hcju/proofwikiqa huggingface dataset:
Let $O$ denote the point $\tuple {0, 0}$. Let $A$ denote the point $\tuple {1, 0}$. The $x$-axis is identified with the straight line through $O$ and $A$. The $y$-axis is constructed as the line perpendicular to $OA$ through $O$. From Construction of Integer Multiple of Line Segment, the point $\tuple {a, 0}$ is constructed. Drawing a circle whose center is at $O$ and whose radius is $OA$ the point $A'$ is constructed on the $y$-axis where $OA' = OA$. Thus $A'$ is the point $\tuple {0, 1}$. From Construction of Integer Multiple of Line Segment, the point $\tuple {0, b}$ is constructed. Using Construction of Parallel Line, a straight line is drawn through $\tuple {a, 0}$ parallel to the $y$-axis. Using Construction of Parallel Line, a straight line is drawn through $\tuple {0, b}$ parallel to the $x$-axis. By definition of Cartesian plane, their intersection is at $\tuple {a, b}$, which is the required point $P$. {{qed}}
Construction of a regular heptagon is the equivalent of constructing the point $\tuple {\cos \dfrac {2 \pi} 7, \sin \dfrac {2 \pi} 7}$ from the points $\tuple {0, 0}$ and $\tuple {1, 0}$ Let $\epsilon = \map \exp {\dfrac {2 \pi} 7}$. Then $\epsilon$ is a root of $x^7 - 1$. We have: {{begin-eqn}} {{eqn | l = x^7 - 1       | r = \paren {x - 1} \paren {x^6 + x^5 + x^4 + x^3 + x^2 + x + 1}       | c =  }} {{eqn | ll= \leadsto       | l = \epsilon^6 + \epsilon^5 + \epsilon^4 + \epsilon^3 + \epsilon^2 + \epsilon + 1       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \epsilon^3 + \epsilon^2 + \epsilon + 1 + \epsilon^{-1} + \epsilon^{-2} + \epsilon^{-3}       | r = 0       | c =  }} {{end-eqn}} But we have: {{begin-eqn}} {{eqn | l = \epsilon       | r = \cos \dfrac {2 \pi} 7 + i \sin \dfrac {2 \pi} 7 }} {{eqn | l = \epsilon^{-1}       | r = \cos \dfrac {2 \pi} 7 - i \sin \dfrac {2 \pi} 7 }} {{eqn | ll= \leadsto       | l = \epsilon + \epsilon^{-1}       | r = 2 c       | c = where $c = \cos \dfrac {2 \pi} 7$ }} {{eqn | ll= \leadsto       | l = \epsilon^2 + \epsilon^{-2} + 2       | r = 4 c^2       | c = squaring }} {{eqn | ll= \leadsto       | l = \epsilon^3 + \epsilon^{-3} + 3 \paren {\epsilon + \epsilon^{-1} }       | r = 8 c^3       | c = cubing }} {{eqn | ll= \leadsto       | l = \epsilon^2 + \epsilon^{-2}       | r = 4 c^2 - 2       | c =  }} {{eqn | ll= \leadsto       | l = \epsilon^3 + \epsilon^{-3}       | r = 8 c^3 - 6 c       | c =  }} {{eqn | ll= \leadsto       | l = \paren {8 c^3 - 6 c} + \paren {4 c^2 - 2} + 2 c - 1       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = 8 c^3 + 4 c^2 - 4 c - 1       | r = 0       | c =  }} {{end-eqn}} Thus $2 c$ is a root of the polynomial $x^3 + x^2 - 2 x - 1$ But from Irreducible Polynomial: $x^3 + x^2 - 2 x - 1$ in Rationals: :$x^3 + x^2 - 2 x - 1$ is irreducible over $\Q$. Thus by Algebraic Element of Degree 3 is not Element of Field Extension of Degree Power of 2, $\cos \dfrac {2 \pi} 7$ is not an element of any extension of $\Q$ of degree $2^m$. The result follows from Point in Plane is Constructible iff Coordinates in Extension of Degree Power of 2. {{qed}}
