hcju/proofwikiqa huggingface dataset:
Performing the calculation using long division: <pre>    0.142857...   ------------  7)1.000000000      7      --      30      28      --       20       14       --        60        56        --         40         35         --          50          49          --           10            7           --           .. </pre> {{qed}} Category:7 Category:Examples of Reciprocals 9f9jealnmyoryjbz628gzjux1untqco
:400px Let $BC$ be produced to meet $A'B'$ and $A'C'$ at $L$ and $M$ respectively. Because $A'$ is the pole of the great circle $LBCM$, the spherical angle $A'$ equals the side of the spherical triangle $ALM$. By construction we have that $B'$ is the pole of $AC$. Thus the length of the arc of the great circle from $B$ to any point on $AC$ is a right angle. Similarly, the length of the arc of the great circle from $A'$ to any point on $BC$ is also a right angle. Hence: :the length of the great circle arc $CA'$ is a right angle :the length of the great circle arc $CB'$ is a right angle and it follows by definition that $C$ is a pole of $A'B'$. In the same way: :$A$ is a pole of $B'C'$ :$B$ is a pole of $A'C'$. Hence, by definition, $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$. {{qed}}
:400px Let $BC$ be produced to meet $A'B'$ and $A'C'$ at $L$ and $M$ respectively. Because $A'$ is the pole of the great circle $LBCM$, the spherical angle $A'$ equals the side of the spherical triangle $A'LM$. That is: :$(1): \quad \sphericalangle A' = LM$ From Spherical Triangle is Polar Triangle of its Polar Triangle, $\triangle ABC'$ is also the polar triangle of $\triangle A'B'C'$. That is, $C$ is a pole of the great circle $A'LB'$. Hence $CL$ is a right angle. Similarly, $BM$ is also a right angle. Thus we have: {{begin-eqn}} {{eqn | l = LM       | r = LB + BM       | c =  }} {{eqn | n = 2       | r = LB + \Box       | c = where $\Box$ denotes a right angle }} {{end-eqn}} By definition, we have that: :$BC = a$ {{begin-eqn}} {{eqn | l = BC       | r = a       | c = by definition of $\triangle ABC$ }} {{eqn | ll= \leadsto       | l = LB + a       | r = LC       | c =  }} {{eqn | n = 3       | ll= \leadsto       | l = LB       | r = \Box - a       | c = as $LC = \Box$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \sphericalangle A'       | r = LM       | c = from $(1)$ }} {{eqn | r = LB + \Box       | c = from $(2)$ }} {{eqn | r = \paren {\Box - a} + \Box       | c = from $(3)$ }} {{eqn | r = \paren {2 \Box} - a       | c =  }} {{end-eqn}} where $2 \Box$ is $2$ right angles, that is, $\pi$ radians. That is, $A'$ is the supplement of $a$: :$A' = \pi - a$ By applying the same analysis to $B'$ and $C'$, it follows similarly that: :$B' = \pi - b$ :$C' = \pi - c$ {{qed}}
From Closed Balls of P-adic Number: :$\map {B^-_1} 0 = \map {B^-_{p^0}} 0 = 0 + p^0 \Z_p = \Z_p$ From Integers are Dense in P-adic Integers, the integers $\Z$ are dense in the closed ball $\map {B^-_1} 0$. {{qed}} Category:P-adic Number Theory dxyfk8n2fni4vmlbp4uyicd5n9r9ayv
