nyuuzyou/wikis huggingface dataset:
Meet Semilattice is Semilattice

== Theorem ==
Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Then $\struct {S, \wedge}$ is a semilattice.
== Proof ==
By definition of meet semilattice, $\wedge$ is closed.
The other three defining properties for a semilattice follow respectively from:
:Meet is Commutative
:Meet is Associative
:Meet is Idempotent
Hence $\struct {S, \wedge}$ is a semilattice.
== Also see ==
* Join Semilattice is Semilattice
Category:Meet Semilattices
Category:Semilattices
Meet in Inclusion Ordered Set

== Theorem ==
Let $P = \struct {X, \subseteq}$ be an inclusion ordered set.
Let $A, B \in X$ such that
:$A \cap B \in X$
Then $A \wedge B = A \cap B$
== Proof ==
By Intersection is Subset:
:$A \cap B \subseteq A$ and $A \cap B \subseteq B$
By definition:
:$A \cap B$ is a lower bound for $\set {A, B}$
We will prove that
:$\forall C \in X: C$ is a lower bound for $\set {A, B} \implies C \subseteq A \cap B$
Let $C \in X$ such that:
:$C$ is a lower bound for $\set {A, B}$.
By definition of lower bound:
:$C \subseteq A$ and $C \subseteq B$
Thus by Intersection is Largest Subset:
:$C \subseteq A \cap B$
By definition of infimum:
:$\inf \set {A, B} = A \cap B$
Thus by definition of meet:
:$A \wedge B = A \cap B$
== Sources ==
* 
Category:Join and Meet
Meet in Set of Ideals

== Theorem ==
Let $\mathscr S = \struct {S, \preceq}$ be a meet semilattice.
Let $\map {\it Ids} {\mathscr S}$ be the set of all ideals in $\mathscr S$.
Let $I_1, I_2$ be ideals in $\mathscr S$.
Then
:$I_1 \wedge I_2 = \set {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}$
where
:$I_1 \wedge I_2$ denotes the meet in $\struct {\map {\it Ids} {\mathscr S}, \subseteq}$
== Proof ==
By Meet is Intersection in Set of Ideals:
:$I_1 \wedge I_2 = I_1 \cap I_2$
=== ($\subseteq$) ===
Let $x \in I_1 \cap I_2$.
By definition of intersection:
:$x \in I_1 \land x \in I_2$
By Meet is Idempotent:
:$x = x \wedge x$
Thus
:$x \in \set {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}$
=== ($\supseteq$) ===
Let $x \in \set {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}$.
Then
:$\exists i_1 \in I_1, i_2 \in I_2: x = i_1 \wedge i_2$
By Meet Precedes Operands:
:$x \preceq i_1 \land x \preceq i_2$
By definition of lower section:
:$x \in I_1 \land x \in I_2$
Thus by definition of intersection:
:$x \in I_1 \cap I_2$
== Sources ==
* 
Category:Join and Meet Semilattices
Category:Join and Meet
Meet is Associative

== Theorem ==
Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Then $\wedge$ is associative.
== Proof ==
Let $a, b, c \in S$ be arbitrary.
Then:
Hence the result.
== Also see ==
* Join is Associative
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.23 \ \text {(a)}$
Category:Meet Operation
Category:Examples of Associative Operations
Meet is Commutative

== Theorem ==
Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Then $\wedge$ is commutative.
== Proof ==
Let $a, b \in S$ be arbitrary.
Then:
Hence the result.
== Also see ==
* Join is Commutative
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.23 \ \text {(a)}$
Category:Meet Operation
Category:Examples of Commutative Operations
Meet is Directed Suprema Preserving implies Meet of Suprema equals Supremum of Meet of Directed Subsets

