hcju/proofwikiqa huggingface dataset:
Define $B := \left\{ {\complement_S\left({x^\succeq}\right): x \in S}\right\}$ By definition of lower topology: :$B$ is an analytic sub-basis. Let $F$ be a filtered subset of $S$ such that :$F$ admits an infimum in $T$. Thus by definition of complete lattice: :$f\left[{F}\right]$ admits an infimum in $Q$. We will prove that :$\forall A \in B: \inf F \in A \implies F \cap A \ne \varnothing$ Let $A \in B$ be such that :$\inf F \in A$ By definition of $B$: :$\exists x \in S: A = \complement_S\left({x^\succeq}\right)$ By definition of relative complement: :$\inf F \notin x^\succeq$ By definition of upper closure of element: :$x \npreceq \inf F$ By definition of infimum: :$x$ is not lower bound for $F$. By definition of lower bound: :$\exists y \in F: x \npreceq y$ By definition of upper closure of element: :$y \notin x^\succeq$ By definition of relative complement: :$y \in A$ By definitions of intersection and non-empty set: :$F \cap A \ne \varnothing$ {{qed|lemma}} Then by If Infimum of Filtered Subset belongs to Element of Sub-Basis then Subset and Element Intersect implies Infimum of Subset belongs to Closure of Subset: :$\inf F \in F^-$ We will prove that :$f$ is an increasing mapping. Let $x, y \in S$ be such that :$x \preceq y$ By definition of reflexivity: :$f\left({x}\right) \preceq' f\left({x}\right)$ By definition of upper closure of element: :$f\left({x}\right) \in \left({f\left({x}\right)}\right)^{\succeq'}$ By definition of preimage of set: :$x \in f^{-1}\left[{\left({f\left({x}\right)}\right)^{\succeq'} }\right]$ By Complement of Upper Closure of Element is Open in Lower Topology: :$\left({f\left({x}\right)}\right)^{\succeq'}$ is closed. By Continuity by Closed Sets: :$f^{-1}\left[{\left({f\left({x}\right)}\right)^{\succeq'} }\right]$ is closed. By Closed Subset is Upper in Lower Topology: :$f^{-1}\left[{\left({f\left({x}\right)}\right)^{\succeq'} }\right]$ is upper. By definition of upper set: :$y \in f^{-1}\left[{\left({f\left({x}\right)}\right)^{\succeq'} }\right]$ By definition of preimage of set: :$f\left({y}\right) \in \left({f\left({x}\right)}\right)^{\succeq'}$ Thus by definition of upper closure of element: :$f\left({x}\right) \preceq' f\left({y}\right)$ {{qed|lemma}} We will prove that :$f\left({\inf F}\right)$ is lower bound for $f\left[{F}\right]$ By definition of infimum: :$f\left({\inf F}\right) \preceq' \inf\left({f\left[{F}\right]}\right)$ We will prove that :$F \subseteq f^{-1}\left[{\left({\inf\left({f\left[{F}\right]}\right)}\right)^{\succeq'} }\right]$ {{qed}}
By Intersection of Ring Ideals is Ideal, $P$ is an ideal of $R$. We show that $P$ is prime. Let $a, b\in R$ with $a, b \notin P$. We show that $a b \notin P$. Because $a \notin P$, there exists $\alpha \in A$ with $a \notin P_\alpha$. Because $b \notin P$, there exists $\beta \in A$ with $b \notin P_\beta$. Because $\set {P_\gamma}_{\gamma \mathop \in A}$ is totally ordered, $P_\alpha \subseteq P_\beta$ or $P_\beta \subseteq P_\alpha$. {{WLOG}}, we can assume $P_\alpha \subseteq P_\beta$, which gives us $a, b \notin P_\alpha$. Since $P_\alpha$ is an prime ideal, $a b \notin P_\alpha$. Thus $a b \notin P$. {{qed}}
