times S \implies \tuple {x, z} \in S \times S$
By definition of Cartesian product, we have that:
:$\forall x, z \in S: \tuple {x, z} \in S \times S$
hence by True Statement is implied by Every Statement, it follows that $\RR$ is transitive.
Having verified all three conditions, we conclude $\RR$ is an equivalence.
== Sources ==
* : $\S 7$: Relations
* : Chapter $\text{I}$: Sets and Functions: Relations
* : Appendix $\text{A}.3$: Equivalence Relations: Problem Set $\text{A}.3$: $15$
Category:Examples of Equivalence Relations
Category:Trivial Relation
Trivial Relation is Largest Equivalence Relation

== Theorem ==
The trivial relation $\TT$ on $S$ is the largest equivalence in $S$, in the sense that:
:$\forall \EE \subseteq S \times S: \EE \subseteq \TT$
where $\EE$ denotes a general equivalence relation.
== Proof ==
The trivial relation $\TT$ on $S$ is defined as:
:$\TT = S \times S$
It is confirmed from Trivial Relation is Equivalence that the trivial relation is in fact an equivalence relation.
Let $\EE$ be an arbitrary equivalence relation on $S$.
By definition of relation, $\EE \subseteq S \times S$ and so (trivially) $\EE \subseteq \TT$.
== Sources ==
* : $\S 7$: Relations
Category:Equivalence Relations
Category:Trivial Relation
Trivial Relation is Universally Congruent

== Theorem ==
The trivial relation $\RR = S \times S$ on a set $S$ is universally congruent on $S$.
== Proof ==
Let $\struct {S, \circ}$ be any algebraic structure which is closed for $\circ$.
By definition of trivial relation:
:$x \in S \land y \in S \implies x \mathrel \RR y$
So:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Example $11.3$
Category:Examples of Congruence Relations
Category:Trivial Relation
Trivial Ring from Abelian Group

== Theorem ==
An abelian group $\struct {G, +}$ may be turned into a trivial ring by defining the ring product to be:
:$\forall x, y \in G: x \circ y = e_G$
== Proof ==
Follows directly from the definition of a trivial ring.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $3$
Category:Trivial Rings
Category:Abelian Groups
Trivial Ring is Commutative Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a trivial ring.
Then $\struct {R, +, \circ}$ is a commutative ring.
== Proof ==
First we need to show that a trivial ring is actually a ring in the first place.
Taking the ring axioms in turn:
=== ===
$\struct {R, +}$ is an abelian group:
This follows from the definition.
=== ===
$\struct {R, \circ}$ is closed:
From Ring Product with Zero:
:$x \circ y = 0_R \in R$
=== ===
$\circ$ is associative on $\struct {R, +, \circ}$:
:$x \circ \paren {y \circ z} = 0_R = \paren {x \circ y} \circ z$
=== ===
$\circ$ distributes over $+$ in $\struct {R, +, \circ}$:
:$x \circ \paren {y + z} = 0_R$
by definition.
Then:
and the same for $\paren {y + z} \circ x$.
=== Commutativity ===
From the definition of trivial ring:
:$\forall x, y \in R: x \circ y = 0_R = y \circ x$
Hence its commutativity. 
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Example $21.5$
Category:Ring Theory
Category:Commutative Rings
Category:Trivial Rings
Trivial Solution of Homogeneous Linear 1st Order ODE

== Theorem ==
The homogeneous linear first order ODE:
:$\dfrac {\d y} {\d x} + \map Q x y = 0$
has the particular solution:
:$\map y x = 0$
that is, the zero constant function.
This particular solution is referred to as the trivial solution.
== Proof ==
We have:
:$\map {\dfrac {\d} {\d x} } 0 = 0$
from which:
:$\dfrac {\d y} {\d x} + \map Q x y = 0$
Hence the result.
Category:Linear First Order ODEs
Trivial Solution of Homogeneous Linear 2nd Order ODE

== Theorem ==
The homogeneous linear second order ODE:
:$\dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
has the particular solution:
:$\map y x = 0$
that is, the zero constant function.
This particular solution is referred to as the trivial solution.
== Proof ==
We have:
:$\map {\dfrac {\d} {\d x} } 0 = 0$
and so:
:$\map {\dfrac {\d^2} {\d x^2} } 0 = 0$
from which:
:$\dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
Hence the result.
== Sources ==
* : $\S 3.14$: Second Order Linear Equations: Introduction
Category:Homogeneous LSOODEs
Trivial Solution to System of Homogeneous Simultaneous Linear Equations is Solution

