hcju/proofwikiqa huggingface dataset:
Note that by Infinite Product of Analytic Functions is Analytic, $f$ is analytic. Let $n_0 \in \N$ and $U\subset D$ an open neighborhood of $z_0$ such that $\displaystyle \prod_{n \mathop = n_0}^\infty f_n \left({z}\right) \ne 0$ for $z\in U$. Let $f$ be identically zero on $U$. Then $\displaystyle \prod_{n \mathop = 1}^{n_0-1} f_n \left({z}\right)$ is identically zero on $U$. Then some $f_k$ with $k<n_0$ is identically zero on $U$. By Uniqueness of Analytic Continuation, $f_k$ is identically zero on $D$. Now let $f$ be not identically zero on $U$. We have: {{begin-eqn}} {{eqn | l = \operatorname {mult}_{z_0} \left({f}\right)       | r = \sum_{n \mathop = 1}^{n_0} \operatorname {mult}_{z_0} \left({f_n}\right) + \operatorname {mult}_{z_0} \left({\prod_{n \mathop = n_0}^\infty f_n}\right)       | c = Multiplicity of Product of Analytic Functions }} {{eqn | r = \sum_{n \mathop = 1}^{n_0} \operatorname {mult}_{z_0} \left({f_n}\right)       | c = $\displaystyle \prod_{n \mathop = n_0}^\infty f_n \left({z_0}\right) \ne 0$ }} {{eqn | l =        | r = \sum_{n \mathop = 1}^\infty \operatorname {mult}_{z_0} \left({f_n}\right)       | c =  $f_n \left({z_0}\right) \ne 0$ for $n \ge n_0$ }} {{end-eqn}} {{qed}}
Follows directly from: :Partial Products of Uniformly Convergent Product Converge Uniformly :Uniform Limit of Analytic Functions is Analytic {{qed}}
Let: :$m = a^2 + b^2$ :$n = c^2 + d^2$ Then: {{begin-eqn}} {{eqn | l = m n       | r = \paren {a^2 + b^2} \paren {c^2 + d^2}       | c =  }} {{eqn | r = \paren {a c + b d}^2 + \paren {a d - b c}^2       | c = Brahmagupta-Fibonacci Identity }} {{eqn | r = \paren {a c - b d}^2 + \paren {a d + b c}^2       | c = Brahmagupta-Fibonacci Identity: Corollary }} {{end-eqn}} It remains to be shown that $a \ne b$ and $c \ne d$, then the four numbers: :$a c + b d, a d - b c, a c - b d, a d + b c$ are distinct. Because $a, b, c, d > 0$, we have: :$a c + b d \ne a c - b d$ :$a d + b c \ne a d - b c$ We also have: {{begin-eqn}} {{eqn | l = a c \pm b d       | r = a d \pm b c }} {{eqn | ll= \leadstoandfrom       | l = a c \mp b c - a d \pm b d       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = c \paren {a \mp b} - d \paren {a \mp b}       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = \paren {a \mp b} \paren {c \mp d}       | r = 0 }} {{end-eqn}} Thus $a \ne b$ and $c \ne d$ implies $a c \pm b d \ne a d \pm b c$. The case for $a c \pm b d \ne a d \mp b c$ is similar. {{qed}}
=== Lower bound === Follows by expanding. {{qed|lemma}} === Upper Bound === ==== Proof 1 ==== By Exponential of x not less than 1+x: :$\displaystyle \prod_{k \mathop = 1}^n \paren {1 + a_k} \le \prod_{k \mathop = 1}^n \exp a_k = \map \exp {\sum_{k \mathop = 1}^n a_k}$ {{qed}} ==== Proof 2 ==== By the AM-GM Inequality: :$\displaystyle \prod_{k \mathop = 1}^n \paren {1 + a_k} \le \paren {\frac {n + \sum_{k \mathop = 1}^n a_k} n}^n$ {{link wanted|link to an article that this is less than $\exp(\sum a_k)$, i.e. that this expression is increasing with $n$}} {{qed}}
