nyuuzyou/wikis huggingface dataset:
Jacobi's Necessary Condition

== Theorem ==
Let $J$ be a functional, such that:
:$J \sqbrk y = \ds \int_a^b \map F {x, y, y'} \rd x$
Let $\map y x$ correspond to the minimum of $J$.
Let:
:$F_{y'y'}>0$
along $\map y x$.
Then the open interval $\openint a b$ contains no points conjugate to $a$.
== Proof ==
By Necessary Condition for Twice Differentiable Functional to have Minimum, $J$ is minimised by $y = \map {\hat y} x$ if:
:$\delta^2 J \sqbrk {\hat y; h} \ge 0$ 
for all admissable real functions $h$.
By lemma 1 of Legendre's Condition,
:$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {P h'^2 + Q h^2} \rd x$
where:
:$P = F_{y' y'}$
By Nonnegative Quadratic Functional implies no Interior Conjugate Points, $\openint a b$ does not contain any conjugate points $J$.
== Sources ==
* : $\S 5.27$: Jacobi's Necessary Condition. More on Conjugate Points
Category:Calculus of Variations
Jacobi's Necessary Condition/Dependent on N Functions

== Theorem ==
Let $J$ be a functional, such that:
:$J \sqbrk {\mathbf y} = \ds \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
where $\mathbf y = \paren {\sequence {y_i}_{1 \le i \le N} }$ is an N-dimensional real vector.
Let $\map {\mathbf y} x$ correspond to the minimum of $J$.
Let the $N\times N$ matrix $\mathbf P = F_{y_i' y_j'}$ be positive definite along $\map {\mathbf y} x$.
Then the open interval $\openint a b$ contains no points conjugate to $a$.
== Proof ==
By Necessary Condition for Twice Differentiable N Function dependent Functional to have Minimum, $J$ is minimised by $y = \map {\mathbf {\hat y} } x$ if:
:$\delta^2 J \sqbrk {\mathbf {\hat y}; \mathbf h} \ge 0$ 
for all admissable real functions $\mathbf h$.
By lemma 1 of Legendre's Condition:
:$\ds \delta^2 J \sqbrk {\mathbf y; \mathbf h} = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h} \rd x$
where:
:$\mathbf P = F_{y_i' y_j'}$
By Nonnegative Quadratic N function dependent Functional implies no Interior Conjugate Points, $\openint a b$ does not contain any conjugate points $J$.
== Sources ==
* : $\S 5.29$: Generalization to n Unknown Functions
Category:Calculus of Variations
Jacobi's Theorem

== Theorem ==
Let $\mathbf y = \sequence {y_i}_{1 \le i \le n}$, $\boldsymbol \alpha = \sequence {\alpha_i}_{1 \le i \le n}$, $\boldsymbol \beta = \sequence {\beta_i}_{1 \le i \le n}$ be vectors, where $\alpha_i$ and $ \beta_i$ are parameters.
Let $S = \map S {x, \mathbf y, \boldsymbol \alpha}$ be a a complete solution of the Hamilton-Jacobi equation.
Let:
:$\begin {vmatrix} \dfrac {\partial^2 S} {\partial \alpha_i \partial y_k} \end{vmatrix} \ne 0$
where $\begin {vmatrix} \cdot \end{vmatrix}$ is a determinant.
Let:
:$\dfrac {\partial S} {\partial \alpha_i} = \beta_i$
Then:
:$p_i = \map {\dfrac {\partial S} {\partial y_i} } {x, \mathbf y, \boldsymbol \alpha}$
:$y_i = \map {y_i} {x, \boldsymbol \alpha, \boldsymbol \beta}$
constitute a general solution of the canonical Euler's equations.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 4.23$: The Hamilton-Jacobi Equation. Jacobi's Theorem 
Category:Calculus of Variations
Category:Jacobi's Theorem
Jacobi's Theorem/Proof 1

== Theorem ==
== Proof ==
Consider the total derivative of $\dfrac {\partial S} {\partial \alpha_i}$ $x$:
Next, consider the total derivative of $p_i$ $x$:
On the other hand, the partial derivative of Hamilton-Jacobi equation yields:
By comparison of this and previous expressions:
:$\dfrac {\d p_i} {\d x} = -\dfrac {\partial H} {\partial y_i}$
== Sources ==
* : $\S 4.23$: The Hamilton-Jacobi Equation. Jacobi's Theorem 
Category:Jacobi's Theorem
Jacobi's Theorem/Proof 2

