hcju/proofwikiqa huggingface dataset:
From the definition of binomial coefficients: :$\dbinom r k = \dfrac {r^{\underline k} } {k!}$ for $k \ge 0$ where $r^{\underline k}$ is the falling factorial. In turn: :$\displaystyle x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \paren {x - j}$ But when $k = 1$, we have: :$\displaystyle \prod_{j \mathop = 0}^0 \paren {x - j} = \paren {x - 0} = x$ So: :$\forall r \in \R: \dbinom r 1 = r$ {{qed}} This is completely compatible with the result for natural numbers: :$\forall n \in \N: \dbinom n 1 = n$ as from the definition: :$\dbinom n 1 = \dfrac {n!} {1! \ \paren {n - 1}!}$ the result following directly, again from the definition of the factorial where $1! = 1$. {{qed}}
From the definition of binomial coefficient: :$\dbinom n n = \dfrac {n!} {n! \ \paren {n - n}!} = \dfrac {n!} {n! \ 0!}$ the result following directly from the definition of the factorial, where $0! = 1$. From N Choose Negative Number is Zero: :$\forall k \in \Z_{<0}: \dbinom n k = 0$ So for $n < 0$: :$\dbinom n n = 0$ {{qed}}
The case where $n = 1$ can be taken separately. From Binomial Coefficient with Zero: :$\dbinom 1 0 = 1$ demonstrating that the result holds for $n = 1$. Let $n \in \N: n > 1$. From the definition of binomial coefficients: :$\dbinom n {n - 1} = \dfrac {n!} {\left({n - 1}\right)! \left({n - \left({n - 1}\right)}\right)!} = \dfrac {n!} {\left({n - 1}\right)! \ 1!}$ the result following directly from the definition of the factorial. {{qed}}
From Cardinality of Set of Subsets, $\dbinom n {n - 1}$ is the number of combination of things taken $n - 1$ at a time. Choosing $n - 1$ things from $n$ is the same thing as choosing which $1$ of the elements to be left out. There are $n$ different choices for that $1$ element. Therefore there are $n$ ways to choose $n - 1$ things from $n$. {{qed}}
From the definition of binomial coefficients: :$\dbinom r k = \dfrac {r^{\underline k}} {k!}$ for $k \ge 0$ where $r^{\underline k}$ is the falling factorial. In turn: :$\displaystyle x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \left({x - j}\right)$ When $k = 2$: :$\displaystyle \prod_{j \mathop = 0}^1 \left({x - j}\right) = \frac {\left({x - 0}\right) \left({x - 1}\right)} {2!}$ where $2! = 1 \times 2 = 2$. So: :$\forall r \in \R: \dbinom r 2 = \dfrac {r \left({r - 1}\right)} 2$ {{qed}}
From the definition of the binomial coefficient, when $i < 0$ and $i > n$ we have $\dbinom n i = 0$. The result follows directly from Sum of Binomial Coefficients over Lower Index. {{qed}} Category:Binomial Coefficients Category:Sum of Binomial Coefficients over Lower Index 0kwxkf33w9yqy7i9kaksyn7yxmjwnb4
{{begin-eqn}} {{eqn | l = \left({r + 1}\right) \binom r {k - 1} + \left({r + 1}\right) \binom r k       | r = \left({r + 1}\right) \binom r {k - 1} + \left({r + 1}\right) \binom r {r - k}       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = k \binom {r + 1} k + \left({r - k + 1}\right) \binom {r + 1} {r - k + 1}       | c = Factors of Binomial Coefficient }} {{eqn | r = k \binom {r + 1} k + \left({r - k + 1}\right) \binom {r + 1} k       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \left({r + 1}\right) \binom {r + 1} k       | c =  }} {{end-eqn}} Dividing by $\left({r + 1}\right)$ yields the result. {{qed}}
