hcju/proofwikiqa huggingface dataset:
==== Necessary Condition ==== Let $\mathscr B$ satisfy the base axiom: {{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1}} $\dots$ It follows that $\mathscr B$ satisfies the base axiom: {{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 3}}{{qed|lemma}} ==== Sufficient Condition ==== By choosing $y = \map \pi x$ in Definition 3, Definition 1 follows immediately.
Let $N_m$ denote the set $\set {1, 2, \ldots, m}$. Let $f: N_m \to S$ be the mapping defined as: :$\forall k \in N_m: \map f t = t_m$ By definition, $f$ corresponds to one of the specific instances of $\sequence T_m$. Hence the number of different instances of $\sequence T_m$ is found from Cardinality of Set of All Mappings: :$\card S^{\card {N_m} }$ The result follows. {{qed}}
From Definite Integral to Infinity of Reciprocal of x Squared plus a Squared: :$\ds \int_0^\infty \dfrac {\d x} {x^2 + a^2} = \frac \pi {2 a}$ which holds for for $a \ne 0$.  The result follows by setting $a = 1$. {{qed}}
=== Norm Axiom $(\text N 1)$ === Let $x \in \ell^p$ with $1 \le p < \infty$. By definition of $p$-norm: :$\ds \norm {\mathbf x}_p = \paren {\sum_{n \mathop = 0}^\infty \size {x_n}^p}^{1/p}$ The complex modulus of $x_n$ is real and non-negative. We have the results: :Sum of Non-Negative Reals is Non-Negative :Power of Positive Real Number is Positive :Zero Raised to Positive Power is Zero Hence, $\norm {\mathbf x}_p \ge 0$. Suppose $\norm {\mathbf x}_p = 0$. Then: {{begin-eqn}} {{eqn | l = \norm {\mathbf x}_p       | r = 0 }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 0}^\infty \size {x_n}^p       | r = 0       | c = raising to power $p > 0$ }} {{eqn | ll= \leadsto       | l = \size {x_n}       | r = 0       | c = Sum of Non-Negatives vanishes iff Summands vanish }} {{eqn | ll= \leadsto       | l = x_n       | r = 0       | c = Complex Modulus equals Zero iff Zero }} {{eqn | ll= \leadsto       | l = \bf x       | r = \sequence 0_{n \mathop \in \N} }} {{end-eqn}} Thus norm axiom $(\text N 1)$ is satisfied. {{qed|lemma}} === Norm Axiom $(\text N 2)$ === Suppose $\alpha \in \set {\R, \C}$. {{begin-eqn}} {{eqn| l = \norm {\alpha \cdot \mathbf x}_p      | r = \paren {\sum_{n \mathop = 0}^\infty \size {\alpha x_n}^p }^{\frac 1 p} }} {{eqn| r = \paren {\sum_{n \mathop = 0}^\infty \size \alpha^p \size {x_n}^p }^{\frac 1 p} }} {{eqn| r = \size {\alpha} \paren {\sum_{n \mathop = 0}^\infty \size {x_n}^p }^{\frac 1 p} }} {{eqn| r = \size {\alpha} \norm {\mathbf x}_p }} {{end-eqn}} {{qed|lemma}} === Norm Axiom $(\text N 3)$ === From P-Norm is Norm we have that: :$\ds \paren {\sum_{n \mathop = 0}^d \size {x_n + y_n}^p}^{\frac 1 p} \le \paren {\sum_{n \mathop = 0}^d \size {x_n}^p }^{\frac 1 p} + \paren {\sum_{n \mathop = 0}^d \size {y_n}^p }^{\frac 1 p}$ $\map f z = z^{\frac 1 p}$ is a continuous function for $z \ge 0$ and $p > 0$. For $\mathbf x \in \ell^p$, changing $d$ is equivalent to changing $z$ in the interval $0 \le z < \infty$.  Take the composite limit $d \to \infty$. Then: :$\norm {\mathbf x + \mathbf  y}_p \le \norm {\mathbf x}_p + \norm {\mathbf y}_p$ {{qed}}
We have that $p$-norm is a norm on complex numbers. Since real numbers are wholy real complex numbers, the same results holds. {{qed}} Category:Examples of Norms Category:P-Norms d0t8j9qm69yeccq7h5c3xktgx0b9b4j
