hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \forall A       | o = :       | r = \paren {A \in U \iff \exists x: \paren {x = A \land x \in U} }       | c = {{Defof|Class Membership}} }} {{eqn | l = \forall x       | o = :       | r = x \in U       | c = Fundamental Law of Universal Class }} {{eqn | ll= \leadsto       | l = \forall A       | o = :       | r = \paren {A \in U \iff \exists x: x = A}       | c = Conjunction with Tautology }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \forall A, B       | o = :       | r = \exists x: \forall y: \left({y \in x \iff y = A \lor y = B}\right)       | c = Axiom of Pairing }} {{eqn | ll= \implies       | l = \forall A, B       | o = :       | r = \exists x: x = \left\{ {y: y = A \lor y = B}\right\}       | c = Definition of Set Equality }} {{eqn | ll= \implies       | l = \forall A, B       | o = :       | r = \left\{ {y: y = A \lor y = B}\right\} \in U       | c = Element of Universe }} {{eqn | ll= \implies       | l = \forall A, B       | o = :       | r = \left\{ {A, B}\right\} \in U       | c = Definition of Doubleton }} {{end-eqn}} {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations. $\begin{array}{|c|ccc||c|} \hline p & p & \land & \top & \top \\ \hline F & F & F & T & T \\ T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations. $\begin{array}{|c|c||ccc|} \hline p & \top & p & \lor & \top \\ \hline F & T & F & T & T \\ T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations. $\begin{array}{|c|ccc||c|ccc|} \hline \bot & p & \land & \bot & p \\ \hline F & F & F & F & F \\ F & T & F & F & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations. $\begin{array}{|c|ccc||c|ccc|} \hline \bot & p & p & \lor & \bot \\ \hline F & F & F & F & F \\ F & T & T & T & F \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values in the appropriate columns match. $\begin{array}{|c||cc|} \hline \top & \neg & \bot \\ \hline T & T & F \\ \hline \end{array}$ {{qed}}
Let $p$ be a propositional formula. Let $v$ be any arbitrary boolean interpretation of $p$. Then $\map v p = F \iff \map v {\neg p} = T$ by the definition of the logical not. Since $v$ is arbitrary, $p$ is false in all interpretations {{iff}} $\neg p$ is true in all interpretations. Hence: :$\bot \dashv \vdash \neg \top$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values in the appropriate columns match. $\begin{array}{|c||cc|} \hline \bot & \neg & \top \\ \hline F & F & T \\ \hline \end{array}$ {{qed}}
Let $p$ be a propositional formula. Let $v$ be an arbitrary boolean interpretation of $p$. Then: :$\map v p = T \iff \map v {\neg p} = F$  by the definition of the logical not. Since $v$ is arbitrary, $p$ is true in all interpretations {{iff}} $\neg p$ is false in all interpretations. Hence: :$\top \dashv \vdash \neg \bot$ {{qed}}
Let the conjugation operator on $A$ be $*$. Let $\left({a, b}\right), \left({c, d}\right) \in A'$. In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by product notation: :$a \oplus b =: a b$ :$x \oplus' y =: x y$ The context will make it clear which is meant. Let $A$ be a nicely normed algebra. Then: {{begin-eqn}} {{eqn | l = \left({a, b}\right) + \left({a, b}\right)^*       | r = \left({a, b}\right) + \left({a^*, -b}\right)       | c =  }} {{eqn | r = \left({a + a^*, b - b}\right)       | c =  }} {{eqn | r = \left({a + a^*, 0}\right)       | c =  }} {{end-eqn}} As $A$ is itself nicely normed, we have that $a + a^*$ is real. Hence it follows that $\left({a + a^*, 0}\right)$ is real. Next: {{begin-eqn}} {{eqn | l = \left({a, b}\right) \left({a, b}\right)^*       | r = \left({a, b}\right) \left({a^*, -b}\right)       | c =  }} {{eqn | r = \left({a a^* - \left({-b b^*}\right), a^* b + \left({-a^* b}\right)}\right)       | c =  }} {{eqn | r = \left({\left \Vert {a} \right \Vert^2 + \left \Vert {b} \right \Vert^2, 0}\right)       | c = as $A$ is nicely normed }} {{end-eqn}} We have by definition of norm that $\left \Vert {a} \right \Vert^2 + \left \Vert {b} \right \Vert^2$ is real. Hence it follows that $\left({\left \Vert {a} \right \Vert^2 + \left \Vert {b} \right \Vert^2, 0}\right)$ is real. It follows from reversing the argument that if $A'$ is not nicely normed then nor will $A$ be. {{qed}}
