hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \tan \paren {a + b i}       | r = \frac {\tan a + i \tanh b} {1 - i \tan a \tanh b}       | c = Tangent of Complex Number: Formulation 2 }} {{eqn | r = \frac {\paren {\tan a + i \tanh b} \paren {1 + i \tan a \tanh b} } {1 + \tan ^2 a \tanh ^2 b}       | c = multiplying denominator and numerator by $1 + i \tan a \tanh b$ }} {{eqn | r = \frac {\tan a + i \tanh b + i \tan ^2 a \tanh b - \tan a \tanh ^2 b} {1 + \tan ^2 a \tanh ^2 b}       | c =  }} {{eqn | r = \frac {\tan a - \tan a \tanh ^2 b} {1 + \tan ^2 a \tanh ^2 b} + \frac {\tanh b + \tan ^2 a \tanh b} {1 + \tan ^2 a \tanh ^2 b} i       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \csc \paren {a + b i}       | r = \frac 1 {\sin \paren {a + b i} }       | c = {{Defof|Complex Cosecant Function}} }} {{eqn | r = \dfrac 1 {\sin a \cosh b + i \cos a \sinh b}       | c = Sine of Complex Number }} {{eqn | r = \dfrac {\sin a \cosh b - i \cos a \sinh b} {\paren {\sin a \cosh b - i \cos a \sinh b} \paren {\sin a \cosh b + i \cos a \sinh b} }       | c = multiplying denominator and numerator by $\sin a \cosh b - i \cos a \sinh b$ }} {{eqn | r = \dfrac {\sin a \cosh b - i \cos a \sinh b} {\sin^2 a \cosh^2 b - i^2 \cos^2 a \sinh^2 b}       | c = Difference of Two Squares }} {{eqn | r = \dfrac {\sin a \cosh b - i \cos a \sinh b} {\sin^2 a \cosh^2 b + \cos^2 a \sinh^2 b}       | c = {{Defof|Imaginary Unit}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cot \paren {a + b i}       | r = \frac {\cos \paren {a + b i} } {\sin \paren {a + b i} }       | c = {{Defof|Complex Cotangent Function}} }} {{eqn | r = \dfrac {\cos a \cosh b - i \sin a \sinh b} {\sin a \cosh b + i \cos a \sinh b}       | c = Sine of Complex Number and Cosine of Complex Number }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \cot {a + b i}       | r = \dfrac {\cos a \cosh b - i \sin a \sinh b} {\sin a \cosh b + i \cos a \sinh b}       | c = Cotangent of Complex Number: Formulation 1 }} {{eqn | r = \dfrac {\cot a \cosh b - i \sinh b} {\cosh b + i \cot a \sinh b}       | c = multiplying denominator and numerator by $\dfrac 1 {\sin a}$ }} {{eqn | r = \dfrac {\cot a \coth b - i} {\coth b + i \cot a}       | c = multiplying denominator and numerator by $\dfrac 1 {\sinh b}$ }} {{eqn | r = \dfrac {i \cot a \coth b - 1} {\cot a - i \coth b}       | c = multiplying denominator and numerator by $-i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cot \paren {a + b i}       | r = \dfrac {i \cot a \coth b - 1} {\cot a - i \coth b}       | c = Cotangent of Complex Number: Formulation 2 }} {{eqn | r = \dfrac {\paren {i \cot a \coth b - 1} \paren {\cot a + i \coth b} } {\paren {\cot a - i \coth b} \paren {\cot a + i \coth b} }       | c = multiplying denominator and numerator by $\cot a + i \coth b$ }} {{eqn | r = \dfrac {\paren {i \cot a \coth b - 1} \paren {\cot a + i \coth b} } {\cot^2 a + \coth^2 b}       | c = Difference of Two Squares }} {{eqn | r = \dfrac {i \cot^2 a \coth b - \cot a - \cot a \coth^2 b - i \coth b} {\cot^2 a + \coth^2 b}       | c =  }} {{eqn | r = \dfrac {- \cot a - \cot a \coth^2 b} {\cot^2 a + \coth^2 b} + \dfrac {\cot^2 a \coth b - \coth b} {\cot^2 a + \coth^2 b} i       | c =  }} {{end-eqn}} {{qed}}
