avewright/proofwiki-math huggingface dataset:
Theorem: Quaternions form Vector Space over Reals

Let $\R$ be the set of real numbers.
Let $\H$ be the set of quaternions.
Then the $\R$-module $\H$ is a vector space.

Proof:
Recall that Real Numbers form Field.
Thus $\R$ is a division ring {{afortiori}}.
Thus we only need to show that $\R$-module $\H$ is a unitary module, by demonstrating the unitary (left) module axioms:
{{begin-axiom}}
{{axiom | n = \text {UM} 1
        | q = x, y \in \H: \forall \lambda \in \R
        | ml= \lambda \paren {x + y}
        | mo= =
        | mr= \paren {\lambda x} + \paren {\lambda y}
}}
{{axiom | n = \text {UM} 2
        | q = x \in \H: \forall \lambda, \mu \in \R
        | ml= \paren {\lambda + \mu} x
        | mo= =
        | mr= \paren {\lambda x} + \paren {\mu x}
}}
{{axiom | n = \text {UM} 3
        | q = x \in \H: \forall \lambda, \mu \in \R
        | ml= \paren {\lambda \mu} x
        | mo= =
        | mr= \lambda \paren {\mu x}
}}
{{axiom | n = \text {UM} 4
        | q = \forall x \in \H
        | ml= 1 x
        | mo= =
        | mr= x
}}
{{end-axiom}}
As $\lambda, \mu \in \R$ it follows that $\lambda, \mu \in \H$.
Thus from Quaternion Multiplication Distributes over Addition, $(1)$ and $(2)$ immediately follow.
$(3)$ follows from Quaternion Multiplication is Associative.
$(4)$ follows from Multiplicative Identity for Quaternions, as $\mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$ is the unity of $\H$.
{{qed}}
Theorem: Quaternions form Vector Space over Themselves

The set of quaternions $\H$, with the operations of addition and multiplication, forms a vector space.

Proof:
Let the set of quaternions be denoted $\struct {\H, +, \times}$.
From Quaternions form Skew Field, the algebraic structure $\struct {\H, +, \times}$ is a skew field.
By definition, a skew field is a division ring.
{{ProofWanted}}
Theorem: Set of Matrices under Entrywise Addition forms Vector Space

Let $m, n \in \Z_{>0}$ be (strictly) positive integers.
Let $\map {\MM_\GF} {m, n}$ be the $m \times n$ matrix space over a field $\GF$.
Let $\struct {\map {\MM_\GF} {m, n}, +}$ denote the abelian group formed from $\map {\MM_\GF} {m, n}$ under matrix entrywise addition.
Let $\struct {\map {\MM_\GF} {m, n}, +, \times}_\GF$ denote the unitary module over $\GF$ where $\times$ denotes the matrix scalar product.
Then $\struct {\map {\MM_\GF} {m, n}, +, \times}_\GF$ forms a vector space.

Proof:
It is to be demonstrated that $\map {\MM_\GF} {m, n}$ satisfies all the vector space axioms.
First we note that from Matrix Entrywise Addition forms Abelian Group, $\struct {\map {\MM_\GF} {m, n}, +}$ is indeed an abelian group.
Hence the underlying group axioms $\text G 0$ to $\text G 3$ and $\text C$ are satisfied.
Theorem: Simple Events are Mutually Exclusive

Let $\EE$ be an experiment.
Let $e_1$ and $e_2$ be distinct simple events in $\EE$.
Then $e_1$ and $e_2$ are mutually exclusive.

Proof:
By definition of simple event:
{{begin-eqn}}
{{eqn | l = e_1
      | r = \set {s_1}
}}
{{eqn | l = e_2
      | r = \set {s_2}
}}
{{end-eqn}}
for some elementary events $s_1$ and $s_2$ of $\EE$ such that $s_1 \ne s_2$.
It follows that:
{{begin-eqn}}
{{eqn | l = e_1 \cap e_2
      | r = \set {s_1} \cap \set {s_2}
      | c = Definition of $e_1$ and $e_2$
}}
{{eqn | r = \O
      | c = {{Defof|Set Intersection}}
}}
{{end-eqn}}
The result follows by definition of mutually exclusive events.
{{qed}}
Theorem: Non-Trivial Event is Union of Simple Events

Let $\EE$ be an experiment.
Let $e$ be an event in $\EE$ such that $e \ne \O$.
That is, such that $e$ is non-trivial.
Then $e$ can be expressed as the union of a set of simple events in $\EE$.

