hcju/proofwikiqa huggingface dataset:
The proof proceeds by strong induction on $n$. For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :$\map {\Phi_n} x$ has integer coefficients === Basis for the Induction === By First Cyclotomic Polynomial: :$\map {\Phi_1} x = x - 1$ Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true. This is the induction hypothesis: :For all $j$ such that $0 \le j \le k$, $\map {\Phi_j} x$ has integer coefficients from which it is to be shown that: :$\map {\Phi_{k + 1} } x$ has integer coefficients === Induction Step === This is the induction step: Suppose $\map P j$ holds for all $j \le k$. From Product of Cyclotomic Polynomials: :$\displaystyle \prod_{d \mathop \divides {k + 1} } \map {\Phi_d} x = x^k - 1$ By the induction hypothesis: :$\displaystyle \prod_{\substack {d \mathop \divides k \\ d \mathop \ne k}} \map {\Phi_d} x$ is a monic polynomial with integer coefficients, and thus primitive. From Content of Polynomial is Multiplicative it follows that $\map {\Phi_{k + 1} } x$ has integer coefficients. {{explain|Determine the exact nature of what the content of the polynomial is in this context}} {{qed}} Category:Cyclotomic Polynomials 2aur7jochp1371xk6wj0jpt8bu97j0q
Both sides are bilinear functions of $f$ and $g$, so it suffices to verify the equality in the case where $f \left({X}\right) = X^n$ and $g \left({X}\right) = X^m$. Then: :$\left({X^n X^m}\right)' = \left({n + m}\right) X^{n + m}$ and: :$\left({X^n}\right)' X^m + X^n \left({X^m}\right)' = n X^n X^m + m X^n X^m$ {{qed}} Category:Polynomial Theory r3gdqv26fn051y9ma6lul15778kqaph
Because $a$ has multiplicity at least $2$, we can write: : $f \left({X}\right) = \left({X - a}\right)^2 g \left({X}\right)$ with $g \left({X}\right) \in R \left[{X}\right]$. From Formal Derivative of Polynomials Satisfies Leibniz's Rule: :$f' \left({X}\right) = 2 \left({X - a}\right) g \left({X}\right) + \left({X - a}\right)^2 g' \left({X}\right)$ and thus: :$f' \left({a}\right) = 0$ {{qed}} Category:Polynomial Theory 1sf2w1ms5ssraoogsc824daz8mq01ov
Let $k$ be the order of $a$ modulo $p$. By Element to Power of Multiple of Order is Identity, $k \divides p - 1$. If $k = n$, the result follows. Let $k < n$. Then by Product of Cyclotomic Polynomials, there exists $d \divides k$ such that $p \divides \map {\Phi_d} a$. Consequently, $a$ is a double root of $\Phi_d \Phi_n$ modulo $p$. Again by Product of Cyclotomic Polynomials, $a$ is a double root of $x^n - 1$ modulo $p$. Thus, by Double Root of Polynomial is Root of Derivative, $a$ is a root of the derivative of $x^n - 1$ modulo $p$, which is the constant polynomial $n$. Thus $n \equiv 0 \pmod p$, for a nonzero constant polynomial has no roots. {{qed}}
Then: {{begin-eqn}} {{eqn | l = r^{\psi \mathop + \tau}       | r = \map \exp {\paren {\psi + \tau} \ln r}       | c = {{Defof|Power (Algebra)/Complex Number/Principal Branch/Positive Real Base|Principal Branch of Positive Real Number raised to Complex Number}} }} {{eqn | r = \map \exp {\psi \ln r + \tau \ln r} }} {{eqn | r = \map \exp {\psi \ln r} \, \map \exp {\tau \ln r}       | c = Exponential of Sum }} {{eqn | r = r^\psi \times r^\tau       | c = {{Defof|Power (Algebra)/Complex Number/Principal Branch/Positive Real Base|Principal Branch of Positive Real Number raised to Complex Number}} }} {{end-eqn}} {{qed}} Category:Powers k3hqknj348ae2u78j6idayayc85fnij
