nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of P-adic Norms

== Theorem ==
Let $p \in \N$ be a prime.
Let $\Q$ denote the rational numbers.
 
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
From Negative Powers of Group Elements, Definition 2 can be rewritten as:
:$\forall r \in \Q: \norm r_p = \begin {cases} 0 & : r = 0 \\ p^{-k} & : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n \end {cases}$
Hence if follows that Definition 1 and Definition 2 are equivalent if it is shown:
:$\forall r \in \Q_{\ne 0}: \map {\nu_p} r = k : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n$
Let $r \in \Q_{\ne 0}$.
Let $r = \dfrac a b : a, b \in \Z_{\ne 0}$
We have:
Let:
:$k_a := \map {\nu_p} a$
:$k_b := \map {\nu_p} b$
=== Lemma 1 ===
From Lemma 1:
:$a = p^{k_a} m$
:$b = p^{k_b} n$
:$p \nmid m, n$
Hence:
Let $k = k_a - k_b$.
It follows that:
:$\map {\nu_p} r = k : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n$
Since $r$ was arbitrary:
:$\forall r \in \Q_{\ne 0}: \map {\nu_p} r = k : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n$
Category:Equivalence of Definitions of P-adic Norms
Category:P-adic Norms
Equivalence of Definitions of P-adic Norms/Lemma 1

== Theorem ==
Let $p \in \N$ be a prime number.
Let $\nu_p: \Z \to \N \cup \set {+\infty}$ be the $p$-adic valuation on the integers.
Then:
:$\forall x \in Z_{\ne 0}: \map {\nu_p} x = k : x = p^k y : p \nmid y$
== Proof ==
Let $x \in \Z_{\ne 0}$.
By definition of the $p$-adic valuation:
:$\map {\nu_p} x = \sup \set {v \in \N: p^v \divides x}$
Let $\map {\nu_p} x = k$.
Then:
:$p^k \nmid x$
By definition of a divisor:
:$\exists y \in Z : x = p^k y$
:
:$p \divides y$
By definition of a divisor:
:$\exists y' \in Z : y = p y'$
Hence:
This contradicts:
:$k = \sup \set {v \in \N: p^v \divides x}$
Hence:
:$p \nmid y$
Since the choice of $x \in \Z_{\ne 0}$ was arbitrary:
:$\forall x \in Z_{\ne 0}: \map {\nu_p} x = k : x = p^k y : p \nmid y$
Category:Equivalence of Definitions of P-adic Norms
Equivalence of Definitions of P-adic Valuation on P-adic Numbers

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $x \in \Q_p \setminus \set 0$.
Let $l$ be the index of the first non-zero coefficient in the $p$-adic expansion:
:$l = \min \set {i: i \ge m \land d_i \ne 0}$
From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient:
:$\norm x_p = p^{-l}$
By definition of real general logarithm:
:$-\log_p \norm x_p = l$
The result follows.
Category:P-adic Numbers
Category:P-adic Valuations
Equivalence of Definitions of Path Component

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
=== Equivalence Class ===
=== Union of Path-Connected Sets ===
=== Maximal Path-Connected Set ===
== Proof ==
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text { is path-connected in } T}$
Let $C = \bigcup \CC_x$.
=== Lemma ===
Let $C'$ be the equivalence class containing $x$ of the equivalence relation $\sim$ defined by:
:$y \sim z$ $y$ and $z$ are connected in $T$.
=== Equivalence Class equals Union of Path-Connected Sets ===
It needs to be shown that $C = C'$.
=== Union of Path-Connected Sets is Maximal Path-Connected Set ===
=== Maximal Path-Connected Set is Union of Path-Connected Sets ===
== Also see ==
* Path-Connectedness is Equivalence Relation
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Path Components
Category:Path-Connected Spaces
Category:Equivalence of Definitions of Path Component
Equivalence of Definitions of Path Component/Equivalence Class equals Union of Path-Connected Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\CC_x = \left\{ {A \subseteq S : x \in A \land A }\right.$ is path-connected in $\left. {T}\right\}$.
Let $C = \bigcup \CC_x$
Let $\sim$ be the equivalence relation defined by:
:$y \sim z$ $y$ and $z$ are path-connected in $T$.
Let $C'$ be the equivalence class of $\sim$ containing $x$.
Then $C = C'$.
== Proof ==
The result follows.
== Also see ==
* Path-Connectedness is Equivalence Relation
Category:Equivalence of Definitions of Path Component
Equivalence of Definitions of Path Component/Lemma 1