We have that: :$P$-sequence space is a vector space :$P$-norm on the $p$-sequence space is a norm By definition, $\struct {\ell^p, \norm {\, \cdot \,}_p}$ is a normed vector space. {{qed}}
Let the Cartesian coordinate system be selected such that $\mathbf u$ and $\mathbf v$ are embedded in the $x$-$y$ plane. Then: {{begin-eqn}} {{eqn | l = \mathbf u       | r = \tuple {u_1, u_2} }} {{eqn | l = \mathbf v       | r = \tuple {v_1, v_2} }} {{end-eqn}} and all other components, if there are any, are then zero. Thus by the component definition: :$\mathbf u + \mathbf v = \tuple {u_1 + v_1, u_2 + v_2}$ {{qed|lemma}} Now we use the triangle law: :500px Let: {{begin-eqn}} {{eqn | l = A       | r = \tuple {A_x, A_y} }} {{eqn | l = B       | r = \tuple {B_x, B_y} }} {{eqn | l = C       | r = \tuple {C_x, C_y} }} {{end-eqn}} Thus we have: {{begin-eqn}} {{eqn | l = \mathbf u + \mathbf v       | r = \vec {AC}       | c =  }} {{eqn | r = \tuple {C_x - A_x, C_y - A_y}       | c =  }} {{eqn | r = \tuple {\paren {C_x - B_x} + \paren {B_x - A_x}, \paren {C_y - B_y} + \paren {B_y - A_y} }       | c =  }} {{eqn | r = \tuple {v_1 + u_1, v_2 + u_2}       | c =  }} {{end-eqn}} It is seen that $\mathbf u + \mathbf v$ calculated using the triangle law is the same as $\mathbf u + \mathbf v$ calculated using the component definition. {{qed}} Category:Vector Addition qh7096tcvw6uxwn5hn084bzommo4tnd
Let $\mathbf 0$ denote a zero vector. {{AimForCont}} $\mathbf 0$ has a direction. Then $\mathbf 0$ can be represented as an arrow in a real vector space $\R^n$ with a Cartesian frame. Let $\mathbf 0$ be so embedded. Thus it consists of a line segment between two points with an initial point $A$ and a terminal point $B$. The initial point and a terminal point are distinct from each other. Let these points be identified as: {{begin-eqn}} {{eqn | l = A       | r = \tuple {a_1, a_2, \ldots, a_n} }} {{eqn | l = B       | r = \tuple {b_1, b_2, \ldots, b_n} }} {{end-eqn}} Hence we have that the length of $\mathbf 0$ is defined as: :$\norm {\mathbf 0} = \ds \sqrt {\sum_{i \mathop = 1}^n \paren {a_i - b_i}^2} > 0$ which means that at least one of $a_i - b_i$ is non-zero. But this contradicts the definition of $\mathbf 0$ being the zero vector. It follows by Proof by Contradiction that our assumption that $\mathbf 0$ has a direction must be false. Hence the result. {{qed}}
Let $\sequence {A_i}$ be a countably infinite sequence of sets in $\powerset S$. Consider an element of the union of all the sets in this $\sequence {A_i}$: :$\ds x \in \bigcup_{i \mathop \in \N} A_i$ By definition of union: :$\exists i \in \N: x \in A_i$ But $A_i \in \powerset S$ and so by definition $A_i \subseteq S$. By definition of subset, it follows that $x \in S$. Hence, again by definition of subset: :$\ds \bigcup_{i \mathop \in \N} A_i \subseteq S$ Finally, by definition of power set: :$\ds \bigcup_{i \mathop \in \N} A_i \in \powerset S$ {{qed}}
Let $r_1, r_2, \ldots r_k$ be a set of $k$ elements of $\set {1, 2, \ldots, m}$. Then: :$\paren {A_{r_1} \cap A_{r_2} \cap \cdots \cap A_{r_k} } \setminus \paren { A_{r_{k + 1} } \cup A_{r_{k + 2} } \cup \cdots \cup A_{r_m} }$ contains exactly those elements of $\Omega$ which are contained in exactly those events $A_{r_1}, A_{r_2}, \ldots, \cap A_{r_k}$. By Elementary Properties of Event Space, that last set is an event of $\EE$. By selecting all combinations of $\set {1, 2, \ldots, m}$, we can form the subsets of $\Sigma$ containing all those elementary events of exactly $k$ events. Each one is an element of $\Sigma$. We then form the union $S_k$ of each of those element of $\Sigma$. By Elementary Properties of Event Space, $S_k$ is an event of $\EE$. By construction, $S_k$ contains all and only the elementary events of $\EE$ which are elements of exactly $k$ of the events $A_1, A_2, \ldots, A_m$. {{qed}}