Combine Free Module is Isomorphic to Free Module Indexed by Set and Universal Property of Free Module on Set. {{handwaving}} Category:Module Theory Category:Universal Properties 7gxxsstllxvoymdm4wgtb13rifn7q8y
From Euler Phi Function of Prime: :$\map \phi 7 = 7 - 1 = 6$ From Euler Phi Function of Prime Power: :$\map \phi 9 = \map \phi {3^2} = 2 \times 3^{2 - 1} = 6 = \map \phi 7$ From the corollary to Euler Phi Function of Prime Power: :$\map \phi 8 = \map \phi {2^3} = 2^{3 - 1} = 4$ From Euler Phi Function of Integer: :$\map \phi {10} = \map \phi {2 \times 5} = 10 \paren {1 - \dfrac 1 2} \paren {1 - \dfrac 1 5} = 10 \times \dfrac 1 2 \times \dfrac 4 5 = 4 = \map \phi 8$ {{qed}}
From Sum of Sequence of Squares: {{:Sum of Sequence of Squares}} Thus: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 7}^{29} i^2       | r = \sum_{i \mathop = 1}^{29} i^2 - \sum_{i \mathop = 1}^6 i^2       | c =  }} {{eqn | r = \frac {29 \left({29 + 1}\right) \left({2 \times 29 + 1}\right)} 6 - \frac {6 \left({6 + 1}\right) \left({2 \times 6 + 1}\right)} 6       | c =  }} {{eqn | r = \frac {29 \times 30 \times 59} 6 - \frac {6 \times 7 \times 13} 6       | c =  }} {{eqn | r = 8 \, 555 -  91       | c =  }} {{eqn | r = 8 \, 464       | c =  }} {{eqn | r = 92^2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 7}^{39} i^2       | r = \sum_{i \mathop = 1}^{39} i^2 - \sum_{i \mathop = 1}^6 i^2       | c =  }} {{eqn | r = \frac {39 \left({39 + 1}\right) \left({2 \times 39 + 1}\right)} 6 - \frac {6 \left({6 + 1}\right) \left({2 \times 6 + 1}\right)} 6       | c =  }} {{eqn | r = \frac {39 \times 40 \times 79} 6 - \frac {6 \times 7 \times 13} 6       | c =  }} {{eqn | r = 20 \, 540 -  91       | c =  }} {{eqn | r = 20 \, 449       | c =  }} {{eqn | r = 143^2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 7}^{56} i^2       | r = \sum_{i \mathop = 1}^{56} i^2 - \sum_{i \mathop = 1}^6 i^2       | c =  }} {{eqn | r = \frac {56 \left({56 + 1}\right) \left({2 \times 56 + 1}\right)} 6 - \frac {6 \left({6 + 1}\right) \left({2 \times 6 + 1}\right)} 6       | c =  }} {{eqn | r = \frac {56 \times 57 \times 113} 6 - \frac {6 \times 7 \times 13} 6       | c =  }} {{eqn | r = 60 \, 116 -  91       | c =  }} {{eqn | r = 60 \, 025       | c =  }} {{eqn | r = 245^2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 7}^{190} i^2       | r = \sum_{i \mathop = 1}^{190} i^2 - \sum_{i \mathop = 1}^6 i^2       | c =  }} {{eqn | r = \frac {190 \left({190 + 1}\right) \left({2 \times 190 + 1}\right)} 6 - \frac {6 \left({6 + 1}\right) \left({2 \times 6 + 1}\right)} 6       | c =  }} {{eqn | r = \frac {190 \times 191 \times 381} 6 - \frac {6 \times 7 \times 13} 6       | c =  }} {{eqn | r = 2 \, 304 \, 415 -  91       | c =  }} {{eqn | r = 2 \, 304 \, 324       | c =  }} {{eqn | r = 1 \, 518^2       | c =  }} {{end-eqn}} {{qed}}
