hcju/proofwikiqa huggingface dataset:
Let $n = r s$. Then: {{begin-eqn}} {{eqn | l = R_n       | r = \sum_{k \mathop = 0}^{n - 1} 10^k       | c = Basis Representation Theorem }} {{eqn | r = \sum_{j \mathop = 0}^{s - 1} \paren {\sum_{k \mathop = 0}^{r - 1} 10^k} 10^{r j}       | c =  }} {{eqn | r = \paren {\sum_{k \mathop = 0}^{r - 1} 10^k} \paren {\sum_{j \mathop = 0}^{s - 1}  10^{r j} }       | c =  }} {{eqn | r = R_r \paren {\sum_{j \mathop = 0}^{s - 1}  10^{r j} }       | c =  }} {{end-eqn}} Similarly: :$R_n = R_s \paren {\displaystyle \sum_{j \mathop = 0}^{r - 1}  10^{s j} }$ {{qed}} Thus, for example: {{begin-eqn}} {{eqn | l = R_6 = 111 \, 111       | r = 11 \times 10 \, 101 }} {{eqn | r = 111 \times 1001 }} {{eqn | l = R_{10} = 1 \, 111 \, 111 \, 111       | r = 11 \times 101 \, 010 \, 101 }} {{eqn | r = 11111 \times 100 \, 001 }} {{end-eqn}} The pattern is clear. Category:Repunits ro5wr7je4f0budpzjm2tq6uvpxg6f87
Let $m$ be a repunit base $9$ with $n$ digits. We have: {{begin-eqn}} {{eqn | l = m       | r = \displaystyle \sum_{k \mathop = 0}^{n - 1} 9^k       | c = Basis Representation Theorem }} {{eqn | r = \dfrac {9^n - 1} 8       | c = Sum of Geometric Sequence }} {{eqn | r = \dfrac {3^{2 n} - 1} 8       | c =  }} {{eqn | r = \dfrac {\paren {3^n - 1} \paren {3^n + 1} } 8       | c = Difference of Two Squares }} {{eqn | r = \dfrac {\paren {\paren {2 s + 1} - 1} \paren {\paren {2 s + 1} + 1} } 8       | c = $3^n$ is odd, and so $3^n = 2 s + 1$ for some integer $s$ }} {{eqn | r = \dfrac {2 s \paren {2 s + 2} } 8       | c = simplifying }} {{eqn | r = \dfrac 1 2 \dfrac {2 s} 2 \dfrac {2 s + 2} 2       | c =  }} {{eqn | r = \dfrac {s \paren {s + 1} } 2       | c =  }} {{end-eqn}} The result follows from Closed Form for Triangular Numbers. {{qed}}
The $n$th power of $10$ has $n + 1$ digits. Hence the $n$th power of $m$ such that $m > 10$ has more than $n$ digits. The $11$th power of $8$ has $10$ digits: :$8^{10} = 8 \, 589 \, 934 \, 592$ and so $8^n$ where $n > 10$ has fewer than $n$ digits. Hence the $n$th power of $m$ such that $m < 9$ and $n < 21$ has fewer than $n$ digits. Finally we note that: :$9^{22} = 984 \, 770 \, 902 \, 183 \, 611 \, 232 \, 881$ has $21$ digits. Hence the $n$th power of $9$ such that $n > 21$ has fewer than $n$ digits. Hence the result. {{qed}}
We have that: :$112 \, 359 \, 550 \, 561 \, 797 \, 732 \, 809 = 7 \times 11 \times 61 \times 87 \, 629 \times 337 \, 411 \times 809 \, 063$ while: {{begin-eqn}} {{eqn | l = 11 \, 123 \, 595 \, 505 \, 617 \, 977 \, 328 \, 091       | r = 3^2 \times 7 \times 11^2 \times 61 \times 87 \, 629 \times 337 \, 411 \times 809 \, 063       | c =  }} {{eqn | r = 3^2 \times 11 \times 112 \, 359 \, 550 \, 561 \, 797 \, 732 \, 809       | c =  }} {{end-eqn}} Let $N$ be the smallest integer satisfying $99 N = \sqbrk {1N1}$ when expressed in decimal notation. Suppose $N$ is $k$ digits long. Then: :$\sqbrk {1 N 1} = 10^{k + 1} + 10 N + 1$ Subtracting $10 N$ from $99 N$ gives: :$89 N = 10^{k + 1} + 1$ One can show, by trial and error, that the smallest $k$ where $10^{k + 1} + 1$ is divisible by $89$ is $21$. Then $N = \dfrac {10^{22} + 1} {89} = 112 \, 359 \, 550 \, 561 \, 797 \, 732 \, 809$. {{qed}}
