hcju/proofwikiqa huggingface dataset:
First, it is to be shown that $+$ and $\circ$ are in fact well-defined operations on $R / J$. === Well-definition of $+$ === {{:Quotient Ring is Ring/Quotient Ring Addition is Well-Defined}} === Well-definition of $\circ$ === {{:Quotient Ring is Ring/Quotient Ring Product is Well-Defined}} Now to prove that $\struct {R / J, +, \circ}$ is a ring, proceed by verifying the ring axioms in turn: === $\text A$: Addition forms a Group === From: :Ideal is Additive Normal Subgroup :The definition of a quotient group :Quotient Group is Group it follows that $\struct {R / J, +}$ is a group. {{qed|lemma}} === $\text M 0$: Closure of Ring Product === By definition of $\circ$ in $R / J$, it follows that $\struct {R / J, \circ}$ is closed. {{qed|lemma}} === $\text M 1$: Associativity of Ring Product === Associativity can be deduced from the fact that $\circ$ is associative on $R$: {{begin-eqn}} {{eqn | lo= \forall x, y, z \in R:       | o =        | r = \paren {x + J} \circ \paren {\paren {y + J} \circ \paren {z + J} } }} {{eqn | r = \paren {x + J} \circ \paren {y \circ z + J} }} {{eqn | r = x \circ y \circ z + J }} {{eqn | r = \paren {x \circ y + J} \circ \paren {z + J} }} {{eqn | r = \paren {\paren {x + J} \circ \paren {y + J} } \circ \paren {z + J} }} {{end-eqn}} {{qed|lemma}} === $\text D$: Distributivity of Ring Product over Addition === Distributivity can be deduced from the fact that $\circ$ is distributive on $R$: {{begin-eqn}} {{eqn | lo= \forall x, y, z \in R:       | o =        | r = \paren {\paren {x + J} + \paren {y + J} } \circ \paren {z + J} }} {{eqn | r = \paren {x + y + J} \circ \paren {z + J} }} {{eqn | r = \paren {\paren {x + y} \circ z} + J }} {{eqn | r = \paren {\paren {x \circ z} + \paren {y \circ z} } + J }} {{eqn | r = \paren {\paren {x \circ z} + J} + \paren {\paren {y \circ z} + J} }} {{eqn | r = \paren {\paren {x + J} \circ \paren {z + J} } + \paren {\paren {y + J} \circ \paren {z + J} } }} {{end-eqn}} {{qed|lemma}} Having verified all of the ring axioms, it follows that $\struct {R / J, +, \circ}$ is a ring. {{qed}}
As $J$ is an ideal, $\left({J, +}\right)$ is a subgroup of $\left({R, +}\right)$. By definition of a ring, $\left({R, +}\right)$ is abelian. The result follows from Subgroup of Abelian Group is Normal. {{qed}} Category:Ideal Theory eas96ddq9bmt95eqmyswfp6se6v6le4
Let $q_\EE$ be the quotient mapping from $\struct {R, +, \circ}$ to $\struct {R / \EE, +_\EE, \circ_\EE}$. From Quotient Mapping on Structure is Canonical Epimorphism: :$q_\EE: \struct {R, +} \to \struct {R / \EE, +_\EE}$ is an epimorphism :$q_\EE: \struct {R, \circ} \to \struct {R / \EE, \circ _\EE}$ is an epimorphism. As the morphism property holds for both $+$ and $\circ$, it follows that $q_\EE: \struct {R, +, \circ} \to \struct {R / \EE, +_\EE, \circ_\EE}$ is also an epimorphism. From Epimorphism Preserves Rings, it follows that $\struct {R / \EE, +_\EE, \circ_\EE}$ is a ring. {{qed}}
Consider the field of real numbers $\left({\R, +, \times}\right)$. We have that a field is by definition a ring, hence so is $\left({\R, +, \times}\right)$. From Rational Numbers form Subfield of Real Numbers and Integers form Subdomain of Rationals, it follows that the integers $\left({\Z, +, \times}\right)$ are a subring of $\left({\R, +, \times}\right)$. Consider $1 \in \Z$, and consider $\dfrac 1 2 \in \R$. We have that $1 \times \dfrac 1 2 = \dfrac 1 2 \notin \Z$. From this counterexample it is seen that $\Z$ is not an ideal of $R$. Hence the result, again by Proof by Counterexample. {{qed}}
