hcju/proofwikiqa huggingface dataset:
By Open Ball of Metric Space is Open Set, every element of $\mathcal B_{\sequence{x_n}}$ is an open neighborhood of $a$. Let $U$ be an open neighborhood of $a$. By the definition of an open set, there exists a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} a \subseteq U$. By definition of a real null sequence: :$\exists N \in \N : \forall n > N : \size {x_n} < \epsilon$ Let $m = N + 1$. Then $\size {x_m} < \epsilon$. Since $x_m > 0$ then $x_m = \size {x_m} < \epsilon$, and so $\map {B_{x_m}} a \subseteq \map {B_\epsilon} a \subseteq U$. The result follows from Subset Relation is Transitive. {{qed}} Category:Metric Spaces p6e164vnwhzpwwhkc02y9l82madmz2z
Let $\sequence {x_n}$ be the sequence in $\R$ defined as: :$x_n = \dfrac 1 n$ From Sequence of Reciprocals is Null Sequence, $\sequence {x_n}$ is a real null sequence. From Null Sequence induces Local Basis in Metric Space: :$\mathcal B = \set {\map {B_{1/n} } a : n \in \N}$ is a local basis at $a$. {{qed}} Category:Metric Spaces 3a732x13wk7s1coxts0ddou0bxq98k8
Let the statement be rewritten as: For all $r \in \Z_{\ge 0}$:  :$3^r \equiv \begin {cases} 1 \pmod 8 & : r = 2 n \\ 3 \pmod 8 & : r = 2 n + 1 \end {cases}$ where $n \in \Z_{\ge 0}$. The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$3^{2 n} \equiv 1 \pmod 8$ $\map P 0$ is the case: :$3^{2 \times 0} = 3^0 = 1 \equiv 1 \pmod 8$ Thus $\map P 0$ is seen to hold. === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = 3^{2 \times 1}       | r = 3^2       | c =  }} {{eqn | l = 3^{2 \times 1}       | r = 9       | c =  }} {{eqn | o = \equiv       | r = 1 \pmod 8       | c =  }} {{end-eqn}} Thus $\map P 1$ 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: :$3^{2 k} \equiv 1 \pmod 8$ from which it is to be shown that: :$3^{2 \paren {k + 1} } \equiv 1 \pmod 8$ === Induction Step === This is the induction step: We have: {{begin-eqn}} {{eqn | l = 3^2       | o = \equiv       | r = 1       | rr= \pmod 8       | c = Basis for the Induction }} {{eqn | l = 3^{2 k}       | o = \equiv       | r = 1       | rr= \pmod 8       | c = Induction Hypothesis }} {{eqn | ll= \leadsto       | l = 3^2 \times 3^{2 k}       | o = \equiv       | r = 1 \times 1       | rr= \pmod 8       | c = Congruence of Product }} {{eqn | ll= \leadsto       | l = 3^{2 \paren {k + 1} }        | o = \equiv       | r = 1       | rr= \pmod 8       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and then it follows by the Principle of Mathematical Induction that: :$\forall n \in \Z_{\ge 0}: 3^{2 n} \equiv 1 \pmod 8$ Then we have: {{begin-eqn}} {{eqn | l = 3^{2 n + 1}       | r = 3 \times 3^{2 n}       | rr=        | c =  }} {{eqn | o = \equiv       | r = 3 \times 1       | rr= \pmod 8       | c = Congruence of Product }} {{eqn | o = \equiv       | r = 3       | rr= \pmod 8 }} {{end-eqn}} and the result follows. {{qed}}