Let $A \in \Sigma$ be one of the events of $\EE$. We have by definition that $\Omega$ is itself an events of $\EE$. Hence by Set Difference of Events is Event, $\Omega \setminus A$ is also an event of $\EE$. As $A$ is arbitrary, the same applies to all events of $\EE$. Thus all events of $\EE$ come in pairs: $A$ and $\Omega \setminus A$. Hence the result. {{qed}}
From the Kolmogorov Axioms: :$\ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$ where $\set {A_1, A_2, \ldots}$ is a countable set of pairwise disjoint events of $\EE$. This applies directly to $\map \Pr {A \cup B}$ where $A \cap B = \O$. {{qed}}
From Power Set of Sample Space is Event Space we have that $\Sigma$ is an event space. {{qed|lemma}} We check the axioms defining a probability measure: {{begin-axiom}} {{axiom | n = \text I         | q = \forall A \in \Sigma         | ml= \map \Pr A         | mo= \ge         | mr= 0 }} {{axiom | n = \text {II}         | ml= \map \Pr \Omega         | mo= =         | mr= 1 }} {{axiom | n = \text {III}         | q = \forall A \in \Sigma         | ml= \map \Pr A         | mo= =         | mr= \sum_{\bigcup \set e \mathop = A} \map \Pr {\set e}         | rc= where $e$ denotes the elementary events of $\EE$ }} {{end-axiom}} Axiom $\text I$ is seen to be satisfied by the observation that the cardinality of a set is never negative. Hence $\map \Pr A \ge 0$. {{qed|lemma}} Then we have: {{begin-eqn}} {{eqn | l = \map \Pr \Omega       | r = \dfrac 1 n \card \Omega       | c =  }} {{eqn | r = \dfrac 1 n \times n       | c = Definition of $\Omega$: it has been defined as having $n$ elements }} {{eqn | r = 1       | c =  }} {{end-eqn}} Axiom $\text {II}$ is thus seen to be satisfied. {{qed|lemma}} Let $A = \set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} }$ where $k = \card A$. Then by Union of Set of Singletons: :$A = \set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} }$ Hence: {{begin-eqn}} {{eqn | l = \map \Pr A       | r = \dfrac 1 n \card A       | c =  }} {{eqn | r = \dfrac 1 n \card {\set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} } }       | c =  }} {{eqn | r = \dfrac 1 n \card {\set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} } }       | c =  }} {{eqn | r = \dfrac 1 n \paren {\underbrace {1 + 1 + \cdots + 1}_{\text {$k$ times} } }       | c =  }} {{eqn | r = \underbrace {\dfrac 1 n + \dfrac 1 n + \cdots + \dfrac 1 n}_{\text {$k$ times} }       | c =  }} {{eqn | r = \map \Pr {\set {\omega_{r_1} } } + \map \Pr {\set {\omega_{r_2} } } + \cdots + \map \Pr {\set {\omega_{r_k} } }       | c =  }} {{eqn | r = \sum_{\bigcup \set e \mathop = A} \map \Pr {\set e}       | c =  }} {{end-eqn}} Hence Axiom $\text {III}$ is thus seen to be satisfied. {{qed|lemma}} All axioms are seen to be satisfied. Hence the result. {{qed}}
