hcju/proofwikiqa huggingface dataset:
Let $\map S t = \map \Phi {t + T} {\map \Phi T}^{-1}$. Then: {{begin-eqn}} {{eqn | l = \map {\frac \d {\d t} } {\map S t}       | r = \map {\Phi'} {t + T} {\map \Phi T}^{-1}       | c =  }} {{eqn | r = \map {\mathbf A} {t + T} \map \Phi {t + T} {\map \Phi T}^{-1}       | c =  }} {{eqn | r = \map {\mathbf A} t \map S t       | c =  }} {{end-eqn}} So $\map S t$ is a fundamental matrix and: :$\map S 0 = Id$ Then: :$\map S t = \map \Phi t$ which means that: :$\map \Phi {t + T} = \map \Phi t \map \Phi T$ Hence by the existence of the matrix logarithm, there exists a matrix $\mathbf B$ such that: :$\map \Phi T = e^{\mathbf B T}$ Defining $\map {\mathbf P} t = \map \Phi t e^{-\mathbf B t}$, it follows that: {{begin-eqn}} {{eqn | l = \map {\mathbf P} {t + T}       | r = \map \Phi {t + T} e^{-\mathbf B t - \mathbf B T}       | c =  }} {{eqn | r = \map \Phi t \map \Phi T e^{-\mathbf B T} e^{-\mathbf B t}       | c =  }} {{eqn | r = \map \Phi t e^{-\mathbf B t}       | c =  }} {{eqn | r = \map {\mathbf P} t       | c =  }} {{end-eqn}} and hence $\map {\mathbf P} t$ is a periodic function with period $T$. As $\map \Phi t = \map {\mathbf P} t e^{\mathbf B t}$, the second implication also holds. {{qed}}
By definition, $\map \Phi t$ is non-singular, and therefore has an inverse $\map {\Phi^{-1} } t$. If $z$ is an arbitrary solution, then $\map \Phi t \, \map {\Phi^{-1} } {t_0} \, \map z {t_0}$ also solves the system and has the same initial condition. Hence by Existence and Uniqueness Theorem for 1st Order IVPs $\map \Phi t \, \map {\Phi^{-1} } {t_0} \, \map z {t_0}$ equals $z$. Letting $c = \map {\Phi^{-1} } {t_0} \, \map z {t_0}$ finishes the proof. {{Qed}} Category:Differential Equations 3r820ua0yknw20g5ljibmg4nxwv2ozf
$\map \Phi t C$ is a fundamental matrix as follows: :$\dfrac \d {\d t} \map \Phi t C = \map {\Phi'} t C = \map A t \, \map \Phi t C$ :$\map \det {\map \Phi t C} = \map \det {\map \Phi t} \, \map \det C \ne 0$ Let $\map \Psi t$ be an arbitrary fundamental matrix. Then from General Vector Solution of Fundamental Matrix $\map \Phi t \map {\Phi^{-1} } {t_0} \, \map \Psi {t_0}$ solves the same matrix equation and has the same initial conditions. Hence by uniqueness: :$\map \Phi t \, \map {\Phi^{-1} } {t_0} \, \map \Psi {t_0}$ is equal to $\map \Psi t$ Letting $C = \map {\Phi^{-1} } {t_0} \, \map \Psi {t_0}$ finishes the proof. {{qed}} Category:Differential Equations 34fd9ivjgm58qxodnj89k4eqhayi1v7
Let the circles $ABC$ and $ADE$ touch externally at $A$. Let $F$ be the center of $ABC$ and let $G$ be the center of $ADE$. We are to show that the straight line joining $F$ to $G$ passes through $A$. :400px Suppose, as in the diagram above, that it does not, and that it were possible for it to pass through $C$ and $D$, as $FCDG$. (It is clear that the diagram does ''not'' have $F$ and $G$ as the ''actual'' centers of these circles - it is the point of this proof to demonstrate that this would ''not'' be possible.) Join $AF$ and $AG$. We have that: * Since $F$ is the center of $ABC$, then $FA$ and $FC$ are both radii of $ABC$, and so $FA = FC$. * Since $G$ is the center of $ADE$, then $GA$ and $GD$ are both radii of $ADE$, and so $GA = GD$. So $FA + AG = FC + GD$. So all of $FCDG$ is greater than $FA + AG$. But from Sum of Two Sides of Triangle Greater than Third Side $FA + AG$ is greater than $FCDG$. Hence we have a contradiction, and so $FG$ has to go through point $A$, the point of contact of the two circles. Hence the result. {{Qed}} {{Euclid Note|12|III}}