== Theorem ==
Let $\struct {S, \preceq}$ be an up-complete meet semilattice.
Let $\struct {S \times S, \precsim}$ be the simple order product of $\struct {S, \preceq}$ and $\struct {S, \preceq}$.
Let $f: S \times S \to S$ be a mapping such that:
:$\forall s, t \in S: \map f {s, t} = s \wedge t$
and:
:$f$ preserves directed suprema.
Let $D_1, D_2$ be directed subsets of $S$.
Then:
:$\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {x \wedge y: x \in D_1, y \in D_2}$
== Proof ==
By Up-Complete Product:
:$\struct {S \times S, \precsim}$ is up-complete.
By Up-Complete Product/Lemma 1:
:$D_1 \times D_2$ is directed subsets of $S \times S$
By definition of mapping preserves directed suprema:
:$f$ preserves the supremum of $D_1 \times D_2$
By definition of up-complete:
:$D_1 \times D_2$ admits a supremum
and:
$D_1$ and $D_2$ admit suprema.
Thus:
== Sources ==
* 
* 
Category:Up-Complete Semilattices
Meet is Idempotent

== Theorem ==
Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Then $\wedge$ is idempotent.
== Proof ==
Let $a \in S$ be arbitrary.
Then:
Hence the result.
== Also see ==
* Join is Idempotent
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.23 \ \text {(a)}$
Category:Meet Operation
Category:Examples of Idempotence‎
Meet is Increasing

== Theorem ==
Let $\struct {S, \preceq}$ be a meet semilattice.
Let $f: S \times S \to S$ be a mapping such that:
:$\forall s, t \in S: \map f {s, t} = s \wedge t$
Then:
:$f$ is increasing as a mapping from the simple order product $\struct {S \times S, \precsim}$ of $\struct {S, \preceq}$ and $\struct {S, \preceq}$ into $\struct {S, \preceq}$.
== Proof ==
Let $\tuple {x, y}, \tuple {z, t} \in S \times S$ such that:
:$\tuple {x, y} \precsim \tuple {z, t}$
By definition of simple order product:
:$x \preceq z$ and $y \preceq t$
By Meet Semilattice is Ordered Structure:
:$x \wedge y \preceq z \wedge t$
By definition of $f$:
:$\map f {x, y} \preceq \map f {z, t}$
Thus by definition:
:$f$ is increasing mapping.
== Sources ==
* 
Category:Meet Operation
Meet is Intersection in Set of Ideals

== Theorem ==
Let $\mathscr S = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $\map {\mathit {Ids} } {\mathscr S}$ be the set of all ideals in $\mathscr S$.
Let $P = \struct {\map {\mathit {Ids} } {\mathscr S}, \precsim}$ be an ordered set where $\mathord \precsim = \mathord \subseteq \restriction_{\map {\mathit {Ids} } {\mathscr S} \times \map {\mathit {Ids} } {\mathscr S} }$
Let $I_1, I_2$ be ideals in $\mathscr S$.
Then
:$I_1 \wedge_P I_2 = I_1 \cap I_2$
== Proof ==
By Intersection of Semilattice Ideals is Ideal:
:$I_1 \cap I_2 \in \map {\mathit {Ids} } {\mathscr S}$
We will prove that
:$I_1 \cap I_2$ is lower bound for $\set {I_1, I_2}$
Let $x \in \set {I_1, I_2}$.
Then by definition of unordered tuple:
:$x = I_1$ or $x = I_2$
By Intersection is Subset
:$I_1 \cap I_2 \subseteq x$
Thus by definition of $\precsim$:
:$I_1 \cap I_2 \precsim x$
We will prove that
:$\forall I \in \map {\mathit {Ids} } {\mathscr S}: I$ is lower bound for $\set {I_1, I_2} \implies I \precsim I_1 \cap I_2$
Let $I \in \map {\mathit {Ids} } {\mathscr S}$ such that
:$I$ is lower bound for $\set {I_1, I_2}$
By definition of lower bound:
:$I \precsim I_1$ and $I \precsim I_2$
By definition of $\precsim$:
:$I \subseteq I_1$ and $I \subseteq I_2$
By Intersection is Largest Subset:
:$I \subseteq I_1 \cap I_2$
Thus by definition of $\precsim$
:$I \precsim I_1 \cap I_2$
By definition of infimum:
:$\inf \set {I_1, I_2} = I_1 \cap I_2$
Hence by definition of meet:
:$I_1 \wedge I_2 = I_1 \cap I_2$
== Sources ==
* 
Category:Join and Meet
Meet of Directed Subsets is Directed