=== Existence === By Morphism from Ring with Unity to Module, for all $i$ there exists a morphism $\psi_i:R\to M$ with $\psi_i(1)=m_i$. By Universal Property of Direct Sum of Modules, there exists a morphism $\Psi:R^{(I)}\to M$ such that $\Psi\circ\iota_i=\psi_i$ for all $i$. Thus $\Psi(e_i)=\Psi(\iota_i(1))=\psi_i(1)=m_i$ for all $i$. We have $(r_i)_{i\in I}=\sum_{i\in I}r_ie_i$, so the expression for $\Psi$ follows by linearity. === Uniqueness === Let $\Psi$ be such a morphism. Then $\Psi\circ\iota_i$ sends $1$ to $m_i$. By Morphism from Ring with Unity to Module, $\Psi\circ\iota_i=\psi_i$, with $\psi_i$ as above. By Universal Property of Direct Sum of Modules, $\Psi$ is determined by $\Psi\circ\iota_i$. Thus $\Psi$ is unique. {{qed}} Category:Module Theory Category:Direct Products Category:Universal Properties p7el9r1q9guxfa9y6efy0lo3ejn7eh2
=== Sufficient Condition === Let $\mathcal R$ be multiplicative. Let $x \in S$. Let $a, b \in {\mathcal R}\left({x}\right)$. By definition of $\mathcal R$-image of element: :$\left({x, a}\right), \left({x, b}\right) \in \mathcal R$ By definition of multiplicative relation: :$\left({x, a \wedge b}\right) \in \mathcal R$ By definition of $\mathcal R$-image of element: :$a \wedge b \in {\mathcal R}\left({x}\right)$ By Meet Precedes Operands: :$a \wedge b \preceq a$ and $a \wedge b \preceq b$ Thus :$\exists c \in {\mathcal R}\left({x}\right): c \preceq a \land c \preceq b$ Hence ${\mathcal R}\left({x}\right)$ is filtered. {{qed|lemma}} === Necessary Condition === Suppose that :for all $x \in S$: ${\mathcal R}\left({x}\right)$ is filtered Let $a, x, y \in S$ such that :$\left({a, x}\right), \left({a, y}\right) \in \mathcal R$ By definition of $\mathcal R$-image of element: :$x, y \in {\mathcal R}\left({a}\right)$ By Auxiliary Relation Image of Element is Upper Set: :${\mathcal R}\left({a}\right)$ is upper set. By assumption: :${\mathcal R}\left({a}\right)$ is filtered. By Filtered in Meet Semilattice: :$x \wedge y \in {\mathcal R}\left({a}\right)$ Thus by definition of $\mathcal R$-image of element: :$\left({a, x \wedge y}\right) \in \mathcal R$ {{qed}}
For $\left\langle{r_i}\right\rangle_{i \mathop \in I} \in R^{\left({I}\right)}$ we have: :$\Psi \left({\left\langle{r_i}\right\rangle_{i \mathop \in I}}\right) = \displaystyle \sum_{i \mathop \in I} m_i r_i$ Thus $\Psi$ is surjective {{iff}} every element of $M$ is a linear combination of $S$. {{qed}} Category:Module Theory pgaxjw8u1428mvui4ro0gn8dhdxm9jr
We have: :$\Psi(\left\langle{r_i}\right\rangle_{i \mathop \in I}) = 0$ {{iff}}: :$\displaystyle \sum_{i \mathop \in I} r_i m_i = 0$ {{explain|Justify the above statement.}} Thus  injectivity and linearly independent are equivalent. {{explain|Justify the above statement.}} {{qed}} Category:Module Theory 2u5c0gexeulic4vtazjnvahb6v1ek06
Follows directly from: : Characterisation of Linearly Independent Set through Free Module Indexed by Set : Characterisation of Spanning Set through Free Module Indexed by Set. {{qed}} Category:Module Theory d3onr5592y2nzbayp16sgunznhp5emy
From Direct Product of Modules is Module, $M$ is an $R$-module. It remains to verify that: :$\forall x \in M: 1 \circ x = x$ We have: {{begin-eqn}} {{eqn | l = 1_R \circ (m_i)_{i \in I}       | r = \left({ 1_R \circ_i m_i }\right)_{i \in I} }} {{eqn | r = (m_i)_{i \in I} }} {{end-eqn}} {{qed}} Category:Module Theory Category:Direct Products 7icx8rqtkyen4nvys9dfq0v8ggic4qi
From Finite Direct Product of Modules is Module we have that $G$ is a module. It remains to be shown that: :$\forall x \in G: 1_R \circ x = x$ Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$. Then: {{begin-eqn}} {{eqn | l = 1_R \circ x       | r = 1_R \circ \tuple {x_1, x_2, \ldots, x_n}       | c =  }} {{eqn | r = \tuple {1_R \circ x_1, 1_R \circ x_2, \ldots, 1_R \circ x_n}       | c =  }} {{eqn | r = \tuple {x_1, x_2, \ldots, x_n}       | c =  }} {{eqn | r = x       | c =  }} {{end-eqn}} Hence the result. {{qed}}
