nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Order Complete Set

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\struct {S, \preceq}$ be an order complete set by definition 1.
Let $H \subseteq S$ have a lower bound.
Let $K$ be the set of all lower bounds of $H$.
Then $K \ne \O$.
By definition of lower bound:
:$\forall x \in K: \forall y \in H: x \le y$
and so all elements of $H$ are upper bounds of $K$.
Thus by hypothesis $K$ admits a supremum.
Let $k = \map \sup K$.
By definition of supremum, $k$ precedes every upper bound of $K$.
In particular, $k$ precedes every element of $H$.
Thus $k$ is a lower bound of $H$.
But we have that $k$ is an upper bound of $K$.
That is, $k$ succeeds every lower bound of $H$.
That is, $k$ is an infimum of $H$.
Thus $\struct {S, \preceq}$ is an order complete set by definition 2.
=== Definition 2 implies Definition 1 ===
Let $\struct {S, \preceq}$ be an order complete set by definition 2.
Let $H \subseteq S$ have an upper bound.
Let $K$ be the set of all upper bounds of $H$.
Then $K \ne \O$.
By definition of upper bound:
:$\forall x \in K: \forall y \in H: y \le x$
and so all elements of $H$ are lower bounds of $K$.
Thus by hypothesis $K$ admits an infimum.
Let $k = \map \inf K$.
By definition of infimum, $k$ succeeds every lower bound of $K$.
In particular, $k$ succeeds every element of $H$.
Thus $k$ is an upper bound of $H$.
But we have that $k$ is a lower bound of $K$.
That is, $k$ precedes every upper bound of $H$.
That is, $k$ is a supremum of $H$.
Thus $\struct {S, \preceq}$ is an order complete set by definition 1.
== Sources ==
* : Chapter $0$: Orderings: Theorem $9$
Category:Order Theory
Equivalence of Definitions of Order Embedding

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: S \to T$ be a mapping.
=== Definition 1 implies Definition 2 ===
Let $\phi$ be an order embedding by definition 1.
Then by definition:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
It follows from Order Embedding is Injection that $\phi$ is an injection.
Thus $\phi$ is an order embedding by definition 2.
=== Definition 2 implies Definition 1 ===
Let $\phi$ be an order embedding by definition 2.
Then condition $(2)$ of that definition directly states:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
Thus $\phi$ is an order embedding by definition 1.
=== Definition 1 implies Definition 3 ===
=== Definition 3 implies Definition 1 ===
=== Definition 2 implies Definition 4 ===
Let $\phi$ be an order embedding by definition 2.
Then by definition:
:$\phi$ is an injection
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
Let $T' = \Img S$ be the image of $S$ under $\phi$.
From Injection to Image is Bijection:
:$\phi {\restriction_{S \times T'} }: S \to T'$ is a bijection.
It follow by definition that $\phi {\restriction_{S \times T'} }: S \to T'$ is an order isomorphism.
Thus $\phi$ is an order embedding by definition 4.
=== Definition 4 implies Definition 1 ===
Let $\phi$ be an order embedding by definition 4.
Then by definition:
:the restriction of $\phi$ to $S \times T'$ is an order isomorphism between $\struct {S, \preceq_1}$ and $\struct {T', \preceq_2 \restriction_{T' \times T'} }$.
By definition of order isomorphism:
:$\forall x, y \in S: x \preceq_1 y \iff \map {\phi {\restriction_{S \times T'} } } x \preceq_2 \map {\phi {\restriction_{S \times T'} } } y$
But by definition of restriction:
:$\forall z \in S: \map {\phi {\restriction_{S \times T'} } } z = \map \phi z$
Thus:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
and so $\phi$ is an order embedding by definition 1.
== Sources ==
* : Chapter $0$: Orderings
Category:Order Embeddings
Category:Equivalence of Definitions of Order Embedding
Equivalence of Definitions of Order Embedding/Definition 1 implies Definition 3