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$. Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively. From Spherical Triangle is Polar Triangle of its Polar Triangle we have that: :not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$ :but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$. We have: {{begin-eqn}} {{eqn | l = \sin a' \cos B'       | r = \cos b' \sin c' - \sin b' \cos c' \cos A'       | c = Analogue Formula for Spherical Law of Cosines }} {{eqn | ll= \leadsto       | l = \map \sin {\pi - A} \, \map \cos {\pi - b}       | r = \map \cos {\pi - B} \, \map \sin {\pi - C} - \map \sin {\pi - B} \, \map \cos {\pi - C} \, \map \cos {\pi - a}       | c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle }} {{eqn | ll= \leadsto       | l = \map \sin {\pi - A} \paren {-\cos b}       | r = \paren {-\cos B} \, \map \sin {\pi - C} - \map \sin {\pi - B} \, \paren {-\cos C} \, \paren {-\cos a}       | c = Cosine of Supplementary Angle }} {{eqn | ll= \leadsto       | l = \sin A \, \paren {-\cos b}       | r = \paren {-\cos B} \sin C - \sin B \, \paren {-\cos C} \, \paren {-\cos a}       | c = Sine of Supplementary Angle }} {{eqn | ll= \leadsto       | l = \sin A \cos b       | r = \cos B \sin C + \sin B \cos C \cos a       | c = simplifying }} {{end-eqn}} {{qed}} and: {{begin-eqn}} {{eqn | l = \sin a' \cos C'       | r = \cos c' \sin b' - \sin c' \cos b' \cos A'       | c = Analogue Formula for Spherical Law of Cosines }} {{eqn | ll= \leadsto       | l = \map \sin {\pi - A} \, \map \cos {\pi - c}       | r = \map \cos {\pi - C} \, \map \sin {\pi - B} - \map \sin {\pi - C} \, \map \cos {\pi - B} \, \map \cos {\pi - a}       | c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle }} {{eqn | ll= \leadsto       | l = \map \sin {\pi - A} \paren {-\cos c}       | r = \paren {-\cos C} \, \map \sin {\pi - B} - \map \sin {\pi - C} \, \paren {-\cos B} \, \paren {-\cos a}       | c = Cosine of Supplementary Angle }} {{eqn | ll= \leadsto       | l = \sin A \, \paren {-\cos c}       | r = \paren {-\cos C} \sin B - \sin C \, \paren {-\cos B} \, \paren {-\cos a}       | c = Sine of Supplementary Angle }} {{eqn | ll= \leadsto       | l = \sin A \cos c       | r = \cos C \sin B + \sin C \cos B \cos a       | c = simplifying }} {{end-eqn}} {{qed}}
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$. Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively. From Spherical Triangle is Polar Triangle of its Polar Triangle we have that: :not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$ :but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$. We have: {{begin-eqn}} {{eqn | l = \cos a' \cos C'       | r = \sin a' \cot b' - \sin C' \cot B'       | c = Four-Parts Formula }} {{eqn | ll= \leadsto       | l = \map \cos {\pi - A} \, \map \cos {\pi - c}       | r = \map \sin {\pi - A} \, \map \cot {\pi - B} - \map \sin {\pi - c} \, \map \cot {\pi - b}       | c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle }} {{eqn | ll= \leadsto       | l = \paren {-\cos A} \, \paren {-\cos c}       | r = \map \sin {\pi - A} \, \map \cot {\pi - B} - \map \sin {\pi - c} \, \map \cot {\pi - b}       | c = Cosine of Supplementary Angle }} {{eqn | ll= \leadsto       | l = \paren {-\cos A} \, \paren {-\cos c}       | r = \sin A \, \map \cot {\pi - B} - \sin c \, \map \cot {\pi - b}       | c = Sine of Supplementary Angle }} {{eqn | ll= \leadsto       | l = \paren {-\cos A} \, \paren {-\cos c}       | r = \sin A \, \paren {-\cot B} - \sin c \, \paren {-\cot b}       | c = Cotangent of Supplementary Angle }} {{eqn | ll= \leadsto       | l = \cos A \cos c       | r = \sin c \cot b - \sin A \cot B        | c = simplifying and rearranging }} {{end-eqn}} {{qed}}