Let $a \in R\left({x}\right), b \in S$ such that :$a \preceq b$ By definition of $R$-image of element: :$\left({x, a}\right) \in R$ By definition of reflexivity: :$x \preceq x$ By definition of auxiliary relation: :$\left({x, b}\right) \in R$ Thus by definition of $R$-image of element: :$b \in R \left({x}\right)$ {{qed}}
=== Sufficient Condition === Let $\mathcal R$ be multiplicative. Let $a, b, x, y \in S$ such that :$\left({a, x}\right), \left({b, y}\right) \in \mathcal R$ By Meet Precedes Operands: :$a \wedge b \preceq a$ and $a \wedge b \preceq b$ By definition of reflexivity: :$x \preceq x$ and $y \preceq y$ By definition of auxiliary relation: :$\left({a \wedge b, x}\right), \left({a \wedge b, y}\right) \in \mathcal R$ Thus by definition of multiplicative relation: :$\left({a \wedge b, x \wedge y}\right) \in \mathcal R$ {{qed|lemma}} === Necessary Condition === Suppose that :$\forall a, b, x, y \in S: \left({a, x}\right), \left({b, y}\right) \in \mathcal R \implies \left({a \wedge b, x \wedge y}\right) \in \mathcal R$ Let $a, x, y \in S$ such that :$\left({a, x}\right), \left({a, y}\right) \in \mathcal R$ By Meet is Idempotent: :$a \wedge a = a$ Thus by assumption: :$\left({a, x \wedge y}\right) \in \mathcal R$ {{qed}}
This is a specific instance of 1 plus Perfect Power is not Prime Power except for 9. {{qed}}
This is a direct consequence of 1 plus Perfect Power is not Prime Power except for 9. {{qed}}
Let $N$ be divisible by $8$. Then: {{begin-eqn}} {{eqn | l = N       | o = \equiv       | r = 0 \pmod 8 }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^n a_k 10^k       | o = \equiv       | r = 0 \pmod 8 }} {{eqn | ll= \leadstoandfrom       | l = 100 a_2 + 10 a_1 + a_0 + 10^3 \sum_{k \mathop = 3}^n a_k 10^{k - 3}       | o = \equiv       | r = 0 \pmod 8 }} {{eqn | ll= \leadstoandfrom       | l = 100 a_2 + 10 a_1 + a_0       | o = \equiv       | r = 0 \pmod 8       | c = as $10^3 \equiv 0 \pmod 8$ }} {{end-eqn}} {{qed}}
=== Definition 1 implies Definition 2 === Let $O_n$ be an octagonal number by definition 1. Let $n = 0$. By definition: :$O_0 = 0$ By vacuous summation: :$\displaystyle O_0 = \sum_{i \mathop = 1}^0 \left({6 i - 5}\right) = 0$ By definition of summation: {{begin-eqn}} {{eqn | l = O_{n - 1}       | r = \sum_{i \mathop = 1}^{n - 1} \left({6 i - 5}\right)       | c =  }} {{eqn | r = 1 + 7 + \cdots + 6 \left({n - 1}\right) - 5       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = O_n       | r = O_{n - 1} + 6 n - 5       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \left({6 i - 5}\right)       | c =  }} {{eqn | r = 1 + 7 + \cdots + \left({6 n - 5}\right)       | c =  }} {{end-eqn}} Thus $O_n$ is an octagonal number by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $O_n$ be an octagonal number by definition 2. Then: {{begin-eqn}} {{eqn | l = O_n       | r = \sum_{i \mathop = 1}^n \left({6 i - 5}\right)       | c =  }} {{eqn | r = 1 + 7 + \cdots + \left({6 \left({n - 1}\right) - 5}\right) + \left({6 n - 5}\right)       | c =  }} {{eqn | r = O_{n - 1} + 6 n - 5       | c =  }} {{end-eqn}} Then: :$\displaystyle O_0 = \sum_{i \mathop = 1}^0 \left({6 i - 5}\right)$ is a vacuous summation and so: :$O_0 = 0$ Thus $O_n$ is an octagonal number by definition 1. {{qed|lemma}} === Definition 1 equivalent to Definition 3 === We have by definition that $O_n = 0 = P \left({8, n}\right)$. Then: {{begin-eqn}} {{eqn | lo= \forall n \in \N_{>0}:       | l = P \left({8, n}\right)       | r = P \left({8, n - 1}\right) + \left({8 - 2}\right) \left({n - 1}\right) + 1       | c =  }} {{eqn | r = P \left({8, n - 1}\right) + 6 \left({n - 1}\right) + 1       | c =  }} {{eqn | r = P \left({8, n - 1}\right) + 6 n - 6 + 1       | c =  }} {{eqn | r = P \left({8, n - 1}\right) + 6 n - 5       | c =  }} {{end-eqn}} Thus $P \left({8, n}\right)$ and $O_n$ are generated by the same recurrence relation. {{qed}} Category:Octagonal Numbers jgqxa1jcb5ntwvz5kx10cdqfzi1uzq9
Octagonal numbers are $k$-gonal numbers where $k = 8$. From Closed Form for Polygonal Numbers we have that: :$P \left({k, n}\right) = \dfrac n 2 \left({\left({k - 2}\right) n - k + 4}\right)$ Hence: {{begin-eqn}} {{eqn | l = O_n       | r = \frac n 2 \left({\left({8 - 2}\right) n - 8 + 4}\right)       | c = Closed Form for Polygonal Numbers }} {{eqn | r = n \left({3 n - 2}\right)       | c =  }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \cmod {t^z}       | r = \cmod {t^{\map \Re z + i \map \Im z} } }} {{eqn | r = \cmod {t^{\map \Re z} t^{i \map \Im z} }       | c = Sum of Complex Indices of Real Number }} {{eqn | r = \cmod {t^{\map \Re z} } \cmod {t^{i \map \Im z} }       | c = Complex Modulus of Product of Complex Numbers }} {{eqn | r = \cmod {t^{\map \Re z} }       | c = Modulus of Exponential of Imaginary Number is One:Corollary }} {{eqn | r = t^{\map \Re z}       | c = Power of Positive Real Number is Positive }}  {{end-eqn}} {{qed}} Category:Complex Modulus Category:Complex Analysis ix6871u9khqhoqtc8e7ei0pqor9yd6y
=== Sufficient Condition === Let $p$ be pseudoprime element. Let $a, b \in S$ such that :$a \wedge b \ll p$ By definition of meet: :$\inf \left\{ {a, b}\right\} \ll p$ By Characterization of Pseudoprime Element by Finite Infima: :$\exists c \in \left\{ {a, b}\right\}: c \preceq p$ Thus :$a \preceq p$ or $b \preceq p$ {{qed|lemma}} === Necessary Condition === Suppose :$\forall a, b \in S: a \wedge b \ll p \implies a \preceq p \lor b \preceq p$ Aiming for a contradiction suppose that :$p$ is not a pseudoprime element. By Prime is Pseudoprime (Order Theory): :$p$ is not a prime element. By definition of prime element: :$\exists x, y \in S: x \wedge y \preceq p$ and $x \npreceq p$ and $y \npreceq p$ By definition of continuous: :$\forall z \in S: z^\ll$ is directed. and :$L$ satisfies axiom of approximation. By Axiom of Approximation in Up-Complete Semilattice: :$\exists u \in S: u \ll x \land u \npreceq p$ and :$\exists v \in S: v \ll y \land v \npreceq p$ By Way Below Relation is Auxiliary Relation: :$\ll$ is auxiliary relation. By Multiplicative Auxiliary Relation iff Congruent: :$u \wedge v \ll x \wedge y$ By definition of transitivity: :$u \wedge v \ll p$ By assumption: :$u \preceq p$ or $v \preceq p$ This contradicts $u \npreceq p$ and $v \npreceq p$ {{qed}}
To any deltahedron, it is possible to attach a regular tetrahedron to any of its faces to create a new deltahedron with $2$ more faces.