We prove the whole cycle of implications: :$(1) \implies (2) \quad$ by definition of $(1)$ :$(2) \implies (3)\quad$ because $\size f =  f^+ + f^-$ and Integral of Positive Measurable Function is Additive  :$(3) \implies (4)\quad$ because $g:= \size f$ exists It remains to demonstrate $(4) \implies (1)$. Let $f \in \MM_{\overline \R}$ and $g$ according to $(4)$. Then: :$f = f^+ - f^-$ where $f^+$ is the positive and $f^-$ is the negative part of $f$. We have that $f^+$ and $f^-$ are positive and measurable. Let $f^0$ stand for either $f^+$ or $f^-$. We have that: :$\size f = f^+ + f^-$ Therefore: :$f^0 \le \size f \le g$ It is to be shown that the Integral of Positive Measurable Function of $f^0$ exists and is finite. Let $\EE^+$ and $\map {I_\mu} h$ be defined as in Integral of Positive Measurable Function. Then: :$\forall h \in \EE^+$: $h \le f^0 \implies h \le g$ Hence: :$\set {h: h \le f^0, h \in \EE^+} \subseteq \set {h: h \le g, h \in \EE^+}$ :$\displaystyle \int f^0 \rd \mu := \sup \set {\map {I_\mu} h: h \le f^0, h \in \EE^+} \le \sup \set {\map {I_\mu} h: h \le g, h \in \EE^+} < \infty$ We have that the integrals for $f^+$ and $f^-$ both are finite. Therefore $f$ is $\mu$-integrable according to definition. {{qed}} {{MissingLinks|The steps in the above need to be demonstrated and explained by linking to appropriate results on {{ProofWiki}} As it stands, none of the above can be followed without knowing all about this subject.}}.
From Ideal is Additive Normal Subgroup that $J$ is a normal subgroup of $R$ under $+$. Thus, the quotient group $\struct {R / J, +}$ is defined, and as a Quotient Group is Group, $+$ is well-defined. <includeonly>{{qed|lemma}}</includeonly>
Both $\chi_A$ and $\chi_B$ take values in $\left\{{0, 1}\right\}$ as they are characteristic functions. So if $\chi_A \left({s}\right) = 0$, then the statement of the theorem is automatically satisfied (since both $0 \le 0$ and $0 \le 1$). Now assume $\chi_A \left({s}\right) = 1$. By definition of $\chi_A$, this happens {{iff}} $s \in A$. But as $A \subseteq B$, this implies $s \in B$ as well. Hence, $\chi_B \left({s}\right) = 1$, and so $\chi_A \left({s}\right) \le \chi_B \left({s}\right)$. The theorem follows from Proof by Cases. {{qed}} Category:Characteristic Functions gdk1etepo4hrntko2sjtwauby8ig3j4
=== $(1)$ implies $(2)$ === Follows directly from the definition of characteristic function. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $s \in S$. Suppose that $s \notin A$. Then by assumption, $\map f s = 0$. Also, by definition of characteristic function, $\map {\chi_A} s = 0$. Next, suppose that $s \in A$. Then $\map f s \ne 0$ by assumption. As $\map f s \in \set {0, 1}$, it follows that $\map f s = 1$. Again, by definition of characteristic function, also have $\map {\chi_A} s = 1$. Hence, for all $s \in S$, have $\map f s = \map {\chi_A} s$. By Equality of Mappings, it follows that $f = \chi_A$. {{qed}} Category:Characteristic Functions ky9w3kxpse1dj9yoddaplxmkehfvgpy
Let $J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$. From Congruence Relation on Ring induces Ideal, we have that $J$ is an ideal of $R$. From Ideal is Additive Normal Subgroup, we have that $\left({J, +}\right)$ is a normal subgroup of $\left({R, +}\right)$. From Normal Subgroup induced by Congruence Relation defines that Congruence, the equivalence defined by the partition $\left({R, +}\right) / \left({J, +}\right)$ is $\mathcal E$. As $\mathcal E$ was the congruence relation on $R$ that was originally posited, we already know that it is compatible with $\circ$. Thus the equivalence defined by $J$ is the same congruence relation on $R$ that gave rise to $J$ to start with. Hence the result. {{qed}}