The quaternions $\Bbb H$ are formed by the Cayley-Dickson Construction from the complex numbers $\C$. From Complex Numbers form Algebra, we have that $\C$ forms: :$(1): \quad$ An associative algebra :$(2): \quad$ A commutative algebra :$(3): \quad$ A normed division algebra :$(4): \quad$ A nicely normed $*$-algebra. From Cayley-Dickson Construction forms Star-Algebra, $\Bbb H$ is a $*$-algebra. From Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed, $\Bbb H$ is a nicely normed $*$-algebra. From Cayley-Dickson Construction from Commutative Associative Algebra is Associative, $\Bbb H$ is an associative algebra. Now suppose $\Bbb H$ formed a commutative algebra. Then from Cayley-Dickson Construction from Real Algebra is Commutative, that would mean $\C$ is a real algebra. But from Complex Numbers form Algebra it is explicitly demonstrated that $\C$ is '''not''' a real algebra. So $\Bbb H$ can not be a commutative algebra. === Proof of Normed Division Algebra === Consider the element $\left({1, 0}\right)$ of $\C^2$. We have: {{begin-eqn}} {{eqn | o =        | r = \left({x_1, x_2}\right) \times \left({1, 0}\right)       | c =  }} {{eqn | r = \left({x_1 \times 1 - 0 \times x_2, x_1 \times 0 + x_2 \times 1}\right)       | c =  }} {{eqn | r = \left({x_1, x_2}\right)       | c =  }} {{end-eqn}} As $\times$ is commutative on $\C$, it follows that $\left({1, 0}\right) \times \left({x_1, x_2}\right) = \left({x_1, x_2}\right)$. So $\left({1, 0}\right) \in \C^2$ functions as a unit. That is, $\left({\C^2, \times}\right)$ is a unitary algebra. <!-- From Inverses for Complex Multiplication, every element of $\left({\C, \times}\right)$ except $0$ has a multiplicative inverse. So $\left({\C, \times}\right)$ is a division algebra and hence a unitary division algebra. --> We define a norm on $\left({\C^2, \times}\right)$ by: :$\forall \mathbf a = \left({a_1, a_2}\right) \in \C^2: \left \Vert {\mathbf a} \right \Vert = \sqrt {a_1^2 + a_2^2}$ This is a norm because: : $(1): \quad \left \Vert \mathbf x \right \Vert = 0 \iff \mathbf x = \mathbf 0$ : $(2): \quad \left \Vert \lambda \mathbf x \right \Vert = \left \vert \lambda \right \vert \left \Vert x \right \Vert$ : $(3): \quad \left \Vert x - y \right \Vert \le \left \Vert x - z \right \Vert + \left \Vert z - y \right \Vert$ It also follows that: : $\left \Vert x \times y \right \Vert = \left \vert x \times y \right \vert = \left \vert x \right \vert \times \left \vert y \right \vert = \left \Vert x \right \Vert \times \left \Vert y \right \Vert$ and so $\left({\C^2, \times}\right)$ is a normed division algebra. {{Proofread|Make sure the above are all proved properly.}} {{qed}}
The octonions $\Bbb O$ are formed by the Cayley-Dickson construction from the quaternions $\Bbb H$. From Quaternions form Algebra, we have that $\Bbb H$ forms: :$(1): \quad$ An associative algebra :$(2): \quad$ A normed division algebra :$(3): \quad$ A nicely normed $*$-algebra. From Cayley-Dickson Construction forms Star-Algebra, $\Bbb O$ is a $*$-algebra. From Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed, $\Bbb O$ is a nicely normed $*$-algebra. From Nicely Normed Cayley-Dickson Construction from Associative Algebra is Alternative, $\Bbb O$ is an alternative algebra. Now suppose $\Bbb O$ formed an associative algebra. Then from Cayley-Dickson Construction from Commutative Associative Algebra is Associative, that would mean $\Bbb H$ is a commutative algebra. But from Quaternions form Algebra it is explicitly demonstrated that $\Bbb H$ is '''not''' a commutative algebra. So $\Bbb O$ can not be a associative algebra. === Proof of Normed Division Algebra === {{finish|Might not need this section. Might find a better way of processing this bit.}} {{qed}}