== Theorem ==
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: S \to T$ be a mapping.
Let $\phi: S \to T$ be an order embedding by Definition 1:
Then $\phi: S \to T$ is an order embedding by Definition 3:
== Proof ==
Let $\phi$ be an order embedding by definition 1.
Then by definition:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
$\phi$ is injective by Order Embedding is Injection.
It remains to be shown that:
: $x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$
Suppose first that $x \prec_1 y$.
Then $x \preceq_1 y$ and $x \ne y$.
Thus by the premise:
:$\map \phi x \preceq_2 \map \phi y$
Since $\phi$ is injective:
:$\map \phi x \ne \map \phi y$
Therefore:
:$\map \phi x \prec_2 \map \phi y$
Suppose instead that $\map \phi x \prec_2 \map \phi y$
Then:
:$\map \phi x \preceq_2 \map \phi y$
and:
:$\map \phi x \ne \map \phi y$
By the premise:
:$x \preceq_1 y$
By the substitutive property of equality:
:$x \ne y$
Thus:
:$x \prec_1 y$
Thus $\phi$ is an order embedding by definition 3.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $20 \ \text {(a)}$
Category:Equivalence of Definitions of Order Embedding
Equivalence of Definitions of Order Embedding/Definition 3 implies Definition 1

== Theorem ==
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: S \to T$ be a mapping.
Let $\phi: S \to T$ be an order embedding by Definition 3:
Then $\phi: S \to T$ is an order embedding by Definition 1:
== Proof ==
Let $\phi$ be an order embedding by definition 3.
Then by definition:
:$(1): \quad \phi$ is injective
:$(2): \quad \forall x, y, \in S: x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$
Let $x \preceq_1 y$.
Then $x \prec_1 y$ or $x = y$.
If $x \prec_1 y$, then by hypothesis:
:$\map \phi x \prec_2 \map \phi y$
Thus:
:$\map \phi x \preceq_2 \map \phi y$
If $x = y$, then:
:$\map \phi x = \map \phi y$
Thus:
:$\map \phi x \preceq_2 \map \phi y$
Thus it has been shown that:
:$x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$
Let $\map \phi x \preceq_2 \map \phi y$.
Then:
:$\map \phi x \prec_2 \map \phi y$
or:
:$\map \phi x = \map \phi y$
Suppose $\map \phi x \prec_2 \map \phi y$.
Then by hypothesis:
:$x \prec_1 y$
and so:
:$x \preceq_1 y$
Suppose $\map \phi x = \map \phi y$.
Then since $\phi$ is injective:
:$x = y$
and so:
:$x \preceq_1 y$
Thus in both cases:
:$x \preceq_1 y$
and so:
:$\map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$
Hence the result:
:$x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
and so $\phi$ is an order embedding by definition 1.
Category:Equivalence of Definitions of Order Embedding
Equivalence of Definitions of Order Isomorphism

== Theorem ==
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\phi: S \to T$ be an order isomorphism by Definition 1.
Then $\phi$ is bijective, and thus trivially surjective.
Let $x, y \in S$.
Then by Definition 1:
:$x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$
Suppose that $\map \phi x \preceq_2 \map \phi y$.
Then by Definition 1:
:$\map {\phi^{-1} } {\map \phi x} \preceq_1 \map {\phi^{-1} } {\map \phi y}$
By the definition of inverse:
:$x \preceq_1 y$
Thus by Rule of Implication:
:$\map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$.
So:
:$x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
As this holds for all $x, y \in S$ and $\phi$ is surjective, $\phi$ is an order isomorphism by Definition 2.
=== Definition 2 implies Definition 1 ===
Let $\phi: S \to T$ be an order isomorphism by Definition 2.
Then $\phi$ is a surjective order embedding.
By Order Embedding is Injection, $\phi$ is injective.
As it is also surjective, $\phi$ is bijective, and thus satisfies the first condition of Definition 1.
Since $\phi$ is an order embedding:
:$\forall x, y \in S: \paren {x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y}$
Thus
:$\forall x, y \in S: \paren {x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y}$
This satisfies the second condition of Definition 1.
Furthermore:
:$\forall x, y \in S: \paren {\map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y}$
Let $p, q \in T$ and let $p \preceq_2 q$.
Then since $\phi$ is bijective, it has an inverse $\phi^{-1}$ such that
:$\map \phi {\map {\phi^{-1} } p} = p$
:$\map \phi {\map {\phi^{-1} } q} = q$
Thus:
:$\map \phi {\map {\phi^{-1} } p} \preceq_2 \map \phi {\map {\phi^{-1} } q}$
Therefore:
:$\map {\phi^{-1} } p \preceq_1 \map {\phi^{-1} } q$
Thus we see that:
:$\forall p, q \in T: \paren {p \preceq_2 q \implies \map {\phi^{-1} } p \preceq_2 \map {\phi^{-1} } q}$
Thus we have satisfied the final condition of Definition 1.
=== Definition 1 iff Definition 3 ===
Let $\phi: S \to T$ be an order isomorphism by Definition 1.
Then by Definition 1:
That is:
:$\forall x, y \in S: \map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$
which together with:
:$\forall x, y \in S: x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$
gives:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
Hence $\phi$ is an order isomorphism by Definition 3.
The above argument reverses.
Category:Order Isomorphisms
Equivalence of Definitions of Order Topology