We have that the Rational Numbers are Countably Infinite. The result follows from Countable Set is Null Set under Lebesgue Measure. {{qed}}
Let $\family {x_i}_{i \mathop \in I} \in \displaystyle \paren{ \prod_{i \mathop \in I} S_i } \cap \paren{ \prod_{i \mathop \in I} T_i }$. By definition of intersection, this is equivalent to the conjunction of: :$\family {x_i}_{i \mathop \in I} \in \displaystyle \prod_{i \mathop \in I} S_i$ :$\family {x_i}_{i \mathop \in I} \in \displaystyle \prod_{i \mathop \in I} T_i$ By definition of Cartesian product, this means, for all $i \in I$: :$x_i \in S_i$ and $x_i \in T_i$ Again by definition of intersection, this is equivalent to, for all $i \in I$: :$x_i \in S_i \cap T_i$ Finally, by definition of Cartesian product, this is equivalent to: :$\family {x_i}_{i \mathop \in I} \in \displaystyle \prod_{i \mathop \in I} \paren{S_i \cap T_i}$ The result follows by definition of set equality. {{qed}} Category:Cartesian Product of Intersections 7933c0rqsliz6350edef2esv35htdqb
Recall the conditions for $\SS \times \TT$ to be a semiring of sets: :$(1): \quad \O \in \SS \times \TT$ :$(2): \quad \SS \times \TT$ is $\cap$-stable :$(3'):\quad$ If $A, B \in \SS \times \TT$, then there exists a finite sequence of pairwise disjoint sets $A_1, A_2, \ldots, A_n \in \SS \times \TT$ such that $\displaystyle A \setminus B = \bigcup_{k \mathop = 1}^n A_k$. === Proof of $(1)$ === From Empty Set is Subset of All Sets: :$\O \in \SS$ and: :$\O \in \TT$ So: :$\O \times \O \in \SS \times \TT$ From Cartesian Product is Empty iff Factor is Empty: :$\O \times \O = \O$ {{qed|lemma}} === Proof of $(2)$ === Let $S_1 \times T_1$ and $S_2 \times T_2$ be in $\SS \times \TT$. Then from Cartesian Product of Intersections: :$\paren {S_1 \times T_1} \cap \paren {S_2 \times T_2} = \paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2}$ Since $\SS$ and $\TT$ are $\cap$-stable, $S_1 \cap S_2 \in \SS$ and $T_1 \cap T_2 \in \TT$. Hence $\paren {S_1 \times T_1} \cap \paren {S_2 \times T_2} \in \SS \times \TT$. {{qed|lemma}} === Proof of $(3')$ === Let $S_1 \times T_1$ and $S_2 \times T_2$ be in $\SS \times \TT$. Let $\sqcup$ signify union of disjoint sets. Then: {{begin-eqn}} {{eqn | l = \paren {S_1 \times T_1} \setminus \paren {S_2 \times T_2}       | r = \paren {S_1 \times \paren {T_1 \setminus T_2} } \cup \paren {\paren {S_1 \setminus S_2} \times T_1}       | c = Set Difference of Cartesian Products }} {{eqn | o = }} {{eqn | l = S_1       | r = \paren {S_1 \setminus S_2} \sqcup \paren {S_1 \cap S_2}       | c = Set Difference Union Intersection }} {{eqn | ll= \leadsto       | l = S_1 \times \paren {T_1 \setminus T_2}       | r = \paren {S_1 \setminus S_2} \times \paren {T_1 \setminus T_2} \ \sqcup \ \paren {S_1 \cap S_2} \times \paren {T_1 \setminus T_2}       | c = Cartesian Product Distributes over Union }} {{eqn | o = }} {{eqn | l = T_1       | r = \paren {T_1 \cap T_2} \sqcup \paren {T_1 \setminus T_2}       | c = Set Difference Union Intersection }} {{eqn | ll= \leadsto       | l = \paren {S_1 \setminus S_2} \times T_2       | r = \paren {S_1 \setminus S_2} \times \paren {T_1 \cap T_2} \ \sqcup \ \paren {S_1 \setminus S_2} \times \paren {T_1 \setminus T_2}       | c = Cartesian Product Distributes over Union }} {{eqn | o = }} {{eqn | ll= \leadsto       | l = \paren {S_1 \times T_1} \setminus \paren {S_2 \times T_2}       | r = \paren {S_1 \setminus S_2} \times \paren {T_1 \setminus T_2} \ \sqcup \ \paren {S_1 \cap S_2} \times \paren {T_1 \setminus T_2} \ \sqcup \ \paren {S_1 \setminus S_2} \times \paren {T_1 \cap T_2} }} {{end-eqn}} Now recall that $\SS$ and $\TT$ are semirings of sets. Thence the expressions $S_1 \setminus S_2$ and $T_1 \setminus T_2$ may be written as finite disjoint unions. Applying Cartesian Product Distributes over Union again, each of the three $\sqcup$-operands in the expression above may thus be written as a finite disjoint union. This yields the same fact for $\paren {S_1 \times T_1} \setminus \paren {S_2 \times T_2}$ as well, completing the proof. {{qed}}
We are given that $\mathcal R$ is a congruence relation for $\circ$. From Congruence Relation iff Compatible with Operation, we have: : $\forall u \in G: x \mathop {\mathcal R} y \implies \left({x \circ u}\right) \mathop {\mathcal R} \left({y \circ u}\right), \left({u \circ x}\right)\mathop {\mathcal R} \left({u \circ y}\right)$ ==== Proof of being a Subgroup ==== We show that $H$ is a subgroup of $G$. First we note that $H$ is not empty: :$e \in H \implies H \ne \varnothing$ Then we show $H$ is closed: {{begin-eqn}} {{eqn | l = x, y       | o = \in       | r = H       | c =  }} {{eqn | ll= \implies       | l = e       | o = \mathcal R       | r = x       | c =  }} {{eqn | lo= \land       | l = e       | o = \mathcal R       | r = y       | c = by definition of $H$ }} {{eqn | ll= \implies       | l = \left({e \circ e}\right)       | o = \mathcal R       | r = \left({x \circ y}\right)       | c = $\mathcal R$ is compatible with $\circ$ }} {{eqn | ll= \implies       | l = x \circ y       | o = \in       | r = H       | c = by definition of $H$ }} {{end-eqn}} Next we show that $x \in H \implies x^{-1} \in H$: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = H       | c =  }} {{eqn | ll= \implies       | l = e       | o = \mathcal R       | r = x       | c = by definition of $H$ }} {{eqn | ll= \implies       | l = \left({x^{-1} \circ e}\right)       | o = \mathcal R       | r = \left({x^{-1} \circ x}\right)       | c = $\mathcal R$ is compatible with $\circ$ }} {{eqn | ll= \implies       | l = x^{-1}       | o = \mathcal R       | r = e       | c = Group properties }} {{eqn | ll= \implies       | l = x^{-1}       | o = \in       | r = H       | c = by definition of $H$ }} {{end-eqn}} Thus by the Two-Step Subgroup Test, $H$ is a subgroup of $G$. {{qed|lemma}} ==== Proof of Normality ==== Next we show that $H$ is normal in $G$. Thus: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = H       | c =  }} {{eqn | ll= \implies       | l = e       | o = \mathcal R       | r = h       | c = for some $h \in H$, by definition of $H$ }} {{eqn | ll= \implies       | l = \left({x \circ e}\right)       | o = \mathcal R       | r = \left({x \circ h}\right)       | c = $\mathcal R$ is compatible with $\circ$ }} {{eqn | ll= \implies       | l = \left({x \circ e \circ x^{-1} }\right)       | o = \mathcal R       | r = \left({x \circ h \circ x^{-1} }\right)       | c = $\mathcal R$ is compatible with $\circ$ }} {{eqn | ll= \implies       | l = e       | o = \mathcal R       | r = \left({x \circ h \circ x^{-1} }\right)       | c = Group properties }} {{eqn | ll= \implies       | l = x \circ h \circ x^{-1}       | o = \in       | r = H       | c = by definition of $H$ }} {{end-eqn}} Thus from Subgroup is Normal iff Contains Conjugate Elements, we have that $H$ is normal. {{qed}}