From Rational Numbers form Integral Domain we have that $\struct {\Q, +, \times}$ is an integral domain. Hence to demonstrate that $\struct {S, +, \times}$ is an integral domain, we can use the Subdomain Test. We have that the unity of $\struct {\Q, +, \times}$ is $1$. Then we note: :$1 = \dfrac 1 1$ and: :$1 = 2^0$ and so $1 \in S$. Thus property $(2)$ of the Subdomain Test is fulfilled. It remains to demonstrate that $\struct {S, +, \times}$ is a subring of $\struct {\Q, +, \times}$, so fulfilling property $(2)$ of the Subdomain Test. Hence we use the Subring Test. We note that $S \ne \O$ as $1 \in S$. This fulfils property $(1)$ of the Subring Test. Let $x, y \in S$. Then: {{begin-eqn}} {{eqn | l = x + \paren {-y}       | r = \dfrac a {2^p} + \dfrac b {2^q}       | c = for some $a, b \in \Z$ and $p, q \in \Z_{\ge 0}$ }} {{eqn | r = \dfrac {a 2^q - b 2^p} {2^p 2^q}       | c = {{Defof|Rational Addition}} }} {{eqn | r = \dfrac {a 2^q - b 2^p} {2^{p + q} }       | c =  }} {{eqn | o = \in       | r = S       | c =  }} {{end-eqn}} This fulfils property $(2)$ of the Subring Test. Then:  {{begin-eqn}} {{eqn | l = x \times y       | r = \dfrac a {2^p} \times \dfrac b {2^q}       | c = for some $a, b \in \Z$ and $p, q \in \Z_{\ge 0}$ }} {{eqn | r = \dfrac {a b} {2^p 2^q}       | c = {{Defof|Rational Multiplication}} }} {{eqn | r = \dfrac {a b} {2^{p + q} }       | c =  }} {{eqn | o = \in       | r = S       | c =  }} {{end-eqn}} This fulfils property $(3)$ of the Subring Test. Hence the result. {{qed}}
Let $r$ be the number of Sylow $p$-subgroup of order $5^2 = 25$ The First Sylow Theorem guarantees existence, so $r \ge 1$. From the Fourth Sylow Theorem: :$r \equiv 1 \pmod 5$ That is: :$r \in \set {1, 6, 11, 16, \ldots}$ From the Fifth Sylow Theorem: :$r \divides \dfrac {100} {25} = 4$ from which it follows that $r = 1$. From Sylow p-Subgroup is Unique iff Normal, this unique Sylow $5$-subgroup must be normal. {{qed}}
Let $P$ be a Sylow $p$-subgroup of $G$. By Second Sylow Theorem, $H$ is a subset of a conjugate of $P$. Then: :$\exists g \in G: H \subseteq g P g^{-1}$. This implies: {{begin-eqn}} {{eqn | l = H       | r = g^{-1} H g       | c = Subgroup equals Conjugate iff Normal }} {{eqn | o = \subseteq       | r = g^{-1} \paren {g P g^{-1} } g       | c = Subset Relation is Compatible with Subset Product/Corollary 2 }} {{eqn | r = P }} {{end-eqn}} Since $P$ is arbitrary, $H$ is a subset of every Sylow $p$-subgroup of $G$. {{qed}}
The order $\order {G / H}$ of the quotient group $G / H$ is $\index G H$. Hence $p^k$ divides $\order {G / H}$. By Group has Subgroups of All Prime Power Factors, $G / H$ has a subgroup of order $p^k$. By Correspondence Theorem, this subgroup is in the form $K / H$ where $H \le K \le G$. Hence the result. {{qed}}
From Group of Order $p^2 q$ has Normal Sylow $p$-Subgroup, $G$ has a normal subgroup of order $p^2$. Hence the result, by definition of simple group. {{qed}}
