nyuuzyou/wikis huggingface dataset:
Condition for Commensurability of Roots of Quadratic Equation

== Theorem ==
Consider the quadratic equation:
:$(1): \quad a x - x^2 = \dfrac {b^2} 4$
Then $x$ and $a - x$ are commensurable $\sqrt{a^2 - b^2}$ and $a$ are commensurable.
=== Lemma ===
== Proof ==
We have that:
Let $a \frown b$ denote that $a$ is commensurable with $b$.
=== Necessary Condition ===
Let $\paren {a - x} \frown x$.
From Commensurability of Sum of Commensurable Magnitudes:
:$a \frown x$
From Magnitudes with Rational Ratio are Commensurable:
:$x \frown 2 x$
So:
=== Sufficient Condition ===
Let $a \frown \sqrt {a^2 - b^2}$.
Then:
From Magnitudes with Rational Ratio are Commensurable:
:$x \frown 2 x$
From Commensurability is Transitive Relation:
:$a \frown x$
From Commensurability of Sum of Commensurable Magnitudes:
:$\paren {a - x} \frown x$
== Sources ==
* : Book $\text X$. Propositions
Category:Euclidean Number Theory
Condition for Commensurability of Roots of Quadratic Equation/Lemma

== Theorem ==
== Proof ==
:
Let $AB$ be a straight line.
Let the parallelogram $AD$ be applied to $AB$ which is deficient by the square $DB$.
As $DB$ is a square:
: $DC = BC$
Thus $AD$ is the rectangle contained by $AC$ and $CB$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Condition for Complete Bipartite Graph to be Edgeless

== Theorem ==
Let $G = \struct {A \mid B, E} =: K_{m, n}$ be the complete bipartite graph with $m$ vertices in $A$ and $n$ vertices in $B$.
Then:
:$K_{m, n}$ is edgeless either $m = 0$ or $n = 0$.
== Proof ==
Consider $K_{m, n}$ where both $m > 0$ and $n > 0$.
Then:
:$\exists u \in A, v \in V: \set {u, v} \in E$
and it follows by definition that $K_{m, n}$ is not edgeless.
Consider $K_{m, n}$ where either $m = 0$ or $n = 0$.
Hence one of the partite sets of $G$ is empty.
Hence there are no vertices for the vertices in the other partite set to be adjacent to.
So $K_{m, 0}$ and $K_{0, n}$ are edgeless.
Hence the result.
Category:Complete Bipartite Graphs
Category:Edgeless Graphs
Condition for Complete Bipartite Graph to be Semi-Hamiltonian

== Theorem ==
Let $K_{m, n}$ be a complete bipartite graph.
$K_{m, n}$ is semi-Hamiltonian either:
:$m = n = 1$
or:
:$m = n + 1$ (or $n = m + 1$).
== Proof ==
Let $K_{m, n}$ be a complete bipartite graph.
If $m = n = 1$ then $K_{m, n} = K_{1, 1}$ is trivially semi-Hamiltonian:
:
If $m = n + 1$, we can construct a Hamiltonian path as follows.
Let $K_{m, n} = \struct {A \mid B, E}$ such that $\card A = m, A = \set {u_1, u_2, \ldots, u_m}, \card B = n, B = \set {v_1, v_2, \ldots, v_m}$.
We start at vertex $u_1 \in A$.
As $K_{m, n}$ is complete, there is an edge connecting $u_1$ to $v_1$.
From $v_1$, there is an edge connecting $v_1$ to $u_2$.
And so on.
There is therefore a Hamiltonian path $\tuple {u_1, v_1, u_2, v_2, \ldots, u_{n - 1}, v_{n - 1}, u_n, v_n, u_m}$ (where of course $u_m = u_{n + 1}$).
So when $m = n + 1$, $K_{m, n}$ has a Hamiltonian path.
However, from Condition for Bipartite Graph to be Hamiltonian, $K_{m, n}$ is only fully Hamiltonian if $m = n$.
Hence when $m = n + 1$, $K_{m, n}$ is semi-Hamiltonian.
Similarly, if $n = m + 1$ the same argument applies, but this time the Hamiltonian path is:
$\tuple {v_1, u_1, v_2, u_2, \ldots, v_{n - 1}, u_{n - 1}, v_m, u_m, v_n}$
where of course $v_n = u_{m + 1}$.
Now, suppose $K_{m, n}$ is such that neither $m = n + 1$ nor $n = m + 1$, and it is not the case that $m = n = 1$.
If $m = n$, then from Complete Hamiltonian Bipartite Graph $K_{m, n}$ is Hamiltonian, and therefore not semi-Hamiltonian.
If $m > n + 1$, then we can once more trace a path $P$ through $K_{m, n} = \struct {A \mid B, E}$ as before.
However, at the end of $P$ there is a spare vertex in $A$ which has not been traversed which you can't get to without going back to one of the vertices in $B$ that you have already visited.
Similarly if $n > m + 1$.
Hence the result.
Category:Semi-Hamiltonian Graphs
Category:Complete Bipartite Graphs
Condition for Complex Number to be in Right Half Plane