There are $20$ ways to make the $1$st move by White: :Each of the $8$ pawns may be moved either $1$ or $2$ squares forward, making $16$ moves :Each of the $2$ knights may be moved to either of $2$ squares before it, making $4$ moves. For each of those $20$ first moves by White, Black has the same $20$ options. Thus there are $20 \times 20$ possible different games after the $2$nd move. To count the $3$rd moves, one needs to consider cases. First note that after the $1$st move, whatever it was, there are $7$ pawns on the $2$nd rank, each of which can be moved $1$ or $2$ squares forward, making $14$ moves for each of those $400$ possibilities. Note that if a knight was one of the pieces to have moved first, the pawn in the square behind where it ends up cannot move -- hence the $7$ pawns on the $2$nd rank that can move. Thus there are $400 \times 14 = 5600$ possible moves involving a so-far unmoved pawn. For each of the $400$ positions, there are exactly $8$ which consist of two pawns in opposition on the $4$th and $5$th rank. There are also another $4 \times 20 = 80$ positions in which white moved a knight. For all other $400 - 88 = 312$ positions, the already-moved pawn has the option of moving another square forward. This gives another $312$ options for the $3$rd move. We now need to take into account the possibility that White may be able to capture a Black pawn. {{finish|Each case needs to be investigated.}}
{{begin-eqn}} {{eqn | l = 11 \, 593       | r = 4 \times 2898 + 1 }} {{eqn | l = 11 \, 597       | r = 4 \times 2899 + 1 }} {{eqn | l = 11 \, 617       | r = 4 \times 2904 + 1 }} {{eqn | l = 11 \, 621       | r = 4 \times 2905 + 1 }} {{eqn | l = 11 \, 633       | r = 4 \times 2908 + 1 }} {{eqn | l = 11 \, 657       | r = 4 \times 2914 + 1 }} {{eqn | l = 11 \, 677       | r = 4 \times 2919 + 1 }} {{eqn | l = 11 \, 681       | r = 4 \times 2920 + 1 }} {{eqn | l = 11 \, 689       | r = 4 \times 2922 + 1 }} {{end-eqn}} It remains to be noted that: :the prime number before $11 \, 593$ is $11 \, 587$ which is $4 \times 2897 - 1$ :the prime number after $11 \, 689$ is $11 \, 699$ which is $4 \times 2925 - 1$ confirming that they are not of the form $4 n + 1$. {{qed}}
To streamline the argument, the term '''$9$-pan''' will be coined to mean '''$9$-digit pandigital number which excludes the $0$ digit'''. Let $n$ be the smallest positive integer whose square is '''$9$-pan'''. First it is noted that the smallest '''$9$-pan''' is $123 \, 456 \, 789$. Hence any square '''$9$-pan''' must be at least as large as that. Thus we can can say that: :$n \ge \ceiling {\sqrt {123 \, 456 \, 789} } = 11 \, 112$ where $\ceiling {\, \cdot \,}$ denotes the ceiling function. It remains to be demonstrated that no positive integer between $11 \, 112$ and $11 \, 826$ has a '''$9$-pan''' square. {{ProofWanted|This is over $700$ numbers. The task can be filtered by, for example, disregarding all $n$ ending in $1$ and $9$ because their squares will both begin and end in $1$ at this low range of the $10000$s, and of course all $n$ ending in $0$ because their squares will end in $0$.}}
Because $T$ is locally compact, every point of $S$ has a neighborhood basis consisting of compact sets. From Space is Neighborhood of all its Points, $S$ is a neighborhood of each point of $S$.   By assumption, each point has a compact neighborhood contained in $S$. Thus $T$ is weakly locally compact. {{qed}}
By the definition of the topology $\tau$, $X$ is an open set. From Set is Open iff Neighborhood of all its Points, $X$ is a neighborhood of $x$.