== Theorem ==
== Proof ==
Consider canonical Euler's equations:
:$\dfrac {\d y_i} {\d x} = \dfrac {\partial H} {\partial p_i}$
:$\dfrac {\d p_i} {\d x} = -\dfrac {\partial H} {\partial y_i}$
Apply a canonical transformation:
:$\tuple {x, \mathbf y, \mathbf p, H} \to \tuple {x, \boldsymbol \alpha, \boldsymbol \beta, H^*}$
where $\Phi = S$.
By Conditions for Transformation to be Canonical:
:$p_i = \dfrac {\partial S} {\partial y_i}$
:$\beta_i = \dfrac {\partial S} {\partial \alpha_i}$
:$H^* = H + \dfrac {\partial S} {\partial x}$
Because $S$ is a solution to the Hamilton-Jacobi equation:
:$H^* = 0$
In these new coordinates canonical Euler's equations are:
:$\dfrac {\d\alpha_i} {\d x} = \dfrac {\partial H^*} {\partial \beta_i}$
:$\dfrac {\d \beta_i} {\d x} = -\dfrac {\partial H^*} {\partial \alpha_i}$
By $H^* = 0$:
:$\dfrac {\d \alpha_i} {\d x} = 0$
:$\dfrac {\d\beta_i} {\d x} = 0$
which imply that $ \alpha_i$ and $\beta_i$ are constant along each extremal.
$\beta_i$ constancy provides with $n$ first integrals:
:$\dfrac {\partial S} {\partial \alpha_i} = \beta_i$
Because $S = \map S {x, \mathbf y, \boldsymbol \alpha}$, the aforementioned set of first integrals is also a system of equations for functions $y_i$.
Thus, functions $y_i$ can be found.
Functions $p_i$ are found by the results of Conditions for Transformation to be Canonical:
:$p_i = \dfrac {\partial} {\partial y_i} \map S {x, \mathbf y, \boldsymbol \alpha}$
Then:
:$\map {y_i} {x, \boldsymbol \alpha, \boldsymbol \beta}$
:$\map {p_i} {x, \boldsymbol \alpha, \boldsymbol \beta}$
are solutions to canonical Euler's equations.
== Sources ==
* : $\S 4.23$: The Hamilton-Jacobi Equation. Jacobi's Theorem 
Category:Jacobi's Theorem
Jadhav Theorem

== Theorem ==
Let $a, b, c$ be real numbers in arithmetic sequence.
Let the common difference of this arithmetic sequence be $d$.
Then:
:$b^2 - a c = d^2$
== Proof ==
== Examples ==
== Also see ==
* Definition:Arithmetic Progression
* Definition:Quadratic Equation
== Historical Note ==
== Sources ==
* https://artofproblemsolving.com/wiki/index.php/Jadhav_Theorem
Category:Arithmetic Sequences
Category:Jadhav Theorem
Jadhav Theorem/Examples/Square of 102

== Examples of Use of Jadhav Theorem ==
Consider the square of $102$.
Consider $102$ as a term in an arithmetic sequence:
:$100$, $102$, $104$
whose common difference is $2$.
Then from the Jadhav Theorem:
:$102^2 - 100 \times 104 = 2^2$
from which we have:
:$102^2 = 100 \times 104 + 4 = 10404$
Category:Jadhav Theorem
Jadhav Theorem/Examples/Square of 406

== Examples of Use of Jadhav Theorem ==
Consider the square of $406$.
Consider $406$ as a term in an arithmetic sequence:
:$400$, $406$, $412$
whose common difference is $6$.
Then from the Jadhav Theorem:
:$406^2 - 400 \times 412 = 36^2$
from which we have:
:$406^2 = 400 \times 412 + 36 = 164 \, 800 + 36 = 164 \, 836$
Category:Jadhav Theorem
Jensen's Formula

== Theorem ==
Let $S$ be an open subset of the complex plane containing the closed disk:
:$D_r = \set {z \in \C : \cmod z \le r}$
of radius $r$ about $0$.
Let $f: S \to \C$ be holomorphic on $S$. 
Let $f$ have no zeroes on the circle $\cmod z = r$.
Let $\map f 0 \ne 0$.
Let $\rho_1, \ldots, \rho_n$ be the zeroes of $f$ in $D_r$, counted with multiplicity.
Then:
:$(1): \quad \ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \cmod {\map f {r e^{i \theta} } } \rd \theta = \ln \cmod {\map f 0} + \sum_{k \mathop = 1}^n \paren {\ln r - \ln \size {\rho_k} }$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Complex Analysis
Category:Jensen's Formula
Jensen's Formula/Proof 1