Let $a$ and $b$ be two arbitrary consecutive positive integers. === Both Integers Powers of $2$ === The only powers of $2$ that differ by $1$ are $2^0$ and $2^1$, which gives us the pair $\tuple {1, 2}$. === Both Integers Powers of $3$ === The powers of $3$ are: :$1, 3, 9, \ldots$ and so trivially there are no two consecutive positive integers which are powers of $3$. === The Remaining Case === There is no need to consider the case where $a$ or $b$ is $1$, as that has already been investigated. So, let $a = 2^m$ and $b = 3^n$ where both $m$ and $n$ are greater than $0$. From Powers of 3 Modulo 8, $b \equiv 1 \pmod 8$ or $b \equiv 3 \pmod 8$. For $m \ge 3$ we have that: :$2^m \equiv 0 \pmod 8$ while: :$2^1 \equiv 2 \pmod 8$ :$2^2 \equiv 4 \pmod 8$ Let $a = b + 1$. Then we have the cases: :$b = 1, a = 2$ which has already been covered, and: :$b = 3, a = 4$ which gives us the pairs $\tuple {3, 4}$. Otherwise, for $a \ge 8$, we have: {{begin-eqn}} {{eqn | l = b + 1       | r = a       | c =  }} {{eqn | ll= \leadsto       | l = b + 1       | o = \equiv       | r = 0       | rr= \pmod 8       | c =  }} {{eqn | ll= \leadsto       | l = b       | o = \equiv       | r = 7       | rr= \pmod 8       | c =  }} {{end-eqn}} and we have demonstrated that there are no such $b$, as $b \equiv 1 \pmod 8$ or $b \equiv 3 \pmod 8$ Let $b = a + 1$. Let $b = 3^n$ where $n$ is odd. Then $b \equiv 3 \pmod 8$ which means: :$b \equiv 2 \pmod 8$ leading to the pair $\tuple {2, 3}$. No further pairs are possible for $b = 3^n$ with $n$ odd. Let $b = 3^n$ where $n$ is even. Thus: {{begin-eqn}} {{eqn | l = a       | r = 2^m       | c =  }} {{eqn | r = b - 1       | c =  }} {{eqn | r = 3^n - 1       | c =  }} {{eqn | r = 3^{2 k} - 1       | c =  }} {{eqn | r = \paren {3^k + 1} \paren {3^k - 1}        | c = Difference of Two Squares }} {{eqn | r = \paren {3^k + 1} \paren {3^k - 1}        | c = Difference of Two Squares }} {{eqn | r = 2^m        | c =  }} {{eqn | r = 2^p \, 2^q        | c = for some $p, q \in \Z_{\ge 0}$ such that $2^p = 2^q + 2$ }} {{end-eqn}} The only such $p$ and $q$ are $2^p = 2$ and $2^q = 4$ This means that the only possible pair fulfilling this condition is $\tuple {8, 9}$. Thus all such possible pairs have been found. {{qed}}
Standard Lagrangian is of the following form: :$\displaystyle L = \sum_{i \mathop = 1}^n \frac {m_i} 2 \dot {\mathbf x}_i^2 - U$ By the Principle of Stationary Action, equations of motion of $P$ are: :$\forall i \in \N: m_i {\ddot {\mathbf x} }_i + \dfrac {\partial U} {\partial \mathbf x_i} = 0$ By definition of velocity: :$\displaystyle {\dot {\mathbf x} }_i = {\mathbf v}_i$ By assumption: :${\mathbf F}_i = -\dfrac {\partial U} {\partial {\mathbf x}_i}$ Then it follows that the equations of motion of $P$ can be rewritten as: :$\dfrac \d {\d t} \paren {m_i {\mathbf v}_i } = {\mathbf F}_i$ By definition, these equations together represent Newton's Second Law of Motion. {{qed}}
:400px Consider the triangles $\triangle ABE$ and $\triangle ACF$. We have that: :$\angle FAC$ and $\angle BAE$ are common :$\angle AFC$ and $\angle AEB$ are both right angles and it follows from Triangles with Two Equal Angles are Similar that $\triangle ABE$ and $\triangle ACF$ are similar. Thus: :$\angle ABE = \angle ACF$ Consider the quadrilateral $\Box BFHD$. We have that $\angle BFH$ and $\angle BDH$ are both right angles. Thus two opposite angles of $\Box BFHD$ sum to two right angles So by Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles $\Box BFHD$ is a cyclic quadrilateral. From Angles in Same Segment of Circle are Equal: :$\angle FBH = \angle FDH$. By similar analysis of quadrilateral $\Box DHEC$, we note that: :$\angle HDE = \angle HCE$ But then we have: :$\angle FBH = \angle ABE$ and: :$\angle HCE = \angle ACF$ Hence it follows that: :$\angle FDH = \angle HDE$ demonstrating that $AD$ is the angle bisector of $\angle FDE$. The same argument applies mutatis mutandis to $\angle FDE$ and $\angle FED$. Hence the result. {{Qed}}
