hcju/proofwikiqa huggingface dataset:
Let: :$\alpha_1 = \displaystyle \limsup_{R \mathop \to \infty} \frac {\displaystyle \ln \ln \max_{\cmod z \mathop \le R} \cmod f} {\ln R}$ :$\alpha_2 = \inf \set {\beta \ge 0: \displaystyle \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } = \map \OO {R^\beta} }$ :$\alpha_3 = \inf \set {\beta \ge 0: \map f z = \map \OO {\map \exp {\cmod z^\beta} } }$ === 3 less than 2 === Let $\beta > \alpha_2$. Then there exists $K > 0$ such that: :$\displaystyle \ln \max_{\cmod z \mathop \le R} \cmod f \le K \cdot R^\beta$ for $R$ sufficiently large. That is: :$\dfrac {\ln \ln \max_{\cmod z \le R} \cmod f} {\ln R} \le \frac K {\ln R} + \beta$ for $R$ sufficiently large. Taking $\limsup$, we get $\alpha_1 \le \beta$. Taking limits, $\alpha_2 \ge \alpha_1$. {{qed|lemma}} === 2 less than 3 === Let $\beta > \alpha_1$. We have $\displaystyle \ln \max_{\cmod z \mathop \le R} \cmod f \le R^\beta$ for $R$ sufficiently large. Thus $\beta \ge \alpha_2$. Taking limits, $\alpha_1 \ge \alpha_2$. {{qed|lemma}} === 1 less than 2 === Let $\beta > \alpha_2$. Let $\epsilon > 0$ be such that $\beta - \epsilon > \alpha_1$. Then: :$\displaystyle \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } = \map \OO {R^{\beta - \epsilon} }$ Then there exists $R_0 > 0$ such that: :$\displaystyle \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } \le R^\beta$ for $R \ge R_0$. By Exponential is Strictly Increasing, this implies: :$\displaystyle \max_{\cmod z \mathop \le R} \cmod {\map f z} \le \map \exp {R^\beta}$ Let $z \in \C$ with $\cmod z \ge R_0$. Then: :$\cmod {\map f z} \le \max_{\cmod w \le \cmod z} \cmod {\map f w} \le \map \exp {\cmod z^\beta}$ Thus: :$\map f z = \map \OO {\map \exp {\cmod z^\beta} }$ Thus $\beta \ge \alpha_3$. Taking the limit $\beta \to \alpha_2$, we obtain: :$\alpha_2 \ge \alpha_3$ {{qed|lemma}} {{ProofWanted}} Category:Entire Functions l8pol3ujopmruaz3voa85cyvlb36j7f
The square of a $2$-digit integer cannot have more than $4$ digits: :$99^2 = 9801$ The square of a $4$-digit integer has at least $7$ digits: :$1000^2 = 1 \, 000 \, 000$ Hence we only need to inspect $3$-digit integers, with a corresponding $6$-digit square. A lower bound is given by $\ceiling {\sqrt {123456}} = 352$, where $123456$ is the smallest $6$-digit integer without repeating digits or $0$. An upper bound is given by $\floor {\sqrt {876543}} = 936$, where $876543$ is the largest $6$-digit integer without repeating digits or $9$. It is seen that it is not necessary to investigate $6$-digit squares beginning with $9$, because its square root would also begin with $9$. It is noted that integers ending in $1$, $5$ or $6$ have squares ending in those same digits. Such numbers can be eliminated from our search, as they will duplicate the appearance of those digits. We also cannot have $0$ as a digit. Moreover, observe that an integer $\bmod 9$ and its square $\bmod 9$ have the following pattern: :$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline n \bmod 9 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline n^2 \bmod 9 & 0 & 1 & 4 & 0 & 7 & 7 & 0 & 4 & 1 \\ \hline \end{array}$ By Congruence of Sum of Digits to Base Less 1, the sum of digits of $n$ and $n^2 \pmod 9$ are congruent to $n$ and $n^2 \pmod 9$ respectively. We must require their sum to be congruent to $0 \bmod 9$ since the sum of all their digits is: :$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 \equiv 0 \pmod 9$ Hence $n \equiv 0$ or $8 \pmod 9$. {{finish|There are $52$ numbers left to check:}} Those divisible by $9$: :$369, 378, 387, 423, 432, 459, 468, 513, 549, 567, 594, 612, 639, 648, 657, 684, 693, 729, 738, 783, 792, 819, 837, 864, 873, 918, 927$ Those not divisible by $9$: :$359, 368, 413, 458, 467, 512, 539, 548, 584, 593, 629, 638, 647, 674, 683, 692, 719, 728, 764, 782, 827, 854, 863, 872, 917$