Let $P = \tuple {x, y}$ and $A = \tuple {d, y}$. We have that: :$\dfrac {OM} {MN} = \dfrac {2 a} x = \dfrac y d$. Also, by Pythagoras's Theorem: :$\paren {a - y}^2 + d^2 = a^2 \implies y \paren {2 a - y} = d^2$ Eliminating $d$ gives us: :$\dfrac {y^2} {y \paren {2 a - y} } = \dfrac {\paren {2 a}^2} {x^2}$ Hence: :$\dfrac y {2 a - y} = \dfrac {4 a^2} {x^2}$ from which: :$y = \dfrac {8 a^3} {x^2 + 4 a^2}$ {{qed}}
{{begin-eqn}} {{eqn | l = \cos a       | r = \cos b \cos c + \sin b \sin c \cos A       | c = Spherical Law of Cosines }} {{eqn | ll= \leadsto       | l = 1 - 2 \hav a       | r = \cos b \cos c + \sin b \sin c \paren {1 - 2 \hav A}       | c = Cosine in Terms of Haversine }} {{eqn | r = \map \cos {b - c} - 2 \sin b \sin c \hav A       | c = Cosine of Difference }} {{eqn | r = 1 - 2 \map \hav {b - c} - 2 \sin b \sin c \hav A       | c = Cosine in Terms of Haversine }} {{eqn | ll= \leadsto       | l = \hav a       | r = \map \hav {b - c} + \sin b \sin c \hav A       | c = simplifying }} {{end-eqn}} {{qed}}
The haversine is conventionally defined on the real numbers only. We have that: :$\forall \theta \in \R: -1 < \cos \theta < 1$ and so: :$\forall \theta \in \R: 0 < 1 - \cos \theta < 2$ from which the result follows by definition of haversine. {{qed}}
{{begin-eqn}} {{eqn | l = \map \hav {-\theta}       | r = \dfrac 1 2 \paren {1 - \map \cos {-\theta} }       | c = {{Defof|Haversine}} }} {{eqn | r = \dfrac 1 2 \paren {1 - \cos \theta}       | c = Cosine Function is Even }} {{eqn | r = \hav \theta       | c = {{Defof|Haversine}} }} {{end-eqn}} {{qed}}
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$. Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively. From Spherical Triangle is Polar Triangle of its Polar Triangle we have that: :not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$ :but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$. Let $s' = \dfrac {a' + b' + c'} 2$. We have: {{begin-eqn}} {{eqn | l = \cos \dfrac {A'} 2       | r = \sqrt {\dfrac {\sin s' \, \map \sin {s' - a'} } {\sin b' \sin c'} }       | c = Cosine of Half Angle for Spherical Triangles }} {{eqn | ll= \leadsto       | l = \cos \dfrac {\pi - a} 2       | r = \sqrt {\dfrac {\sin s' \, \map \sin {s' - a'} } {\map \sin {\pi - B} \, \map \sin {\pi - C} } }       | c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle }} {{eqn | ll= \leadsto       | l = \map \cos {\dfrac \pi 2 - \dfrac a 2}       | r = \sqrt {\dfrac {\sin s' \, \map \sin {s' - a'} } {\sin B \sin C} }       | c = Sine of Supplementary Angle }} {{eqn | ll= \leadsto       | l = \sin \dfrac a 2       | r = \sqrt {\dfrac {\sin s'  \, \map \sin {s' - a'} } {\sin B \sin C} }       | c = Sine of Complement equals Cosine }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = s' - a'       | r = \dfrac {\paren {\pi - A} + \paren {\pi - B} + \paren {\pi - C} } 2 - \paren {\pi - A}       | c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle  }} {{eqn | r = \dfrac \pi 2 - \dfrac {A + B + C} 2 + A       | c = simplifying }} {{eqn | r = \dfrac \pi 2 - \paren {S - A}       | c = where $S = \dfrac {A + B + C} 2$ }} {{eqn | ll= \leadsto       | l = \map \sin {s' - a'}       | r = \map \sin {\dfrac \pi 2 - \paren {S - A} }       | c =  }} {{eqn | r = \map \cos {S - A}       | c = Sine of Complement equals Cosine }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = s'       | r = \dfrac {\paren {\pi - A} + \paren {\pi - B} + \paren {\pi - C} } 2       | c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle  }} {{eqn | r = \dfrac {3 \pi} 2 - \dfrac {A + B + C} 2       | c = simplifying }} {{eqn | r = \dfrac {3 \pi} 2 - S       | c = where $S = \dfrac {A + B + C} 2$ }} {{eqn | ll= \leadsto       | l = \sin s'       | r = \map \sin {\dfrac {3 \pi} 2 - S}       | c =  }} {{eqn | ll= \leadsto       | l = \sin s'       | r = -\map \sin {\dfrac \pi 2 - S}       | c = Sine of Angle plus Straight Angle }} {{eqn | r = -\cos S       | c = Sine of Complement equals Cosine }} {{end-eqn}} The result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \tan \dfrac a 2       | r = \dfrac {\sqrt {\dfrac {-\cos S \, \map \cos {S - A} } {\sin B \sin C} } } {\sqrt {\dfrac {\map \cos {S - B} \, \map \cos {S - C} } {\sin B \sin C} } }       | c = Sine of Half Angle for Spherical Triangles, Cosine of Half Angle for Spherical Triangles }} {{eqn | r = \sqrt {\dfrac {-\cos S \, \map \cos {S - A} } {\map \cos {S - B} \, \map \cos {S - C} } }       | c = simplification }} {{end-eqn}} Hence the result. {{qed}}
The vertical circle through $X$ is defined as the great circle that passes through $Z$. By definition, the angle of the arc from $Z$ to the horizon is a right angle. Hence $z + a = 90 \degrees$. The result follows. {{qed}}
Let $d_p$ be the $p$-adic metric on $\Q_p$: :$\forall x, y \in \Q_p: \map {d_p} {x, y} = \norm {x - y}_p$ From P-adic Metric on P-adic Numbers is Non-Archimedean Metric, $d_p$ is a non-Archimedean norm. By definition of a non-Archimedean norm: :$\forall x, y, z \in R: \norm {x - y}_p \le \max \set {\norm {x - z}_p, \norm {y - z}_p}$ {{qed}} Category:Normed Division Rings Category:Non-Archimedean Norms 1gv6hs4a6f9mz3ak4qmxkpzjaafetxk
Let $Z$ denote the zenith. The zenith distance $z$ of $P$ is by definition the length of the arc $PZ$ of the prime vertical. This in turn is defined as the angle $\angle POZ$ subtended by $PZ$ at $O$. This is equivalent to the angle between the radius of Earth through $O$ and Earth's axis. This is by definition the (terrestrial) colatitude of $O$. Hence the result. {{qed}}
Let $z$ denote the zenith distance of $P$. Let $\psi$ denote the (terrestrial) colatitude of $O$. By definition we have: :$a = 90 \degrees - z$ :$\phi - 90 \degrees - \psi$ Then: {{begin-eqn}} {{eqn | l = z       | r = \psi       | c = Zenith Distance of North Celestial Pole equals Colatitude of Observer }} {{eqn | ll= \leadsto       | l = 90 \degrees - z       | r = 90 \degrees - \psi       | c =  }} {{eqn | ll= \leadsto       | l = a       | r = \phi       | c =  }} {{end-eqn}} Hence the result. {{qed}}
Let $P$ denote the north celestial pole. The celestial equator is by definition perpendicular to $OP$. Also by definition, $OP$ is parallel to Earth's axis. Also by definition, Earth's axis is perpendicular to the geographical equator. Hence $OP$ is also perpendicular to the geographical equator. The result follows from Planes Perpendicular to same Straight Line are Parallel. {{qed}}
The apparent movement of $C$ is caused by the rotation of Earth about its axis. This rotation takes $1$ sidereal day, by definition, for the principal vertical circle to be in the same position again. As the Earth rotates, the observer $O$ moves in an easterly direction. Hence the celestial sphere $C$ appears to move in the opposite direction, so from north to east and so on. $B$ remains at the same distance from the north celestial pole as it did. Hence its declination $\delta$ stays the same. Hence the result. {{qed}}