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
By definition, an open ray of $S$ is defined as a subset of $S$ of one of the two forms:
:An upward-pointing ray:
:: $\left\{{x \in S: s \prec x}\right\}$ (the strict upper closure of $s$), denoted $s^\succ$
:A downward-pointing ray:
:: $\left\{{x \in S: x \prec s}\right\}$ (the strict lower closure of $s$), denoted $s^\prec$
from a given point $s \in S$.
Thus the set of open rays of $S$ can be expressed as:
:the set of all subsets of $S$ of the form:
::$\left\{ {s^\succ: s \in S}\right\}$
and:
:the set of all subsets of $S$ of the form:
::$\left\{ {s^\prec: s \in S}\right\}$
This is precisely the set ${\Uparrow} \left({S}\right) \cup {\Downarrow} \left({S}\right)$, where:
:${\Uparrow} \left({S}\right) = \left\{{s^\succ: s \in S}\right\}$
:${\Downarrow} \left({S}\right) = \left\{{s^\prec: s \in S}\right\}$
Hence the result, by definition of the topology on $S$ generated by a sub-basis of $S$.
Category:Order Topology
Equivalence of Definitions of Order of Entire Function

== Theorem ==
Let $f: \C \to \C$ be an entire function.
Let $\ln$ denote the natural logarithm.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
Let:
:$\alpha_1 = \ds \limsup_{R \mathop \to \infty} \frac {\ds \ln \ln \max_{\cmod z \mathop \le R} \cmod f} {\ln R}$
:$\alpha_2 = \inf \set {\beta \ge 0: \ds \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } = \map \OO {R^\beta} }$
:$\alpha_3 = \inf \set {\beta \ge 0: \map f z = \map \OO {\map \exp {\cmod z^\beta} } }$
=== $3$ less than $2$ ===
Let $\beta > \alpha_2$.
Then there exists $K > 0$ such that:
:$\ds \ln \max_{\cmod z \mathop \le R} \cmod f \le K \cdot R^\beta$
for $R$ sufficiently large.
That is:
:$\dfrac {\ds \ln \ln \max_{\cmod z \le R} \cmod f} {\ln R} \le \dfrac K {\ln R} + \beta$
for $R$ sufficiently large.
Taking $\limsup$, we get $\alpha_1 \le \beta$.
Taking limits, $\alpha_2 \ge \alpha_1$.
=== $2$ less than $3$ ===
Let $\beta > \alpha_1$.
We have $\ds \ln \max_{\cmod z \mathop \le R} \cmod f \le R^\beta$ for $R$ sufficiently large.
Thus $\beta \ge \alpha_2$.
Taking limits, $\alpha_1 \ge \alpha_2$.
=== $1$ less than $2$ ===
Let $\beta > \alpha_2$.
Let $\epsilon > 0$ be such that $\beta - \epsilon > \alpha_1$.
Then:
:$\ds \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } = \map \OO {R^{\beta - \epsilon} }$
Then there exists $R_0 > 0$ such that:
:$\ds \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } \le R^\beta$
for $R \ge R_0$.
By Exponential is Strictly Increasing, this implies:
:$\ds \max_{\cmod z \mathop \le R} \cmod {\map f z} \le \map \exp {R^\beta}$
Let $z \in \C$ with $\cmod z \ge R_0$.
Then:
:$\ds \cmod {\map f z} \le \max_{\cmod w \mathop \le \cmod z} \cmod {\map f w} \le \map \exp {\cmod z^\beta}$
Thus:
:$\map f z = \map \OO {\map \exp {\cmod z^\beta} }$
Thus $\beta \ge \alpha_3$.
Taking the limit $\beta \to \alpha_2$, we obtain:
:$\alpha_2 \ge \alpha_3$
Category:Order of Entire Function
Equivalence of Definitions of Order of Group Element

