hcju/proofwikiqa huggingface dataset:
Let: :$\mathbf A = \begin{bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 r} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 r} & \cdots & a_{2 n} \\  \vdots &  \vdots & \ddots &  \vdots & \ddots &  \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n r} & \cdots & a_{n n} \\ \end{bmatrix}$ :$\mathbf B = \begin{bmatrix} b_{1 1} & b_{1 2} & \cdots & b_{1 r} & \cdots & b_{1 n} \\ b_{2 1} & b_{2 2} & \cdots & b_{2 r} & \cdots & b_{1 n} \\  \vdots &  \vdots & \ddots &  \vdots & \ddots &  \vdots \\ b_{n 1} & b_{n 2} & \cdots & b_{n r} & \cdots & b_{n n} \\ \end{bmatrix} = \begin{bmatrix} a_{1 1} & a_{1 2} & \cdots & c a_{1 r} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & c a_{2 r} & \cdots & a_{1 n} \\  \vdots &  \vdots & \ddots &  \vdots & \ddots &  \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n r} & \cdots & a_{n n} \\ \end{bmatrix}$ We have that: :$\mathbf A^\intercal = \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{r 1} & a_{r 2} & \cdots & a_{r n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$ where $\mathbf A^\intercal$ denotes the transpose of $\mathbf A$. Similarly, we have that: :$\mathbf B^\intercal = \begin{bmatrix} a_{1 1} & a_{1 2} & \ldots & a_{1 n} \\ a_{2 1} & a_{2 2} & \ldots & a_{2 n} \\  \vdots &  \vdots & \ddots &  \vdots \\ c a_{r 1} & c a_{r 2} & \cdots & c a_{r n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$ From Determinant with Row Multiplied by Constant: :$\map \det {\mathbf B^\intercal} = c \map \det {\mathbf A^\intercal}$ From from Determinant of Transpose: :$\map \det {\mathbf B^\intercal} = \map \det {\mathbf B}$ :$\map \det {\mathbf A^\intercal} = \map \det {\mathbf A}$ and the result follows. {{qed}}
Let $\mathbf I$ denote the unit matrix of arbitrary order $n$. By Determinant of Unit Matrix: :$\map \det {\mathbf I} = 1$ Let $\rho$ be the permutation on $\tuple {1, 2, \ldots, n}$ which transposes $i$ and $j$. From Parity of K-Cycle, $\map \sgn \rho = -1$. By definition we have that $\mathbf E_3$ is $\mathbf I$ with columns $i$ and $j$ transposed. By the definition of a determinant: :$\displaystyle \map \det {\mathbf I} = \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k k} }$ By Permutation of Determinant Indices: :$\displaystyle \map \det {\mathbf E_3} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k \map \rho k} }$ We can take $\map \sgn \rho = -1$ outside the summation because it is constant, and so we get: {{begin-eqn}} {{eqn | l = \map \det {\mathbf E_3}       | r = \map \sgn \rho \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k \map \rho k} }       | c =  }} {{eqn | r = -\sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k k} }       | c =  }} {{eqn | r = -\map \det {\mathbf I}       | c =  }} {{end-eqn}} Hence the result. {{qed}}
For $1 \le k \le n$, let $e_k$ be the elementary row operation that multiplies row $k$ of $\mathbf A$ by $\lambda$. By definition of the scalar product, $\lambda \mathbf A$ is obtained by multiplying every row of $\mathbf A$ by $\lambda$. That is the same as applying $e_k$ to $\mathbf A$ for each of $k \in \set {1, 2, \ldots, n}$. Let $\mathbf E_k$ denote the elementary row matrix corresponding to $e_k$. By Determinant of Elementary Row Matrix: Scale Row: :$\map \det {\mathbf E_k} = \lambda$ Then we have: {{begin-eqn}} {{eqn | l = \lambda \mathbf A       | r = \prod_{k \mathop = 1}^n \mathbf E_k \mathbf A       | c =  }} {{eqn | ll= \leadsto       | l = \map \det {\lambda \mathbf A}       | r = \map \det {\prod_{k \mathop = 1}^n \mathbf E_k \mathbf A}       | c =  }} {{eqn | r = \paren {\prod_{k \mathop = 1}^n \map \det {\mathbf E_k} } \map \det {\mathbf A}       | c = Determinant of Matrix Product }} {{eqn | r = \paren {\prod_{k \mathop = 1}^n \lambda} \map \det {\mathbf A}       | c = Determinant of Elementary Row Matrix: Scale Row }} {{eqn | r = \lambda^n \map \det {\mathbf A}       | c =  }} {{end-eqn}} {{qed}}
Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the finite sequence of elementary row operations that compose $\Gamma_1$. Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding finite sequence of the elementary row matrices. Let $\sequence {f_i}_{1 \mathop \le i \mathop \le l}$ be the finite sequence of elementary row operations that compose $\Gamma_2$. Let $\sequence {\mathbf F_i}_{1 \mathop \le i \mathop \le l}$ be the corresponding finite sequence of the elementary row matrices. From Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, we have: :$\mathbf R_1 \mathbf A = \mathbf B$ where $\mathbf R$ is the product of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$: :$\mathbf R_1 = \mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1$ Also from Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, we have: :$\mathbf R_2 \mathbf B = \mathbf C$ where $\mathbf R_2$ is the product of $\sequence {\mathbf F_i}_{1 \mathop \le i \mathop \le l}$: :$\mathbf R_2 = \mathbf F_l \mathbf F_{l - 1} \dotsb \mathbf F_2 \mathbf F_1$ Hence we have: :$\mathbf R_2 \mathbf R_1 \mathbf A = \mathbf C$ where $\mathbf R := \mathbf R_2 \mathbf R_1$ is the product: :$\mathbf F_l \mathbf F_{l - 1} \dotsb \mathbf F_2 \mathbf F_1 \mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1$ Let $\Gamma$ be the row operation composed of the finite sequence of elementary row operations $\tuple {e_1, e_2, \ldots, e_{k - 1}, e_k, f_1, f_2, \ldots, f_{l - 1}, f_l}$. Thus $\Gamma$ is a row operation which transforms $\mathbf A$ into $\mathbf C$. Hence the result. {{qed}}
Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the finite sequence of elementary column operations that compose $\Gamma_1$. Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding finite sequence of the elementary column matrices. Let $\sequence {f_i}_{1 \mathop \le i \mathop \le l}$ be the finite sequence of elementary column operations that compose $\Gamma_2$. Let $\sequence {\mathbf F_i}_{1 \mathop \le i \mathop \le l}$ be the corresponding finite sequence of the elementary column matrices. From Column Operation is Equivalent to Post-Multiplication by Product of Elementary Matrices, we have: :$\mathbf A \mathbf R_1 = \mathbf B$ where $\mathbf R$ is the product of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$: :$\mathbf R_1 = \mathbf E_1 \mathbf E_2 \dotsb \mathbf E_{k - 1} \mathbf E_k$ Also from Column Operation is Equivalent to Post-Multiplication by Product of Elementary Matrices, we have: :$\mathbf B \mathbf R_2 = \mathbf C$ where $\mathbf R_2$ is the product of $\sequence {\mathbf F_i}_{1 \mathop \le i \mathop \le l}$: :$\mathbf R_2 = \mathbf F_1 \mathbf F_2 \dotsb \mathbf F_{l - 1} \mathbf F_l$ Hence we have: :$\mathbf A \mathbf R_1 \mathbf R_2 = \mathbf C$ where $\mathbf R := \mathbf R_1 \mathbf R_2$ is the product: :$\mathbf E_1 \mathbf E_2 \dotsb \mathbf E_{k - 1} \mathbf E_k \mathbf F_1 \mathbf F_2 \dotsb \mathbf F_{l - 1} \mathbf F_l$ Let $\Gamma$ be the column operation composed of the finite sequence of elementary column operations $\tuple {e_1, e_2, \ldots, e_{k - 1}, e_k, f_1, f_2, \ldots, f_{l - 1}, f_l}$. Thus $\Gamma$ is a column operation which transforms $\mathbf A$ into $\mathbf C$. Hence the result. {{qed}}
