hcju/proofwikiqa huggingface dataset:
Proof by Counterexample: Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology. Let $H_1 := \set 0 \cup \closedint 1 2 \cup \set 3$, where $\closedint 1 2$ denotes the closed interval from $1$ to $2$. Let $H_2 := \closedint 0 1 \cup \set 2 \cup \set 3$. $H_1$ and $H_2$ are homeomorphic, as can be demonstrated by the mapping $\phi: H_1 \to \H_2$ defined as: :$\forall x \in H_1: \map \phi x = \begin {cases} 2 & : x = 0 \\ 3 & : x = 3 \\ x - 1 & : x \in \closedint 1 2 \end {cases}$ which is trivially a homeomorphism. So, let $g$ be a homeomorphism from $H_1$ and $H_2$. Each of the singletons in $H_1$ has to map to one of the singletons in $H_2$. As a result, $g$ is not monotone. Let $f: \R \to \R$ be a homeomorphism such that $g$ is a restriction of $f$. From Continuous Real Function on Closed Interval is Bijective iff Strictly Monotone, $f$ is monotone. {{explain|We need another result for the above over the whole range $\R$ -- for some reason this does not exist. We have Surjective Monotone Function is Continuous but not in the other direction.}} Hence $f$ cannot be a bijection. Hence $f$ cannot be a homeomorphism. {{qed}}
The expectation of a continuous random variable $X$ with sample space $\Omega_X$ is given by: :$\displaystyle \expect X := \int_{x \mathop \in \Omega_X} x \map {f_X} x \rd x$ For the exponential distribution, we have $\Omega_X = \hointr 0 \infty$, and:  :$\displaystyle \map {f_X} x = \frac 1 \beta \map \exp {- \frac x \beta}$ So: :$\displaystyle \expect X = \int_0^\infty x \frac 1 \beta \map \exp {- \frac x \beta} \rd x$ where $\exp$ is the exponential function.  Substituting $u = \dfrac x \beta$, we have:: :$\displaystyle \expect X = \beta \int_0^\infty u \map \exp {-u} \rd u$ The integral evaluates to: :$\displaystyle \expect X = \bigintlimits {-\beta \paren {u + 1} \map \exp {-u} } 0 \infty$ So: :$\displaystyle \expect X = \beta - \beta \lim_{u \mathop \to \infty} \frac {u + 1} {\exp u}$ {{MissingLinks|The link below should be to the real counterpart of this result, but it's not up yet}} By Limit at Infinity of Polynomial over Complex Exponential, it follows that this limit is zero, so that: :$\expect X = \beta$ {{qed}}
From Moment Generating Function of Exponential Distribution, the moment generating function $M_X$ of $X$, is given by: :$\map {M_X} t = \dfrac 1 {1 - \beta t}$ By Moment in terms of Moment Generating Function: :$\expect X = \map {M_X'} 0$ We have:  {{begin-eqn}} {{eqn	| l = \map {M_X'} t  	| r = \map {\frac \d {\d t} } {\frac 1 {1 - \beta t} } }} {{eqn	| r = \frac {-\beta} {-1} \frac 1 {\paren {1 - \beta t}^2} 	| c = Chain Rule for Derivatives, Derivative of Power }} {{eqn	| r = \frac \beta {\paren {1 - \beta t}^2} }} {{end-eqn}}  Setting $t = 0$ gives: {{begin-eqn}} {{eqn	| l = \expect X 	| r = \frac \beta {\paren {1 - 0 \beta}^2} }} {{eqn	| r = \beta }} {{end-eqn}} {{qed}}
=== Necessary Condition === Let $G$ be a finite group whose order is $p^n$ for some $n \in \Z_{>0}$. Let $g \in G$. From Order of Element Divides Order of Finite Group, the order of $g$ is a divisor of $p^n$. That is, $x$ is a $p$-element by definition. As $x$ is arbitrary, it follows that all elements of $G$ are $p$-elements. Thus $G$ is a $p$-group. {{qed|lemma}} === Sufficient Condition === Let $G$ be a finite $p$-group. By definition, every element of $G$ is a $p$-element. From Order of Finite $p$-Group is Power of $p$: :$G$ is a finite group whose order is $p^n$ for some $n \in \Z_{\ge 0}$. Hence the result. {{qed}}