{{begin-eqn}} {{eqn | o =        | r = 587       | c = is prime }} {{eqn | l = 3 \times 587 + 16       | r = 1777       | c = which is prime }} {{eqn | l = 3 \times 1777 + 16       | r = 5347       | c = which is prime }} {{eqn | l = 3 \times 5347 + 16       | r = 16 \, 057       | c = which is prime }} {{eqn | l = 3 \times 16 \, 057 + 16       | r = 48 \, 187       | c = which is prime }} {{eqn | l = 3 \times 48 \, 187 + 16       | r = 144 \, 577       | c = which is prime }} {{eqn | l = 3 \times 144 \, 577 + 16       | r = 433 \, 747       | c = which is prime }} {{eqn | l = 3 \times 433 \, 747 + 16       | r = 1 \, 301 \, 257       | c = which is prime }} {{eqn | l = 3 \times 1 \, 301 \, 257 + 16       | r = 3 \, 903 \, 787       | c = which is prime }} {{eqn | l = 3 \times 3 \, 903 \, 787 + 16       | r = 11 \, 711 \, 377       | c = which is prime }} {{eqn | l = 3 \times 11 \, 711 \, 377 + 16       | r = 35 \, 134 \, 147       | c = which is prime }} {{eqn | l = 3 \times 35 \, 134 \, 147 + 16       | r = 105 \, 402 \, 457       | c = which is not prime : $105 \, 402 \, 457 = 67 \times 137 \times 11 \, 483$ }} {{end-eqn}} {{qed}}
Let $\epsilon > 0$. Then by definition of little-o notation: :$\exists n_1 \in \N: \paren {n \ge n_1 \implies \size {a_n} \le \epsilon \cdot \size {b_n}}$ :$\exists n_2 \in \N: \paren {n \ge n_2 \implies \size {c_n} \le \epsilon \cdot \size {d_n}}$ For $n \ge \max \set {n_1, n_2}$: {{begin-eqn}} {{eqn | l = \size {a_n + c_n}       | o = \le       | r = \size {a_n} + \size {c_n}       | c = Triangle Inequality }} {{eqn | o = \le       | r = \epsilon \cdot \size {b_n} + \epsilon \cdot \size {d_n} }} {{eqn | r = \epsilon \cdot \size {\size {b_n} + \size {d_n} }       | c = Absolute value is positive }} {{end-eqn}} Hence by definition of little-o notation: :$a_n + c_n = \map o {\size {b_n} + \size {d_n} }$ {{qed}} Category:Asymptotic Notation kospwasornr8yzsr5oo9di0ssd1ev2u
=== 2 iff 3 === Let $a_n - b_n = \map o {b_n}$. Let $0 < \epsilon < 1/2$. Then {{begin-eqn}} {{eqn | l = \epsilon \cdot \size {b_n}       | o = \ge       | r = \size {a_n - b_n}       | c = For $n$ sufficiently large }} {{eqn | r = \paren {1 - \epsilon} \size {a_n - b_n} + \epsilon \cdot \size {a_n - b_n}       | c =  }} {{eqn | o = \ge       | r = \paren {1 - \epsilon} \size {a_n - b_n} - \epsilon \cdot \size {a_n} + \epsilon \cdot \size {b_n}       | c = Triangle Inequality }} {{end-eqn}} So $\size {a_n - b_n} \le \dfrac {\epsilon \cdot \size {a_n} } {1 - \epsilon} \le 2 \epsilon \cdot \size {a_n}$. Thus $a_n - b_n = \map o {a_n}$. The other implication follows by symmetry. {{qed|lemma}} === 1 iff 2 === {{begin-eqn}} {{eqn | l = \lim_{n \to \infty} \dfrac {a_n} {b_n}       | o = \to       | r = 1 }} {{eqn | ll = \leadstoandfrom       | l = \lim_{n \to \infty} \paren {\dfrac {a_n} {b_n} - \dfrac {b_n} {b_n} }       | o = \to       | r = 0 }} {{eqn | ll = \leadstoandfrom       | l = \lim_{n \to \infty} \dfrac {a_n - b_n} {b_n}       | o = \to       | r = 0 }} {{eqn | ll = \leadstoandfrom       | l = a_n - b_n       | r = \map o {b_n} }} {{end-eqn}} {{qed}} Category:Asymptotic Notation d46q51xbla75ikhfwlgi0fbzo6j4d6p