== Theorem ==
== Proof ==
Write:
:$\map f z = \paren {z - \rho_1} \dotsm \paren {z - \rho_n} \map g z$
so $\map g z \ne 0$ for $z \in D_r$.
It is sufficient to check the equality for each factor of $f$ in this expansion.
First let:
:$\map h z = z - \rho_k$
for some $k \in \set {1, \ldots, n}$.
Making use of the substitution $u = r e^{i \theta} - \rho_k$ we find that:
:$\ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map h {r e^{i \theta} } } \rd \theta = \frac 1 {2 \pi i} \int_\gamma \frac {\ln \size u} {u + \rho_k} \rd u$
where $\gamma$ is a circle of radius $r$ centred at $-\rho_k$, traversed anticlockwise.
On this circle, $\ln \size u = \ln r$ is constant, and we have that:
:$\ds \int_\gamma \frac 1 {u + \rho_k} \rd u = \int_{\size z \mathop = r} \frac {\d u} u = 2 \pi i$
Therefore the of $(1)$ is $\ln r$ as required.
To show equality for $\map g z$, first observe that by Cauchy's Residue Theorem:
:$\ds \int_{\size z = r} \frac {\ln \map g z} z \rd z = 2 \pi i \ln \map g 0$
Therefore substituting $z = r e^{i \theta}$ we have
:$\ds 2 \pi i \ln \map g 0 = i \int_0^{2 \pi} \ln \map g {r e^{i \theta} } \rd \theta$
Comparing the imaginary parts of this equality we see that:
:$\ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map g {r e^{i \theta} } } \rd \theta = \ln \size {\map g 0}$
as required.
Category:Jensen's Formula
Jensen's Formula/Proof 2

== Theorem ==
== Proof ==
Write
:$\map f z = \dfrac {r^2 - z \overline {\rho_1} } {r \paren {z - \rho_1} } \cdots \dfrac {r^2 - z \overline {\rho_n} } {r \paren {z - \rho_n} } \map g z$
so $\map g z \ne 0$ for $z \in D_r$.
It is sufficient to check equality for each factor of $f$ in this expansion.
When $\cmod z = r$, we have:
:$\dfrac 1 z = \dfrac {\overline z} {r^2}$
and:
:$\cmod {\dfrac z r} = 1$
where $\overline z$ denotes the complex conjugate of $z$.
So:
so the is $0$.
Moreover:
:$\\ln \cmod {\dfrac {r^2 - 0 \overline {\rho_k} } {r \paren {0 - \rho_k} } } = \ln \cmod {\rho_k} - \ln r$
so the is $0$.
Therefore, the formula holds for $\dfrac {r^2 - z \overline {\rho_k} } {r \paren {z - \rho_k} }$.
Since:
:$r^2 - z \overline {\rho_i} = 0 \implies \cmod z = r^2 / \cmod {\overline {\rho_i} } > r$
it follows that $\map g z$ is holomorphic without zeroes on $D_r$.
So the is $\ln \size {\map g 0}$.
On the other hand, by the mean value property:
:$\ds \dfrac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map g {r e^{i \theta} } } \rd \theta = \ln \map g 0$
as required.
Category:Jensen's Formula
Jensen's Formula/Proof 3

== Theorem ==
== Proof ==
, assume that $r = 0$. 
Write:
:$\map f z = \map {B_{\rho_1} } z \dotsm \map {B_{\rho_n} } z \map g z$
where $\map {B_\rho} z := \dfrac {z - \rho} {1 - z \overline \rho}$ is the Blaschke factor at $\rho$.
So:
:$\map g z$ is holomorphic and nonzero in $D_r$
and:
:each $\map {B_{\rho_i} } z$ is holomorphic and has a simple zero $\rho_i$ inside $D_r$.
It is sufficient to check the equality for each factor of $f$ in this expansion.
First let:
:$\map h z = \map {B_{\rho_k} } z = \dfrac {z - \rho_k} {1 - z \overline {\rho_k} }$
for some $k \in \set {1, \ldots, n}$.
Note that:
:$\size {\map h z} \equiv 1$ for $\size z = 1$
so the integral is $0$.
Also:
:$\map {B_{\rho_k} } 0 = \rho_k$
so the equality holds in this case.
To show equality for $\map g z$, first observe that by the Cauchy's Residue Theorem:
:$\ds \int_{\size z \mathop = r} \frac {\ln \map g z} z \rd z = 2 \pi i \ln \map g 0$
Therefore substituting $z = r e^{i \theta}$ we have:
:$\ds 2 \pi i \ln \map g 0 = i \int_0^{2 \pi} \ln \map g {r e^{i \theta} } \rd \theta$
Comparing the imaginary parts of this equality we see that:
:$\ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map g {r e^{i \theta} } } \rd \theta = \ln \size {\map g 0}$
as required.
Category:Jensen's Formula
Jensen's Inequality (Complex Analysis)

