hcju/proofwikiqa huggingface dataset:
Let $x \in \R$ such that $x > 0$. Then for all $k \in \Z$ such that $0 \le k < n$: :$x + \dfrac k n \in \Z_{> 0}$ and so from Membership of Set of Integers is Replicative Function: :$\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z_{> 0} } = \sqbrk {n x \in \Z_{> 0} }$ Let $x \le 0$. Then for all $k \in \Z$ such that $0 \le k < n$: :$x + \dfrac k n < 1$ and so: :$\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z_{> 0} } = 0 = \sqbrk {n x \in \Z_{> 0} }$ {{qed}}
Let $\map f x = 1$. Then: :$\exists r \in \Q, \exists m \in \Z: x = r \pi + m$ and: {{begin-eqn}} {{eqn | l = x       | r = r \pi + m       | c =  }} {{eqn | ll= \leadsto       | l = n x       | r = \paren {r n} \pi + m n       | c =  }} {{end-eqn}} But $r n \in Q$ and $m n \in \Z$. So: :$\map f x = 1 \implies \map f {n x} = 1$ Let $n \in \Z_{>0}$. Let $k \in \Z: 0 \le k < n$. Then: {{begin-eqn}} {{eqn | l = x + \frac k n       | r = r \pi + m + \frac k n        | c =  }} {{eqn | ll= \leadsto       | l = x + \frac k n       | r = \paren {r + \frac {k \pi} n} \pi + m       | c =  }} {{end-eqn}} Thus, when $0 \le k < n$ it follows that the coefficient of $\pi$ cannot be rational. Thus: :$\map f x = 1 \implies \map f {x + \dfrac k n} = 0$ and so: :$\displaystyle \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n} = \map f {n x}$ {{qed|lemma}} Now let $\map f {n x} = 1$. Then: :$r_n \pi + m_n = n x$ for some rational $r_n$ and integer $m_n$. Thus exactly one of $m_n + k$ where $0 \le n < k$ is a multiple of $n$. Therefore there exists exactly one $k \in \Z$ such that $0 \le n < k$ such that: :$\map f {x + \dfrac k n} = 1$ Therefore: :$\displaystyle \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n} = 1 = \map f {n x}$ Hence the result by definition of replicative function. {{qed}}
We have that: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n}       | r = \sum_{k \mathop = 0}^{n - 1} \log \, \size {2 \sin \pi \paren {x + \frac k n} }       | c =  }} {{eqn | r = \log \prod_{k \mathop = 0}^{n - 1} \size {2 \sin \pi \paren {x + \frac k n} }       | c = Sum of Logarithms }} {{end-eqn}} Thus to demonstrate that $f$ is replicative, it is sufficient to demonstrate that: :$\displaystyle \prod_{k \mathop = 0}^{n - 1} \left({2 \sin \pi \left({x + \frac k n}\right)}\right) = 2 \sin \pi n x$ By Product Formula for Sine: {{begin-eqn}} {{eqn | l = 2 \sin \pi n x       | r = 2^n \prod_{k \mathop = 0}^{n - 1} \paren {\map \sin {\pi x + \frac {k \pi} n} } }} {{eqn | r = \prod_{k \mathop = 0}^{n - 1} \paren {2 \sin \pi \paren {x + \frac k n} } }} {{end-eqn}} Hence the result. {{qed}}