== Theorem ==
Let $\struct {S, \preceq}$ be a meet semilattice.
Let $D_1, D_2$ be directed subset of $S$.
Then:
:$\set {x \wedge y: x \in D_1, y \in D_2}$ is a directed subset of $S$.
== Proof ==
Let $a, b \in \set {x \wedge y: x \in D_1, y \in D_2}$.
Then:
:$\exists x \in D_1, y \in D_2: a = x \wedge y$
and
:$\exists z \in D_1, t \in D_2: b = z \wedge t$
By definition of directed subset:
:$\exists g \in D_1: x \preceq g \land z \preceq g$
and
:$\exists h \in D_2: y \preceq h \land t \preceq h$
By Meet Semilattice is Ordered Structure:
:$x \wedge y \preceq g \wedge h$ and $z \wedge t \preceq g \wedge h$
:$g \wedge h \in \set {x \wedge y: x \in D_1, y \in D_2}$
Thus:
:$\exists c \in \set {x \wedge y: x \in D_1, y \in D_2}: a \preceq c \land b \preceq c$
Hence by definition:
:$\set {x \wedge y: x \in D_1, y \in D_2}$ is directed.
== Sources ==
* 
Category:Meet Semilattices
Category:Directed Preorderings
Meet of Suprema equals Supremum of Meet of Ideals implies Ideal Supremum is Meet Preserving

== Theorem ==
Let $\mathscr S = \struct {S, \wedge, \preceq}$ be an up-complete meet semilattice.
Let $f: \map {\it Ids} {\mathscr S} \to S$ be a mapping such that:
:$\forall I \in \map {\it Ids} {\mathscr S}: \map f I = \sup_{\mathscr S} I$
where 
:$\map {\it Ids} {\mathscr S}$ denotes the set of all ideals in $\mathscr S$
Let
:$\forall I_1, I_2 \in \map {\it Ids} {\mathscr S}: \paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {i \wedge j: i \in I_1, j \in I_2}$
Then:
$f$ preserves meet as a mapping from $\struct {\map {\it Ids} {\mathscr S}, \subseteq}$ into $\mathscr S$
== Proof ==
Let $I, J \in \map {\it Ids} {\mathscr S}$ such that
:$\set {I, J}$ admits an infimum in $\struct {\map {\it Ids} {\mathscr S}, \subseteq}$.
By definition of image of set:
:$\map {f^\to} {\set {I, J} } = \set {\map f I, \map f J}$
Thus by definition of meet semilattice:
:$\map {f^\to} {\set {I, J} }$ admits an infimum in $\mathscr S$.
Thus:
== Sources ==
* 
* 
Category:Up-Complete Semilattices
Meet with Complement is Bottom

== Theorem ==
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 2.
Then:
:$\exists \bot \in S: \forall a \in S: a \wedge \neg a = \bot$
where $\wedge$ denotes the meet operation in $S$.
This element $\bot$ is unique for any given $S$, and is named bottom.
== Proof ==
Let $\exists r, s \in S: r \wedge \neg r = a, \ s \wedge \neg s = b$
Then:
Thus, whatever $r$ and $s$ may be:
:$r \wedge \neg r = s \wedge \neg s$
This unique element can be assigned the symbol $\bot$ and named bottom as required.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.8$: Problems: $1 \ \text{A}$
Category:Boolean Algebras
Mellin Transform of Dirac Delta Function