== Theorem ==
Let $\C$ be the complex plane.
Let $P$ be the half-plane of $\C$ to the right of the infinite straight line $x = \lambda$.
The points in $P$ can be defined by:
:$\map \Re z > \lambda$
where $\map \Re z$ denotes the real part of $z$.
== Proof ==
== Sources ==
* : $\S 2$. Geometrical Representations: Example $2$.
Category:Geometry of Complex Plane
Condition for Complex Root of Unity to be Primitive

== Theorem ==
Let $n, k \in \N$.
Then $\alpha_k = \map \exp {\dfrac {2 \pi i k} n}$ is a primitive $n$th root of unity $\gcd \set {n, k} = 1$.
== Proof ==
Let $U_n = \set {\map \exp {\dfrac {2 \pi i k} n}: 0 \le k \le n - 1}$.
Let $V = \set {1, \dotsc, {\alpha_k}^{n - 1} }$.
By Complex Roots of Unity in Exponential Form it is sufficient to show that $U_n = V$ $\gcd \set {n, k} = 1$.
Let $\gcd \set {n, k} = d > 1$.
Then there are $n', k' \in \N$ such that:
:$n' = d n$
and:
:$k' = d k$
Then we have:
:$\alpha_k = \map \exp {\dfrac {2 \pi i k'} {n'} }$
and:
:$\alpha_k^{n'} = \map \exp {2 \pi i k'} = 1$
Therefore:
:$V = \set {1, \dotsc, \alpha^{n' - 1} }$
such that $n' < n$.
So:
:$\cmod V = n' < n = \cmod {U_n}$
and $U_n \ne V$.
Let $\gcd \set {n, k} = 1$.
Let:
:$\map \exp {\dfrac {2 \pi i k} n}^d = \map \exp {\dfrac {2 \pi i k} n} = 1$
Then it must be the case that $\dfrac {k d} n \in \Z$.
Since $\gcd \set {n, k}\ = 1$ it follows that:
:$n \divides d$
and so:
:$d \ge n$
Therefore $\set {1, \dotsc, \alpha^{n - 1} }$ are distinct
Hence $\card V = \card {U_n}$.
Moreover each element of $V$ can be written in the form:
:$\map \exp {\dfrac {2 \pi i k} n}$
with $0 \le k \le n - 1$.
It follows that $V = U_n$.
== Also see ==
* Complex Roots of Unity in Exponential Form
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 44$
Category:Complex Roots of Unity
Condition for Composite Mapping on Left