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $x \in G$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
Let $k$ be the order of $x$ in $G$ according to Definition 1.
Let $l$ be the order of $x$ in $G$ according to Definition 3.
=== Definition $1$ is equivalent to Definition $3$ ===
 $k \ne l$.
Then from Ordering on 1-Based Natural Numbers is Trichotomy either $k < l$ or $k > l$.
According to Definition 3:
:$\forall i, j \in \Z: 0 \le i < j < l \implies x^i \ne x^j$
If $k < l$, then letting $i = 0$ and $j = k$ yields a contradiction by Element to Power of Zero is Identity.
If $k > l$, then:
:$\exists i,j \in \Z: 0 \le i < j < k: x^i = x^j$
But then $x^{j - i} = e$ and $j - i < k$, contradicting the fact that $k \in \Z_{> 0}$ is the smallest such that $x^k = e$.
Therefore $k = l$.
=== Definition $1$ is equivalent to Definition $2$ ===
It follows straight away from List of Elements in Finite Cyclic Group that $\order {\gen a} = k$:
:$\gen a = \set {a^0, a^1, a^2, \ldots, a^{k - 1} }$
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 41$
* : $\S 1.9$: Subgroups: Example $28$
* : $\S 38.4$ Period of an element: $\text{(iv)}$
Category:Subgroups
Category:Order of Group Elements
Equivalence of Definitions of Ordered Integral Domain

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\struct {D, +, \times}$ be a integral domain whose zero is $0_D$ and whose unity is $1_D$.
=== $(1)$ implies $(2)$ ===
Let $\struct {D, +, \times \le}$ be a ordered integral domain by definition 1.
By Strict Positivity Property induces Total Ordering, $P$ induces this total ordering $\le$ on $D$.
Thus $\struct {D, +, \times \le}$ is a ordered integral domain by definition 2.
=== $(2)$ implies $(1)$ ===
Let $\struct {D, +, \times \le}$ be a ordered integral domain by definition 2.
That is, $\struct {D, +, \times}$ has a relation $\le$ which is compatible with the ring structure of $D$:
Let $P$ be the set of elements of $D$ which fulfil the conditions:
:$P = \set {x \in D: 0_D \le x \land 0_D \ne x}$
We check the (strict) positivity property axioms as follows.
Let $x, y \in P$.
That is:
:$0_D \le x \land 0_D \ne x$
:$0_D \le y \land 0_D \ne y$
$(\text P 1)$:
Because $\le$ is an ordering, it is a fortiori a preordering.
We have:
But as $x, y \ne 0_D$ it follows that:
:$0_D \ne x + y$
That is:
:$x + y \in P$
So it is seen that $P$ fulfils (strict) positivity property $\text P 1$.
$(\text P 2)$:
From $\text {OR} 2$: Product of Positive Elements is Positive:
:$0_D \le x \times y$
We have that:
:$0_D \ne x$
and:
:$0_D \ne y$
As $\struct {D, +, \times}$ is an integral domain, it has no proper zero divisors by definition.
It follows that:
:$0_D \ne x \times y$
and so:
:$x \times y \in P$
So it is seen that $P$ fulfils (strict) positivity property $\text P 2$.
$(\text P 3)$:
Thus $\struct {D, +, \times \le}$ is a ordered integral domain by definition 1.
Category:Ordered Integral Domains
Equivalence of Definitions of Ordered Pair

== Theorem ==
=== Informal Definition ===
=== Kuratowski Formalization ===
== Proof ==
=== Equality of Ordered Pairs ===
From Equality of Ordered Pairs, we have that:
:$\set {\set a, \set {a, b} } = \set {\set c, \set {c, d} } \iff a = c, b = d$
hence showing that the Kuratowski formalization fulfils the requirement for equality.
=== Existence of Cartesian Product ===
By Binary Cartesian Product in Kuratowski Formalization contained in Power Set of Power Set of Union:
:$A \times B \subseteq \powerset {\powerset {A \cup B} }$
By Axiom of Specification, there is a set $C$ where:
:$C = \set {x \in \powerset {\powerset {A \cup B} } : x \in A \times B}$
Thus, the cartesian product exists.
By Cartesian Product is Unique, the cartesian product is unique.
== Sources ==
* : Exercise $1.2: \ 12$
* : $\S 2$: Sets and functions: Graphs and functions
Category:Cartesian Product
Category:Axiomatic Set Theory
Equivalence of Definitions of Ordering