By definitions of image of set and composition of mappings: :$f\left[{N\left[{\precsim \left({j}\right)}\right]}\right] = \left({f \circ N}\right)\left[{\precsim \left({j}\right)}\right]$ By definition of complete lattice: :$f\left[{N\left[{\precsim \left({j}\right)}\right]}\right]$ and $N\left[{\precsim \left({j}\right)}\right]$ admit infima. Thus by Subset and Image Admit Infima and Mapping is Increasing implies Infimum of Image Succeeds Mapping at Infimum: :$f\left({\inf\left({N\left[{\precsim \left({j}\right)}\right]}\right)}\right) \preceq_2 \inf \left({\left({f \circ N}\right)\left[{\precsim \left({j}\right)}\right]}\right)$ {{qed}}
By definition of non-empty set: :$\exists j: j \in A$ By definition of $D$: :$\inf \left({Z\left[{\precsim \left({j}\right)}\right]}\right) \in D$ Hence by definition: :$D$ is a non-empty set. Let $x, y \in D$. By definition of $D$: :$\exists j_1 \in A: x = \inf \left({Z\left[{\precsim \left({j_1}\right)}\right]}\right)$ and :$\exists j_2 \in A: y = \inf \left({Z\left[{\precsim \left({j_2}\right)}\right]}\right)$ By definition of directed set: :$\exists j \in A: j_1 \precsim j \land j_2 \precsim j$ By Preceding implies Image is Subset of Image: :$\precsim\left({j}\right) \subseteq \mathord\precsim\left({j_1}\right)$ and $\precsim\left({j}\right) \subseteq \mathord\precsim\left({j_2}\right)$ By Image of Subset under Relation is Subset of Image/Corollary 2 :$Z\left[{\precsim\left({j}\right)}\right] \subseteq Z\left[{\precsim\left({j_1}\right)}\right]$ and $Z\left[{\precsim\left({j}\right)}\right] \subseteq Z\left[{\precsim\left({j_2}\right)}\right]$ Thus by definition of $D$: :$z := \inf \left({Z\left[{\precsim \left({j}\right)}\right]}\right) \in D$ Thus by Infimum of Subset: :$x \preceq z$ and $y \preceq z$ {{qed}}
Let $z \in \mathord\precsim\left({y}\right)$ By definition of image of element: :$y \precsim z$ By definition of transitivity: :$x \precsim z$ Thus by definition of image of element: :$z \in \mathord\precsim\left({x}\right)$ {{qed}}
=== Sufficient Condition === Assume that :$f$ is continuous. By Continuous iff Mapping at Limit Inferior Precedes Limit Inferior of Composition of Mapping and Sequence: :for all directed set $\left({D, \precsim}\right)$ and Moore-Smith sequence $N:D \to S$ in $S$: $f\left({\liminf N}\right) \preceq_2 \liminf\left({f \circ N}\right)$ Thus by Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Mapping Preserves Directed Suprema: :$f$ preserves directed suprema. {{qed|lemma}} === Necessary Condition === Thus by Mapping Preserves Directed Suprema implies Mapping is Continuous: :if $f$ preserves directed suprema, then $f$ is continuous. {{qed}}