First it is demonstrated that $357 \, 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137$ is indeed a left-truncatable prime: {{begin-eqn}} {{eqn | o =        | l = 357 \, 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 57 \, 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 7 \, 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 86 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 6 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 312 \, 646 \, 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 12 \, 646 \, 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 2 \, 646 \, 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 646 \, 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 46 \, 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 6 \, 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 216 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 16 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 6 \, 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 567 \, 629 \, 137       | c = is prime }} {{eqn | o =        | l = 67 \, 629 \, 137       | c = is the $3 \, 986 \, 726$th prime }} {{eqn | o =        | l = 7 \, 629 \, 137       | c = is the $516 \, 434$th prime }} {{eqn | o =        | l = 629 \, 137       | c = is the $51 \, 275$th prime }} {{eqn | o =        | l = 29 \, 137       | c = is the $3167$th prime }} {{eqn | o =        | l = 9 \, 137       | c = is the $1133$rd prime }} {{eqn | o =        | l = 137       | c = is the $33$rd prime }} {{eqn | o =        | l = 37       | c = is the $12$th prime }} {{eqn | o =        | l = 7       | c = is the $4$th prime }} {{end-eqn}} {{ProofWanted|It remains to be shown that it is the largest such.}}
{{begin-eqn}} {{eqn | l = 3       | r = 1 \times 3 }} {{eqn | l = 36       | r = 2 \times 18 }} {{eqn | l = 360       | r = 3 \times 120 }} {{eqn | l = 3608       | r = 4 \times 902 }} {{eqn | l = 36 \, 085       | r = 5 \times 7217 }} {{eqn | l = 360 \, 852       | r = 6 \times 60 \, 142 }} {{eqn | l = 3 \, 608 \, 528       | r = 7 \times 515 \, 504 }} {{eqn | l = 36 \, 085 \, 288       | r = 8 \times 4 \, 510 \, 661 }} {{eqn | l = 360 \, 852 \, 885       | r = 9 \times 40 \, 094 \, 765 }} {{eqn | l = 3 \, 608 \, 528 \, 850       | r = 10 \times 360 \, 852 \, 885 }} {{eqn | l = 36 \, 085 \, 288 \, 503       | r = 11 \times 3 \, 280 \, 480 \, 773 }} {{eqn | l = 360 \, 852 \, 885 \, 036       | r = 12 \times 30 \, 071 \, 073 \, 753 }} {{eqn | l = 3 \, 608 \, 528 \, 850 \, 368       | r = 13 \times 277 \, 579 \, 142 \, 336 }} {{eqn | l = 36 \, 085 \, 288 \, 503 \, 684       | r = 14 \times 2 \, 577 \, 520 \, 607 \, 406 }} {{eqn | l = 360 \, 852 \, 885 \, 036 \, 840       | r = 15 \times 24 \, 056 \, 859 \, 002 \, 456 }} {{eqn | l = 3 \, 608 \, 528 \, 850 \, 368 \, 400       | r = 16 \times 225 \, 533 \, 053 \, 148 \, 025 }} {{eqn | l = 36 \, 085 \, 288 \, 503 \, 684 \, 007       | r = 17 \times 2 \, 122 \, 664 \, 029 \, 628 \, 471 }} {{eqn | l = 360 \, 852 \, 885 \, 036 \, 840 \, 078       | r = 18 \times 20 \, 047 \, 382 \, 502 \, 046 \, 671 }} {{eqn | l = 3 \, 608 \, 528 \, 850 \, 368 \, 400 \, 786       | r = 19 \times 189 \, 922 \, 571 \, 072 \, 021 \, 094 }} {{eqn | l = 36 \, 085 \, 288 \, 503 \, 684 \, 007 \, 860       | r = 20 \times 1 \, 804 \, 264 \, 425 \, 184 \, 200 \, 393 }} {{eqn | l = 360 \, 852 \, 885 \, 036 \, 840 \, 078 \, 603       | r = 21 \times 17 \, 183 \, 470 \, 716 \, 040 \, 003 \, 743 }} {{eqn | l = 3 \, 608 \, 528 \, 850 \, 368 \, 400 \, 786 \, 036       | r = 22 \times 164 \, 024 \, 038 \, 653 \, 109 \, 126 \, 638 }} {{eqn | l = 36 \, 085 \, 288 \, 503 \, 684 \, 007 \, 860 \, 367       | r = 23 \times 1 \, 568 \, 925 \, 587 \, 116 \, 695 \, 993 \, 929 }} {{eqn | l = 360 \, 852 \, 885 \, 036 \, 840 \, 078 \, 603 \, 672       | r = 24 \times 15 \, 035 \, 536 \, 876 \, 535 \, 003 \, 275 \, 153 }} {{eqn | l = 3 \, 608 \, 528 \, 850 \, 368 \, 400 \, 786 \, 036 \, 725       | r = 25 \times 144 \, 341 \, 154 \, 014 \, 736 \, 031 \, 441 \, 469 }} {{end-eqn}} From No Polydivisible Number with 26 Digits Exists, there are no polydivisible numbers with more digits. {{ProofWanted|It remains to be shown there is no larger one with $25$ digits.}}
Let $p q = 10^n$ for some $n \in \Z_{>0}$. Then $p q = 2^n 5^n$. {{WLOG}}, suppose $p = 2^r 5^s$ for $r, s \ge 1$. Then $2 \times 5 = 10$ is a divisor of $p$ and so $p$ ends with a zero. Thus for $p$ and $q$ to be zero-free, it must be the case that $p = 2^n$ and $q = 5^n$ (or the other way around). The result follows from Powers of 2 and 5 without Zeroes. {{qed}}
By direct calculation: {{begin-eqn}} {{eqn | o =       | r = 191 \, 918 \, 080 \, 818 \, 091 \, 909 \, 090 \, 909 \, 190 \, 818 \, 080 \, 819 \, 191       | c = is palindromic and prime }} {{eqn | o =        | r = 2 \times 191 \, 918 \, 080 \, 818 \, 091 \, 909 \, 090 \, 909 \, 190 \, 818 \, 080 \, 819 \, 191 + 1       | c =  }} {{eqn | r = 383 \, 836 \, 161 \, 636 \, 183 \, 818 \, 181 \, 818 \, 381 \, 636 \, 161 \, 638 \, 383       | c = is palindromic and prime }} {{eqn | o =        | r = 2 \times 383 \, 836 \, 161 \, 636 \, 183 \, 818 \, 181 \, 818 \, 381 \, 636 \, 161 \, 638 \, 383 + 1       | c =  }} {{eqn | r = 767 \, 672 \, 323 \, 272 \, 367 \, 636 \, 363 \, 636 \, 763 \, 272 \, 323 \, 276 \, 767       | c = is palindromic and prime }} {{eqn | o =        | r = 2 \times 767 \, 672 \, 323 \, 272 \, 367 \, 636 \, 363 \, 636 \, 763 \, 272 \, 323 \, 276 \, 767 + 1       | c =  }} {{eqn | r = 1 \, 535 \, 344 \, 646 \, 544 \, 735 \, 272 \, 727 \, 273 \, 526 \, 544 \, 646 \, 553 \, 535       | c = is neither prime nor palindromic }} {{end-eqn}} {{qed}}
Let $N$ be divisible by each of the integers from $1$ to $100$. Each prime number between $2$ and $97$ must be a divisor of $N$. Also: :$2^6 = 64 \divides N$ :$3^4 = 81 \divides N$ :$5^2 = 25 \divides N$ :$7^2 = 49 \divides N$ Every other integer between $1$ and $100$ is the product of a subset of all of these. Hence by Euclid's Lemma: {{begin-eqn}} {{eqn | l = N       | o = \ge       | r = 2^6 \times 3^4 \times 5^2 \times 7^2 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47 \times 53 \times 59 \times 61 \times 67 \times 71 \times 73 \times 79 \times 83 \times 89 \times 97       | c =  }} {{eqn | r = 69 \, 720 \, 375 \, 229 \, 712 \, 477 \, 164 \, 533 \, 808 \, 935 \, 312 \, 303 \, 556 \, 800       | c =  }} {{end-eqn}} {{qed}}