== Theorem ==
Let $S$ be a system of homogeneous simultaneous linear equations:
:$\ds \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \alpha_{i j} x_j = 0$
Consider the trivial solution to $A$:
:$\tuple {x_1, x_2, \ldots, x_n}$
such that:
:$\forall j \in \set {1, 2, \ldots, n}: x_j = 0$
Then the trivial solution is indeed a solution to $S$.
== Proof ==
Let $i \in \set {1, 2, \ldots, m}$.
We have:
This holds for all $i \in \set {1, 2, \ldots, m}$.
Hence:
:$\ds \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \alpha_{i j} x_j = 0$
and the result follows.
Category:Simultaneous Linear Equations
Trivial Subgroup is Normal

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then the trivial subgroup $\struct {\set e, \circ}$ of $G$ is a normal subgroup in $G$.
== Proof ==
First, by Trivial Subgroup is Subgroup, $\struct {\set e, \circ}$ is a subgroup of $G$.
To show $\struct {\set e, \circ}$ is normal in $G$:
:$\forall a \in G: a \circ e \circ a^{-1} = a \circ a^{-1} = e$
Hence the result.
== Sources ==
* : $\S 6.6$. Normal subgroups: Example $123$
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures
* : $\S 1.10$: Example $35$
* : $1$: Introduction to Finite Group Theory: $1.7$
* : $\S 49.1$ Normal subgroups
* : Chapter $7$: Normal subgroups and quotient groups: Definition $7.3$: Remark
Category:Normal Subgroups
Trivial Subgroup is Subgroup

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then the trivial subgroup $\struct {\set e, \circ}$ is indeed a subgroup of $\struct {G, \circ}$.
== Proof ==
Using the One-Step Subgroup Test:
: $(1): \quad e \in \set e \leadsto \set e \ne \O$
: $(2): \quad e \in \set e \leadsto e \circ e^{-1} = e \in \set e$
== Sources ==
* : $\S 5.2$. Subgroups: Example $90$
* : $\S 1.9$: Subgroups: Example $25$
* : Chapter $\text{II}$: Groups: Subgroups
* : Chapter $2$: Subgroups and Cosets: $\S 35$
* : $\S 36.1$: Subgroups
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(ii)}$
* : Chapter $4$: Subgroups: Definition $4.1$
Category:Subgroups
Trivial Topological Space is Indiscrete

== Theorem ==
Let $T = \struct {S, \tau}$ be a trivial topological space.
Then $\tau$ is the indiscrete topology.
== Proof ==
By definition of trivial topological space, $S$ is a singleton.
That is, $S$ is a set containing exactly one element.
Suppose $S = \set x$ for some object $x$. 
Then the power set of $S$ is the set:
:$\powerset S = \set {\O, \set x}$
That is:
:$\powerset S = \set {\O, S}$
Let $\tau$ be a topology on $S$.
We have that $\tau$ is a subset of $\powerset S$.
Hence $\tau$ must equal one of the following sets:
:$\tau_1 = \O$
:$\tau_2 = \set \O$
:$\tau_3 = \set S$
or
:$\tau_4 = \set {\O, S}$.
By definition of a topology, $S \in \tau$.
Thus $\tau \ne \tau_1$ and $\tau \ne \tau_2$.
By Empty Set is Element of Topology, also $\O \in \tau$. 
Therefore $\tau \ne \tau_3$.
By Indiscrete Topology is Topology, $\tau_4$ is a topology on $S$.
Hence if $S$ is a singleton, the only possible topology on $S$ is the indiscrete topology. 
Category:Indiscrete Topology
Category:Trivial Topological Space
Trivial Topological Space is Irreducible

== Theorem ==
Let $T = \struct {\set s, \tau}$ be a trivial topological space.
Then $T$ is irreducible.
== Proof ==
Follows from:
:Trivial Topological Space is Indiscrete
:Indiscrete Space is Irreducible
Category:Irreducible Spaces
Category:Trivial Topological Space
Trivial Topological Space is Non-Meager

== Theorem ==
Let $T = \struct {S, \tau}$ be a trivial topological space.
Then $T$ is non-meager.
== Proof ==
As $T$ is a trivial topological space, by definition $S$ is a singleton: $S = \set s$, say.
Then $\set s$ is an open set.
That is, $s$ is an open point.
The result follows from Space with Open Point is Non-Meager.
== Sources ==
* : Problems: Section $1: \ 4$
Category:Non-Meager Spaces
Category:Trivial Topological Space
Trivial Vector Space iff Zero Dimension