First it is noted that Prime Group is Simple. We also note from Alternating Group is Simple that the alternating group $A_5$, which is of order $60$, is simple. Hence the motivation for the result. It remains to be shown that all groups of composite order such that $\order G < 60$ are not simple. Let $S$ be the set: :$S = \set {n \in \Z: 0 < n < 60: \text { there exists a simple group of order $n$ such that $n$ is composite} }$ The aim is to show that $S$ is empty. From Abelian Group is Simple iff Prime, all abelian groups of composite order are not simple. Thus any simple group must be non-abelian. Let $p$ and $q$ be prime. From Prime Power Group has Non-Trivial Proper Normal Subgroup, no group of order $p^n$ is simple. Thus: :$\forall n \in \Z_{>0}: p^n \notin S$ Thus, with the primes and prime powers eliminated, we have: :$S \subseteq {6, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 30, 33, 34, 35, 36, 38, 40, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58}$ From Group of Order $p q$ has Normal Sylow $p$-Subgroup: :$p q \notin S$ The set of non-square semiprimes less than $60$ is: :$\set {6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58}$ Thus we have so far: :$S \subseteq {12, 18, 20, 24, 28, 30, 36, 40, 42, 44, 45, 48, 50, 52, 54, 56}$ From Group of Order $p^2 q$ is not Simple: :$p^2 q \notin S$ This eliminates: {{begin-eqn}} {{eqn | l = 12       | r = 2^2 \times 3 }} {{eqn | l = 18       | r = 3^2 \times 2 }} {{eqn | l = 20       | r = 2^2 \times 5 }} {{eqn | l = 28       | r = 2^2 \times 7 }} {{eqn | l = 44       | r = 2^2 \times 11 }} {{eqn | l = 45       | r = 3^2 \times 5 }} {{eqn | l = 50       | r = 5^2 \times 2 }} {{eqn | l = 52       | r = 2^2 \times 13 }} {{end-eqn}} Thus we are left with: :$S \subseteq {24, 30, 36, 40, 42, 48, 54, 56}$ From Normal Subgroup of Group of Order 24, a group of order $24$ has a normal subgroup either of order $4$ or order $8$. Hence $24 \notin S$. From Group of Order 56 has Unique Sylow 2-Subgroup or Unique Sylow 7-Subgroup, a group of order $56$ has at least one normal subgroup. Hence $56 \notin S$. From Group of Order 30 is not Simple: :$30 \notin S$ Thus we are left with: :$S \subseteq {36, 40, 42, 48, 54}$ $40$ is eliminated by Group of Order 40 has Normal Subgroup of Order 5. $42$ is eliminated by Group of Order 42 has Normal Subgroup of Order 7. $54$ is eliminated by Group of Order 54 has Normal Subgroup of Order 27. Thus remains: :$S \subseteq {36, 48}$ {{finish}}
We have that: :$42 = 2 \times 3 \times 7$ From the First Sylow Theorem, $G$ has at least one Sylow $7$-subgroup, which is of order $7$. Let $n_7$ denote the number of Sylow $7$-subgroups of $G$. From the Fourth Sylow Theorem: :$n_7 \equiv 1 \pmod 7$ and from the Fifth Sylow Theorem: :$n_7 \divides 6$ where $\divides$ denotes divisibility. It follows that $n_7 = 1$. From Sylow $p$-Subgroup is Unique iff Normal, this Sylow $7$-subgroup is normal. {{qed}} Category:Groups of Order 42 qzh3u1kuc9cl8kk2l8oa9yookmzd6a5
We have that: :$54 = 2 \times 3^3$ From the First Sylow Theorem, $G$ has at least one Sylow $3$-subgroup, which is of order $3^3 = 27$. Let $n_3$ denote the number of Sylow $3$-subgroups of $G$. From the Fourth Sylow Theorem: :$n_3 \equiv 1 \pmod 3$ and from the Fifth Sylow Theorem: :$n_3 \divides 2$ where $\divides$ denotes divisibility. It follows that $n_3 = 1$. From Sylow $p$-Subgroup is Unique iff Normal, this Sylow $3$-subgroup is normal. {{qed}} Category:Groups of Order 54 49bi4fhqoq3ujnnbiungpqlspagce6y
