hcju/proofwikiqa huggingface dataset:
{{AimForCont}} the contrary. If the prime reciprocal series converges then there must exist some $k \in \N$ such that: :$\displaystyle \sum_{n \mathop = k \mathop + 1}^{\infty} \frac 1 {p_n} < \frac 1 2$ Let: :$\displaystyle Q = \prod_{i \mathop = 1}^k {p_i}$ and: :$\displaystyle \map S r = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q}$ Let $\map S {r, j}$ be the sum of all of the terms from $\map S r$ for which $1 + i Q$ has exactly $j$ prime factors. Notice that $1 + i Q$ is coprime with every prime factor in $Q$. Thus every prime factor of $1 + i Q$ where $i = 1, \ldots, r$ falls into some finite sequence of consecutive primes: :$\map P r = \set {p_{k + 1}, p_{k + 2}, \ldots, p_{\map m r} }$ {{explain|What exactly is $\map m r$?}} Notice again that each term of $\map S {r, j}$ occurs at least once in the expansion of: :$\displaystyle \paren {\sum_{n \mathop = k \mathop + 1}^{\map m r} \frac 1 {p_n} }^j < \paren {\sum_{n \mathop = k \mathop + 1}^\infty \frac 1 {p_n} }^j < \paren {\frac 1 2}^j$ and also by Sum of Infinite Geometric Sequence: :$\displaystyle \map S r = \sum_{j \mathop = 1}^r \map S {r, j} < \sum_{j \mathop = 1}^r \paren {\frac 1 2}^j < 1$ for every $r$. Finally notice that: :$\displaystyle \map S r = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q} > \frac 1 {1 + Q} \sum_{n \mathop = 1}^r \frac 1 n$ which implies that $\map S r$ diverges towards $+\infty$ by Harmonic Series is Divergent, a contradiction. {{qed}}
=== Necessary Condition === Let $C$ be closed and bounded in $\R$. Then, by Closed Bounded Subset of Real Numbers is Compact, $C$ is compact. {{qed|lemma}} === Sufficient Condition === Let $C$ be compact in $\R$. Then, by Compact Subspace of Real Numbers is Closed and Bounded, $C$ is closed and bounded in $\R$. {{qed}}
Let the domain of $f$ be $D$. We have that Measurable Sets form Algebra of Sets. First we note that, from Properties of Algebras of Sets, the difference of two measurable sets is measurable. So: : $\left\{{x: f \left({x}\right) \le \alpha}\right\} = D - \left\{{x: f \left({x}\right) > \alpha}\right\}$ and so $(1) \iff (4)$. Similarly, $(2) \iff (3)$. Next we note that, also from Properties of Algebras of Sets, the intersection of a sequence of measurable sets is measurable. Now: :$\displaystyle \left\{{x: f \left({x}\right) \ge \alpha}\right\} = \bigcap_{n=1}^\infty \left\{{x: f \left({x}\right) > \alpha -\dfrac 1 n}\right\}$ because if: :$x \in \left\{{x: f \left({x}\right) \ge \alpha}\right\}$ that is: :$f \left({x}\right) \ge \alpha$ and since: :$\forall n \in \N: n > 0: \alpha > \alpha - \dfrac 1 n$ then: :$\forall n \in \N: n > 0: f \left({x}\right) > \alpha - \dfrac 1 n$ That is: :$\forall n \in \N: n > 0: x \in  \left\{{x: f \left({x}\right) > \alpha - \dfrac 1 n}\right\}$ Hence: :$\displaystyle x \in \bigcap_{n \mathop = 1}^\infty \left\{{x: f \left({x}\right) > \alpha - \dfrac 1 n}\right\}$ Conversely, suppose: :$\displaystyle x \in \bigcap_{n \mathop = 1}^\infty \left\{{x: f \left({x}\right) > \alpha - \dfrac 1 n}\right\}$ that is: :$\forall n \in \N: n > 0: x \in  \left\{{x: f \left({x}\right) > \alpha - \dfrac 1 n}\right\}$ Claim $f \left({x}\right) \ge \alpha $. Otherwise $f \left({x}\right) < \alpha$, say for example $f \left({x}\right) = \alpha - \left\vert{\epsilon}\right\vert$. Choose $N = \left\lceil{\dfrac 1 {\left\vert{\epsilon}\right\vert} }\right\rceil + 1 \in \N$. Therefore: :$N > \left\lceil{\dfrac 1 {\left\vert{\epsilon}\right\vert} }\right\rceil \ge \dfrac 1 {\left\vert{\epsilon}\right\vert}$ and so: :$\alpha - \dfrac 1 N > \alpha - \left\vert{\epsilon}\right\vert$ By hypothesis: :$\forall N \in \N: f \left({x}\right) > \alpha - \dfrac 1 N$ and therefore by the just previous: :$f \left({x}\right) > \alpha - \left\vert{\epsilon}\right\vert$ But we had $f \left({x}\right) = \alpha - \left\vert{\epsilon}\right\vert$, a contradiction. Therefore: :$f \left({x}\right) \ge \alpha $ that is: $x \in \left\{{x: f \left({x}\right) \ge \alpha}\right\}$ which was to be shown. So $(1) \implies (2)$. Similarly: :$\displaystyle \left\{{x: f \left({x}\right) > \alpha}\right\} = \bigcup_{n \mathop = 1}^\infty \left\{{x: f \left({x}\right) \ge \alpha + \dfrac 1 n}\right\}$ and so $(2) \implies (1)$. This shows that $(1) \iff (2) \iff (3) \iff (4)$. For the fifth statement, we have: : $\left\{{x: f \left({x}\right) = \alpha}\right\} = \left\{{x: f \left({x}\right) \ge \alpha}\right\} \cap \left\{{x: f \left({x}\right) \le \alpha}\right\}$ and so $(3) \land (4) \implies (5)$ for $\alpha \in \R$. Since: :$\displaystyle \left\{{x: f \left({x}\right) = +\infty }\right\} = \bigcap_{n \mathop = 1}^\infty \left\{{x: f \left({x}\right) \ge n}\right\}$ we have that $(2) \implies (5)$ for $\alpha = +\infty$. Similarly $(4) \implies (5)$ for $\alpha = - \infty$. {{qed}} Category:Measure Theory 4x9u1du0b8e1x5fwhum72uy3los1pwm
Since every step function is also a simple function, we have :$\displaystyle \map L P \le \sup_{\phi \mathop \le f} \int_a^b \map \phi x \rd x \le \inf_{\psi \mathop \ge f} \int_a^b \map \psi x \rd x \le \map U P$ where $\map L P$ and $\map U P$ are the lower sum and upper sum as defined in the definition of definite integral. Since $f$ is Darboux integrable, the inequalities are all equalities and $f$ is measurable by basic properties of measurable functions. {{questionable|all reasoning is behind red links, and it's incomplete}} {{ProofWanted}} {{qed}}
The Euclidean metric on $\R^n$ is a special case of the $p$-product metric. The result follows from $p$-Product Metric on Real Vector Space is Metric. {{qed}}
Consider the Euclidean space $M = \struct {\R^n, d_2}$ where $d_2$ is the distance function given by: :$\displaystyle \map {d_2} {x, y} = \paren {\sum_{i \mathop = 1}^n \paren {x_i - y_i}^2}^{\frac 1 2}$ where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$. === Proof of $\text M 1$ === {{begin-eqn}} {{eqn | l = \map {d_2} {x, x}       | r = \paren {\sum_{i \mathop = 1}^n \paren {x_i - x_i}^2}^{\frac 1 2}       | c = Definition of $d_2$ }} {{eqn | r = \paren {\sum_{i \mathop = 1}^n 0^2}^{\frac 1 2}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $\text M 1$ holds for $d_2$. {{qed|lemma}} === Proof of $\text M 2$ === It is required to be shown: :$\map {d_2} {x, y} + \map {d_2} {y, z} \ge \map {d_2} {x, z}$ for all $x, y, z \in \R^n$. Let: :$(1): \quad z = \tuple {z_1, z_2, \ldots, z_n}$ :$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$ :$(3): \quad x_i - y_i = r_i$ :$(4): \quad y_i - z_i = s_i$. Thus we need to show that: :$\displaystyle \paren {\sum \paren {x_i - y_i}^2}^{\frac 1 2} + \paren {\sum \paren {y_i - z_i}^2}^{\frac 1 2} \ge \paren {\sum \paren {x_i - z_i}^2}^{\frac 1 2}$ We have: {{begin-eqn}} {{eqn | l = \map {d_2} {x, y} + \map {d_2} {y, z}       | r = \paren {\sum \paren {x_i - y_i}^2}^{\frac 1 2} + \paren {\sum \paren {y_i - z_i}^2}^{\frac 1 2}       | c = Definition of $d_2$ }} {{eqn | r = \paren {\sum r_i^2}^{\frac 1 2} + \paren {\sum s_i^2}^{\frac 1 2}       | c =  }} {{eqn | o = \ge       | r = \paren {\sum \paren {r_i + s_i}^2}^{\frac 1 2}       | c = Minkowski's Inequality for Sums: index $2$ }} {{eqn | r = \paren {\sum \paren {x_i - y_i + y_i - z_i}^2}^{\frac 1 2}       | c = Definition of $r_i$ and $s_i$ }} {{eqn | r = \paren {\sum \paren {x_i - z_i}^2}^{\frac 1 2}       | c =  }} {{eqn | r = \map {d_2} {x, z}       | c = Definition of $d_2$ }} {{end-eqn}} So axiom $\text M 2$ holds for $d_2$. {{qed|lemma}} === Proof of $\text M 3$ === {{begin-eqn}} {{eqn | l = \map {d_2} {x, y}       | r = \paren {\sum \paren {x_i - y_i}^2}^{\frac 1 2}       | c = Definition of $d_2$ }} {{eqn | r = \paren {\sum \paren {y_i - x_i}^2}^{\frac 1 2}       | c =  }} {{eqn | r = \map {d_2} {y, x}       | c = Definition of $d_2$ }} {{end-eqn}} So axiom $\text M 3$ holds for $d_2$. {{qed|lemma}} === Proof of $\text M 4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = \exists k \in \closedint 1 n: x_k       | o = \ne       | r = y_k       | c =  }} {{eqn | ll= \leadsto       | l = x_k - y_k       | o = \ne       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\sum \paren {x_k - y_k}^2}^{\frac 1 2}       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \map {d_2} {x, y}       | o = >       | r = 0       | c = Definition of $d_2$ }} {{end-eqn}} So axiom $\text M 4$ holds for $d_2$. {{qed}}
We examine each of the criteria for being a topology separately. :$(1): \quad$ By Union of Open Sets of Metric Space is Open, the union of any collection of open sets of a metric space is open. :$(2): \quad$ By Finite Intersection of Open Sets of Metric Space is Open, a finite intersection of open sets of a metric space is open. :$(3): \quad$ By Open Sets in Metric Space, $\O \in \tau$ and $A \in \tau$. Hence the result. {{qed}}
Let: :$X$ be a set :$\mathfrak A$ be an algebra of sets on $X$ :$A, B \in \mathfrak A$ By the definition of algebra of sets, we have that: :$A \cup B \in \mathfrak A$ :$\relcomp X A \in \mathfrak A$ Thus: {{begin-eqn}} {{eqn | o =        | r = A, B \in \mathfrak A       | c =  }} {{eqn | o = \leadsto       | r = \relcomp X A \cup \relcomp X B \in \mathfrak A       | c = {{Defof|Algebra of Sets}} }} {{eqn | o = \leadsto       | r = \relcomp X {A \cap B} \in \mathfrak A       | c = De Morgan's Laws: Complement of Intersection }} {{eqn | o = \leadsto       | r = A \cap B \in \mathfrak A       | c = {{Defof|Algebra of Sets}} }} {{end-eqn}} and so we have that the intersection of two sets in $\mathfrak A$ is in $\mathfrak A$. Next: {{begin-eqn}} {{eqn | o =        | r = A, B \in \mathfrak A       | c =  }} {{eqn | o = \leadsto       | r = A \cap \relcomp X B \in \mathfrak A       | c = from above }} {{eqn | o = \leadsto       | r = A \setminus B \in \mathfrak A       | c = Set Difference as Intersection with Relative Complement }} {{end-eqn}} and so we have that the difference of two sets in $\mathfrak A$ is in $\mathfrak A$. We have that $\mathfrak A \ne \O$ and so $\exists A \subseteq X: A \in \mathfrak A$. Then: {{begin-eqn}} {{eqn | l = \relcomp X A       | o = \in       | r = \mathfrak A       | c = {{Defof|Algebra of Sets}} }} {{eqn | ll= \leadsto       | l = \relcomp X A \cup A       | o = \in       | r = \mathfrak A       | c = {{Defof|Algebra of Sets}} }} {{eqn | ll= \leadsto       | l = X       | o = \in       | r = \mathfrak A       | c = Union with Relative Complement }} {{end-eqn}} Also, $\relcomp X A \cap A \in \mathfrak A$ from above. So by Intersection with Relative Complement is Empty: :$\O \in \mathfrak A$ {{qed}} Category:Algebras of Sets hq8mr16m39bsgf2dmbi1np8ux46w1mp
