hcju/proofwikiqa huggingface dataset:
=== Definition 1 implies Definition 2 === Let $M$ be totally bounded by definition 1. That is: :$\forall \epsilon \in \R_{>0}: \left({A, d}\right)$ has a finite $\epsilon$-net. So, let $\epsilon \in \R_{>0}$. Let $A' = \left\{{x_0, x_1, \ldots, x_n}\right\}$ be such a finite $\epsilon$-net of $A$. By definition: : $\displaystyle A \subseteq \bigcup_{x_i \mathop \in A'} B_\epsilon \left({x_i}\right)$ Now let $x \in A$, and so: : $\displaystyle x \in \bigcup_{x_i \mathop \in A'} B_\epsilon \left({x_i}\right)$ Thus: : $\exists i: 0 \le i \le n: x \in B_\epsilon \left({x_i}\right)$ and so: : $d \left({x_i, x}\right) < \epsilon$ But: : $\displaystyle \inf_{0 \mathop \le i \mathop \le n} d \left({x_i, x}\right) \le d \left({x_i, x}\right)$ Thus it follows that there exist finitely many points $x_0, \dots, x_n \in x$ such that: : $\displaystyle \inf_{0 \mathop \le i \mathop \le n} d \left({x_i, x}\right) \le \epsilon$ for all $x \in A$. Thus $M$ is totally bounded by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $M$ be totally bounded by definition 2. That is: : for every $\epsilon > 0$ there exist finitely many points $x_0, \dots, x_n \in A$ such that $\displaystyle \inf_{0 \mathop \le i \mathop \le n} d \left({x_i, x}\right) \le \epsilon$ for all $x \in A$. So, let $x \in A$. Let $\epsilon' = \dfrac \epsilon 2$. Then by definition $\exists n \in \N: A' = \left\{{x_0, x_1, \ldots, x_n}\right\}$ such that $\forall x \in A: \exists x_i \in A': d \left({x_i, x}\right) \le \epsilon'$. Hence: : $x \in B_{\epsilon'} \left({x_i}\right)$ where $B_{\epsilon'} \left({x_i}\right)$ is the open $\epsilon'$-ball of $x_i$. So $\displaystyle x \in \bigcup_{x_i \mathop \in A'} B_{\epsilon'} \left({x_i}\right)$ and hence $\displaystyle A \subseteq \bigcup_{x_i \mathop \in A'} B_{\epsilon'} \left({x_i}\right)$. Thus by definition, $A'$ is a finite $\epsilon'$-net of $A$. Thus $M$ is totally bounded by definition 1. {{qed}} Category:Totally Bounded Metric Spaces 9uknbpjzn8v1gunocdt0rrbmrgbsyqt
From Convergent Sequence in Hausdorff Space has Unique Limit, if $\left \langle {x_n} \right \rangle$ is convergent in a Hausdorff space it has exactly one limit. From Limit of Subsequence equals Limit of Sequence, any subsequence of a convergent sequence must have the same limit. So, if a sequence has two convergent subsequences with different limit, it must in fact be divergent. {{qed}} Category:Topology Category:Named Theorems jj2p654n8qyhaihrpdkx6h6iiapuese
Follows directly from the result that a Convergent Sequence is Bounded. {{qed}} Category:Metric Spaces Category:Named Theorems p9axnabeombo5itr4ffpe77zr9q76ha
Let $x \in U \setminus \left\{{\alpha}\right\}$. Let $\delta = d \left({x, \alpha}\right)$. Let $B_\epsilon \left({x}\right) \subseteq U$ be an open $\epsilon$-ball of $x$ in $U$. Let $\zeta = \min \left\{{\epsilon, \delta}\right\}$. Then $B_\epsilon \left({x}\right) \subseteq U \setminus \left\{{\alpha}\right\}$. The result follows. {{qed}} Category:Open Sets 5j7ijxim5g7g06792qw2kusgit96v2s
{{begin-eqn}} {{eqn | l = X       | r = 10^0 X_0 + 10^1 X_1 + \cdots + 10^{n - 1} X_{n-1} }} {{eqn | r = \sum_{i \mathop = 0}^{n - 1} 10^i X_i }} {{eqn | r = \sum_{i \mathop = 0}^{n - 1} \paren {10^i - 3^i + 3^i} X_i }} {{eqn | r = \sum_{i \mathop = 0}^{n - 1} \paren {\paren {10^i - 3^i} X_i} + \sum_{i \mathop = 0}^{n - 1} \paren {3^i X_i} }} {{end-eqn}} From Difference of Two Powers, $10^i - 3^i$ always produces a number divisible by $7$. Thus the first addend is always divisible by $7$.   So $X$ will be divisible by $7$ {{iff}} the second addend is divisible by $7$. Hence the result. {{qed}}