Recall the Kolmogorov axioms: {{:Axiom:Kolmogorov Axioms}} First we determine that $\Pr$ as defined is actually a probability measure. By definition, we have that $\map \Pr A$ is the sum of some subset of $\set {p_1, p_2, \ldots, p_n}$. Thus $0 \le \map \Pr A \le 1$ and Axiom $(1)$ is fulfilled trivially by definition. Let $A \in \Sigma$ be such that: :$A = \set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} }$ We have that: :$A = \set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} }$ From Simple Events are Mutually Exclusive, $\set {\set {\omega_{r_1} }, \set {\omega_{r_2} }, \ldots, \set {\omega_{r_k} } }$ constitutes a set of pairwise disjoint events. Hence: :$\map \Pr {\set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} } } = \ds \sum_{i \mathop = 1}^k \map \Pr {\omega_{r_1} }$ and it is seen that axiom $(3)$ is fulfilled. Then we have that: {{begin-eqn}} {{eqn | l = \map \Pr \Omega       | r = \map \Pr {\bigcup_{i \mathop = 1}^n \omega_i}       | c =  }} {{eqn | r = \ds \sum_{i \mathop = 1}^n \map \Pr {\omega_i}       | c =  }} {{eqn | r = \ds \sum_{i \mathop = 1}^n p_i       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} Hence axiom $(2)$ is satisfied. {{Qed}}
From Event Space from Single Subset of Sample Space, we have that $\Sigma$ is an event space. Recall the Kolmogorov axioms: {{:Axiom:Kolmogorov Axioms}} First we determine that $\Pr$ as defined is actually a probability measure. Axiom $(1)$ and axiom $(2)$ are fulfilled trivially by definition. Then we note that, apart from $\O$, $\set {A, \Omega \setminus A}$ are the only pairwise disjoint events whose union is $\Omega$. Hence by definition of $\Pr$ we see that axiom $(3)$ degenerates to: {{begin-eqn}} {{eqn | l = \map \Pr {A \cup \paren {\Omega \setminus A} }       | r = \map \Pr \Omega       | c =  }} {{eqn | r = 1       | c = $\text {Pr 4}$ }} {{eqn | r = p + \paren {1 - p}       | c = Definition of $p$ }} {{eqn | r = \map \Pr A + \map \Pr {\Omega \setminus A}       | c = $\text {Pr 2}$ and $\text {Pr 3}$ }} {{end-eqn}} and thus axiom $(3)$ is fulfilled. Thus $\Pr$ is indeed a probability measure. {{qed|lemma}} Now it is established that a probability measure on $\Sigma$ is of the form of $\Pr$ as defined. First we note that from Probability of Empty Event is Zero: :$\map \Pr \O  = 0$ Hence $\text {Pr 1}$ is satisfied. From Axiom $(2)$ we have that $\map \Pr \Omega = 1$ Hence $\text {Pr 4}$ is satisfied. By definition of probability measure, $\map \Pr A$ is between $0$ and $1$ inclusive. That is: :$\map \Pr A = p$ for some $p \in \R$ such that $0 \le p \le 1$. This is exactly $\text {Pr 2}$, which is hence seen to be satisfied. Then we note that from Probability of Event not Occurring: :$\map \Pr {\Omega \setminus A} = 1 - \map \Pr A$ That is: :$\map \Pr {\Omega \setminus A} = 1 - p$ Hence $\text {Pr 3}$ is satisfied. Thus it has been proved that a probability measure on such an an event space $\Sigma$ has to be of the form $\Pr$ as defined. {{Qed}}
From Set Difference and Intersection form Partition: :$A$ is the union of the two disjoint sets $A \setminus B$ and $A \cap B$ So, by the definition of probability measure: :$\map \Pr A = \map \Pr {A \setminus B} + \map \Pr {A \cap B}$ Hence the result. {{qed}}
Let $s$ be the length of a curve between $2$ points $A$ and $B$. The problem becomes one of finding the curve for which $\ds \int_a^B \rd s$ is a minimum. {{ProofWanted|In due course as the work progresses}} Hence such a curve has the equation: :$y = m x + c$ which defines a straight line.