Let $\displaystyle \set {x_i}_{i \mathop = 1}^\infty$ be an enumeration of the elements of $S$. For any (strictly) positive real number $\epsilon$, define: :$A_i = \paren {x_i - 2^{-i} \epsilon, x_i + 2^{-i} \epsilon}$ {{explain|The notation of the above seems to be of an open real interval. This needs to be clarified.}} Then: :$\displaystyle S \subseteq \bigcup_{i \mathop = 1}^\infty A_i$ and: :$\displaystyle \map m {\bigcup A_i} \le \sum_{i \mathop = 1}^\infty 2^{1 - i} \epsilon = 2 \epsilon$ Since our choice of $\epsilon$ was arbitrary, for any positive real $z$ we can construct a set $T$ such that $S \subseteq T$ and $\map m T \le z$. Hence $X$ has zero measure. {{qed}} Category:Analysis l9se6p00ujy712vxsgsx3xbbo2ikxro
By the Mean Value Theorem: :$\exists \eta \in \openint x \xi: \map {f'} \eta = \dfrac {\map f x - \map f \xi} {x - \xi}$ From Real Function is Convex iff Derivative is Increasing, the derivative of $f$ is increasing. Thus: :$x > \xi \implies \map {f'} \eta \ge \map {f'} \xi$ :$x < \xi \implies \map {f'} \eta \le \map {f'} \xi$ Hence: :$\map f x - \map f \xi \ge \map {f'} \xi \paren {x - \xi}$ {{qed}}
From Logarithm is Strictly Concave: :$\ln$ is (strictly) concave. From Mean Value of Concave Real Function: :$\ln y - \ln 1 \le \left({D \ln 1}\right) \left({y - 1}\right)$ From Derivative of Natural Logarithm: :$D \ln 1 = \dfrac 1 1 = 1$ So: :$\ln y - \ln 1 \le \left({y - 1}\right)$ But from Logarithm of 1 is 0: :$\ln 1 = 0$ Hence the result. {{qed}}
Let $Y = \exp y$. From Exponential of Natural Logarithm: :$\map \ln {\exp y} = y$ From Logarithms of Powers, we have: :$\ln Y^x = x \ln Y = x \, \map \ln {\exp y} = x y$ Thus: :$\map \exp {x y} = \map \exp {\ln Y^x} = Y^x = \paren {\exp y}^x$ {{qed}}
{{begin-eqn}} {{eqn | l = \paren {\exp y}^x       | r = \map \exp {x \, \map \ln {\exp y} }       | c = {{Defof|Power to Real Number}} }} {{eqn | r = \map \exp {x y}       | c = Exponential of Natural Logarithm }} {{end-eqn}} {{qed}}
This proof assumes the Exponential of Sum property. First, for $n \in \Z_{\ge 0}$: {{begin-eqn}} {{eqn | l = \map \exp {n y}       | r = \map \exp {\sum_{k \mathop = 1}^n y} }} {{eqn | r = \prod_{k \mathop = 1}^n \exp y       | c = Exponential of Sum }} {{eqn | n = 1       | r = \paren {\exp y}^n }} {{end-eqn}} That is: :$\forall n \in \Z_{\ge 0}: \map \exp {n y} = \paren {\exp y}^n$ Now let $n \in \Z_{<0}$. It follows that $-n \in \Z_{>0}$, so: {{begin-eqn}} {{eqn | l = \exp n       | r = \map \exp {-\paren {-n y} } }} {{eqn | r = \frac 1 {\map \exp {-n y} }       | c = Reciprocal of Real Exponential }} {{eqn | r = \frac 1 {\paren {\exp y}^{-n} }       | c = from $(1)$ }} {{eqn | r = \paren {\exp y}^n       | c = Real Number to Negative Power: Positive Integer }} {{end-eqn}} Thus: :$(2): \quad \forall m \in \Z: \map \exp {m y} = \paren {\exp y}^m$ Next, for $n \in \Z_{>0}$: {{begin-eqn}} {{eqn | l = \exp y       | r = \map \exp {n \frac y n} }} {{eqn | r = \paren {\map \exp {\frac y n} }^n       | c = from $(1)$ }} {{eqn | n = 3       | ll= \leadsto       | l = \sqrt [n] {\exp y}       | r = \map \exp {\frac y n} }} {{end-eqn}} So fix $r \in \Q$. Let $r = \dfrac m n$, where $m \in \Z$ is an integer and $n \in \Z_{>0}$ is a strictly positive integer. From the above: {{begin-eqn}} {{eqn | l = \map \exp {r y}       | r = \map \exp {\frac m n y} }} {{eqn | r = \sqrt [n] {\map \exp {m y} }       | c = from $(3$ }} {{eqn | r = \sqrt [n] {\paren {\exp y}^m}       | c = from $(2)$ }} {{eqn | r = \paren {\exp y}^r }} {{end-eqn}} Thus, from the definition of $\paren {\exp y}^x$ as the unique continuous extension of $r \mapsto \paren {\exp y}^r$ from $\Q$ to $\R$: :$\map \exp {x y} = \paren {\exp y}^x$ {{qed}}
For all $x \in \R$: :$\exp 0 - 1 = 0$ from Exponential of Zero :$\map {D_x} {\exp x - 1} = \exp x$ from Sum Rule for Derivatives :$D_x x = 1$ from Derivative of Identity Function. Its prerequisites having been verified, Corollary 1 to L'Hôpital's Rule yields immediately: :$\displaystyle \lim_{x \mathop \to 0} \frac {\exp x - 1} x = \lim_{x \mathop \to 0} \frac {\exp x} 1 = \exp 0 = 1$ {{qed}}
Note that this proof does not presuppose Derivative of Exponential Function. We use the definition of the exponential as a limit of a sequence: {{begin-eqn}} {{eqn | l = \frac {\exp h - 1} h       | r = \frac {\lim_{n \mathop \to \infty} \paren {1 + \dfrac h n}^n - 1} h       | c = {{Defof|Exponential Function/Real|subdef = Limit of Sequence|Exponential Function}} }} {{eqn | r = \frac {\displaystyle \lim_{n \mathop \to \infty} \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac h n}^k - 1} h       | c = Binomial Theorem }} {{eqn | r = \lim_{n \mathop \to \infty} \frac {\displaystyle \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac h n}^k - 1} h       | c = as $h$ is constant }} {{eqn | r = \lim_{n \mathop \to \infty} \paren { {n \choose 0} 1 - 1 + {n \choose 1} \paren {\frac h n} \frac 1 h + \sum_{k \mathop = 2}^n {n \choose k} \paren {\frac h n}^k \frac 1 h} }} {{eqn | r = \lim_{n \mathop \to \infty}1 + \lim_{n \mathop \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k - 1} }{n^k}       | c = Powers of Group Elements }} {{eqn | r = 1 + h \lim_{n \mathop \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k - 2} } {n^k}       | c = Powers of Group Elements }} {{end-eqn}} The right summand converges to zero as $h \to 0$, and so: :$\displaystyle \lim_{h \mathop \to 0} \frac {\exp h - 1} h = 1$ {{qed}}
{{begin-eqn}} {{eqn | l = \frac {e^x - 1} x       | r = \frac {e^x - e^0} x       | c = Exponential of Zero }} {{eqn | o = \to       | r = \intlimits {\dfrac \d {\d x} e^x} {x \mathop = 0} {}       | c = {{Defof|Derivative of Real Function at Point}}, as $x \to 0$ }} {{eqn | r = \bigintlimits {e^x} {x \mathop = 0} {}       | c = Derivative of Exponential Function }} {{eqn | r = 1       | c = Exponential of Zero }} {{end-eqn}} {{qed}}
