hcju/proofwikiqa huggingface dataset:
In the summation: :$\displaystyle \sum_{j \mathop = 0}^2 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 2 \\ \forall p \ge 1: k_p \mathop \ge 0} } 2! \prod_{m \mathop = 1}^2 \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$ we need to consider $j = 0, j = 1$ and $j = 2$. Note that when $k_m = 0$: :$\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } = 1$ by definition of zeroth derivative and factorial of $0$. Thus any contribution to the summation where $k_m = 0$ can be disregarded. Let $j = 0$. Consider the set of $k_p$ such that: :$k_1 + k_2 + \cdots = 0$ :$1 \times k_1 + 2 k_2 + \cdots = 2$ :$k_1, k_2, \ldots \ge 0$ It is apparent by inspection that no set of $k_p$ can fulfil these conditions. Therefore when $j = 0$ the summation is vacuous Let $j = 1$. Consider the set of $k_p$ such that: :$k_1 + k_2 + \cdots = 1$ :$1 \times k_1 + 2 k_2 + \cdots = 2$ :$k_1, k_2, \ldots \ge 0$ By inspection, it is seen that these can be satisfied only by: :$k_2 = 1$ and all other $k_p = 0$. Let $j = 2$. Consider the set of $k_p$ such that: :$k_1 + k_2 + \cdots = 2$ :$1 \times k_1 + 2 k_2 + \cdots = 2$ :$k_1, k_2, \ldots \ge 0$ By inspection, it is seen that these can be satisfied only by: :$k_1 = 2$ and all other $k_p = 0$. Thus we have: {{begin-eqn}} {{eqn | o =        | r = \sum_{j \mathop = 0}^2 D_u^j w \sum_{\substack {k_1 \mathop + k_2 \mathop = j \\ k_1 \mathop + 2 k_2 \mathop = 2 \\ k_1, k_2 \mathop \ge 0} } 2! \dfrac {\left({D_x^1 u}\right)^{k_1} \left({D_x^2 u}\right)^{k_2} } {k_1! \left({1!}\right)^{k_1} k_2! \left({2!}\right)^{k_2} }        | c =  }} {{eqn | r = D_u^1 w \dfrac {2! \left({D_x^1 u}\right)^0 \left({D_x^2 u}\right)^1} {0! \left({1!}\right)^0 1! 2^1} + D_u^2 w \dfrac {2! \left({D_x^1 u}\right)^2 \left({D_x^2 u}\right)^0} {2! \left({1!}\right)^2 0! \left({2!}\right)^0}        | c = substituting for $k_1$ and $k_2$ for each case }} {{eqn | r = D_u^1 w \left({D_x^2 u}\right) + D_u^2 w \left({D_x^1 u}\right)^2       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = D_x \left({\left({D_x^m u}\right)^{k_m} }\right)       | r = \left({k_m \left({D_x^m u}\right)^{k_m - 1} }\right) D_x \left({D_x^m u}\right)       | c = Derivative of Power of Function }} {{eqn | r = k_m \left({\left({D_x^m u}\right)^{k_m - 1} }\right) D_x^{m + 1} u       | c = {{Defof|Higher Derivative}} }} {{end-eqn}} {{qed}} Category:Faà di Bruno's Formula lxvuaw0ammntxkyzm0qexwv6cw3pn3e
{{begin-eqn}} {{eqn | r = D_x \left({\prod_{m \mathop = 1}^r \left({\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } }\right) }\right)       | o =        | c =  }} {{eqn | r = \prod_{m \mathop = 1}^r \left({\dfrac 1 {k_m! \left({m!}\right)^{k_m} } }\right) D_x \left({\prod_{m \mathop = 1}^r \left({\left({D_x^m u}\right)^{k_m} }\right) }\right)       | c =  }} {{eqn | r = \prod_{m \mathop = 1}^r \left({\dfrac 1 {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r \left({D_x \left({\left({D_x^m u}\right)^{k_m} }\right) \prod_{s \mathop \ne m} \left({\left({D_x^s u}\right)^{k_s} }\right) }\right)       | c = Product Rule for Derivatives: General Result }} {{eqn | r = \prod_{m \mathop = 1}^r \left({\dfrac 1 {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r \left({\left({k_m \left({D_x^m u}\right)^{k_m - 1} D_x^{m + 1} u}\right) \prod_{s \mathop \ne m} \left({\left({D_x^s u}\right)^{k_s} }\right)}\right)       | c = Faà di Bruno's Formula: Lemma 1 }} {{eqn | r = \prod_{m \mathop = 1}^r \left({\dfrac 1 {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u} \prod_{m \mathop = 1}^r \left({\left({D_x^m u}\right)^{k_m} }\right)       | c =  }} {{eqn | r = \prod_{m \mathop = 1}^r \left({\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u}       | c =  }} {{end-eqn}} {{qed}} Category:Faà di Bruno's Formula 56df0zm1pg9z1et79yyuorg5w41mfb0