Proof:
By definition of event, $e$ is a subset of the sample space $\Omega$ of $\EE$.
By hypothesis:
:$e \ne \O$
and so:
:$\exists s \in \Omega: s \in e$
Let $S$ be the set defined as:
:$S = \set {\set s: s \in e}$
By Union is Smallest Superset: Set of Sets it follows that:
:$\ds \bigcup S \subseteq e$
Let $x \in e$.
Then by Singleton of Element is Subset:
:$\set x \subseteq e$
and by definition of $S$ it follows that:
:$\set x \in S$
and so by definition of set union:
:$x \in \ds \bigcup S$
Thus we have:
:$e \subseteq \ds \bigcup S$
The result follows by definition of set equality.
{{qed}}
Theorem: Sample Space is Union of All Distinct Simple Events

Let $\EE$ be an experiment.
Let $\Omega$ denote the sample space of $\EE$.
Then $\Omega$ is the union of the set of simple events in $\EE$.

Proof:
By Set is Subset of Itself:
:$\Omega \subseteq \Omega$
That is, $\Omega$ is itself an event in $\EE$.
The result as an application of Non-Trivial Event is Union of Simple Events.
{{qed}}
Theorem: Equivalence of Definitions of Probability Measure

Let $\EE$ be an experiment.
{{TFAE|def = Probability Measure}}

Proof:
{{ProofWanted}}
Category:Probability Measures
3h2zo3dlivx9v7189tggd120brw05hd
Theorem: Total Probability Theorem/Conditional Probabilities

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\set {B_1, B_2, \ldots}$ be a partition of $\Omega$ such that $\forall i: \map \Pr {B_i} > 0$.
Let $C \in \Sigma$ be an event independent to any of the $B_i$.
Let $\map \Pr C > 0$.
Then:
:$\ds \forall A \in \Sigma: \condprob A C = \sum_i \condprob A {C \cap B_i} \, \map \Pr {B_i}$

Proof:
First define $Q_C := \condprob {\, \cdot} C$.
Then, from Conditional Probability Defines Probability Space, $\struct {\Omega, \Sigma, Q_C}$ is a probability space.
Moreover:
{{begin-eqn}}
{{eqn | q = \forall i
      | l = \map {Q_C} {B_i}
      | r = \map \Pr {B_i \mid C}
}}
{{eqn | r = \map \Pr {B_i}
      | c = $C$ and $B_i$ are independent
}}
{{eqn | o = >
      | r = 0
}}
{{end-eqn}}
Therefore the Total Probability Theorem also holds true.
Hence we have:
{{begin-eqn}}
{{eqn | l = \map {Q_C} A
      | r = \sum_i \map {Q_C} {A \mid B_i} \, \map {Q_C} {B_i}
      | c = Total Probability Theorem
}}
{{eqn | r = \sum_i \frac {\map {Q_C} {A \cap B_i} } {\map {Q_C} {B_i} } \, \map {Q_C} {B_i} 
      | c = {{Defof|Conditional Probability}}
}}
{{eqn | r = \sum_i \frac {\condprob {\paren {A \cap B_i} } C } {\condprob {B_i} C} \, \condprob {B_i} C
      | c = Definition of $Q_C$
}}
{{eqn | r = \sum_i \frac {\map \Pr {A \cap B_i \cap C} } {\map \Pr C} \frac {\map \Pr C} {\map \Pr {B_i \cap C} } \, \condprob {B_i} C
      | c = {{Defof|Conditional Probability}} for $\condprob {A \cap B_i} C$
}}
{{eqn | r = \sum_i \frac {\map \Pr {A \cap B_i \cap C} } {\map \Pr {B_i \cap C} } \, \condprob {B_i} C
      | c = cancelling $\map \Pr C$ from top and bottom
}}
{{eqn | r = \sum_i \condprob A {B_i \cap C} \, \condprob {B_i} C
      | c = {{Defof|Conditional Probability}} for $\condprob A {B_i \cap C}$
}}
{{eqn | r = \sum_i \condprob A {B_i \cap C} \, \map \Pr {B_i}
      | c = $C$ and $B_i$ are independent
}}
{{eqn | r = \sum_i \condprob A {C \cap B_i} \, \map \Pr {B_i}
      | c = Intersection is Commutative
}}
{{end-eqn}}
{{qed}}
Category:Total Probability Theorem
994upsy8n5pfj591jg8z2foql8089nd
Theorem: Product Space is Product in Category of Topological Spaces