Let $A$ be a division algebra, in the sense that: :$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$ Suppose that $\exists a, b \in A_F \setminus \left\{{\mathbf 0_A}\right\}: \mathbf 0_A = b \oplus a$. Then by definition of the zero vector we also have that $\mathbf 0_A = b \oplus \mathbf 0_A$. So there are two elements $x$ of $A_F$ such that $\mathbf 0_A = b \oplus x$, that is, $a$ and $\mathbf 0_A$. Similarly, suppose that $\exists a, b \in A_F \setminus \left\{{\mathbf 0_A}\right\}: \mathbf 0_A = a \oplus b$. Then by definition of the zero vector we also have that $\mathbf 0_A = \mathbf 0_A \oplus b$. So there are two elements $y$ of $A_F$ such that $\mathbf 0_A = y \oplus b$, that is, $a$ and $\mathbf 0_A$. So $A$ can not be a division algebra. So it follows that if: :$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$ then: :$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$ {{qed|lemma}} Now suppose that: :$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$ Suppose $\exists x_1, x_2 \in A_F, x_1 \ne x_2$ such that: :$a = b \oplus x_1$ :$a = b \oplus x_2$ We have that $x_1 \ne x_2$ and so $x_1 - x_2 = z \ne \mathbf 0_A$. Thus $b \oplus x_1 - b \oplus x_2 = \mathbf 0_A$. As $\oplus$ is a bilinear mapping, it follows that $b \oplus \left({x_1 - x_2}\right) = \mathbf 0_A$ and so $b \oplus z = \mathbf 0_A$. But $z \ne \mathbf 0_A$ and so it is not the case that $\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$. Similarly it can be shown that if $\exists y_1, y_2 \in A_F, y_1 \ne y_2$ such that: :$a = y_1 \oplus b$ :$a = y_2 \oplus b$ then it is not the case that $\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$. Now suppose that: :$\exists a, b \in A_F, b \ne \mathbf 0_A: \not \exists_1 x \in A_F: a = b \oplus x$ or :$\exists a, b \in A_F, b \ne \mathbf 0_A: \not \exists_1 y \in A_F: a = y \oplus b$ {{finish|Haven't found an approach yet ...}} Thus it is not the case that $\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$. So it follows that if: :$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$ then: :$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$ {{qed}} Category:Algebras ezmzalib7kvhrt517srmx91mhyslu5d
Let $A = \struct {A_F, \oplus}$ be a normed divison algebra as defined in the hypothesis. The fact that $A$ is a unitary algebra is a consequence of the definition of normed divison algebra. From the definition of a norm, we have that: :$\forall a \in A: \norm a = 0 \iff a = 0_A$ So, let $a, b \in A \setminus \set {0_A}$. We have: {{begin-eqn}} {{eqn | l = \norm a \norm b       | o = \ne       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \norm {a \oplus b}       | o = \ne       | r = 0       | c = {{Defof|Normed Division Algebra}}: $\norm {a \oplus b} = \norm a \norm b$ }} {{eqn | ll= \leadsto       | l = a \oplus b       | o = \ne       | r = 0_A       | c = {{Defof|Norm on Vector Space}} }} {{end-eqn}} Thus for any arbitrary $a, b \ne 0_A$ we have shown that $a \oplus b \ne 0_A$. Thus $A$ is a division algebra. {{qed|lemma}} Next: {{begin-eqn}} {{eqn | o =        | r = \norm a \norm {1_A}       | c =  }} {{eqn | r = \norm {a \oplus 1_A}       | c = {{Defof|Normed Division Algebra}}: $\norm {a \oplus b} = \norm a \norm b$ }} {{eqn | r = \norm a       | c = {{Defof|Norm on Vector Space}} }} {{end-eqn}} demonstrating that $\norm {1_A} = 1$. {{qed}}