By definition of complex exponential: :$\exp z = \displaystyle \sum_{n \mathop = 1}^\infty \frac {z^n} {n!}$ Thus {{begin-eqn}} {{eqn  | l = \cmod {\exp z - 1 - z}        | r = \cmod {\sum_{n \mathop = 2}^\infty \frac {z^n} {n!} }        | c = Linear Combination of Convergent Series }} {{eqn  | o = \le        | r = \sum_{n \mathop = 2}^\infty \cmod {\frac {z^n} {n!} }        | c = Triangle Inequality for Series }} {{eqn  | o = \le        | r = \sum_{n \mathop = 2}^\infty \frac {\cmod z^n} 2        | c = as $n \ge 2$ }} {{eqn  | o = =        | r = \frac {\cmod z^2 / 2} {1 - \cmod z}        | c = Sum of Geometric Sequence }} {{eqn  | o = \le        | r = \frac 1 2 \cmod z        | c = as $\cmod z \le \dfrac 1 2$ }} {{end-eqn}} By the Triangle Inequality: :$\dfrac 1 2 \cmod z \le \cmod {\exp z - 1} \le \dfrac 3 2 \cmod z$ {{qed}}
The smallest two positive integers which can be expressed as the sum of two distinct square numbers are: {{begin-eqn}} {{eqn | l = 5       | r = 1^2 + 2^2 }} {{eqn | l = 10       | r = 1^2 + 3^2 }} {{end-eqn}} We have that: :$50 = 5 \times 10$ Thus: {{begin-eqn}} {{eqn | r = \paren {1^2 + 2^2} \paren {1^2 + 3^2}       | c =  }} {{eqn | r = \paren {1 \times 1 + 2 \times 3}^2 + \paren {1 \times 3 - 2 \times 1}^2       | c = Brahmagupta-Fibonacci Identity }} {{eqn | r = \paren {1 + 6}^2 + \paren {3 - 2}^2       | c =  }} {{eqn | r = 7^2 + 1^2       | c =  }} {{eqn | r = 49 + 1       | c =  }} {{eqn | r = 50       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | r = \paren {1^2 + 2^2} \paren {1^2 + 3^2}       | c =  }} {{eqn | r = \paren {1 \times 1 - 2 \times 3}^2 + \paren {1 \times 3 + 2 \times 1}^2       | c = Brahmagupta-Fibonacci Identity: Corollary }} {{eqn | r = \paren {1 - 6}^2 + \paren {3 + 2}^2       | c =  }} {{eqn | r = \paren {6 - 1}^2 + \paren {3 + 2}^2       | c =  }} {{eqn | r = 5^2 + 5^2       | c =  }} {{eqn | r = 25 + 25       | c =  }} {{eqn | r = 50       | c =  }} {{end-eqn}} {{qed}}
We have that: :$65 = 5 \times 13$ Both $5$ and $13$ can be expressed as the sum of two distinct square numbers: {{begin-eqn}} {{eqn | l = 5       | r = 1^2 + 2^2 }} {{eqn | l = 13       | r = 2^2 + 3^2 }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | r = \paren {1^2 + 2^2} \paren {2^2 + 3^2}       | c =  }} {{eqn | r = \paren {1 \times 2 + 2 \times 3}^2 + \paren {1 \times 3 - 2 \times 2}^2       | c = Brahmagupta-Fibonacci Identity }} {{eqn | r = \paren {2 + 6}^2 + \paren {3 - 4}^2       | c =  }} {{eqn | r = \paren {2 + 6}^2 + \paren {4 - 3}^2       | c =  }} {{eqn | r = 8^2 + 1^2       | c =  }} {{eqn | r = 64 + 1       | c =  }} {{eqn | r = 65       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | r = \paren {1^2 + 2^2} \paren {2^2 + 3^2}       | c =  }} {{eqn | r = \paren {1 \times 2 - 2 \times 3}^2 + \paren {1 \times 3 + 2 \times 2}^2       | c = Brahmagupta-Fibonacci Identity: Corollary }} {{eqn | r = \paren {2 - 6}^2 + \paren {3 + 4}^2       | c =  }} {{eqn | r = \paren {6 - 2}^2 + \paren {3 + 4}^2       | c =  }} {{eqn | r = 4^2 + 7^2       | c =  }} {{eqn | r = 16 + 49       | c =  }} {{eqn | r = 65       | c =  }} {{end-eqn}} {{qed}}
Let $x \in X$. Let $U = \map {f^{-1} } {\map B {\map f x, 1} }$. {{explain|$\map B {\map f x, 1}$}} By continuity, $U$ is a neighborhood of $x$. Because $\map f U \subset \map B {\map f x, 1}$, $f$ is bounded on $U$. Thus $f$ is locally bounded. {{qed}} Category:Boundedness 032x9q3dgsc1qdh2h7vr31bk99m0l6f
Let $t \in T$. By assumption: :$\map d t = \min \set {g^{-1} \sqbrk {\set t} }$ By definition of min operation: :$\map d t = \map \inf {g^{-1} \sqbrk {\set t} }$ and $\map d t \in g^{-1} \sqbrk {\set t}$ By definition of image of set: :$\map g {\map d t} \in \set t$ By definition of singleton: :$\map g {\map d t} = t$ Thus by definition of composition of mappings: :$\map {\paren {g \circ d} } t = t$ {{qed}}