== Theorem ==
Let $c \in \R_{>0}$ be a (strictly) positive real number.
Let $\map {\delta_c} t$ be the Dirac delta function.
Let $\MM$ be the Mellin transform.
Then:
:$\map {\MM \set {\map {\delta_c} t} } s = c^{s - 1}$
== Proof ==
== Also see ==
* Mellin Transform of Dirac Delta Function by Function
Category:Mellin Transforms
Category:Dirac Delta Function
Mellin Transform of Dirac Delta Function by Function

== Theorem ==
Let $f: \R \to \R$ be a function.
Let $c \in \R_{>0}$ be a positive constant real number.
Let $\map {\delta_c} t$ be the Dirac delta function.
Let $\MM$ be the Mellin transform.
Then:
:$\map {\MM \set {\map {\delta_c} t \map f t} } s = c^{s - 1} \map f c$
== Proof ==
Category:Mellin Transforms
Category:Dirac Delta Function
Mellin Transform of Exponential

== Theorem ==
Let $a$ be a complex constant and $e^t$ be the complex exponential.
Let $\MM$ be the Mellin transform.
Then:
:$\map {\MM \set {e^{-a t} } } s = a^{-s} \, \map \Gamma s$
where $\map \Re a, \map \Re s > 0$
== Proof ==
== Also see ==
* Mellin Transform of Higher Order Exponential
Category:Mellin Transforms
Mellin Transform of Heaviside Step Function

== Theorem ==
Let $c$ be a constant real number.
Let $\map {u_c} t$ be the Heaviside step function.
Let $\MM$ be the Mellin transform.
Then:
:$\map {\MM \set {\map {u_c} t} } s = -\dfrac {c^s} s$
for $c > 0, \map \Re s < 0$.
=== Corollary ===
== Proof ==
=== Lemma ===
Category:Mellin Transforms
Category:Heaviside Step Function
Mellin Transform of Heaviside Step Function/Corollary

== Theorem ==
Let $c$ be a constant real number.
Let $\map {u_c} t$ be the Heaviside step function.
Let $\MM$ be the Mellin transform.
Then:
:$\map {\MM \set {\map u {c - t} } } s = \dfrac {c^s} s$
for $c > 0, \map \Re s > 0$
== Proof ==
Category:Mellin Transforms
Category:Heaviside Step Function
Mellin Transform of Heaviside Step Function/Lemma

== Theorem ==
Let $t \in \R$.
Let $s \in \C$ with $\map \Re s < 0$.
Then:
:$\ds \lim_{t \mathop \to +\infty} t^s = 0$
== Proof ==
Let $s = a + b i$, where $a, b \in \R$, $a<0$.
Category:Heaviside Step Function
Mellin Transform of Higher Order Exponential

== Theorem ==
Let $a$ be a complex constant.
Let $n$ be a natural number.
Let $e^t$ be the complex exponential of $t$.
Let $\MM$ be the Mellin transform.
Then:
:$\map {\MM \set {e^{-a t^n} } } s = \dfrac {a^{-s/n} } n \map \Gamma {\dfrac s n}$
where $\map \Gamma z$ is the Gamma function and $\map \Re a$, $\map \Re s > 0$.
== Proof ==
Category:Mellin Transforms
Mellin Transform of Power Times Function

== Theorem ==
Let $t^n: \R \to \R$ be $t$ to the $n$th power for some $n \in \N_{\ge 0}$.
Let $\MM$ be the Mellin transform.
Then:
:$\map {\MM \set {t^n \map f t} } s = \map {\MM \set {\map f t} } {s + n}$
given that both transforms exist.
== Proof ==
Category:Mellin Transforms
Membership Rank Inequality