By Largest Number not Expressible as Sum of Multiples of Coprime Integers, the largest such number is: :$\paren {23 - 1} \times \paren {28 - 1} - 1 = 593$ {{ProofWanted|This is a specific "historical" example of a general result which I read in Polya and Szego some time back, which still needs to be added into {{ProofWiki}}. All in due course.  I may already have it on here, I misremember.}}
By definition of continuous ordered set: :$p^\ll$ is directed and :$L$ satisfies axiom of approximation. By axiom of approximation: :$p = \sup \left({p^\ll}\right)$ By definition of Scott topology: :$A$ is inaccessible by directed suprema. By definition of inaccessible by directed suprema: :$A \cap p^\ll \ne \varnothing$ By definition of non-empty set: :$\exists q: q \in A \cap p^\ll$ Thus by definition of intersection: :$q \in A$ By definition of intersection: :$q \in p^\ll$ By definition of way below closure: :$q \ll p$ {{qed}}
We have: :$\left\{ {G \in \left\{ {g^\gg: g \in S}\right\}: G \subseteq X}\right\} = \left\{ {x^\gg: x \in S \land x^\gg \subseteq X}\right\}$ By Way Above Closures Form Basis: :$\left\{ {x^\gg: x \in S}\right\}$ is basis of $\left({S, \tau}\right)$. By Interior is Union of Elements of Basis: :$X^\circ = \bigcup \left\{ {x^\gg: x \in S \land x^\gg \subseteq X}\right\}$ {{qed}}
Define $B = \left\{ {x^\gg: x \in S}\right\}$ Thus by Way Above Closure is Open: :$B \subseteq \tau$ We will prove that :for all $x \in S$: there exists local basis $Q$ of $x$: $Q \subseteq B$ Let $x \in S$. By Way Above Closures that Way Below Form Local Basis: :$Q := \left\{ {g^\gg: g \in S \land g \ll x}\right\}$ is local basis at $x$. Thus by definition of subset: :$Q \subseteq B$ {{qed|lemma}} Thus by Characterization of Analytic Basis by Local Bases: :$B$ is basis of $L$. {{qed}}
First it is noted that $3$ and $5$ twin primes which are both Sophie Germain. Prime numbers greater than $3$ are of the form $6 n - 1$ and $6 n + 1$. Thus a pair of twin primes is of the form $\left({6 n - 1, 6 n + 1}\right)$. The result follows from Sophie Germain Prime cannot be 6n+1. {{qed}} Category:Twin Primes Category:Sophie Germain Primes snxstobc2zqlmxfufczkwl31v2642ye
Let $p$ be a Sophie Germain prime. Then, by definition, $2 p + 1$ is prime. {{AimForCont}} $p = 6 n + 1$ for some $n \in \Z_{>0}$. Then: {{begin-eqn}} {{eqn | l = 2 p + 1       | r = 2 \paren {6 n + 1} + 1       | c =  }} {{eqn | r = 12 n + 3       | c =  }} {{eqn | r = 3 \paren {4 n + 1}       | c =  }} {{end-eqn}} and so $2 p + 1$ is not prime. The result follows by Proof by Contradiction. {{qed}} Category:Sophie Germain Primes runb03zh6lpwhgzstdneqn105w7lp87
By Way Above Closure is Upper: :$x^\gg$ is upper. We will prove that :$x^\gg$ is inaccessible by directed suprema. Let $D$ be a directed subset of $S$ such that :$\sup D \in x^\gg$ By definition of way above closure: :$x \ll \sup D$ By Way Below iff Second Operand Preceding Supremum of Directed Set There Exists Element of Directed Set First Operand Way Below Element: :$\exists d \in D: x \ll d$ By definition of way above closure: :$d \in x^\gg$ By definitions of intersection and non-empty set: :$x^\gg \cap D \ne \varnothing$ {{qed|lemma}} Thus by definition of Scott topology: :$x^\gg$ is open. {{qed}}