This is proved in Expansion Theorem for Determinants. {{qed}} Category:Determinants ejpg48vysa61ntozt8m9l4q41r80xl6
{{begin-eqn}} {{eqn | o =       | r = \set x \cap S = \O       | c = }} {{eqn | ll = \leadstoandfrom       | o =       | r = \forall y : \lnot \paren{ y \in \set x \cap S}       | c = {{Defof|Empty Set}} }} {{eqn | ll = \leadstoandfrom       | o =       | r = \forall y : \lnot \paren{ y \in \set x \land y \in S}       | c = {{Defof|Set Intersection}} }} {{eqn | ll = \leadstoandfrom       | o =       | r = \forall y : \lnot \paren{ y \in \set x \land \lnot \lnot \paren{y \in S} }       | c = Double Negation Introduction }} {{eqn | ll = \leadstoandfrom       | o =       | r = \forall y : y \in \set x \implies \lnot \paren{y \in S}       | c = Implication Equivalent to Negation of Conjunction with Negative }} {{eqn | ll = \leadstoandfrom       | o =       | r = \forall y : y \in \set x \implies y \notin S       | c = {{Defof|Element|Not Element}} }} {{eqn | ll = \leadstoandfrom       | o =       | r = \forall y : y = x \implies y \notin S       | c = {{Defof|Singleton}} }} {{eqn | ll = \leadstoandfrom       | o =       | r = x \notin S       | c = {{Defof|Equality}} }} {{end-eqn}} {{qed}} Category:Singletons Category:Disjoint Sets osfxprp7vvy9sl3nsfye0u7x1a4su30
From Transpose of Upper Triangular Matrix is Lower Triangular, the transpose $\mathbf T_n^\intercal$ of $\mathbf T_n$ is an upper triangular matrix. From Determinant of Upper Triangular Matrix, the determinant of $\mathbf T_n^\intercal$ is equal to the product of all the diagonal elements of $\mathbf T_n^\intercal$. From Determinant of Transpose, the determinant of $\mathbf T_n^\intercal$ equals the determinant of $\mathbf T_n$. {{qed}}
Consider the ordered set $\struct {\mathscr I, \subseteq}$. From Element of Finite Ordered Set is Between Maximal and Minimal Elements: :$\exists B \in \mathscr I : A \subseteq B$ and $B$ is maximal in $\struct {\mathscr I, \subseteq}$. By definition of a base: :$B \in \mathscr B$ {{qed}} Category:Matroid Theory bct1wemdzy528q031kddmjegtcudbpp
Let $T = \set{y : x \preceq y}$. By the reflexivity of the ordering $\preceq$: :$x \preceq x$ So $x \in T$ and $T$ is non-empty. From Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements: :$\struct{T, \preceq}$ has a maximal element $M \in T$ We now show that $M$ is a maximal element in $\struct{S, \preceq}$.  Let $y \in S$ such that: :$M \preceq y$ By the transitiviy of the ordering $\preceq$: :$x \prec y$ So $y \in T$. By the definition of a maximal element: :$y = M$ Similarly for $T' = \set{y : y \preceq x}$: :$\struct{T', \preceq}$ has a minimal element $m \in T'$ and $m$ is a minimal element in $\struct{S, \preceq}$ {{qed}} Category:Order Theory qptimv3q5zr0p4ls4ge0z0tr9gluzdq