== Theorem ==
Let $S$ and $T$ be sets.
Let $\map {\operatorname{rank} } S$ denote the rank of $S$.
Then:
:$S \in T \implies \map {\operatorname{rank} } S < \map {\operatorname{rank} } T$
== Proof ==
By Ordinal Equal to Rank:
:$T \in \map V {\map {\operatorname{rank} } T + 1}$
By the definition of rank:
:$T \subseteq \map V {\map {\operatorname{rank} } T}$
Since $S \in T$:
:$S \in \map V {\map {\operatorname{rank} } T}$
By Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy:
:$\map {\operatorname{rank} } T \nsubseteq \map {\operatorname{rank} } S$
Therefore by Ordinal Membership is Trichotomy and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal:
:$\map {\operatorname{rank} } S < \map {\operatorname{rank} } T$
== Sources ==
* : $\S 9.16$
Category:Von Neumann Hierarchy
Membership Relation is Antisymmetric

== Theorem ==
Let $\Bbb S$ be a set of sets in the context of pure set theory
Let $\RR$ denote the membership relation on $\Bbb S$:
:$\forall \tuple {a, b} \in \Bbb S \times \Bbb S: \tuple {a, b} \in \RR \iff a \in b$
$\RR$ is an antisymmetric relation.
== Proof ==
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.5$: Relations: Exercise $1.5.1$
Category:Membership Relation
Membership Relation is Not Reflexive

== Theorem ==
Let $\Bbb S$ be a set of sets in the context of pure set theory
Let $\RR$ denote the membership relation on $\Bbb S$:
:$\forall \tuple {a, b} \in \Bbb S \times \Bbb S: \tuple {a, b} \in \RR \iff a \in b$
$\RR$ is not in general a reflexive relation.
== Proof ==
In the extreme pathological edge case:
:$S = \set S$
it is seen that:
:$S \in S$
and so:
:$\forall x \in S: x \in x$
demonstrating that $\RR$ is reflexive in this specific case.
However, in this case $\set S$ is a set on which the Axiom of Foundation does not apply.
This is seen in Set is Not Element of Itself.
Hence this set is not supported by Zermelo-Fraenkel set theory.
Consider the set:
:$T = \set {\O, \set \O}$
Then we immediately see that:
:$\O \notin \O$
and so $\RR$ is seen to be not reflexive.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.5$: Relations: Exercise $1.5.1$
Category:Membership Relation
Membership Relation is Not Symmetric

== Theorem ==
Let $\Bbb S$ be a set of sets in the context of pure set theory
Let $\RR$ denote the membership relation on $\Bbb S$:
:$\forall \tuple {a, b} \in \Bbb S \times \Bbb S: \tuple {a, b} \in \RR \iff a \in b$
$\RR$ is not in general a symmetric relation.
== Proof ==
In the extreme pathological edge case:
:$S = \set S$
it is seen that:
:$S \in S$
and so:
:$\forall x \in S: \tuple {a, b} \in \RR \implies \tuple {b, a} \in \RR$
demonstrating that $\RR$ is symmetric in this specific case.
However, in this case $\set S$ is a set on which the Axiom of Foundation does not apply.
This is seen in Set is Not Element of Itself.
Hence this set is not supported by Zermelo-Fraenkel set theory.
Consider the set:
:$T = \set {\O, \set \O}$
Then we immediately see that while:
:$\O \in \set \O$
we have that:
:$\set \O \notin \O$
and so $\RR$ is seen to be not symmetric.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.5$: Relations: Exercise $1.5.1$
Category:Membership Relation
Membership is Left Compatible with Ordinal Addition

== Theorem ==
Let $x$, $y$, and $z$ be ordinals.
Let $<$ denote membership relation $\in$, since $\in$ is a strict well-ordering on the ordinals.
Then:
:$x < y \implies \paren {z + x} < \paren {z + y}$
== Proof ==
The proof proceeds by transfinite induction on $y$.
In the proof, we shall use $\in$, $\subsetneq$, and $<$ interchangeably.
We are justified in this by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
=== Base Case ===
The conclusion:
:$x < \O \implies \paren {z + x} < \paren {z + \O}$
follows from propositional logic.
=== Inductive Case ===
=== Limit Case ===
== Sources ==
* : $\S 8.4$
Category:Ordinal Arithmetic
Membership is Left Compatible with Ordinal Exponentiation