=== Norm Axiom $(\text N 1)$ === Let $x \in \ell^\infty$. By definition of supremum norm: :$\ds \norm {\mathbf x}_\infty = \sup_{n \mathop \in \N} \size {x_n}$ The complex modulus of $x_n$ is real and non-negative. Hence, $\norm {\mathbf x}_\infty \ge 0$. Suppose $\norm {\mathbf x}_\infty = 0$. Then: {{begin-eqn}} {{eqn | l = \norm {\mathbf x}_\infty       | r = 0 }} {{eqn | ll= \leadsto       | l = \sup_{n \mathop \in \N} \size {x_n}       | r = 0 }} {{eqn | ll= \leadsto       | l = x_n       | r = 0       | c = Complex Modulus equals Zero iff Zero }} {{eqn | ll= \leadsto       | l = \bf x       | r = \sequence 0_{n \mathop \in \N} }} {{end-eqn}} Thus norm axiom $(\text N 1)$ is satisfied. {{qed|lemma}} === Norm Axiom $(\text N 2)$ === Suppose $\alpha \in \C$. {{begin-eqn}} {{eqn| l = \norm {\alpha \cdot \mathbf x}_\infty      | r = \sup_{n \mathop \in \N} \size {\alpha x_n} }} {{eqn| r = \size {\alpha} \sup_{n \mathop \in \N} \size {x_n} }} {{eqn| r = \size {\alpha} \norm {\mathbf x}_\infty }} {{end-eqn}} {{qed|lemma}} === Norm Axiom $(\text N 3)$ === Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N}, \mathbf y = \sequence {y_n}_{n \mathop \in \N} \in \ell^\infty$. By Triangle Inequality for Complex Numbers: :$\forall n \in \N : \size {x_n + y_n} \le \size {x_n} + \size {y_n}$ Then: {{begin-eqn}} {{eqn | l = \norm {\mathbf x + \mathbf y}_\infty       | r = \sup_{n \mathop \in \N} \size {x_n + y_n}       | c = {{defof|Supremum Norm}} }} {{eqn | o = \le       | r = \sup_{n \mathop \in \N} \size {x_n} + \sup_{n \mathop \in \N} \size {y_n} }} {{eqn | r = \norm {\mathbf x}_\infty + \norm {\mathbf y}_\infty       | c = {{defof|Supremum Norm}} }} {{end-eqn}} {{qed}}
We have that: :Space of bounded sequences is a vector space :Supremum norm on the space of bounded sequences is a norm By definition, $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ is a normed vector space. {{qed}}
Since $Y < X$: :$X \setminus Y$ is non-empty.  Since $Y$ is closed: :$X \setminus Y$ is open.  Let $x \in X \setminus Y$, then there exists $\epsilon > 0$ such that:  :$\map {B_\epsilon} x \subset X \setminus Y$ So, for all $y \in Y$, we must have:  :$\norm {x - y} \ge \epsilon$ That is:  :$\inf \left\{\norm {x - y} \colon y \in Y\right\} \ge \epsilon$ For brevity, let: :$d = \inf \left\{\norm {x - y} \colon y \in Y\right\}$ Since $\alpha^{-1} > 1$, there exists $z \in Y$ with:  :$\norm {x - z} < d \alpha^{-1}$ otherwise the infimum would be at least $d \alpha^{-1}$, a contradiction. Since $x \in X \setminus Y$ and $z \in Y$, we clearly have $x \ne z$. So, we can set:  :$\displaystyle x_\alpha = \frac {x - z} {\norm {x - z} }$ Clearly: :$\displaystyle \norm {x_\alpha} = \frac {\norm {x - z} } {\norm {x - z} } = 1$ Now, for any $y \in Y$ we have:  {{begin-eqn}} {{eqn	| l = \norm {x_\alpha - y} 	| r = \norm {\frac {x - z} {\norm {x - z} } - y} }} {{eqn	| r = \frac {\norm {x - \paren {z + \norm {x - z} y} } } {\norm {x - z} } }} {{end-eqn}} We've already seen that:  :$\norm {x - z} < d \alpha^{-1}$ Since $Y$ is closed under linear combination, we have: :$z + \norm {x - z} y \in Y$ and so:  :$\norm {x - \paren {z + \norm {x - z} y} } \ge d$ We conclude that:  {{begin-eqn}} {{eqn	| l = \norm {x_\alpha - y} 	| r = \frac {\norm {x - \paren {z + \norm {x - z} y} } } {\norm {x - z} } }} {{eqn	| o = > 	| r = \frac d {d \alpha^{-1} }  }} {{eqn	| r = \alpha }} {{end-eqn}} {{qed}} {{Namedfor|Frigyes Riesz|cat = Riesz F}} Category:Functional Analysis 2dtquazxgxfx8hgz6crolrqva0ijaoz