We will prove that :$\forall x \in S: \exists I \in \mathit{Ids}\left({L}\right): x \preceq \sup I \land \forall J \in \mathit{Ids}\left({L}\right): x \preceq \sup J \implies I \subseteq J$  Let $x \in S$. Define $I := \inf \left({f^{-1}\left[{x^\succeq}\right]}\right)$. By definition of $P$: :$I \in \mathit{Ids}\left({L}\right)$ We will prove that :$\forall J \in \mathit{Ids}\left({L}\right): x \preceq \sup J \implies I \subseteq J$ Let $J \in \mathit{Ids}\left({L}\right)$ such that :$x \preceq \sup J$ By definition of $f$: :$x \preceq f\left({J}\right)$ By definition of upper closure of element: :$f\left({J}\right) \in x^\succeq$ By definition of image of set: :$J \in f^{-1}\left[{x^\succeq}\right]$ By definition of infimum: :$I \precsim J$ Hence by definition of $\precsim$: :$I \subseteq J$ {{qed|lemma}} By definition of upper adjoint of a Galois connection: :there exists a mapping $d: S \to \mathit{Ids}\left({L}\right)$: $\left({f, d}\right)$ is a Galois connection. By Galois Connection is Expressed by Minimum :$d\left({x}\right) = \min \left({f^{-1} \left[{x^\succeq}\right]}\right)$ By definition of smallest element: :$I \in f^{-1} \left[{x^\succeq}\right]$ By definition of image of set: :$f\left({I}\right) \in x^\succeq$ By definition of upper closure of element: :$x \preceq f\left({I}\right)$ Thus by definition of $f$: :$x \preceq \sup I$ Hence :$\exists I \in \mathit{Ids}\left({L}\right): x \preceq \sup I \land \forall J \in \mathit{Ids}\left({L}\right): x \preceq \sup J \implies I \subseteq J$  {{qed|lemma}} Hence by Continuous iff For Every Element There Exists Ideal Element Precedes Supremum: :$L$ is continuous. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n - 1} \left({f + g}\right) \left({x + \frac k n}\right)       | r = \sum_{k \mathop = 0}^{n - 1} \left({f \left({x + \frac k n}\right) + g \left({x + \frac k n}\right)}\right)       | c = {{Defof|Pointwise Sum}} }} {{eqn | r = \sum_{k \mathop = 0}^{n - 1} f \left({x + \frac k n}\right) + \sum_{k \mathop = 0}^{n - 1} g \left({x + \frac k n}\right)       | c =  }} {{eqn | r = f \left({n x}\right) + g \left({n x}\right)       | c = {{Defof|Replicative Function}} }} {{eqn | r = \left({f + g}\right) \left({n x}\right)       | c = {{Defof|Pointwise Sum}} }} {{end-eqn}} Hence the result by definition of replicative function. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n - 1} g \left({x + \frac k n}\right)       | r = \sum_{k \mathop = 0}^{n - 1} c f \left({x + \frac k n}\right)       | c =  }} {{eqn | r = c \sum_{k \mathop = 0}^{n - 1} f \left({x + \frac k n}\right)       | c =  }} {{eqn | r = c f \left({n x}\right)       | c = {{Defof|Replicative Function}} }} {{eqn | r = g \left({n x}\right)       | c =  }} {{end-eqn}} Hence the result by definition of replicative function. {{qed}}
From Closed Form for Triangular Numbers, for $n = 1, 3, 6, 10, \ldots$: :$n = \dfrac {a_n \left({a_n + 1}\right)} 2$ Thus by the Quadratic Formula: $a_n = \dfrac {-1 \pm \sqrt {1 + 8 n} } 2$ In this context it is the positive root that is required. The result follows by definition of ceiling function. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n a_k       | r = \sum_{k \mathop = 1}^n \left({k - \left({k - 1}\right)}\right) a_k       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n k a_k - \sum_{k \mathop = 1}^n \left({k - 1}\right) a_k       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n k a_k - \sum_{k \mathop = 0}^{n - 1} k a_{k + 1}       | c = Translation of Index Variable of Summation }} {{eqn | r = n a_n + \sum_{k \mathop = 1}^{n - 1} k a_k - \sum_{k \mathop = 1}^{n - 1} k a_{k + 1} + 0 \times a_0       | c = extracting the end terms }} {{eqn | r = n a_n + \sum_{k \mathop = 1}^{n - 1} k \left({a_k - a_{k + 1} }\right)       | c =  }} {{eqn | r = n a_n - \sum_{k \mathop = 1}^{n - 1} k \left({a_{k + 1} - a_k}\right)       | c =  }} {{end-eqn}} {{qed}}
From Abel's Lemma: Formulation 2, after renaming and reassigning variables: :$\displaystyle \sum_{k \mathop = 1}^n a_k b_k = \sum_{k \mathop = 1}^{n - 1} A_k \left({a_k - a_{k + 1} }\right) + A_n a_n$ where: :$\left \langle {a} \right \rangle$ and $\left \langle {b} \right \rangle$ be sequences in $\R$ :$\displaystyle A_n = \sum_{k \mathop = 1}^n {b_k}$ be the partial sum of $\left \langle {b} \right \rangle$ from $1$ to $n$. Let $\left \langle{b}\right \rangle$ be defined as: :$\forall k: b_k = 1$ Thus: :$\displaystyle A_n = \sum_{k \mathop = 1}^n 1 = n$ and so: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n a_k       | r = \sum_{k \mathop = 1}^{n - 1} k \left({a_k - a_{k + 1} }\right) + n a_n       | c =  }} {{eqn | r = n a_n - \sum_{k \mathop = 1}^{n - 1} k \left({a_{k + 1} - a_k}\right)       | c =  }} {{end-eqn}} {{qed}}
By definition of way below closure: :$\forall c \in a^\ll: c \preceq b$ By definition: :$b$ is an upper bound for $a^\ll$ By definition of supremum: :$\sup \left({a^\ll}\right) \preceq b$ By definition of continuous: :$L$ satisfies axiom of approximation. Thus by axiom of approximation: :$a = \sup \left({a^\ll}\right) \preceq b$ {{qed}}
From Sum of Sequence as Summation of Difference of Adjacent Terms: :$\displaystyle \sum_{k \mathop = 1}^n \left \lfloor{\log_b k}\right \rfloor = n \left \lfloor{\log_b n}\right \rfloor - \sum_{k \mathop = 1}^{n - 1} k \left({\left \lfloor{\log_b {k + 1} }\right \rfloor - \left \lfloor{\log_b k}\right \rfloor}\right)$ Let $S$ be defined as: :$\displaystyle S := \sum_{k \mathop = 1}^{n - 1} k \left({\left \lfloor{\log_b {k + 1} }\right \rfloor - \left \lfloor{\log_b k}\right \rfloor}\right)$ As $b \ge 2$, we have that: :$\log_b {k + 1} - \log_b k < 1$ As $b$ is an integer: :$\left \lfloor{\log_b {k + 1} }\right \rfloor - \left \lfloor{\log_b k}\right \rfloor = 1$ {{iff}} $k + 1$ is a power of $b$. So: {{begin-eqn}} {{eqn | l = S       | r = \sum_{k \mathop = 1}^{n - 1} k \left[{k + 1 \text{ is a power of } b}\right]       | c =  }} {{eqn | r = \sum_{1 \mathop \le t \mathop < \log_b n} \left({b^t - 1}\right)       | c =  }} {{eqn | r = \sum_{1 \mathop \le t \mathop < \log_b n} b^t - \left \lfloor{\log_b n}\right \rfloor       | c =  }} {{eqn | r = \frac {b^{\left \lfloor{\log_b n}\right \rfloor + 1} - 1} {b - 1} - \left \lfloor{\log_b n}\right \rfloor       | c =  }} {{end-eqn}} Hence the result. {{qed}}