Let $U = \set {x_0, x_1, \ldots, x_n}$. Aiming for a contradiction, suppose that $U$ is open. Let: :$\displaystyle D = \min_{i \mathop \ne j} \map d {x_i, x_j}$ Let $x_j \in U$. Consider the open ball $\map {B_{D/2}} {x_j}$ of $x_j$. From Open Ball of Metric Space is Open Set, $\map {B_{D/2}} {x_j}$ is an open set. Then from Finite Intersection of Open Sets of Metric Space is Open, we can see that $U \cap B_{D/2}$ is also an open set. However, we have that: :$D/2 < \displaystyle \min_{i \mathop \ne j} \map d {x_i, x_j}$ Thus the open ball $\map {B_{D/2}} {x_j}$ is simply the singleton $\set {x_j}$. But then $x_j$ is an isolated point of $M$. This contradicts the fact that $M$ is dense-in-itself. So in fact, $U$ cannot be open. The result follows. {{qed}} Category:Metric Spaces Category:Denseness 48qkhhugbc34gcz825h2r6j2kqu0az4
From the definition, a region is a non-empty, open, path-connected subset of $M$. First, note that as $R$ is open it can not be a singleton from Finite Subspace of Dense-in-itself Metric Space is not Open. Therefore $R \setminus \left\{{x}\right\}$ is not empty. Next, we see that from Open Set Less One Point is Open that $R$ is open. Now, let $\alpha, \beta \in R$. As $R$ is path-connected, we can join $\alpha$ and $\beta$ with a path $\Gamma$. If $x \notin \Gamma$, then $\Gamma$ is also a path in $R \setminus \left\{{x}\right\}$, and we are done. If $x \in \Gamma$, then we consider the open $\epsilon$-ball $B_\epsilon \left({x}\right)$ of $x$ for some $\epsilon$ such that $B_\epsilon \left({x}\right) \subseteq R$. {{wtd|Needs more work done to prove that a deleted neighborhood in a connected open set is path-connected. Obvious intuitively, but there's some groundwork to be done.<br>Also see Euclidean Space without Origin is Path-Connected which is effectively the same result.}} Category:Metric Spaces sqwgw3p5n965wxjxb6sl9u1exly60ml
=== Sufficient Condition === {{:Theorem of Even Perfect Numbers/Sufficient Condition}} {{qed|lemma}} === Necessary Condition === {{:Theorem of Even Perfect Numbers/Necessary Condition}}
Let $n$ be composite such that $n \ge 0$. From Composite Number has Two Divisors Less Than It, we can write $n = a b$ where $a, b \in \Z$ and $1 < a, b < n$.  {{WLOG}}, suppose that $a \le b$. Let $a > \sqrt n$. Then $b \ge a > \sqrt n$.  However, if $b \ge a > \sqrt n$ is true, then: :$n = a b > \sqrt n \sqrt n > n$ This is clearly a contradiction. So: :$a \le \sqrt n$ From Positive Integer Greater than 1 has Prime Divisor it follows that there is some prime $p$ which divides $a$. From Absolute Value of Integer is not less than Divisors, we have that $p \le a$ and so: :$p \le \sqrt n$ From Divisor Relation on Positive Integers is Partial Ordering: : $p \divides n$ Hence the result. {{qed}}
If $d$ is a divisor of $n$ then $d \times \dfrac n d = n$ and so $\dfrac n d$ is also a divisor of $n$. Therefore if $d_1, d_2, \ldots, d_r$ are all the divisors of $n$, then so are $\dfrac n {d_1}, \dfrac n {d_2}, \ldots, \dfrac n {d_r}$, except in a different order. Hence: {{begin-eqn}} {{eqn | l = \sum_{d \mathop \divides n} \map f {\frac n d}       | r = \map f {\frac n {d_1} } + \map f {\frac n {d_2} } + \cdots + \map f {\frac n {d_r} }       | c =  }} {{eqn | r = \map f {d_1} + \map f {d_2} + \cdots + \map f {d_r}       | c =  }} {{eqn | r = \sum_{d \mathop \divides n} \map f d       | c =  }} {{end-eqn}} {{qed}} Category:Number Theory Category:Divisors 10kv4fgjkff1q5detzj7a4ie7o0jpkk