Let $S_1$ and $S_2$ be sets which are not empty. Suppose there exists an injection $f: S_1 \to S_2$ and another injection $g: S_2 \to S_1$. Then by the Cantor-Bernstein-Schröder Theorem there exists a bijection between $S_1$ and $S_2$ and by definition $S_1$ is equivalent to $S_2$. Let $\mathcal A$ be the set of invertible mappings $\phi: A \to B$ where $A \subseteq S_1$ and $B \subseteq S_2$. Since $S_1$ and $S_2$ are not empty, $\exists s_1 \in S_1$ and $\exists s_2 \in S_2$. Thus we can construct the mapping $\alpha: \set {s_1} \to \set {s_2}$ such that $\map \alpha {s_1} = s_2$. This is trivially an invertible mapping, so $\mathcal A$ is not empty. We can impose an ordering $\le$ on $\mathcal A$ by letting $\phi_1 \le \phi_2 \iff \phi_1 \subseteq \phi_2$, that is, if $\phi_2$ is an extension of $\phi_1$. Let $\mathcal C$ be a chain in $\struct {\mathcal A, \le}$. Then $\displaystyle \bigcup \set {\phi \in \mathcal C}$ is an upper bound of every $\phi \in \mathcal C$, and it lies in $\mathcal A$. The conditions of Zorn's Lemma are satisfied, so we can find a maximal element $M$ in $\mathcal A$. Let: :$M_1 = \set {s_1: \tuple {s_1, s_2} \in M}$ :$M_2 = \set {s_2: \tuple {s_1, s_2} \in M}$ We have that $M_1 \subsetneq S_1$ and $M_2 \subsetneq S_2$ both together contradict the fact that $M$ is maximal element. Thus either $M_1 = S_1$ or $M_2 = S_2$, and possibly both. Thus either: : $M$ is an injection of $S_1$ into $S_2$ or: : $M^{-1}$ is an injection of $S_2$ into $S_1$. Thus either $S_1 \le S_2$ or $S_2 \le S_1$, and the result follows. {{qed}} {{AoC||Zorn's Lemma}}
Follows directly from: :The definition of a cardinal as $S \sim T \iff \map \Card S = \map \Card T$ :Set Equivalence is Equivalence Relation :Relation Partitions Set iff Equivalence. {{qed}} Category:Cardinals hh5nfciyzjy5f3jv4q3qua739f4wgzt
We have that: :$\struct {\Z, +, \times}$ form a commutative ring. :$\struct {\Z, +, \times}$ has a unity, and the unity is $1$. {{Qed}}
From Matrix Entrywise Addition forms Abelian Group we have that $\struct {\map {\MM_R} n, +}$ is an abelian group, because $\struct {R, +}$ is itself an abelian group. Similarly, it is clear that $\struct {\map {\MM_R} n, \times}$ is a semigroup, as Matrix Multiplication is Closed and Matrix Multiplication is Associative. Finally, we note that Matrix Multiplication Distributes over Matrix Addition. {{qed}}
From Ring of Square Matrices over Ring is Ring we have that $\struct {\map {\MM_R} n, +, \times}$ is a ring. As $R$ has a unity, the unit matrix can be formed. The unity of $\struct {\map {\MM_R} n, +, \times}$ is this unit matrix. {{qed}}
We have by definition that a field is a division ring which is also commutative. Hence $F$ is a commutative ring with unity. So, from Ring of Square Matrices over Commutative Ring with Unity we have that $\struct {\map {\MM_F} n, +, \times}$ is a ring with unity. From Matrix Multiplication is not Commutative, we have that $\struct {\map {\MM_F} n, +, \times}$ is not a commutative ring. Hence the result. {{qed}}
Recall that Real Numbers form Field. The result follows directly from Ring of Square Matrices over Field is Ring with Unity. {{qed}}
As $R$ is a ring with unity, we have that: {{begin-eqn}} {{eqn | l = 0_R       | o = \ne       | r = 1_R }} {{eqn | l = 0_R \times 0_R       | r = 0_R       | c =  }} {{eqn | l = 0_R \times 1_R       | r = 0_R = 1_R \times 0_R       | c =  }} {{eqn | l = 1_R \times 1_R       | r = 1_R       | c =  }} {{end-eqn}} Now let: {{begin-eqn}} {{eqn | l = \mathbf A       | r = \begin {pmatrix} 0_R & 1_R \\ 0_R & 0_R \end {pmatrix} }} {{eqn | l = \mathbf B       | r = \begin {pmatrix} 0_R & 0_R \\ 1_R & 0_R \end {pmatrix} }} {{end-eqn}} By definition, both $\mathbf A$ and $\mathbf B$ are elements of $\map {\MM_R} 2$. It will be demonstrated that $\mathbf A$ and $\mathbf