=== Condition $(1)$ === {{:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Union of Closed Balls}}{{qed|lemma}} === Condition $(2)$ === {{:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Disjoint Closed Balls}}{{qed}} Category:Topology of P-adic Numbers Category:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers mvxihpzrbvi3tzh1op2y0hjxavlfjfw
==== Lemma ==== {{:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Lemma 1}}{{qed|lemma}} Let $0 \le i \le p^\paren{m - n} - 1$. Let $x \in \map {B^{\,-}_{p^{-m} } } {a + i p^{-n} }$ By definition of a closed ball: :$\norm {x - a - i p^{-n} } \le p^{-m}$ From Lemma: :$\norm {x - a}_p \le p^{-n}$ By definition of a closed ball: :$x \in \map {B^-_{p^{-n} } } a$ Since $x$ was arbitrary: :$\map {B^-_{p^{-m} } } {a + i p^{-n} } \subseteq \map {B^-_{p^{-n} } } a$ Since $i$ was arbitrary:  :$\displaystyle \bigcup_{i \mathop = 0}^{p^\paren {m - n} - 1} \map {B^-_{p^{-m} } } {a + i p^{-n} } \subseteq \map {B^-_{p^{-n} } } a$ Let $x \in \map {B^-_{p^{-n} } } a$. By definition of a closed ball: :$\norm {x - a}_p \le p^{-n}$ From Lemma: :$\exists i \in \N : 0 \le i \le p^\paren {m - n} - 1: \norm {x - a - i p^{-n} } \le p^{-m}$ By definition of a closed ball: :$\exists i \in \N : 0 \le i \le p^\paren {m - n} - 1 : x \in \map {B^-_{p^{-m} } } {a + i p^{-n} }$ Hence: :$\map {B^-_{p^{-n} } } a \subseteq \displaystyle \bigcup_{i \mathop = 0}^{p^\paren {m - n} - 1} \map {B^-_{p^{-m} } } {a + i p^{-n} }$ It follows that: :$\map {B^-_{p^{-n} } } a = \displaystyle \bigcup_{i \mathop = 0}^{p^\paren {m - n} - 1} \map {B^-_{p^{-m} } } {a + i p^{-n} }$
Let $0 \le i, j \le p^\paren {m - n} - 1$. Let $x \in \map {B^-_{p^{-m} } } {a + i p^n} \cap \map {B^-_{p^{-m} } } {a + j p^n}$ From Characterization of Open Ball in P-adic Numbers: :$\norm {\paren {x -a} - i p^n}_p \le p^{-m}$ and: :$\norm {\paren {x -a} - j p^n}_p \le p^{-m}$ Then: {{begin-eqn}} {{eqn | l = \norm {i p^n - j p^n}_p       | o = \le       | r = p^{-m}       | c = Corollary to P-adic Metric on P-adic Numbers is Non-Archimedean Metric }} {{eqn | ll= \leadsto       | l = \norm {p^n}_p \norm {i - j}_p       | o = \le       | r = p^{-m}       | c = P-adic Norm satisfies Non-Archimedean Norm Axioms: {{NormAxiom|2}} }} {{eqn | ll= \leadsto       | l = p^{-n} \norm {i - j}_p       | o = \le       | r = p^{-m}       | c = {{Defof|P-adic Norm|$p$-adic norm}} }} {{eqn | ll= \leadsto       | l = \norm {i - j}_p       | o = \le       | r = p^{n - m}       | c = multiplying both sides by $p^n$. }} {{eqn | ll= \leadsto       | l = p^\paren {m - n}       | o = \divides       | r = \paren {i - j}       | c = {{Defof|P-adic Norm|$p$-adic norm}} }} {{eqn | ll= \leadsto       | l = j       | o = \equiv       | r = i \mod p^\paren {m - n}       | c = {{Defof|Congruence Modulo Integer|Congruence Modulo $p$}} }} {{eqn | ll= \leadsto       | l = i       | r = j        | c = Integer is Congruent to Integer less than Modulus }} {{eqn | ll= \leadsto       | l = \map {B^-_{p^{-m} } } {a + i p^n}       | r = \map {B^-_{p^{-m} } } {a + j p^n}        | c = }} {{end-eqn}} The result follows.