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.27 \ \text {(a)}$
Category:Orderings
Category:Equivalence of Definitions of Ordering
Equivalence of Definitions of Ordering/Proof 1

== Theorem ==
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\RR$ be a relation on $S$ satisfying:
By Reflexive and Transitive Relation is Idempotent:
:$\RR \circ \RR = \RR$
By Relation is Antisymmetric and Reflexive iff Intersection with Inverse equals Diagonal Relation:
:$\RR \cap \RR^{-1} = \Delta_S$
Thus $\RR$ is an ordering by definition 2.
=== Definition 2 implies Definition 1 ===
Let $\RR$ be a relation which fulfils the conditions:
:$(1): \quad \RR \circ \RR = \RR$
:$(2): \quad \RR \cap \RR^{-1} = \Delta_S$
By definition of set equality, it follows from $(1)$:
:$\RR \circ \RR \subseteq \RR$
Thus, by definition, $\RR$ is transitive.
By Relation is Antisymmetric and Reflexive iff Intersection with Inverse equals Diagonal Relation, it follows from $(2)$ that:
:$\RR$ is reflexive
:$\RR$ is antisymmetric.
Thus $\RR$ is an ordering by definition 1.
Category:Equivalence of Definitions of Ordering
Equivalence of Definitions of Ordering/Proof 2

== Theorem ==
The following definitions of ordering are equivalent:
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\RR$ be a relation on $S$ satisfying:
==== Condition $(1)$ ====
Let $\tuple {x, y} \in \RR \circ \RR$.
Then there exists a $z \in \RR$ such that:
:$\tuple {x, z}, \tuple {z, y} \in \RR$
By $\RR$ being transitive:
:$\tuple {x, y} \in \RR$
Hence:
:$\RR \circ \RR \subseteq \RR$
Now let $\tuple {x, y} \in \RR$.
By $\RR$ being reflexive:
:$\tuple {y, y} \in \RR$
Hence by the definition of relation composition:
:$\tuple {x, y} \in \RR \circ \RR$
Hence:
:$\RR \subseteq \RR \circ \RR$
==== Condition $(2)$ ====
Follows immediately from Relation is Antisymmetric iff Intersection with Inverse is Coreflexive and $\RR$ being reflexive.
Thus $\RR$ is an ordering by definition 2.
=== Definition 2 implies Definition 1 ===
Let $\RR$ be a relation which fulfils the conditions:
:$(1): \quad \RR \circ \RR = \RR$
:$(2): \quad \RR \cap \RR^{-1} = \Delta_S$
==== Reflexivity ====
By Intersection is Subset the condition:
:$\RR \cap \RR^{-1} = \Delta_S$
implies:
:$\Delta_S \subseteq \RR$
Thus $\RR$ is reflexive by definition.
==== Antisymmetry ====
By Relation is Antisymmetric iff Intersection with Inverse is Coreflexive the condition:
:$\RR \cap \RR^{-1} = \Delta_S$
implies that $\RR$ is antisymmetric.
==== Transitivity ====
Let $\tuple {x, y}, \tuple {y, z} \in \RR$.
Then by the definition of relation composition:
:$\tuple {x, z} \in \RR \circ \RR$
But by the condition:
:$\RR \circ \RR = \RR$
It follows that:
:$\tuple {x, z} \in \RR$
Hence $\RR$ is transitive.
Thus $\RR$ is an ordering by definition 1.
Category:Equivalence of Definitions of Ordering
Equivalence of Definitions of Ordering on Integers