=== Definition 1 implies Definition 2 === Let $G$ be an abelian group by definition 1. Then by definition: : $\forall a \in G: \forall x \in G: a x = x a$ Thus: : $\forall a \in G: a \in \map Z G$ By definition of subset: :$G \subseteq \map Z G$ By definition of center: :$\map Z G \subseteq G$ So by definition of set equality: :$G = \map Z G$ Thus $G$ is an abelian group by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $G$ be an abelian group by definition 2. Then by definition: : $G = \map Z G$ So by the definition of center: : $\forall a \in G: \forall x \in G: a x = x a$ Thus $G$ is an abelian group by definition 1. {{qed}}
=== Definition 1 implies Definition 2 === Let $\mathbf A, \mathbf B$ be equivalent according to definition 1. Let $v$ be a boolean interpretation. By definition. either $v \left({\mathbf A}\right) = T$ or $v \left({\mathbf A}\right) = F$. In the first case, it follows by hypothesis that $v \left({\mathbf B}\right) = T$. In particular, then: :$v \left({\mathbf A}\right) = v \left({\mathbf B}\right)$ In the second case, it must be that $v \left({\mathbf B}\right) \ne T$. That is, $v \left({\mathbf B}\right) = F$, so that: :$v \left({\mathbf A}\right) = v \left({\mathbf B}\right)$ Hence $\mathbf A$ and $\mathbf B$ are also equivalent in the sense of definition 2. {{qed|lemma}} === Definition 2 implies Definition 3 === Let $\mathbf A, \mathbf B$ be equivalent according to definition 2. By definition of the boolean interpretation of $\iff$: :$v \left({\mathbf A \iff \mathbf B}\right)= T$ {{iff}} $v \left({\mathbf A}\right) = v \left({\mathbf B}\right)$ Therefore, by hypothesis and definition of tautology: :$\mathbf A \iff \mathbf B$ is a tautology. {{qed|lemma}} === Definition 3 implies Definition 1 === Let $\mathbf A, \mathbf B$ be equivalent according to definition 3. That is, let $\mathbf A \iff \mathbf B$ be a tautology. From the boolean interpretation of $\iff$, we have: :$v \left({\mathbf A}\right) = v \left({\mathbf B}\right)$ for every boolean interpretation $v$. Therefore it immediately follows that: :$v \left({\mathbf A}\right) = T$ {{iff}} $v \left({\mathbf B}\right) = T$ i.e. $\mathbf A$ and $\mathbf B$ are equivalent in the sense of definition 1. {{qed}}
Observe: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^N \Lambda \left({n}\right)       | r = \Lambda \left({1}\right) + \Lambda \left({2}\right) + \cdots + \Lambda \left({N}\right)       | c= }} {{eqn | r = 0 + \ln \left({2}\right) + \ln \left({3}\right) + \ln \left({2}\right) + \ln \left({5}\right) + 0 + \ln \left({7}\right) + \ln \left({2}\right) + \ln \left({3}\right) + 0 + \cdots       | c = }} {{end-eqn}} Notice this sum will have: : as many $\ln \left({2}\right)$ terms as there are powers of $2$ less than or equal to $N$ : as many $\ln \left({3}\right)$ terms as there are powers of $3$ less than or equal to $N$ and in general, if $p$ is a prime less than $N$, $\ln \left({p}\right)$ will occur in this sum $\left \lfloor {\log_p \left({N}\right)}\right \rfloor$ times. But: :$\displaystyle \ln \left({p}\right) \left \lfloor {\log_p \left({N}\right)} \right \rfloor \sim \ln \left({p}\right) \log_p \left({N}\right) = \ln \left({N}\right)$ so: :$\displaystyle \sum_{p \text{ prime} \mathop \le N} \ln \left({p}\right) \left \lfloor {\log_p \left({N}\right)} \right \rfloor \sim \sum_{p \text{ prime} \mathop \le N} \ln \left({N}\right) = \pi \left({N}\right) \ln \left({N}\right)$ Therefore: :$\displaystyle \sum_{n \mathop = 1}^N \Lambda \left({n}\right) \sim \pi \left({N}\right) \ln \left({N}\right)$ and so if: :$\displaystyle \lim_{N \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \Lambda \left({n}\right) = 1$  then: :$\displaystyle \lim_{N \to \infty} \frac 1 N \pi \left({N}\right) \ln \left({N}\right) = 1$ and vice versa. But this last equation is precisely the Prime Number Theorem. Hence our statement regarding the von Mangoldt function is logically equivalent to the Prime Number Theorem. {{qed}} {{namedfor|Hans Carl Friedrich von Mangoldt|cat=von Mangoldt}} Category:Prime Numbers 0540rt6ylzhmz1bwdz29u41jdhj8243
The Von Mangoldt Equivalence is equivalent (clearly) to the statement that the average of the coefficients of the function of $z$ defined as: :$(1): \quad \displaystyle \sum_{n \mathop = 1}^\infty \frac {\map \Lambda n} {n^z}$ tend to $1$. {{handwaving|Needs to be explained in more detail.}} Let $ \set {p_1, p_2, p_3, \dots}$ be an enumeration of the prime numbers: :$\set { 2, 3, 5, 7, 11, \dots}$ In the proof of the Von Mangoldt Equivalence, in the sum of von Mangoldt function, the $\map \ln p$ term will appear once for each power of $p$. So, we expand out $(1)$ as: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \frac{\map \Lambda n} {n^z}       | r = \map \ln {p_1} \paren {\frac 1 {p_1^z} + \frac 1 {p_1^{2 z} } + \frac 1 {p_1^{3 z} } + \cdots} + \map \ln {p_2} \paren {\frac 1 {p_2^z} + \frac 1 {p_2^{2 z} } + \cdots} + \cdots       | c =  }} {{eqn | r = \map \ln {p_1} \sum_{n \mathop = 1}^\infty \paren {\paren {p_1^{-z} }^n} + \map \ln {p_2} \sum_{n \mathop = 1}^\infty \paren {\paren {p_2^{-z} }^n} + \cdots       | c =  }} {{eqn | r = \map \ln {p_1} \frac {p_1^{-z} } {1 - p_1^{-z} } + \map \ln {p_2} \frac {p_2^{-z} } {1 - p_2^{-z} } + \cdots       | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \sum_{p \text{ prime} } \map \ln p \frac {p^{-z} } {1 - p^{-z} }       | c =  }} {{end-eqn}} This function of $z$ can be recognized as: {{begin-eqn}} {{eqn | l = \sum_{p \text{ prime} } \map \ln p \frac {p^{-z} } {1 - p^{-z} }       | r = \sum_{p \text{ prime} } \paren {1 - p^{-z} } \frac {-\paren {0 - \map \ln p p^{-z} } } {\paren {1 - p^{-z} }^2}       | c =  }} {{eqn | r = \sum_{p \text{ prime} } \frac \d {\d z} \map \ln {\frac {-1} {1 - p^{-z} } }       | c =  }} {{eqn | r = \map {\frac \d {\d z} } {\sum_{p \text{ prime} } \map \ln {\frac {-1} {1 - p^{-z} } } }       | c =  }} {{eqn | r = \map {\frac \d {\d z} } {\ln \prod_{p \text{ prime} } \frac {-1} {1 - p^{-z} } }       | c =  }} {{eqn | r = -\frac \d {\d z} \map \ln {\map \zeta z}       | c = $\displaystyle \prod_{p \text{ prime} } \frac 1 {1 - p^{-z} }$ is the Riemann zeta function }} {{eqn | r = -\frac {\map {\zeta'} z} {\map \zeta z}       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Prime Numbers ffru57tk03klkfodswx7sp9kwozdsiy
Proof by Counterexample: Let $c = 6, a = 3, b = 4$. Then $6 \times 2 = 12$ so $c \divides a b$. But neither $6 \divides 4$ nor $6 \divides 3$. {{qed}}