=== Necessary Condition === Let $y \in \Q_p$. Let $\norm y_p \le p^{-n}$. {{:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Lemma 1/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === Let $y\in \Q_p$. Let there exist $i \in \Z$ such that: :$(1) \quad 0 \le i \le p^\paren {m - n} - 1$ :$(2) \quad \norm {y - i p^n}_p \le p^{-m}$ {{:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Lemma 1/Sufficient Condition}}{{qed}} Category:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers sld6g03brfc43dvpv2t8pgs4rfk6010
We have: {{begin-eqn}} {{eqn | l = \norm y_p         | o = \le       | r = p^{-n} }} {{eqn | ll= \leadsto       | l = p^n \norm{y}_p        | o = \le       | r = 1       | c = Multiply both sides by $p^n$ }} {{eqn | ll= \leadsto       | l = \norm{p^{-n} }_p \norm{y}_p       | o = \le       | r = 1       | c = {{Defof|P-adic Norm|$p$-adic norm}} }} {{eqn | ll= \leadsto       | l = \norm{p^{-n} y}_p       | o = \le       | r = 1       | c = P-adic Norm satisfies Non-Archimedean Norm Axioms: {{NormAxiom|2}} }} {{eqn | ll= \leadsto       | l = \map {B_1^-} {p^{-n}y}       | r = \map {B_1^-} 0       | c = Characterization of Closed Ball in P-adic Numbers }} {{end-eqn}} From Integers are Dense in Unit Ball of P-adic Numbers: :$\exists \mathop k \in \Z : \norm{p^{-n} y - k}_p \le p^\paren {n - m}$ From Residue Classes form Partition of Integers: :$\exists \mathop 0 \le i \le p^\paren {m - n} - 1: p^\paren {m - n} \divides k - i$ By definition of the $p$-adic norm: $\norm {k - i}_p  \le p^\paren {n - m}$ It follows that: {{begin-eqn}} {{eqn | l = \norm {p^{-n} y - i}_p       | o = \le        | r = \max \set {\norm {p^{-n} y - k}_p, \norm {i - k}_p}       | c = Corollary to P-adic Metric on P-adic Numbers is Non-Archimedean Metric }} {{eqn | o = \le       | r = p^\paren {n - m}       | c =  }} {{eqn | ll= \leadsto        | l = \norm {p^{-n} }_p \norm {y - i p^n}_p        | o = \le       | r = p^\paren {n - m}       | c = P-adic Norm satisfies Non-Archimedean Norm Axioms: {{NormAxiom|2}} }} {{eqn | ll= \leadsto        | l = p^n \norm {y - i p^n}_p       | o = \le        | r = p^\paren {n - m}       | c = {{Defof|P-adic Norm|$p$-adic norm}} }} {{eqn | ll= \leadsto        | l = \norm {y - i p^n}_p       | o = \le       | r = p^{-m}       | c = Divide both sides by $p^{-n}$ }} {{end-eqn}}
We have that: {{begin-eqn}} {{eqn | l = \norm y_p       | r = \norm {y - i p^n + i p^n}_p }} {{eqn | o = \le       | r = \max \set {\norm {y - i p^n}_p, \norm {i p^n}_p}       | c = P-adic Norm satisfies Non-Archimedean Norm Axioms: {{NormAxiom|4}} }} {{end-eqn}} By assumption: :$\norm{ y - i p^n} \le p^{-m} \le p^{-n}$ and: {{begin-eqn}} {{eqn | l = \norm{i p^n}_p       | r = \norm{i}_p \norm{p^n}_p       | c = P-adic Norm satisfies Non-Archimedean Norm Axioms: {{NormAxiom|2}} }} {{eqn | o = \le       | r = 1 \cdot p^{-n}       | c = As $i \in \Z \subseteq \Z_p$ }} {{eqn | r = p^{-n} }} {{end-eqn}} Hence: :$\max \set {\norm{ y - i p^n}_p, \norm{i p^n}_p} \le p^{-n}$ So: :$\norm{ y }_p \le p^{-n}$
We begin by proving the theorem for the closed ball $\map {B^-_{p^{-n} } } a$. From Open Ball in P-adic Numbers is Closed Ball then the theorem will be proved. Let $d$ denote the subspace metric induced by the norm $\norm {\,\cdot\,}_p$ on $\map {B^-_{p^{-n} } } a$. That is, $d: \map {B^-_{p^{-n} } } a \times \map {B^-_{p^{-n} } } a \to \R_{>0}$ is the metric defined by: ::$\forall x, y \in \map {B^-_{p^{-n} } } a: \map d {x,y} = \norm {x - y}_p$ By the definition of totally bounded it needs to be shown that: :for every $\epsilon \in \R_{>0}$ there exist finitely many points $x_0, \dots, x_k \in \map {B^-_{p^{-n} } } a$ such that: ::$\displaystyle \inf_{0 \mathop \le i \mathop \le k} \norm {x_i - x}_p \le \epsilon$ :for all $x \in \map {B^-_{p^{-n} } } a$.  