From Sum of Sequence as Summation of Difference of Adjacent Terms: :$\displaystyle \sum_{k \mathop = 1}^n \left \lfloor{\sqrt k}\right \rfloor = n \left \lfloor{\sqrt n}\right \rfloor - \sum_{k \mathop = 1}^{n - 1} k \left({\left \lfloor{\sqrt {k + 1} }\right \rfloor - \left \lfloor{\sqrt k}\right \rfloor}\right)$ Let $S$ be defined as: :$\displaystyle S := \sum_{k \mathop = 1}^{n - 1} k \left({\left \lfloor{\sqrt {k + 1} }\right \rfloor - \left \lfloor{\sqrt k}\right \rfloor}\right)$ We have that: :$\sqrt {k + 1} - \sqrt k < 1$ and so: :$\left \lfloor{\sqrt {k + 1} }\right \rfloor - \left \lfloor{\sqrt k}\right \rfloor = 1$ {{iff}} $k + 1$ is a square number. So: {{begin-eqn}} {{eqn | l = S       | r = \sum_{k \mathop = 1}^{n - 1} k \left[{k + 1 \text{ is a square number} }\right]       | c =  }} {{eqn | l = S       | r = \sum_{k \mathop = 2}^{n} (k - 1) \left[{k \text{ is a square number} }\right]       | c =  }} {{eqn | r = \sum_{2 \mathop \le t \mathop \le \sqrt n} \left({t^2 - 1}\right)       | c = $1$ is not included in this summation }} {{eqn | r = \sum_{1 \mathop \le t \mathop \le \sqrt n} t^2 - 1 - \sum_{2 \mathop \le t \mathop \le \sqrt n} 1       | c =  }} {{eqn | r = \sum_{1 \mathop \le t \mathop \le \sqrt n} t^2 - \sum_{1 \mathop \le t \mathop \le \sqrt n} 1       | c =  }} {{eqn | r = \frac {\left \lfloor{\sqrt n}\right \rfloor \left({\left \lfloor{\sqrt n}\right \rfloor + 1}\right) \left({2 \left \lfloor{\sqrt n}\right \rfloor + 1}\right)} 6 - \left \lfloor{\sqrt n}\right \rfloor       | c = Sum of Sequence of Squares  }} {{eqn | r = \frac {\left \lfloor{\sqrt n}\right \rfloor \left({\left \lfloor{\sqrt n}\right \rfloor + 1}\right) \left({2 \left \lfloor{\sqrt n}\right \rfloor + 1}\right) - 6 \left \lfloor{\sqrt n}\right \rfloor} 6       | c =  }} {{eqn | r = \left \lfloor{\sqrt n}\right \rfloor \frac {\left({\left \lfloor{\sqrt n}\right \rfloor + 1}\right) \left({2 \left \lfloor{\sqrt n}\right \rfloor + 1}\right) - 6} 6       | c =  }} {{eqn | r = \left \lfloor{\sqrt n}\right \rfloor \frac {2 \left \lfloor{\sqrt n}\right \rfloor^2 + 3 \left \lfloor{\sqrt n}\right \rfloor - 5} 6       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 1}^n \left \lfloor{\sqrt k}\right \rfloor       | r = n \left \lfloor{\sqrt n}\right \rfloor - \left \lfloor{\sqrt n}\right \rfloor \frac {\left({2 \left \lfloor{\sqrt n}\right \rfloor + 5}\right) \left({\left \lfloor{\sqrt n}\right \rfloor - 1}\right)} 6       | c =  }} {{eqn | r = \left \lfloor{\sqrt n}\right \rfloor \left({n - \dfrac {\left({2 \left \lfloor{\sqrt n}\right \rfloor + 5}\right) \left({\left \lfloor{\sqrt n}\right \rfloor - 1}\right)} 6 }\right)       | c =  }} {{end-eqn}} {{qed}}
Let $x, y \in S$ such that :$\top = x \wedge y$ By Meet Precedes Operands :$\top \preceq x$ and $\top \preceq y$ By definition of greatest element: :$x \preceq \top$ Thus by definition of antisymmetry: :$\top = x$ {{qed}}