Let $\mathcal R_N$ be the equivalence defined by $N$. Then: {{begin-eqn}} {{eqn | l = x       | o = \mathcal R       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = e       | o = \mathcal R       | r = \paren {x^{-1} \circ y}       | c = $\mathcal R$ is compatible with $\circ$ }} {{eqn | ll= \leadsto       | l = \paren {e \circ e}       | o = \mathcal R       | r = \paren {x^{-1} \circ y}       | c = Group properties }} {{eqn | ll= \leadsto       | l = x^{-1} \circ y       | o = \in       | r = N       | c = Definition of $N$ }} {{end-eqn}} But from Congruence Class Modulo Subgroup is Coset: :$x \mathop {\mathcal R_N} y \iff x^{-1} \circ y \in N$ Thus: :$\mathcal R = \mathcal R_N$ {{Qed}}
Let $\eqclass x {\mathcal R} \in G / \mathcal R$. By Congruence Relation on Group induces Normal Subgroup: :$\eqclass x {\mathcal R} = x N$ where $x N$ is the (left) coset of $N$ in $G$. Similarly, let $y N \in G / N$. Then from Normal Subgroup induced by Congruence Relation defines that Congruence: :$y N = \eqclass x {\mathcal R}$ where: :$\eqclass x {\mathcal R}$ is the equivalence class of $y$ under $\mathcal R$ :$\mathcal R$ is the equivalence relation defined by $N$. Hence the result. {{qed}}
{{AimForCont}} that: :$\exists \lambda \in F: \exists \mathbf v \in \mathbf V: \lambda \circ \mathbf v = \bszero \land \lambda \ne 0_F \land \mathbf v \ne \bszero$ which is the negation of the exposition of the theorem. Utilizing the vector space axioms: {{begin-eqn}} {{eqn | l = \lambda \circ \mathbf v       | r = \bszero }} {{eqn | ll= \leadsto       | l = \lambda^{-1} \circ \paren {\lambda \circ \mathbf v}       | r = \lambda^{-1} \circ \mathbf 0       | c = multiplying both sides by $\lambda^{-1}$ }} {{eqn | ll= \leadsto       | l = \bszero       | r = \lambda^{-1} \circ \paren {\lambda \circ \mathbf v}       | c = Zero Vector Scaled is Zero Vector }}  {{eqn | r = \paren {\lambda^{-1} \cdot \lambda} \circ \mathbf v }} {{eqn | r = 1_F \circ \mathbf v }} {{eqn | r = \mathbf v }} {{end-eqn}} which contradicts the assumption that $\mathbf v \ne \mathbf 0$. {{qed}}
Let $d$ be a metric on $Y$ generating its topology. Using the oscillation we have, following the convention that we omit the metric when writing the oscillation, that: :$\displaystyle D \left({f}\right) = \bigcup_{n \mathop = 1}^\infty \left\{{x \in X: \omega_f \left({x}\right) \ge \frac 1 n}\right\}$ which is a countable union of closed sets. Since we have this expression for $D \left({f}\right)$, the claim follows if we can prove that for all $\epsilon \in \R_{>0}$ the closed set: : $F_\epsilon = \left\{{x \in X : \omega_f \left({x}\right) \ge 5 \epsilon}\right\}$ is nowhere dense. Let $\epsilon \in \R_{>0}$ be given and consider the sets: :$\displaystyle A_n = \bigcap_{i, j \mathop \ge n} \left\{{x \in X: d \left({f_i \left({x}\right), f_j \left({x}\right)}\right) \le \epsilon}\right\}$ which are closed because $d$ and the $f_i$ are continuous. Because $\left \langle{f_n}\right\rangle$ is pointwise convergent, it is pointwise Cauchy with respect to any metric generating the topology on $Y$, so $\displaystyle \bigcup_{n \mathop = 1}^\infty A_n = X$. Given a nonempty open $U \subseteq X$ we wish to show that $U \nsubseteq F_\epsilon$. Consider the sequence $\left\langle{A_n \cap U}\right\rangle$ of closed subsets of $U$. The union of these is all of $U$. As $U$ is an open subspace of a Baire space, it is a Baire space. So one of the elements of $\left\langle{A_n \cap U}\right\rangle$, say $A_k$, must have an interior point, so there is an open $V \subseteq A_k \cap U$. Because $U$ is open in $X$, $V$ is open in $X$ as well.   