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\CC_x = \leftset {A \subseteq S : x \in A \land A}$ is path-connected in $\rightset T$
Let $C = \bigcup \CC_x$
Then:
:$C$ is path-connected in $T$ and $C \in \CC_x$.
== Proof ==
From Point is Path-Connected to Itself, $\set x$ is a path-connected subset of $T$ containing $x$.
It follows that $x \in C$.
From Union of Path-Connected Sets with Common Point is Path-Connected, $C$ is a path-connected subset of $T$.
Hence $C \in \CC_x$.
Category:Equivalence of Definitions of Path Component
Equivalence of Definitions of Path Component/Maximal Path-Connected Set is Union of Path-Connected Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\tilde C$ be a maximal path-connected set of $T$ that contains $x$.
Then:
:$\tilde C = \bigcup \set {A \subseteq S : x \in A \land A \text{ is path-connected in } T}$.
== Proof ==
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text{ is path-connected in } T}$
Let $C = \bigcup \CC_x$
=== Lemma ===
By definition:
:$\tilde C \in \CC_x$
From Set is Subset of Union:
:$\tilde C \subseteq C$
By maximality of $\tilde C$:
:$\tilde C = C$
== Also see ==
* Union of Path-Connected Sets is Path-Maximal Connected Set
Category:Equivalence of Definitions of Path Component
Equivalence of Definitions of Path Component/Union of Path-Connected Sets is Maximal Path-Connected Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\CC_x = \left\{ {A \subseteq S : x \in A \land A } \right.$ is path-connected in $\left. {T} \right\}$
Let $C = \bigcup \CC_x$
Then $C$ is a maximal path-connected set of $T$.
== Proof ==
=== Lemma ===
Let $\tilde C$ be any path-connected set such that:
:$C \subseteq \tilde C$
Then $x \in \tilde C$.
Hence $\tilde C \in \CC_x$.
From Set is Subset of Union, 
:$\tilde C \subseteq C$.
Hence $\tilde C = C$.
It follows that $C$ is a maximal path-connected set of $T$ by definition.
== Also see ==
* Maximal Path-Connected Set is Union of Path-Connected Sets
Category:Equivalence of Definitions of Path Component
Equivalence of Definitions of Pentagonal Number

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $P_n$ be a pentagonal number by definition 1.
Let $n = 0$.
By definition:
:$P_0 = 0$
By vacuous summation:
:$P_0 = \ds \sum_{i \mathop = 1}^0 \paren {3 i - 2} = 0$
By definition of summation:
and so:
Thus $P_n$ is a pentagonal number by definition 2.
=== Definition 2 implies Definition 1 ===
Let $P_n$ be a pentagonal number by definition 2.
Then:
Then:
:$P_0 = \ds \sum_{i \mathop = 1}^0 \paren {3 n - 2}$
is a vacuous summation and so:
:$P_0 = 0$
Thus $P_n$ is a pentagonal number by definition 1.
=== Definition 1 equivalent to Definition 3 ===
We have by definition that $P_n = 0 = \map P {5, n}$.
Then:
Thus $\map P {5, n}$ and $P_n$ are generated by the same recurrence relation.
Category:Pentagonal Numbers
Equivalence of Definitions of Perfect Number