== Theorem ==
Let $A, B, C$ be sets.
Suppose that $C$ is non-empty.
Let $f: A \to B$ and $g: A \to C$ be mappings.
Let $\RR: B \to C$ be a relation such that $g = \RR \circ f$ is the composite of $f$ and $\RR$.
Then $\RR$ may be a mapping :
:$\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$
That is:
:$\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$
:
:$\exists h: B \to C$ such that $h$ is a mapping and $h \circ f = g$
== Proof ==
=== Sufficient Condition ===
Suppose $\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$.
Consider the subset $G \subseteq \Img f \times C$ defined by:
:$G = \set {\tuple {y, z}: \exists x \in A: y = \map f x, z =\map g x}$
Clearly $G \ne \O$ because for any $x \in A$ we have $\tuple {\map f x, \map g x} \in G$.
What we need to show is that $G$ is the graph of a mapping $t: \Img f \to C$.
To do that, we need to show that for every $y \in \Img f$, there is a unique $z \in C$ such that $\tuple {y, z} \in G$.
It is clear that there is at least one such $z$: choose any $x \in A$ such that $y = \map f x$ and set $z = \map g x$.
Now we need to show that such a $z$ is unique.
Suppose we have $\tuple {y, z} \in G$ and $\tuple {y, z'} \in G$.
Then by the definition of $G$:
:$\exists x, x' \in A: y = \map f x = \map f {x'}, z = \map g x, z' = \map g {x'}$
We have taken as a hypothesis that:
:$\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$
So $\map g x = \map g {x'}$ and so $z = z'$.
So $G$ is the graph of a mapping which we can denote:
:$t: \Img f \to C$
Also, since:
:$\forall x \in A: \tuple {\map f x, \map g x} \in G$
it follows that:
:$\forall x \in A: \map f x = \map t {\map f x}$
Since $C$ is non-empty, it has an element $c$.
By Law of Excluded Middle, we can now construct a mapping $h$ as follows:
:$\map h x = \begin{cases}
 \map t x & : x \in \Img f \\
 c & : x \in B \setminus \Img f
\end{cases}$
So:
:$\forall x \in A: \map {\paren {h \circ f} } x = \map h {\map f x} = \map t {\map f x} = \map g x$
Thus we have constructed a mapping $h$ such that $h \circ f = g$, as required.
=== Necessary Condition ===
Suppose there exists some mapping $h: B \to C$ such that $h \circ f = g$.
Let $\map f x = \map f y$.
Then we have:
== Comment ==
Hence we have a necessary and sufficient condition for determining whether the composition of mappings actually exists as a mapping.
Note that this is different from being given two mappings and creating their composition.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.3$
Category:Composite Mappings
Condition for Composite Mapping on Right

== Theorem ==
Let $A, B, C$ be sets.
Let $f: B \to A$ and $g: C \to A$ be mappings.
Let $\RR: C \to B$ be a relation such that $g = f \circ \RR$ is the composite of $\RR$ and $f$.
Then $\RR$ may be a mapping :
:$\Img g \subseteq \Img f$
That is:
:$\Img g \subseteq \Img f$
:
:$\exists h: C \to B$ such that $h$ is a mapping and $f \circ h = g$
== Proof ==
=== Sufficient Condition ===
Suppose $\Img g \subseteq \Img f$.
That is:
:$\forall x \in C: \map g x \in \Img f$
and so:
:$\forall x \in C: \exists y \in B: \map g x = \map f y$
Take any $x \in C$.
Consider the set $Y_x = \set {y \in B: \map g x = \map f y}$.
We know from above that $Y_x \ne \O$.
So, using the Axiom of Choice, for each $x$ we may select some $y_x \in Y_x$.
Then we may define the mapping $h: C \to B$ such that:
:$\forall x \in C: \map h x = y_x$
We then see that:
Thus we have constructed a mapping $h$ such that $f \circ h = g$, as required.
=== Necessary Condition ===
Suppose there exists some mapping $h: C \to B$ such that $f \circ h = g$.
Let $y \in \Img g$.
Then we have:
Hence by definition of subset, $\Img g \subseteq \Img f$.
== Comment ==
Hence we have a necessary and sufficient condition for determining whether the composition of mappings actually exists as a mapping.
Note that this is different from being given two mappings and creating their composition.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.5$
Category:Composite Mappings
Condition for Composite Mapping to be Identity