We will show that: :$V \subseteq F_\epsilon^c = \left\{{x \in X: \omega_f \left({x}\right) < 5 \epsilon}\right\}$ This will show that: :$V \nsubseteq F_\epsilon$ and thus that $U \nsubseteq F_\epsilon$ Since $V \subseteq A_k$: :$d \left({f_i \left({x}\right), f_j \left({x}\right)}\right) \le \epsilon$ for all $x \in V$ and all $i, j \ge k$. Pointwise convergence of $\left \langle{f_n}\right\rangle$ gives that: : $d \left({f \left({x}\right), f_k \left({x}\right)}\right) \le \epsilon$ for all $x \in V$. By continuity of $f_k$ we have for every $x_0 \in V$ an open $V_{x_0} \subseteq V$ such that: : $d \left({f_k \left({x}\right), f_k \left({x_0}\right)}\right) \le \epsilon$ for all $x \in V_{x_0}$. By the triangle inequality: :$d \left({f \left({x}\right), f_k \left({x_0}\right)}\right) \le 2 \epsilon$  for all $x \in V_{x_0}$. Applying the triangle inequality again: :$d \left({f \left({x}\right), f \left({y}\right)}\right) \le 4 \epsilon$ for all $x, y \in V_{x_0}$. Thus we have the bound: :$\omega_f \left({x_0}\right) \le \omega_f \left({V_{x_0}}\right) \le  4 \epsilon$ showing that $x_0 \notin F_\epsilon$ as desired. {{qed}} {{Namedfor|René-Louis Baire|name2 = William Fogg Osgood|cat = Baire|cat2 = Osgood}}
From Null Ring and Ring Itself Subrings, $\set {0_R}$ is a subring of $\struct {R, +, \circ}$. Also: :$\forall x \in \struct {R, +, \circ}: x \circ 0_R = 0_R = 0_R \circ x \in \set {0_R}$ thus fulfilling the condition for $\set {0_R}$ to be an ideal of $R$. {{qed}}
A direct application of Ring Homomorphism Preserves Subrings. {{qed}}
As an ideal is a subring, the result Preimage of Image of Subring under Ring Homomorphism applies directly. {{Qed}}
As an ideal is a subring, the result Image of Preimage of Subring under Ring Epimorphism applies directly. {{Qed}}
Utilizing the vector space axioms: {{begin-eqn}} {{eqn | l = \mathbf a + \mathbf c       | r = \mathbf b + \mathbf c }} {{eqn | ll= \leadsto       | l = \paren {\mathbf a + \mathbf c} - \mathbf c       | r = \paren {\mathbf b + \mathbf c} - \mathbf c }} {{eqn | ll= \leadsto       | l = \mathbf a + \paren {\mathbf c - \mathbf c}       | r = \mathbf b + \paren {\mathbf c - \mathbf c} }} {{eqn | ll= \leadsto       | l = \mathbf a + \mathbf 0       | r = \mathbf b + \mathbf 0 }} {{eqn | ll= \leadsto       | l = \mathbf a       | r = \mathbf b }} {{end-eqn}} {{qed}}
Utilizing the vector space axioms: {{begin-eqn}} {{eqn | l = \mathbf c + \mathbf a       | r = \mathbf c + \mathbf b }} {{eqn | ll = \implies       | l = \mathbf a + \mathbf c       | r = \mathbf b + \mathbf c }} {{eqn | ll = \implies       | l = \mathbf a       | r = \mathbf b       | c = Vectors are Right Cancellable }} {{end-eqn}} {{qed}} Category:Vector Algebra 99qrr8db0lqjny0a9ktic3ydue0x6ux
=== Proof of Existence === Follows from the vector space axioms. {{qed|lemma}} === Proof of Uniqueness === Let $\mathbf 0$, $\mathbf 0'$ be zero vectors. Utilizing the vector space axioms: {{begin-eqn}} {{eqn | l = \mathbf 0       | r = \mathbf 0 + \mathbf 0' }} {{eqn | r = \mathbf 0' + \mathbf 0 }} {{eqn | r = \mathbf 0' }} {{end-eqn}} {{qed}}