=== Basis for the Induction === For $n = 0$ we have: :$\displaystyle \paren {x + y}^0 = 1 = \binom 0 0 x^{0 - 0} y^0 = \sum_{k \mathop = 0}^0 \binom 0 k x^{0 - k} y^k$ This is the basis for the induction. === Induction Hypothesis === This is our induction hypothesis: :$\displaystyle \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {x + y}^{n + 1}       | r = \paren {x + y} \paren {x + y}^n       | c =  }} {{eqn | r = x \sum_{k \mathop = 0}^n \binom n k x^{n - k}y^k + y \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k       | c = Inductive Hypothesis }} {{eqn | r = \sum_{k \mathop = 0}^n \binom n k x^{n + 1 - k} y^k + \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^{k + 1}       | c =  }} {{eqn | r = \binom n 0 x^{n + 1} + \sum_{k \mathop = 1}^n \binom n k x^{n + 1 - k} y^k + \binom n n y^{n + 1} + \sum_{k \mathop = 0}^{n - 1} \binom n k x^{n - k} y^{k + 1}       | c =  }} {{eqn | r = x^{n + 1} + y^{n + 1} + \sum_{k \mathop = 1}^n \binom n k x^{n + 1 - k} y^k + \sum_{k \mathop = 0}^{n - 1} \binom n k x^{n - k} y^{k + 1}       | c =  }} {{eqn | r = \binom {n + 1} 0 x^{n + 1} + \binom {n + 1} {n + 1} y^{n + 1} + \sum_{k \mathop = 1}^n \binom n k x^{n + 1 - k} y^k + \sum_{k \mathop = 1}^n \binom n {k - 1} x^{n + 1 - k} y^k       | c =  }} {{eqn | r = \binom {n + 1} 0 x^{n + 1} + \binom {n + 1} {n + 1} y^{n + 1} + \sum_{k \mathop = 1}^n \paren {\binom n k + \binom n {k - 1} } x^{n + 1 - k} y^k       | c =  }} {{eqn | r = \binom {n + 1} 0 x^{n + 1} + \binom {n + 1} {n + 1} y^{n + 1} + \sum_{k \mathop = 1}^n \binom {n + 1} k x^{n + 1 - k} y^k       | c = Pascal's Rule }} {{eqn | r = \sum_{k \mathop = 0}^{n + 1} \binom {n + 1} k x^{n + 1 - k} y^k       | c =  }} {{end-eqn}} The result follows by the Principle of Mathematical Induction. {{qed}}
{{begin-eqn}} {{eqn | l = \frac \d {\d x} \size x       | r = \frac \d {\d x} \sqrt{x^2}       | c = Square of Real Number is Non-Negative }} {{eqn | r = \frac \d {\d x} \paren {x^2}^{\frac 1 2} }} {{eqn | r = \frac 1 2 \paren {x^2}^{-\frac 1 2} \cdot 2 x       | c = Chain Rule for Derivatives }} {{eqn | r = \frac x {\sqrt{x^2} } }} {{eqn | r = \frac x {\size x} }} {{end-eqn}} {{qed|lemma}} Now consider $x = 0$. From the definition of derivative: {{begin-eqn}} {{eqn | l = \valueat {\dfrac {\d \size x} {\d x} } {x \mathop = 0}       | r = \lim_{x \mathop \to 0}\frac {\size x - 0} {x - 0} }} {{eqn | r = \begin {cases} \lim_{x \mathop \to 0^+} \dfrac x x & : x > 0 \\ \lim_{x \mathop \to 0^-} \dfrac {-x} x & : x < 0 \end {cases}       | c = {{Defof|Absolute Value}} }} {{eqn | r = \begin {cases} 1 & : x > 0 \\ -1 & : x < 0 \end{cases} }} {{end-eqn}} From Limit iff Limits from Left and Right, the limit does not exist. {{qed}} Category:Derivatives Category:Absolute Value Function kwp3633hffzgpfq9nnhi7ctylso5m5w
Let the identity of $\left({S, \circ}\right)$ be $e$. Let $z$ be the inverse of $y$ for $\circ$: :$z \circ y = e$ :$y \circ z = e$ From Identity of Inverse Completion of Commutative Monoid: :$z \circ' y = e$ :$y \circ' z = e$ Hence $z$ is the inverse of $y$ for $\circ'$. {{qed}}
From Semigroup is Subsemigroup of Itself, $\left({S, \circ}\right)$ is a subsemigroup of $\left({S, \circ}\right)$. Also from Semigroup is Subsemigroup of Itself, $\left({C, \circ {\restriction_C}}\right)$ is a subsemigroup of $\left({C, \circ {\restriction_C}}\right)$. The result follows from Cross-Relation is Congruence Relation. {{qed}}
From Cross-Relation is Equivalence Relation we have that $\boxtimes$ is an equivalence relation. Hence the equivalence class of $\tuple {x, y}$ under $\boxtimes$ is defined for all $\tuple {x, y} \in S \times C$. From Semigroup is Subsemigroup of Itself, $\struct {S, \circ}$ is a subsemigroup of $\struct {S, \circ}$. Also from Semigroup is Subsemigroup of Itself, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {C, \circ {\restriction_C} }$. The result follows from Elements of Cross-Relation Equivalence Class. {{qed}}