{{begin-eqn}} {{eqn | l = \sum_{d \mathop \divides n} \frac 1 d       | r = \sum_{d \mathop \divides n} \frac 1 {\paren {\frac n d} }       | c = Sum Over Divisors Equals Sum Over Quotients }} {{eqn | r = \frac 1 n \sum_{d \mathop \divides n} d       | c =  }} {{eqn | r = \frac {\map \sigma n} n       | c = {{Defof|Sigma Function}} }} {{end-eqn}} {{qed}} Category:Number Theory Category:Abundancy helnafv3kiz33xfbhbhdk8go3uubv6a
From Rule of Transposition, we may replace the ''only if'' statement by its contrapositive. Therefore, the following suffices: === Implication === Suppose $n$ is a prime. By definition, the only positive divisors of $n$ are $1$ and $n$ itself. Therefore $\map \sigma n$, defined as the sum of the divisors of $n$, equals $n + 1$. {{qed|lemma}} === Contrapositive Implication === Suppose $n$ is not a prime. From One Divides all Integers and Integer Divides Itself, both $1$ and $n$ are divisors of $n$. As $n$ is composite: :$\exists r, s \in \N: r, s > 1: r s = n$ Trivially, both $r$ and $s$ are divisors of $n$. Hence: :$\map \sigma n \ge n + 1 + r + s > n + 1$ {{qed}}
From Divisors of Power of Prime, the divisors of $n = p^k$ are $1, p, p^2, \ldots, p^{k - 1}, p^k$. Hence from Sum of Geometric Sequence: :$\map \sigma {p^k} = 1 + p + p^2 + \cdots + p^{k - 1} + p^k = \dfrac {p^{k + 1} - 1} {p - 1}$ {{qed}}
From Divisors of Power of Prime, the divisors of $n = p^k$ are $1, p, p^2, \ldots, p^{k-1}, p^k$. There are $k + 1$ of them. Hence the result. {{qed}} Category:Tau Function Category:Prime Numbers gsb8iojqu8nzvhj21g55mmbzuh6vyxj
Let $\displaystyle \map F n = \sum_{d \mathop \divides n} \map f d$. Let $m, n \in \Z_{>0}: m \perp n$. Then by definition: :$\displaystyle \map F {m n} = \sum_{d \mathop \divides m n} \map f d$ The divisors of $m n$ are of the form $d = r s$ where $r$ and $s$ are divisors of $m$ and $n$ respectively, from Divisors of Product of Coprime Integers. It is noted that $r \perp s$, otherwise any common divisor of $r$ and $s$ would be a common divisor of $m$ and $n$. Therefore: :$\displaystyle \map F {m n} = \sum_{r \mathop \divides m, \ s \mathop \divides n} \map f {r s}$ So, as $f$ is multiplicative: :$\displaystyle \map F {m n} = \sum_{r \mathop \divides m, \ s \mathop \divides n} \map f r \map f s$ But at the same time: :$\displaystyle \map F m \map F n = \paren {\sum_{r \mathop \divides m} \map f r} \paren {\sum_{s \mathop \divides n} \map f s}$ Multiplying out the product on the {{RHS}}, $\map F {m n}$ and $\map F m \map F n$ are seen to be the same. {{qed}}
By Divisor of Product, there exists $\tuple {r, s}$ satisfying: :$r \divides b$ :$s \divides c$ :$r s = a$ We have: :$r, s \divides a$ By definition of GCD: :$\gcd \set {a, b} \divides r$ :$\gcd \set {a, c} \divides s$ By Absolute Value of Integer is not less than Divisors: :$\gcd \set {a, b} \le \size r$ :$\gcd \set {a, c} \le \size s$ We also have: {{begin-eqn}} {{eqn | l = \size {r s}       | o = \ge       | r = \gcd \set {a, b} \gcd \set {a, c} }} {{eqn | r = \gcd \set {a, b c}       | c = GCD with One Fixed Argument is Multiplicative Function }} {{eqn | r = \size a       | c = GCD of Integer and Divisor }} {{eqn | r = \size {r s} }} {{end-eqn}} This forces both inequalities to be equalities, i.e.: :$\size r = \gcd \set {a, b}$ :$\size s = \gcd \set {a, c}$ Hence $\tuple {r, s}$ is unique up to absolute value. {{qed}} Category:Coprime Integers eviwamuiv88g3epnkp1xh547qre566o