Let $p$ be a pseudoprime element. Aiming for a contradiction suppose that :$p$ is not a prime element. By definition of prime element: :$\exists x, y \in S: x \wedge y \preceq p$ and $x \npreceq p$ and $y \npreceq p$ By definition of continuous: :$\forall z \in S: z^\ll$ is directed. and :$L$ satisfies axiom of approximation. By Axiom of Approximation in Up-Complete Semilattice: :$\exists u \in S: u \ll x \land u \npreceq p$ and :$\exists v \in S: v \ll y \land v \npreceq p$ By Way Below Relation is Auxiliary Relation: :$\ll$ is auxiliary relation. By Multiplicative Auxiliary Relation iff Congruent: :$u \wedge v \ll x \wedge y$ By definition of transitivity: :$u \wedge v \ll p$ By Characterization of Pseudoprime Element when Way Below Relation is Multiplicative: :$u \preceq p$ or $v \preceq p$ This contradicts $u \npreceq p$ and $v \npreceq p$ {{qed}}
Let $a, x, y \in S$ such that :$a \ll x$ and $a \ll y$ Aiming for a contradiction suppose that :$a \not\ll x \wedge y$ We will prove that :$\forall z \in S: z \in \left({x \wedge y}\right)^\ll \implies z \notin a^\succeq$ Let $z \in S$ such that :$z \in \left({x \wedge y}\right)^\ll$ By definition of way below closure: :$z \ll x \wedge y$ Aiming for a contradiction suppose that :$z \in a^\succeq$ By definition of upper closure of element: :$a \preceq z$ By Preceding and Way Below implies Way Below :$a \ll x \wedge y$ This contradicts $a \not\ll x \wedge y$ {{qed|lemma}} By definitions of empty set and intersection: :$\left({x \wedge y}\right)^\ll \cap a^\succeq = \varnothing$ By Way Below Closure is Ideal in Bounded Below Join Semilattice: :$\left({x \wedge y}\right)^\ll$ is ideal in $L$. By Upper Closure of Element is Filter: :$a^\succeq$ is filter on $L$. By If Ideal and Filter are Disjoint then There Exists Prime Ideal Including Ideal and Disjoint from Filter: :there exists prime ideal $P$ in $L$: :$\left({x \wedge y}\right)^\ll \subseteq P$ and $P \cap a^\succeq = \varnothing$ By definition of pseudoprime element: :$\sup P$ is pseudoprime. By assumption: :$\sup P$ is prime. By definition of reflexivity: :$a \preceq a$ By definition of upper closure of element: :$a \in a^\succeq$ By definition of axiom of approximation: :$\sup \left({x \wedge y}\right)^\ll = x \wedge y$ By definition of up-complete: :$\left({x \wedge y}\right)^\ll$ admits a supremum and :$P$ admits a supremum. By Supremum of Subset: :$x \wedge y \preceq \sup P$ By definition of prime element: :$x \preceq \sup P$ or $y \preceq \sup P$ By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal: :$a \in P$ By definition of intersection: :$a \in P \cap a^\succeq$ This contradicts $P \cap a^\succeq = \varnothing$ {{qed}}
=== Sufficient Condition === Suppose : $x^\succeq$ is a way below open filter on $L$ By definitions of upper closure of element and reflexivity: :$x \in x^\succeq$ By definition of way below open: :$\exists y \in x^\succeq: y \ll x$ By definition of upper closure of element: :$x \preceq y$ By Preceding and Way Below implies Way Below and definition of reflexivity: :$x \ll x$ By definition: :$x$ is compact. {{qed|lemma}} === Necessary Condition === Let $x$ be compact. By Upper Closure of Element is Filter: :$x^\succeq$ is filter on $L$. It remains to show that :$x^\succeq$ is way below open. Let $u \in x^\succeq$. By definition of upper closure of element: :$x \preceq u$ and by definition of reflexivity: :$x \in x^\succeq$ By definition of compact: :$x \ll x$ By Preceding and Way Below implies Way Below: :$x \ll u$ Thus :$\exists z \in x^\succeq: z \ll u$ {{qed}}