From Semigroup is Subsemigroup of Itself, $\left({S, \circ}\right)$ is a subsemigroup of $\left({S, \circ}\right)$. Also from Semigroup is Subsemigroup of Itself, $\left({C, \circ {\restriction_C}}\right)$ is a subsemigroup of $\left({C, \circ {\restriction_C}}\right)$. The result follows from Equivalence Class of Equal Elements of Cross-Relation. {{qed}}
The canonical epimorphism from $\left({S \times C, \oplus}\right)$ onto $\left({T', \oplus'}\right)$ is given by: :$q_\boxtimes: \left({S \times C, \oplus}\right) \to \left({T', \oplus'}\right): q_\boxtimes \left({x, y}\right) = \left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes$ where, by definition: {{begin-eqn}} {{eqn | l = \forall \left({x_1, y_1}\right), \left({x_2, y_2}\right) \in S \times C:       | o =        | r = q_\boxtimes \left({\left({x_1, y_1}\right) \oplus \left({x_2, y_2}\right)}\right)       | c =  }} {{eqn | r = q_\boxtimes \left({\left({x_1, y_1}\right)}\right) \oplus' q_\boxtimes \left({\left({x_2, y_2}\right)}\right)       | c =  }} {{end-eqn}} By Morphism Property Preserves Closure, as $\oplus$ is closed, then so is $\oplus'$. By Epimorphism Preserves Associativity, as $\oplus$ is associative, then so is $\oplus'$. By Epimorphism Preserves Commutativity, as $\oplus$ is commutative, then so is $\oplus'$. Thus $\left({T', \oplus'}\right)$  is closed, associative and commutative, and therefore a commutative semigroup. {{Qed}}
{{begin-eqn}} {{eqn | l = \psi \left({x}\right)       | r = \psi \left({y}\right)       | c =  }} {{eqn | ll= \implies       | l = \forall a \in C: \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes       | r = \left[\!\left[{\left({y \circ a, a}\right)}\right]\!\right]_\boxtimes       | c = Definition of $\left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes$ }} {{eqn | ll= \implies       | l = x       | r = y       | c = from $(1)$ above }} {{end-eqn}} The result follows by the definition of injection. {{Qed}}
We have that this quotient mapping $\psi: S \to T'$ is an injection. Let $x, y \in S$. Then: {{begin-eqn}} {{eqn | l = \psi \left({x}\right) \oplus' \psi \left({y}\right)       | r = \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes \oplus' \left[\!\left[{\left({y \circ a, a}\right)}\right]\!\right]_\boxtimes       | c = Definition of $\psi$ }} {{eqn | r = \left[\!\left[{\left({x \circ a, a}\right) \oplus' \left({y \circ a, a}\right)}\right]\!\right]_\boxtimes       | c = Definition of $\oplus'$ }} {{eqn | r = \left[\!\left[{\left({x \circ a \circ y \circ a, a \circ a}\right)}\right]\!\right]_\boxtimes       | c = Definition of $\boxtimes$ }} {{eqn | r = \left[\!\left[{\left({\left({x \circ y}\right) \circ \left({a \circ a}\right), a \circ a}\right)}\right]\!\right]_\boxtimes       | c = Commutativity of $\circ$ }} {{eqn | r = \psi \left({x \circ y}\right)       | c = as $a \circ a \in C$ }} {{end-eqn}} So $\psi \left({x \circ y}\right) = \psi \left({x}\right) \oplus' \psi \left({y}\right)$, and the morphism property is proven. Thus $\psi$ is an injective homomorphism, and so by definition a monomorphism. {{Qed}}