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ and $g: T \to S$ be mappings such that:
: $g \circ f = I_S$
where $I_S$ is the identity mapping on $S$.
Then $f$ is an injection and $g$ is a surjection.
== Proof ==
From Identity Mapping is Bijection, $I_S$ is a bijection.
From Identity Mapping is Injection, $I_S$ is an injection.
From Injection if Composite is Injection it follows that $f$ is an injection.
From Identity Mapping is Surjection, $I_S$ is a surjection.
From Surjection if Composite is Surjection it follows that $g$ is a surjection.
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Lemma $1$
Category:Identity Mappings
Category:Injections
Category:Surjections
Condition for Composite Relation with Inverse to be Identity

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $\RR \circ \RR^{-1}$ be the composite of $\RR$ with its inverse.
Let $I_T$ be the identity mapping on $T$.
Then:
:$\RR \circ \RR^{-1} = I_T$
:
:$\RR$ is many-to-one
and:
:$\RR$ is right-total.
== Proof ==
=== Sufficient Condition ===
Let $\RR \circ \RR^{-1} = I_T$.
:
:$\exists t \in T: t \notin \Img \RR$
Then:
:$t \notin \Img {\RR \circ \RR^{-1} }$
But:
:$t \in \Img {I_T}$
by definition of the identity mapping on $T$.
Hence:
:$\RR \circ \RR^{-1} \ne I_T$
From this contradiction we deduce that:
:$\RR \circ \RR^{-1} = I_T \implies T \setminus \Img \RR = \O$
where $T \setminus \Img \RR$ denotes set difference.
So from Set Difference with Superset is Empty Set‎:
:$T \subseteq \Img \RR$
But from Image is Subset of Codomain we have:
:$T \supseteq \Img \RR$
and so:
:$\Img \RR = T$
which means $\RR$ is right-total.
Suppose $\RR$ is not many-to-one.
Then:
:$\exists s \in S: \exists t_1, t_2 \in T, t_1 \ne t_2: \tuple {s, t_1} \in \RR \land \tuple {s, t_2} \in \RR$
By definition of inverse relation:
:$\exists s \in S: \exists t_1, t_2 \in T: \tuple {t_1, s} \in \RR^{-1} \land \tuple {t_2, s} \in \RR^{-1}$
The composite of $\RR^{-1}$ and $\RR$ is defined as:
:$\RR \circ \RR^{-1} = \set {\tuple {x, z} \in T \times T: \exists y \in S: \tuple {x, y} \in \RR^{-1} \land \tuple {y, z} \in \RR}$
Thus:
So, by definition of identity mapping:
:$\RR \circ \RR^{-1} \ne I_T$
From this contradiction we deduce that $\RR$ must be many-to-one.
So it has been demonstrated that if:
:$\RR \circ \RR^{-1} = I_T$
then:
:$\RR$ is many-to-one
and
:$\RR$ is right-total.
=== Necessary Condition ===
Let:
:$\RR$ be many-to-one
and
:$\RR$ be right-total.
Let $\tuple {t_1, t_2} \in \RR \circ \RR^{-1}$.
The composite of $\RR^{-1}$ and $\RR$ is defined as:
:$\RR \circ \RR^{-1} = \set {\tuple {t_1, t_2} \in T \times T: \exists s \in S: \tuple {t_1, s} \in \RR^{-1} \land \tuple {s, t_2} \in \RR}$
By definition of inverse:
:$\RR \circ \RR^{-1} = \set {\tuple {t_1, t_2} \in T \times T: \exists s \in S: \tuple {s, t_1} \in \RR \land \tuple {s, t_2} \in \RR}$
But $\RR$ is many-to-one, and so:
:$t_1 = t_2$
So:
:$\forall \tuple {t_1, t_2} \in \RR \circ \RR^{-1}: t_1 = t_2$
and so:
:$\RR \circ \RR^{-1} \subseteq I_T$
Now let $t \in T$.
By definition of identity mapping on $T$:
:$\tuple {t, t} \in I_T$
As $\RR$ is right-total:
:$\Img \RR = T$
and so:
:$\exists s \in S: \tuple {s, t} \in \RR$
and so:
:$\exists s \in S: \tuple {t, s} \in \RR^{-1}$
Hence by definition of the composite of $\RR^{-1}$ and $\RR$:
:$\tuple {t, t} \in \RR \circ \RR^{-1}$
So:
:$\RR \circ \RR^{-1} \supseteq I_T$
and so:
:$\RR \circ \RR^{-1} = I_T$
Hence the result.
== Examples ==
== Sources ==
* : $\S 10$: Inverses and Composites
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Exercise $5.8 \ \text{(e)}$
Category:Inverse Relations
Category:Identity Mappings
Category:Composite Relations
Category:Condition for Composite Relation with Inverse to be Identity
Condition for Composite Relation with Inverse to be Identity/Examples/Arbitrary Example 1