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
Consider a strictly positive integer $n$.
By Definition 1, $n$ is a perfect number $n$ equals the sum of its aliquot parts.
By definition of divisor sum function, $\map {\sigma_1} n$ equals the sum of all the divisors of $n$.
Thus $\map {\sigma_1} n - n$ equals the aliquot sum of $n$.
So by Definition 2, $n$ is a perfect number $\map {\sigma_1} n - n = n$.
Hence the definitions are equivalent.
The equivalence of definition 4 to definition 2 follows directly.
The equivalence of definition 3 to definition 2 follows from the definition of abundance.
== Sources ==
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
Category:Perfect Numbers
Equivalence of Definitions of Perfect Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a subset of $S$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
Let $T = \struct {S, \tau}$ be a topological space and let $H \subseteq S$.
=== Definition 1 implies Definition 2 ===
Suppose that $H = H'$ where $H'$ is the derived set of $H$.
By definition of derived set, $H'$ is the set of all limit points of $H$.
So $H$ contains all its limit points and so by definition is closed.
Now we also have that any point not in $H'$ is an isolated point.
But there are no points in $H$ which are not in $H'$, so $H$ has no isolated points.
Therefore $H = H'$ implies that $H$ is closed and has no isolated points.
=== Definition 2 implies Definition 1 ===
Suppose $H$ is closed and has no isolated points.
By Closed Set iff Contains all its Limit Points we have that $H' \subseteq H$ where $H'$ is the derived set of $H$.
As $S$ has no isolated points, all its elements are elements of its derived set.
Therefore $H \subseteq H'$.
So by definition of set equality it follows that $H = H'$.
=== Definition 2 iff Definition 3 ===
Let $H$ be closed with no isolated points.
By definition, $H$ is closed it contains all its limit points.
Also by definition, $H$ is dense-in-itself it has no isolated points.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Limit Points
Category:Perfect Sets
Equivalence of Definitions of Points Separated by Neighborhoods

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $x, y \in S$.
From Singleton of Element is Subset:
:$x$ and $y$ are separated as points by neighborhoods the singletons $\set x$ and $\set y$ are separated as sets by neighborhoods.
From Equivalence of Definitions of Sets Separated by Neighborhoods:
:the singletons $\set x$ and $\set y$ are separated as sets by neighborhoods the singletons $\set x$ and $\set y$ are separated as sets by open sets.
From Singleton of Element is Subset:
:the singletons $\set x$ and $\set y$ are separated as sets by open sets $x$ and $y$ are separated as points by open sets.
Category:Separation Axioms
Equivalence of Definitions of Polynomial Function on Subset of Ring

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $S \subset R$ be a subset.
=== Definition 1 ===
=== Definition 2 ===
== Outline of proof ==
This follows by interpreting both definitions as $f$ being a polynomial in the element $\iota$.
== Proof ==
=== 1 implies 2 ===
Let $\map P X \in R \sqbrk X$ be the polynomial:
:$P = \ds \sum_{k \mathop = 0}^n a_k \cdot X^k$
where $\sum$ denotes indexed summation.
We show that $\map P \iota = f$.
=== 2 implies 1 ===
Let $P \in R \sqbrk X$ be a polynomial and $f = \map P \iota \in R^S$.
By Polynomial is Linear Combination of Monomials, there exist:
:$n \in \N$
:$a_0, \ldots, a_n \in R$
such that:
:$P = \ds \sum_{k \mathop = 0}^n a_k \cdot X^k$
where $\sum$ denotes indexed summation.
Let $\operatorname{ev}_{\iota}$ denote the evaluation homomorphism at $\iota$.
Then $\map {\operatorname{ev}_{\iota} } P = f$.
We have:
Category:Polynomial Theory
Equivalence of Definitions of Polynomial Ring in Multiple Variables

== Theorem ==
Let $R$ be a commutative ring with unity.
The following definitions of polynomial ring are equivalent in the following sense:
: For every two constructions, there exists an $R$-isomorphism which sends indeterminates to indeterminates.
=== Definition 1: As the monoid ring on a free monoid on a set ===
== Proof ==
== Also see ==
* Equivalence of Definitions of Polynomial Ring
Category:Polynomial Theory
Equivalence of Definitions of Polynomial Ring in One Variable

== Theorem ==
Let $R$ be a commutative ring with unity.
The following definitions of polynomial ring are equivalent in the following sense:
: For every two constructions, there exists an $R$-isomorphism which sends indeterminates to indeterminates.
=== Definition 1: As a Ring of Sequences ===
=== Definition 2: As a Monoid Ring on the Natural Numbers ===
== Outline of Proof ==
We show that they all satisfy the same universal property.
== Proof ==
Use Polynomial Ring of Sequences Satisfies Universal Property
== Also see ==
* Uniqueness of Polynomial Ring in One Variable
* Equivalence of Definitions of Polynomial Ring
Category:Polynomial Theory
Equivalence of Definitions of Positive Definite Matrix