From Quotient Mapping is Monomorphism, $\psi: \left({S, \circ}\right) \to \left({S', \oplus'}\right)$ is a monomorphism. Therefore by definition: :$\psi$ is a homomorphism :$\psi$ is an injection. Because $S'$ is the image of $\psi \left({S}\right)$, by Surjection by Restriction of Codomain $\psi$ is a surjection. Therefore by definition $\psi: S \to S'$ is a bijection. A bijective homomorphism is an isomorphism. {{Qed}}
We have Morphism Property Preserves Cancellability. Thus: :$c \in C \implies \psi \left({c}\right) \in C'$ So by Image of Subset under Relation is Subset of Image: Corollary 2: :$\psi \left[{C}\right] \subseteq C'$ From above, $\psi$ is an isomorphism. {{explain|Above where?}} Hence, also from Morphism Property Preserves Cancellability: :$c' \in C' \implies \psi^{-1} \left({c'}\right) \in C$ So by Image of Subset under Relation is Subset of Image: Corollary 3: : $\psi^{-1} \left[{C'}\right] \subseteq C$ Hence by definition of set equality: :$\psi \left[{C}\right] = C'$ {{Qed}}
We have that $S'$ is the image $\psi \left({S}\right)$ of $S$. For $\left({S', \oplus'}\right)$ to be a subsemigroup of $\left({T', \oplus'}\right)$, by Subsemigroup Closure Test we need to show that $\left({S', \oplus'}\right)$ is closed. Let $x, y \in S'$. Then $x = \phi \left({x'}\right), y = \phi \left({y'}\right)$ for some $x', y' \in S$. But as $\phi$ is an isomorphism, it obeys the morphism property. So $x \oplus' y = \phi \left({x'}\right) \oplus' \phi \left({y'}\right) = \phi \left({x' \circ y'}\right)$. Hence $x \oplus' y$ is the image of $x' \circ y' \in S$ and hence $x \oplus' y \in S'$. Thus by the Subsemigroup Closure Test, $\left({S', \oplus'}\right)$ is a subsemigroup of $\left({T', \oplus'}\right)$ {{Qed}}
{{begin-eqn}} {{eqn | l = \paren {x \circ c} \circ y       | r = x \circ \paren {y \circ c}       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass {\tuple {x, y} } \boxtimes \oplus' \eqclass {\tuple {c, c} } \boxtimes       | r = \eqclass {\tuple {x \circ c, y \circ c} } \boxtimes       | c = Definition of $\oplus'$ }} {{eqn | r = \eqclass {\tuple {x, y} } \boxtimes       | c = Cancellability of elements of $C$ }} {{end-eqn}} Hence the result, by definition of identity element. {{Qed}}
From Identity of Quotient Structure, $\struct {T', \oplus'}$ has an identity, and it is $\eqclass {\tuple {c, c} } \boxtimes$ for any $c \in C$. Call this identity $e_{T'}$. Let the mapping $\psi: S \to T'$ be defined as: :$\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$ From Image of Cancellable Elements in Quotient Mapping: :$C' = \psi \sqbrk C$ So: {{begin-eqn}} {{eqn | l = x'       | o = \in       | r = C'       | c =  }} {{eqn | ll= \leadsto       | l = \exists x \in C: x'       | r = \map \psi x       | c = as $\psi$ is a surjection }} {{eqn | ll= \leadsto       | l = \forall a \in C: x'       | r = \eqclass {\tuple {x \circ a, a} } \boxtimes       | c = Definition of $\psi$ }} {{end-eqn}} The inverse of $x'$ is $\eqclass {\tuple {a, a \circ x} } \boxtimes$, as follows: {{begin-eqn}} {{eqn | l = a \circ x       | o = \in       | r = C       | c = Cancellable Elements of Semigroup form Subsemigroup }} {{eqn | l = a \circ a \circ x       | o = \in       | r = C       | c = Cancellable Elements of Semigroup form Subsemigroup }} {{eqn | ll= \leadsto       | l = \eqclass {\tuple {x \circ a, a} } \boxtimes \oplus' \eqclass {\tuple {a, a \circ x} } \boxtimes       | r = \eqclass {\tuple {x \circ a \circ a, a \circ a \circ x} } \boxtimes       | c = by definition of $\oplus'$ }} {{eqn | r = \eqclass {\tuple {a \circ a \circ x, a \circ a \circ x} } \boxtimes       | c = Commutativity of $\circ$ }} {{eqn | r = e_{T'}       | c = Identity of Quotient Structure }} {{end-eqn}} thus showing that the inverse of $\eqclass {\tuple {x \circ a, a} } \boxtimes$ is $\eqclass {\tuple {a, a \circ x} } \boxtimes$. {{Qed}}