=== Proof of some bound === Let $\cmod z \le \dfrac 1 2$. We may assume $p \ge 1$. We have: :$\map {E_p} z = \map \exp {\map \log {1 - z} + \displaystyle \sum_{k \mathop = 1}^p \frac {z^k} k}$ Then: {{begin-eqn}} {{eqn | l = \cmod {\map \log {1 - z} + \sum_{k \mathop = 1}^p \frac {z^k} k}       | r = \cmod {-\sum_{k \mathop = p + 1}^\infty \frac{z^k} k}       | c = Series Expansion of Complex Logarithm }} {{eqn | o = \le       | r = \sum_{k \mathop = p + 1}^\infty \frac {\cmod z^k} k       | c = Triangle Inequality for Series }} {{eqn | o = \le       | r = \sum_{k \mathop = p + 1}^\infty \cmod z^k       | c = because $k \ge 1$ }} {{eqn | r = \frac {\cmod z^{p + 1} } {1 - \cmod z}       | c = Sum of Geometric Sequence }} {{eqn | o = \le       | r = 2 \cmod z^{p + 1}       | c = because $\cmod z \le \dfrac 1 2$ }} {{end-eqn}} Because $p \ge 1$: :$2 \cmod z^{p + 1} \le \dfrac 1 2$ By Bounds for Complex Exponential: :$\cmod {\map {E_p} z - 1} \le 3 \cmod z^{p + 1}$ {{qed}} === Proof of another bound === {{ProofWanted}}
By Order of Product of Entire Functions and Polynomial has Order Zero, $f\cdot P$ has order at most $\omega$. By Limit at Infinity of Polynomial, there exist $r,\delta>0$ such that $|P(z)| \geq \delta$ for $|z|\geq r$. Suppose $\displaystyle \log\left(\max_{|z|\leq R}|f(z)P(z)|\right) = O(R^\beta)$ for some $\beta < \omega$. By the Maximum Modulus Principle $\max_{|z|\leq R}|f(z)| \leq \frac1\delta \max_{|z|\leq R}|f(z)P(z)|$ for $R\geq r$. Thus $\displaystyle \log\left(\max_{|z|\leq R}|f(z)|\right) = O(R^\beta)$. This is a contradiction. Thus $f\cdot P$ has order $\omega$. {{qed}}
This is an immediate consequence of Hadamard Factorization Theorem. {{qed}} Category:Entire Functions fv2qke2k3ajg5o0nqogidmbn8so234u
We may assume $\map f 0 \ne 0$. Let $f$ have finitely many zeroes. Then: :$\tau = 0 \le \omega$ {{qed|lemma}} Let $f$ have infinitely many zeroes. Let $\sequence {a_n}$ be the sequence of nonzero zeroes of $f$, repeated according to multiplicity and ordered by increasing modulus. Let $r_n = \size {a_n}$ and $R_n = 2 \size {a_n}$. By Jensen's Inequality: :$n \le \dfrac {\map \log {\max_{\size z \le R_n} \size f} - \log \size {\map f 0} } {\log 2}$ for all $n \in \N$. Let $\epsilon > 0$. Because $f$ has order $\omega$: :$n \ll_\epsilon \size {a_n}^{\omega + \epsilon}$ Thus: :$\size {a_n}^{-\paren {\omega + \epsilon} } \ll_\epsilon n^{-1}$ By Convergence of P-Series: :$\omega + \epsilon \ge \tau$ Because $\epsilon$ is arbitrary: :$\omega \ge \tau$ {{qed}}
{{NotZFC}} From Biconditional is Transitive and Class is Proper iff Bijection from Class to Proper Class, it suffices to show that: :There exists a bijection from $A$ to $\mathrm P$ iff there exists a bijection from $\mathrm P$ to $A$. The rest follows from Inverse of Bijection is Bijection. {{qed}} Category:Gödel-Bernays Class Theory Category:Class Mappings 309a4dy28gwmdpy3bfpxv5rnvpwu80i
{{NotZFC}} Aiming for contradiction, suppose that $A$ is not proper. Then it must be a set. By Injection to Image is Bijection, it follows that the restriction $f \restriction_{\mathrm P \times f \left({\mathrm P}\right)}$ is a bijection. By the corollary of Class is Proper iff Bijection from Class to Proper Class, $f \left({\mathrm P}\right)$ is proper. But since $f \left({\mathrm P}\right) \subseteq A$, this contradicts Subclass of Set is Set. And so by contradiction, $A$ cannot be a set. Therefore $A$ is proper. {{qed}} Category:Gödel-Bernays Class Theory Category:Class Mappings fztqgk4v4mg4qqcuavzhh0sbuzsfhsj