Because $N\subset \operatorname{ker} f$, $f$ is constant on the cosets of $N$. Thus $f$ is invariant under congruence modulo $N$. By Universal Property of Quotient Set, there exists a unique mapping $\overline f : G/N \to H$ such that $f = \overline f \circ \pi$. It suffices to verify that it is a continuous group homomorphism. By Universal Property of Quotient Space, there exists a continuous mapping $\overline g : G/N \to H$ such that $f = \overline g \circ \pi$. By uniqueness of $f$, $g=f$. By Universal Property of Quotient Group, there exists a group homomorphism $\overline h : G/N \to H$ such that $f = \overline h \circ \pi$. By uniqueness of $f$, $h=f$. {{qed}} Category:Topological Groups Category:Universal Properties k5zu4pdsabc95m0js903eifs2c2vpgq
=== Existence === For $x \in X$, let $[x] \in X/\sim$ denote its equivalence class under $\sim$. Define the relation $\overline f : X/\sim \to Y$ by: :$([x], y) \in \overline f \iff f(x) = y$. We check that this is a well-defined mapping. By Quotient Mapping is Surjection, $f$ is left-total. Let $[x] \in X/\sim$ and $y, z \in Y$ with: :$([x], y) \in \overline f$ :$([x], z) \in \overline f$ We show that $y = z$. By definition of $\overline f$, there exist $u, v \in X$ with $[u] = [v] = [x]$ and $f(u) = y$ and $f(v) = z$. By definition of equivalence class, $u \sim v$. Because $f$ is $\sim$-invariant, $y = z$. Thus $\overline f$ is many-to-one. Thus $\overline f$ is a well-defined mapping. {{qed|lemma}} === Uniqueness === Follows from: * Quotient Mapping is Surjection * Surjection iff Epimorphism in Category of Sets {{qed}} Category:Equivalence Relations Category:Universal Properties 0pajq8g4foujgbs0tcgtmdtt5vow2hf
=== Sufficient Condition === Assume that :$f$ is continuous. By Continuous iff Directed Suprema Preserving: :$f$ preserves directed suprema. By Directed Suprema Preserving Mapping at Element is Supremum: :$\forall x \in S: \map f x = \sup \set {\map f w: w \in S \land w \ll x}$ Thus by Mapping at Element is Supremum implies Way Below iff There Exists Element that Way Below and Way Below: :$\forall x \in S, y \in T: y \ll \map f x \iff \exists w \in S: w \ll x \land y \ll \map f w$ {{qed|lemma}} === Necessary Condition === Assume that :$\forall x \in S, y \in T: y \ll \map f x \iff \exists w \in S: w \ll x \land y \ll \map f w$ Let $U$ be an open subset of $T$. By definition of Scott topology: :$U$ is upper. By definition of interior: :$\paren {f^{-1} \sqbrk U}^\circ \subseteq f^{-1} \sqbrk U$ We will prove that :$f^{-1} \sqbrk U \subseteq \paren {f^{-1} \sqbrk U}^\circ$ Let $x \in f^{-1} \sqbrk U$ By definition of preimage of set: :$p := \map f x \in U$ By Open implies There Exists Way Below Element: :$\exists u \in T: u \ll p \land u \in U$ By assumption: :$\exists w \in S: w \ll x \land u \ll \map f w$ We will prove that :$f \sqbrk {w^\gg} \subseteq U$ Let $h \in f \sqbrk {w^\gg}$ By definition of image of set: :$\exists z \in w^\gg: h = \map f z$ By definition of way above closure: :$w \ll z$ By assumption: :$u \ll h$ By Way Below implies Preceding: :$u \preceq_2 h$ Thus by definition of upper set: :$h \in U$ {{qed|lemma}} By Image of Subset under Relation is Subset of Image/Corollary 3: :$f^{-1} \sqbrk {f \sqbrk {w^\gg} } \subseteq f^{-1} \sqbrk U$ By Preimage of Image under Left-Total Relation is Superset: :$w^\gg \subseteq \paren {f^{-1} \circ f} \sqbrk {w^\gg}$ By definition of composition of relations: :$w^\gg \subseteq f^{-1} \sqbrk {f \sqbrk {w^\gg} }$ By Subset Relation is Transitive: :$w^\gg \subseteq f^{-1} \sqbrk U$ By Interior is Union of Way Above Closures: :$\paren {f^{-1} \sqbrk U}^\circ = \bigcup \set {g^\gg: g \in S \land g^\gg \subseteq f^{-1} \sqbrk U}$ By definition of way above closure: :$x \in w^\gg$ We have: :$w^\gg \in \set {g^\gg: g \in S \land g^\gg \subseteq f^{-1} \sqbrk U}$ Thus by definition of union: :$x \in \paren {f^{-1} \sqbrk U}^\circ$ {{qed|lemma}} By definition of set equality: :$\paren {f^{-1} \sqbrk U}^\circ = f^{-1} \sqbrk U$ Thus by definition of interior: :$f^{-1} \sqbrk U$ is open. Hence $f$ is continuous. {{qed}}
The Conjugacy Classes of Symmetric Group are determined uniquely by the cycle type of the elements. Since any element in $S_n$ is of the same cycle type with its inverse, they are in the same conjugacy class. Hence they are conjugates of each other. This implies that $S_n$ is ambivalent. {{qed}} Category:Ambivalent Groups 8dh89ilrc2bqnqj6sgtu3z57lhhvwpt
Because $f = \map \OO g$ and $g = \map \OO h$, there exist neighborhoods $U$ and $V$ of $x_0$ and real numbers $c, d \ge 0$ such that: :$\norm {\map f x} \le c \cdot \norm {\map g x}$ for all $x \in U$ :$\norm {\map g x} \le d \cdot \norm {\map h x}$ for all $x \in V$. By Intersection of Neighborhoods in Topological Space is Neighborhood, $U\cap V$ is a neighborhood of $x_0$. For $x \in U \cap V$, we have: :$\norm {\map f x} \le c \cdot \norm {\map g x} \le c d \cdot \norm {\map h x}$ Thus $f = \map \OO h$ for $x \to x_0$. {{qed}} Category:Asymptotic Notation 2q0vllwo8bhm4dwet2s8v688obgsd8e
From the definition of little-O notation: :there exists a neighborhood $U$ of $x_0$ such that $\norm {\map f x} \le \norm {\map g x}$ for all $x \in U$. By definition of big-$\OO$ notation, $f = \map \OO g$ as $x \to x_0$. {{qed}} Category:Asymptotic Notation Category:Order Notation m96cmklqesjtl62cy9wwf2wxtn5d0wc
It will be shown that the two matrices defined are equal column-wise. Let $\displaystyle b_i = \sum_{j \mathop = 1}^n c_{i j} a_j$ for $i$ ranging from $1$ to $n$, where $c_{i j}$'s are scalars. The uniqueness of the above expression is justified by Expression of Vector as Linear Combination from Basis is Unique. Then by definition of coordinate vector, the $i$th column of the matrix defined in definition 1 is: :$\begin {bmatrix} c_{i 1} & c_{i 2} & \ldots & c_{in} \end {bmatrix}^T$ We also have: :$\displaystyle \map {I_G} {b_i} = b_i = \sum_{j \mathop = 1}^n c_{i j} a_j$ So by definition of relative matrix, the $i$th column of the matrix defined in definition 2 is: :$\begin {bmatrix} c_{i 1} & c_{i 2} & \ldots & c_{i n} \end {bmatrix}^T$ The two matrices are equal, so the two definitions are equivalent. {{qed}} Category:Linear Algebra Category:Matrix Theory 18lwx8an6rap6pmfy0ne495dys719ed