=== Sufficient Condition === Suppose $\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$. Consider the subset $G \subseteq \Img f \times C$ defined by: :$G = \set {\tuple {y, z}: \exists x \in A: y = \map f x, z =\map g x}$ Clearly $G \ne \O$ because for any $x \in A$ we have $\tuple {\map f x, \map g x} \in G$. What we need to show is that $G$ is the graph of a mapping $t: \Img f \to C$. To do that, we need to show that for every $y \in \Img f$, there is a unique $z \in C$ such that $\tuple {y, z} \in G$. It is clear that there is at least one such $z$: choose any $x \in A$ such that $y = \map f x$ and set $z = \map g x$. Now we need to show that such a $z$ is unique. Suppose we have $\tuple {y, z} \in G$ and $\tuple {y, z'} \in G$. Then by the definition of $G$: :$\exists x, x' \in A: y = \map f x = \map f {x'}, z = \map g x, z' = \map g {x'}$ We have taken as a hypothesis that: :$\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$ So $\map g x = \map g {x'}$ and so $z = z'$. So $G$ is the graph of a mapping which we can denote: :$t: \Img f \to C$ Also, since: :$\forall x \in A: \tuple {\map f x, \map g x} \in G$ it follows that: :$\forall x \in A: \map f x = \map t {\map f x}$ Since $C$ is non-empty, it has an element $c$. By Law of Excluded Middle, we can now construct a mapping $h$ as follows: :$\map h x = \begin{cases}    \map t x  & : x \in \Img f \\    c         & : x \in B \setminus \Img f \end{cases}$ So: :$\forall x \in A: \map {\paren {h \circ f} } x = \map h {\map f x} = \map t {\map f x} = \map g x$ Thus we have constructed a mapping $h$ such that $h \circ f = g$, as required. {{qed|lemma}} === Necessary Condition === Suppose there exists some mapping $h: B \to C$ such that $h \circ f = g$. Let $\map f x = \map f y$. Then we have: {{begin-eqn}} {{eqn | l = \map f x       | r = \map f y       | c =  }} {{eqn | ll= \leadsto       | l = \map h {\map f x}       | r = \map h {\map f y}       | c = {{Defof|Mapping}} }} {{eqn | ll= \leadsto       | l = \map {h \circ f} x       | r = \map {h \circ f} y       | c = {{Defof|Composition of Mappings}} }} {{eqn | ll= \leadsto       | l = \map g x       | r = \map g y       | c = Definition of $g$ }} {{end-eqn}} {{qed}} {{LEM}}
=== Sufficient Condition === Suppose $\Img g \subseteq \Img f$. That is: :$\forall x \in C: \map g x \in \Img f$ and so: :$\forall x \in C: \exists y \in B: \map g x = \map f y$ Take any $x \in C$. Consider the set $Y_x = \set {y \in B: \map g x = \map f y}$. We know from above that $Y_x \ne \O$. So, using the Axiom of Choice, for each $x$ we may select some $y_x \in Y_x$. Then we may define the mapping $h: C \to B$ such that: :$\forall x \in C: \map h x = y_x$ We then see that: {{begin-eqn}} {{eqn | l = \forall x \in C: \map {\paren {f \circ h} } x       | r = \map f {\map h x}       | c =  }} {{eqn | r = \map f {y_x}       | c =  }} {{eqn | r = \map g x       | c =  }} {{end-eqn}} Thus we have constructed a mapping $h$ such that $f \circ h = g$, as required. === Necessary Condition === Suppose there exists some mapping $h: C \to B$ such that $f \circ h = g$. Let $y \in \Img g$. Then we have: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \Img g       | c = }} {{eqn | ll= \leadsto       | l = \exists x \in C: \map g x       | r = y       | c = {{Defof|Image of Mapping}} }} {{eqn | ll= \leadsto       | l = \map {\paren {f \circ h} } x       | r = y       | c = Definition of $g$ }} {{eqn | ll= \leadsto       | l = \map f {\map h x}       | r = y       | c = {{Defof|Composition of Mappings}} }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = \Img f       | c = {{Defof|Image of Mapping}} }} {{end-eqn}} Hence by definition of subset, $\Img g \subseteq \Img f$. {{qed}}