Let $\mathbf {Top}$ be the category of topological spaces.
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\struct {\XX, \tau}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Then $\struct {\XX, \tau}$ is the product of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ in $\mathbf{Top}$.

Proof:
{{ProofWanted|Use Continuous Mapping to Product Space/General Result}}
Category:Category of Topological Spaces
Category:Product Topology
Category:Products (Category Theory)
auofr2hdeswk67x1womd9iyj6pl1yki
Theorem: Strictly Positive Rational Numbers are Closed under Addition

Let $\Q_{>0}$ denote the set of strictly positive rational numbers:
:$\Q_{>0} := \set {x \in \Q: x > 0}$
where $\Q$ denotes the set of rational numbers.
The algebraic structure $\struct {\Q_{>0}, +}$ is closed in the sense that:
:$\forall a, b \in \Q_{>0}: a + b \in \Q_{>0}$
where $+$ denotes rational addition.

Proof:
Let $a$ and $b$ be expressed in canonical form:
:$a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$
where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$.
As $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$.
By definition of rational addition:
:$\dfrac {p_1} {q_1} + \dfrac {p_2} {q_2} = \dfrac {p_1 q_2 + p_2 q_1} {q_1 q_2}$
From Integers form Ordered Integral Domain, it follows that:
{{begin-eqn}}
{{eqn | l = p_1 q_2
      | o = >
      | r = 0
}}
{{eqn | l = p_2 q_1
      | o = >
      | r = 0
}}
{{eqn | l = q_1 q_2
      | o = >
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = p_1 q_2 + p_2 q_1
      | o = >
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \dfrac {p_1 q_2 + p_2 q_1} {q_1 q_2}
      | o = >
      | r = 0
}}
{{end-eqn}}
{{qed}}
Theorem: Strictly Positive Rational Numbers are Closed under Multiplication

:$\forall a, b \in \Q_{>0}: a b \in \Q_{>0}$

Proof:
Let $a$ and $b$ be expressed in canonical form:
:$a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$
where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$.
As $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$.
By definition of rational multiplication:
:$\dfrac {p_1} {q_1} \times \dfrac {p_2} {q_2} = \dfrac {p_1 \times p_2} {q_1 \times q_2}$
From Integers form Ordered Integral Domain, it follows that:
{{begin-eqn}}
{{eqn | l = p_1 \times p_2
      | o = >
      | r = 0
}}
{{eqn | l = q_1 \times q_2
      | o = >
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \dfrac {p_1} {q_1} \times \dfrac {p_2} {q_2}
      | o = >
      | r = 0
}}
{{end-eqn}}
{{qed}}
Theorem: Set of Rationals Less than Root 2 has no Greatest Element

Let $A$ be the set of all positive rational numbers $p$ such that $p^2 < 2$.
Then $A$ has no greatest element.