Let $y \in x^\gg$, $z \in S$ such that :$y \preceq z$ By definition of way above closure: :$x \ll y$ By Preceding and Way Below implies Way Below: :$x \ll z$ Thus by definition of way above closure: :$z \in x^\gg$ {{qed}}
Define $B := \left\{ {q^\gg: q \in S \land q \ll p}\right\}$ By Way Above Closure is Open: :$B \subseteq \tau$ By definition of way above closure: :$\forall X \in B: p \in X$ Thus by definition: :$B$ is set of open neighborhoods. Let $U$ be an open subset of $S$ such that :$p \in U$ By Open implies There Exists Way Below Element: :$\exists u \in U: u \ll p$ Thus by definition of $B$: :$u^\gg \in B$ By definition of Scott topology: :$U$ is upper. We will prove that :$u^\gg \subseteq U$ Let $z \in u^\gg$ By definition of way above closure: :$u \ll z$ By Way Below implies Preceding: :$u \preceq z$ Thus by definition of upper set: :$z \in U$ {{qed}}
By assumption: :$P \subseteq \tau$ Let $U$ be an open subset of $S$. Define: :$X := \set {V \in P: V \subseteq U}$ By definition of subset: :$X \subseteq P$ We will prove that: :$\forall u \in S: u \in U \iff \exists Z \in X: u \in Z$ Let $u \in S$. We will prove that: :$u \in U \implies \exists Z \in X: u \in Z$ Assume that: :$u \in U$ By assumption: :there exists local basis $B$ at $u: B \subseteq P$. By definition of local basis: :$\exists V \in B: V \subseteq U$ Thus by definitions of subset and $X$: :$V \in X$ Thus by definition of local basis: :$u \in V$ {{qed|lemma}} Assume that: :$\exists Z \in X: u \in Z$ By definition of $X$: :$Z \subseteq U$ Thus by definition of subset: :$u \in U$ {{qed|lemma}} Thus by definition of union: :$U = \bigcup X$ Hence $P$ is basis of $L$. {{qed}}
From Euler Phi Function of Integer: :$\displaystyle \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$ where $p \divides n$ denotes the primes which divide $n$. We have that: :$666 = 2 \times 3^2 \times 37$ Thus: {{begin-eqn}} {{eqn | l = \map \phi {666}       | r = 666 \paren {1 - \dfrac 1 2} \paren {1 - \dfrac 1 3} \paren {1 - \dfrac 1 {37} }       | c =  }} {{eqn | r = 666 \times \frac 1 2 \times \frac 2 3 \frac {36} {37}       | c =  }} {{eqn | r = 3 \times 1 \times 2 \times 36       | c =  }} {{eqn | r = 3 \times 2 \times \paren {2^2 \times 3^2}       | c =  }} {{eqn | r = 216       | c =  }} {{eqn | r = 6 \times 6 \times 6       | c =  }} {{end-eqn}} {{qed}}
By definition of interior: :$\left\{ {G \in B: G \subseteq V}\right\} = \left\{ {G \in B: G \subseteq V^\circ}\right\}$ and :$V^\circ$ is open. Thus by Open Set is Union of Elements of Basis: :$V^\circ = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$ {{qed}}
By investigating all square numbers which are palindromic. {{begin-eqn}} {{eqn | l = 676       | r = 26^2 }} {{eqn | l = 69 \, 696       | r = 264^2 }} {{eqn | l = 94 \, 249       | r = 307^2 }} {{eqn | l = 698 \, 896       | r = 836^2 }} {{eqn | l = 5 \, 221 \, 225       | r = 2285^2 }} {{eqn | l = 6 \, 948 \, 496       | r = 2636^2 }} {{eqn | l = 522 \, 808 \, 225       | r = 22 \, 865^2 }} {{end-eqn}} and so on. {{qed}}
{{begin-eqn}} {{eqn | o =       | r = 680       | c = the $15$th tetrahedral number }} {{eqn | r = 120       | c = the $8$th tetrahedral number }} {{eqn | o =       | ro= +       | r = 560       | c = the $14$th tetrahedral number }} {{end-eqn}} {{ProofWanted|Demonstrate cases}}