Because $g = \map O h$, there exists a neighborhood $V$ of $y_0$ and a real number $c$ such that: :$\norm {\map g x} \le c \cdot \norm {\map h x}$ for all $y \in V$. By definition of continuity, there exists a neighborhood $U$ of $x_0$ with $\map f U \subset V$. For $x \in U$, we have: :$\norm {\map g {\map f x} } \le c \cdot \norm {\map h {\map f x} }$ Thus $g \circ f = \map O {h \circ f}$ as $x \to x_0$. {{qed}} Category:Asymptotic Notation 9s3z4pj7o5jlquyp13g5r1u6ap5193l
Because $a_n = O(b_n)$, there exists $M\geq0$ and $n_0 \in\N$ such that $|a_n| \leq M \cdot |b_n|$ for $n\geq n_0$. Because $n_k$ diverges, there exists $k_0\in\N$ such that $n_k\geq n_0$ for $k\geq k_0$. Then $|a_{n_k}| \leq M\cdot |b_{n_k}|$ for $k\geq k_0$. Thus $a_{n_k} = O(b_{n_k})$. {{qed}} Category:Asymptotic Notation q917p4cyyh16q4kjx1g18nd1hkq82oy
{{begin-eqn}} {{eqn | l = \map \phi {523}       | r = 522       | c = Euler Phi Function of Prime }} {{eqn | l = \map \phi {524}       | r = 260       | c = {{EulerPhiLink|524}} }} {{eqn | l = \map \phi {525}       | r = 240       | c = {{EulerPhiLink|525}} }} {{end-eqn}} {{ProofWanted|It remains to be shown that this is the first such.}}
By definition, a basis is a generator. By Basis of Free Module is No Greater than Generator, there exist: :an injection $f : B \to C$ :an injection $g : C \to B$ By the Cantor-Bernstein-Schröder Theorem, $B$ and $C$ are equivalent. {{qed}}
By Vector Space has Basis between Linearly Independent Set and Spanning Set, there exists a basis $C \subset G$. By Bases of Vector Space have Equal Cardinality, there exists a bijection between $B$ and $C$. By Composite of Injections is Injection, composing this bijection with the inclusion of $C$ in $G$, we obtain an injection from $B$ to $G$. {{qed}} Category:Vector Spaces o50qamytl06tbrbu0zgnt3j68uncmuj
Let $\displaystyle S = \bigcup_{i \mathop \in \N} S_i$. From Increasing Union of Subrings is Subring, we have that $S$ is a subring of $R$. Now we need to show that it is an ideal of $R$. Let $a \in S$. Then $\exists i \in \N: a \in S_i$. Let $b \in R$. Then $a b \in S_i$ and $b a \in S_i$, as $S_i$ is an ideal of $R$. Thus $a b \in S$ and $b a \in S$. So $S$ is an ideal of $R$. {{qed}}
===== Property 1: $0 \in I$ ===== Since $\left\{{I_\alpha}\right\}_{\alpha \in A}$ is non-empty chain, it must contain some ideal $I_\beta$ Since $I_\beta$ is an ideal, $0 \in I_\beta$.  Thus $0 \in I$. ===== Property 2: $x \in I \implies -x \in I$ ===== If $x \in I$, then $x \in I_\beta$ for some $\beta\in A$. Since $I_\beta$ is an ideal, $-x \in I_\beta$.  Thus $-x \in I$. ===== Property 3: $x, y \in I \implies x + y \in I$ ===== If $x,y \in I$, then $x \in I_\beta$ for some $\beta\in A$, and $y \in I_\gamma$ for some $\gamma \in A$.  