Define $d: S \to \mathit{Ids}\left({L}\right)$ :$\forall t \in S: d\left({t}\right) =  \inf \left({f^{-1}\left[{t^\succeq}\right]}\right)$ where :$t^\succeq$ denotes the upper closure of $t$, :$f^{-1}\left[{t^\succeq}\right]$ denotes the image of $t^\succeq$ over $f^{-1}$. We will prove that :$\forall t \in S: d\left({t}\right) =  \min \left({f^{-1}\left[{t^\succeq}\right]}\right)$ Let $t \in S$. By Continuous iff For Every Element There Exists Ideal Element Precedes Supremum: :there exists an ideal $J$ in $L$ such that ::$t \preceq \sup J$ and for every ideal $K$ in $L$: $t \preceq \sup K \implies J \subseteq K$ We will prove that :$\forall K \in \mathit{Ids}\left({L}\right): K$ is lower bound for $f^{-1}\left[{t^\succeq}\right] \implies K \precsim J$ Let $K \in \mathit{Ids}\left({L}\right)$ such that :$K$ is lower bound for $f^{-1}\left[{t^\succeq}\right]$ By definition of $f$: :$t \preceq f\left({J}\right)$ By definition of upper closure of element: :$f\left({J}\right) \in t^\succeq$ By definition of image of set: :$J \in f^{-1}\left[{t^\succeq}\right]$ Thus by definition of lower bound: :$K \precsim J$ {{qed|lemma}} We will prove that :$J$ is lower bound for $f^{-1}\left[{t^\succeq}\right]$ Let $K \in \mathit{Ids}\left({L}\right)$ such that :$K \in f^{-1}\left[{t^\succeq}\right]$ By definition of image of set: :$f\left({K}\right) \in t^\succeq$ By definition of upper closure of element: :$t \preceq f\left({K}\right)$ By definition of $f$: :$t \preceq \sup K$ Then :$J \subseteq K$ Thus by definition of $\precsim$: :$J \precsim K$ {{qed|lemma}} By definition of supremum: :$t \preceq \sup \left({\inf \left({f^{-1}\left[{t^\succeq}\right]}\right)}\right)$ By definition of $f$: :$t \preceq f\left({\inf \left({f^{-1}\left[{t^\succeq}\right]}\right)}\right)$ By definition of upper closure of element: :$f\left({\inf \left({f^{-1}\left[{t^\succeq}\right]}\right)}\right) \in t^\succeq$ By definition of image of set: :$\inf \left({f^{-1}\left[{t^\succeq}\right]}\right) \in f^{-1}\left[{t^\succeq}\right]$ Thus by definition of smallest element: :$d\left({t}\right) = \min \left({f^{-1}\left[{t^\succeq}\right]}\right)$ {{qed|lemma}} By Supremum of Ideals is Increasing: :$f$ is an increasing mapping. By Galois Connection is Expressed by Minimum: :$\left({f, d}\right)$ is a Galois connection. Hence $f$ is an upper adjoint of Galois connection. {{qed}}
Let $I, J \in \mathit{Ids}\left({L}\right)$ such that :$I \precsim J$ By definition of $\precsim$: :$I \subseteq J$ By definition of up-complete: :$I$ and $J$ admit suprema in $L$. By Supremum of Subset: :$\sup I \preceq \sup J$ Thus by definition of $f$: :$f\left({I}\right) \preceq f\left({J}\right)$ Hence $f$ is an increasing mapping. {{qed}}