Let $x$ be arbitrary. We will prove that :$x \in V \implies \exists Y \in \left\{ {G \in B: G \subseteq V}\right\}: x \in Y$ Assume that :$x \in V$ By definition of basis: :$\exists F \subseteq B: V = \bigcup F$ By definition of union: :$\exists Y \in F: x \in Y$ By Set is Subset of Union/General Result: :$Y \subseteq V$ Thus by definition of subset: :$Y \in \left\{ {G \in B: G \subseteq V}\right\}$ Thus $x \in Y$ {{qed|lemma}} Assume that :$\exists Y \in \left\{ {G \in B: G \subseteq V}\right\}: x \in Y$ By assumption: :$Y \subseteq V$ Thus by definition of subset: :$x \in V$ Hence by definition of union: :$V = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$ {{qed}}
Let $x, y \in S$ such that :$x \preceq_1 y$ By Compact Closure is Increasing: :$x^{\mathrm{compact} } \subseteq y^{\mathrm{compact} }$ By Image of Subset under Relation is Subset of Image/Corollary 2: :$f\left[{x^{\mathrm{compact} } }\right] \subseteq f\left[{y^{\mathrm{compact} } }\right]$ By assumption: :$f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \preceq_1 x \land w}\right.$ is compact$\left.{}\right\}$ and :$f\left({y}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \preceq_1 y \land w}\right.$ is compact$\left.{}\right\}$ By definitions of image of set and compact closure: :$f\left({x}\right) = \sup \left({f\left[{x^{\mathrm{compact} } }\right]}\right)$ and :$f\left({y}\right) = \sup \left({f\left[{y^{\mathrm{compact} } }\right]}\right)$ Thus by Supremum of Subset and definition of complete lattice: :$f\left({x}\right) \preceq_2 f\left({y}\right)$ {{qed}}
=== 1 implies 2 === Follows immediately from the definition of $C^k$-atlas. {{qed|lemma}} === 2 implies 1 === Let $(U,\phi)$ and $(V,\psi)$ be charts in $\mathscr F \cup \mathscr G$. If they are both in $\mathscr F$, they are $C^k$-compatible because $\mathscr F$ is a $C^k$-atlas. If they are both in $\mathscr G$, they are $C^k$-compatible because $\mathscr G$ is a $C^k$-atlas. If $(U,\phi) \in \mathscr F$ and $(V,\psi) \in \mathscr G$, they are $C^k$-compatible by assumption. Thus $\mathscr F \cup \mathscr G$ is a $C^k$-atlas. {{qed}} Category:Manifolds kbyy48ho66z0akiplglybgcbh9iadz9
For every $t\in N$, the mapping: :$f_t: M \to M : x \mapsto f \left({x, t}\right)$ is a contraction. By the Banach Fixed-Point Theorem, there exists a unique $g \left({t}\right) \in M$ such that $f_t \left({g \left({t}\right)}\right) = g \left({t}\right)$. We show that $g$ is continuous. Let $K < 1$ be a uniform Lipschitz constant for $f$. Let $s, t \in N$. Then {{begin-eqn}} {{eqn | l = d \left({g \left({s}\right), g \left({t}\right)}\right)       | r = d \left({f \left({g \left({s}\right), s}\right), f \left({g \left({t}\right), t}\right)}\right)       | c = Definition of $g$ }} {{eqn | o = \le       | r = d \left({f \left({g \left({s}\right), s}\right), f \left({g \left({t}\right), s}\right)}\right) + d \left({f \left({g \left({t}\right), s}\right), f \left({g \left({t}\right), t}\right)}\right)        | c = {{Defof|Metric}} }} {{eqn | o = \le       | r = K \cdot d \left({g \left({s}\right), g \left({t}\right)}\right) + d \left({f \left({g \left({t}\right), s}\right), f \left({g \left({t}\right), t}\right)}\right)       | c = $f$ is a uniform contraction }} {{end-eqn}} and thus: :$d \left({g \left({s}\right), g \left({t}\right)}\right) \le \dfrac 1 {1 - K} d  \left({f \left({g \left({t}\right), s}\right), f \left({g \left({t}\right), t}\right)}\right)$ The continuity of $g$ now follows from that of $f$ using the definition of product metric {{qed}}