First we demonstrate that this number has this property: {{begin-eqn}} {{eqn | o =        | r = 10 \, 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 808 \, 988 \, 764 \, 044 \, 943 \, 820 \, 224 \, 719 \times 9       | c =  }} {{eqn | r = 91 \, 011 \, 235 \, 955 \, 056 \, 179 \, 775 \, 280 \, 898 \, 876 \, 404 \, 494 \, 382 \, 022 \, 471       | c =  }} {{end-eqn}} A way to find this positive integer is given below. Suppose $N = \sqbrk {a_k a_{k - 1} \dots a_1}$ satisfies the property above. Consider the rational number represented by $q = 0.\dot {a_k} a_{k - 1} \dots \dot {a_1}$. Since $a_k \ne 0$, we require $q \ge 0.1$. By the property we have $9q = 0.\dot {a_1} a_k \dots \dot {a_2}$. Then $90 q = a_1.\dot {a_k} a_{k - 1} \dots \dot {a_1}$. Notice that the recurring part of $90 q$ is the same as the recurring part of $q$. Subtracting, we get $89 q = a_1$. We have $a_1 \ge 89 \times 0.1 = 8.9$, so $a_1 = 9$, which gives $q = \dfrac 9 {89}$. In decimal form: :$q = 0. \dot 1011235955056179775280898876404494382022471 \dot 9$. which gives us $N = 10 \, 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 808 \, 988 \, 764 \, 044 \, 943 \, 820 \, 224 \, 719$ as the smallest positive integer satisfying the property. {{qed}}
The existence and uniqueness of $f$ follows from the universal property. Likewise, there is a unique ring homomorphism $g: R \sqbrk Y \to R \sqbrk X$ such that: :$g \circ \kappa = \iota$ :$\map g Y = X$ and a unique ring homomorphism $h: R \sqbrk X \to R \sqbrk X$ such that: :$h \circ \iota = \iota$ :$\map X = X$ By uniqueness and Identity Mapping is Ring Homomorphism, $h = \operatorname{id}$ is the identity mapping on $R \sqbrk X$. Again by uniqueness and Composition of Ring Homomorphisms is Ring Homomorphism, $g\circ f = \operatorname{id}_{R \sqbrk X}$. By symmetry, $f \circ g = \operatorname{id}_{R \sqbrk Y}$. Thus $f$ is an isomorphism. {{qed}} Category:Polynomial Theory 7oevp9mhpny0q2646gcng8tg9nsbvwj
=== 1 iff 2 === This is by definition of coefficients and degree. {{qed|lemma}} === 1 iff 3 === {{proof wanted|use Definition:Module Structure of Polynomial Ring}} Category:Constant Polynomials bblquu7tqa8syoo2ej80tsu3klpwgew
=== 1 equals 2 === By Minimal Polynomial is Irreducible, it follows that the two are equal. {{qed|lemma}} === 1 equals 3 === Let $f \in K \sqbrk x$ be the monic polynomial of smallest degree such that $\map f \alpha = 0$. Let $g \in K \sqbrk x$ be a polynomial. If $f \divides g$, then $g = q f$ for some $q \in K \sqbrk x$. Thus $\map g \alpha = \map f \alpha \, \map q \alpha = 0$. Conversely, suppose $\map g \alpha = 0$. By the Division Theorem for Polynomial Forms over Field, there exists $q, r \in K \sqbrk x$ such that: :$g = q f + r$ and: :$r = 0$ or $\deg r < \deg f$. Evaluating this expression at $\alpha$ we find that: :$\map g \alpha = \map q \alpha \, \map f \alpha + \map r \alpha \implies \map r \alpha = 0$ since $\map f \alpha = \map g \alpha = 0$. But $f$ has minimal degree among the non-zero polynomials that are zero at $\alpha$. Therefore $r = 0$. Therefore: :$g = q f$ That is, $f$ divides $g$. {{qed}} Category:Minimal Polynomials 43r03ieejpk5uv4gr3hkohcxitv89fj
=== Existence === Follows from Well-Ordering Principle and because $K$ is a field. {{explain|link to some result why we may assume $f$ is monic}} {{qed|lemma}} === Uniqueness === Let $f$ and $g$ be two such monic polynomial. Then: :$\map f \alpha - \map g \alpha = 0$ {{AimForCont}} $f - g \ne 0$. Let $a$ be the leading coefficient of $f-g$. Then $\dfrac 1 {a} \paren {f - g}$ is another such polynomial of strictly smaller degree. This is a contradiction. Thus $f - g = 0$. {{qed}} Category:Minimal Polynomials go17sjwcnpz95u5shizfzxrprqw3oii