Let $f_1: \Z_{>0} \to \Z_{>0}$ be the constant function: :$\forall n \in \Z_{>0}: \map {f_1} n = 1$ Thus we have: :$\displaystyle \map \tau n = \sum_{d \mathop \divides n} 1 = \sum_{d \mathop \divides n} \map {f_1} d$ But from Unity Function is Completely Multiplicative, $f_1$ is multiplicative. The result follows from Sum Over Divisors of Multiplicative Function. {{qed}} Category:Multiplicative Functions Category:Tau Function lo34yjiju2pxjv7407tw8mj35vs52kr
Let $I_{\Z_{>0}}: \Z_{>0} \to \Z_{>0}$ be the identity function: :$\forall n \in \Z_{>0}: \map {I_{\Z_{>0} } } n = n$ Thus we have: :$\displaystyle \map \sigma n = \sum_{d \mathop \divides n} d = \sum_{d \mathop \divides n} \map {I_{\Z_{>0} } } d$ But from Identity Function is Completely Multiplicative, $I_{\Z_{>0} }$ is multiplicative. The result follows from Sum Over Divisors of Multiplicative Function. {{qed}}
Let $a, b$ be coprime integers. Because $a$ and $b$ have no common divisor, the divisors of $a b$ are integers of the form $a_i b_j$, where $a_i$ is a divisor of $a$ and $b_j$ is a divisor of $b$. That is, any divisor $d$ of $a b$ is in the form: :$d = a_i b_j$ in a unique way, where $a_i \divides a$ and $b_j \divides b$. We can list the divisors of $a$ and $b$ as  :$1, a_1, a_2, \ldots, a$ and: :$1, b_1, b_2, \ldots, b$ and so their $\sigma$ function values are: :$\displaystyle \map \sigma a = \sum_{i \mathop = 1}^r a_i$ :$\displaystyle \map \sigma b = \sum_{j \mathop = 1}^s b_j$ Consider all divisors of $a b$ with the same $a_i$. Their sum is: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^s a_i b_j       | r = a_i \sum_{j \mathop = 1}^s b_j       | c =  }} {{eqn | r = a_i \map \sigma b       | c =  }} {{end-eqn}} Summing for all $a_i$: {{begin-eqn}} {{eqn | l = \map \sigma {a b}       | r = \sum_{i \mathop = 1}^r \paren {a_i \map \sigma b}       | c =  }} {{eqn | r = \paren {\sum_{i \mathop = 1}^r a_i} \map \sigma b       | c =  }} {{eqn | r = \map \sigma a \map \sigma b       | c =  }} {{end-eqn}} {{qed}}
Suppose $n$ has an odd divisor apart from $1$. Then $n$ can be expressed as $n = \paren {2 r + 1} s$. So: {{begin-eqn}} {{eqn | l = 2^n + 1       | r = 2^{\paren {2 r + 1} s} + 1       | c =  }} {{eqn | r = \paren {2^s}^{\paren {2 r + 1} } + 1^{\paren {2 r + 1} }       | c =  }} {{eqn | r = \paren {2^s + 1} \paren {2^{2 r s} - 2^{\paren {2 r - 1} s} + 2^{\paren {2 r - 2} s} - \cdots - 2^s + 1}       | c = Sum of Odd Positive Powers }} {{end-eqn}} and so $2^n + 1$ is not prime. Hence $2^n + 1$ can be prime only if $n$ has only even divisors. That is, if $n = 2^k$ for some natural number $k$. {{qed}}
A specific instance of Primes of form Power plus One: $q^n + 1$ is prime only if: :$(1): \quad q$ is even and :$(2): \quad n$ is of the form $2^k$ for some positive integer $k$. As $2$ is even, the result applies. {{qed}}
=== Necessary Condition === Suppose $c \in \Z_{>0}$ is the multiplicative order of $a$ modulo $n$. Then by definition: :$a^c \equiv 1 \pmod n$ Hence, by definition, $a^c = k n + 1$ for some $k \in \Z$. Thus: :$a r + n s = 1$ where $r = a^{c-1}$ and $s = -k$. It follows from Integer Combination of Coprime Integers that $a$ and $n$ are coprime. === Sufficient Condition === Suppose $a \perp n$. Then by Euler's Theorem: : $a^{\phi \left({n}\right)} \equiv 1 \pmod n$ where $\phi \left({n}\right)$ is the Euler Phi Function of $n$. Hence the multiplicative order of $a$ modulo $n$ exists, by taking $c = \phi \left({n}\right)$. {{qed}} Category:Coprime Integers Category:Modulo Arithmetic ciiwxi45ftzbw816mx9aob2x1wfd3x4