==== Reduction to $\tuple {a, b} = \tuple {0, 0}$ ==== We may assume $\tuple {a, b} = \tuple {0, 0}$. {{explain|why}} Define: :$F : \Omega \to \R^k: \map F {x, y} = y - \map {D_2 \map f {a, b}^{-1} } {\map f {x, y} }$ By Linear Function is Continuous, $D_2 \map f {a, b}^{-1}$ is continuous on $\R^k$. Thus $F$ is continuous on $\Omega$. ==== $F$ is locally a uniform contraction ==== Let $r_1 > 0$ such that the open ball $\map B {0, r_1} \subseteq \Omega$. We have, for $\tuple {x, y_1}, \tuple {x, y_2} \in \map B {0, r_1}$, by the Mean Value Inequality: :$\norm {\map F {x, y_2} - \map F {x, y_1} } \le \displaystyle \sup_{y \mathop \in \closedint {y_1} {y_2} } \norm {D_2 \map F {x, y} } \cdot \norm {y_2 - y_1}$ We have, for $\tuple {x, y} \in \Omega$: :$D_2 \map F {x, y} = I - \paren {D_2 \map F {0, 0} }^{-1} \circ D_2 \map F {x, y}$ where $I$ is the identity mapping. Thus $D_2 F$ is continuous in $\Omega$. By definition, $D_2 \map F {0, 0} = 0$. By continuity, there exists $r_2 > 0$ such that: :$\norm {D_2 \map F {x, y} } \le \dfrac 1 2$ for $\norm {\tuple {x, y} } \le r_2$ From the above inequality, $F$ is a uniform contraction mapping on the open ball $\map B {0, r_2}$ for the Lipschitz Constant $\dfrac 1 2$. ==== Constructing a stable neighborhood ==== We have, for $x, y \in \map {\overline B} {0, r_2}$: {{begin-eqn}} {{eqn | l = \norm {\map F {x, y} }       | o = \le       | r = \norm {\map F {x, y} - \map F {x, 0} } + \norm {\map F {x, 0} }       | c = Triangle Inequality }} {{eqn | o = \le       | r = \frac 1 2 \norm y + \norm {\map F {x, 0} }       | c = Definition of $r_2$ }} {{end-eqn}} Because $\map F {0, 0} = 0$ and $F$ is continuous, there exists $r_3>0$ such that $\norm {\map F {x, 0} } \le \dfrac {r_2} 2$. Then: :$\norm {\map F {x, y} } \le r_2$ for $\norm y \le r_2$ and: :$\norm x \le r_3$ Thus the restriction of $F$ to $\map B {0, r_3} \times \map {\overline B} {0, r_2}$ is a uniform contraction: :$F: \map B {0, r_3} \times \map {\overline B} {0, r_2} \to \map {\overline B} {0, r_2}$ ==== Applying the Uniform Contraction Theorem ==== By Subspace of Complete Metric Space is Closed iff Complete, $\map {\overline B} {0, r_2}$ is complete. By the Uniform Contraction Mapping Theorem, there exists a unique mapping $g: \map B {0, r_3} \to \map {\overline B} {0, r_2}$ such that $\map F {x, \map g x} = \map g x$. Moreover, $g$ is continuous. A mapping $h: \map B {0, r_3} \to \map {\overline B} {0, r_2}$ satisfies $\map F {x, \map h x} = \map h x$ {{iff}} it satisfies $\map {D_2 \map f {a, b}^{-1} } {\map f {x, \map h x} } = 0$ {{iff}} it satisfies $\map f {x, \map h x} = 0$, because $D_2 \map f {a, b}^{-1}$ is invertible. Thus $g$ is the unique mapping $g: \map B {0, r_3} \to \map {\overline B} {0, r_2}$ such that $\map f {x, \map g x} = 0$. {{qed}}