Suppose that $\dim W = n$. Let:  :$\set {e_1, e_2, \ldots, e_n}$ be a basis for $W$. {{AimForCont}} that $W$ is not a closed set. Then there exists a convergent sequence $\sequence {w_k}$ in $W$ such that:  :$w_k \to w$ where $w \in V \setminus W$. {{explain|What does the notation $w_k \to w$ actually mean? As it stands, we are asked to accept at the end of this that $0 \to 1$ is "impossible", but that cannot be recognised as such unless we know what the notation means.}} Note that: :$\set {e_1, e_2, \ldots, e_n}$ is linearly independent in $W$, and hence $V$. Note that since $w \not \in W$, $w$ cannot be written as a linear combination of elements of $\set {e_1, e_2, \ldots, e_n}$. So: :$\set {e_1, e_2, \ldots, e_n, w}$ is linearly independent in $V$. So consider the subspace: :$W^* = \mathrm {span} \set {e_1, e_2, \ldots, e_n, w}$. Using the sequence $\sequence {w_n}$ from before, write:  :$w_k = \tuple {w_k^{\paren 1}, w_k^{\paren 2}, \ldots, w_k^{\paren n}, 0} \in W^*$ and: :$w = \tuple {0, 0, \ldots, 0, 1} \in W^*$ We necessarily have:  :$w_k^{\paren j} \to 0$ for each $1 \le j \le n$. However, we would also have:  :$0 \to 1$ Clearly this is impossible, so we have derived a contradiction. So $W$ is necessarily closed. {{qed}} Category:Functional Analysis 2opja2acq5674ufrwguwbahy9fh4u7e
:500px By definition, the force $\mathbf F_{21}$ and $\mathbf F_{31}$ are vector quantities. Hence their resultant can be found by using the Parallelogram Law. The result follows from Coulomb's Law of Electrostatics. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition: :$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$ $\map P 2$ is the case: $\mathbf F_1 = \dfrac 1 {4 \pi \varepsilon_0} \dfrac {q_1 q_2} {r_{2 1}^3} \mathbf r_{2 1}$ which is Coulomb's Law of Electrostatics. Thus $\map P 0$ is seen to hold. === Basis for the Induction === $\map P 3$ is the case: :$\mathbf F_1 = \dfrac {q_2 q_1} {4 \pi \varepsilon_0 r_{2 1}^3} \mathbf r_{2 1} + \dfrac {q_3 q_1} {4 \pi \varepsilon_0 r_{3 1}^3} \mathbf r_{3 1}$ which is proved in Total Force on Charged Particle from 2 Charged Particles. Thus $\map P 3$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le k  \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$ from which it is to be shown that: :$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le {k + 1}  \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \mathbf F_i       | r = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le k \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i} + \dfrac {q_i q_{k + 1} } {4 \pi \varepsilon_0 r_{\paren {k + 1} i}^3}  \mathbf r_{\paren {k + 1} i}       | c = Basis for the Induction }} {{eqn | r = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le {k + 1} \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}       | c = {{Defof|Summation}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 2}: \ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$ </onlyinclude> {{qed}}