== Example of Use of Condition for Composite Relation with Inverse to be Identity ==
:
In the above we see that:
:$\RR$ is many-to-one
:$\RR$ is right-total
:$\RR \circ \RR^{-1} = I_T$.
Note, however, that $\RR^{-1}$ is neither many-to-one nor right-total, and does not need to be for $\RR \circ \RR^{-1} = I_T$.
Category:Condition for Composite Relation with Inverse to be Identity
Condition for Composition Series

== Theorem ==
Let $G$ be a finite group.
Then:
:a normal series $\HH$ for $G$ is a composition series for $G$
:
:every factor group of $\HH$ is a simple group.
== Proof ==
Let $G$ be a finite group whose identity is $e$.
Let:
:$(1): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{n - 1} \lhd G_n = G$
be a normal series for $G$.
=== Necessary Condition ===
Suppose there exists $k$ such that $G_{k + 1} / G_k$ is not a simple group.
Then there exists a normal subgroup $G'$ such that:
:$\set e \lhd G' \lhd G_{k + 1} / G_k$
It follows that:
:$G_k \lhd G \lhd G_{k + 1}$
where:
:$G$ is a normal subgroup of $G_{k + 1}$
and:
:$G' = G / G_{k + 1}$
Thus $(1)$ has a proper refinement and so is not a composition series.
By the Rule of Transposition it follows that if normal series is a composition series, every factor group of that normal series is a simple group.
=== Sufficient Condition ===
Suppose $(1)$ is not a composition series for $G$.
Then a proper refinement of $(1)$ can be constructed by inserting a group $G$ into the series somewhere, for example:
:$G_k \lhd G \lhd G_{k + 1}$
It follows that $G / G_k$ is a normal subgroup of $G_{k + 1} / G$.
Thus, by definition, $G_{k + 1} / G_k$ is not a simple group.
Thus it has been shown that if a normal series is not a composition series, then it contains at least one factor group which is not a simple group
By the Rule of Transposition it follows that if every factor group of a normal series is a simple group, then that normal series is a composition series.
Hence the result.
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 74$
* : $1$: Introduction to Finite Group Theory: $1.10$
Category:Composition Series
Condition for Composition of Linear Real Functions to be Commutative

== Theorem ==
Let $a, b, c, d \in \R$ be real numbers.
Let $\theta_{a, b}: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map {\theta_{a, b} } x = a x + b$
Let $\theta_{c, d} \circ \theta_{a, b}$ denote the composition of $\theta_{c, d}$ with $\theta_{a, b}$.
Then:
:$\theta_{c, d} \circ \theta_{a, b} = \theta_{a, b} \circ \theta_{c, d}$
:
:$b c + d = a d + b$
== Proof ==
== Sources ==
* : Chapter $3$. Mappings: Exercise $5$
Category:Linear Algebra
Condition for Conchoid of Nicomedes to have Cusp