Let $x \in S$. Define $X = \left\{ {f\left({w}\right): w \in S \land w \preceq_1 x \land w}\right.$ is compact$\left.{}\right\}$ Define $Y = \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$ We will prove that :$X \subseteq Y$ Let $b \in X$. By definition of $X$: :$\exists w \in S: b = f\left({w}\right) \land w \preceq_1 x \land w$ is compact. By definition of compact element: :$w \ll w$ By Preceding and Way Below implies Way Below: :$w \ll x$ Thus by definition of $Y$: :$b \in Y$ {{qed|lemma}} We will prove that :$f\left({x}\right)$ is upper bound for $Y$. Let $b \in Y$. By definition of $Y$: :$\exists w \in S: b = f\left({w}\right) \land w \ll x$ By Way Below implies Preceding: :$w \preceq_1 x$ By Mapping at Element is Supremum implies Mapping is Increasing: :$f$ is an increasing mapping. Thus by definition of increasing mapping: :$b \preceq_2 f\left({x}\right)$ {{qed|lemma}} We will prove that :$\forall b \in T: b$ is upper bound for $Y \implies f\left({x}\right) \preceq_2 b$ Let $b \in T$ such that :$b$ is upper bound for $Y$. By Upper Bound for Subset: :$b$ is upper bound for $X$. By assumption: :$f\left({x}\right) = \sup X$ Thus by definition of supremum: :$f\left({x}\right) \preceq_2 b$ {{qed|lemma}} Thus by definition of supremum: :$f\left({x}\right) = \sup Y$ {{qed}}
We have that: :$714 + 715 = 1429$ Hence we investigate its anagrams. We bother only to check those which do not end in either $2$ or $4$, as those are even. {{begin-eqn}} {{eqn | l = 1429       | o =        | c = is prime }} {{eqn | l = 1249       | o =        | c = is prime }} {{eqn | l = 4129       | o =        | c = is prime }} {{eqn | l = 4219       | o =        | c = is prime }} {{eqn | l = 2149       | r = 7 \times 307       | c = and so is not prime }} {{eqn | l = 2419       | r = 41 \times 59       | c = and so is not prime }} {{eqn | l = 9241       | o =        | c = is prime }} {{eqn | l = 9421       | o =        | c = is prime }} {{eqn | l = 2941       | r = 17 \times 173       | c = and so is not prime }} {{eqn | l = 2491       | r = 47 \times 53       | c = and so is not prime }} {{eqn | l = 4291       | r = 7 \times 613       | c = and so is not prime }} {{eqn | l = 4921       | r = 7 \times 19 \times 37       | c = and so is not prime }} {{end-eqn}} Of the above, $6$ are seen to be prime. {{qed}}
Follows from Factorial as Product of Two Factorials: :$10! = 6! \times 7!$ and so: :$\dfrac {10!} {7!} = 10 \times 9 \times 8 = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ Hence the result. {{qed}}
We show that for $n \ge 7$, $n!$ cannot be expressed as the sum of two squares. By refining the result in Interval containing Prime Number of forms 4n - 1, 4n + 1, 6n - 1, 6n + 1, one can show that: :There exists a prime of the form $4 k + 3$ strictly between $m$ and $2 m$ whenever $m \ge 4$. Let $n \ge 7$. Then $\ceiling {\dfrac n 2} \ge 4$. Using the result above, there is a prime $p$ of the form $4 k + 3$ such that: :$\ceiling {\dfrac n 2} < p < 2 \ceiling {\dfrac n 2}$ We then have, by multiplying the inequality by $2$: :$2 \ceiling {\dfrac n 2} < 2 p < 4 \ceiling {\dfrac n 2}$ This gives: :$p < 2 \ceiling {\dfrac n 2} < 2 p$ Which implies: :$p \le n < 2 p$ From Integer as Sum of Two Squares: :$n!$ can be expressed as the sum of two squares {{iff}} each of its prime divisors of the form $4 k + 3$ (if any) occur to an even power. The inequality above shows that there are no multiples of $p$ which are not greater than $n$ except $p$ itself. Hence $p$ occurs to an odd power, $1$, in $n!$. This shows that for $n \ge 7$, $n!$ cannot be expressed as the sum of two squares. Checking the rest of the factorials we see that the only ones satisfying the criteria are: {{begin-eqn}} {{eqn | l = 0! = 1!       | r = 0^2 + 1^2 }} {{eqn | l = 2!       | r = 1^2 + 1^2 }} {{eqn | l = 6!       | r = 12^2 + 24^2 }} {{end-eqn}} Hence the result. {{qed}}