{{begin-eqn}} {{eqn | l = x \in \map {B_\epsilon} {a; \norm {\,\cdot\,} }       | o = \leadstoandfrom       | r = \norm {x - a} < \epsilon       | c = {{Defof|Open Ball of Normed Division Ring|open ball}} in $\struct {R, \norm {\,\cdot\,} }$ }} {{eqn | o = \leadstoandfrom       | r = \map d {x, a} < \epsilon       | c = {{Defof|Metric Induced by Norm on Division Ring|metric induced}} by $\norm {\,\cdot\,}$ }} {{eqn | o = \leadstoandfrom       | r = x \in \map {B_\epsilon} {a; d }       | c = {{Defof|Open Ball|open ball}} in $\struct {R, d}$ }} {{end-eqn}} The result follows from Equality of Sets. {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \map { {B_\epsilon}^-} {a; \norm {\,\cdot\,} }       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \norm {x - a}       | o = \le       | r = \epsilon       | c = {{Defof|Closed Ball of Normed Division Ring}} }} {{eqn | ll= \leadstoandfrom       | l = \map d {x, a}       | o = \le       | r = \epsilon       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \map { {B_\epsilon}^-} {a; d }       | c = {{Defof|Closed Ball}} }} {{end-eqn}} The result follows from Equality of Sets. {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \map {S_\epsilon} {a; \norm {\,\cdot\,} }       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \norm {x - a}       | r = \epsilon       | c = {{Defof|Sphere in Normed Division Ring}} }} {{eqn | ll= \leadstoandfrom       | l = \map d {x, a}       | r = \epsilon       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \map {S_\epsilon} {a; d }       | c = {{Defof|Sphere}} in $\struct {R, d}$ }} {{end-eqn}} The result follows from Equality of Sets. {{qed}}
The result follows directly from: :Closed Ball in Normed Division Ring is Closed Ball in Induced Metric :Open Ball in Normed Division Ring is Open Ball in Induced Metric :Sphere in Normed Division Ring is Sphere in Induced Metric :Sphere is Set Difference of Closed Ball with Open Ball {{qed}} Category:Normed Division Rings 6vn02xt3symwmexuwwu1ze2ydr4u7br
By assumption, $S$ is invariant under the following family of transformations: :$T = t + \epsilon$ :$\mathbf X = \mathbf x$ By Noether's Theorem: :$\nabla_{\dot {\mathbf x} } L \cdot \boldsymbol \psi + \paren {L - \dot {\mathbf x} \cdot \nabla_{\dot {\mathbf x} } L} \phi = C$ where $\phi = 1$, $\boldsymbol \psi = \mathbf 0$ and $C$ is an arbitrary constant. Then it follows that: {{begin-eqn}} {{eqn | l = L - \dot {\mathbf x} \cdot \nabla_{\dot {\mathbf x} } L       | r = T - U - \dot {\mathbf x} \cdot \nabla_{\dot{\mathbf x} } \paren {T - U} }} {{eqn | r = T - U - \dot {\mathbf x} \cdot \nabla_{\dot{\mathbf x} } \paren{\frac m 2 \dot {\mathbf x}^2} }} {{eqn | r = T - U - m \dot {\mathbf x}^2 }} {{eqn | r = T - U - 2 T }} {{eqn | r = - \paren {T + U} }} {{end-eqn}} Since the last term is the total energy of $P$, we conclude that it is conserved. {{qed}}