Follows from a slightly stronger result than Limit at Infinity of Polynomial over Complex Exponential. {{ProofWanted}}
Performing the calculation using long division: <pre>    0.018867924528301...   --------------------- 53)1.000000000000000000      53    212                --    ---      470    280      424    265      ---    ---       460    150       424    106       ---    ---        360    440        318    424        ---    ---         420    160         371    159         ---    ---          490     100          477      53          ---     ---           130    ...           106           ---            240            212            --- </pre> {{qed}}
By Exponent of Convergence is Less Than Order: :$\tau \le \omega$ By Hadamard Factorization Theorem: :$h \le \omega$ Thus: :$\map \max {\tau, h} \le \omega$ By Order is Less Than Maximum of Exponent of Convergence and Degree: :$\omega \le \map \max {\tau, h}$ {{qed}}
By Zerofree Analytic Function on Simply Connected Set has Logarithm, there exists an entire function $g$ with $f = \exp g$. {{ProofWanted}}
By Necessary Condition for Twice Differentiable N Function dependent Functional to have Minimum, $J$ is minimised by $y = \map {\mathbf {\hat y} } x$ if: :$\delta^2 J \sqbrk {\mathbf {\hat y}; \mathbf h} \ge 0$  for all admissable real functions $\mathbf h$. By lemma 1 of Legendre's Condition: :$\ds \delta^2 J \sqbrk {\mathbf y; \mathbf h} = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h} \rd x$ where: :$\mathbf P = F_{y_i' y_j'}$ By Nonnegative Quadratic N function dependent Functional implies no Interior Conjugate Points, $\openint a b$ does not contain any conjugate points {{WRT}} $J$. {{qed}}
{{Refactor|The analyis has already been performed in Birthday Paradox. Extract that general case and make it a theorem, and then introduce Birthday Paradox and this page as examples.}} Let there be $n$ people in the room. Let $\map p n$ be the probability that no two people in the room have the same birthday. For simplicity, let us ignore leap years and assume there are $365$ days in the year. Let the birthday of person $1$ be established. The probability that person $2$ shares person $1$'s birthday is $\dfrac 1 {365}$. Thus, the probability that person $2$ does not share person $1$'s birthday is $\dfrac {364} {365}$. Similarly, the probability that person $3$ does not share the birthday of either person $1$ or person $2$ is $\dfrac {363} {365}$. And further, the probability that person $n$ does not share the birthday of any of the people indexed $1$ to $n - 1$ is $\dfrac {365 - \paren {n - 1} } {365}$. Hence the total probability that none of the $n$ people share a birthday is given by: :$\map p n = \dfrac {364} {365} \dfrac {363} {365} \dfrac {362} {365} \cdots \dfrac {365 - n + 1} {365}$ {{begin-eqn}} {{eqn | l = \map p n       | r = \dfrac {364} {365} \dfrac {363} {365} \dfrac {362} {365} \cdots \dfrac {365 - n + 1} {365}       | c =  }} {{eqn | r = \dfrac {365!} {365^n} \binom {365} n       | c =  }} {{end-eqn}} Setting $n = 53$ and evaluating the above gives: :$\map p {53} \approx 0.01887$ or: :$\map p {53} \approx \dfrac 1 {53.01697}$ {{qed}}