=== Condition $(1)$ iff Condition $(2)$ === Follows immediately from Element is Loop iff Member of Closure of Empty Set. {{qed|lemma}}  === Condition $(1)$ iff Condition $(3)$ === Follows immediately from Element is Loop iff Rank is Zero. {{qed|lemma}}  === Condition $(1)$ iff Condition $(4)$ === Follows immediately from Element is Loop iff Singleton is Circuit. {{qed|lemma}}  === Condition $(1)$ iff Condition $(5)$ === Follows immediately from the contrapositive statement of Element is Member of Base iff Not Loop. {{qed}} Category:Matroid Theory 84i77rbmvcca6a8jdcv7zwft3kdctww
By definition of a subset: :$\set x , \set y, \set {x, y} \subseteq \set{x, y}$ Let $A \subseteq \set {x, y}$: :$A \ne \set x, \set y, \set {x,y}$ From set equality: :$\set {x,y} \not \subseteq A$ From Doubleton of Elements is Subset: :either $x \notin A$ or $y \notin A$. {{WLOG}} assume that $x \notin A$. From Intersection With Singleton is Disjoint if Not Element: :$A \cap \set x = \O$ From Subset of Set Difference iff Disjoint Set: :$A \subseteq \set{x, y} \setminus \set x$ From Set Difference of Doubleton and Singleton is Singleton: :$A \subseteq \set y$ From set equality: :$\set y \not \subseteq A$ From Singleton of Element is Subset: :$y \notin A$. From Intersection With Singleton is Disjoint if Not Element: :$A \cap \set y = \O$ From Subset of Set Difference iff Disjoint Set: :$A \subseteq \set y \setminus \set y$ From Set Difference with Self is Empty Set: :$A \subseteq \O$ From Empty Set is Subset of All Sets: :$\O \subseteq A$ From set equality: :$A = \O$ It follows that: :$\powerset {\set {x, y}} = \big \{ \O, \set x, \set y, \set {x,y} \big \}$ {{qed}} Category:Power Set Category:Doubletons mcqoyqldh5icrnt4mrvnznotubp8h9r
=== Necessary Condition === Let $x, y \in S$. From Singleton of Element is Subset: :$\set x \subseteq S$ :$\set y \subseteq S$ From Union of Subsets is Subset: :$\set x \cup \set y \subseteq S$ From Union of Disjoint Singletons is Doubleton: : $\set x \cup \set y = \set {x, y}$ Hence: :$\set {x,y} \subseteq S$ {{qed|lemma}} === Sufficient Condition === Let $\set {x,y} \subseteq S$. From the definition of a subset: :$x \in \set {x,y} \implies x \in S$ :$y \in \set {x,y} \implies y \in S$ {{qed}} Category:Subsets Category:Doubletons snl24lnninfwoacfvoj1ighxevol7be
Let $d \divides n$. By Divisors of Power of Prime, $d = p^a$ for some positive integer $a \le k$. We have $\dfrac n d = p^{k - a}$. Suppose $d$ is a unitary divisor of $n$. Then $d$ and $\dfrac n d$ are coprime. If both $a, k - a \ne 0$, $p^a$ and $p^{k - a}$ have a common divisor: $p$. Hence either $a = 0$ or $k - a = 0$. This leads to $d = 1$ or $p^k$. Hence the sum of all positive unitary divisors of $n$ is: :$1 + p^k = 1 + n$ {{qed}} Category:Prime Numbers Category:Sum of Unitary Divisors ebj0fv5t3kwe6rz3pfhvp8mk2uyf2lt
Let $a, b$ be coprime integers. Because $a$ and $b$ have no common divisor, the divisors of $a b$ are integers of the form $a_i b_j$, where $a_i$ is a divisor of $a$ and $b_j$ is a divisor of $b$. That is, any divisor $d$ of $a b$ is in the form: :$d = a_i b_j$ in a unique way, where $a_i \divides a$ and $b_j \divides b$. First we show that: :$d$ is an unitary divisor of $a b$ {{iff}} $a_i, b_j$ are unitary divisors of $a, b$ respectively In the forward implication we are given $d \perp \dfrac {a b} d$. By Divisor of One of Coprime Numbers is Coprime to Other: :$a_i, b_j \perp \dfrac {a b} d$ By Divisor of One of Coprime Numbers is Coprime to Other again: :$a_i \perp \dfrac a {a_i} \land b_j \perp \dfrac b {b_j}$ In the backward implication we are given $a_i \perp \dfrac a {a_i} \land b_j \perp \dfrac b {b_j}$. By Divisor of