== Theorem ==
=== Definition $1$ ===
=== Definition $2$ ===
== Proof ==
=== Definition $(1)$ implies Definition $(2)$ ===
Let $\mathbf A$ be a positive definite matrix by definition $1$.
Thus $\mathbf A$ is a positive definite matrix by definition $2$.
=== Definition $(2)$ implies Definition $(1)$ ===
Let $\mathbf A$ be a positive definite matrix by definition $2$.
Thus $\mathbf A$ is a positive definite matrix by definition $1$.
== Sources ==
* 
* 
Category:Positive Definite Matrices
Equivalence of Definitions of Positive Semidefinite Matrix

== Theorem ==
=== Definition $1$ ===
=== Definition $2$ ===
== Proof ==
=== Definition $(1)$ implies Definition $(2)$ ===
Let $\mathbf A$ be a positive semidefinite matrix by definition $1$.
Thus $\mathbf A$ is a positive semidefinite matrix by definition $2$.
=== Definition $(2)$ implies Definition $(1)$ ===
Let $\mathbf A$ be a positive semidefinite matrix by definition $2$.
Thus $\mathbf A$ is a positive semidefinite matrix by definition $1$.
== Sources ==
* 
Category:Positive Semidefinite Matrices
Equivalence of Definitions of Preimage of Subset under Mapping

== Theorem ==
Let $f: S \to T$ be a mapping from a set $S$ to a set $T$.
Let $Y \subseteq T$ be a subset of $T$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
The difference in definitions is no more than a difference in notations.
Let $X$ be a preimage by definition $2$.
Then by definition:
:$X := \map {f^\gets} Y$
By definition of inverse image mapping of mapping:
:$\forall Y \in \powerset T: \map {f^\gets} Y = \set {s \in S: \exists t \in Y: \map f s = t}$
Thus $X$ is a preimage by definition $1$.
Category:Preimage of Subset under Mapping
Equivalence of Definitions of Preordering

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\RR$ be a relation on $S$ satisfying:
By Reflexive and Transitive Relation is Idempotent:
:$\RR \circ \RR = \RR$
which is condition $(1)$ of definition 2.
By definition of reflexive relation:
:$\RR$ is reflexive it is a superset of the diagonal relation: 
::$\Delta_S \subseteq \RR$
which is condition $(2)$ of definition 2.
Thus $\RR$ is an ordering by definition 2.
=== Definition 2 implies Definition 1 ===
Let $\RR$ be a relation which fulfils the conditions:
:$(1): \quad \RR \circ \RR = \RR$
:$(2): \quad \Delta_S \subseteq \RR$
By definition of set equality, it follows from $(1)$:
:$\RR \circ \RR \subseteq \RR$
Thus, by definition, $\RR$ is transitive.
By definition of reflexive relation:
:$\RR$ is reflexive it is a superset of the diagonal relation: 
::$\Delta_S \subseteq \RR$
Thus $\RR$ is an ordering by definition 1.
Category:Preorderings
Equivalence of Definitions of Primary Ideal

== Theorem ==
Let $R$ be a commutative ring with unity.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $x + \mathfrak q$ be a zero-divisor of $R / \mathfrak q$.
That is, there is a $y \not \in \mathfrak q$ such that:
:$\paren {x + \mathfrak q} \paren {y + \mathfrak q} = 0 + \mathfrak q$
Thus:
:$xy + \mathfrak q = 0 + \mathfrak q$
which means:
:$xy \in \mathfrak q$
Then, :
:$\exists n \in \N_{>0} : x^n \in \mathfrak q$
so that:
:$\paren {x + \mathfrak q}^n = x^n + \mathfrak q = 0 + \mathfrak q$
Therefore $x + \mathfrak q$ is nilpotent.
=== Definition 2 implies Definition 1 ===
Let $xy \in \mathfrak q$ but $x \not \in \mathfrak q$.
That is:
:$\paren {x + \mathfrak q} \paren {y + \mathfrak q} = 0 + \mathfrak q$
Then, , $\paren {y + \mathfrak q}$ is nilpotent in $R \ \mathfrak q$.
Thus:
so that:
:$y^n \in \mathfrak q$
Category:Primary Ideals
Equivalence of Definitions of Prime Ideal of Commutative and Unitary Ring