Let $\left({x, y}\right) \in S \times C$. Then: {{begin-eqn}} {{eqn | l = \psi \left({x}\right) \oplus' \left({\psi \left({y}\right)}\right)^{-1}       | r = \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes \oplus' \left[\!\left[{\left({a, a \circ y}\right)}\right]\!\right]_\boxtimes       | c = Invertible Elements in Quotient Structure above }} {{eqn | r = \left[\!\left[{\left({x \circ a \circ a, a \circ a \circ y}\right)}\right]\!\right]_\boxtimes       | c = Definition of $\oplus'$ }} {{eqn | r = \left[\!\left[{\left({x \circ a \circ a, y \circ a \circ a}\right)}\right]\!\right]_\boxtimes       | c = Commutativity of $\circ$ }} {{eqn | r = \left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes       | c = Cancellability of $a \in C$ }} {{end-eqn}} {{Qed}}
Every cancellable element of $S'$ is invertible in $T'$, from Invertible Elements in Quotient Structure. $T' = S' \cup \left({C'}\right)^{-1}$ is a generator for the semigroup $T'$, from Generator for Quotient Structure. Hence the result, by definition of inverse completion {{Qed}}
Now suppose $m \in \Z: m \ge 2$ be composite. Then: : $\exists k, l \in \N_{> 0}: 1 < k < m, 1 < l < m: m = k \times l$ Thus: {{begin-eqn}} {{eqn | l = \eqclass 0 m       | r = \eqclass m m       | c =  }} {{eqn | r = \eqclass {k l} m       | c =  }} {{eqn | r = \eqclass k m \times \\eqclass l m       | c =  }} {{end-eqn}} So $\struct {\Z_m, +, \times}$ is a ring with zero divisors. So by definition $\struct {\Z_m, +, \times}$ is not an integral domain. {{qed}}
{{AimForCont}} $0$ is cancellable. Let $a, b \in R$ such that $a \ne b$. By definition of ring zero: :$0 \circ a = 0 = 0 \circ b$ By our supposition that $0$ is cancellable: :$a = b$ The result follows by Proof by Contradiction. {{qed}}
Let $J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$. By Congruence Relation induces Normal Subgroup, $\left({J, +}\right)$ is a normal subgroup of $\left({R, +}\right)$. Thus the elements of $\left({R, +}\right) / \left({J, +}\right)$ are the cosets of $\left[\!\left[{0_R}\right]\!\right]_\mathcal E$ by $+$. We have that $\mathcal E$ is also compatible with $\circ$. Thus from Quotient Structure is Well-Defined, and so: :$\left[\!\left[{x_1}\right]\!\right]_\mathcal E = \left[\!\left[{x_2}\right]\!\right]_\mathcal E \land \left[\!\left[{y_1}\right]\!\right]_\mathcal E = \left[\!\left[{y_2}\right]\!\right]_\mathcal E \implies \left[\!\left[{x_1 \circ y_1}\right]\!\right]_\mathcal E = \left[\!\left[{x_2 \circ y_2}\right]\!\right]_\mathcal E$ Putting $x_1 = 0_R, x_2 = x, y_1 = y_2 = y$, so that $\left[\!\left[{y_1}\right]\!\right]_\mathcal E = \left[\!\left[{y_2}\right]\!\right]_\mathcal E$ by definition: :$\left[\!\left[{0}\right]\!\right]_\mathcal E = \left[\!\left[{x}\right]\!\right]_\mathcal E  \implies \left[\!\left[{0 \circ y}\right]\!\right]_\mathcal E = \left[\!\left[{x \circ y}\right]\!\right]_\mathcal E$ Hence: : $\forall y \in R: \left[\!\left[{y}\right]\!\right]_\mathcal E \circ \left[\!\left[{0_R}\right]\!\right]_\mathcal E = \left[\!\left[{0_R}\right]\!\right]_\mathcal E = \left[\!\left[{0_R}\right]\!\right]_\mathcal E \circ \left[\!\left[{y}\right]\!\right]_\mathcal E$ That is: : $\forall x \in J, y \in R: y \circ x \in J, x \circ y \in J$ demonstrating that $J$ is an ideal of $R$. {{qed}}