{{begin-eqn}} {{eqn	| l = \cosh^2 x 	| r = \cos^2 i x 	| c = Hyperbolic Cosine in terms of Cosine }} {{eqn	| r = \frac {\map \cos {2 i x} + 1} 2 	| c = Square of Cosine }} {{eqn	| r = \frac {\cosh 2 x + 1} 2 	| c = Hyperbolic Cosine in terms of Cosine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \cosh^3 x 	| r = \cos^3 i x 	| c = Hyperbolic Cosine in terms of Cosine }} {{eqn	| r = \frac {\map \cos {3 i x} + 3 \cos i x} 4 	| c = Cube of Cosine }} {{eqn	| r = \frac {\cosh 3 x + 3 \cosh x} 4 	| c = Hyperbolic Cosine in terms of Cosine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \cosh^4 x 	| r = \cos^4 i x 	| c = Hyperbolic Cosine in terms of Cosine }} {{eqn	| r = \frac {3 + 4 \cos \paren {2 i x} + \cos \paren {4 i x} } 8 	| c = Fourth Power of Cosine }} {{eqn	| r = \frac {3 + 4 \cosh 2 x + \cosh 4 x} 8 	| c = Hyperbolic Cosine in terms of Cosine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \map {\frac \d {\d x} } {\sech x} 	| r = 2 \map {\frac \d {\d x} } {\frac {e^x} {e^{2 x} + 1} } 	| c = {{Defof|Hyperbolic Secant}} }} {{eqn	| r = \frac 2 {\paren {e^{2 x} + 1}^2} \paren {\map {\frac \d {\d x} } {e^x} \paren {e^{2 x} + 1} - e^x \map {\frac \d {\d x} } {e^{2 x} + 1} } 	| c = Quotient Rule for Derivatives }} {{eqn	| r = -\frac 2 {\paren {e^{2 x} + 1}^2} \paren {2 e^{2 x} \cdot e^x - e^x \cdot e^{2 x} - e^x} 	| c = Derivative of Exponential Function }} {{eqn	| r = -\frac {2 \paren {e^{3 x} - e^x} } {\paren {e^{2 x} + 1}^2} }} {{eqn	| r = -\frac {2 e^x} {\paren {e^{2 x} + 1} } \cdot \frac {e^{2 x} - 1} {e^{2 x} + 1}  }} {{eqn	| r = -\sech x \tanh x 	| c = {{Defof|Hyperbolic Secant}}, {{Defof|Hyperbolic Tangent}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\sech x}       | r = \map {\frac \d {\d x} } {\frac 1 {\cosh x} }       | c = {{Defof|Hyperbolic Secant}} }} {{eqn | r = \map {\frac \d {\d x} } {\paren {\cosh x}^{-1} }       | c = Exponent Laws }} {{eqn | r = -\paren {\cosh x}^{-2} \sinh x       | c = Derivative of Hyperbolic Cosine, Power Rule for Derivatives, Chain Rule for Derivatives }} {{eqn | r = \frac {-1} {\cosh x} \frac {\sinh x} {\cosh x}       | c = Exponent Combination Laws }} {{eqn | r = -\sech z \tanh z       | c = {{Defof|Hyperbolic Secant}} and {{Defof|Hyperbolic Tangent}} }} {{end-eqn}} {{qed}}
Let $x \in S$. Consider the set $\set {p, x}$. From the definition of particular point topology, $\set {p, x}$ is open in $T$. By Finite Topological Space is Compact, $\set {p, x}$ is compact. Let $N$ be a neighborhood of $x$. Then: :$\exists U \in \tau_p: x \in U \subseteq N$. From the definition of particular point topology, since $U \ne \O$, we must have $p \in U$. Therefore $\set {p, x} \subseteq U \subseteq N$. Since $N$ is arbitrary, $\set {\set {p, x}}$ is a neighborhood basis for $x$. The result follows from definition of a locally compact space. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1134}^{2133} k^3       | r = \sum_{k \mathop = 1}^{2133} k^3 - \sum_{k \mathop = 1}^{1133} k^3       | c =  }} {{eqn | r = \frac {2133^2 \paren {2133 + 1}^2} 4 - \frac {1133^2 \paren {1133 + 1}^2} 4       | c = Sum of Sequence of Cubes }} {{eqn | r = \frac {\paren {3^3 \times 79}^2 \paren {2 \times 11 \times 97}^2} 4 - \frac {\paren {11 \times 103}^2 \paren {2 \times 3^4 \times 7}^2} 4       | c = extracting Prime Decomposition }} {{eqn | r = 3^6 \times 11^2 \paren {\paren {79 \times 97}^2 - \paren {103 \times 3 \times 7}^2}       | c = factorising }} {{eqn | r = 3^6 \times 11^2 \paren {7663^2 - 2163^2}       | c = factorising }} {{eqn | r = 3^6 \times 11^2 \paren {7663 + 2163} \times \paren {7663 - 2163}       | c = Difference of Two Squares }} {{eqn | r = 3^6 \times 11^2 \times 9826 \times 5500       | c =  }} {{eqn | r = 3^6 \times 11^2 \times \paren {2 \times 17^3} \times \paren {2^2 \times 5^3 \times 11}       | c =  }} {{eqn | r = 2^3 \times 3^6 \times 5^3 \times 11^3 \times 17^3       | c =  }} {{eqn | r = \paren {2 \times 3^2 \times 5 \times 11 \times 17}^3       | c =  }} {{eqn | r = 16 \, 830^3       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 1       | r = \dfrac {1 \paren {1 + 1} \paren {1 + 2} } 6       | c = Closed Form for Tetrahedral Numbers }} {{eqn | r = 1^2       | c = {{Defof|Square Number}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 4       | r = \dfrac {2 \paren {2 + 1} \paren {2 + 2} } 6       | c = Closed Form for Tetrahedral Numbers }} {{eqn | r = 2^2       | c = {{Defof|Square Number}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 19 \, 600       | r = \dfrac {48 \paren {48 + 1} \paren {48 + 2} } 6       | c = Closed Form for Tetrahedral Numbers }} {{eqn | r = 140^2       | c = {{Defof|Square Number}} }} {{end-eqn}} {{ProofWanted|It remains to be shown that these are the only such instances.}}
First we show that for $1 < k < 90$, $315 k$ is abundant. If $k$ is divisible by $3, 5, 7$, note that: :$945, 1575, 2205$ are all abundant, and $315 k$ is a multiple of at least one of them. Hence $315 k$ is abundant by Multiple of Abundant Number is Abundant. If $k$ is not divisible by $3, 5, 7$: Let $p$ be a prime such that $p \divides k$. Then: {{begin-eqn}} {{eqn | l = \frac {\map \sigma {315 p} } {315 p}       | r = \frac 1 {315 p}  \paren {1 + 3 + 3^2} \paren {1 + 5} \paren {1 + 7}  \paren {1 + p} }} {{eqn | r = \frac {208} {105} \paren {1 + \frac 1 p} }} {{eqn | o = >       | r = \frac {208} {105} \paren {1 + \frac 1 {90} }       | c = $p < 90$ }} {{eqn | o = >       | r = 2 }} {{end-eqn}} hence $315 p$ and $315 k$ are abundant. Since $88$ and $315$ are coprime: :$88 = 2^3 \times 11$ :$315 = 3^2 \times 5 \times 7$ By Largest Number not Expressible as Sum of Multiples of Coprime Integers, all numbers greater than or equal to: :$\paren {88 - 1} \paren {315 - 1} = 27 \, 318$ can be expressed as a sum of multiples of $88$ and $315$. Hence for $n \ge 27 \, 318 + 315 \times 2 = 27 \, 948$: :$\exists s, t \in \N: 90 > t \ge 2: n = 88 s + 315 t$ and both $88 s$ and $315 t$ are abundant for $s > 0$. For $s = 0$, $t \ge \dfrac {27 \, 948} {315} > 2 = \paren {2 - 1} \paren {3 - 1}$. By Largest Number not Expressible as Sum of Multiples of Coprime Integers, $t - 5$ can be expressed as a sum of multiples of $2$ and $3$. Hence: :$\exists a, b \in \Z_{> 0}: 2 a + 3 b = t$ This gives: :$n = 630 a + 945 b$ and both $630 a$ and $945 b$ are abundant. We still need to find representations for $20 \, 162 < n < 27 \, 948$. We can check this via brute force. Using Largest Number not Expressible as Sum of Multiples of Coprime Integers/Generalization, we can narrow down our search to numbers that are