Let $A \subseteq B$ and $S \subseteq T$. Then: {{begin-eqn}} {{eqn | l = x \in A       | o = \leadsto       | r = x \in B       | c = {{Defof|Subset}} }} {{eqn | l = x \in S       | o = \leadsto       | r = x \in T       | c = {{Defof|Subset}} }} {{end-eqn}} Now we invoke the Constructive Dilemma of propositional logic: :$p \implies q, \ r \implies s \vdash p \lor r \implies q \lor s$ applying it as: :$\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \implies \paren {x \in A \lor x \in S \implies x \in B \lor x \in T}$ The result follows directly from the definition of set union: :$\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \implies \paren {x \in A \cup S \implies x \in B \cup T}$ and from the definition of subset: :$A \subseteq B, \ S \subseteq T \implies A \cup S \subseteq B \cup T$ {{qed}}
By Subset Relation is Transitive, $\subseteq$ is a transitive relation. By the corollary to Set Union Preserves Subsets (Proof 2), $\subseteq$ is compatible with $\cup$. Thus the theorem holds by Operating on Transitive Relationships Compatible with Operation. {{qed}}
{{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = T       | c =  }} {{eqn | ll= \leadstoandfrom       | l = S \cap T       | r = S       | c = Intersection with Subset is Subset }} {{eqn | ll= \leadstoandfrom       | l = \map \complement {S \cap T}       | r = \map \complement S       | c = Complement of Complement }} {{eqn | ll= \leadstoandfrom       | l = \map \complement S \cup \map \complement T       | r = \map \complement S       | c = De Morgan's Laws: Complement of Intersection }} {{eqn | ll= \leadstoandfrom       | l = \map \complement T       | o = \subseteq       | r = \map \complement S       | c = Union with Superset is Superset }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = T       | c =  }} {{eqn | ll= \leadstoandfrom       | l = (x \in S        | o = \implies       | r = x \in T)       | c = {{Defof|Subset}} }} {{eqn | ll= \leadstoandfrom       | l = (x \notin T       | o = \implies       | r = x \notin S)       | c = Rule of Transposition }} {{eqn | ll= \leadstoandfrom       | l = (x \in \map \complement T       | o = \implies       | r = x \in \map \complement S)       | c = {{Defof|Set Complement}} }} {{eqn | l = \map \complement T       | o = \subseteq       | r = \map \complement S       | c = {{Defof|Subset}} }} {{end-eqn}} {{qed}}
By definition of set complement: :$\map \complement T := \relcomp {\mathbb U} T$ where: :$\mathbb U$ is the universe :$\relcomp {\mathbb U} T$ denotes the complement of $T$ relative to $\mathbb U$. Thus the statement can be expressed as: :$S \subseteq T \iff \relcomp {\mathbb U} T \subseteq \relcomp {\mathbb U} S$ The result then follows from Relative Complement inverts Subsets. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = S \setminus T = \O }} {{eqn | o = \leadstoandfrom       | r = \neg \paren {\exists x: x \in S \land x \notin T}       | c = {{Defof|Empty Set}} }} {{eqn | o = \leadstoandfrom       | r = \forall x: \neg \paren {x \in S \land x \notin T}       | c = De Morgan's Laws (Predicate Logic) }} {{eqn | o = \leadstoandfrom       | r = \forall x: x \notin S \lor x \in T       | c = De Morgan's Laws: Disjunction of Negations }} {{eqn | o = \leadstoandfrom       | r = \forall x: x \in S \implies x \in T       | c = Rule of Material Implication }} {{eqn | o = \leadstoandfrom       | r = S \subseteq T       | c = {{Defof|Subset}} }} {{end-eqn}} {{qed}}