== Theorem ==
Let $D \subset \C$ be an open set with $0 \in D$.
Let $R > 0$ be such that $\map B {0, R} \subset D$.
Let $f: D \to \C$ be analytic with $\map f 0 \ne 0$.
Let $\cmod {\map f z} \le M$ for $\cmod z \le R$.
Let $0 < r <R$.
Then the number of zeroes of $f$, counted with multiplicity, for $\cmod z \le r$, is at most:
:$\dfrac {\map \log {M / \cmod {\map f 0} } } {\map \log {R / r} }$
== Proof ==
== Sources ==
* $6$: The Prime Number Theorem: $\S 6.1$: A zero-free region: Lemma $6.1$
Category:Complex Analysis
Jensen's Inequality (Measure Theory)

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \R$ be a $\mu$-integrable function such that $f \ge 0$ pointwise.
=== Convex Functions ===
=== Concave Functions ===
== Also see ==
* Jensen's Inequality (Real Analysis), a similar result in the context of real analysis.
== Sources ==
* : $12.14$
Category:Measure Theory
Jensen's Inequality (Measure Theory)/Concave Functions

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \R$ be a $\mu$-integrable function such that $f \ge 0$ pointwise.
Let $\Lambda: \hointr 0 \infty \to \hointr 0 \infty$ be a concave function.
Then for all positive measurable functions $g: X \to \R$, $g \in \map {\MM^+} \Sigma$:
:$\dfrac {\int \paren {\Lambda \circ g} \cdot f \rd \mu} {\int f \rd \mu} \le \map \Lambda {\dfrac {\int g \cdot f \rd \mu} {\int f \rd \mu} }$
where $\circ$ denotes composition, and $\cdot$ denotes pointwise multiplication.
== Proof ==
Category:Measure Theory
Jensen's Inequality (Measure Theory)/Convex Functions

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \R$ be a $\mu$-integrable function such that $f \ge 0$ pointwise.
Let $V: \hointr 0 \infty \to \hointr 0 \infty$ be a convex function.
Then for all positive measurable functions $g: X \to \R$, $g \in \map {\MM^+} \Sigma$:
:$\map V {\dfrac {\ds \int g \cdot f \rd \mu} {\ds \int f \rd \mu} } \le \dfrac {\ds \int \paren {V \circ g} \cdot f \rd \mu} {\ds\int f \rd \mu}$
where $\circ$ denotes composition, and $\cdot$ denotes pointwise multiplication.
== Proof ==
Let $\d \map \nu x := \dfrac {\map f x} {\ds \int \map f s \rd \map \mu s} \rd \map \mu x$ be a probability measure.
Let $\ds x_0 := \int \map g s \rd \map \nu s$.
Then by convexity there exists constants $a, b$ such that:
In other words, there is a tangent line at $\tuple {x_0, V_0}$ that falls below the graph of $V$.
Therefore:
Category:Measure Theory
Jensen's Inequality (Real Analysis)