Let $N = p! + 1$. If $N$ is prime, then the proof is complete. Otherwise, from Positive Integer Greater than 1 has Prime Divisor, $N$ has a prime divisor $q$. From Absolute Value of Integer is not less than Divisors: :$q < N$ {{AimForCont}} $q \le p$. Then by the definition of factorial: :$q \divides p!$ But $q$ was chosen so that :$q \divides N$. Let $p! = q m_1, N = q m_2$. Then: :$1 = N - p! = q \paren {m_2 - m_1}$ and so: :$q \divides 1$ From Absolute Value of Integer is not less than Divisors: :$q \le 1$ and so $q$ cannot be a prime number. It follows by Proof by Contradiction that $q > p$. {{qed}}
By definition of Kronecker delta: :$\delta_{i j} = \begin {cases} 1 & : i = j \\ 0 & : i \ne j \end {cases}$ Thus: :$\delta_{i j} a_{j k} = \begin {cases} a_{i k} & : i = j \\ 0 & : i \ne j \end {cases}$ and: :$\delta_{i j} a_{k j} = \begin {cases} a_{k i} & : i = j \\ 0 & : i \ne j \end {cases}$ from which the result follows. {{qed}}
By definition: :$\mathbf I_n := \sqbrk a_n: a_{i j} = \delta_{i j}$ That is: each of the elements on the main diagonal is equal to $1$. There are $n$ such elements. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = S \cup \paren {S \cap T} }} {{eqn | o = \leadstoandfrom       | r = \paren {S \cup S} \cap \paren {S \cup T}       | c = Union Distributes over Intersection }} {{eqn | o = \leadstoandfrom       | r = S \cap \paren {S \cup T}       | c = Union is Idempotent }} {{end-eqn}} {{qed}} Category:Set Intersection Category:Set Union Category:Absorption Laws 3su6gpzcntao36l5ekfopv7x983honp
Let $\mathbf A$ be invertible. Let $1_R$ denote the unity of $R$. Let $\mathbf I_n$ denote the unit matrix of order $n$. Then: {{begin-eqn}} {{eqn | l = 1_R       | r = \map \det {\mathbf I_n}       | c = Determinant of Unit Matrix }} {{eqn | r = \map \det {\mathbf A \mathbf B}       | c = {{Defof|Inverse Matrix}} }} {{eqn | r = \map \det {\mathbf A} \, \map \det {\mathbf B}       | c = Determinant of Matrix Product }} {{end-eqn}}  This shows that: :$\map \det {\mathbf B} = \dfrac 1 {\map \det {\mathbf A} }$ {{qed|lemma}}
By definition, an orthogonal matrix is one such that: :$\mathbf Q^\intercal = \mathbf Q^{-1}$ and so the result follows by definition of inverse. {{qed}}
By Determinant of Transpose: :$\det \mathbf Q^\intercal = \det \mathbf Q$ Then: {{begin-eqn}} {{eqn | l = \mathbf Q \mathbf Q^\intercal       | r = \mathbf I       | c = Product of Orthogonal Matrix with Transpose is Identity }} {{eqn | ll= \leadsto       | l = \map \det {\mathbf Q \mathbf Q^\intercal}       | r = \det \mathbf I       | c =  }} {{eqn | ll= \leadsto       | l = \map \det {\mathbf Q \mathbf Q^\intercal}       | r = 1       | c = Determinant of Unit Matrix }} {{eqn | ll= \leadsto       | l = \det \mathbf Q \det \mathbf Q^\intercal       | r = 1       | c = Determinant of Matrix Product }} {{end-eqn}} Hence the result. {{qed}}
By definition, $\mathbf P \mathbf Q$ is a proper orthogonal matrix {{iff}} it is an orthogonal matrix with a determinant of $1$. From Product of Orthogonal Matrices is Orthogonal Matrix, $\mathbf P \mathbf Q$ is an orthogonal matrix. By definition, $\mathbf P$ and $\mathbf Q$ both have a determinant of $1$. From Determinant of Matrix Product: :$\det \left({\mathbf P \mathbf Q}\right) = \det \left({\mathbf P}\right) \det \left({\mathbf Q}\right)$ Thus: :$\det \left({\mathbf P \mathbf Q}\right) = 1$ Hence the result. {{qed}} Category:Orthogonal Matrices renl7fwzyrwo2rltv5hg4vjq8pbpvku