Let $S \cap T = S$. Then by the definition of set equality, $S \subseteq S \cap T$. Thus: {{begin-eqn}}      {{eqn | l = S \cap T       | o = \subseteq       | r = T       | c = Intersection is Subset }} {{eqn | ll= \leadsto       | l = S       | o = \subseteq       | r = T       | c = Subset Relation is Transitive }} {{end-eqn}} Now let $S \subseteq T$. From Intersection is Subset we have $S \supseteq S \cap T$. We also have: {{begin-eqn}}      {{eqn | l = S       | o = \subseteq       | r = T       | c =  }} {{eqn | ll= \leadsto       | l = S \cap S       | o = \subseteq       | r = T \cap S       | c = Set Intersection Preserves Subsets }} {{eqn | ll= \leadsto       | l = S       | o = \subseteq       | r = S \cap T       | c = Intersection is Idempotent and Intersection is Commutative }} {{end-eqn}} So as we have: {{begin-eqn}}      {{eqn | l = S \cap T       | o = \subseteq       | r = S }} {{eqn | l = S \cap T       | o = \supseteq       | r = S }} {{end-eqn}} it follows from the definition of set equality that: :$S \cap T = S$ So we have: {{begin-eqn}}      {{eqn | l = S \cap T = S       | o = \implies       | r = S \subseteq T }} {{eqn | l = S \subseteq T       | o = \implies       | r = S \cap T = S }} {{end-eqn}} and so: :$S \subseteq T \iff S \cap T = S$ from the definition of equivalence. {{Qed}}
Let $S \cup T = T$. Then by definition of set equality: :$S \cup T \subseteq T$ Thus: {{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = S \cup T       | c = Subset of Union }} {{eqn | ll= \leadsto       | l = S       | o = \subseteq       | r = T       | c = Subset Relation is Transitive }} {{end-eqn}} Now let $S \subseteq T$. From Subset of Union, we have $S \cup T \supseteq T$. We also have: {{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = T       | c = }} {{eqn | ll= \leadsto       | l = S \cup T       | o = \subseteq       | r = T \cup T       | c = Set Union Preserves Subsets }} {{eqn | ll= \leadsto       | l = S \cup T       | o = \subseteq       | r = T       | c = Union is Idempotent }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = S \cup T       | o = \subseteq       | r = T       | c =  }} {{eqn | l = S \cup T       | o = \supseteq       | r = T       | c =  }} {{end-eqn}} By definition of set equality: :$S \cup T = T$ So: {{begin-eqn}} {{eqn | l = S \cup T = T       | o = \implies       | r = S \subseteq T       | c =  }} {{eqn | l = S \subseteq T       | o = \implies       | r = S \cup T = T       | c =  }} {{end-eqn}} and so: :$S \subseteq T \iff S \cup T = T$ from the definition of equivalence. {{Qed}}