== Theorem ==
Let $I$ be a real interval.
Let $\phi: I \to \R$ be a convex function.
Let $x_1, x_2, \ldots, x_n \in I$.
Let $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$ be real numbers, at least one of which is non-zero.
Then:
:$\ds \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
For $\phi$ strictly convex, equality holds $x_1 = x_2 = \cdots = x_n$.
=== Corollary ===
== Proof ==
The proof proceeds by mathematical induction on $n$.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
$\map P 1$ is true, as this just says:
:$\ds \map \phi {\frac {\lambda_1 x_1} {\lambda_1} } \le \frac {\lambda_1 \map \phi {x_1} } {\lambda_1}$
trivially.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.
So this is our induction hypothesis:
:$\ds \map \phi {\frac {\sum_{k \mathop = 1}^r \lambda_k x_k} {\sum_{k \mathop = 1}^r \lambda_k} } \le \frac {\sum_{k \mathop = 1}^r \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^r \lambda_k}$
where $\lambda_1, \lambda_2, \ldots, \lambda_r \ge 0$ are real numbers, at least one of which is non-zero.
Then we need to show:
:$\ds \map \phi {\frac {\sum_{k \mathop = 1}^{r + 1} \lambda_k x_k} {\sum_{k \mathop = 1}^{r + 1} \lambda_k} } \le \frac {\sum_{k \mathop = 1}^{r + 1} \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^{r + 1} \lambda_k}$
where $\lambda_1, \lambda_2, \ldots, \lambda_{r + 1} \ge 0$ are real numbers, at least one of which is non-zero.
=== Induction Step ===
This is our induction step:
, assume that $\lambda_r$ and $\lambda_{r + 1}$ are non-zero.
Otherwise, the statement reduces to the induction hypothesis.
Define:
:$y := \dfrac {\lambda_r x_r + \lambda_{r + 1} x_{r + 1} } {\lambda_r + \lambda_{r + 1} }$
Then:
Note that in the case that $\phi$ is strictly convex, equality holds $x_1 = x_2 = \cdots = x_r = x_{r + 1}$.
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N_{>0}: \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
Category:Real Analysis
Category:Inequalities
Category:Jensen's Inequality (Real Analysis)
Jensen's Inequality (Real Analysis)/Corollary

== Corollary to Jensen's Inequality: Real Analysis ==
Let $I$ be a real interval.
Let $\phi: I \to \R$ be a concave function.
Let $x_1, x_2, \ldots, x_n \in I$.
Let $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$ be real numbers, at least one of which is non-zero.
Then:
:$\ds \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \ge \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
For $\phi$ strictly concave, equality holds $x_1 = x_2 = \cdots = x_n$.
== Proof ==
By Real Function is Concave iff its Negative is Convex, $-\phi: I \to \R$ is a convex function.
Therefore, we can apply Jensen's Inequality: Real Analysis with $-\phi$ to obtain:
:$\ds -\map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le -\frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
with equality for $-\phi$ strictly convex $x_1 = x_2 = \cdots = x_n$.
That is, for $\phi$ strictly concave, equality holds $x_1 = x_2 = \cdots = x_n$.
With that, we have established the equality case.
Multiplying through $-1$ in our inequality gives: 
:$\ds \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \ge \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
as required.
Category:Jensen's Inequality (Real Analysis)
Joachimsthal's Section-Formulae

== Theorem ==
Let $P = \tuple {x_1, y_1}$ and $Q = \tuple {x_2, y_2}$ be points in the Cartesian plane.
Let $R = \tuple {x, y}$ be a point on $PQ$ dividing $PQ$ in the ratio:
:$PR : RQ = l : m$
Then:
== Proof ==
:
Let the ordinates $PL$, $QM$ and $RN$ be constructed for $P$, $Q$ and $R$ respectively.
Then we have:
== Historical Note ==
== Sources ==
* : Chapter $\text I$. Coordinates: $10$. Joachimsthal's section-formulae
Category:Analytic Geometry
Join Absorbs Meet

== Theorem ==
Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $\vee$ denote join.
Then $\vee$ absorbs $\wedge$.
That is, for all $a, b \in S$:
:$a \vee \paren {a \wedge b} = a$
== Proof ==
By Ordering in terms of Join, we have that:
:$a \vee \paren {a \wedge b} = a$ $a \wedge b \preceq a$
The result thus follows from Meet Precedes Operands.
== Duality ==
The dual to this theorem is Meet Absorbs Join.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.23 \ \text {(a)}$
Category:Join and Meet
Category:Absorption Laws
Join Irreducible Element is Dual of Meet Irreducible Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $z \in S$.
The following are dual statements:
:$z$ is a meet irreducible element of the meet semilattice $\struct {S, \wedge, \preceq}$
:$z$ is a join irreducible element of the join semilattice $\struct {S, \vee, \preceq}$
== Proof ==
By definition of meet irreducible element:
:$z$ is the meet irreducible element of the meet semilattice $\struct {S, \wedge, \preceq}$
:
:$\forall x, y \in S: \paren{z = x \wedge y} \implies \leftparen{z = x}$ or $\rightparen{z = y}$
By the duality principle, the dual of this statement is:
:$\forall x, y \in S: \paren{z = x \vee y} \implies \leftparen{z = x}$ or $\rightparen{z = y}$
By definition of join irreducible element:
:$z$ is the join irreducible element of the join semilattice $\struct {S, \vee, \preceq}$
== Also see ==
* Duality Principle (Order Theory)
Category:Meet Irreducible Elements
Category:Join Irreducible Elements
Category:Dual Pairs (Order Theory)
Join Operation on Ordered Set such that Every Doubleton admits Supremum is Entropic