Proof:
{{AimForCont}} $p \in A$ is the greatest element of $A$.
Then by definition of $A$:
:$p^2 < 2$
Let $h \in \Q$ be a rational number such that $0 < h < 1$ such that:
:$h < \dfrac {2 - p^2} {2 p + 1}$
This is always possible, because by definition $2 - p^2 > 0$ and $2 p + 1 > 0$.
Let $q = p + h$.
Then $q > p$, and:
{{begin-eqn}}
{{eqn | l = q^2
      | r = p^2 + \paren {2 p + h} h
      | c = 
}}
{{eqn | o = <
      | r = p^2 + \paren {2 p + 1} h
      | c = 
}}
{{eqn | o = <
      | r = p^2 + \paren {2 - p^2}
      | c = 
}}
{{eqn | r = 2
      | c = 
}}
{{end-eqn}}
That means $q \in A$.
This contradicts our assertion that $p$ is the greatest element of $A$.
It follows that $A$ can have no greatest element.
{{qed}}
Theorem: Set of Rationals Greater than Root 2 has no Smallest Element

Let $B$ be the set of all positive rational numbers $p$ such that $p^2 > 2$.
Then $B$ has no smallest element.

Proof:
{{AimForCont}} $p \in B$ is the smallest element of $B$.
Then by definition of $B$:
:$p^2 > 2$
Let $q = p - \dfrac {p_2 - 2} {2 p}$.
Then $q > p$, and:
{{begin-eqn}}
{{eqn | l = q
      | r = p - \dfrac {p_2 - 2} {2 p}
      | c = 
}}
{{eqn | r = \dfrac p 2 + \dfrac 1 p
      | c = 
}}
{{end-eqn}}
Hence:
:$0 < q < p$
and so:
{{begin-eqn}}
{{eqn | l = q^2
      | r = p^2 - \dfrac {p^2 - 2} + \paren {\dfrac {p^2 - 2} {2 p} }^2
      | c = 
}}
{{eqn | o = >
      | r = p^2 - \paren {p^2 - 2}
      | c = 
}}
{{eqn | r = 2
      | c = 
}}
{{end-eqn}}
That means $q \in B$.
This contradicts our assertion that $p$ is the smallest element of $B$.
It follows that $B$ can have no smallest element.
{{qed}}
Theorem: Rational Number Not in Cut is Greater than Element of Cut

Let $\alpha$ be a cut.
Let $p \in \alpha$.
Let $q \in \Q$ such that $q \notin \alpha$.
Then $q > p$.

Proof:
We have been given that $p \in \alpha$ while $q \notin \alpha$.
{{AimForCont}} $q \le p$.
If $q = p$ then $q \in \alpha$ immediately.
If $q < p$ then $q \in \alpha$ by property $(2)$ of the definition of cut.
In either case $q \in \alpha$ which contradicts $q \notin \alpha$.
Hence by Proof by Contradiction it follows that $q > p$.
{{Qed}}
Theorem: Rational Cut has Smallest Upper Number

Let $r \in \Q$ be rational.
Let $\alpha$ be the rational cut consisting of all rational numbers $p$ such that $p < r$.
Then $\alpha$ is indeed a cut, and has a smallest upper number that is $r$.

Proof:
That $\alpha$ fulfils conditions $(1)$ and $(2)$ of the definition of a cut follows directly from that definition.
Then it is noted that if $p \in \alpha$ then from Mediant is Between:
:$p < \dfrac {p + r} 2 < r$
and so $p$ cannot be the greatest element of $\alpha$.
Hence it is seen that $\alpha$ fulfils all conditions to be a cut.
Because it is not the case that $r < r$, it follows that $r \notin \alpha$.
Thus $r$ is an upper number of $\alpha$.
{{AimForCont}} $q \in \Q$ is an upper number of $\alpha$ such that $q < r$.
But then by definition of a cut, $q \in \alpha$.
That is, $q$ is not an upper number of $\alpha$.
Hence by Proof by Contradiction it follows that there is no upper number of $\alpha$ which is smaller than $r$.
That is, $r$ is the smallest upper number of $\alpha$.
{{Qed}}
Theorem: Natural Basis of Product Topology

Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
:$\ds X := \prod_{i \mathop \in I} X_i$
Then the natural basis on $X$ is the set $\BB$ of cartesian products of the form $\ds \prod_{i \mathop \in I} U_i$ where:
:for all $i \in I : U_i \in \tau_i$
:for all but finitely many indices $i : U_i = X_i$