The Abelian cases are the direct result of the Fundamental Theorem of Finite Abelian Groups. The non-Abelian cases follow from seven separate theorems: :$(1): \quad$ Trivial Group is Cyclic Group - determines theorem for order $1$ :$(2): \quad$ Prime Group is Cyclic - determines theorem for orders $2$, $3$, $5$, $7$, $11$, and $13$ :$(3): \quad$ Group of Order Prime Squared is Abelian - determines theorem for orders $4$ and $9$ :$(4): \quad$ Cyclic Groups of Order p q - determines theorem for order $15$ :$(5): \quad$ Groups of Order Twice a Prime - determines theorem for orders $6$, $10$, $14$ :$(6): \quad$ Groups of Order 8 - determines theorem for order $8$ :$(7): \quad$ Groups of Order 12 - determines theorem for order $12$ {{qed}} Category:Order of Groups pp24q0v3ow83txltn6grzbjuoqymggr
Let $f: \R \setminus \set 0 \to \R: \map f x = \dfrac 1 x$. Clearly $f$ is continuous and positive on $\hointr 1 {+\infty}$. From Reciprocal Sequence is Strictly Decreasing, $f$ is decreasing on $\hointr 1 {+\infty}$. Therefore the conditions of the Integral Test hold. Thus the sequence $\sequence {\Delta_n}$ defined as: :$\displaystyle \Delta_n = \sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x$ is decreasing and bounded below by zero. But from the definition of the natural logarithm: :$\displaystyle \int_1^n \frac {\d x} x = \ln n$ Hence the result. {{qed}}
Let $H$ be a Sylow $p$-subgroup of $G$. Let $K$ be a Sylow $q$-subgroup of $G$.   By the Fourth Sylow Theorem, the number of Sylow $p$-subgroups of $G$ is of the form $1 + k p$ and divides $p q$. We have that $1 + k p$ cannot divide $p$. Then $1 + k p$ must divide $q$. But as $q$ is prime, either: :$1 + k p = 1$ or: :$1 + k p = q$ But: {{begin-eqn}} {{eqn | l = 1 + k p       | r = q       | c =  }} {{eqn | ll= \leadsto       | l = k p       | r = q - 1       | c =  }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = q - 1       | c =  }} {{end-eqn}} which contradicts our condition that $p$ does not divide $q - 1$. Hence $1 + k p = 1$. Thus there is only one Sylow $p$-subgroup of $G$. Similarly, there is only one Sylow $q$-subgroup of $G$. Thus, by Sylow $p$-Subgroup is Unique iff Normal, $H$ and $K$ are normal subgroups of $G$. Let $H = \gen x$ and $K = \gen y$. To show $G$ is cyclic, it is sufficient to show that $x$ and $y$ commute, because then: :$\order {x y} = \order x \order y = p q$ where $\order x$ denotes the order of $x$ in $G$. {{explain|Why does it follow that $\order {x y} {{=}} \order x \order y {{=}} p q$?}} Since $H$ and $K$ are normal: :$x y x^{-1} y^{-1} = \paren {x y x^{-1} } y^{-1} \in K y^{-1} = K$ and  :$x y x^{-1} y^{-1} = x \paren {y x ^{-1} y^{-1} } \in x H = H$ Now suppose $a \in H \cap K$. Then: {{begin-eqn}} {{eqn | l = \order a       | o = \divides       | r = p       | c =  }} {{eqn | lo= \land       | l = \order a       | o = \divides       | r = q       | c =  }} {{eqn | ll= \leadsto       | l = \order a       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = a       | r = e       | c =  }} {{end-eqn}} where $e$ is the identity of $G$. Thus: :$x y x^{-1}y^{-1} \in K \cap H = e$ Hence $x y = y x$ and the result follows. {{Qed}}
Let $H$ be a Sylow $p$-subgroup of $G$. Let $K$ be a Sylow $q$-subgroup of $G$.   By the Fourth Sylow Theorem, the number of Sylow $p$-subgroups of $G$ is of the form $1 + k p$ and divides $p q$. We have that $1 + k p$ cannot divide $p$. Then $1 + k p$ must divide $q$. But as $q$ is prime, either: :$1 + k p = 1$ or: :$1 + k p = q$ But: {{begin-eqn}} {{eqn | l = 1 + k p       | r = q       | c =  }} {{eqn | ll= \leadsto       | l = k p       | r = q - 1       | c =  }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = q - 1       | c =  }} {{end-eqn}} which contradicts our condition that $p$ does not divide $q - 1$. Hence $1 + k p = 1$. Thus there is only one Sylow $p$-subgroup of $G$. Similarly, there is only one Sylow $q$-subgroup of $G$. Let the Sylow $p$-subgroup of $G$ be denoted $P$. Let the Sylow $q$-subgroup of $G$ be denoted $Q$. We have that: :$P \cap Q = \set e$ where $e$ is the identity element of $G$. Hence in $P \cup Q$ there are $q + p - 1$ elements. As $p q \ge 2 q > q + p - 1$, there exists a non- identity element in $G$ that is not in $H$ or $K$. Its order must be $p q$. Hence, by definition, $G$ is cyclic. {{qed}}
L'Hôpital's Rule gives: :$\displaystyle \lim_{x \mathop \to c} \frac {\map f x} {\map g x} = \lim_{x \mathop \to c} \frac {\map {f'} x} {\map {g'} x}$ (provided the appropriate conditions are fulfilled). Here we have: {{begin-eqn}} {{eqn | l = \map \ln {1 + 0}       | r = 0 }} {{eqn | l = \map {D_x} {\map \ln {1 + x} }       | r = \dfrac 1 {1 + x}       | c = Chain Rule for Derivatives }} {{eqn | l = \map {D_x} x       | r = 1       | c = Derivative of Identity Function }} {{end-eqn}} Thus: :$\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = \lim_{x \mathop \to 0} \frac {\paren {1 + x}^{-1} } 1 = \frac 1 {1 + 0} = 1$ {{qed}}
{{begin-eqn}} {{eqn | l = \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x       | r = \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} - \ln 1} x       | c = subtract $\ln 1 = 0$ from the numerator, from Logarithm of 1 is 0 }} {{eqn | r = \intlimits {\dfrac {\d} {\d x} \ln x} {x \mathop = 1} {}       | c = {{Defof|Derivative of Real Function at Point}} }} {{eqn | r = \frac 1 1       | c = Derivative of Natural Logarithm Function }} {{eqn | r = 1 }} {{end-eqn}} {{qed}}
Note that this proof does not presuppose Derivative of Natural Logarithm Function. {{begin-eqn}} {{eqn | l = \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x       | r = \lim_{n \mathop \to \infty} \frac {\map \ln {1 + \frac 1 n} } {\frac 1 n}       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} n \, \map \ln {1 + \frac 1 n}       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} \map \ln {\paren {1 + \frac 1 n}^n}       | c =  }} {{eqn | r = \ln e       | c = {{Defof|Euler's Number|subdef = Limit of Sequence|Euler's Number as Limit of Sequence}} }} {{eqn | r = 1       | c = Natural Logarithm of e is 1 }} {{end-eqn}} {{qed}}
Suppose $\map S x$ converges when $x = y$. We need to show that it converges for all $x$ which satisfy $\size {x - \xi} < \size {y - \xi}$. So, let $\map S x$ converge when $x = y$. Then from Terms in Convergent Series Converge to Zero: :$a_n \paren {y - \xi}^n \to 0$ as $n \to \infty$ Hence, from Convergent Sequence is Bounded: :$\sequence {a_n \paren {y - \xi}^n}$ is bounded Thus: :$\exists H \in \R: \forall n \in \N_{>0}: \size {a_n \paren {y - \xi}^n} \le H$ Now suppose $\size {x - \xi} < \size {y - \xi}$. Then: :$\rho = \dfrac {\size {x - \xi} } {\size {y - \xi} } < 1$ (Note that if $\size {x - \xi} < \size {y - \xi}$ then $\size {y - \xi} > 0$ and the above fraction always exists.) Hence: :$\forall n \in \N_{>0} \size {a_n \paren {x - \xi}^n} = \rho^n \size {a_n \paren {y - \xi}^n} \le H \rho^n$ By Sequence of Powers of Number less than One: :$\displaystyle \sum_{n \mathop = 1}^\infty \rho^n$ converges Thus $\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ converges by the Comparison Test. The result follows. {{qed}}