B$ do not commute. We have: {{begin-eqn}} {{eqn | l = \mathbf A \mathbf B       | r = \begin {pmatrix} 0_R & 1_R \\ 0_R & 0_R \end {pmatrix} \begin {pmatrix} 0_R & 0_R \\ 1_R & 0_R \end {pmatrix}       | c =  }} {{eqn | r = \begin {pmatrix} 0_R \times 0_R + 1_R \times 1_R & 0_R \times 0_R + 1_R \times 0_R \\ 0_R \times 0_R + 0_R \times 1_R & 0_R \times 0_R + 0_R \times 0_R \end{pmatrix}       | c =  }} {{eqn | r = \begin {pmatrix} 1_R & 0_R \\ 0_R & 0_R \end{pmatrix}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \mathbf B \mathbf A       | r = \begin {pmatrix} 0_R & 0_R \\ 1_R & 0_R \end {pmatrix} \begin {pmatrix} 0_R & 1_R \\ 0_R & 0_R \end {pmatrix}       | c =  }} {{eqn | r = \begin {pmatrix} 0_R \times 0_R + 0_R \times 0_R & 0_R \times 1_R + 0_R \times 0_R \\ 1_R \times 0_R + 0_R \times 0_R & 1_R \times 1_R + 0_R \times 0_R \end{pmatrix}       | c =  }} {{eqn | r = \begin {pmatrix} 0_R & 0_R \\ 0_R & 1_R \end{pmatrix}       | c =  }} {{end-eqn}} and it is seen that: :$\mathbf A \mathbf B \ne \mathbf B \mathbf A$ Thus, whatever the nature of the ring with unity $R$, it is never the case that matrix multiplication is commutative over $\map {\MM_R} 2$. {{qed}} Category:Matrix Multiplication is not Commutative rui1jhhw7alanyt33ovxir9qxs8daf5
The ring zero is, by definition of a ring, the identity element of the additive group $\struct {R, +}$. The result then follows from Identity of Group is Unique. {{qed}}
From Ring Product with Zero we have that the ring zero of $R$ is indeed a zero element, as suggested by its name. The result then follows from Zero Element is Unique. {{qed}}
Suppose $0$ and $0'$ are both ring zeroes of $\struct {R, +, \circ}$. Then by Ring Product with Zero: :$0' \circ 0 = 0$ by dint of $0$ being a ring zero :$0' \circ 0 = 0'$ by dint of $0'$ being a ring zero. So $0 = 0' \circ 0 = 0'$. So $0 = 0'$ and there is only one ring zero of $\struct {R, +, \circ}$ after all. {{qed}}
The ring negative is, by definition of a ring, the inverse element of $a$ in the additive group $\struct {R, +}$. The result then follows from Inverse in Group is Unique. {{qed}}
The ring negative is, by definition of a ring, the inverse element of $a$ in the additive group $\struct {R, +}$. The result then follows from Inverse of Group Inverse. {{qed}}
Follows from: : Even Perfect Number is Hexagonal : Hexagonal Number is Triangular Number {{qed}}
{{begin-eqn}} {{eqn | l = a       | o = \ne       | r = 0_F       | c =  }} {{eqn | ll= \leadsto       | l = a^{-1} \times \paren {a \times b}       | r = a^{-1} \times 0_F       | c = $a^{-1}$ exists because $a \ne 0_F$ }} {{eqn | ll= \leadsto       | l = \paren {a^{-1} \times a} \times b       | r = a^{-1} \times 0_F       | c = {{Field-axiom|M1}} }} {{eqn | ll= \leadsto       | l = 1_F \times b       | r = a^{-1} \times 0_F       | c = {{Field-axiom|M4}} }} {{eqn | ll= \leadsto       | l = b       | r = a^{-1} \times 0_F       | c = {{Field-axiom|M3}} }} {{eqn | ll= \leadsto       | l = b       | r = 0_F       | c = Field Product with Zero }} {{end-eqn}} {{qed}}
From the definition of multiplicative identity, $1_F$ is the identity element of the multiplicative group $\struct {F^*, \times}$. The result follows from Identity of Group is Unique. {{qed}}
From the definition of multiplicative inverse, $a^{-1}$ is the inverse element of the multiplicative group $\struct {F^*, \times}$. The result follows from Inverse in Group is Unique. {{qed}}
From the definition of a field as a division ring, every element of $F^*$ is a unit. The result follows from Product Inverse in Ring is Unique. {{qed}}