Proof:
Let $\NN$ denote the natural basis on $X$.
By definition of the natural basis, $\NN$ is the basis generated by:
:$\SS = \set {\pr_i^{-1} \sqbrk U: i \in I, \, U \in \tau_i}$
where for each $i \in I$, $\pr_i: X \to X_i$ denotes the $i$th projection on $X$:
:$\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\family {x_j}_{j \mathop \in I} } = x_i$
Theorem: Natural Basis of Product Topology/Finite Product

Let $n \in \N$.
For all $k \in \set {1, \ldots, n}$, let $\struct {X_k, \tau_k}$ be topological spaces.
Let $\ds X = \prod_{k \mathop = 1}^n X_k$ be the (finite) cartesian product of $X_1, \ldots, X_n$.
Then the natural basis on $X$ is:
:$\BB = \set {\ds \prod_{k \mathop = 1}^n U_k : \forall k : U_k \in \tau_k}$

Proof:
From Natural Basis of Product Topology, the natural basis on $X$ is the set $\BB$ of cartesian products of the form:
:$\ds \prod_{i \mathop \in I} U_i$
where:
:for all $k = 1, \dots, n: U_k \in \tau_i$
:for all but finitely many indices $k : U_k = X_k$
Let:
:$\BB' = \set {\ds \prod_{k \mathop = 1}^n U_k : \forall k : U_k \in \tau_k}$
By definition of $\BB$:
:$\BB \subseteq \BB'$
Let $B \in \BB'$, that is: 
:$B = \ds \prod_{k \mathop = 1}^n U_k$
where $\forall k: U_k \in \tau_k$.
Let $J = \set {k: U_k \neq X_k}$.
Since $J \subseteq \set{ 1, \dots, n}$ is finite then:
:for all but finitely many indices $k : U_k = X_k$
Thus:
:$B \in \BB$
Since $B$ was arbitrary, then:
:$\BB' \subseteq \BB$
It follows that:
:$\BB = \BB'$
from the definition of set equality.
{{qed}}
Category:Natural Basis of Product Topology
ri2bqusw9lk7rcfi3i2kz351wl0gp0b
Theorem: Box Topology may not be Coarsest Topology such that Projections are Continuous

Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$, that is:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the box topology on $X$.
For each $i \in I$, let $\pr_i: X \to X_i$ denote the $i$th projection on $X$:
:$\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\family {x_j}_{j \mathop \in I} } = x_i$
Then $\tau$ may not be the coarsest topology on $X$ for which the projections $\family{\pr_i} _{i \mathop \in I}$ are continuous.

Proof:
From Projection from Box Topology is Continuous, the projections $\family{\pr_i} _{i \mathop \in I}$ are continuous.
From Product Topology is Coarsest Topology such that Projections are Continuous, the product topology is the coarsest topology such that the projections $\family{\pr_i} _{i \mathop \in I}$ are continuous.
Therefore, it is sufficient to show that in some cases the product topology is strictly coarser than the box topology.
That is, it is sufficient to show that in some cases the product topology is not equal to the box topology.
Theorem: Box Topology may not form Categorical Product in the Category of Topological Spaces

Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$, that is:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the box topology on $X$.
Then $\struct{X, \tau}$ may not be the categorical product in the category of topological spaces.

Proof:
There exists such a $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ that the box topology $\tau$ on $X$ differs from the product topology, since:
:Box Topology may not be Coarsest Topology such that Projections are Continuous
:Product Topology is Coarsest Topology such that Projections are Continuous
Then $\struct{X, \tau}$ is not the categorical product, as Product Space is Product in Category of Topological Spaces.
{{qed}}
Category:Box Topology
Category:Products (Category Theory)
1krtkp70b7rfqfwm47nsr3pbnz17z65
Theorem: Ordering on Cuts satisfies Trichotomy Law

Let $\alpha$ and $\beta$ be cuts.
Then exactly one of the following applies:
{{begin-eqn}}
{{eqn | n = 1
      | l = \alpha
      | o = <
      | r = \beta
}}
{{eqn | n = 2
      | l = \alpha
      | r = \beta
}}
{{eqn | n = 3
      | l = \alpha
      | o = >
      | r = \beta
}}
{{end-eqn}}
where $<$ and so $>$ denote the strict ordering of cuts:
:$\alpha < \beta \iff \exists p \in \Q: p \in \alpha, p \notin \beta$
Hence the ordering of cuts $\le$ is a total ordering.