=== Proof of Existence === Follows from the vector space axioms. {{qed|lemma}} === Proof of Uniqueness === Let $\mathbf v$ have inverses $\mathbf x$ and $\mathbf y$. Then: {{begin-eqn}} {{eqn | l = \mathbf v + \mathbf x       | r = \mathbf 0 }} {{eqn | lo= \land       | l = \mathbf v + \mathbf y       | r = \mathbf 0 }} {{eqn | ll= \leadsto       | l = \mathbf v + \mathbf x       | r = \mathbf v + \mathbf y }} {{eqn | ll= \leadsto       | l = \mathbf x       | r = \mathbf y       | c = Vectors are Left Cancellable }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf v + \paren {-1_F \circ \mathbf v}       | r = \paren {1_F \circ \mathbf v} + \paren {-1_F \circ \mathbf v}       | c = {{Field-axiom|M3}} }} {{eqn | r = \paren {1_F + \paren {- 1_F} } \circ \mathbf v       | c = {{Vector-space-axiom|5}} }} {{eqn | r = 0_F \circ \mathbf v       | c = {{Field-axiom|A4}} }} {{eqn | r = \mathbf 0       | c = Vector Scaled by Zero is Zero Vector }} {{end-eqn}} so $-1_F \circ \mathbf v$ is an additive inverse of $\mathbf v$. From Additive Inverse in Vector Space is Unique: :$-1_F \circ \mathbf v = -\mathbf v$ {{qed}}
Suppose $\struct {R, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative. By definition $\struct {R, +}$ is an abelian group. Thus field axioms $(\text A 0)$ to $(\text A 4)$ are satisfied. Also by definition, $\struct {R, \circ}$ is a semigroup such that $\circ$ is commutative. Thus field axioms $(\text M 0)$ to $(\text M 2)$ are satisfied. As $\struct {R, +, \circ}$ is a ring with unity, $(\text M 3)$ is satisfied. Field axiom $(\text D)$ is satisfied by dint of $\struct {R, +, \circ}$ being a ring. Finally note that by definition of division ring, field axiom $(\text M 4)$ is satisfied. Thus all the field axioms are satisfied, and so $\struct {R, +, \circ}$ is a field. {{qed|lemma}} Suppose $\struct {F, +, \circ}$ is a field. Then by definition $\struct {F, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative. {{qed}}
Let $\left({R, +, \circ}\right)$ be a division ring whose zero is $\left\{{0_R}\right\}$ and whose unity is $1_R$. By definition of division ring, every element $x$ of $R^* = R \setminus \left\{{0_R}\right\}$ has an element $y$ such that: :$y \circ x = x \circ y = 1_R$ That is, by definition, every element of $R^*$ is a unit of $R$. The result follows from Unit Not Zero Divisor. {{qed}}
Proof by Counterexample: Let $S = \set {a, b, c}$. Let $\preccurlyeq$ be defined as: :$x \preccurlyeq y \iff \tuple {x, y} \in \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {a, c} }$ A straightforward but laborious process determines that $\preccurlyeq$ is a partial ordering on $S$. We have that: :$c \preccurlyeq x \implies c = x$ and: :$b \preccurlyeq x \implies b = x$ and so by definition, both $b$ and $c$ are maximal elements of $S$. Suppose $b$ is the greatest element of $S$. Then from Greatest Element is Unique it follows that $c$ can not be the greatest element of $S$. Hence the result. In fact, from the definition of the greatest element of $S$: :$x$ is the greatest element of $S$ {{iff}} $\forall y \in S: y \preccurlyeq x$ it can be seen directly that neither $b$ nor $c$ is the greatest element of $S$. {{qed}}
From Ideals of Division Ring, the only ideals of a Division Ring $\left({D, +, \circ}\right)$ are $\left({D, +, \circ}\right)$ and $\left({\left\{{0}\right\}, +, \circ}\right)$. Hence the result by definition of maximal ideal. {{qed}}