From Ideal is Additive Normal Subgroup, we have that $\struct {J, +}$ is a normal subgroup of $\struct {R, +}$. Let $x \mathop {\EE_J} y$ denote that $x$ and $y$ are in the same coset, that is: :$x \mathop {\EE_J} y \iff x + N = y + N$ From Congruence Modulo Normal Subgroup is Congruence Relation, $\EE_J$ is a congruence relation for $+$. Now let $x \mathop {\EE_J} x', y \mathop {\EE_J} y'$. By definition of congruence modulo $J$: :$x + \paren {-x'} \in J$ :$y + \paren {-y'} \in J$ Then: :$x \circ y + \paren {-x' \circ y'} = \paren {x + \paren {-x'} } \circ y + x' \circ \paren {y + \paren {-y'} } \in J$ demonstrating that $\EE_J$ is a congruence relation for $\circ$. Hence the result by definition of quotient ring. {{qed}}
Let $\mathscr I$ be the set of linearly independent subsets of $S$ that contain $L$, ordered by inclusion. Note that $L \in \mathscr I$, so $\mathscr I \ne \varnothing$. Let $\mathscr C$ be a nest in $\mathscr I$. Let $C = \bigcup \mathscr C$. {{AimForCont}} that $C$ is linearly dependent. Then there exist $v_1, v_2, \ldots, v_n \in C$ and $r_1, r_2, \ldots, r_n \in F$ such that $r_1 \ne 0$: :$\displaystyle \sum_{k \mathop = 1}^n r_k v_k = 0$ Then there are $C_1, C_2, \ldots, C_n \in \mathscr C$ such that $v_k \in C_k$ for each $k \in \set {1, 2, \ldots, n}$. Since $\mathscr C$ is a nest, $C_1 \cup C_2 \cup \cdots \cup C_n$ must equal $C_k$ for some $k \in \set {1, 2, \ldots, n}$. But then $C_k \in \mathscr C$ and $C_k$ is linearly dependent, which is a contradiction. Thus $C$ is linearly independent. By Zorn's Lemma, $\mathscr I$ has a maximal element $M$ (one that is not contained in any other element). Since $M \in \mathscr I$, $M$ is linearly independent. All that remains is to show that $M$ spans $V$. Suppose, to the contrary, that there exists a $v \in V \setminus \map {\operatorname {span} } M$. Then, since $S$ spans $V$, there must be an element $s$ of $S$ such that $s \notin \map {\operatorname {span} } M$. Then $M \cup \set s$ is linearly independent. Thus $M \cup \set s \supsetneq M$, contradicting the maximality of $M$. Thus $M$ is a linearly independent subset of $V$ that spans $V$. Therefore, by definition, $M$ is a basis for $V$. {{qed}}
Note that the trivial relation $\TT = S \times S$ on $S$ contains $\RR$, by the definition of a relation on $S$. Further, $\TT$ is transitive by Trivial Relation is Equivalence. Thus there is at least one transitive relation on $S$ that contains $\RR$. Now define $\RR^+$ as the intersection of all transitive relations on $S$ that contain $\RR$: :$\displaystyle \RR^+ := \bigcap \set {\RR': \text {$\RR'$ is transitive and $\RR \subseteq \RR'$} }$ By Intersection of Transitive Relations is Transitive, $\RR^+$ is also a transitive relation on $S$. By Set Intersection Preserves Subsets, it also holds that $\RR \subseteq \RR^+$. Lastly, by Intersection is Subset, for any transitive relation $\RR'$ containing $\RR$, it must be that $\RR^+ \subseteq \RR'$. Thus $\RR^+$ is indeed the smallest transitive relation on $S$ containing $\RR$. {{qed}}
Let $\RR^+$ be the intersection of all transitive relations containing $\RR$. By Trivial Relation is Equivalence: :the trivial relation $S \times S$ is transitive. By the definition of endorelation: :$\RR \subseteq S \times S$. Next we have that the Intersection of Transitive Relations is Transitive. {{handwaving}} Thus $\RR^+$ is the smallest transitive relation on $S$ containing $\RR$.