=== Sufficient Condition === Let $x$ be meet irreducible. We will prove the result by induction on cardinality of $A$. ==== Base case ==== :for every non-empty subset $A$ of $S$: $\left\vert{A}\right\vert = 1 \land x = \inf A \implies x \in A$ Let $A$ be a non-empty subset of $S$ such that :$\left\vert{A}\right\vert = 1 \land x = \inf A$ Then :$A = \left\{ {a}\right\}$ By Infimum of Singleton: :$x = a$ Thus by definition of singleton: :$x \in A$ {{qed|lemma}} ==== Induction Hypothesis ==== :$n \ge 1$ and for every non-empty subset $A$ of $S$: $\left\vert{A}\right\vert = n \land x = \inf A \implies x \in A$ ==== Induction Step ==== :$n \ge 1$ and for every non-empty subset $A$ of $S$: $\left\vert{A}\right\vert = n+1 \land x = \inf A \implies x \in A$ Let $A$ be a non-empty subset of $S$ such that :$\left\vert{A}\right\vert = n+1 \land x = \inf A$ Then :$A = \left\{ {a_1, \dots, a_n, a_{n+1} }\right\}$ By definition of infimum: :$\inf A = \left({a_1 \wedge \dots \wedge a_n}\right) \wedge a_{n+1}$ By definition of meet irreducible: :$x = a_1 \wedge \dots \wedge a_n$ or $x = a_{n+1}$ By Induction Hypothesis: :$x \in \left\{ {a_1, \dots, a_n}\right\}$ or $x = a_{n+1}$ Thus by definition of unordered tuple: :$x \in A$ {{qed|lemma}} === Necessary Condition === Assume that :for every non-empty finite subset $A$ of $S$: $x = \inf A \implies x \in A$ Let $y, z \in S$ such that :$x = y \wedge z$ By definition of meet: :$x = \inf \left\{ {y, z}\right\}$ By assumption: :$x \in \left\{ {y, z}\right\}$ Thus by definition of unordered tuple: :$x = y$ or $x = z$ {{qed}}
We have that $\left \lfloor{\dfrac {n + j b^k} {b^{k + 1} } }\right \rfloor$ is in the form $\left \lfloor{\dfrac {m k + x} n}\right \rfloor$ so that: {{begin-eqn}} {{eqn | l = \sum_{1 \mathop \le j \mathop < b} \left \lfloor{\dfrac {n + j b^k} {b^{k + 1} } }\right \rfloor       | r = \sum_{1 \mathop \le j \mathop < b} \left \lfloor{\dfrac {j + \frac n {b^k} } b}\right \rfloor       | c =  }} {{eqn | r = \sum_{0 \mathop \le j \mathop < b} \left \lfloor{\dfrac {j + \frac n {b^k} } b}\right \rfloor - \left \lfloor{\dfrac n {b^{k + 1} } }\right \rfloor       | c =  }} {{eqn | r = \dfrac {\left({1 - 1}\right) \left({b - 1}\right)} 2 + \dfrac {\left({1 - 1}\right)} 2 + 1 \left \lfloor{\dfrac n {b^k} }\right \rfloor - \left \lfloor{\dfrac n {b^{k + 1} } }\right \rfloor       | c = (find link) }} {{eqn | r = \left \lfloor{\dfrac n {b^k} }\right \rfloor - \left \lfloor{\dfrac n {b^{k + 1} } }\right \rfloor       | c =  }} {{end-eqn}} {{LinkWanted}} Thus: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \left \lfloor{\dfrac {n + j b^k} {b^{k + 1} } }\right \rfloor       | r = \sum_{k \mathop \ge 0} \left({\left \lfloor{\dfrac n {b^k} }\right \rfloor - \left \lfloor{\dfrac n {b^{k + 1} } }\right \rfloor}\right)       | c =  }} {{eqn | r = \lim_{k \mathop \to \infty} \left \lfloor{\dfrac n 1}\right \rfloor - \left \lfloor{\dfrac n {b^{k + 1} } }\right \rfloor       | c = Telescoping Series }} {{eqn | r = n       | c =  }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = r       | m = \floor{\dfrac {m j} n}       | c =  }} {{eqn | ll= \leadsto       | l = r       | o = \le       | m = \dfrac {m j} n       | mo= <       | r = r + 1       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {r n} m       | o = \le       | m = j       | mo= <       | r = \dfrac {\paren {r + 1} } m       | c =  }} {{eqn | ll= \leadsto       | l = \ceiling {\dfrac {r n} m}       | o = \le       | m = j       | mo= <       | r = \ceiling {\dfrac {\paren {r + 1} } m}       | c = as $j$ is an integer }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | o =        | r = \sum_{0 \mathop \le j \mathop < n} \map f {\floor {\dfrac {m j} n} }       | c =  }} {{eqn | r = \sum_{0 \mathop \le \ceiling {\frac {r n} m} \mathop < n} \map f r       | c =  }} {{eqn | r = \sum_{0 \mathop \le r \mathop < m} \map f r \paren {\ceiling {\dfrac {\paren {r + 1} } m} - \ceiling {\dfrac {r n} m} }       | c =  }} {{eqn | r = \map f 0 \ceiling {\dfrac n m} + \map f 1 \paren {\ceiling {\dfrac {2 n} m} - \ceiling {\dfrac n m} } + \cdots + \map f {m - 1} \paren {\ceiling {\dfrac {m n} m} - \ceiling {\dfrac {\paren {m - 1} n} m} }       | c =  }} {{eqn | r = \ceiling {\dfrac n m} \paren {\map f 0 + \map f 1} + \ceiling {\dfrac {2 n} m} \paren {\map f 0 + \map f 1} + \cdots + \ceiling {\dfrac {\paren {m - 1} n} m} \paren {\map f {m - 2} + \map f {m - 1} } + n \map f {m - 1}       | c =  }} {{eqn | r = \sum_{0 \mathop \le r \mathop < m} \ceiling {\dfrac {r n} m} \paren {\map f {r - 1} + \map f r} + n \map f {m - 1}       | c =  }} {{end-eqn}} {{qed}}
From Integer to Power of $\dfrac {p - 1} 2$ Modulo $p$, one of the following cases holds: :$(1): \quad a^{\frac{(p - 1)} 2} \bmod p = 0$ :$(2): \quad a^{\frac{(p - 1)} 2} \bmod p = 1$ :$(3): \quad a^{\frac{(p - 1)} 2} \bmod p = p - 1$ Let $\paren {\dfrac a p}$ be defined as the Legendre symbol by definition 1: {{:Definition:Legendre Symbol/Definition 1}} By Negative Number is Congruent to Modulus minus Number we have: :$p - 1 \pmod p = -1$ Thus $\paren {\dfrac a p}$ is the Legendre symbol by definition 2: {{:Definition:Legendre Symbol/Definition 2}} {{qed}} Category:Legendre Symbol 6mmfykrvmn3tw8fytzql9zcxr16oxlz
Let $a, b \in S$. Aiming for a contradiction suppose that :$x = a \wedge b$ and $x \ne a$ and $x \ne b$ By Meet Precedes Operands: :$x \preceq b$ and $x \preceq a$ By definition of upper closure of element: :$b, a \in x^\succeq$ By definitions of singleton and difference: :$b, a \in x^\succeq \setminus \left\{ {x}\right\}$ By definition of filtered: :$\exists z \in x^\succeq \setminus \left\{ {x}\right\}: z \preceq a \land z \preceq b$ By definition of infimum: :$z \preceq x$ By definition of upper set: :$x \in x^\succeq \setminus \left\{ {x}\right\}$ Thus this contradicts $x \in \left\{ {x}\right\}$ by definition of singleton. {{qed}}
Let $x, y \in S$. Aiming for a contradiction suppose that :$p = x \wedge y$ and $p \ne x$ and $p \ne y$ By Meet Precedes Operands: :$p \preceq x$ and $p \preceq y$ By definition of $\prec$: :$p \prec x$ and $p \prec y$ We will prove that :$x \notin F$ or $y \notin F$ Aiming for a contradiction suppose that :$x \in F \land y \in F$ By definition of filtered: :$\exists z \in F: z \preceq x \land z \preceq y$ By definition of infimum: :$z \preceq p$ By definition of upper set: :$p \in F$ Thus this contradicts $p \in \complement_S\left({F}\right)$ by definition of greatest element. {{qed|lemma}} By definition of relative complement: :$x \in \complement_S\left({F}\right)$ or $y \in \complement_S\left({F}\right)$ Thus by definition of greatest element: this contradicts $p \prec x$ and $p \prec y$ {{qed}}