== Theorem ==
Let $\KK$ be a conchoid of Nicomedes such that:
:the perpendicular distance from the directrix $\CC$ of $\KK$ to the focus point $P$ of $\KK$ is $a$
:the constant length of the line segment between $\KK$ and $\CC$ passing through $P$ is $b$.
Then:
:$\KK$ has a cusp at $P$
:
:$a = b$
== Proof ==
== Sources ==
* 
* 
Category:Conchoid of Nicomedes
Condition for Conchoid of Nicomedes to have Loop

== Theorem ==
Let $\KK$ be a conchoid of Nicomedes such that:
:the perpendicular distance from the directrix $\CC$ of $\KK$ to the focus point $P$ of $\KK$ is $a$
:the constant length of the line segment between $\KK$ and $\CC$ through $P$ is $b$.
Then:
:$\KK$ has a loop
:
:$a < b$
== Proof ==
== Sources ==
* 
* 
Category:Conchoid of Nicomedes
Condition for Concurrency of Three Straight Lines

== Theorem ==
Let $3$ straight lines $\LL_1$, $\LL_2$ and $\LL_3$ be embedded in a cartesian plane $\CC$, expressed using the general equations:
Then $\LL_1$, $\LL_2$ and $\LL_3$ are concurrent :
:$\begin {vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end {vmatrix} = 0$
where $\begin {vmatrix} \cdot \end {vmatrix}$ denotes a determinant.
== Proof ==
=== Necessary Condition ===
Let $\LL_1$, $\LL_2$ and $\LL_3$ be concurrent.
From Equation of Straight Line through Intersection of Two Straight Lines, $\LL_1$, $\LL_2$ and $\LL_3$ are concurrent $\LL_3$ has an equation of the form:
:$\paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2} = 0$
That is:
:$\paren {l_1 - k l_2} x + \paren {m_1 - k m_2} y + \paren {n_1 - k n_2} = 0$
Hence:
=== Sufficient Condition ===
Suppose $\begin {vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end {vmatrix} = 0$
Then the rows of this matrix are linearly dependent.
That is, there exists $a_1, a_2, a_3 \in \R_{\ne 0}$ such that:
$\LL_3: l_3 x + m_3 y + n_3 = 0$ is equivalent to $- \dfrac {a_3} {a_1} l_3 x - \dfrac {a_3} {a_1} m_3 y - \dfrac {a_3} {a_1} n_3 = 0$,
which, from above, is in turn equivalent to:
:$\paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2} = 0$
with $k = -\dfrac {a_2} {a_1}$.
Hence from Equation of Straight Line through Intersection of Two Straight Lines, $\LL_1$, $\LL_2$ and $\LL_3$ are concurrent.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $11$. Condition for concurrency
Category:Straight Lines
Category:Concurrency
Condition for Conditional Expectation to be Almost Surely Non-Negative