By definition of scalar product: :$\size {0 \mathbf a} = 0 \size {\mathbf a}$ where $\size {\mathbf a}$ denotes the magnitude of $\mathbf a$. Thus: :$\size {0 \mathbf a} = 0$ That is: $0 \mathbf a$ is a vector quantity whose magnitude is zero. Hence, by definition, $0 \mathbf a$ is the zero vector. {{qed}}
Let $I := \closedint a b$ be a closed real interval. We have that: :Space of Continuously Differentiable on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication form Vector Space :$\map {C^1} I$ norm on the space of continuously differentiable on closed interval real-valued functions is a norm By definition, $\struct {\map {\CC^1} I, \norm {\, \cdot \,}_{1, \infty} }$ is a normed vector space. {{qed}}
By Structure Induced by Group Operation is Group, $\struct {\mathbb G, \oplus}$ is an abelian group. By Set of Homomorphisms is Subgroup of All Mappings, it follows that $\struct {\mathbb G, \oplus}$ is a subgroup of $\struct {G^G, \oplus}$. Next, we establish that $*$ is associative. By definition, $\forall u, v \in \mathbb G: u * v = u \circ v$ where $u \circ v$ is defined as composition of mappings. Associativity of $*$ follows directly from Composition of Mappings is Associative. Next, we establish that $*$ is distributive over $\oplus$. Let $u, v, w \in \mathbb G$. Then: :$\paren {u \oplus v} * w = \paren {u \oplus v} \circ w$ :$u * \paren {v \oplus w} = u \circ \paren {v \oplus w}$ So let $x \in G$. Then: {{begin-eqn}} {{eqn | l = \map {\paren {\paren {u \oplus v} * w} } x       | r = \map {\paren {\paren {u \oplus v} \circ w} } x       | c =  }} {{eqn | r = \map {\paren {u \oplus v} } {\map w x}       | c =  }} {{eqn | r = \map u {\map w x} \oplus \map v {\map w x}       | c =  }} {{eqn | r = \map {\paren {u \circ w} } x \oplus \map {\paren {v \circ w} } x       | c =  }} {{eqn | r = \map {\paren {u * w} } x \oplus \map {\paren {v * w} } x       | c =  }} {{end-eqn}} So $\paren {u \oplus v} * w = \paren {u * w} \oplus \paren {v * w}$. Similarly: {{begin-eqn}} {{eqn | l = \map {\paren {u * \paren {v \oplus w} } } x       | r = \map {\paren {u \circ \paren {v \oplus w} } } x       | c =  }} {{eqn | r = \map u {\map {\paren {v \oplus w} } x}       | c =  }} {{eqn | r = \map u {\map v x \oplus \map w x}       | c =  }} {{eqn | r = \map u {\map v x} \oplus \map u {\map w x}       | c = $u$ has the morphism property }} {{eqn | r = \map {\paren {u \circ v} } x \oplus \map {\paren {u \circ w} } x       | c =  }} {{eqn | r = \map {\paren {u * v} } x \oplus \map {\paren {u * w} } x       | c =  }} {{end-eqn}} So: :$u * \paren {v \oplus w} = \paren {u * v} \oplus \paren {u * w}$ Thus $*$ is distributive over $\oplus$. The ring axioms are satisfied, and $\struct {\mathbb G, \oplus, *}$ is a ring. The zero is easily checked to be the mapping which takes everything to the identity: :$e: G \to \set {e_G}: \map e x = e_G$ The unity is easily checked to be the identity mapping, which is known to be an automorphism. {{Qed}}
We have that $\struct {\mathbb G, \oplus, *}$ is a ring with unity. It remains to show that the operation $*$ is not necessarily commutative. {{ProofWanted}} Category:Rings of Endomorphisms 5h75pwdez5hb75bqi74jytghgcmbg3v
In order to be a ring, it is necessary that the additive operation $\oplus$ is commutative. However, as $\struct {G, \oplus}$ is specifically defined as being non-abelian, a fortiori $\oplus$ is not commutative. Hence the result. {{qed}}