From Determinant of Orthogonal Matrix is Plus or Minus One and Matrix is Invertible iff Determinant has Multiplicative Inverse it follows that both $\mathbf P$ and $\mathbf Q$ are invertible. Thus: {{begin-eqn}} {{eqn | l = \paren {\mathbf P \mathbf Q}^{-1}       | r = \mathbf Q^{-1} \mathbf P^{-1}       | c = Inverse of Matrix Product }} {{eqn | r = \mathbf Q^\intercal \mathbf P^\intercal       | c = {{Defof|Orthogonal Matrix}} }} {{eqn | r = \paren {\mathbf P \mathbf Q}^\intercal       | c = Transpose of Matrix Product }} {{end-eqn}} Hence the result, by definition of orthogonal matrix. {{qed}}
Let $\mathbf C := \mathbf A \mathbf B$. By definition of matrix product: :$\ds c_{i k} = \sum_{j \mathop = 1}^n a_{i j} b_{j k}$ Thus for the diagonal elements: :$\ds c_{i i} = \sum_{j \mathop = 1}^n a_{i j} b_{j i}$ By definition of trace: :$\ds \map \tr {\mathbf C} = \sum_{i \mathop = 1}^n c_{i i}$ Hence the result. {{qed}}
Let $\mathbf C = \mathbf A_1 \mathbf A_2 \cdots \mathbf A_m$ From Product of Finite Sequence of Matrices, the general element of $\mathbf C$ is given in  the Einstein summation convention by: :$\map c {i_1, j} = \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \cdots \map {a_{m - 1} } {i_{m - 1}, i_m} \map {a_m} {i_m, j}$ Thus for the diagonal elements: :$\ds \map c {i_1, i_1} = \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \cdots \map {a_{m - 1} } {i_{m - 1}, i_m} \map {a_m} {i_m, i_1}$ which is the Einstein summation convention for the trace of $\mathbf C$. {{qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \map c {i_1, i_{n + 1} } = \sum_{i_n \mathop = 1}^{d_n} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_{n - 1} } {i_{n - 1}, i_n} \map {a_n} {i_n, i_{n + 1} }$ $\map P 1$ is trivially true, as this just says: :$\map c {i_1, i_2} = \map {a_1} {i_1, i_2}$ === Basis for the Induction === $\map P 2$ is the case: :$\displaystyle \map c {i_1, i_3} = \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3}$ which is the definition of (conventional) matrix product. The order of $\mathbf C$ is $d_1 \times d_3$, while the orders of $\mathbf A_1$ and $\mathbf A_2$ are $d_1 \times d_2$ and $d_2 \times d_3$ respectively. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \map c {i_1, i_{k + 1} } = \sum_{i_k \mathop = 1}^{d_k} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_k} {i_k, i_{k + 1} }$ Then we need to show: :$\displaystyle \map c {i_1, i_{k + 2} } = \sum_{i_{k + 1} \mathop = 1}^{d_{k + 1} } \sum_{i_k \mathop = 1}^{d_k} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_k} {i_k, i_{k + 1} } \, \map {a_{k + 1} } {i_{k + 1}, i_{k + 2} }$ === Induction Step === This is our induction step: Let $\displaystyle \mathbf C_1 := \prod_{j \mathop = 1}^k \mathbf A_j$. Let $\displaystyle \mathbf C := \prod_{j \mathop = 1}^{k + 1} \mathbf A_j = \mathbf C_1 \mathbf A_{k + 1}$. {{begin-eqn}} {{eqn | l = \map c {i_1, i_{k + 2} }       | r = \sum_{i_{k + 1} \mathop = 1}^{d_{k + 1} } \map {c_1} {i_1, i_{k + 1} } \map {a_{k + 1} } {i_{k + 1}, i_{k + 2} }       | c = {{Defof|Matrix Product (Conventional)|Matrix Product}} }} {{eqn | r = \sum_{i_{k + 1} \mathop = 1}^{d_{k + 1} } \paren {\sum_{i_k \mathop = 1}^{d_k} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_k} {i_k, i_{k + 1} } } \map {a_{k + 1} } {i_{k + 1}, i_{k + 2} }       | c = Induction Hypothesis }} {{eqn | r = \sum_{i_{k + 1} \mathop = 1}^{d_{k + 1} } \sum_{i_k \mathop = 1}^{d_k} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_k} {i_k, i_{k + 1} } \map {a_{k + 1} } {i_{k + 1}, i_{k + 2} }       | c = Associative Law of Addition }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore for all $n \in \N_{>0}$: :$\displaystyle \map c {i_1, i_{n + 1} } = \sum_{i_n \mathop = 1}^{d_n} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_{n - 1} } {i_{n - 1}, i_n} \map {a_n} {i_n, i_{n + 1} }$ {{qed}} Category:Conventional Matrix Multiplication k54na6jenrbhsne4qu3bxh3gvy39zz0
By Characteristic of Field is Zero or Prime, it follows that $\Char \GF$ is $0$ or a prime number.  By Finite Field has Non-Zero Characteristic: :$\Char \GF \ne 0$ Thus $\Char \GF$ is a prime number. {{qed}} Category:Galois Fields j1df1dpfgmn7ywm6w4um79xpevopykt
As $R$ is non-null, there is at least one nonzero element in $R$.  Consider the two maps from $R$ to itself, for each nonzero $a \in R$: :$\varphi_R: x \mapsto a \circ x$ :$\varphi_L: x \mapsto x \circ a$ By Ring Element is Zero Divisor iff not Cancellable, all nonzero elements in $R$ are cancellable. Thus: :$a \circ x = a \circ y \implies x=y$ :$x \circ a = y \circ a \implies x=y$ Therefore, both maps are by definition injective. By Equivalence of Mappings between Sets of Same Cardinality, the maps are then also surjective. First, we show that $R$ has a unity.  Since $\varphi_R$ is surjective, any element $b \in R$ can be written in the form: :$b = a \circ x_b$  for some $x_b \in R$. As $\varphi_L$ is surjective, there is some $x_L \in R$ such that: :$a = x_L \circ a$ We now have: {{begin-eqn}} {{eqn | l = b       | r = a \circ x_b }} {{eqn | r = x_L \circ a \circ x_b       | c = as $a = x_L \circ a$ }} {{eqn | r = x_L \circ b       | c = as $b = a \circ x_b$ }} {{end-eqn}} Thus, $x_L$ is a multiplicative left identity.  By similar arguments, $R$ also has a multiplicative right identity, which can be denoted $x_R$. By Left and Right Identity are the Same, it follows that $x_L = x_R$, so they are the unity of $R$.  Now we show that $R$ is a division ring, i.e. that each nonzero element of $R$ has an inverse. Let $1_R$ denote the unity of $R$. Since $\varphi_R$ is surjective, it follows that: :$a \circ y_R = 1_R$ for some $y_R \in R$.  Since $\varphi_L$ is surjective, it follows that: :$y_L \circ a = 1_R$  for some $y_L \in R$. Recalling that the maps were defined for each nonzero $a \in R$, this means that every nonzero element of $R$ has both a left inverse and a right inverse.  By Left Inverse and Right Inverse is Inverse, each nonzero element of $R$ has an inverse, so $R$ is by definition a division ring.  It now follows by Wedderburn's Theorem that $R$ is a field.  {{qed}} Category:Ring Theory Category:Field Theory Category:Galois Fields oelmpme9wovskd87f0oix14lqd2ojsq