{{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = T }} {{eqn | ll= \leadstoandfrom       | l = S \setminus T       | r = \O       | c = Set Difference with Superset is Empty Set }} {{eqn | ll= \leadstoandfrom       | l = S \cap \map \complement T       | r = \O       | c = Set Difference as Intersection with Complement }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = T }} {{eqn | ll= \leadstoandfrom       | l = S \cap \map \complement T       | r = \O       | c = Intersection with Complement is Empty iff Subset }} {{eqn | ll= \leadstoandfrom       | l = \map \complement {S \cap \map \complement T}       | r = \mathbb U       | c = Complement of Empty Set is Universe }} {{eqn | ll= \leadstoandfrom       | l = \map \complement S \cup \map \complement {\map \complement T}       | r = \mathbb U       | c = De Morgan's Laws: Complement of Intersection }} {{eqn | ll= \leadstoandfrom       | l = \map \complement S \cup T       | r = \mathbb U       | c = Complement of Complement }} {{end-eqn}} {{qed}}
We assume a universe $\mathbb U$ such that $R, S, T \subseteq \mathbb U$. We have the identity Set Difference as Intersection with Complement: :$R \setminus S = R \cap \overline S$ where $\overline S$ is the set complement of $S$: :$\overline S = \relcomp {\Bbb U} S$ Thus we can represent the two expressions as follows: {{begin-eqn}} {{eqn | l = \paren {R \setminus S} \setminus T       | r = \paren {R \cap \overline S} \cap \overline T       | c = Set Difference as Intersection with Complement }} {{eqn | r = R \cap \overline S \cap \overline T       | c = Intersection is Associative }} {{end-eqn}} For the second: {{begin-eqn}} {{eqn | o =        | r = R \setminus \paren {S \setminus T}       | c =  }} {{eqn | r = R \cap \overline {\paren {S \cap \overline T} }       | c = Set Difference as Intersection with Complement }} {{eqn | r = R \cap \paren {\overline S \cup \overline {\paren {\overline T} } }       | c = De Morgan's Laws: Complement of Intersection }} {{eqn | r = R \cap \paren {\overline S \cup T}       | c = Complement of Complement }} {{eqn | r = \paren {R \cap \overline S} \cup \paren {R \cap T}       | c = Intersection Distributes over Union }} {{eqn | r = \paren {R \cap \overline S \cap \mathbb U} \cup \paren {R \cap \mathbb U \cap T}       | c = Intersection with Universe }} {{eqn | r = \paren {R \cap \overline S \cap \paren {T \cup \overline T} } \cup \paren {R \cap \paren {S \cup \overline S} \cap T}       | c = Union with Complement }} {{eqn | r = \paren {R \cap \overline S \cap T} \cup \paren {R \cap \overline S \cap \overline T}       | c =  }} {{eqn | o =        | lo= \cup       | r = \paren {R \cap S \cap T} \cup \paren {R \cap \overline S \cap T}       | c = Intersection Distributes over Union }} {{eqn | r = \paren {R \cap \overline S \cap T} \cup \paren {R \cap \overline S \cap \overline T} \cup \paren {R \cap S \cap T}       | c = Union is Idempotent }} {{eqn | r = \paren {R \cap \overline S \cap \overline T} \cup \paren {R \cap \paren {S \cup \overline S} \cap T}       | c = Intersection Distributes over Union }} {{eqn | r = \paren {R \cap \overline S \cap \overline T} \cup \paren {R \cap \mathbb U \cap T}       | c = Union with Complement }} {{eqn | r = \paren {R \cap \overline S \cap \overline T} \cup \paren {R \cap T}       | c = Intersection with Universe }} {{end-eqn}} As can be seen, this full expansion of the second expression can be expressed: :$R \setminus \paren {S \setminus T} = \paren {\paren {R \setminus S} \setminus T} \cup \paren {R \cap T}$ It directly follows from Intersection with Empty Set that: :$R \setminus \paren {S \setminus T} = \paren {R \setminus S} \setminus T \iff R \cap T = \O$ {{qed}}