{{begin-eqn}} {{eqn | l = 0       | r = \dfrac {0 \times 1} 2 }} {{eqn | l = 1       | r = \dfrac {1 \times 2} 2 }} {{eqn | l = 3       | r = \dfrac {2 \times 3} 2       | rr= = 1 + 2 }} {{eqn | l = 21       | r = \dfrac {6 \times 7} 2       | rr= = 8 + 13 }} {{eqn | l = 55       | r = \dfrac {10 \times 11} 2       | rr= = 21 + 34 }} {{end-eqn}} It remains to be shown that these are the only ones. Let $F_n$ be the $n$th Fibonacci number. From Odd Square is Eight Triangles Plus One, $F_n$ is triangular {{iff}} $8 F_n + 1$ is square. It remains to be demonstrated that $8 F_n + 1$ is square {{iff}}: :$n \in \set{\pm 1, 0, 2, 4, 8, 10}$ So, let $8 F_n + 1$ be square. Then: :$n \equiv \begin{cases} \pm 1 \pmod {2^5 \times 5} & : n \text { odd} \\ 0, 2, 4, 8, 10 \pmod {2^5 \times 5^2 \times 11} & : n \text { even} \end{cases}$ {{ProofWanted|The proof needs to be studied.}}
{{begin-eqn}} {{eqn | l = 55       | r = \dfrac {10 \times 11} 2 }} {{eqn | l = 66       | r = \dfrac {11 \times 12} 2 }} {{eqn | l = 666       | r = \dfrac {36 \times 37} 2 }} {{end-eqn}} Let $\dfrac {d \paren {10^j - 1}} 9$ be a $j$-digit repdigit number. Suppose it is triangular. For $d = 1$, write: {{begin-eqn}} {{eqn | l = \frac {k \paren {k + 1} } 2       | r = \frac {10^j - 1} 9 }} {{eqn | l = 9 k \paren {k + 1}       | r = 2 \times 10^j - 2 }} {{eqn | l = \paren {3 k + 1} \paren {3 k + 2}       | r = 2^{j + 1} \times 5^j }} {{end-eqn}} Since $3 k + 1 \perp 3 k + 2$, one of them is equal to $2^{j + 1}$ while the other is equal to $5^j$. This gives $2^{j + 1} \pm 1 = 5^j$. Since $j \ge 1$, taking modulo $4$: :$0 \pm 1 \equiv 1^j \equiv 1 \pmod 4$ so we have: :$2^{j + 1} + 1 = 5^j$ By 1 plus Power of 2 is not Perfect Power except 9, the above equation has no solutions for $j > 1$. For $j = 1$ we have equality. Hence $1$ is the only triangular repunit. {{qed|lemma}} By Odd Square is Eight Triangles Plus One: :$x = 1 + \dfrac {8 d \paren {10^j - 1}} 9$ is a perfect square. For $d = 2, 4, 7, 9$, $x$ ends in $2$ or $3$, which is not a perfect square by Square Modulo 5/Corollary. {{qed|lemma}} For $d = 3, 8$, $x$ ends in $5$. Since it is a perfect square, $x$ must end in $25$. This is satisfied only when $d = 3, j = 1$. For $d = 3, j > 1$ and $d = 8, j = 1$, $x$ ends in $65$. For $d = 8, j > 1$, $x$ ends in $05$. This corresponds to $\dfrac {x - 1} 8 = 3$. {{qed|lemma}} For $d = 5$, $x$ is in the form $44 \dots 41$. We show that $x$ is square only when $x = 441$. {{finish|see talk}} This corresponds to $\dfrac {x - 1} 8 = 55$. {{qed|lemma}} For $d = 6$, $x$ is either: :$49$ for $j = 1$ :$53 \dots 329$ for $j > 1$, with $j - 1$ $3$'s we show that $x$ is square only when $x = 49, 529, 5329$. {{finish|see talk}} These correspond to $\dfrac {x - 1} 8 = 6$, $66$ and $666$. {{qed|lemma}} We have hence demonstrated that the only triangular repdigit numbers are: :$1, 3, 6, 55, 66, 666$