not divisible by small primes: Since $\gcd \set {18, 20} = 2$, the largest multiple of $2$ not expressible as a sum of multiples of $18$ and $20$ is: :$\dfrac {18 \times 20} 2 - 18 - 20 = 142 < 20161$ Since $\gcd \set {12, 945} = 3$, the largest multiple of $3$ not expressible as a sum of multiples of $12$ and $945$ is: :$\dfrac {12 \times 945} 3 - 12 - 945 = 2823 < 20161$ Since $\gcd \set {20, 945} = 5$, the largest multiple of $5$ not expressible as a sum of multiples of $20$ and $945$ is: :$\dfrac {20 \times 945} 5 - 20 - 945 = 2815 < 20161$ Since $\gcd \set {56, 945} = 7$, the largest multiple of $7$ not expressible as a sum of multiples of $56$ and $945$ is: :$\dfrac {56 \times 945} 7 - 56 - 945 = 6559 < 20161$ All numbers involved above are abundant. Hence we only need to consider $n$ not divisible by $2, 3, 5, 7$. {{finish|Brute force by computer}}
All integers up to $100$ ('''one hundred''') use either $1$ or $2$ words: :'''one''' :'''sixty''' :'''seventeen''' :'''ninety-eight''' All integers of the form $100 n$ for $n = 1, 2, \ldots 9$ use exactly $2$ words: :'''one hundred''' :'''seven hundred''' :'''nine hundred''' In British English, the technique for describing integers from $101$ to $199$, and $201$ to $299$ and so on, is to use '''and''' between the number of hundreds and the rest: :'''one hundred and one''' :'''three hundred and thirteen''' :'''four hundred and twenty-six''' :'''seven hundred and seventy''' thus using either $4$ or $5$ words. All integers of the form $1000 n$ for $n = 1, 2, \ldots 10$ use exactly $2$ words: :'''two thousand''' :'''five thousand''' :'''eight thousand''' :'''twelve thousand''' :'''nineteen thousand''' :'''twenty thousand''' Similarly with hundreds, the technique for describing integers of the form $1000 m + n$ for $1 \le n \le 99$ is to use '''and''' between the number of thousands and the rest: :'''five thousand and eighteen''' :'''sixteen thousand and forty-eight''' :'''thirty-seven thousand and sixty''' thus using either $4$ or $5$ words. All other numbers between $1100$ and $20 \, 999$ trivially use more than $3$ words: :'''four thousand, eight hundred''' :'''sixteen thousand, one hundred and seventy-seven''' :'''twenty thousand, nine hundred and ninety-nine''' and so on. The smallest integer to use exactly $2$ words is $21$: :'''twenty-one''' Hence: :'''twenty-one thousand''' {{qed}}
:800px The $4$ right triangles are inspected: {{begin-eqn}} {{eqn | l = 25^2 + 60^2       | r = 625 + 3600 }} {{eqn | r = 4225       | c =  }} {{eqn | r = 65^2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 91^2 + 60^2       | r = 8281 + 3600 }} {{eqn | r = 11 \, 881       | c =  }} {{eqn | r = 109^2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 91^2 + 312^2       | r = 8281 + 97 \, 344 }} {{eqn | r = 105 \, 625       | c =  }} {{eqn | r = 325^2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 25^2 + 312^2       | r = 625 + 97 \, 344 }} {{eqn | r = 97 \, 969       | c =  }} {{eqn | r = 313^2       | c =  }} {{end-eqn}} The area of each right triangle is calculated: {{begin-eqn}} {{eqn | l = \dfrac {25 \times 60} 2       | r = 750 }} {{eqn | l = \dfrac {91 \times 60} 2       | r = 2730 }} {{eqn | l = \dfrac {91 \times 312} 2       | r = 14 \, 196 }} {{eqn | l = \dfrac {25 \times 312} 2       | r = 3900 }} {{end-eqn}} Thus the total area is: :$750 + 2730 + 14 \, 196 + 3900 = 21 \, 576$ {{ProofWanted|It needs to be proved that this is the smallest.}}