Since $\left\{{I_\alpha}\right\}_{\alpha \in A}$ is totally ordered, $I_\beta \subseteq I_\gamma$ or $I_\gamma\subseteq I_\beta$. {{WLOG}}, we can assume $I_\beta \subseteq I_\gamma$, which gives us $x,y \in I_\gamma$. Since $I_\gamma$ is an ideal, $x + y \in I_\gamma$. Thus $x + y \in I$. ===== Property 4: $x \in I \land r \in R \implies rx, xr \in I$ ===== If $x \in I$, then $x \in I_\beta$ for some $\beta\in A$.  If $r \in R$, then since $I_\beta$ is an ideal, $rx, xr \in I_\beta$. Thus $rx, xr \in I$. Since $I$ satisfies these 4 properties, it is a ideal of $R$. {{qed}}
By Union of Chain of Ideals is Ideal, $I$ is an ideal. It remains to show that $I \subsetneq R$. The ideals $I_\alpha$ are all proper, so none of them contain the unity. Thus $I$ does not contain $1$, which means $I \subsetneq R$. {{qed}}
Follows from: :Canonical Basis of Free Module on Set is Basis :Cardinality of Canonical Basis of Free Module on Set {{qed}} Category:Vector Spaces tfppnzuta22hqh19b9ep3xu1k0mgyru
On a case-by-case basis: From Smallest Number not Expressible as Sum of Fewer than 19 Fourth Powers: {{:Smallest Number not Expressible as Sum of Fewer than 19 Fourth Powers}} From 159 is not Expressible as Sum of Fewer than 19 Fourth Powers: {{:159 is not Expressible as Sum of Fewer than 19 Fourth Powers}} From 239 is not Expressible as Sum of Fewer than 19 Fourth Powers: {{:239 is not Expressible as Sum of Fewer than 19 Fourth Powers}} From 319 is not Expressible as Sum of Fewer than 19 Fourth Powers {{:319 is not Expressible as Sum of Fewer than 19 Fourth Powers}} From 399 is not Expressible as Sum of Fewer than 19 Fourth Powers {{:399 is not Expressible as Sum of Fewer than 19 Fourth Powers}} From 479 is not Expressible as Sum of Fewer than 19 Fourth Powers {{:479 is not Expressible as Sum of Fewer than 19 Fourth Powers}} From 559 is not Expressible as Sum of Fewer than 19 Fourth Powers {{:559 is not Expressible as Sum of Fewer than 19 Fourth Powers}} {{ProofWanted|It needs to be shown that there are no other numbers known with this property.}}
First note that $4^4 = 256 > 239$. Then note that $3 \times 3^4 = 243 > 239$. Hence any expression of $239$ as fourth powers uses no $n^4$ for $n \ge 4$, and uses not more than $2$ instances $3^4$. For the remainder, using $2^4$ uses fewer fourth powers than $16$ instances $1^4$ does. Now we have: {{begin-eqn}} {{eqn | l = 239       | r = 2 \times 3^4 + 4 \times 2^4 + 13 \times 1^4 }} {{eqn | r = 1 \times 3^4 + 9 \times 2^4 + 14 \times 1^4 }} {{eqn | r = 0 \times 3^4 + 13 \times 2^4 + 15 \times 1^4 }} {{end-eqn}} and it is apparent the first one uses the least number of fourth powers, at $19$. {{qed}}