We have that: :$40 = 2^3 \times 5$ From the First Sylow Theorem, $G$ has at least one Sylow $5$-subgroup, which is of order $5$. Let $n_5$ denote the number of Sylow $5$-subgroups of $G$. From the Fourth Sylow Theorem: :$n_5 \equiv 1 \pmod 5$ and from the Fifth Sylow Theorem: :$n_5 \divides 8$ where $\divides$ denotes divisibility. It follows that $n_5 = 1$. From Sylow $p$-Subgroup is Unique iff Normal, this Sylow $5$-subgroup is normal. {{qed}} Category:Groups of Order 40 djhna8lndx7fauabsnev82th3in9f1m
If each of the Sylow $p$-subgroups are unique, they are all normal. As the order of each one is coprime to each of the others, their intersection is $\set e$. {{finish|It remains to be shown that the direct product is what is is}}
{{ProofWanted|Similar to the First Sylow Theorem at first glance. The book suggests this proof is due to Frobenius, and is long.}}
For any $n \in \N$ then: {{begin-eqn}} {{eqn | l = \norm {\alpha_{n + 1} - \alpha_n }_p       | r = \norm {\paren {\alpha_{n + 1} - x} + \paren {x - \alpha_n } }_p }} {{eqn | o = \le       | r = \max \set {\norm {\alpha_{n + 1} - x}_p, \: \norm {x - \alpha_n }_p }       | c = {{NormAxiom|4}} }} {{eqn | o = \le       | r = \max \set {\norm {x - \alpha_{n + 1} }_p, \: \norm {x - \alpha_n }_p}       | c = Norm of negative }} {{eqn | o = \le       | r = \max \set {p^{-\paren{n + 2} } , p^{-\paren{n + 1} } } }} {{eqn | r = p^{-\paren{n + 1} }       | c = Since $p^{-n - 1} < p^{-n}$ }} {{end-eqn}} Hence: :$p^{n + 1} \divides \paren {\alpha_{n + 1} - \alpha_n} $ or equivalently: :$\alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1} }$ The result follows. {{qed}}
Suppose that there exists a sequence $\sequence {\alpha'_n}$ with: :$(1'): \quad \forall n \in \N: \alpha'_n \in \Z$ and $0 \le \alpha'_n \le p^{n + 1} - 1$ :$(2'): \quad \forall n \in \N: \alpha'_{n + 1} \equiv \alpha'_n \pmod {p^{n + 1} }$ :$(3'): \quad \displaystyle \lim_{n \mathop \to \infty} \alpha'_n = x$ {{AimForCont}}: :$\alpha'_N \ne \alpha_N$ for some $N \in \N$ By Initial Segment of Natural Numbers forms Complete Residue System: :$\alpha'_N \not \equiv \alpha_N \pmod {p^{N + 1}}$ Then for all $n > N$: {{begin-eqn}} {{eqn | l = \alpha'_n       | o = \equiv       | r = \alpha'_N \pmod {p^{N + 1} }       | c = by $(2)$ above }} {{eqn | o = \not \equiv       | r = \alpha_N \pmod {p^{N + 1} } }} {{eqn | o = \equiv       | r = \alpha_n \pmod {p^{N + 1} }       | c = by $(2)$ above }} {{end-eqn}} That is, for all $n > N$: :$\alpha'_n \not \equiv \alpha_n \pmod {p^{N + 1}}$ Hence for all $n > N$: :$\norm {\alpha'_n - \alpha_n}_p > p^{-\paren{N + 1}}$ By $(3)$ the limit of $\sequence{\alpha_n}$ is $x$: :$\exists N_1 \in \N: \forall n \ge N_1: \norm {x - \alpha_n}_p \le p^{-\paren{N + 1}}$ Similarly for $\sequence{\alpha'_n}$: :$\exists N_2 \in \N: \forall n \ge N_2: \norm {x - \alpha'_n}_p \le p^{-\paren{N + 1}}$ Let $M = \max \set {N+1, N_1, N_2}$. Then: :$\norm {\alpha'_M - \alpha_M}_p > p^{-\paren{N + 1}}$ :$\norm {x - \alpha_M} _p\le p^{-\paren{N + 1}}$ :$\norm {x - \alpha'_M}_p \le p^{-\paren{N + 1}}$ But: {{begin-eqn}} {{eqn | l = \norm {\alpha'_M - \alpha_M}_p       | r = \norm {\paren {\alpha'_M - x} + \paren {x - \alpha_M} }_p }} {{eqn | o = \le       | r = \max \set{\norm{\alpha'_M - x}_p, \: \norm {x - \alpha_M}_p}       | c = Norm axiom (N4) (Ultrametric Inequality) }} {{eqn | r = \max \set{\norm{x - \alpha'_M}_p, \: \norm {x - \alpha_M}_p}       | c = Norm of negative }} {{eqn | o = \le       | r = \max \set{p^{-\paren{N + 1} } , p^{-\paren{N + 1} } } }} {{eqn | r = p^{-\paren{N + 1} } }} {{end-eqn}} This contradicts the previous assertion that: :$\norm {\alpha'_M - \alpha_M}_p > p^{-\paren{N + 1}}$ Hence: :$\sequence{\alpha'_n} = \sequence{\alpha_n}$. The result follows. {{qed}}
From Sequence of Powers of Number less than One: :$\displaystyle \lim_{n \mathop \to \infty} p^{-n} = 0$ From Multiple Rule for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} p^{-\paren{n + 1}} = 0$ By the Squeeze Theorem for Real Sequences : :$\displaystyle \lim_{n \mathop \to \infty} \norm{x - \alpha_n}_p = 0$. Hence the limit of $\sequence {\alpha_n}$ is by definition: :$\displaystyle \lim_{n \mathop \to \infty} \alpha_n = x$ {{qed}}
From Group of Order Prime Squared is Abelian, $G$ is an abelian group. From Abelian Group of Prime-power Order is Product of Cyclic Groups, $G$ is either: :the cyclic group of order $p^2$ :the direct product of the cyclic group  of order $p$ with itself. The result follows from Finite Cyclic Group is Isomorphic to Integers under Modulo Addition. {{Qed}}
=== Necessary Condition === Let $S$ be everywhere dense in $X$. By Open Set Characterization of Denseness then every non-empty open set contains an element of $S$. Every non-empty set of an analytic basis is an open set by definition. Hence every non-empty set of non-empty open set of $\mathcal B$ contains an element of $S$. {{qed|lemma}} === Sufficient Condition === Suppose that every non-empty open set of $\mathcal B$ contains an element of $S$. Let $U$ be any non-empty open set. Let $x \in U$. By the definition of an analytic basis then: :$\exists B \in \mathcal B: x \in B \subseteq U$ By assumption $B$ contains an element $s$ of $S$. Hence $s \in U$. Since $U$ was an arbitrary non-empty open set then every non-empty open set contains an element of $S$ By Open Set Characterization of Denseness, $S$ is everywhere dense in $X$. {{qed}} Category:Denseness 9ml9xc4pcclv46blz9n2x6euv1yrene
By Open Balls form Basis for Open Sets of Metric Space, the set of open balls are an analytic basis for the topology $\tau_d$. By Analytic Basis Characterization of Denseness then: :$S$ is (everywhere) dense in $\struct{X, \tau_d}$ {{iff}} every open ball contains an element of $S$. {{qed}} Category:Denseness 5c156wij1viyrf4bkv3dd0c590r9jv0
The integers $\Z$ are a subring of the $p$-adic integers $Z_p$ by Integers form Subring of P-adic Integers. Hence $\Z \subseteq \Z_p$. The set of $p$-adic integers $\Z_p$ is closed in the $p$-adic metric by Set of P-adic Integers is Clopen in P-adic Numbers. By Closure of Subset of Closed Set of Metric Space is Subset then the closure of $\Z$ is contained in $\Z_p$: :$\map \cl \Z \subseteq \Z_p$ By P-adic Integer is Limit of Unique Coherent Sequence of Integers then: :for all $x \in \Z_p$ there exists a sequence $\sequence {x_n}$ in $\Z$ such that $\displaystyle \lim_{n \to \infty} x_n = x$ By Closure of Subset of Metric Space by Convergent Sequence then: :$\Z_p \subseteq \map \cl \Z$  Hence $\map \cl \Z = \Z_p$. By Metric Subspace Induces Subspace Topology then the topology induced by $d$ on $\Z_p$ is the subspace topology. By Closure in Subspace then $\map \cl \Z = \Z_p$ in the subspace metric $d$ on $\Z_p$. It follows that $\struct {\Z_p, d}$ is the metric completion of $\Z$. {{qed}}