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
Let $\struct {R, +, \circ}$ be a commutative and unitary ring throughout.
=== $(1)$ implies $(2)$ ===
Let $P$ be a prime ideal of $R$ by definition 1.
Then by definition:
:$\forall a, b \in R : a \circ b \in P \implies a \in P$ or $b \in P$
Let $I \circ J \subseteq P$.
 that both $I \nsubseteq P$ and $J \nsubseteq P$.
Then by definition of subset:
:$\exists a \in I \setminus P, b \in J \setminus P$
But by definition of subset product
:$a \circ b \in P$ as $I \circ J \subseteq P$
Thus we have $a, b \in P$ such that:
:$a \circ b \in P$ where $a \notin P$ and $b \notin P$
But this contradicts the criterion for $P$ to be a prime ideal of $R$ by definition 1
Thus by Proof by Contradiction, either $I \subseteq P$ or $J \subseteq P$,
Thus $P$ is a prime ideal of $R$ by definition 2.
=== $(2)$ implies $(1)$ ===
Let $P$ be a prime ideal of $R$ by definition 2.
Then by definition:
:$I \circ J \subseteq P \implies I \subseteq P \text { or } J \subseteq P$
for all ideals $I$ and $J$ of $R$.
Let $a, b \in R$ such that $a \circ b \in P$.
Consider $\ideal a$ and $\ideal b$, the ideals generated by $a$ and $b$.
Let $r \in \ideal a, s \in \ideal b$. 
Then:
:$\exists m, n \in R: r = m \circ a, s = n \circ b$
Therefore:
This shows that $\ideal a \circ \ideal b \subseteq P$.
By definition 2, $\ideal a \subseteq P$ or $\ideal b \subseteq P$.
This implies $a \in P$ or $b \in P$.
Thus $P$ is a prime ideal of $R$ by definition 1.
=== $(1)$ implies $(3)$ ===
Let $P$ be a prime ideal of $R$ by definition 1.
Then by definition:
:$\forall a, b \in R : a \circ b \in P \implies a \in P$ or $b \in P$
 $R \setminus P$ is not multiplicatively closed.
That is:
But this contradicts the assertion that $a, b \in R \setminus P$.
Thus by Proof by Contradiction $R \setminus P$ is multiplicatively closed.
Thus $P$ is a prime ideal of $R$ by definition 3.
=== $(3)$ implies $(1)$ ===
Let $P$ be a prime ideal of $R$ by definition 3.
Then by definition:
:the complement $R \setminus P$ of $P$ in $R$ is closed under the ring product $\circ$.
 let $a \circ b \in P$ such that $a \notin P$ and $b \notin P$.
Then:
:$a, b \in R \setminus P$
by definition of relative complement.
But $R \setminus P$ is closed under the ring product $\circ$.
That means:
:$\forall a, b \in R \setminus P \implies a \circ b \in R \setminus P $
But this contradicts the assertion that $a \circ b \in P$.
Thus by Proof by Contradiction either $a \in P$ or $b \in P$ (or both).
Thus $P$ is a prime ideal of $R$ by definition 1.
Category:Prime Ideals of Commutative and Unitary Rings
Equivalence of Definitions of Prime Ideal of Ring

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\struct {R, +, \circ}$ be a ring throughout.
=== $(1)$ implies $(2)$ ===
Let $P$ be a prime ideal of $R$ by definition 1.
Let $I \circ J \subseteq P$.
 that both $I \nsubseteq P$ and $J \nsubseteq P$.
Then by definition of subset:
:$\exists a \in I \setminus P, b \in J \setminus P$
But by definition of subset product
:$a \circ b \in P$ as $I \circ J \subseteq P$
Thus we have $a, b \in P$ such that:
:$a \circ b \in P$ where $a \notin P$ and $b \notin P$
But this contradicts the criterion for $P$ to be a prime ideal of $R$ by definition 1
Thus by Proof by Contradiction, either $I \subseteq P$ or $J \subseteq P$,
Thus $P$ is a prime ideal of $R$ by definition 2.
=== $(2)$ implies $(1)$ ===
Let $P$ be a prime ideal of $R$ by definition 2.
Let $a, b \in R$ such that $a \circ b \in P$.
Consider $\ideal a$ and $\ideal b$, the ideals generated by $a$ and $b$.
Let $r \in \ideal a, s \in \ideal b$. 
Then:
:$\exists m, n \in R: r = m \circ a, s = n \circ b$
Therefore:
This shows that $\ideal a \circ \ideal b \subseteq P$.
By definition 2, $\ideal a \subseteq P$ or $\ideal b \subseteq P$.
This implies $a \in P$ or $b \in P$.
Thus $P$ is a prime ideal of $R$ by definition 1.
Category:Prime Ideals of Rings
Equivalence of Definitions of Prime Number