We have that $R$ is the null ring. That is, by definition it has a single element, which can be denoted $0_R$, such that: :$R := \struct {\set {0_R}, +, \circ}$ where ring addition and the ring product are defined as: :$0_R + 0_R = 0_R$ :$0_R \circ 0_R = 0_R$ Consider the operation $+$. By definition, the algebraic structure $\struct {\set {0_R}, +}$ is a trivial group. Then: :$\forall a, b \in R: a \circ b = 0_R$ Thus by definition, $R$ is a trivial ring. The fact that $R$ is a commutative ring follows from Trivial Ring is Commutative Ring. {{qed}} Category:Trivial Rings Category:Null Ring tt65y4tpz8lxtj6f61x8ekhmnman5k1
A null ring is a trivial ring. So, by Trivial Ring is Commutative Ring, the result follows. {{qed}} Category:Examples of Rings qux1837zmt7hpbdfrx5a964j2w469mi
In Trivial Group is Group it is shown that the algebraic structure $\struct {\set e, \circ}$ such that $e \circ e = c$ is in fact a group. It remains to be shown that it is cyclic. In order for $G$ to be a cyclic group, every element $x$ of $G$ has to be expressible in the form $x = g^n$ for some $g \in G$ and some $n \in \Z$. In this case, for every integer $n$, every element of $G$ can be expressed in the form $e^n$. Thus $G$ is trivially a cyclic group. Thus $G$ is abelian from Cyclic Group is Abelian. {{qed}}
First we need to show that a trivial ring is actually a ring in the first place. Taking the ring axioms in turn: === $(\text A)$: Ring Addition forms Group === $\struct {R, +}$ is a group: This follows from the definition. {{qed|lemma}} === $(\text M 0)$: Closure of Ring Product === $\struct {R, \circ}$ is closed: From Ring Product with Zero, we have $x \circ y = 0_R \in R$. {{qed|lemma}} === $(\text M 1)$: Associativity of Ring Product === $\circ$ is associative on $\struct {R, +, \circ}$: :$x \circ \paren {y \circ z} = 0_R = \paren {x \circ y} \circ z$ {{qed|lemma}} === $(\text D)$: Distributivity of Ring Product over Addition === $\circ$ distributes over $+$ in $\struct {R, +, \circ}$: :$x \circ \paren {y + z} = 0_R$ by definition. Then: {{begin-eqn}} {{eqn | l = x \circ y + x \circ z       | r = 0_R + 0_R       | c =  }} {{eqn | r = 0_R       | c = {{Defof|Ring Zero}} }} {{end-eqn}} and the same for $\paren {y + z} \circ x$. {{qed|lemma}} === Commutative === From the definition of trivial ring: :$\forall x, y \in R: x \circ y = 0_R = y \circ x$ Hence its commutativity.  {{qed}}
=== Part $(1a)$ === This is shown on Congruence Relation on Ring induces Ideal. {{qed|lemma}} === Part $(2a)$ === This is shown on Ideal induced by Congruence Relation defines that Congruence. {{qed|lemma}} === Part $(1b)$ === This is shown on Ideal induces Congruence Relation on Ring. {{qed|lemma}} === Part $(2b)$ === {{finish|Argument previously here didn't prove anything. This part is genuinely different from the other three}} Category:Ideal Theory Category:Congruence Relations 87sxahizg95u7ell6r5i6eiy63r1vs9