The expression for the $n$th square number is: :$n^2$ for $n \in \Z$. The expression for the $n$th cube number is: :$n^3$ again, for $n \in \Z$. Therefore the closed-form expression for the $n$th sum of two consecutive cubes is: :$n^3 + \paren {n + 1}^3$ This simplifies to: :$2 n^3 + 3 n^2 + 3 n + 1$ Equate the two expressions with a variable replacing $n$: {{begin-eqn}} {{eqn | l = y^2       | r = 2 x^3 + 3 x^2 + 3 x + 1       | c = }} {{eqn | ll= \leadsto       | l = 4 y^2       | r = 8 x^3 + 12 x^2 + 12 x + 4       | c = }} {{eqn | ll= \leadsto       | l = \paren {2 y}^2       | r = \paren {2 x}^3 + 3 \paren {4 x^2 + 4 x + 1} + 1       | c = }} {{eqn | ll= \leadsto       | l = \paren {2 y}^2       | r = \paren {2 x}^3 + 1^3 + 3 \paren {2 x + 1}^2       | c = }} {{eqn | ll= \leadsto       | l = \paren {2 y}^2       | r = \paren {2 x + 1} \paren { \paren {2 x}^2 -2 x + 1} + 3 \paren {2 x + 1}^2       | c = }} {{eqn | ll= \leadsto       | l = \paren {2 y}^2       | r = \paren {2 x + 1} \paren { \paren {2 x}^2 -2 x + 1 + 3 \paren {2 x + 1} }       | c = }} {{eqn | ll= \leadsto       | l = \paren {2 y}^2       | r = \paren {2 x + 1} \paren {4 x^2 + 4x + 1 + 3 }       | c = }} {{eqn | ll= \leadsto       | l = \paren {2 y}^2       | r = \paren {2 x + 1} \paren {\paren {2x + 1}^2 + 3}       | c = }} {{end-eqn}} Substituting: :$u = 2 x + 1$ :$v = 2 y$ The equation becomes: :$v^2 = u^3 + 3 u$ We then apply Integral Points of Elliptic Curve $y^2 = x^3 + 3 x$: {{:Integral Points of Elliptic Curve y^2 = x^3+3x}} We then write them in terms of $x$ and $y$: :$\tuple {-\frac 1 2, 0}, \tuple {0, \pm 1}, \tuple {1, \pm 3}, \tuple {\frac {11} 2, \pm 21}$ Due to the restrictions on the variables, solutions with non-integer inputs are invalid. This leaves $4$ solutions: :$\tuple {0, \pm 1}, \tuple {1, \pm 3}$ as follows: :$\paren {\pm1}^2 = 1 = 0^3 + 1^3$ :$\paren {\pm3}^2 = 9  = 1^3 + 2^3$ {{qed}}
By definition of compact subset: :$k$ is compact. By definition of compact: :$k \ll k$ Thus by Preceding and Way Below implies Way Below: :$x \ll y$ {{qed}}
From Five Platonic Solids, there exist $5$ regular polyhedra which are convex: :the regular tetrahedron :the cube :the regular octahedron :the regular dodecahedron :the regular icosahedron. From Four Kepler-Poinsot Polyhedra: {{:Four Kepler-Poinsot Polyhedra}} All $4$ of the above are regular polyhedra which are non-convex. making the total $9$.
Let $a, b, x, y \in S$ such that $a \ll x$ and $b \ll y$ By Join Succeeds Operands: :$x \preceq x \vee y$ and $y \preceq x \vee y$ By Preceding and Way Below implies Way Below and definition of reflexivity: :$a \ll x \vee y$ and $b \ll x \vee y$ Thus by Join is Way Below if Operands are Way Below: :$a \vee b \ll x \vee y$ {{qed}}
=== Lemma === {{:Lifting The Exponent Lemma/Lemma}} ---- Let $k = \nu_p \left({n}\right)$. Then $n = p^k m$ with $p \nmid m$. By P-adic Valuation of Difference of Powers with Coprime Exponent: :$\nu_p \left({x^n - y^n}\right) = \nu_p \left({x^{p^k} - y^{p^k} }\right)$ By repeatedly applying the /Lemma/: :$\nu_p \left({x^{p^k} - y^{p^k} }\right) = \nu_p \left({x - y}\right) + k$ {{qed}}
Let $x, y \in \map K L$. By definition of compact subset: :$x$ and $y$ are compact. By definition of compact: :$x \ll x$ and $y \ll y$ By Way Below is Congruent for Join: :$x \vee y \ll x \vee y$ By definition: :$x \vee y$ is compact. Thus by definition compact subset: :$x \vee y \in \map K L$ {{qed}}
By the definition of total ordering, $\preceq$ is connected. As $x \ne 0_R$, one of the following is true, but not both: :$(1): \quad 0_R \prec x$ :$(2): \quad x \prec 0_R$ If $(2)$, because $\prec$ is compatible with $+$: {{begin-eqn}} {{eqn | l = x       | o = \prec        | r = 0_R }} {{eqn | ll= \leadsto       | l = x + \paren {-x}       | o = \prec       | r = 0_R + \paren {-x} }} {{eqn | ll= \leadsto       | l = 0_R       | o = \prec       | r = -x       | c = {{Defof|Ring Zero}}, {{Defof|Ring Negative}} }} {{end-eqn}} {{qed}} Category:Ordered Rings 8wp8cg2du1ssfeaen6ry67n4xlytf9l