From Floor of $\dfrac {x + m} n$: Corollary: :$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$ which is valid for all integers $m, n$ such that $n > 0$. In this instance, $m = 0$ and $n = p$, while $x = \dfrac n {p^k}$. {{qed}}
By definition of continuous: :$L$ satisfies axiom of approximation and :$\forall x \in S: x^\ll$ is directed. By Axiom of Approximation in Up-Complete Semilattice :$\exists u \in S: u \ll y \land u \npreceq x$ By Way Below implies There Exists Way Below Open Filter Subset of Way Above Closure: :there exists way below open filter $F$ in $L$: $y \in F \subseteq u^\gg$ By Way Above Closure is Subset of Upper Closure of Element: :$u^\gg \subseteq u^\succeq$ By definition of upper set: :$x \notin F$ By definition of relative complement: :$x \in \complement_S\left({F}\right)$ By Upper Way Below Open Subset Complement is Non Empty implies There Exists Maximal Element of Complement: :$\exists m \in \complement_S\left({F}\right): x \preceq m \land m = \max \complement_S\left({F}\right)$ Thus by Maximal Element of Complement of Filter is Meet Irreducible: :$m$ is meet irreducible. Thus :$x \preceq m$ Aiming for a contradiction suppose that :$y \preceq m$ By definition of upper set: :$m \in F$ This contradicts $m \in \complement_S\left({F}\right)$ by definition of greatest element. {{qed}}
From the definition of the termial function on the integers: :$\displaystyle n? = \sum_{k \mathop = 1}^n k = 1 + 2 + \cdots + n$ From Closed Form for Triangular Numbers, we have that: :$\displaystyle \forall n \in \Z_{> 0}: \sum_{k \mathop = 1}^n k = \dfrac {n \paren {n + 1} } 2$ This agrees with the definition of the termial function on the real numbers. Hence the result, by definition of extension. {{qed}}
From $(1)$, by induction we have: :$\forall n \in \N: \map f {x + n} = \map f x + n$ and :$\forall n \in\N: \map f {x - n} = \map f x - n$ and therefore, in particular: :$(3): \quad \forall n \in \Z: \map f n = \map f 0 + n$ From $(2)$, we get {{begin-eqn}} {{eqn | l = \map f 0       | r = \map f {\map f 0} }} {{eqn | r = \map f 0 + \map f 0       | c = using $(3)$ and $f$ being integer valued }} {{end-eqn}} Hence :$\map f 0 = 0$ Thus from $(3)$ it follows that: :$\forall n \in \Z: \map f n = n$ Suppose that $\map f {\dfrac 1 2} = k \le 0$. Then: {{begin-eqn}} {{eqn | l = k       | r = \map f {\dfrac 1 {1 - 2 k} \map f {\dfrac 1 2 - k} }       | c =  }} {{eqn | r = \map f {\dfrac 1 {1 - 2 k} \paren {\map f {\dfrac 1 2} - k} }       | c =  }} {{eqn | r = \map f 0       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} We show that in this case, by induction: :$\map f {\dfrac 1 n} = 0$ for all $n \in \N$ Induction hypothesis: :$\map f {\dfrac 1 {n - 1} } = 0$ Then from $(1)$: :$\map f {\dfrac n {n - 1} } = \map f {\dfrac 1 {n - 1} + 1} = 0 + 1 = 1$ so: {{begin-eqn}} {{eqn | l = \map f {\dfrac 1 n}       | r = \map f {\dfrac 1 n \map f {\dfrac n {n - 1} } }       | c =  }} {{eqn | r = \map f {\dfrac 1 {n - 1} }       | c = using $(2)$ }} {{eqn | r = 0       | c =  }} {{end-eqn}} Finally, we show by induction on $m$ that even: :$\map f {\dfrac m n} = 0$ for $m \in \set {1, \ldots, n - 1}$. Above we have shown this for $m = 1$. Let $1 \le m < n$.  If $m \divides n$, then :$\map f {\dfrac m n} = \map f {\dfrac 1 {n / m} } = 0$ Otherwise, write: :$n = \paren {k - 1} m + r$ Then: :$k > 1$ and: :$1 \le r \le m - 1$ We therefore have: :$k m = n + m - r$ so by the induction hypothesis: :$\map f {\dfrac {k m} n} = \map f {1 + \dfrac {m - r} n} = 1 + \map f {\dfrac {m - r} n} = 1$ Then: {{begin-eqn}} {{eqn | l = \map f {\dfrac m n}       | r = \map f {\dfrac 1 m \map f {\dfrac {m k} n} }       | c =  }} {{eqn | r = \map f {\dfrac 1 m}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} By $(1)$, this shows that :$\map f {\dfrac 1 2} \le 0 \implies \map f x = \floor x$ for all rational $x$. Suppose that $\map f {\dfrac 1 2} = k > 0$. Then the integer-valued function $g: \R \to \Z$ satisfies: :$\map g x = -\map f {-x}$ satisfies $(1)$ and $(2)$, and also: {{begin-eqn}} {{eqn | l = \map g {\dfrac 1 2}       | r = 1 - \map f {\dfrac 1 2}       | c =  }} {{eqn | o = \ge       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \map f x       | r = -\map g {-x}       | c =  }} {{eqn | r = -\floor {-x}       | c =  }} {{eqn | r = \ceiling x       | c = Floor of Negative equals Negative of Ceiling }} {{end-eqn}} Thus: :$\map f {\dfrac 1 2} > 0 \implies \map f x = \ceiling x$ for all rational $x$. {{qed}}
Let $f$ be a replicative function. Then: :$\forall n > 0: f \left({n x + 1}\right) - f \left({n x}\right) = f \left({x + 1}\right) - f \left({x}\right)$ If $f$ is then also continuous: :$\forall x \in \R: f \left({x + 1}\right) - f \left({x}\right)$ and so: :$g \left({x}\right) = f \left({x}\right) - c \left \lfloor{x}\right \rfloor$ is both replicative and periodic. We have: :$\displaystyle \int_0^1 e^{2 \pi i n x} g \left({x}\right) \mathrm d x = \dfrac 1 n \int_0^1 e^{2 \pi y} g \left({y}\right) \mathrm d y$ Expanding in a Fourier series shows: :$g \left({x}\right) = \left({x - \dfrac 1 2}\right) a$ for $0 < x < 1$. Thus it follows that: :$f \left({x}\right) = \left({x - \dfrac 1 2}\right) a$ {{qed}} {{MissingLinks|The above needs to be linked to the results used.}}
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Sum of Indices: :$\forall m, n \in \Z: g^m \circ g^n = g^{m + n}$ where in this context the group operation is $+$ and $n$th power of $g$ is denoted $n g$. {{qed}}
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Sum of Indices: :$\forall n \in \Z: \paren {g^n}^{-1} = g^{-n} = \paren {g^{-1} }^n$ where in this context: :the group operation is $+$ :the $n$th power of $g$ is denoted $n g$ :the inverse of $g$ is $-g$. {{qed}}