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an integrable random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra of $\Sigma$.
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.
Then we have: 
:$\expect {X \mid \GG} \ge 0$ almost everywhere
:
:$\expect {X \cdot \chi_A} \ge 0$ for each $A \in \GG$
where $\chi_A$ is the characteristic function of $A$. 
== Proof ==
=== Sufficient Condition ===
Suppose that:
:$\expect {X \chi_A} \ge 0$ for each $A \in \GG$.
Then we have, by the definition of the conditional expectation of $X$ given $\GG$:
:$\expect {\expect {X \mid \GG} \chi_A} \ge 0$ for each $A \in \GG$.
Set: 
:$A = \set {\omega \in \Omega : \map {\paren {\expect {X \mid \GG} } } \omega < 0}$
Since $\expect {X \mid \GG}$ is $\GG$-measurable, we have $A \in \GG$ from Characterization of Measurable Functions.
From Expectation is Monotone, we have:
:$\expect {\expect {X \mid \GG} \chi_A} \le 0$
So, we have:
:$\expect {\expect {X \mid \GG} \chi_A} = 0$
Then, from Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we have:
:$\expect {X \mid \GG} \chi_A = 0$ almost everywhere.
But, for all $\omega \in A$, we have:
:$\expect {X \mid \GG} \chi_A < 0$ 
Hence we have $\map \Pr A = 0$. 
So $\expect {X \mid \GG} \ge 0$ almost everywhere.
=== Necessary Condition ===
Suppose that:
:$\expect {X \mid \GG} \ge 0$ almost everywhere.
Then for each $A \in \GG$, we have:
:$\expect {X \mid \GG} \cdot \chi_A \ge 0$ almost everywhere.
So, from Expectation is Monotone:
:$\expect {\expect {X \mid \GG} \cdot \chi_A} \ge 0$ for each $A \in \GG$.
From the definition of the conditional expectation of $X$ given $\GG$, we have:
:$\expect {\expect {X \mid \GG} \cdot \chi_A} = \expect {X \cdot \chi_A}$
So:
:$\expect {X \cdot \chi_A} \ge 0$ for each $A \in \GG$.
Category:Conditional Expectation
Condition for Constant Operation to be Distributive over Another Operation

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\sqbrk c$ denote the constant operation for some $c \in S$.
Then:
:$\sqbrk c$ is distributive over $\circ$
:
:$c \circ c = c$
== Proof ==
=== Sufficient Condition ===
Let $\sqbrk c$ be distributive over $\circ$
:$c \circ c = c$
=== Necessary Condition ===
Let $\forall c \circ c = c$.
Then:
and:
That is, $\sqbrk c$ is distributive over $\circ$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.24$
Category:Constant Operation
Category:Distributive Operations
Condition for Continuity on Interval

== Theorem ==
Let $f$ be a real function defined on an interval $\mathbb I$.
Then $f$ is continuous on $\mathbb I$ :
:$\forall x \in \mathbb I: \forall \epsilon > 0: \exists \delta > 0: y \in \mathbb I \land \size {x - y} < \delta \implies \size {\map f x - \map f y} < \epsilon$
== Proof ==
Let $x \in \mathbb I$ such that $x$ is not an end point.
Then the condition $y \in \mathbb I \land \size {x - y} < \delta$ is the same as $\size {x - y} < \delta$ provided $\delta$ is small enough.
The criterion given therefore becomes the same as the statement $\ds \lim_{y \mathop \to x} \map f y = \map f x$, that is, that $f$ is continuous at $x$.
Now suppose $x \in \mathbb I$ and $x$ is a left hand end point of $\mathbb I$.
Then the condition $y \in \mathbb I \land \size {x - y} < \delta$ reduces to $x \le y < x + \delta$ provided $\delta$ is small enough.
The criterion given therefore becomes the same as the statement $\ds \lim_{y \mathop \to x^+} \map f y = \map f x$, that is, that $f$ is continuous on the right at $x$.
Similarly, if $x \in \mathbb I$ and $x$ is a right hand end point of $\mathbb I$, then the criterion reduces to the statement that $f$ is continuous on the left at $x$.
Thus the assertions are equivalent to the statement that $f$ is continuous at all points in $\mathbb I$, that is, that $f$ is continuous on $\mathbb I$.
== Sources ==
* : $\S 9.3$
* : Appendix: $\S 18.7$: Continuity
Category:Real Intervals
Category:Continuous Functions
Condition for Coordinate Axes to have Same Orientation

== Theorem ==
Let $\CC_1$ and $\CC_2$ be Cartesian coordinate systems of the same dimension.
Let $\EE_1$ and $\EE_2$ denote an ordered basis of $\CC_1$ and $\CC_2$ respectively.
Then:
:$\CC_1$ and $\CC_2$ have the same orientation
:
:the matrix representing $\EE_1$ in terms of $\EE_2$ (and $\EE_2$ in terms of $\EE_1$) has a (strictly) positive determinant .
== Proof ==
== Sources ==
* 
* 
Category:Cartesian Coordinate Systems