First note from Integer has Multiplicative Order Modulo n iff Coprime to n that unless $a \perp n$ the multiplicative order of $a$ modulo $n$ does not exist. === Necessary Condition === Let $k$ be a multiple of $c$. Then $k = c r$, say. Then: :$a^k = a^{c r} = \paren {a^c}^r \equiv 1^r \equiv 1 \pmod n$ {{qed|lemma}} === Sufficient Condition === Let $a^k \equiv 1 \pmod n$. Then: :$1 \equiv a^k \equiv a^{q c + r} \equiv \paren {a^c}^q a^r \equiv 1^q a^r \equiv a^r \pmod n$ So $r = 0$ or else (from the Division Theorem) $0 < r < c$ and this contradicts $c$ being the smallest integer such that $a^c \equiv 1 \pmod n$. Hence $k$ is a multiple of $c$. {{qed}} Category:Integer to Power of Multiple of Order Category:Number Theory f4jrsgq26v9thhuvvix3e6r998ru3rl
From Sum of Geometric Sequence, we have: :$\displaystyle s_N = \sum_{n \mathop = 0}^N z^n = \frac {1 - z^{N + 1} } {1 - z}$ We have that $\size z < 1$. So by Sequence of Powers of Number less than One: :$z^{N + 1} \to 0$ as $N \to \infty$ Hence $s_N \to \dfrac 1 {1 - z}$ as $N \to \infty$. The result follows. {{qed|lemma}} It remains to demonstrate absolute convergence: The absolute value of $\size z$ is just $\size z$. By assumption: :$\size z < 1$ So $\size z$ fulfils the same condition for convergence as $z$. Hence: :$\displaystyle \sum_{n \mathop = 0}^\infty \size z^n = \frac 1 {1 - \size z}$ {{qed}}
By the Chain Rule for Derivatives and the corollary to $n$th Derivative of Reciprocal of $m$th Power: :$\dfrac {\d^n} {\d z^n} \dfrac 1 {1 - z} = \dfrac {n!} {\paren {1 - z}^{n + 1}}$ Thus the Maclaurin series expansion of $\dfrac 1 {1 - z}$ is: :$\displaystyle \sum_{n \mathop = 0}^\infty \frac {z^n} {n!} \dfrac {n!} {\paren {1 - 0}^{n + 1}}$ whence the result. {{qed}}
Let $S = \displaystyle \sum_{n \mathop = 0}^\infty z^n$. Then: {{begin-eqn}} {{eqn | l = z S       | r = z \sum_{n \mathop = 0}^\infty z^n       | c =  }} {{eqn | r = \sum_{n \mathop = 0}^\infty z^{n + 1}       | c =  }} {{eqn | r = \sum_{n \mathop = 1}^\infty z^n       | c = Translation of Index Variable of Summation }} {{eqn | r = S - z^0       | c =  }} {{eqn | r = S - 1       | c =  }} {{eqn | ll= \leadsto       | l = 1       | r = \paren {1 - z} S       | c =  }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \frac {\d \zeta} {\d z}       | r = \map {\frac \d {\d z} } {\sum_{n \mathop = 1}^\infty n^{-z} }       | c =  }} {{eqn | n = 1       | r = \sum_{n \mathop = 1}^\infty \map {\frac \d {\d z} } {n^{-z} }       | c =  }} {{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-\map \ln n n^{-z} }       | c = Derivative of Exponential Function }} {{eqn | r = -\sum_{n \mathop = 1}^\infty \frac {\map \ln n} {n^z}       | c =  }} {{eqn | r = -\sum_{n \mathop = 2}^\infty \frac {\map \ln n} {n^z}       | c = as $\ln 1 = 0$ }} {{end-eqn}} {{qed}} {{MissingLinks|The step at $(1)$ requires justification.  Need to show that the sum converges uniformly and also that the sum of the derivatives converge uniformly. Note that this result may already exist somewhere in the database and so would just need hunting down.}} Category:Derivatives Category:Riemann Zeta Function dojyxdoiccjkn6c2fz356ajsgte6w3u