== Theorem ==
Let $x$, $y$, and $z$ be ordinals.
Suppose $1 < z$.
Then:
:$x < y \iff z^x < z^y$
== Proof ==
=== Sufficient Condition ===
The proof shall proceed by Transfinite Induction on $y$.
=== Basis for the Induction ===
It is a contradiction that $x < 0$ by the definition of empty set.
This proves the basis for the induction.
=== Induction Step ===
The inductive step states that $x < y \implies z^x < z^y$.
Then, if $x < y^+$, then $x = y \lor x < y$.
==== Lemma ====
$z^y < z^y \times y$ by Membership is Left Compatible with Ordinal Multiplication.
So $z^y < z^{y^+}$ by the definition of ordinal exponentiation.
Suppose $x = y$. Then $z^x < z^{y^+}$.
Suppose $x < y$. Then:
In either case, the inductive step holds.
This proves the induction step.
=== Limit Case ===
The inductive hypothesis states that:
:$\forall w \in y: z^x < z^w$ for $x < w$
Take any $x < y$.
By Limit Ordinal Equals its Union, it follows that $x < w$ for some $w \in y$.
By the inductive hypothesis, $z^x < z^w$.
Since $w \in y$, it follows that:
This proves the limit case.
=== Necessary Condition ===
Note that the last part proves that:
:$x \in y \implies z^x \in z^y$
Moreover:
:$x = y \implies z^x = z^y$
Therefore, by the fact that $\in$ is a strong ordering:
:$x \notin y \implies z^x \notin z^y$
By the Rule of Transposition:
:$z^x < z^y \implies x < y$
== Sources ==
* : $\S 8.33$, $\S 8.34$
Category:Ordinal Arithmetic
Membership is Left Compatible with Ordinal Multiplication

== Theorem ==
Let $x$, $y$, and $z$ be ordinals.
Then:
:$\paren {x < y \land z > 0} \iff \paren {z \cdot x} < \paren {z \cdot y}$
== Proof ==
=== Sufficient Condition ===
The proof of the sufficient condition shall proceed by Transfinite Induction on $y$.
=== Basis for the Induction ===
Both $x < 0$ and $\paren {x \cdot z} < \paren {0 \cdot z}$ are contradictory, so the statement holds for the condition that $y = 0$.
This proves the basis for the induction.
=== Induction Step ===
Suppose the biconditional statement holds for $y$. Then:
In either case:
:$\paren {z \cdot x} < \paren {z \cdot y^+}$
This proves the induction step.
=== Limit Case ===
Suppose $y$ is a limit ordinal:
This proves the limit case.
=== Necessary Condition ===
Conversely, suppose $\paren {z \cdot x} < \paren {z \cdot y^+}$.
Then $z \ne 0$ because if it were equal, both sides of the inequality would be $0$.
So $z > 0$.
Furthermore:
:$y < x \implies \paren {z \cdot y} < \paren {z \cdot x}$
:$y = x \implies \paren {z \cdot y} = \paren {z \cdot x}$
So if $\paren {z \cdot x} < \paren {z \cdot y}$, then $y \ne x$ and $y \not < x$, so $x < y$ by Ordinal Membership is Trichotomy.
== Sources ==
* : $\S 8.19$
Category:Ordinal Arithmetic
Membership of Equivalence Class of m mod pi is Replicative Function