Let $t \in T$. By definition of ordering on mappings: :$I_T\left({t}\right) \precsim \left({g \circ d}\right)\left({t}\right)$ By definition of identity mapping: :$t \precsim \left({g \circ d}\right)\left({t}\right)$ By definition of increasing mapping: :$d\left({t}\right) \preceq d\left({\left({g \circ d}\right)\left({t}\right)}\right)$ By definition of composition of mappings: :$d\left({t}\right) \preceq \left({d \circ \left({g \circ d}\right)}\right)\left({t}\right)$ By definition of ordering on mappings: :$\left({d \circ g}\right)\left({d\left({t}\right)}\right) \preceq I_S\left({d\left({t}\right)}\right)$ By definition of identity mapping: :$\left({d \circ g}\right)\left({d\left({t}\right)}\right) \preceq d\left({t}\right)$ By definition of composition of mappings: :$\left({\left({d \circ g}\right) \circ d}\right)\left({t}\right) \preceq d\left({t}\right)$ By Composition of Mappings is Associative: :$\left({d \circ \left({g \circ d}\right)}\right)\left({t}\right) \preceq d\left({t}\right)$ Thus by definition of antisymmetry: :$d\left({t}\right) = \left({d \circ \left({g \circ d}\right)}\right)\left({t}\right)$ {{qed|lemma}} Thus $g = \left({g \circ d}\right) \circ g$ holds by mutatis mutandis. {{qed}}
{{begin-eqn}} {{eqn | l = S_n       | r = \sum_{k \mathop = 1}^n k^2       | c = {{Defof|Square Pyramidal Number}} }} {{eqn | r = \dfrac {n \paren {n + 1} \paren {2 n + 1} } 6       | c = Sum of Sequence of Squares }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = Q_n       | r = \sum_{k \mathop = 1}^n P_k       | c = {{Defof|Pentagonal Pyramidal Number}} }} {{eqn | r = \sum_{k \mathop = 1}^n \dfrac {k \paren {3 k - 1} } 2       | c = Closed Form for Pentagonal Numbers }} {{eqn | r = \dfrac 1 2 \paren {3 \sum_{k \mathop = 1}^n k^2 - \sum_{k \mathop = 1}^n k}       | c =  }} {{eqn | r = \dfrac 1 2 \paren {3 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - \sum_{k \mathop = 1}^n k}       | c = Sum of Sequence of Squares }} {{eqn | r = \dfrac 1 2 \paren {3 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - \dfrac {n \paren {n + 1} } 2}       | c = Closed Form for Triangular Numbers }} {{eqn | r = \dfrac 1 2 \paren {\frac {n \paren {n + 1} \paren {2 n + 1} - n \paren {n + 1} } 2}       | c =  }} {{eqn | r = \dfrac 1 2 \paren {\frac {n \paren {n + 1} \paren {2 n} } 2}       | c =  }} {{eqn | r = \frac {n^2 \paren {n + 1} } 2       | c =  }} {{end-eqn}} {{qed}}
{{NotZFC}} {{AimForCont}} $A$ is not proper. Then $A$ must be a set. By the Axiom of Powers, $\powerset A$ is also a set. Let $g: f \sqbrk A \to \powerset A$ be defined as: :$\map g {\map f a} = f^{-1} \sqbrk {\set {\map f a} }$ It should be noted that: :$\forall a \in A: \map {\paren {g \circ f} } a \ne \O$: Suppose that $\map g {\map f a} = \map g {\map f b}$. Let $x \in f^{-1} \sqbrk {\set {\map f a} }$. Then by the definition of a singleton: :$x \in f^{-1} \sqbrk {\set {\map f a} } \implies \map f x \in \set {\map f a} \implies \map f x = \map f a$ By the same argument: :$x \in f^{-1} \sqbrk {\set {\map f b} } \implies \map f x = \map f b$ Hhence by the properties of equality: :$\map f a = \map f b$ It has been shown that $g$ is an injection. Because $f$ is a surjection, it follows by definition of surjection that: :$f \sqbrk A = \mathrm P$ But this contradicts Injection from Proper Class to Class. Thus by contradiction $A$ must be proper. Hence the result. {{qed}} Category:Gödel-Bernays Class Theory Category:Class Mappings 2uu2rcmfr75ya31wcu0fnuzvv5xm95i
{{NotZFC}} Aiming for contradiction, suppose that $f \sqbrk S$ is not a set. Then $f \sqbrk S$ must be proper. By Restriction of Mapping to Image is Surjection, the restriction $f \restriction_{S \times f \sqbrk S}$ is a surjection. But this contradicts Surjection from Class to Proper Class. Thus by contradiction, $f \sqbrk S$ is a set. Hence the result. {{qed}} Category:Gödel-Bernays Class Theory Category:Class Mappings fdsyztfl8n7ytacauy3trwyj6gk3y16