Proof:
Let $\alpha = \beta$.
By definition of equality of cuts:
:$p \in \alpha \iff p \in \beta$
By definition of strict ordering of cuts it follows that neither $\alpha < \beta$ or $\alpha > \beta$.
{{AimForCont}} both $\alpha < \beta$ and $\alpha > \beta$.
Because $\alpha < \beta$ there exists $p \in \Q$ such that:
:$p \in \beta, p \notin \alpha$
Because $\alpha > \beta$ there exists $q \in \Q$ such that:
:$q \in \alpha, q \notin \beta$
By Rational Number Not in Cut is Greater than Element of Cut:
:$p \in \beta$ and $q \notin \beta$ implies $p < q$.
:$q \in \alpha$ and $p \notin \alpha$ implies $p > q$.
But from Rational Numbers form Totally Ordered Field, $p < q$ and $p > q$ is not possible.
Hence by Proof by Contradiction, it is not possible for both $\alpha < \beta$ and $\alpha > \beta$.
Suppose either $\alpha < \beta$ or $\alpha > \beta$.
Then by definition of strict ordering of cuts it follows that $\alpha \ne \beta$.
Thus mutual exclusivity of the three conditions has been demonstrated.
That is, at most one of the three conditions holds.
{{qed|lemma}}
It remains to be shown that at least one of the three conditions holds.
Suppose $\alpha \ne \beta$.
Then by definition of equality of cuts, either:
:$\exists p \in \alpha: p \notin \beta$
in which case $\alpha > \beta$, or:
:$\exists q \in \beta: q \notin \alpha$
in which case $\alpha < \beta$.
Hence the result.
{{Qed}}
Theorem: Ordering on Cuts is Transitive

Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let:
{{begin-eqn}}
{{eqn | n = 1
      | l = \alpha
      | o = <
      | r = \beta
}}
{{eqn | n = 2
      | l = \beta
      | o = <
      | r = \gamma
}}
{{end-eqn}}
where $<$ denotes the strict ordering of cuts:
:$\alpha < \beta \iff \exists p \in \Q: p \in \alpha, p \notin \beta$
Then:
:$\alpha < \gamma$
Hence the ordering of cuts $\le$ is a transitive relation.

Proof:
We have that:
:$\alpha < \beta$
By definition of strict ordering of cuts:
:$\exists p \in \Q: p \in \beta, p \notin \alpha$
Similarly, we have that:
:$\beta < \gamma$
and so:
:$\exists q \in \Q: q \in \gamma, q \notin \beta$
We have by definition of a cut that:
:$p \in \beta$ and $q \notin \beta$ implies that $p < q$
Together with $p \notin \alpha$, this implies that $q \notin \alpha$.
Thus we have:
:$q \in \gamma$ and $q \notin \alpha$
from which by definition of strict ordering of cuts:
:$\alpha < \gamma$
{{Qed}}
Theorem: Ordering on Cuts is Total

Let $\CC$ denote the set of cuts.
Let $<$ denote the strict ordering on cuts defined as:
:$\forall \alpha, \beta \in \CC: \alpha < \beta \iff \exists p \in \Q: p \in \alpha, p \notin \beta$
Then $<$ is a (strict) total ordering on $\CC$.