From Sequence of Powers of Reciprocals is Null Sequence: :$\exists N \in \N: \forall k > N: p^{-k} \le \epsilon$. Let $m = \max \set {n + 1, N}$. Then $m > n$. From Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers: :$\map {B^-_{p^{-n} } } a = \displaystyle \bigcup_{i \mathop = 0}^{p^\paren {m - n} - 1} \map {B^-_{p^{-m} } } {a + i p^n}$ {{explain|Ambiguously formed $\LaTeX$: should the $1$ in the above upper limit of the union be inside the $p$ superscript? And again below. If not, then best to put the paren in braces, so as to ensure no possibility of misinterpretation.}} Let $x \in \map {B^-_{p^{-n} } } a$. Let $k = p^\paren {m - n} - 1$. Then there exists $0 \le j \le k$ such that: :$x \in \map {B^-_{p^{-m} } } {a + j p^n}$ That is: :$\norm {x - a + j p^n}_p \le p^{-m} \le \epsilon$ Hence: :$\displaystyle \inf_{0 \mathop \le i \mathop \le k} \norm {a + i p^n - x}_p \le \norm{x - a + j p^n}_p \le \epsilon$ Since $x \in \map {B^-_{p^{-n} } } a$ was arbitrary, then:  :$\forall x \in \map {B^-_{p^{-n} } } a : \displaystyle \inf_{0 \mathop \le i \mathop \le k} \norm {a + i p^n - x}_p \le \epsilon$ The result follows. {{qed}}
=== Lemma === Let$\mathbf J_n$ be the $n \times n$ square ones matrix. Let $\mathbf A$ be an $n \times n$ matrix. Let $\mathbf A_{ij}$ denote the cofactor of element $a_{ij}$ in $\map \det {\mathbf A}$, $1 \le i, j \le n$. Then: {{begin-eqn}} {{eqn | l = \map \det {\mathbf A - \mathbf J_n}       | r = \map \det {\mathbf A} - \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \mathbf A_{i j}       | c = Cofactor Sum Identity }} {{end-eqn}} {{qed|lemma}} Let $\mathbf A = \mathbf B^{-1}$. Then: {{begin-eqn}} {{eqn | l = \mathbf A \mathbf B       | r = \mathbf I       | c = {{Defof|Invertible Matrix}} }} {{eqn | l = \mathbf B        | r = \dfrac {\adj {\mathbf A} } {\map \det {\mathbf A} }       | c = Matrix Product with Adjugate Matrix with $\mathbf B = \mathbf A^{-1}$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{ij}       | r = \dfrac 1 {\map \det {\mathbf A} } \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \mathbf A_{ij} }} {{eqn | r = \map \det {\mathbf B} \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \mathbf A_{ij}       | c = Determinant of Matrix Product and $A B = I$ }} {{eqn | r = \map \det {\mathbf B} \paren {\map \det {\mathbf A} - \map \det {\mathbf A - \mathbf J_n} }       | c = Apply the '''Lemma''' }} {{eqn | r = \map \det {\mathbf B} \map \det {\mathbf A} - \map \det {\mathbf B} \map \det {\mathbf B^{-1} - \mathbf J_n}       | c = Definition $A = B^{-1}$ }} {{eqn | r = 1 - \map \det {\mathbf B} \map \det {\mathbf B^{-1} - \mathbf J_n}       | c = Determinant of Matrix Product and $\mathbf B \mathbf A = \mathbf I$ }} {{end-eqn}} {{qed}} Category:Matrix Theory 4e6rv083tspz41uzh9ugbnfccilb15a
We begin by proving the theorem for the closed ball $\map {B^-_{p^{-n}}} a$. From Open Ball in P-adic Numbers is Closed Ball then the theorem will be proved. By definition the $p$-adic numbers are a normed division ring. By definition the $p$-adic closed ball $\map {B^-_{p^{-n} } } a$ is a closed ball in a normed division ring.   By definition the $p$-adic norm $\norm {\,\cdot\,}_p$ is a non-Archimedean division ring norm. From Closed Ball of Non-Archimedean Division Ring is Clopen, $\map {B^-_{p^{-n} } } a$ is clopen in the $p$-adic metric. The result follows. {{qed}}