=== Existence === Follows from Annihilating Polynomial of Minimal Degree is Irreducible. {{qed|lemma}} === Uniqueness === Let $f$ and $g$ be irreducible monic polynomials in $K[x]$ with $f(\alpha) = g(\alpha) = 0$. Suppose $f$ and $g$ are distinct. Then $f$ and $g$ are coprime. Thus there exist polynomials $a, b \in k[x]$ with $af+bg = 1$. Taking the evaluation homomorphism in $\alpha$, we obtain the contradiction that $0=1$. {{qed}} Category:Minimal Polynomials bqyswfeiuxfiaa79gu20wqdstqkn8jj
{{AimForCont}} that $f$ is not irreducible. Then there exist non-constant polynomials $g, h \in K \sqbrk x$ such that $f = g h$. We have: :$0 = \map f \alpha = \map g \alpha \map h \alpha$ Since $L$ is a field, it is an integral domain. Therefore, as $\map g \alpha, \map h \alpha \in L$, either $\map g \alpha = 0$ or $\map h \alpha = 0$. This contradicts the minimality of the degree of $f$. Hence the result, by Proof by Contradiction. {{qed}}
{{proof wanted|see Equivalence of Definitions of Integral Dependence}} Category:Definitions/Commutative Algebra mohc7qdkq03gf281yvw0cyqjaeb8zgy
First note that: {{begin-eqn}} {{eqn | l = \prod_{i \mathop \in I} C_i       | r = \set {x \in S : \forall i \in I, x_i \in C_i}       | c = {{Defof|Cartesian Product}} }} {{eqn | r = \bigcap_{i \mathop \in I} \set {x \in S: x_i \in C_i} }} {{end-eqn}} Thus by Intersection of Closed Sets is Closed in Topological Space, our result is proven if we can show that $\forall i \in I: \set {x \in S: x_i \in C_i}$ is closed in $T$. Let $y \in I$. We see that: {{begin-eqn}} {{eqn | l = S \setminus \set {x \in S: x_y \in C_y}       | r = \set {x \in S: x_y \notin C_y} }} {{eqn | r = \set {x \in S: x_y \in \paren {S_y \setminus C_y} }       | c = {{Defof|Set Complement}} }} {{eqn | r = \pr_y^{-1} \sqbrk {S_y \setminus C_y}       | c = {{Defof|Projection (Mapping Theory)}} }}  {{end-eqn}} Since $C_y$ is closed, $\paren {S_y \setminus C_y}$ is open in $\struct {S_y, \tau_y}$ by definition. Thus: :$\pr_y^{-1} \sqbrk {S_y \setminus C_y} \in \BB$ where $\BB$ is the natural sub-basis of $T$: :$\BB = \set {\pr_i^{-1} \sqbrk U: i \in I, U \in \tau_i}$ Therefore $\pr_y^{-1} \sqbrk {S_y \setminus C_y} = S \setminus \set {x \in S: x_y \in C_y}$ is open in $T$. This implies that $\set {x \in S: x_y \in C_y}$ must be closed in $T$ by definition. {{qed}}
By the Bijection in Yoneda Lemma for Covariant Functors, $\Phi_{(A, F)}$ is a bijection for all $(A,F)$. Let $(f, \xi) : (A, F) \to (B, G)$ be a morphism in $C \times [C, \mathbf{Set}]$. To prove that $\Phi$ is a natural isomorphism, it remains to prove that the following diagram commutes: :$\xymatrix{ \operatorname{Nat}(h^A, F) \ar[d]^{\Phi_{(A, F)}} \ar[r]^{\operatorname{Nat}(h^f, \xi)} & \operatorname{Nat}(h^B, G) \ar[d]^{\Phi_{(B, G)}} \\ F(A) \ar[r]^{\operatorname{ev}(f, \xi)}          & G(B) }$ Let $\eta \in \operatorname{Nat}(h^A, F)$. We have  {{begin-eqn}} {{eqn | l = (\operatorname{ev}(f, \xi) \circ \Phi_{(A, F)})(\eta)       | r = \operatorname{ev}(f, \xi)(\eta_A(\operatorname{id}_A))       | c =  }} {{eqn | r = (\xi_B\circ F(f))(\eta_A(\operatorname{id}_A))       | c = {{Defof|Functor Evaluation Bifunctor}} }} {{eqn | r = (\xi_B\circ \eta_B \circ h^A(f))(\operatorname{id}_A)       | c = {{Defof|Natural Transformation}}, applied to $\eta : h^A \to F$ }} {{eqn | r = (\xi_B\circ \eta_B)(f \circ \operatorname{id}_A)       | c = {{Defof|Covariant Hom Functor}}, applied to $h^A$ }} {{eqn | r = (\xi_B \circ \eta_B)(f)       | c =  }} {{end-eqn}} and {{begin-eqn}} {{eqn | l = (\Phi_{(B, G)} \circ \operatorname{Nat}(h^f, \xi))(\eta)       | r = \Phi_{(B, G)}(\xi \circ \eta \circ h^f)       | c = {{Defof|Hom Bifunctor}} }} {{eqn | r = (\xi \circ \eta \circ h^f)_B(\operatorname{id}_B)       | c = }} {{eqn | r = (\xi_B \circ \eta_B)((h^f)_B(\operatorname{id}_B))       | c = {{Defof|Vertical Composition of Natural Transformations}} }} {{eqn | r = (\xi_B \circ \eta_B)(\operatorname{id}_B \circ f)       | c = {{Defof|Contravariant Yoneda Functor}}, {{Defof|Precomposition Natural Transformation}} }} {{eqn | r = (\xi_B \circ \eta_B)(f)       | c =  }} {{end-eqn}} {{qed}}
{{ProofWanted|Needs to be demonstrated that this is prime and also that it is the smallest such palindromic prime greater than $10^{1000}$.}}
{{proof wanted|use Yoneda Lemma for Contravariant Functors}} Category:Category Theory 1z483gc8if6qzjsud30twh3w2exryrg
The fact that $\map {\operatorname {Nat} } {h^A, F}$ is a small class follows when we prove that the mappings are bijections. === Injectivity === Let $\eta \in \map {\operatorname {Nat} } {h^A, F}$ and $B \in C$ an object. Let $f \in \map {h^A} B = \map {\operatorname {Hom} } {A, B}$ be a morphism. Then: {{begin-eqn}} {{eqn | l = \map {\eta_B} f       | r = \map {\eta_B} {f \circ I_A}       | c = {{Defof|Identity Morphism}} }} {{eqn | r = \map {\eta_B} {\map {\paren {\map {h^A} f} } {I_A} }       | c = {{Defof|Covariant Hom Functor}} }} {{eqn | r = \map {\map F f} {\map {\eta_A} {I_A} }       | c = {{Defof|Natural Transformation}} }} {{end-eqn}} which shows that $\eta$ is determined by $\map {\eta_A} {I_A} = \map \alpha \eta$. {{qed|lemma}} === Surjectivity === Let $u \in \map F A$. Inspired by the result above, we define $\map \beta u = \eta$ by $\map {\eta_B} f = \map {\map F f} u$ for $B \in C$ and $f \in \map {\operatorname {Hom} } {A, B}$. Let $B, D \in C$ and $g: B \to D$ be a morphism. We have to show that the following diagram commutes: :$\xymatrix{ \map {\operatorname {Hom} } {A, B} \ar[d]^{\eta_B} \ar[r]^{g_*} & \map {\operatorname {Hom} } {A, D} \ar[d]^{\eta_D} \\ F(B) \ar[r]^{\map F g}          & \map F D}$ where $g_* = \map {\operatorname {Hom} } {A, g}$ is the postcomposition mapping. Let $f \in \map {\operatorname {Hom} } {A, B}$. We have: {{begin-eqn}} {{eqn | l = \map {\eta_D} {\map {g_*} f}       | r = \map {\eta_D} {g \circ f}       | c = {{Defof|Postcomposition Mapping}} }} {{eqn | r = \map {\map F {g \circ f} } u       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map F g (\map {\eta_B} f)       | r = \map {\map F g} {\map {\map F f} u}       | c =  }} {{eqn | r = \map {\map F {g \circ f} } u       | c = {{Defof|Covariant Functor}}, applied to $F$ }} {{end-eqn}} {{qed|lemma}} === Reverse bijections === By construction, $\map \beta {\map \alpha \eta} = \eta$ for all $\eta \in \map {\operatorname {Nat} } {h^A, F}$. It remains to show that $\map \alpha {\map \beta u} = u$ for all $u \in \map F A$. Let $\eta = \map \beta u$. Then {{begin-eqn}} {{eqn | l = \map \alpha \eta       | r = \map {\eta_A} {I_A} }} {{eqn | r = \map {\map F {I_A} } u       | c =  }} {{eqn | r = \map {I_{\map F A} } u       | c = {{Defof|Functor}} }} {{eqn | r = u       | c = {{Defof|Identity Mapping}} }} {{end-eqn}} {{qed}}