By assumption, $S$ is invariant under the following family of transformations: :$T = t$ :$X_j = x_j + \epsilon$ :$X_{i \ne j} = x_{i \ne j}$ By Noether's Theorem: :$\nabla_{\mathbf x} L \cdot \boldsymbol \psi + \paren {L - \dot {\mathbf x} \cdot \nabla_{\dot {\mathbf x} } L } \phi = C$ where $\phi = 0$, $\psi_i = 1$,  $\psi_{j \mathop \ne i} = 0$  and $C$ is an arbitrary constant. Then it follows that: :$\dfrac {\partial L} {\partial x_j} = C$ Since the last term is the momentum of $P$, we conclude that it is conserved. {{qed}}
By assumption, $S$ is invariant under the following family of transformations: :$T = t$ :$X_i = x_i \cos \epsilon + y_i \sin \epsilon$ :$Y_i = - x_i \sin \epsilon + y_i \cos \epsilon$ :$Z_i = z_i$ where $\epsilon \in \R$. By Noether's Theorem: :$\nabla_{\dot {\mathbf x} } L \cdot \boldsymbol \psi + \paren {L - \dot {\mathbf x} \cdot \nabla_{\dot {\mathbf x} } L } \phi = C$ where: :$\phi = 0$ :$\psi_{ix} = \intlimits {\dfrac {\partial X_i} {\partial \epsilon} } {\epsilon \mathop = 0} {} = y_i$ :$\psi_{iy} = \intlimits {\dfrac {\partial Y_i} {\partial \epsilon} } {\epsilon \mathop = 0} {} = -x_i$     :$\psi_{iz} = 0$ and $C$ is an arbitrary constant. Then it follows that: :$\displaystyle \sum_i \paren {\dfrac {\partial L} {\partial {\dot x}_i} y_i - \dfrac {\partial L} {\partial {\dot y}_i} x_i} = C$ Since the last term is the  $z$-th component of angular momentum of $P$ along $z$-axis, we conclude that it is conserved. {{qed}}
A direct application of Leibniz's law. {{Namedfor|Gottfried Wilhelm von Leibniz|cat = Leibniz}}
:400px The area of $A$ is seen to be: :the area of the outer circle with the area of the inner circle removed. From Area of Circle: :the area of the outer circle is $\pi R^2$ :the area of the inner circle is $\pi r^2$ The result follows. {{qed}}
:400px {{begin-eqn}} {{eqn | l = \map \Area A       | r = \pi \paren {R^2 - r^2}       | c = Area of Annulus }} {{eqn | r = \pi \paren {\paren {r + w}^2 - r^2}       | c = {{Defof|Width of Annulus}} }} {{eqn | r = \pi \paren {r^2 + 2 r w + w^2 - r^2}       | c = Square of Sum }} {{eqn | r = \pi \paren {2 r + w} w        | c = simplifying }} {{eqn | r = 2 \pi \paren {r + \frac w 2} \times w        | c = extracting $2$ }} {{end-eqn}} The result follows from Area of Rectangle. {{Qed}}
{{begin-eqn}} {{eqn| l = H       | r = -L + \sum_{i \mathop = 1}^n \dot {x_i} L_{\dot {x_i} } }} {{eqn| r = - \paren{T - V} + \sum_{i \mathop = 1}^n \dot {x_i} \dfrac {\partial} {\partial {\dot x}_i }\paren {\map T {\dot x} - \map V {t, x} }      | c = Definition of standard Lagrangian }} {{eqn| r = - \paren{T - V} + \sum_{i \mathop = 1}^n m_i \dot {x_i}^2      | c = Definition of kinetic energy of classical particles }} {{eqn| r = - T + V + 2 T      | c = Definition of kinetic energy of classical particles }} {{eqn| r = T + V }} {{end-eqn}} By definition, the last expression is the total energy of $P$. {{qed}}
By definition of the $p$-adic numbers, $\norm {\,\cdot\,}_p$ is a non-Archimedean norm. === Condition $(1)$ iff Condition $(2)$ === This follows directly from the definition of a closed ball in the $p$-adic numbers. {{qed|lemma}} === Condition $(1)$ iff Condition $(3)$ === By definition, $\map {B^{\,-}_{p^{-n}}} y$ is a closed ball in a non-Archimedean norm $\norm {\,\cdot\,}_p$. From Centers of Closed Balls in Non-Archimedean Division Rings: :$x \in \map {B^{\,-}_{p^{-n}}} y \leadsto \map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y$ From Center is Element of Closed Ball in P-adic Numbers: :$\map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y \leadsto x \in \map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y$ {{qed|lemma}} === Condition $(2)$ iff Condition $(4)$ === {{begin-eqn}} {{eqn | l = \norm{x - y}_p \le p^{-n}        | o = \leadstoandfrom       | r = \norm{x - y}_p \le \norm{p^n}_p       | c = {{Defof|P-adic Norm|$p$-adic norm}} on integers }} {{eqn | o = \leadstoandfrom       | r = \dfrac {\norm{x - y}_p} {\norm{p^n}_p} \le 1       | c = Dividing both sides of equation by $p^{-n}$ }} {{eqn | o = \leadstoandfrom       | r = \norm{p^{-n} \paren{x-y} }_p \le 1       | c = Norm of Quotient in Division Ring }} {{eqn | o = \leadstoandfrom       | r = p^{-n} \paren{x-y} \in \Z_p       | c = {{Defof|P-adic Integers|$p$-adic integers}} }} {{eqn | o = \leadstoandfrom       | r = x-y \in p^n\Z_p       | c =  }} {{end-eqn}} {{qed|lemma}} === Condition $(3)$ iff Condition $(5)$ === From Closed Balls of P-adic Number, :$\map {B^{\,-}_{p^{-n}}} x = x + p^n \Z_p$ and  :$\map {B^{\,-}_{p^{-n}}} y = y + p^n \Z_p$ Hence: :$\map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y$ {{iff}} $x + p^n \Z_p = y + p^n \Z_p$ {{qed}}
By definition of the $p$-adic numbers, $\norm {\,\cdot\,}_p$ is a non-Archimedean norm. === Condition $(1)$ iff Condition $(2)$ === This follows directly from the definition of a open ball in the $p$-adic numbers. {{qed|lemma}} === Condition $(1)$ iff Condition $(3)$ === By definition, $\map {B_{p^{-n}}} y$ is an open ball in a non-Archimedean norm $\norm {\,\cdot\,}_p$. From Centers of Open Balls in Non-Archimedean Division Rings: :$x \in \map {B_{p^{-n}}} y \leadsto \map {B_{p^{-n}}} x = \map {B_{p^{-n}}} y$ From Center is Element of Open Ball in P-adic Numbers: :$\map {B_{p^{-n}}} x = \map {B_{p^{-n}}} y \leadsto x \in \map {B_{p^{-n}}} x = \map {B_{p^{-n}}} y$ {{qed|lemma}} === Condition $(2)$ iff Condition $(4)$ === {{begin-eqn}} {{eqn | l = \norm {x - y}_p        | o = <       | r = p^{-n}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \norm {x - y}_p       | o = \le       | r = p^{-\paren {n + 1} }       | c = $p$-adic Norm of $p$-adic Number is Power of $p$ }} {{eqn | ll= \leadstoandfrom       | l = \norm {x - y}_p       | o = \le       | r = \norm {p^{n + 1} }_p       | c = {{Defof|P-adic Norm|$p$-adic Norm}} on Integers }} {{eqn | ll= \leadstoandfrom       | l = \dfrac {\norm {x - y}_p} {\norm {p^{-\paren {n + 1} } }_p}       | o = \le       | r = 1       | c = dividing both sides by $\norm {p^{-\paren {n + 1} } }$ }} {{eqn | ll= \leadstoandfrom       | l = \norm {p^{-\paren {n + 1} } \paren {x - y} }_p       | o = \le       | r = 1       | c = Norm of Quotient in Division Ring }} {{eqn | ll= \leadstoandfrom       | l = p^{-\paren {n + 1} } \paren {x - y}       | o = \in       | r = \Z_p       | c = {{Defof|P-adic Integers|$p$-adic Integers}} }} {{eqn | ll= \leadstoandfrom       | l = x - y       | o = \in       | r = p^{n + 1} \Z_p       | c =  }} {{end-eqn}} {{qed|lemma}} === Condition $(3)$ iff Condition $(5)$ === From Open Balls of P-adic Number, :$\map {B_{p^{-n} } } x = x + p^{n + 1} \Z_p$ and  :$\map {B_{p^{-n} } } y = y + p^{n + 1} \Z_p$ Hence: :$\map {B_{p^{-n} } } x = \map {B_{p^{-n} } } y$ {{iff}} $x + p^{n + 1} \Z_p = y + p^{n + 1} \Z_p$ {{qed}}