Suppose $N = \map \tau {N^3}$. The case $N = 1$ is trivial. Suppose $N$ is a prime power. Write $N = p^n$. By Tau of Power of Prime: :$N = \map \tau {p^{3 n} } = 3 n + 1$ By Bernoulli's Inequality: :$N = p^n \ge 1 + n \paren {p - 1}$ This gives us the inequality: :$3 n + 1 \ge 1 + n \paren {p - 1}$ which can be simplified to: :$3 \ge p - 1$ The only primes satisfying the inequality are $2$ and $3$. We have: :$\map \tau {2^3} = 4 > 2^1$ :$\map \tau {2^6} = 7 > 2^2$ :$\map \tau {2^9} = 10 > 2^3$ :$\map \tau {2^{3 n} } = 3 n + 1 < 2^n$ for $n > 3$ :$\map \tau {3^3} = 4 > 3^1$ :$\map \tau {3^{3 n} } = 3 n + 1 < 3^n$ for $n > 1$ :$\map \tau {p^{3 n} } = 3 n + 1 < p^n$ for all $p > 3$ Hence no prime powers satisfy the property. Note that Tau Function is Multiplicative. To form an integer $N$ with our property, we must choose and multiply prime powers from the list above. If we chose any $\tuple {p, n}$ with $\map \tau {p^{4 n} } < p^n$, we must choose $2^m$ or $3^1$ in order for equality to possibly hold. If $\tuple {2, 1}$ was chosen, $2^2 \nmid N$. But $\map \tau {2^3} = 4 \divides N$, which is a contradiction. Suppose $\tuple {2, 2}$ was chosen. Then $\map \tau {2^6} = 7 \divides N$. Then we must choose some $\tuple {7, n}$. For $n = 1$, $\map \tau {7^3} = 4$. :$\map \tau {2^6 \times 7^3} = 4 \times 7 = 28 = 2^2 \times 3$ :$\map \tau {2^6 \times 7^3 \times p^{3 m} } = 28 \paren {3 m + 1} < 28 \times p^m$ for all $p \ne 2, 7$ For $n > 1$, $\map \tau {2^6 \times 7^{3 n} } = 7 \paren {3 n + 1} < 4 \times 7^n$, a contradiction. Suppose $\tuple {2, 3}$ was chosen. Then: :$\map \tau {2^9} = 10 \divides N$ Then we must choose some $\tuple {5, n}$. For $n = 1$, $\map \tau {5^3} = 4$. :$\map \tau {2^9 \times 5^3} = 10 \times 4 = 40 = 2^3 \times 5$ :$\map \tau {2^9 \times 5^3 \times p^{3 m} } = 40 \paren {3 m + 1} < 40 \times p^m$ for all $p \ne 2, 5$ For $n > 1$, $\map \tau {2^9 \times 5^{3 n} } = 10 \paren {3 n + 1} < 2^3 \times 5^n$, a contradiction. Suppose $\tuple {3, 1}$ was chosen. Then $\map \tau {3^3} = 4 \divides N$. Then this case coincides the cases above. Thus we have exhausted all cases. {{qed}}
{{ProofWanted|We have the list of Poulet numbers and Fermat pseudoprimes base $3$, but not of bases $5$ and $7$. Once we get those lists, we can find the numbers on the lists for both.}}
From Sigma Function of Integer: :$\displaystyle \map \sigma n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ Suppose $k_i = 1$. Then we have: {{begin-eqn}} {{eqn | l = \frac {p_i^{k_i + 1} - 1} {p_i - 1}       | r = \frac {p_i^2 - 1} {p_i - 1}       | c =  }} {{eqn | r = \frac {\paren {p_i + 1} \paren {p_i - 1} } {p_i - 1}       | c = Difference of Two Squares }} {{eqn | r = p_i + 1       | c =  }} {{end-eqn}} Thus the contribution from  a prime factor which is square-free can be expressed in the simpler form $p_i + 1$ instead of the more unwieldy form $\dfrac {p_i^2 - 1} {p_i - 1}$. Hence the result. {{qed}}