== Theorem ==
Let $V$ be a vector space.
Then $V = \set \bszero$ $\map \dim V = 0$, where $\dim$ signifies dimension of vector space.
== Proof ==
=== Necessary Condition ===
Suppose $V = \set \bszero$.
We have that $V$ has no $\mathbf v \in V, \mathbf v \ne \bszero$.
Thus there exists no $\set {\mathbf v} \subseteq V$ such that $\mathbf v \ne \bszero$.
Such a $\set {\mathbf v}$ would be a linearly independent set.
Hence $\O$ is the only possible basis for $V$.
Hence $\map \dim V = \card \O = 0$.
=== Sufficient Condition ===
Suppose $\map \dim V = 0$.
Then by definition of dimension, $0 = \map \dim V = \card B$, where $B$ is a basis for $V$.
Thus $B = \O$.
Suppose there are sets $\set {\mathbf v}$ such that $\mathbf v \in V, \mathbf v \ne \bszero$.
By Singleton is Linearly Independent, any such $\set {\mathbf v}$ would be a linearly independent set.
So $V$ has no such $\mathbf v \ne \bszero$.
Thus $V = \set \bszero$.
Category:Vector Spaces
Category:Dimension of Vector Space
Trivial Zeroes of Riemann Zeta Function are Even Negative Integers

== Theorem ==
Let $\rho = \sigma + i t$ be a zero of the Riemann zeta function not contained in the critical strip:
:$0 \le \map \Re s \le 1$
Then:
:$s \in \set {-2, -4, -6, \ldots}$
These are called the trivial zeros of $\zeta$.
== Proof ==
First we note that by Zeroes of Gamma Function, $\Gamma$ has no zeroes on $\C$.
Therefore, the completed Riemann zeta function:
:$\map \xi s = \dfrac 1 2 s \paren {s - 1} \pi^{-s/2} \map \Gamma {\dfrac s 2} \, \map \zeta s$
has the same zeroes as $\zeta$.
Additionally by Functional Equation for Riemann Zeta Function, we have:
:$\map \xi s = \map \xi {1 - s}$
for all $s \in \C$.
Therefore if:
:$\map \zeta s \ne 0$
for all $s$ with $\map \Re s > 1$, then also:
:$\map \zeta s \ne 0$
for all $s$ with $\map \Re s < 0$.
Let us consider $\map \Re s > 1$.
We have:
:$\ds \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
where here and in the following $p$ ranges over the primes.
Therefore, we have:
:$\ds \map \zeta s \prod_p \paren {1 - p^{-s} } = 1$
All of the factors of this infinite product can be found in the product:
:$\ds \prod_{n \mathop = 2}^\infty \paren {1 - n^{-s} }$
which converges absolutely.
This follows because from Convergence of P-Series:
:$\ds \sum_{k \mathop = 1}^\infty k^{-s}$ converges absolutely.
Hence:
:$\ds \prod_p \paren {1 - p^{-s} }$ converges absolutely.
So by the fact that:
:$\ds \map \zeta s \prod_p \paren {1 - p^{-s} } = 1$
we know $\map \zeta s$ cannot possibly be zero for any point in the region in question.
== Also see ==
* Riemann Hypothesis: all zeroes in the critical strip fall on the critical line $\map \Re s = \dfrac 1 2$.
* Critical Line Theorem
== Sources ==
* : $0,5$
* 
* 
Category:Riemann Zeta Function
True Statement is implied by Every Statement

== Theorem ==
If something is true, then anything implies it.
==== Formulation 1 ====
==== Formulation 2 ====
== Also see ==
* Paradoxes of Material Implication, in which category this result is grouped
== Sources ==
* : $\S 2.3$: Basic Truth-Tables of the Propositional Calculus
* 
* 
Category:True Statement is implied by Every Statement
Category:Paradoxes of Material Implication
Category:Conditional
True Statement is implied by Every Statement/Formulation 1

== Theorem ==
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables: Exercise $\text{II} \ \mathbf 6.$
* : $\S 1.14$: Exercise $17 \ \text{(ii)}$
Category:True Statement is implied by Every Statement
True Statement is implied by Every Statement/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, where the truth value in the relevant column on the is $\T$, that under the one on the is also $\T$:
$\begin{array}{|c||ccc|} \hline
p & q & \implies & p \\
\hline
\F & \F & \T & \F \\
\F & \T & \F & \F \\
\T & \F & \T & \T \\
\T & \T & \T & \T \\
\hline
\end{array}$
Category:True Statement is implied by Every Statement
Category:Truth Table Proofs
True Statement is implied by Every Statement/Formulation 2