The standard Lagrangian of $P$ is: :$L = \dfrac 1 2 \paren {m {\dot x}^2 - k x^2}$. The canonical momentum is: :$p = \dfrac {\partial L} {\partial \dot x} = m \dot x$ The Hamiltonian associated to $L$ in canonical coordinates reads: :$H = \dfrac {p^2} {2 m} + \dfrac k 2 x^2$ Then: :$\sqbrk {x, p} = \dfrac {\partial x} {\partial x} \dfrac {\partial p} {\partial p} - \dfrac {\partial p} {\partial x} \dfrac {\partial x} {\partial p} = 1$ :$\sqbrk {x, H} = \dfrac {\partial x} {\partial x} \dfrac {\map \partial {\frac {p^2} {2 m} + \frac {k x^2} 2} } {\partial p} - \dfrac {\map \partial {\frac {p^2} {2 m} + \frac {k x^2} 2} } {\partial x} \dfrac {\partial x} {\partial p} = \dfrac p m$ :$\sqbrk {p, H} = \dfrac {\partial p} {\partial x} \dfrac {\map \partial {\frac {p^2} {2 m} + \frac {k x^2} 2} } {\partial p} - \dfrac {\map \partial {\frac {p^2} {2 m} + \frac {k x^2} 2} } {\partial x} \dfrac {\partial p} {\partial p} = - k x$ {{qed}}
By definition, the $p$-adic norm is the norm on a division ring. By definition, the $p$-adic metric is the metric induced by the $p$-adic norm. From Equivalence of Definitions of Convergence in Normed Division Rings, it follows that  Definition 2, Definition 3 and Definition 4 are equivalent. By definition of convergence in a normed division ring, $\sequence {x_n}$ converges to $x$ in the $p$-adic norm {{iff}}: :$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N: n > N \implies \norm {x_n - x}_p < \epsilon$ So Definition 1 is equivalent to Definition 2. {{qed}} Category:Sequences Category:Convergence Category:P-adic Number Theory gy59fniyr7y8py0l8rklx9x7qqpz5a2
=== Definition 1 iff Definition 2 === By definition, the metric induced by the norm $\norm {\, \cdot \,}$ is the mapping $d: R \times R \to \R_{\ge 0}$ defined as: :$\map d {x, y} = \norm {x - y}$ From Metric Induced by Norm on Normed Division Ring is Metric, $d$ is a metric. By definition of a convergent sequence in a metric space, $\sequence{x_n}$ converges to $x \in R$ {{iff}}: :$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N: n > N \implies \map d {x_n, x} < \epsilon$ {{iff}}: :$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N: n > N \implies \norm {x_n - x} < \epsilon$ The result follows. {{qed|lemma}} === Definition 1 iff Definition 3 === Let $x \in R$. By definition of norm on a division ring, $\norm {\, \cdot \,}$ is a mapping $\norm {\, \cdot \,}:R \to \R_{\ge 0}$. Then: :$\forall n \in \N: \size {\norm{x_n - x} - 0} = \size {\norm{x_n - x}} = \norm{x_n - x}$ By definition of convergence of a real sequence, the real sequence $\sequence {\norm {x_n - x} }$ converges to $0$ in the reals $\R$ {{iff}} :$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: n > N \implies \size {\norm{x_n - x} - 0} < \epsilon$ {{iff}}: :$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: n > N \implies \norm{x_n - x} < \epsilon$ The result follows. {{qed}}
We have: :Metrizable Space is Hausdorff :Finite Complement Space is not $T_2$ (Hausdorff) {{qed}}