{{proof wanted|from Naturality of Yoneda Lemma for Covariant Functors, by duality}}
The reciprocal of a repunit $R_n$ is of the form: :$\dfrac 1 {R_n} = 0 \cdotp \underbrace{\dot 000 \ldots 000}_{n - 1} \dot 9$ Thus $\dfrac 1 {R_{1031}}$ has a period of $1031$. From Period of Reciprocal of Prime, for prime numbers such that: :$p \nmid 10$ we have that the period of such a prime is the order of $10$ modulo $p$. That is, the smallest integer $d$ such that: :$10^d \equiv 1 \pmod p$ The only other possible primes $p$ whose reciprocals might have a period of $1031$ must also satisfy: :$10^{1031} \equiv 1 \pmod p$ that is: :$p \divides \paren {10^{1031} - 1} = 9 \times R_{1031}$ Therefore the only other possible prime whose reciprocal might have a period of $1031$ is $3$. Trivially: :$\dfrac 1 3 = 0 \cdotp \dot 3$ which has a period of $1$. Hence the result. {{qed}}
It is clear that the digits are instances of $7$, $5$, $3$ and $2$, and so are all prime. It is also noted that it has $275 \times 4 + 4 = 1104$ digits, making it titanic {{ProofWanted|It remains to be shown that it is prime.}}
It is clear that the digits are instances of $0$ and $1$. It is also noted that it has $640 \times 2 + 1 = 1281$ digits, making it titanic. {{ProofWanted|It remains to be shown that it is prime.}}
At $1812$ digits, it is clear by definition that this prime is titanic. {{ProofWanted|It remains to be shown that it is also a Sophie Germain prime.}}
It is clear that the digits are instances of $9$ except for the first $1$. It is also noted that it has $3020 + 1 = 3021$ digits, making it titanic. {{ProofWanted|It remains to be shown that it is prime.}}
It is noted that it has $111 \times 9 + 2284 + 1 = 3284$ digits, making it titanic. {{ProofWanted|It remains to be shown that it is prime.}}
It is clear that the digits are all instances of $9$ except for the initial $1357$, all of which are odd. It is also noted that it has $4 + 3821 = 3825$ digits, making it titanic. {{ProofWanted|It remains to be shown that it is prime.}}
It is noted that these integers have $9 + 5820 = 5829$ digits, making them titanic. {{ProofWanted|It remains to be shown that they are prime.}}
It is clear that this number is palindromic. It is also noted that it has $1 + 5901 + 7 + 5901 + 1 = 11 \, 811$ digits, making it gigantic. {{ProofWanted|It remains to be shown that it is prime.}}
Immediate by definition of the successor function $s: \Z_{\ge 0} \to  \Z_{\ge 0}$: :$\forall y \in \Z_{\ge 0}: s \left({y}\right) = y + 1$ and for the $n$th hyperoperation: $\forall n, x, y \in \Z_{\ge 0}: H_n \left({x, y}\right) = \begin{cases} y + 1 & : n = 0 \\ x & : n = 1, y = 0 \\ 0 & : n = 2, y = 0 \\ 1 & : n > 2, y = 0 \\ H_{n - 1} \left({x, H_n \left({x, y - 1}\right)}\right) & : n > 0, y > 0 \end{cases}$ setting $n = 0$. Thus the zeroth hyperoperation degenerates to a mapping with a single operand. {{qed}} Category:Hyperoperation sva9ij9r42p3rhawixy5b3t0l4cgaw6