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $\struct {S, \preccurlyeq}$ be such that every doubleton subset of $S$ admits a supremum.
Let $\vee$ be the join operation on $S$, defined as:
:$\forall a, b \in S: a \vee b = \sup_\preccurlyeq \set {a, b}$
Then $\vee$ is an entropic operation.
== Proof ==
By definition, $\struct {S, \vee, \preccurlyeq}$ is a join semilattice. 
From Join Semilattice is Semilattice, $\struct {S, \vee, \preccurlyeq}$ is indeed a semilattice.
Then:
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.24$
Category:Entropic Operations
Category:Ordered Structures
Category:Join Operation
Join Prime Element Iff Join Irreducible In Distributive Lattice

== Theorem ==
Let $\struct{S, \vee, \wedge, \preceq}$ be a distributive lattice.
Let $z \in S$.
Then:
:$z$ is join-irreducible
:$z$ is join-prime
== Proof ==
By Dual Pairs (Order Theory):
* join prime element is dual to meet prime element.
* join irreducible element is dual to meet irreducible element.
Thus the theorem holds by the duality principle applied to Characterization of Meet Irreducible Element.
Category:Distributive Lattices
Category:Join Irreducible Elements
Category:Join Prime Elements
Join Prime Element is Dual of Meet Prime Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $z \in S$.
The following are dual statements:
:$z$ is the meet prime element of the meet semilattice $\struct {S, \wedge, \preceq}$
:$z$ is the join prime element of the join semilattice $\struct {S, \vee, \preceq}$
== Proof ==
By definition of meet prime element:
:$z$ is the meet prime element of the meet semilattice $\struct {S, \wedge, \preceq}$
:
:$\forall x, y \in S: \paren{x \wedge y \preceq z} \implies \leftparen{x \preceq z}$ or $\rightparen{y \preceq z}$
By the duality principle, the dual of this statement is:
:$\forall x, y \in S: \paren{z \preceq x \vee y} \implies \leftparen{z \preceq z}$ or $\rightparen{z \preceq y}$
By definition of join prime element:
:$z$ is the join prime element of the join semilattice $\struct {S, \vee, \preceq}$
== Also see ==
* Duality Principle (Order Theory)
Category:Prime Elements
Category:Join Prime Elements
Category:Dual Pairs (Order Theory)
Join Prime Element is Join Irreducible

== Theorem ==
Let $\struct{S, \vee, \preceq}$ be a join semilattice.
Let $z \in S$ be a join-prime element.
Then:
:$z$ is join-irreducible
== Proof ==
By Dual Pairs (Order Theory):
* join semilattice is dual to meet semilattice.
* join prime element is dual to meet prime element.
* join irreducible element is dual to meet irreducible element.
Thus the theorem holds by the duality principle applied to Characterization of Meet Irreducible Element.
Category:Join Prime Elements
Category:Join Irreducible Elements
Join Semilattice Ideal iff Ordered Set Ideal

== Theorem ==
Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Let $I \subseteq S$ be a non-empty subset of $S$.
Then:
:$I$ is a join semilattice ideal of $\struct {S, \vee, \preceq}$ $I$ is an ordered set ideal of $\struct {S, \preceq}$.
== Proof ==
=== Necessary Condition ===
Let $I$ be a join semilattice ideal of $\struct {S, \vee, \preceq}$.
To show that $I$ is an ordered set ideal of $\struct {S, \preceq}$ it is sufficient to show:
Let $x, y \in I$.
Let $z = x \vee y$.
By definition of join semilattice ideal, $I$ is a subsemilattice, so:
:$z \in I$
By definition of join:
:$x \preceq z \text{ and } y \preceq z$
The result follows.
=== Sufficient Condition ===
Let $I$ be an ordered set ideal of $\struct {S, \preceq}$.
To show that $I$ is a join semilattice ideal of $\struct {S, \vee, \preceq}$ it is sufficient to show:
Let $x, y \in I$.
By definition of ordered set ideal, $I$ is a directed subset, so:
:$\exists z \in I : x \preceq z \text{ and } y \preceq z$
By definition of join:
:$x \vee y \preceq z$
By definition of ordered set ideal, $I$ is a lower section, so:
:$x \vee y \in I$
The result follows.
Category:Join Semilattices