For ease of understanding, let Leibniz's notation be used: :$\dfrac {\d^k u} {\d x^k} := D_x^k u$ Then we have: {{begin-eqn}} {{eqn | l = D_x^3 w       | r = \map {\dfrac \d {\d x} } {\dfrac {\d^2 w} {\d x^2} }       | c =  }} {{eqn | r = \map {\dfrac \d {\d x} } {\dfrac {\d^2 w} {\d u^2} \paren {\dfrac {\d u} {\d x} }^2 + \dfrac {\d w} {\d u} \dfrac {\d^2 u} {\d x^2} }       | c = Derivative of Composite Function: Second Derivative }} {{eqn | r = \map {\dfrac \d {\d x} } {\dfrac {\d^2 w} {\d u^2} } \paren {\dfrac {\d u} {\d x} }^2 + \dfrac {\d^2 w} {\d u^2} \map {\dfrac \d {\d x} } {\paren {\dfrac {\d u} {\d x} }^2}       | c = Product Rule for Derivatives }} {{eqn | o =        | ro= +       | r = \map {\dfrac \d {\d x} } {\dfrac {\d w} {\d u} } \dfrac {\d^2 u} {\d x^2} + \dfrac {\d w} {\d u} \map {\dfrac \d {\d x} } {\dfrac {\d^2 u} {\d x^2} }       | c =  }} {{eqn | r = \dfrac {\d^3 w} {\d u^3} \dfrac {\d u} {\d x} \paren {\dfrac {\d u} {\d x} }^2 + \dfrac {\d^2 w} {\d u^2} 2 \dfrac {\d u} {\d x} \dfrac {\d^2 u} {\d x^2}       | c = Derivative of Composite Function }} {{eqn | o =        | ro= +       | r = \dfrac {\d^2 w} {\d u^2} \dfrac {\d u} {\d x} \dfrac {\d^2 u} {\d x^2} + \dfrac {\d w} {\d u} \paren {\dfrac {\d^3 u} {\d x^3} }       | c =  }} {{eqn | r = \dfrac {\d^3 w} {\d u^3} \paren {\dfrac {\d u} {\d x} }^3 + 2 \dfrac {\d^2 w} {\d u^2} \dfrac {\d^2 u} {\d x^2} \dfrac {\d u} {\d x}       | c = simplifying }} {{eqn | o =        | ro= +       | r = \dfrac {\d^2 w} {\d u^2} \dfrac {\d^2 u} {\d x^2} \dfrac {\d u} {\d x} + \dfrac {\d w} {\d u} \paren {\dfrac {\d^3 u} {\d x^3} }       | c =  }} {{eqn | r = \dfrac {\d^3 w} {\d u^3} \paren {\dfrac {\d u} {\d x} }^3 + 3 \dfrac {\d^2 w} {\d u^2} \dfrac {\d^2 u} {\d x^2} \dfrac {\d u} {\d x} + \dfrac {\d w} {\d u} \paren {\dfrac {\d^3 u} {\d x^3} }       | c = simplifying }} {{eqn | r = D_u^3 w \paren {D_x^1 u}^3 + 3 D_u^2 w D_x^2 u D_x^1 u + D_u^1 w D_x^3 u       | c = Leibniz's Notation }} {{end-eqn}} {{qed}} Category:Differential Calculus kiwj42mfwm8i3qy33qkqwbvx9nxljff
The proof proceeds by induction on $n$. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \map f {n, k} = {n \brace k}$ $\map P 0$ is the degenerate case: :$\displaystyle \map f {0, k} = \delta_{0 k} = {0 \brace k}$ That is: the empty set can be partitioned one and only one way: into $0$ subsets. Thus $\map P 0$ is seen to hold. The remainder of the proof considers $n \in \Z_{> 0}$. First we note that when $k < 1$ or $k > n$: :$\displaystyle \map f {n, k} = 0 = {n \brace k}$ Hence, throughout, we consider only such $k$ as $1 \le k \le n$. We define the representative set of cardinality $n$ to be: :$S_n := \set {1, 2, \ldots, n}$ === Basis for the Induction === $\map P 1$ is the case $\map f {1, 1}$. There is exactly one way to partition $\set 1$, and that is: :$\set {\set 1}$ From Stirling Number of the Second Kind of Number with Self: :$\displaystyle {1 \brace 1} = 1$ 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 $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \map f {m, k} = {m \brace k}$ from which it is to be shown that: :$\displaystyle \map f {m + 1, k} = {m + 1 \brace k}$ === Induction Step === This is the induction step: By definition, the number of partitions of $S_m$ into $k$ subsets is $\map f {m, k}$. A partition of $S_{m + 1}$ can be generated by adding element $m + 1$ into one of the existing partitions of $S_m$. There are two ways this can be done: :$(1): \quad$ The subset $\set {m + 1}$ may be added, in one way, to one of the partitions of $S_m$ into $k - 1$ subsets. :$(2): \quad$ The element $m + 1$ may be added to any one of the $k$ subsets in one of the partitions of $S_m$ into $k$ subsets. Option $(1)$ gives $1$ partition of $S_{m + 1}$ for each partition of $S_m$ into $k - 1$ subsets, that is: $\map f {m, k - 1}$. Option $(2)$ gives $k$ partitions of $S_{m + 1}$ for each partition of $S_m$ into $k$ subsets, that is: $k \map f {m, k}$. Thus: :$\map f {m + 1, k} = k \map f {m, k} + \map f {m, k - 1}$ By the induction hypothesis: :$\displaystyle \map f {m + 1, k} = k {m \brace k} + {m  \brace k - 1}$ So by definition of Stirling numbers of the second kind: :$\displaystyle \map f {m + 1, k} = {m + 1 \brace k}$ So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n, k \in \Z_{\ge 0}: \map f {n, k} = {n \brace k}$ {{qed}}
From Exchange of Order of Summation with Dependency on Both Indices: {{:Exchange of Order of Summation with Dependency on Both Indices}} We have that: :$\map R d$ is the propositional function: ::$d \divides n$ :$\map S {d, c}$ is the propositional function: ::$c \divides d$ Thus $\map {R'} {d, c}$ is the propositional function: ::Both $d \divides n$ and $c \divides d$ This is the same as: ::$c \divides n$ and $\dfrac d c \divides \dfrac n c$ Similarly, $\map {S'} c$ is the propositional function: ::$\exists d$ such that both $d \divides n$ and $c \divides d$ This is the same as: ::$c \divides n$ This gives: :$\displaystyle \sum_{d \mathop \divides n} \sum_{c \mathop \divides d} \map f {c, d} = \sum_{c \mathop \divides n} \sum_{\paren {d / c} \mathrel \divides \paren {n / c} } \map f {c, d}$ Replacing $d / c$ with $d$: :$\displaystyle \sum_{d \mathop \divides n} \sum_{c \mathop \divides d} \map f {c, d} = \sum_{c \mathop \divides n} \sum_{d \mathop \divides \paren {n / c} } \map f {c, c d}$ {{qed}}
Either $n + m$ or $n - m + 1$ is even. Thus: :$\dfrac {n + m} 2 \bmod 1 + \dfrac {n - m + 1} 2 \bmod 1 = \dfrac 1 2 < 1$ and so: {{begin-eqn}} {{eqn | l = \floor {\dfrac {n + m} 2} + \floor {\dfrac {n - m + 1} 2}       | r = \floor {\dfrac {n + m} 2 + \dfrac {n - m + 1} 2}       | c = Sum of Floors not greater than Floor of Sum }} {{eqn | r = \floor {\dfrac {n + m + n - m + 1} 2}       | c =  }} {{eqn | r = \floor {n + \dfrac 1 2}       | c =  }} {{eqn | r = n       | c =  }} {{end-eqn}} {{qed}}
Either $n + m$ or $n - m + 1$ is even. Thus either $\dfrac {n + m} 2$ or $\dfrac {n - m + 1} 2$ is an integer. So: {{begin-eqn}} {{eqn | l = \left\lceil{\dfrac {n + m} 2}\right\rceil + \left\lceil{\dfrac {n - m + 1} 2}\right\rceil       | r = \left\lceil{\dfrac {n + m} 2 + \dfrac {n - m + 1} 2}\right\rceil       | c = Sum of Ceilings not less than Ceiling of Sum }} {{eqn | r = \left\lceil{\dfrac {n + m + n - m + 1} 2}\right\rceil       | c =  }} {{eqn | r = \left\lceil{n + \dfrac 1 2}\right\rceil       | c =  }} {{eqn | r = n + 1       | c =  }} {{end-eqn}} {{qed}}
From Floor is between Number and One Less: :$\left\lfloor{x}\right\rfloor \le x$ From Real Number is Integer iff equals Floor: :$x = \left \lfloor {x} \right \rfloor \iff x \in \Z$ But we have $x \notin \Z$. So: :$\left \lfloor {x} \right \rfloor \ne x$ and so: :$\left \lfloor {x} \right \rfloor < x$ {{qed}}