One of Coprime Numbers is Coprime to Other: :$a \perp b \implies \paren {a_i \perp b_j \land \dfrac a {a_i} \perp \dfrac b {b_j} }$ By Product of Coprime Pairs is Coprime: :$d = a_i b_j \perp \dfrac a {a_i} \dfrac b {b_j} = \dfrac {a b} d$ {{qed|lemma}} We can list the unitary divisors of $a$ and $b$ as  :$1, a_1, a_2, \ldots, a$ and: :$1, b_1, b_2, \ldots, b$ and thus the sum of their unitary divisors are: :$\displaystyle \map {\sigma^*} a = \sum_{i \mathop = 1}^r a_i$ :$\displaystyle \map {\sigma^*} b = \sum_{j \mathop = 1}^s b_j$ Consider all unitary divisors of $a b$ with the same $a_i$. Their sum is: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^s a_i b_j       | r = a_i \sum_{j \mathop = 1}^s b_j       | c =  }} {{eqn | r = a_i \map {\sigma^*} b       | c =  }} {{end-eqn}} Summing over all $a_i$: {{begin-eqn}} {{eqn | l = \map {\sigma^*} {a b}       | r = \sum_{i \mathop = 1}^r \paren {a_i \map {\sigma^*} b}       | c =  }} {{eqn | r = \paren {\sum_{i \mathop = 1}^r a_i} \map {\sigma^*} b       | c =  }} {{eqn | r = \map {\sigma^*} a \map {\sigma^*} b       | c =  }} {{end-eqn}} {{qed}} Category:Sum of Unitary Divisors Category:Multiplicative Functions c6yun19lmtgq08xdz2qamz8jtidjz7i
We have that the Sum of Unitary Divisors is Multiplicative. From Value of Multiplicative Function is Product of Values of Prime Power Factors, we have: :$\map {\sigma^*} n = \map {\sigma^*} {p_1^{k_1} } \map {\sigma^*} {p_2^{k_2} } \ldots \map {\sigma^*} {p_r^{k_r} }$ From Sum of Unitary Divisors of Power of Prime, we have: :$\displaystyle \map {\sigma^*} {p_i^{k_i} } = \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ Hence the result. {{qed}} Category:Sum of Unitary Divisors 4ga8cqg48g3xpy1pbq6s76o1kdvqcya
Consider the infinite sequence $\sequence {x_n}_{n \mathop \in \N}$ defined as: :$\forall n \in \N: x_n = \dfrac {1 + \paren {-1}^n} 2$ Thus: :$\sequence {x_n}_{n \mathop \in \N} = 1, 0, 1, 0, \dotsc$ Hence the range of $\sequence {x_n}$ is $\set {0, 1}$, which is finite. {{qed}}
Proof by induction: Let $\map P n$ be the proposition: :$\dfrac {n!} {p^\mu} \equiv \paren {-1}^\mu a_0! a_1! \dotsm a_k! \pmod p$ where $p, a_0, \dots, a_k, \mu$ are as defined above. === Basis for the Induction === For $n = 1$: :$a_0 = 1, \mu = 0$ $\map P 1$ reduces to: :$\dfrac {1!} {p^0} \equiv \paren {-1}^0 1! \pmod p$ which holds trivially. This is the basis for the induction. === Induction Hypothesis === This is our induction hypothesis: :$\dfrac {r!} {p^\mu} \equiv \paren {-1}^\mu a_0! a_1! \dots a_k! \pmod p$ Now we need to show true for $n = r + 1$: :$\dfrac {\paren {r + 1}!} {p^{\mu'}} \equiv \paren {-1}^{\mu'} b_0! b_1! \dots b_k! \pmod p$ where $\displaystyle r + 1 = \sum_{j \mathop = 0}^k b_j p^j$ is the base $p$ presentation of $r + 1$, and: :$p^{\mu'} \divides \paren {r + 1}!$ :$p^{\mu' + 1} \nmid \paren {r + 1}!$ === Induction Step === This is our induction step: Let $p^m$ be the largest power of $p$ which divides $r + 1$, that is: :$p^m \divides r + 1$ :$p^{m + 1} \nmid r + 1$ Then we must have: :$b_m \ne 0$ :$b_j = 0$ for $0 \le j < m$ :$\mu' = \mu + m$ and consequently: :$a_j = b_j$ for $j > m$ :$a_m + 1 = b_m$ :$a_j = p - 1$ for $0 \le j < m$ Thus: {{begin-eqn}} {{eqn | l = \frac {\paren {r + 1}!} {p^{\mu'} }       | r = \frac {r!