== Theorem ==
:$\vdash q \implies \paren {p \implies q}$
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : $\S \text{II}.13$: Symbolism of sentential calculus
* : $2$ Arguments Containing Compound Statements: $2.4$: Statement Forms
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(2) \ \text{(i)}$
* 
Category:True Statement is implied by Every Statement
True Statement is implied by Every Statement/Formulation 2/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth value under the main connective, the first instance of $\implies$, is $\T$ for each boolean interpretation.
$\begin{array}{|ccccc|} \hline
q & \implies & ( p & \implies & q ) \\
\hline
\F & \T & \T & \F & \F \\
\F & \T & \F & \T & \F \\
\T & \T & \T & \T & \T \\
\T & \T & \F & \T & \T \\
\hline
\end{array}$
Category:True Statement is implied by Every Statement
Category:Truth Table Proofs
True Weight from False Balance/Imbalanced Pans

== Theorem ==
Let $B$ be a body whose weight is $W$.
Let $B$ be weighed in a false balance with imbalanced pans.
Let the readings of the weight of $B$ be $a$ and $b$ when placed in opposite pans.
Then:
:$W = \dfrac {a + b} 2$
== Proof ==
We have that the false balance has pans such that one weighs more than the other.
Let the lengths of the arms of the false balance be $x$.
Let one of the pans weigh $m$ more than the other.
Placing $B$ in the lighter pan gives:
:$W x = \paren {a + m} x$
and placing $B$ in the heavier pan gives:
:$\paren {W + m} x = b x$
Then:
Category:False Balances
True Weight from False Balance/Unequal Arms

== Theorem ==
Let $B$ be a body whose weight is $W$.
Let $B$ be weighed in a false balance with unequal arms.
Let the readings of the weight of $B$ be $a$ and $b$ when placed in opposite pans.
Then:
:$W = \sqrt {a b}$
== Proof ==
We have that the false balance has arms of different lengths.
Let the lengths of the arms of the false balance be $x$ and $y$.
, placing $B$ in the pan at the end of $x$ gives:
:$W x = a y$
and placing $B$ in the pan at the end of $y$ gives:
:$W y = b x$
Then:
Category:False Balances
Truncated Distribution/Examples/Manufacturing Process

== Examples of Truncated Distribution ==
Consider a manufacturing process which makes many items of a particular type.
Suppose there is a quality process in place which causes all items under a certain length to be scrapped, leaving the rest available for sale.
The probability distribution of the lengths of those saleable items will follow a truncated normal distribution.
== Sources ==
* 
* 
Category:Examples of Truncated Distributions
Truncation Error/Examples/Trapezium Rule

== Example of Truncation Error ==
Let the trapezium rule be used to integrate $\cos x$ over the closed interval $\closedint 0 {0.8}$ using $8$ subintervals of length $0.1$.
Because $\size {\map {f' '} x} \le 1$, the truncation error cannot exceed a value given by:
:$E \le \size {\dfrac {\paren {0.1}^2 \times 0.8} {12} } = 0.00067$
== Proof ==
Follows from Truncation Error from Trapezium Rule.
== Sources ==
* Truncation errors
* Truncation errors
Category:Trapezium Rule for Definite Integrals
Category:Examples of Truncation Errors
Truncation Error from Trapezium Rule

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let the Trapezium Rule for Definite Integrals be used to integrate $f$ over the closed interval $\closedint a b$ using $n$ subintervals of length $h$.
Then the truncation error $E$ is given by:
:$E = \dfrac {-h^2 \paren {b - a} \map {f' '} \alpha} {12}$
where the second derivative $f' '$ is evaluated at some unspecified point $\alpha \in \closedint a b$.
== Proof ==
== Sources ==
* Truncation errors
* Truncation errors
Category:Trapezium Rule for Definite Integrals
Category:Truncation Errors
Truth Table/Examples

== Examples of Truth Tables ==
=== Example: $p \implies \paren {q \lor r}$ ===
=== Example: $\paren {\lnot p} \land \paren {\lnot q}$ ===
=== Example: $\lnot \paren {\paren {p \implies q} \implies \paren {\lnot \paren {q \implies p} } }$ ===
=== Example: $p \implies \paren {q \implies r}$ ===
=== Example: $\paren {p \land q} \implies r$ ===
=== Example: $\paren {p \iff \paren {\lnot q} } \lor q$ ===
=== Example: $\paren {p \land q} \lor \paren {r \land s}$ ===
=== Example: $\paren {\paren {\lnot p} \land q} \implies \paren {\paren {\lnot q} \land r}$ ===
Category:Truth Tables