{{begin-eqn}} {{eqn | l = 1       | r = \dfrac {1 \paren {1 + 1} \paren {2 \times 1 + 1} } 6       | c = Closed Form for Square Pyramidal Numbers }} {{eqn | r = \dfrac {1 \times \paren {1 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 55       | r = \dfrac {5 \paren {5 + 1} \paren {2 \times 5 + 1} } 6       | c = Closed Form for Square Pyramidal Numbers }} {{eqn | r = \dfrac {10 \times \paren {10 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 91       | r = \dfrac {6 \paren {6 + 1} \paren {2 \times 6 + 1} } 6       | c = Closed Form for Square Pyramidal Numbers }} {{eqn | r = \dfrac {13 \times \paren {13 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 208 \, 335       | r = \dfrac {85 \paren {85 + 1} \paren {2 \times 85 + 1} } 6       | c = Closed Form for Square Pyramidal Numbers }} {{eqn | r = \dfrac {645 \times \paren {645 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{ProofWanted|It remains to be shown that these are the only such instances.}}
From the definition of the Dirichlet convolution: :$\displaystyle \left({f * g}\right) \left({n}\right) = \sum_{a b \mathop = n} f \left({a}\right) g \left({b}\right)$ By definition, arithmetic functions are mappings from the natural numbers $\N$ to the complex numbers $\C$. Thus $f \left({a}\right), g \left({b}\right) \in \C$ and commutativity follows from commutativity of multiplication of complex numbers. {{qed}}
By definition of Galois connection: :$g$ is an increasing mapping. Let $t \in T$. By definition of surjection: :$\Img g = T$ By Image of Preimage under Mapping/Corollary: :$g \sqbrk {g^{-1} \sqbrk {t^\succeq} } = t^\succeq$ By Galois Connection is Expressed by Minimum: :$\map d t = \min \set {g^{-1} \sqbrk {t^\succeq} }$ By definition of min operation: :$\map d t = \inf \set {g^{-1} \sqbrk {t^\succeq} }$ and $\map d t \in g^{-1} \sqbrk {t^\succeq}$ By definition of image of set: :$\map g {\map d t} \in g \sqbrk {g^{-1} \sqbrk {t^\succeq} }$ By definition of upper closure of element: :$t \precsim \map g {\map d t}$ By definition of minimum element: :$g^{-1} \sqbrk {t^\succeq}$ admits an infimum. By definition of infimum: :$\map d t$ is lower bound for $g^{-1} \sqbrk {t^\succeq}$ By definition of surjection: :$\exists s \in S: t = \map g s$ By definition of singleton: :$t \in \set t$ By Set is Subset of Upper Closure :$\set t \subseteq \set t^\succeq$ By Upper Closure of Singleton: :$\set t^\succeq = t^\succeq$ By definition of image of set: :$s \in g^{-1} \sqbrk {t^\succeq}$ By definition of lower bound: :$\map d t \preceq s$ By definition of increasing mapping: :$\map g {\map d t} \precsim t$ By definition of antisymmetry: :$\map g {\map d t} = t$ By definition of preimage of set: :$\map d t \in g^{-1} \sqbrk {\set t}$ By Image of Subset under Relation is Subset of Image/Corollary 3: :$g^{-1} \sqbrk {\set t} \subseteq g^{-1} \sqbrk {t^\succeq}$ We will prove that :$\map d t$ is an infimum of $g^{-1} \sqbrk {\set t}$ Thus by Lower Bound is Lower Bound for Subset: :$\map d t$ is lower bound for $g^{-1} \sqbrk {\set t}$ Thus by definition: :$\forall s \in S: s$ is lower bound for $g^{-1} \sqbrk {\set t} \implies s \preceq \map d t$ {{qed|lemma}} Thus by definition of min operation: :$\map d t = \min \set {g^{-1} \sqbrk {\set t} }$ {{qed}}