== Theorem ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \sqbrk {\exists r \in \Q, \exists m \in \Z: x = r \pi + m}$
where $\sqbrk {\cdots}$ is Iverson's convention.
Then $f$ is a replicative function.
== Proof ==
Let $\map f x = 1$.
Then:
:$\exists r \in \Q, \exists m \in \Z: x = r \pi + m$
and:
But $r n \in \Q$ and $m n \in \Z$.
So:
:$\map f x = 1 \implies \map f {n x} = 1$
Let $n \in \Z_{>0}$.
Let $k \in \Z: 0 \le k < n$.
Then:
Thus, when $0 \le k < n$ it follows that the coefficient of $\pi$ cannot be rational.
Thus:
:$\map f x = 1 \implies \map f {x + \dfrac k n} = 0$
and so:
:$\ds \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n} = \map f {n x}$
Now let $\map f {n x} = 1$.
Then:
:$r_n \pi + m_n = n x$
for some rational $r_n$ and integer $m_n$.
Thus exactly one of $m_n + k$ where $0 \le n < k$ is a multiple of $n$.
Therefore there exists exactly one $k \in \Z$ such that $0 \le n < k$ such that:
:$\map f {x + \dfrac k n} = 1$
Therefore:
:$\ds \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n} = 1 = \map f {n x}$
Hence the result by definition of replicative function.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $39 \ \text{d)}$
Category:Replicative Functions
Membership of Set of Integers is Replicative Function

== Theorem ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \sqbrk {x \in \Z}$
where $\sqbrk {\cdots}$ is Iverson's convention.
Then $f$ is a replicative function.
== Proof ==
First note that the interval between $x$ and $x + \dfrac {n - 1} n$ is less than $1$.
Thus there can be no more than one $k$ such that $0 \le k < n$ such that:
:$x + \dfrac k n \in \Z$
Hence:
:$\ds \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} \le 1$
First let:
:$\ds \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} = 1$
Then:
Let:
:$\ds \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} = 0$
 $\sqbrk {n x \in \Z} = 1$.
Then:
But at least one $n x + k$ such that $0 \le k < n$ is a multiple of $n$.
Hence:
:$\exists k \in \Z: 0 \le k < n: \sqbrk {x + \dfrac k n \in \Z} = 1$
So:
:$\ds \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} = 1$
which contradicts the supposition that $\sqbrk {n x \in \Z} = 1$.
Hence the result by definition of replicative function.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $39 \ \text{b)}$
Category:Replicative Functions
Membership of Set of Strictly Positive Integers is Replicative Function

== Theorem ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \sqbrk {x \in \Z_{> 0} }$
where $\sqbrk \cdots$ is Iverson's convention.
Then $f$ is a replicative function.
== Proof ==
Let $x \in \R$ such that $x > 0$.
Then for all $k \in \Z$ such that $0 \le k < n$:
:$x + \dfrac k n \in \Z_{> 0}$
and so from Membership of Set of Integers is Replicative Function:
:$\ds \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z_{> 0} } = \sqbrk {n x \in \Z_{> 0} }$
Let $x \le 0$.
Then for all $k \in \Z$ such that $0 \le k < n$:
:$x + \dfrac k n < 1$
and so:
:$\ds \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z_{> 0} } = 0 = \sqbrk {n x \in \Z_{> 0} }$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $39 \ \text{c)}$
Category:Replicative Functions
Men Ascending Vertical Ladders yet Moving Apart

== Problem ==
$2$ men each ascend a vertical ladder, at the same speed, and yet they get further apart.
Explain.
== Solution 1 ==
== Solution 2 ==
== Historical Note ==
== Sources ==
* : Henry van Etten: $118$
Category:Geography Puzzles
Category:Classic Problems
Category:Men Ascending Vertical Ladders yet Moving Apart
Men Ascending Vertical Ladders yet Moving Apart/Solution 2

== Problem ==
== Solution ==
To a first approximation, Earth is a sphere.
So, anywhere you place $2$ vertical ladders, their upper ends will be further apart from their lower ends.
Hence the ladders may be anywhere on the surface of the Earth.
== Historical Note ==
== Sources ==
* : Henry van Etten: $118$
Category:Men Ascending Vertical Ladders yet Moving Apart