Let $n$ be expressed in decimal notation as: :$n = \sqbrk {a_r a_{r - 1} \dotso a_1 a_0 \cdotp a_{-1} a_{-2} \dotso a_{-s} a_{-s - 1} \dotso}$ That is: :$n = \displaystyle \sum_{k \mathop \in \Z} a_k 10^k$ Then: {{begin-eqn}} {{eqn | l = n \times 10^d       | r = 10^d \times \sum_{k \mathop \in \Z} a_k 10^k       | c =  }} {{eqn | r = \sum_{k \mathop \in \Z} 10^d \times a_k 10^k       | c = Multiplication of Numbers Distributes over Addition }} {{eqn | r = \sum_{k \mathop \in \Z} a_k 10^{k + d}       | c = Product of Powers }} {{eqn | r = \sum_{k \mathop \in \Z} a_{k - d} 10^k       | c = Translation of Index Variable of Summation }} {{end-eqn}} The effect of presenting digit $a_{k - d}$ in position $k$ of $n$ is the same as what you get having moved the decimal point from between $a_0$ and $a_{-1}$ to between $a_{-d}$ and $a_{-d - 1}$. Thus: :if $d$ is positive, that is equivalent to moving the decimal point $d$ places to the right and: :if $d$ is negative, that is equivalent to moving the decimal point $d$ places to the left. {{qed}}
We have: {{begin-eqn}} {{eqn | l = 1       | r = 10^0       | c = {{Defof|Integer Power}} }} {{eqn | l = 10       | r = 10^1       | c = {{Defof|Integer Power}} }} {{eqn | ll= \leadsto       | l = \log_{10} 1       | r = 0       | c =  }} {{eqn | l = \log_{10} 10       | r = 1       | c =  }} {{end-eqn}} The result follows from Logarithm is Strictly Increasing. {{explain|Strictly speaking we need another step in here, the above is just for the natural logarithm.}}
{{begin-eqn}} {{eqn | l = n       | r = a \times 10^d       | c = by definition }} {{eqn | ll= \leadsto       | l = \log_{10} n       | r = \map {\log_{10} } {a \times 10^d}       | c =  }} {{eqn | r = \log_{10} a + \log_{10} 10^d       | c = Logarithm of Product }} {{eqn | r = \log_{10} a + d       | c = {{Defof|Common Logarithm}} }} {{end-eqn}} We are given that: :$1 \le a < 10$ It follows from Range of Common Logarithm of Number between 1 and 10 that: :$0 \le \log_{10} a < 1$ {{qed}} Category:Common Logarithms n5x32ns0jovsk9ok39zysxn29aq41l4
Let $x$ be expressed in scientific notation: :$x = a \times 10^e$ where: :$1 \le a < 10$ :$e \in \Z_{\ge 0}$ From Range of Common Logarithm of Number between 1 and 10: :$0 \le \log_{10} a < 1$ The characteristic of $\log_{10} x$ equals $\map {\log_{10} } {10^e} = e$. Thus the characteristic of $\log_{10} x$ is equal to the exponent of $x$. By Multiplication by Power of 10 by Moving Decimal Point, multiplication by $10^e$ is the same as moving the decimal point $e$ places to the right. Hence $a \times 10^e$ has $e$ more digits to the left of the decimal point than $a$ does. That is: $e + 1$. Hence the result. {{qed}}
Let $x$ be expressed in scientific notation: :$x = a \times 10^{-e}$ where: :$1 \le a < 10$ :$e \in \Z_{>0}$ From Range of Common Logarithm of Number between 1 and 10: :$0 \le \log_{10} a < 1$ The characteristic of $\log_{10} x$ equals $\map {\log_{10} } {10^{-e} } = \overline e$. Thus the characteristic of $\log_{10} x$ is equal to the exponent of $x$. When $e = -1$, we have that: :$0 \cdotp 1 \le x < 1$ and so $x$ is in the form: :$x = 0 \cdotp rstu \cdotp$ where $rstu \ldots$ are the digits of $x$ following the decimal point such that $r > 0$. Thus when $e = -1$ there are no zero digits to the immediate right of the decimal point of $x$ That is $e$ is one less than the number of zero digits to the immediate right of the decimal point of $x$. By Multiplication by Power of 10 by Moving Decimal Point, multiplication by $10^{-e}$ is the same as moving the decimal point $e$ places to the right. Hence $a \times 10^{-e}$ has $e - 1$ more digits to the left of the decimal point than $a$ does. That is: $e - 1$. Hence the result. {{qed}}