Let $a \in \R_{>0}$ such that $a \ne 1$. Then: {{begin-eqn}} {{eqn | l = \log_b b       | r = \dfrac {\log_a b} {\log_a b}       | c = Change of Base of Logarithm }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}} Category:Logarithms 8p8v7a9h3w1b8jvsqgyq3scehh3d4y0
=== Necessary Condition === Let: :$\forall x \in \R_{\ge 1}: \left\lfloor{\log_b x}\right\rfloor = \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor$ Let $x = b$. Then: :$\left\lfloor{\log_b b}\right\rfloor = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor$ {{begin-eqn}} {{eqn | l = \left\lfloor{\log_b b}\right\rfloor       | r = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor       | c =  }} {{eqn | ll= \leadsto       | l = \left\lfloor{1}\right\rfloor       | r = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor       | c = Logarithm of Base }} {{eqn | ll= \leadsto       | l = 1       | r = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor       | c = Real Number is Integer iff equals Floor }} {{end-eqn}} {{AimForCont}} $b \notin \Z$. {{begin-eqn}} {{eqn | l = \left\lfloor{b}\right\rfloor       | o = <       | r = b       | c = Floor of Non-Integer }} {{eqn | ll= \leadsto       | l = \log_b \left\lfloor{b}\right\rfloor       | o = <       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor       | o = <       | r = \left\lfloor{\log_b b}\right\rfloor       | c =  }} {{eqn | ll= \leadsto       | l = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor       | o = \ne       | r = \left\lfloor{\log_b b}\right\rfloor       | c = which contradicts $\left\lfloor{\log_b b}\right\rfloor = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor$ }} {{end-eqn}} Thus by Proof by Contradiction: :$b \in \Z$ But for $\log_b$ to be defined, $b > 0$ and $b \ne 1$. Hence: :$b \in \Z_{> 1}$ {{qed|lemma}} === Sufficient Condition === Let $b \in \Z_{> 1}$. Let $\left\lfloor{\log_b x}\right\rfloor = n$. Then: {{begin-eqn}} {{eqn | l = n       | m = \left\lfloor{\log_b x}\right\rfloor       | c =  }} {{eqn | ll= \iff       | l = n       | o = \le       | m = \log_b x       | mo= <       | r = n + 1       | c = Integer equals Floor iff Number between Integer and One More }} {{eqn | ll= \iff       | l = b^n       | o = \le       | m = x       | mo= <       | r = b^{n + 1}       | c = Power Function is Strictly Increasing over Positive Reals }} {{eqn | ll= \iff       | l = b^n       | o = \le       | m = \left \lfloor {x}\right \rfloor       | mo= <       | r = b^{n + 1}       | c = Number not less than Integer iff Floor not less than Integer }} {{eqn | ll= \iff       | l = n       | o = \le       | m = \log_b \left\lfloor{x}\right\rfloor       | mo= <       | r = n + 1       | c = Power Function is Strictly Increasing over Positive Reals }} {{eqn | ll= \iff       | l = n       | m = \left \lfloor {\log_b \left\lfloor{x}\right\rfloor}\right \rfloor       | c = Integer equals Floor iff Number between Integer and One More }} {{eqn | ll= \iff       | l = \left\lfloor{\log_b x}\right\rfloor       | m = \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor        | c = Definition of $n$ }} {{end-eqn}} {{qed}}
We have that: :Logarithm is Strictly Increasing and: :Natural Logarithm Function is Continuous Suppose that $\log_b x \in \Z$: let $\log_b x = n$, say. Then: :$x = b^n$ It follows that $x \in \Z \iff b \in \Z$. Thus by McEliece's Theorem (Integer Functions): :$\forall x \in \R_{\ge 1}: \left\lfloor{\log_b x}\right\rfloor = \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor \iff b \in \Z_{> 1}$ {{qed}}