} {p^\mu} \times \frac {r + 1} {p^m} }} {{eqn | o = \equiv       | r = \paren {-1}^\mu a_0! \dots a_{m - 1}! a_m! \dots a_k! \times \frac {r + 1} {p^m}       | rr = \pmod p       | c = Induction Hypothesis }} {{eqn | o = \equiv       | r = \paren {-1}^\mu \paren {\paren {p - 1}!}^m \paren {b_m - 1}! b_{m + 1}! \dots b_k! \times \frac 1 {p^m} \sum_{j \mathop = m}^k b_j p^j       | rr = \pmod p       | c = from above }} {{eqn | o = \equiv       | r = \paren {-1}^\mu \paren {-1}^m \paren {b_m - 1}! b_{m + 1}! \dots b_k! \times \sum_{j \mathop = m}^k b_j p^{j - m}       | rr = \pmod p       | c = Wilson's Theorem }} {{eqn | o = \equiv       | r = \paren {-1}^{\mu'} \paren {b_m - 1}! b_{m + 1}! \dots b_k! b_m       | rr = \pmod p       | c = because $b_j p^{j - m} \equiv 0 \pmod p$ for $j - m > 0$ }} {{eqn | o = \equiv       | r = \paren {-1}^{\mu'} b_m! b_{m + 1}! \dots b_k!       | rr = \pmod p       | c = {{Defof|Factorial}} }} {{eqn | o = \equiv       | r = \paren {-1}^{\mu'} b_0! \dots b_k!       | rr = \pmod p       | c = Factorial of Zero: $b_j = 0$ for $0 \le j < m$ }} {{end-eqn}} By the Principle of Mathematical Induction, $\map P n$ is true for all $n \in \Z_{>0}$. {{qed}}
Consider the numbers of $\set {1, 2, \ldots, n}$ which are not multiples of $p$. There are $\floor {\dfrac n p}$ complete sets of $p - 1$ such consecutive elements of $\set {1, 2, \ldots, n}$. Each one of these has a product which is congruent to $-1 \pmod p$ by Wilson's Theorem. There are also $a_0$ left over which are congruent to $a_0! \pmod p$. Thus: :the contributions of the divisors which are not multiples of $p$ is $\paren {-1}^{\floor {n / p} } a_0!$ :the contributions of the divisors which are multiples of $p$ is the same as the contribution in $\floor {\dfrac n p}!$ Thus the argument can be repeated on $\floor {\dfrac n p}!$ until the formula is complete. {{qed}}
We have that $\struct {\map \CC I, +, \, \cdot \,}_\R$ is a vector space. By Differentiable Function is Continuous, $\map {\CC^1} I \subset \map \CC I$. Let $f, g \in \map {\CC^1} I$. Let $\alpha \in \R$. Let $\map 0 x$ be a real-valued function such that: :$\map 0 x : I \to 0$ Restrict $\paren +$ to $\map {\CC^1} I \times \map {\CC^1} I$. Restrict $\paren {\, \cdot \,}$ to $\R \times \map {\CC^1} I$. === Closure under vector addition === By Sum Rule for Derivatives: :$f + g \in \map {\CC^1} I$ {{qed|lemma}} === Closure under scalar multiplication === By Derivative of Constant Multiple: :$\alpha \cdot f \in \map {\CC^1} I$ {{qed|lemma}} === Nonemptiness === By Derivative of Constant, a constant mapping is differentiable. Hence, $\map 0 x \in \map {\CC^1} I$. {{qed|lemma}} We have that $\map {\CC^1} I$ is closed under restrictions of $\paren +$ and $\paren {\, \cdot \,}$ to $\map {\CC^1} I$. Also, $\map {\CC^1} I$ is non-empty. By definition, $\struct {\map {\CC^1} I, +, \, \cdot \,}_\R$ is a vector subspace of $\struct {\map \CC I, +, \, \cdot \,}_\R$. Since $\struct {\map {\CC^1} I, +, \, \cdot \,}_\R$ satisfies vector space axioms under given restrictions, it is a vector space. {{qed}}