Let $\map \Re z$ be the real part of a complex variable $z$. By Dirichlet Series is Analytic, the Riemann zeta function is analytic. By Trivial Zeroes of Riemann Zeta Function are Even Negative Integers, the Riemann zeta function has no zeroes in $\map \Re z > 1$. Thus the reciprocal of the Riemann zeta function is analytic in $\map \Re z > 1$. By Reciprocal of Riemann Zeta Function, this means that $\displaystyle \sum_{n \mathop = 1}^\infty \map \mu n n^{-z}$ converges to an analytic function in $\map \Re z > 1$. By taking $a_n = \mu \left({n}\right)$ in Ingham's Theorem on Convergent Dirichlet Series: :$\displaystyle \sum_{n \mathop = 1}^\infty \frac {\map \mu n} {n^z}$ converges for $\map \Re z \ge 1$. Taking $z = 1$, we are given a convergent sum: :$\displaystyle \sum_{n \mathop = 1}^\infty \frac {\map \mu n} n$ Clearly: :$\displaystyle \sum_{n \mathop = 1}^N \frac {\map \mu n} n \ge \sum_{n \mathop = 1}^N \frac {\map \mu n} N$ but: :$\displaystyle \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N \frac {\map \mu n} n = \lim_{z \mathop \to 1} \frac 1 {\map \zeta z}$ Since Harmonic Series is Divergent, $\dfrac 1 {\map \zeta z}$ goes to $0$ by Reciprocal of Null Sequence. Hence also: :$\displaystyle \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N \frac {\map \mu n} N = 0$ {{qed}} Category:Möbius Function Category:Analytic Number Theory mjytexxydnlfsj8qtuz0lv2x7nxlwfw
Let $r, s \in K$. Then: {{begin-eqn}} {{eqn | l = f_z \left({r s}\right)       | r = \left({r s}\right)^z       | c =  }} {{eqn | r = \left({r}\right)^z \left({s}\right)^z       | c = Power of Product in Abelian Group }} {{eqn | r = f_z \left({r}\right) f_z \left({s}\right)       | c =  }} {{end-eqn}} {{Qed}} Category:Field Theory Category:Completely Multiplicative Functions hkrbesxhwh2r3p02h5xduuvbmjohnhx
First suppose that $p \nmid n$. Then by Prime not Divisor implies Coprime, $p \perp n$. So by Euler Phi Function is Multiplicative, $\phi \left({p n}\right) = \phi \left({p}\right) \phi \left({n}\right)$. It follows from Euler Phi Function of Prime that $\phi \left({p n}\right) = \left({p - 1}\right) \phi \left({n}\right)$. Now suppose that $p \mathrel \backslash n$. Then $n = p^k m$ for some $k, m \in \Z: k, m \ge 1$ such that $p \perp m$. Then: {{begin-eqn}} {{eqn | l = \phi \left({p n}\right)       | r = \phi \left({p^{k + 1} m}\right)       | c =  }} {{eqn | r = \phi \left({p^{k + 1} }\right) \phi \left({m}\right)       | c = Euler Phi Function is Multiplicative }} {{eqn | r = p^{k + 1} \left({1 - \frac 1 p}\right) \phi \left({m}\right)       | c = Euler Phi Function of Prime Power }} {{end-eqn}} At the same time: {{begin-eqn}} {{eqn | l = p \phi \left({n}\right)       | r = p \phi \left({p^k m}\right)       | c =  }} {{eqn | r = p \phi \left({p^k}\right) \phi \left({m}\right)       | c = Euler Phi Function is Multiplicative }} {{eqn | r = p p^k \left({1 - \frac 1 p}\right) \phi \left({m}\right)       | c = Euler Phi Function of Prime Power }} {{eqn | r = p^{k + 1} \left({1 - \frac 1 p}\right) \phi \left({m}\right)       | c =  }} {{end-eqn}} {{qed}} Category:Euler Phi Function g8jzolur7l4nul1np3pzpd6e01rdkye
We have from the definition of Euler $\phi$ function: :$\map \phi 1 = 1$ and from Euler Phi Function of Prime Power: Corollary: :$\map \phi 2 = 1$ Now let $n \ge 3$. There are two possibilities: === Odd Prime Divisor === $n$ has (at least one) odd prime factor: $p$, say. From the corollary to Euler Phi Function of Integer, it follows that: :$p - 1$ divides $\map \phi n$ But as $p$ is odd, $p - 1$ is even and hence: :$2 \divides \paren {p - 1} \divides \map \phi n$ and so $\map \phi n$ is even. === No Odd Prime Divisor === Now suppose $n$ has no odd prime factors. Then its only prime factor must be $2$. Thus: :$n = 2^k$ where $k > 1$. Then from Euler Phi Function of Prime Power: Corollary: :$\map \phi n = 2^k \paren {1 - \frac 1 2} = 2^{k - 1}$ where $k-1 > 0$. Hence $\map \phi n$ is even. {{qed}} Category:Euler Phi Function tf6k17kvzjqks2fylassj5b5dvwjp3u