Proof:
Let $\struct {\CC, <}$ denote the relational structure defined from the above.
From Ordering on Cuts is Transitive, $<$ is a transitive relation on $\CC$.
From Ordering on Cuts satisfies Trichotomy Law, we have that:
:$\alpha < \beta \implies \lnot \paren {\beta < \alpha}$
demonstrating that $<$ is asymmetric.
Hence, by definition, $<$ is a strict ordering.
Also by Ordering on Cuts satisfies Trichotomy Law, $\struct {\CC, <}$ has no non-comparable pairs:
:$\forall \alpha, \beta \in \CC: \alpha \ne \beta \implies \alpha < \beta \lor \beta < \alpha$
The result follows by definition of strict total ordering.
{{Qed}}
Theorem: Sum of Cuts is Cut

Let $\alpha$ and $\beta$ be cuts.
Let $\gamma$ be the set of all rational numbers $r$ such that:
:$\exists p \in \alpha, q \in \beta: r = p + q$
Then $\gamma$ is also a cut.
Thus the operation of addition on the set of cuts is closed.

Proof:
By definition of cut, neither $\alpha$ nor $\beta$ are empty.
Hence there exist $p \in \alpha$ and $q \in \beta$.
Hence there exists $r = p + q$ and so $\gamma$ is likewise not empty.
Let $s, t \in \Q$ such that $s \notin \alpha$ and $t \notin \beta$, where $\Q$ denotes the set of rational numbers.
Such $s$ and $t$ are bound to exist because by definition of cut, neither $\alpha$ nor $\beta$ equal $\Q$.
We have:
{{begin-eqn}}
{{eqn | l = p
      | o = <
      | r = s
      | c = {{Defof|Cut (Analysis)|Cut}}: $s \notin \alpha$
}}
{{eqn | l = q
      | o = <
      | r = t
      | c = {{Defof|Cut (Analysis)|Cut}}: $t \notin \beta$
}}
{{eqn | ll= \leadsto
      | q = \forall p \in \alpha, q \in \beta
      | l = p + q
      | o = <
      | r = s + t
      | c = Rational Numbers form Totally Ordered Field
}}
{{eqn | ll= \leadsto
      | l = s + t
      | o = \notin
      | r = \gamma
      | c = {{Defof|Cut (Analysis)|Cut}}
}}
{{end-eqn}}
Thus it is demonstrated that $\gamma$ does not contain every rational number.
Thus condition $(1)$ of the definition of a cut is fulfilled.
{{qed|lemma}}
Let $r \in \gamma$.
Let $s \in \Q$ such that $s < r$.
Then $r = p + q$ for some $p \in \alpha, q \in \beta$.
Let $t \in \Q$ such that $s = t + q$.
Then $t < p$.
Hence $t \in \alpha$.
Hence by definition of $\gamma$, $t + q = s \in \gamma$.
Thus we have that $r \in \gamma$ and $s < r$ implies that $s \in \gamma$.
Thus condition $(2)$ of the definition of a cut is fulfilled.
{{qed|lemma}}
{{AimForCont}} $r \in \gamma$ is the greatest element of $\gamma$.
Then $r = p + q$ for some $p \in \alpha, q \in \beta$.
By definition of a cut, $\alpha$ has no greatest element:
Hence:
:$\exists s \in \Q: s > p: s \in \alpha$
But then $s + q \in \gamma$ while $s + q > r$.
This contradicts the supposition that $r$ is the greatest element of $\gamma$.
Hence $\gamma$ itself can have no greatest element.
Thus condition $(3)$ of the definition of a cut is fulfilled.
{{qed|lemma}}
Thus it is seen that all the conditions are fulfilled for $\gamma$ to be a cut.
{{qed}}
Theorem: Addition of Cuts is Commutative

Let $\alpha$ and $\beta$ be cuts.
Let the operation of $\alpha + \beta$ be the sum of $\alpha$ and $\beta$.
Then:
:$\alpha + \beta = \beta + \alpha$

Proof:
$\alpha + \beta$ is the set of all rational numbers of the form $p + q$ such that $p \in \alpha$ and $q \in \beta$.
Similarly, $\beta + \alpha$ is the set of all rational numbers of the form $q + p$ such that $p \in \alpha$ and $q \in \beta$.
From Rational Addition is Commutative we have that:
:$p + q = q + p$
The result follows.
{{qed}}