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
=== Definition 5 ===
=== Definition 6 ===
=== Definition 7 ===
== Proof ==
=== Definition 1 iff Definition 2 ===
This is proved in Prime Number has 4 Integral Divisors:
=== Definition 1 iff Definition 3 ===
This is proved in Divisor Count Function of Prime Number:
=== Definition 1 iff Definition 4 ===
From these two results:
:$1$ Divides all Integers
:Integer Divides Itself
it follows that if $p$ has exactly two positive integer divisors then those are $1$ and $p$.
By the same coin, if the only positive integer divisors of $p$ are $1$ and $p$, then $p$ has exactly two positive integer divisors.
=== Definition 4 iff Definition 5 ===
Let the only two positive integer divisors of $p$ be $1$ and $p$.
Then the only divisor of $p$ strictly less than $p$ is $1$.
Conversely, let the only divisor of $p$ strictly less than $p$ be $1$
From Integer Divides Itself we also have that $p$ is a divisor of $p$.
From Absolute Value of Integer is not less than Divisors it follows that any positive integer greater than $p$ is not a divisor of $p$.
Thus the only positive integer divisors of $p$ are $1$ and $p$.
=== Definition 2 iff Definition 6 ===
This is proved in Prime iff Equal to Product.
=== Definition 4 iff Definition 7 ===
Let the only two positive integer divisors of $p$ be $1$ and $p$.
Then the only positive integer less than $p$ which divisors of $p$ is $1$
So if $a b = p$ then either $a$ or $b$ is $p$ and so it is not the case that both $a$ and $b$ are less than $p$.
Now suppose that there do not exist $a$ and $b$ less than $p$ such that $a b = p$.
Suppose $a \divides p$ such that $a < p$.
That means that $\exists b \in \Z_{>0}: a b = p$.
That means $b \divides p$.
But $b \not < p$ by hypothesis.
It follows that $b = p$ and so $a = 1$
Hence $1$ and $p$ are the only positive integer divisors of $p$.
== Sources ==
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes: Footnote $1$
Category:Prime Numbers
Equivalence of Definitions of Primitive Root of Unity

== Theorem ==
Let $n \in \Z_{> 0}$ be a strictly positive integer.
Let $F$ be a field.
Let $U_n$ denote the set of all $n$-th roots of unity.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\alpha \in U_n$ such that:
:$U_n = \set{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}}$
By :
:$\alpha = \alpha^1$
Hence:
:$U_n = \set{1, \alpha^1, \alpha^2, \ldots, \alpha^{n-1}}$
By :
:$\forall m \in \N: 0 < m < n : \alpha^m \ne 1$ 
=== Definition 2 implies Definition 1 ===
Let $\alpha \in U_n$ such that:
:$\forall m \in \N : 0 < m < n: \alpha^m \ne 1$
Since $\alpha$ is a root of unity:
:$\alpha \ne 0$
:
:$\exists m, k \in \N : 0 \le m < k < n : \alpha^k = \alpha^m$
Then:
Since $0 < k-m < n$, this contradicts the hypothesis:
:$\forall m \in \N : 0 < m < n: \alpha^m \ne 1$
Hence:
:$\forall m, k \in \N: 0 \le m < k < n : \alpha^k \ne \alpha^m$
From Integer Power of Root of Unity is Root of Unity:
:$\set{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}} \subseteq U_n$
From Cardinality of Subset of Finite Set:
:$n = \card {\set{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}}} \le \card {U_n}$
Consider the polynomial $X^n - 1$ over $F$.
By and :
:$\beta \in F$ is a $n$-th root of unity $\beta$ is a root of the polynomial $X^n - 1$
From Polynomial over Field has Finitely Many Roots:
:The polynomial $X^n - 1$ has at most $n$ roots.
So:
:$\card {U_n} \le n$
Hence:
:$\card {U_n} = n = \card {\set{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}}}$
From Tests for Finite Set Equality:
:$U_n = \set{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}}$
Category:Primitive Roots of Unity