Begin by rewriting the function $f$ and differential $\rd z$ in terms of their real and complex parts: :$f = u + iv$ :$\d z = \d x + i \rd y$ Then we have: :$\displaystyle \oint_{\partial D} \map f z \rd z = \oint_{\partial D} \paren {u + iv} \paren {\d x + i \rd y}$ Expanding the result and again separating into real and complex parts yields two integrals of real variables: :$\displaystyle \oint_{\partial D} \paren {u \rd x - v \rd y} + i \oint_{\partial D} \paren {v \rd x + u \rd y}$ We next apply Green's Theorem to each integral term to convert the contour integrals to surface integrals over $D$: :$\displaystyle \iint_D \paren {-\dfrac {\partial v} {\partial x} - \dfrac {\partial u} {\partial y} } \rd x \rd y + \iint_D \paren {\dfrac {\partial u} {\partial x} - \dfrac {\partial v} {\partial y} } \rd x \rd y$ By the assumption that $f$ is holomorphic, it satisfies the Cauchy-Riemann Equations :$\dfrac {\partial v} {\partial x} + \dfrac {\partial u} {\partial y} = 0$ :$\dfrac {\partial u} {\partial x} - \dfrac {\partial v} {\partial y} = 0$ The integrands are therefore zero and hence the integral is zero. {{qed}}
We have that the Sigma Function is Multiplicative. From Value of Multiplicative Function is Product of Values of Prime Power Factors, we have: :$\map f n = \map f {p_1^{k_1} } \map f {p_2^{k_2} } \ldots \map f {p_r^{k_r} }$ From Sigma Function of Power of Prime, we have: :$\displaystyle \map \sigma {p_i^{k_i} } = \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ Hence the result. {{qed}}
{{improve|Better to start with $\displaystyle \phi \left({n}\right) {{=}} n \prod_{p \mathrel \backslash n} \left({1 - \frac 1 p}\right)$ and then a lot of the gnarliness goes away}} From Euler Phi Function of Integer: : $\displaystyle \phi \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i - 1} \left({p_i - 1}\right)$ From Sigma Function of Integer: : $\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ So: : $\displaystyle \sigma \left({n}\right) \phi \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \left({\frac {p_i^{k_i + 1} - 1} {p_i - 1}}\right) p_i^{k_i - 1} \left({p_i - 1}\right)$ Taking a general factor of this product: {{begin-eqn}} {{eqn | l = \left({\frac {p_i^{k_i + 1} - 1} {p_i - 1} }\right) p_i^{k_i - 1} \left({p_i - 1}\right)       | r = \left({p_i^{k_i + 1} - 1}\right) p_i^{k_i - 1}       | c = cancelling $p_i - 1$ top and bottom }} {{eqn | r = p_i^{2 k_i} - p_i^{k_i - 1}       | c = multiplying out the bracket }} {{eqn | r = p_i^{2 k_i} \left({1 - \frac 1 {p_i^{k_i + 1} } }\right)       | c = extracting $p_i^{2 k_i}$ as a factor }} {{end-eqn}} So: :$\displaystyle \sigma \left({n}\right) \phi \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} p_i^{2 k_i} \left({1 - \frac 1 {p_i^{k_i + 1} } }\right)$ Hence: :$\displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{2 k_i} = \left({\prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}}\right)^2 = n^2$ and the result follows. {{qed}} Category:Sigma Function Category:Euler Phi Function g6g2u4a6atp7fl8kcvqvdae7qzugm6t