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $x, y \in \Z$ such that $x \le y$.
Let $x$ and $y$ be defined as from the formal definition of integers:
:$x = \eqclass {x_1, x_2} {}$ and $y = \eqclass {y_1, y_2} {}$ where $x_1, x_2, y_1, y_2 \in \N$.
=== $(1)$ implies $(2)$ ===
Let $\le$ be an ordering on integers by definition $1$.
Then by definition:
:$y - x$ is non-negative 
That is:
:$\map \PP {y - x}$
where $\PP$ is the positivity property.
Thus:
:$\eqclass {y_1, y_2} {} - \eqclass {x_1, x_2} {}$ is non-negative
By definition of integer subtraction:
:$\eqclass {y_1, y_2} {} + \eqclass {x_2, x_1} {}$ is non-negative
and by the formal definition of integers:
:$\eqclass {y_1 + x_2, y_2 + x_1} {}$ is non-negative
We have that $y_1 + x_2$ and $y_2 + x_1$ are natural numbers.
Thus by definition of natural number ordering:
:$y_1 + x_2 \ge y_2 + x_1$
Thus $\le$ is an ordering on integers by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $\le$ be an ordering on integers by definition $2$.
Then by definition:
:$x_1 + y_2 \le x_2 + y_1$
That is:
:$x_2 + y_1 \ge x_1 + y_2$
Hence by the formal definition of integers:
:$\eqclass {y_1 + x_2, y_2 + x_1} {}$ is non-negative
By definition of integer addition:
:$\eqclass {y_1, y_2} {} + \eqclass {x_2, x_1} {}$ is non-negative
By definition of integer subtraction:
:$\eqclass {y_1, y_2} {} - \eqclass {x_1, x_2} {}$ is non-negative
That is:
:$\map \PP {y - x}$
where $\PP$ is the positivity property.
Thus $\le$ is an ordering on integers by definition $1$.
Category:Orderings on Integers
Equivalence of Definitions of Ordinal

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== Definition 1 is equivalent to Definition 2 ===
This follows immediately from the definition of a strict well-ordering.
=== Definition 1 implies Definition 3 ===
Let $\alpha$ be an ordinal according to Definition 1.
Let $\beta \in \alpha$.
Then:
=== Definition 3 implies Definition 1 ===
Let $\struct {\alpha, \prec}$ be an ordinal according to Definition 3.
Let $\beta \in \alpha$.
Then $\beta = \alpha_\beta \subseteq \alpha$ and so $\alpha$ is transitive.
Also, by the definition of set equality:
It has been shown that if $x, \beta \in \alpha$ then:
:$x \in \beta \iff x \prec \beta$
Therefore, ${\prec} = \struct {\alpha, \alpha, \RR}$ where:
Hence ${\prec} = \Epsilon {\restriction_\alpha}$.
=== Definition 3 is equivalent to Definition 4 ===
== Sources ==
* : $\S 7.3$, $\S 7.4$
Category:Equivalence of Definitions of Ordinal
Category:Ordinals
Equivalence of Definitions of P-adic Integer

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
=== Definition 2 implies Definition 1 ===
== Sources ==
* : $\S 1.4$ The field of $p$-adic numbers $\Q_p$
Category:P-adic Integers
Category:Equivalence of Definitions of P-adic Integer
Equivalence of Definitions of P-adic Integer/Definition 1 Implies Definition 2

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $x \in \Q_p$ such that $\norm x_p \le 1$.
Then:
:the canonical expansion of $x$ contains only positive powers of $p$
== Proof ==
Let $x \in \Q_p$ such that $\norm x_p \le 1$.
From P-adic Integer is Limit of Unique P-adic Expansion, there exists a $p$-adic expansion of the form:
:$\ds \sum_{n \mathop = 0}^\infty d_n p^n$
By definition of the canonical expansion:
:$\ds \sum_{n \mathop = 0}^\infty d_n p^n$ is the canonical expansion of $x$
It follows that the canonical expansion of $x$ contains only positive powers of $p$.
Category:Equivalence of Definitions of P-adic Integer
Equivalence of Definitions of P-adic Integer/Definition 2 Implies Definition 1

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the valued field of $p$-adic numbers for some prime $p$. 
That is, such that:
:$\Q_p$ is the field of $p$-adic numbers
:$\norm {\,\cdot\,}_p$ is the $p$-adic norm on $\Q_p$.
Let $x \in \Q_p$ such that the canonical expansion of $x$ contains only positive powers of $p$.
Then:
:$\norm x_p \le 1$
== Proof ==
Let the canonical expansion of $x$ contain only positive powers of $p$.
That is:
:$x = \ds \sum_{n \mathop = 0}^\infty d_n p^n : \forall n \in \N : 0 \le d_n < p$
==== Case 1 : $\forall n \in \N : d_n = 0$ ====
Let: 
:$\forall n \in \N : d_n = 0$
Then:
Hence:
==== Case 2 : $\exists n \in \N : d_n > 0$ ====
Let:
:$\exists n \in \N : d_n > 0$
Let:
:$l = \min \set {i: i \ge 0 \land d_i \ne 0}$
Hence:
:$l \ge 0$
Thus:
Category:Equivalence of Definitions of P-adic Integer