Let $f: \R \to \R$ be the real function defined as: :$\forall x \in \R: f \left({x}\right) = \dfrac {x + m} n$ It is clear that $f$ is both strictly increasing and continuous on the whole of $\R$. Let $\dfrac {x + m} n \in \Z$. Then: {{begin-eqn}} {{eqn | l = \exists s \in \Z: \dfrac {x + m} n       | r = s       | c =  }} {{eqn | ll= \leadsto       | l = x + m       | r = n s       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = n s - m       | c =  }} {{eqn | o = \in       | r = \Z       | c =  }} {{end-eqn}} Thus: :$\forall x \in \R: f \left({x}\right) \in \Z \implies x \in \Z$ So the conditions are fulfilled for McEliece's Theorem (Integer Functions) to be applied: :$\left \lfloor{f \left({x}\right)}\right \rfloor = \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor \iff \left(f \left({x}\right) \in \Z \implies x \in \Z)\right)$ Hence the result. {{qed}}
From Ceiling is between Number and One More: :$\left \lceil{x}\right \rceil \ge x$ From Real Number is Integer iff equals Ceiling: :$x = \left \lceil {x} \right \rceil \iff x \in \Z$ But we have $x \notin \Z$. So: :$\left \lceil {x} \right \rceil \ne x$ and so: :$\left \lceil {x} \right \rceil > x$ {{qed}}
{{begin-eqn}} {{eqn | l = \left \lceil{\dfrac {x + m} n}\right \rceil       | r = -\left \lfloor{-\dfrac {x + m} n}\right \rfloor       | c = Floor of Negative equals Negative of Ceiling }} {{eqn | r = -\left \lfloor{\dfrac {- x - m} n}\right \rfloor       | c =  }} {{eqn | r = -\left \lfloor{\dfrac {\left \lfloor{- x}\right \rfloor - m} n}\right \rfloor       | c = Floor of $\dfrac{x + m} n$ }} {{eqn | r = -\left \lfloor{\dfrac {-\left \lceil{x}\right \rceil - m} n}\right \rfloor       | c = Floor of Negative equals Negative of Ceiling }} {{eqn | r = -\left \lfloor{-\dfrac {\left \lceil{x}\right \rceil + m} n}\right \rfloor       | c =  }} {{eqn | r = \left \lceil{\dfrac {\left \lceil{x}\right \rceil + m} n}\right \rceil        | c = Floor of Negative equals Negative of Ceiling }} {{end-eqn}} {{qed}}
Let $f: \R \to \R$ be the real function defined as: :$\forall x \in \R: f \left({x}\right) = \dfrac {x + m} n$ It is clear that $f$ is both strictly increasing and continuous on the whole of $\R$. Let $\dfrac {x + m} n \in \Z$. Then: {{begin-eqn}} {{eqn | l = \exists s \in \Z: \dfrac {x + m} n       | r = s       | c =  }} {{eqn | ll= \leadsto       | l = x + m       | r = n s       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = n s - m       | c =  }} {{eqn | o = \in       | r = \Z       | c =  }} {{end-eqn}} Thus: :$\forall x \in \R: f \left({x}\right) \in \Z \implies x \in \Z$ So the conditions are fulfilled for McEliece's Theorem (Integer Functions) to be applied: :$\left \lceil{f \left({x}\right)}\right \rceil = \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil \iff \left(f \left({x}\right) \in \Z \implies x \in \Z)\right)$ Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{0 \mathop \le k \mathop < n} \left \lceil{\dfrac {m k + x} n}\right \rceil       | r = \sum_{0 \mathop \le k \mathop < n} \left({-\left \lfloor{-\dfrac {m k + x} n}\right \rfloor}\right)       | c = Floor of Negative equals Negative of Ceiling }} {{eqn | r = -\sum_{0 \mathop \le k \mathop < n} \left \lfloor{\dfrac {-m k - x} n}\right \rfloor       | c =  }} {{eqn | r = -\left({\dfrac {\left({\left({-m}\right) - 1}\right) \left({n - 1}\right)} 2 + \dfrac {d - 1} 2 + d \left \lfloor{\dfrac {\left({-x}\right)} d}\right \rfloor}\right)       | c = Summation over $k$ of $\left \lfloor{\dfrac {m k + x} n}\right \rfloor$ }} {{eqn | r = \dfrac {\left({m + 1}\right) \left({n - 1}\right)} 2 - \dfrac {d - 1} 2 - d \left \lfloor{- \dfrac x d}\right \rfloor       | c =  }} {{eqn | r = \dfrac {\left({m + 1}\right) \left({n - 1}\right)} 2 - \dfrac {d - 1} 2 + d \left \lceil{\dfrac x d}\right \rceil       | c = Floor of Negative equals Negative of Ceiling }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n - 1} \left \lfloor{x + \dfrac k n}\right \rfloor       | r = \left \lfloor{x n + \left \lfloor{x + 1}\right \rfloor \left({\left \lceil{n}\right \rceil - n}\right)}\right \rfloor       | c =  Summation over $k$ of $\left \lfloor{x + \dfrac k y}\right \rfloor$, setting $y = n$ }} {{eqn | r = \left \lfloor{x n + \left \lfloor{x + 1}\right \rfloor \left({n - n}\right)}\right \rfloor       | c = Real Number is Integer iff equals Ceiling }} {{eqn | r = \left \lfloor{n x + \left \lfloor{x + 1}\right \rfloor \times 0}\right \rfloor       | c =  }} {{eqn | r = \left \lfloor{n x}\right \rfloor       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n}       | r = \sum_{k \mathop = 0}^{n - 1} \paren {x - \frac 1 2 + \frac k n}       | c =  }} {{eqn | r = n x - \frac n 2 + \frac 1 n \sum_{k \mathop = 0}^{n - 1} k       | c =  }} {{eqn | r = n x - \frac n 2 + \frac 1 n \frac {n \paren {n - 1} } 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = n x - \frac n 2 + \frac n 2 - \frac 1 2       | c =  }} {{eqn | r = n x - \frac 1 2       | c =  }} {{eqn | r = \map f {n x}       | c =  }} {{end-eqn}} Hence the result by definition of replicative function. {{qed}}
First note that the interval between $x$ and $x + \dfrac {n - 1} n$ is less than $1$. Thus there can be no more than one $k$ such that $0 \le k < n$ such that: :$x + \dfrac k n \in \Z$ Hence: :$\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} \le 1$ First let: :$\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} = 1$ Then: {{begin-eqn}} {{eqn | l = x + \frac k n       | o = \in        | r = \Z       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {x + \frac k n}       | o = \in        | r = \Z       | c =  }} {{eqn | ll= \leadsto       | l = n x + k       | o = \in        | r = \Z       | c =  }} {{eqn | ll= \leadsto       | l = n x       | o = \in        | r = \Z       | c =  }} {{eqn | ll= \leadsto       | l = \sqbrk {n x \in \Z}       | r = 1       | c =  }} {{end-eqn}} {{qed|lemma}} Let: :$\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} = 0$ {{AimForCont}} $\sqbrk {n x \in \Z} = 1$. Then: {{begin-eqn}} {{eqn | l = n x       | o = \in       | r = \Z       | c =  }} {{eqn | lll= \leadsto       | ll= \forall k \in \Z:       | l = n x + k       | o = \in       | r = \Z       | c =  }} {{end-eqn}} But at least one $n x + k$ such that $0 \le k < n$ is a multiple of $n$. Hence: :$\exists k \in \Z: 0 \le k < n: \sqbrk {x + \dfrac k n \in \Z} = 1$ So: :$\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} = 1$ which contradicts the supposition that $\sqbrk {n x \in \Z} = 1$. Hence the result by definition of replicative function. {{qed}}