To show that these are the only ones, we observe that since $p$ is an odd prime, write: :$p = 2 n + 1$ for $n \ge 1$. Let $\map {q_p} 2$ be a perfect power. Then $2^{p - 1} - 1 = p x^y$ for some integers $x, y$. Note that: :$2^{p - 1} - 1 = 2^{2 n} - 1 = \paren {2^n - 1} \paren {2^n + 1}$ and we have: :$\gcd \set {2^n - 1, 2^n + 1} = \gcd \set {2^n - 1, 2} = 1$ so $2^n - 1$ and $2^n + 1$ are coprime. Hence there are $2$ cases: === Case $1$: $p \divides 2^n - 1$ === By Divisor of One of Coprime Numbers is Coprime to Other: :$\gcd \set {\dfrac {2^n - 1} p, 2^n + 1} = 1$ Hence both the numbers are perfect $y$th powers. In particular we have: :$\exists k \in \Z: 2^n + 1 = k^y$ by 1 plus Power of 2 is not Perfect Power except 9, the only solution to the equation above is: :$n = k = 3, y = 2$ This gives $p = 2 n + 1 = 7$. {{qed|lemma}} === Case $2$: $p \divides 2^n + 1$ === By Divisor of One of Coprime Numbers is Coprime to Other: :$\gcd \set {\dfrac {2^n + 1} p, 2^n - 1} = 1$ Hence both the numbers are perfect $y$th powers. In particular we have: :$\exists k \in \Z: 2^n - 1 = k^y$ By 1 plus Perfect Power is not Prime Power except for 9, this equation has no solution for $n, k, y > 1$. Hence we must have $n = 1$. This gives $p = 2 n + 1 = 3$. {{qed|lemma}} In the main theorem we have already shown that $p = 3, 7$ gives perfect power Fermat quotients. Hence the result. {{qed}} Category:Fermat Quotients rqzhs2gdzagzjs1p0zkge6ndnlx5vig
By P-Norm is Norm, $\norm {\, \cdot \,}_p$ is a norm. By definition, the taxicab norm is $\norm {\, \cdot \,}_1$. Therefore, the taxicab norm is a norm. {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac 1 {2 n} + \dfrac 1 {6 n}       | r = \dfrac 3 {6 n} + \dfrac 1 {6 n}       | c =  }} {{eqn | r = \dfrac {3 + 1} {6 n}       | c =  }} {{eqn | r = \dfrac 2 {3 n}       | c =  }} {{eqn | r = \dfrac 1 n \times \dfrac 2 3       | c =  }} {{end-eqn}} Note the case where we multiply $\dfrac 2 3$ by $\dfrac 2 3$ itself: {{begin-eqn}} {{eqn | l = \dfrac 2 3 \times \dfrac 2 3       | r = \dfrac 4 9       | c =  }} {{eqn | r = \dfrac 3 9 + \dfrac 1 9       | c =  }} {{eqn | r = \dfrac 1 3 + \dfrac 1 9       | c = which is in Egyptian form }} {{end-eqn}} {{qed}}
Let $\dfrac {x_k} {y_k}$ and $\dfrac {x_{k + 1} } {y_{k + 1} }$ be consecutive stages of the calculation of the unit fractions accordingly: :$\dfrac {x_k} {y_k} - \dfrac 1 {\ceiling {y_n / x_n} } = \dfrac {x_{k + 1} } {y_{k + 1} }$ By definition of Fibonacci's Greedy Algorithm: :$\dfrac {x_{k + 1} } {y_{k + 1} } = \dfrac {\paren {-y_k} \bmod {x_k} } {y_k \ceiling {y_k / x_k} }$ It is established during the processing of Fibonacci's Greedy Algorithm that: :$\paren {-y_k} \bmod {x_k} < x_k$ Hence successive numerators decrease by at least $1$. Hence there can be no more unit fractions than there are natural numbers between $1$ and $p$. Hence the result. {{Qed}}
We have that: {{begin-eqn}} {{eqn | l = \frac 3 {25}       | r = \frac 1 9 + \frac 2 {225}       | c = as $\ceiling {25 / 3} = \ceiling {8.333\ldots} = 9$ }} {{eqn | r = \frac 1 9 + \frac 1 {113} + \frac 1 {25 \, 425}       | c = as $\ceiling {225 / 2} = \ceiling {112.5} = 113$ }} {{end-eqn}} But then we have: {{begin-eqn}} {{eqn | l = \frac 3 {25}       | r = \frac 6 {50}       | c =  }} {{eqn | r = \frac 5 {50} + \frac 1 {50}       | c =  }} {{eqn | r = \frac 1 {10} + \frac 1 {50}       | c =  }} {{end-eqn}} By Condition for 3 over n producing 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient, we are to find the smallest $n$ such that: :$n \equiv 1 \pmod 6$ :$\exists d: d \divides n$ and $d \equiv 2 \pmod 3$ The first few $n \ge 4$ which satisfies $n \equiv 1 \pmod 6$ are: :$7, 13, 19, 25$ of which $7, 13, 19$ are primes, so they do not have a divisor of the form $d \equiv 2 \pmod 3$. We see that $5 \divides 25$ and $5 \equiv 2 \pmod 3$. Hence the result. {{qed}}