The $p$-adic metric on $\Q_p$ is defined as the metric induced by the $p$-adic norm on $\Q_p$. It follows from Metric Induced by Norm is Metric that $d_p$ is a metric. By definition of the $p$-adic norm on $\Q_p$, $\norm{\,\cdot\,}_p$ is a non-Archimedean norm. From Non-Archimedean Norm iff Non-Archimedean Metric, then $d_p$ is a non-Archimedean metric. The $p$-adic norm $\norm{\,\cdot\,}_p$ on the $p$-adic numbers $\Q_p$ is defined to be an extension of the $p$-adic norm $\norm{\,\cdot\,}_p$ on $\Q$. By the definition of the metric induced by the $p$-adic norm, it follows that $d_p$ is an extension of the $p$-adic metric on the rationals $\Q$. {{qed}} Category:P-adic Metrics la84i9e4l0koi4phd8626sogono0596
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$. === $(1)$ implies $(2)$ === Let $x$ be an irreducible element of $\struct {D, +, \circ}$ by definition 1. By definition: :$x$ has no non-trivial factorization in $D$. Let $x = y \circ z$ for some $y, z \in D$. By definition, it cannot be the case that neither $y$ nor $z$ are units of $D$. So either $y$ or $z$ is a unit of $D$. {{WLOG}}, suppose $y$ is a unit of $D$. Then by definition $z$ is an associate of $x$. Contrariwise, suppose $z$ is a unit of $D$. Then by definition $y$ is an associate of $x$. Thus both $y$ and $z$ are either a unit of $D$ or an associate of $x$. $x = y \circ z$ is an arbitrary factorization of $x$ in $D$. It follows that the only divisors of $x$ are its associates and the units of $D$. Thus $x$ is an irreducible element of $\struct {D, +, \circ}$ by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $x$ be an irreducible element of $\struct {D, +, \circ}$ by definition 2. Then by definition: :the only divisors of $x$ are its associates and the units of $D$. Let $x = y \circ z$. Then either: :$y$ is an associate of $x$ and $z$ is a unit of $D$ or: :$z$ is an associate of $x$ and $y$ is a unit of $D$. {{explain|Have we ruled out the possibility of $x {{=}} y \circ z$ where both $y$ and $z$ are associates of $x$ but neither are units?}} In either case, $y \circ z$ is a trivial factorization of $x$. Thus $x$ is an irreducible element of $\struct {D, +, \circ}$ by definition 1. {{qed}} Category:Irreducible Elements of Rings qei69jydym7qnq0d0a8fox7am92d2z3
Let $g$ be a generator of $C_n$. Let $\varphi$ be an automorphism on $C_n$. By Homomorphic Image of Cyclic Group is Cyclic Group, $\map \varphi g$ is a generator of $C_n$. By Homomorphism of Generated Group, $\varphi$ is uniquely determined by $\map \varphi g$. By Finite Cyclic Group has Euler Phi Generators, there are $\map \phi n$ possible values for $\map \varphi g$. Therefore there are $\map \phi n$ automorphisms on $C_n$: :$\order {\Aut {C_n} } = \map \phi n$ {{qed}}