As $\struct {S, \circ}$ is a semigroup: : it is closed : $\circ$ is associative From Parenthesization of Word of $4$ Elements, there are exactly $5$ different ways of inserting brackets in the expression $a \circ b \circ c \circ d$. As $\circ$ is associative, we have that: :$\forall s_1, s_2, s_3 \in S: \paren {s_1 \circ s_2} \circ s_3 = s_1 \circ \paren {s_2 \circ s_3}$ As $\struct {S, \circ}$ is closed, we know that all products of elements from $\set {a, b, c, d}$ are in $S$, and are likewise bound by the associativity of $S$. So: {{begin-eqn}} {{eqn | l = \paren {\paren {a \circ b} \circ c} \circ d       | r = \paren {a \circ \paren {b \circ c} } \circ d       | c =  }} {{eqn | r = a \circ \paren {\paren {b \circ c} \circ d}       | c =  }} {{eqn | r = a \circ \paren {b \circ \paren {c \circ d} }       | c =  }} {{eqn | r = \paren {a \circ b} \circ \paren {c \circ d}       | c =  }} {{end-eqn}} {{qed}}
An group isomorphism is by definition a group epimorphism. The result follows from Epimorphism Preserves Identity. {{Qed}}
{{begin-eqn}} {{eqn | l = \map \phi {e_G}       | r = \map \phi {e_G \circ e_G}       | c = {{Defof|Identity Element}} }} {{eqn | r = \map \phi {e_G} * \map \phi {e_G}       | c = {{Defof|Group Isomorphism}} }} {{eqn | r = \map \phi {e_G} * \map \phi {e_G}       | c = {{Defof|Group Isomorphism}} }} {{end-eqn}} It follows from Identity is only Idempotent Element in Group that $\map \phi {e_G}$ is the identity of $H$. That is: :$\map \phi {e_G} = e_H$ {{qed}}
From Strictly Positive Rational Numbers under Multiplication form Subgroup of Non-Zero Rational Numbers we have that $\struct {\Q_{> 0}, \times}$ is a subgroup of $\struct {\Q_{\ne 0}, \times}$, where $\Q_{\ne 0}$ is the set of rational numbers without zero: $\Q_{\ne 0} = \Q \setminus \set 0$. From Subgroup of Abelian Group is Abelian it follows that $\struct {\Q_{> 0}, \times}$ is an abelian group. From Positive Rational Numbers are Countably Infinite, it follows that $\struct {\Q_{> 0}, \times}$ is a countably infinite group. {{qed}}
=== Necessary Condition === Suppose $\phi$ is a constant mapping. Then: : $\forall x, y \in S: \phi \left({x}\right) = \phi \left({y}\right)$ So: : $\forall x, y \in S: \phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$ : $\forall x, y \in S: \phi \left({y}\right) \mathop{\preceq_2} \phi \left({x}\right)$ and so $\phi$ is both increasing and decreasing. {{qed|lemma}} === Sufficient Condition === Suppose $\phi$ is both increasing and decreasing. Let $x, y \in S$. Then: : $\phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$ : $\phi \left({y}\right) \mathop{\preceq_2} \phi \left({x}\right)$ As $\preceq$ is an ordering, by definition $\preceq$ is antisymmetric. This means thtat: : $\phi \left({y}\right) = \phi \left({x}\right)$ As this holds for any $x, y \in S$ it follows for ''all'' $x, y \in S$ by Universal Generalisation. Thus: : $\forall x, y \in S: \phi \left({x}\right) = \phi \left({y}\right)$ and so $\phi$ is a constant mapping. {{qed}}