=== Definition 1 implies Definition 2 === Let $\mathbf a \cdot \mathbf b$ be a dot product by definition 1. From Cosine Formula for Dot Product: :$\mathbf a \cdot \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \cos \theta$ where $\theta$ is the angle between $\mathbf v$ and $\mathbf w$. Thus $\cdot$ is a dot product by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\mathbf a \cdot \mathbf b$ be a dot product by definition 2. Let $\mathbf a$ and $\mathbf b$ be expressed in the form: {{begin-eqn}} {{eqn | l = \mathbf a       | r = a_1 \mathbf e_1 + a_2 \mathbf e_2 + \cdots + a_n \mathbf e_n }} {{eqn | r = \sum_{i \mathop = 1}^n a_i \mathbf e_i }} {{eqn | l = \mathbf b       | r = b_1 \mathbf e_1 + b_2 \mathbf e_2 + \cdots + b_n \mathbf e_n }} {{eqn | r = \sum_{j \mathop = 1}^n b_j \mathbf e_j }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | l = \mathbf a \cdot \mathbf b       | r = \paren {\sum_{i \mathop = 1}^n a_i \mathbf e_i} \cdot \paren {\sum_{j \mathop = 1}^n b_j \mathbf e_j}       | c =  }} {{eqn | r = \sum_{i, j \mathop = 1}^n a_i b_j \mathbf e_i \mathbf e_j       | c = Dot Product Distributes over Addition }} {{eqn-intertext | noting at this point that Dot Product Distributes over Addition has been derived from {{Defof|Dot Product|index = 2}} }} {{eqn | r = \sum_{i, j \mathop = 1}^n a_i b_j \delta_{i j}       | c = Dot Product of Orthogonal Basis Vectors }} {{eqn-intertext | noting at this point that Dot Product of Orthogonal Basis Vectors has also ultimately been derived from {{Defof|Dot Product|index = 2}} }} {{eqn | r = \sum_{i \mathop = 1}^n a_i b_i       | c = simplifying }} {{end-eqn}} Thus $\cdot$ is a dot product by definition 1. {{qed}}
Follows from Row Equivalent Matrix for Homogeneous System has same Solutions and from Matrix is Row Equivalent to Reduced Echelon Matrix. {{qed}} Category:Matrix Theory Category:Linear Algebra dla471omeqku92w6a8j8ujatufa5qe7
This follows immediately from the definitions: :$(1): \quad$ A mapping is sequentially continuous everywhere in $X$ {{iff}} if it is sequentially continuous at each point :$(2): \quad$ A mapping is continuous everywhere on $X$ {{iff}} it is continuous at each point. {{Qed}} Category:Continuous Mappings knf1qdbtxukn1rzdw4x6v5lt9y6r4d5
{{begin-eqn}} {{eqn | l = z_1 z_2       | r = r_1 \paren {\cos \theta_1 + i \sin \theta_1} r_2 \paren {\cos \theta_2 + i \sin \theta_2}       | c = {{Defof|Polar Form of Complex Number}} }} {{eqn | r = r_1 r_2 \paren {\paren {\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2} + i \paren {\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2} }       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = r_1 r_2 \paren {\map \cos {\theta_1 + \theta_2} + i \paren {\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2} }       | c = Cosine of Sum }} {{eqn | r = r_1 r_2 \paren {\map \cos {\theta_1 + \theta_2} + i \map \sin {\theta_1 + \theta_2} }       | c = Sine of Sum }} {{end-eqn}} {{qed}}
By definition of a complex number, $z_1$ and $z_2$ can be expressed in the form: :$z_1 = \tuple {a_1, b_1}$ :$z_2 = \tuple {a_2, b_2}$ where $\tuple {a, b}$ denotes an ordered pair. The result follows from Equality of Ordered Pairs.
{{begin-eqn}} {{eqn | l = z_1 - z_2       | r = z_1 + \paren {- z_2}       | c = {{Defof|Complex Subtraction}} }} {{eqn | r = a_1 + \paren {-a_2} + i \paren {b_1 + \paren {-b_2} }       | c = Inverse for Complex Addition }} {{eqn | r = \paren {a_1 - a_2} + i \paren {b_1 - b_2}       | c = {{Defof|Complex Subtraction}} }} {{end-eqn}} {{qed}}