First note that $5^4 = 625 > 479$. Then note that $2 \times 4^4 = 512 > 479$. Hence any expression of $479$ as fourth powers uses no $n^4$ for $n \ge 5$, and uses not more than $1$ instance of $4^4$. For the remainder, using $2^4$ uses fewer fourth powers than $16$ instances of $1^4$ does. Now we have: {{begin-eqn}} {{eqn | l = 479       | r = 4^4 + 2 \times 3^4 + 3 \times 2^4 + 13 \times 1^4 }} {{eqn | r = 4^4 + 1 \times 3^4 + 8 \times 2^4 + 14 \times 1^4 }} {{eqn | r = 4^4 + 0 \times 3^4 + 13 \times 2^4 + 15 \times 1^4 }} {{eqn | r = 5 \times 3^4 + 4 \times 2^4 + 10 \times 1^4 }} {{eqn | r = 4 \times 3^4 + 9 \times 2^4 + 11 \times 1^4 }} {{eqn | r = 3 \times 3^4 + 14 \times 2^4 + 12 \times 1^4 }} {{eqn | r = 2 \times 3^4 + 19 \times 2^4 + 13 \times 1^4 }} {{eqn | r = 1 \times 3^4 + 24 \times 2^4 + 14 \times 1^4 }} {{eqn | r = 0 \times 3^4 + 29 \times 2^4 + 15 \times 1^4 }} {{end-eqn}} and it can be seen that the first and the fourth uses the least number of fourth powers, at $19$. {{qed}}
First note that $5^4 = 625 > 559$. Then note that $3 \times 4^4 = 768 > 559$. Hence any expression of $559$ as fourth powers uses no $n^4$ for $n \ge 5$, and uses not more than $2$ instances of $4^4$. For the remainder, using $2^4$ uses fewer fourth powers than $16$ instances of $1^4$ does. Now we have: {{begin-eqn}} {{eqn | l = 559       | r = 2 \times 4^4 + 0 \times 3^4 + 2 \times 2^4 + 15 \times 1^4 }} {{eqn | r = 1 \times 4^4 + 3 \times 3^4 + 3 \times 2^4 + 12 \times 1^4 }} {{eqn | r = 1 \times 4^4 + 2 \times 3^4 + 8 \times 2^4 + 13 \times 1^4 }} {{eqn | r = 1 \times 4^4 + 1 \times 3^4 + 13 \times 2^4 + 14 \times 1^4 }} {{eqn | r = 1 \times 4^4 + 0 \times 3^4 + 18 \times 2^4 + 15 \times 1^4 }} {{eqn | r = 6 \times 3^4 + 4 \times 2^4 + 9 \times 1^4 }} {{eqn | r = 5 \times 3^4 + 9 \times 2^4 + 10 \times 1^4 }} {{eqn | r = 4 \times 3^4 + 14 \times 2^4 + 11 \times 1^4 }} {{eqn | r = 3 \times 3^4 + 19 \times 2^4 + 12 \times 1^4 }} {{eqn | r = 2 \times 3^4 + 24 \times 2^4 + 13 \times 1^4 }} {{eqn | r = 1 \times 3^4 + 29 \times 2^4 + 14 \times 1^4 }} {{eqn | r = 0 \times 3^4 + 34 \times 2^4 + 15 \times 1^4 }} {{end-eqn}} and it can be seen that the first and the sixth use the least number of fourth powers, at $19$. {{qed}}
From Canonical Basis of Free Module on Set is Basis, $R^{\paren I}$ has a basis. {{qed}} Category:Free Modules 09a67r1n3513hjmtc6sbhi8db56nhxo
=== Sufficient Condition === Let $n$ be a Carmichael number: :$(4): \quad \forall a \in \Z: a \perp n: a^n \equiv a \pmod n$ where $\perp$ denotes coprimality. Suppose $n$ is even. Set $a = -1$ in $(4)$. Then $\paren {-1}^n = 1$ and so: :$1 \equiv -1 \pmod n$ resulting in $n = 2$. But as $2$ is not a Carmichael number, it follows that for $n$ to be Carmichael it must be odd. Thus: : $(1): n$ is odd holds. From the Fundamental Theorem of Arithmetic, let $n$ be expressed as: :$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ where: :the $p_i$'s are distinct odd primes :for all $i$, $k_i \ge 1$. By Integers with Primitive Roots there exists a primitive root $a_i$ modulo $p_i^{k_i}$ for each $i$. In particular, we have that $a_i \perp p_i$. By the Chinese Remainder Theorem: :$\exists a: \forall i: a \equiv a_i \pmod {p_i^{k_i} }, a \perp n$ Consider a particular $i$. We have: {{begin-eqn}} {{eqn | l = a^n       | o = \equiv       | r = a \pmod n       | c = from $(4)$ }} {{eqn | ll= \leadsto       | l = a^{n - 1}       | o = \equiv       | r = 1 \pmod n       | c =  }} {{eqn | ll= \leadsto       | l = a^{n - 1}       | o = \equiv       | r = 1  \pmod {p_i^{k_i} }       | c =  }} {{eqn | ll= \leadsto       | l = {a_i}^{n - 1}       | o = \equiv       | r = 1 \pmod {p_i^{k_i} }       | c =  }} {{end-eqn}} We have that $a_i$ is a primitive root modulo $p_i^{k_i}$. Thus: {{begin-eqn}} {{eqn | l = \map {\operatorname {ord}_{p_i^{k_i} } } {a_i}       | r = \map \phi {p_i^{k_i} }       | c = where $\phi$ is the Euler $\phi$ function }} {{eqn | r = p_i^{k_i - 1} \paren {p_i - 1}       | c =  }} {{eqn | o = \divides       | r = n - 1       | c =  }} {{end-eqn}} But because $p_i \divides n$: :$p_i \perp n - 1$ Thus: :$k_i = 1$ giving that $n$ is square-free, and: :$p_i - 1 \divides n - 1$ As $i$ is arbitrary, both conditions $(1)$ and $(2)$ of the statement of the theorem are thus fulfilled: That is: : $(1): n$ is odd and for all primes $p$ such that $p \divides n$: : $(2): \quad p^2 \nmid n$ : $(3): \quad \paren {p - 1} \divides \paren {n - 1}$ {{qed|lemma}} === Necessary Condition === Suppose $n$ is such that: : $(1): n$ is odd and for all primes $p$ such that $p \divides n$: : $(2): \quad p^2 \nmid n$ : $(3): \quad \paren {p - 1} \divides \paren {n - 1}$ Thus: : $n = p_1 p_2 \cdots p_r$ is the product of $r$ distinct odd primes such that: :$\paren {p_i - 1} \divides \paren {n - 1}$ for all $i$. Let $a \in \Z$. Let $a \perp p_i$. Then by Fermat's Little Theorem: :$a^{p_i - 1} \equiv 1 \pmod {p_i}$ Thus: :$a^{n - 1} \equiv 1 \pmod {p_i}$ and so: :$a^n \equiv a \pmod {p_i}$ Let $\gcd \set {a, p_i} > 1$. Then as $p_i$ is prime: :$p_i \divides a$ and so: :$a^n \equiv a \equiv 0 \pmod {p_i}$ Thus for all $i$ and for all $a \in \Z$: :$a^n \equiv a \pmod {p_i}$ From the Chinese Remainder Theorem: :$a^n \equiv a \pmod n$ for all $a$. Thus, by definition, $n$ is a Carmichael number. {{qed}} {{Proofread}}
By Korselt's Theorem, $n$ is odd. Therefore $n$ has at least $1$ odd prime factor. By Korselt's Theorem, for each prime factor of $n$: :$p^2 \nmid n$ :$\paren {p - 1} \divides \paren {n - 1}$ Suppose $n = p^k$ for some odd prime $p$. By Korselt's Theorem, $k = 1$. However by definition of a Carmichael Number, $n$ cannot be prime. Therefore $n$ has at least $2$ distinct odd prime factors. Suppose $n = p^a q^b$ for distinct odd primes $p$ and $q$. By Korselt's Theorem, the following holds: :$a = b = 1$ :$n = p q$ :$\paren {p - 1} \divides \paren {n - 1}$ :$\paren {q - 1} \divides \paren {n - 1}$ Hence: {{begin-eqn}} {{eqn | l = \paren {p - 1}       | o = \divides       | r = \paren {n - 1 - q \paren {p - 1} } }} {{eqn | r = p q - 1 - p q + q }} {{eqn | r = q - 1 }} {{end-eqn}} Swapping $p$ and $q$ yields $\paren {q - 1} \divides \paren {p - 1}$. Hence $p - 1 = q - 1$ and $p = q$, which is a contradiction. Therefore $n$ has at least $3$ distinct odd prime factors. {{qed}}