By definition, a (group) isomorphism is a (group) homomorphism which is a bijection. Recall the definition of underlying set of $1$-dimensional affine group on $\R$: :$S = \set {f_{a b}: x \mapsto a x + b : a \in \R_{\ne 0}, b \in \R}$ So the bijection $\psi: \map {\operatorname {Af}_1} \R \to \R^+ \rtimes_\phi \R^\times$ defined by $\map \psi {f_{a b} } = \paren {b, a}$ arises naturally. It remains to show that $\psi$ is a (group) homomorphism: Let $f_{a b}, f_{c d} \in \map {\operatorname {Af}_1} \R$. Then: {{begin-eqn}} {{eqn | l = \map {f_{a b} \circ f_{c d} } x       | r = a \paren {c x + d} + b       | c = {{Defof|Affine Group of One Dimension}} }} {{eqn | r = a c x + a d + b       | c =  }} {{eqn | r = f_{\paren {a c} \paren {a d + b} }       | c =  }} {{end-eqn}} Let $\tuple {b, a}, \tuple {d, c} \in \R^+ \rtimes_\phi \R^\times$. Then: {{begin-eqn}} {{eqn | l = \tuple {b, a} \tuple {d, c}       | r = \paren {b + \map {\map \phi a} d, a c}       | c = {{Defof|Semidirect Product}} }} {{eqn | r = \paren {b + a d, a c}       | c = Definition of $\phi$ }} {{eqn | r = \paren {a d + b, a c}       | c = Real Addition is Commutative }} {{end-eqn}} So: :$\map \psi {f_{a b} } \, \map \psi {f_{c d} } = \map \psi {f_{a b} \circ f_{c d} }$ So the bijection $\psi$ is a (group) homomorphism, and thus a (group) isomorphism. {{qed}}
{{begin-eqn}} {{eqn | l = 1       | r = \dfrac {1 \paren {3 \times 1 - 1} } 2       | c = Closed Form for Pentagonal Numbers }} {{eqn | r = 1 \paren {2 \times 1 - 1}       | c = Closed Form for Hexagonal Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 40 \, 755       | r = \dfrac {165 \paren {3 \times 165 - 1} } 2       | c = Closed Form for Pentagonal Numbers }} {{eqn | r = 143 \paren {2 \times 143 - 1}       | c = Closed Form for Hexagonal Numbers }} {{end-eqn}} {{ProofWanted|It remains to be shown that these are the only such instances.}}
From Sequence of Carmichael Numbers: {{:Definition:Carmichael Number/Sequence}} The sequence continues: :$\ldots, 29 \, 341, 41 \, 041$ We now investigate their prime factors: {{begin-eqn}} {{eqn | l = 561       | r = 3 \times 11 \times 17 }} {{eqn | l = 1105       | r = 5 \times 13 \times 17 }} {{eqn | l = 1729       | r = 7 \times 13 \times 19 }} {{eqn | l = 2465       | r = 5 \times 17 \times 29 }} {{eqn | l = 2821       | r = 7 \times 13 \times 31 }} {{eqn | l = 6601       | r = 7 \times 23 \times 41 }} {{eqn | l = 8911       | r = 7 \times 19 \times 67 }} {{eqn | l = 10 \, 585       | r = 5 \times 29 \times 73 }} {{eqn | l = 15 \, 841       | r = 7 \times 31 \times 73 }} {{eqn | l = 29 \, 341       | r = 13 \times 37 \times 61 }} {{eqn | l = 41 \, 041       | r = 7 \times 11 \times 13 \times 41 }} {{end-eqn}} {{qed}}
Performing the calculation using long division: <pre>    0.04761901...   -------------- 21)1.00000000000      84      --      160      147      ---       130       126       ---         40         21        ---         190         189         ---           100            84           ---           ... </pre> {{qed}}
By Language of Propositional Logic has Unique Parsability, $\LL_0$ is uniquely parsable. Therefore, the Principle of Definition by Structural Induction can be applied to $\LL_0$. By inspection, we see that the definition of the boolean interpretation $v$ follows the bottom-up specification of propositional logic. Hence the Principle of Definition by Structural Induction implies that $v$ is well-defined. {{qed}}