We have that the additive group $\struct {R, +}$ is an abelian group. $(1): \quad \paren {m + n} \cdot a = \paren {m \cdot a} + \paren {n \cdot a}$: This is an instance of Powers of Group Elements: Sum of Indices when expressed in additive notation: :$\forall n, m \in \Z: \forall a \in R: m a + n a = \paren {m + n} a$ {{qed|lemma}} $(2): \quad m \cdot \paren {a + b} = \paren {m \cdot a} + \paren {m \cdot b}$: This is an instance of Power of Product in Abelian Group when expressed in additive notation: :$\forall n \in \Z: \forall a, b \in R: n \paren {a + b} = n a + n b$ {{qed}}
We have that $\struct {F^*, \times}$ is the multiplicative group of $\struct {F, +, \times}$. Let $a \in F^*$, that is, $a \in F: a \ne 0_F$, where $0_F$ is the zero of $F$. This is an instance of Powers of Group Elements when expressed in additive notation: :$\forall m, n \in \Z: \paren {m n} a = m \paren {n a}$ {{qed|lemma}} Now suppose $a = 0_F$. Then by definition of the zero element of $F$, we have that: :$\paren {m n} \cdot a = 0_F = m \cdot \paren {n \cdot a}$ {{qed}}
{{begin-eqn}} {{eqn|l = \sin \left({- \theta}\right)      |r = - \sin \theta      |c = Sine Function is Odd }} {{eqn|l = \cos \left({- \theta}\right)      |r = \cos \theta      |c = Cosine Function is Even }} {{eqn|l = \tan \left({- \theta}\right)      |r = - \tan \theta      |c = Tangent Function is Odd }} {{end-eqn}} {{qed}}
We have by definition of integral multiple that: :$\forall r \in R: 0 \cdot r = 0_R$ where $0_R$ is the zero of $R$. Hence the result by definition of annihilator. {{qed}} Category:Ring Theory sky2zptmfq5vwwjsru78klub6slkzlg
Let the zero of $R$ be $0_R$ and the unity of $R$ be $1_R$. First we note that: {{begin-eqn}} {{eqn | l = a       | o = \in       | r = \operatorname{Ann} \left({R}\right)       | c =  }} {{eqn | ll= \leadsto       | l = a \cdot 1_R       | r = 0_R       | c =  }} {{eqn | ll= \leadsto       | l = \left({-a}\right) \cdot 1_R       | r = 0_R       | c =  }} {{eqn | ll= \leadsto       | l = -\left({a \cdot 1_R}\right)       | r = 0_R       | c =  }} {{end-eqn}} So: : $a \in A \implies -a \in A$ But if $a \ne 0$ then either $a > 0$ or $-a > 0$. That is, either $a$ or $-a$ is positive. So, if $\operatorname{Ann} \left({F}\right)$ contains at least one non-zero element, it contains at least one strictly positive integer. {{qed}} Category:Rings with Unity bg2avar95sakmqjanxs4npzpr0q0vfa
{{questionable|To be replaced with something rigorous, based on calculus.}} From Area of Circle, the area of $\CC$ is $\pi r^2$. From Full Angle measures $2 \pi$ Radians, the angle within $\CC$ is $2 \pi$. The fraction of the area of $\CC$ within the sector $BAC$ is therefore $\pi r^2 \times \dfrac \theta {2 \pi}$. Hence the result. {{qed}}
Suppose $F$ is a subfield of $\struct {\Q, +, \times}$. Let $a \in F$. Then $a^{-1} \in F$ and so $a \times a^{-1} = 1$ and so $1 \in F$. So, we have that $1 \in F$ and so $-1 \in F$ and thus $1 + \paren {-1} = 0 \in F$. Suppose $n \in \Z$ such that $n \in F$. Then $n + 1 \in F$ and so by the Principle of Mathematical Induction that all positive integers are in $F$. But then if $m \in F$ we have that $-m \in F$ and so all negative integers are in $F$. Next, as we have seen, $m \in F \implies m^{-1} \in F$. So: :$\forall p \in \Z, q \in \Z_{\ne 0}: p \times q^{-1} \in F$ and it follows by the Subfield Test that $F = \Q$. {{qed}}
If $\struct {F, +, \times}$ is a subfield of $\struct {\Z_p, +, \times}$, then $\struct {F, +}$ is a subgroup of $\struct {\Z_p, +}$. But from Prime Group has no Proper Subgroups, $\struct {\Z_p, +}$ has no proper subgroup except the trivial group. Hence $F = \Z_p$ and so follows the result. {{qed}}
By definition of field, $F$ is a division ring where $\times$ is commutative. Therefore all division subrings of $F$ are in fact subfields of $F$. By Intersection of All Division Subrings is Prime Subfield, the intersection of all subfields of $F$ is a prime field which is a subfield of $F$. {{qed}}