From Identity is in Kernel of Group Homomorphism we have that: :$e_G \in \map \ker \phi$ Hence the result. {{Qed}}
By Valuation Ring of P-adic Norm on Rationals, the induced valuation ring $\OO$ is the set: :$\OO = \Z_{\paren p} = \set {\dfrac a b \in \Q : p \nmid b}$ Since $p \nmid 1$ then for all $a \in \Z$, $a = \dfrac a 1 \in \OO$. Hence $\Z \subseteq \OO$. By Valuation Ring of Non-Archimedean Division Ring is Subring then $\OO$ is a subring of $\Q$. By Integers form Subdomain of Rationals then $\Z$ is a subring of $\Q$. By Intersection of Subrings is Largest Subring Contained in all Subrings then $\Z \cap \OO = \Z$ is a subring of $\OO$.  {{qed}}
The $p$-adic integers is defined as: :$\Z_p = \set {x \in \Q_p: \norm x_p \le 1}$ The induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$ is defined as: :$\Z_{\ideal p} = \set {x \in \Q: \norm x_p \le 1}$ By definition, the $p$-adic norm $\norm {\,\cdot\,}_p$ on $\Q_p$ is an extension of the $p$-adic norm $\norm {\,\cdot\,}_p$ on $\Q$. It follows that $\Z_{\ideal p} = \Q \cap \Z_p$. This proves $(1)$ above. By Valuation Ring of Non-Archimedean Division Ring is Subring then $\Z_p$ is a subring of $\Q_p$. By definition of $p$-adic integers then $\Q$ is a subring of $\Q_p$. By Intersection of Subrings is Largest Subring Contained in all Subrings then $\Z_{\paren p}$ is a subring of $\Z_p$. This proves $(2)$ above. {{qed}}
Let $\Z_{\paren p}$ be the valuation ring induced by $\norm {\,\cdot\,}_p$ on $\Q$. By Integers form Subring of Valuation Ring of P-adic Norm on Rationals then: :$\Z$ is a subring of $\Z_{\paren p}$ By Valuation Ring of P-adic Norm is Subring of P-adic Integers then: :$\Z_{\paren p}$ is a subring of $\Z_p$ The result follows. {{qed}}
Let $\map f t := \map \Ei t = \displaystyle \int_t^\infty \dfrac {e^{-u} } u \rd u$. Then: {{begin-eqn}} {{eqn | l = \map {f'} t       | r = -\dfrac {e^{-t} } t       | c =  }} {{eqn | ll= \leadsto       | l = t \map {f'} t       | r = -e^{-t}       | c =  }} {{eqn | ll= \leadsto       | l = \laptrans {t \map {f'} t}       | r = -\laptrans {e^{-t} }       | c =  }} {{eqn | r = -\dfrac 1 {s + 1}       | c = Laplace Transform of Exponential }} {{eqn | ll= \leadsto       | l = -\dfrac \d {\d s} \laptrans {\map {f'} t}       | r = -\dfrac 1 {s + 1}       | c = Derivative of Laplace Transform }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d s} } {s \laptrans {\map f t} - \map f 0}       | r = \dfrac 1 {s + 1}       | c = Laplace Transform of Derivative }} {{eqn | ll= \leadsto       | l = s \laptrans {\map f t}       | r = \int \dfrac 1 {s + 1} \rd s       | c = $\map f 0 = 0$, and integrating both sides {{WRT|Integration}} $s$ }} {{eqn | ll= \leadsto       | l = s \laptrans {\map f t}       | r = \map \ln {s + 1} + C       | c = Primitive of $\dfrac 1 {a x + b}$ }} {{end-eqn}} By the Initial Value Theorem of Laplace Transform: :$\displaystyle \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$ which leads to: :$c = 0$ Thus: {{begin-eqn}} {{eqn | l = s \laptrans {\map f t}       | r = \map \ln {s + 1}       | c =  }} {{eqn | ll= \leadsto       | l = \laptrans {\map f t}       | r = \dfrac {\map \ln {s + 1} } s       | c =  }} {{end-eqn}} {{qed}}