See Equivalence of Definitions of Real Exponential Function: Inverse of Natural Logarithm implies Limit of Sequence for how $\displaystyle \lim_{n \mathop \to \infty} \left({1 + \frac 1 n}\right)^n = e$ follows from the definition of $e$ as the number satisfied by $\ln e = 1$. See Euler's Number: Limit of Sequence implies Limit of Series for how $\displaystyle e = \sum_{n \mathop = 0}^\infty \frac 1 {n!}$ follows from $\displaystyle \lim_{n \mathop \to \infty} \left({1 + \frac 1 n}\right)^n = e$. Now suppose $e$ is defined as $\displaystyle e = \sum_{n \mathop = 0}^\infty \frac 1 {n!}$. Let us consider the series $\displaystyle f \left({x}\right) = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$. From Series of Power over Factorial Converges, this is convergent for all $x$. We differentiate $f \left({x}\right)$ {{WRT|Differentiation}} $x$ term by term (justified by Power Series is Differentiable on Interval of Convergence), and get: {{begin-eqn}} {{eqn | l = D_x \left({f \left({x}\right)}\right)       | r = D_x \left({1}\right) + D_x \left({\frac {x} {1!} }\right) + D_x \left({\frac {x^2} {2!} }\right) + D_x \left({\frac {x^3} {3!} }\right) + \cdots + D_x \left({\frac {x^n} {n!} }\right) + D_x \left({\frac {x^{n+1} } {\left({n+1}\right)!} }\right) + \cdots       | c =  }} {{eqn | r = 0 + 1 + \frac {2 x} {2!} + \frac {3 x^2} {3!} + \cdots + \frac {n x^{n-1} } {n!} + \frac {\left({n+1}\right) x^n} {\left({n+1}\right)!} + \cdots       | c =  }} {{eqn | r = 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n-1} } {\left({n-1}\right)!} + \frac {x^n} {n!} \cdots       | c =  }} {{eqn | r = f \left({x}\right)       | c =  }} {{end-eqn}} Thus we have: : $D_x \left({f \left({x}\right)}\right) = f \left({x}\right)$ From Derivative of Exponential Function: : $f \left({x}\right) = e^x$ From Derivative of Inverse Function: : $D_x \left({f^{-1} \left({x}\right)}\right) = \dfrac 1 {f^{-1} \left({x}\right)}$ Hence from Derivative of Natural Logarithm Function: : $f^{-1} \left({x}\right) = \ln x$ It follows that $e$ can be defined as that number such that $\ln e = 1$. Hence all the definitions of $e$ as given here are equivalent. {{qed}}
=== 1 implies 2 === See Euler's Number: Limit of Sequence implies Limit of Series. {{qed|lemma}} === 2 implies 3 === See Euler's Number: Limit of Sequence implies Base of Logarithm. {{qed|lemma}} === 3 implies 4 === Let $e$ be the unique solution to the equation $\ln x = 1$. We want to show that $\exp 1 = e$, where $\exp$ is the  exponential function. {{begin-eqn}} {{eqn | l = \exp 1 = 0       | o = \leadstoandfrom       | r = \map \ln {\exp 1} = \ln e       | c =  Logarithm is Injective }} {{eqn | o = \leadstoandfrom       | r = 1 = \ln e       | c = Exponential is Inverse of Logarithm and Inverse of Inverse }} {{end-eqn}} where the final equation holds by hypothesis. Hence the result. {{qed|lemma}} === 4 implies 1 === Let $e = \exp 1$, where $\exp$ denotes the exponential function. We want to show that: :$\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {n!} = e$ By definition of $\exp$: :$\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {n!} = \exp 1$ And $\exp 1 = e$ by hypothesis. Hence the result. {{qed}}
From Upper Bound of Natural Logarithm: When $x > 1$: : $\forall s \in \R: s > 0: \ln x \le \dfrac {x^s} s$ Given that $r > 0$, we can plug $s = \dfrac r 2$ in: {{begin-eqn}} {{eqn | l = x^{-r} \ln x       | r = x^{-r/2} \left({x^{-s} \ln x}\right)       | c =  }} {{eqn | o = \le       | r = \frac {x^{-r/2} } s       | c =  }} {{eqn | r = s \frac 1 {x^{r/2} }       | c =  }} {{end-eqn}} From Sequence of Powers of Reciprocals is Null Sequence: :$\displaystyle \lim_{x \mathop \to \infty} x^{-r} \frac 1 {x^{r/2}} = 0$ and so: :$\displaystyle \lim_{x \mathop \to \infty} x^{-r} \ln x = 0$ by the Squeeze Theorem for Real Sequences. {{qed}}