avewright/proofwiki-math huggingface dataset:
Theorem: Real Multiplication Distributes over Addition

The operation of multiplication on the set of real numbers $\R$ is distributive over the operation of addition:
:$\forall x, y, z \in \R:$
::$x \times \paren {y + z} = x \times y + x \times z$
::$\paren {y + z} \times x = y \times x + z \times x$

Proof:
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers.
Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}, z = \eqclass {\sequence {z_n} } {}$, where $\eqclass {\sequence {x_n} } {}$, $\eqclass {\sequence {y_n} } {}$ and $\eqclass {\sequence {z_n} } {}$ are such equivalence classes.
From the definition of real multiplication, $x \times y$ is defined as:
:$\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$
From the definition of real addition, $x + y$ is defined as:
:$\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n + y_n} } {}$
Thus:
{{begin-eqn}}
{{eqn | l = x \times \paren {y + z}
      | r = \eqclass {\sequence {x_n} } {} \times \paren {\eqclass {\sequence {y_n} } {} + \eqclass {\sequence {z_n} } {} }
      | c = 
}}
{{eqn | r = \eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n + z_n} } {}
      | c = 
}}
{{eqn | r = \eqclass {\sequence {x_n \times \paren {y_n + z_n} } } {}
      | c = 
}}
{{eqn | r = \eqclass {\sequence {\paren {x_n \times y_n} + \paren {x_n \times z_n} } } {}
      | c = Rational Multiplication Distributes over Addition
}}
{{eqn | r = \eqclass {\sequence {x_n \times y_n} } {} + \eqclass {\sequence {x_n \times z_n} } {}
      | c = 
}}
{{eqn | r = \paren {\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} } + \paren {\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {z_n} } {} }
      | c = 
}}
{{eqn | r = \paren {x \times y} + \paren {x \times z}
      | c = 
}}
{{end-eqn}}
By Real Addition is Commutative and Real Multiplication is Commutative, it follows that:
:$\paren {y + z} \times x = \paren {y \times x} + \paren {z \times x}$
{{qed}}
Theorem: Real Multiplication Distributes over Addition

The operation of multiplication on the set of real numbers $\R$ is distributive over the operation of addition:
:$\forall x, y, z \in \R:$
::$x \times \paren {y + z} = x \times y + x \times z$
::$\paren {y + z} \times x = y \times x + z \times x$

Proof:
{{:Euclid:Proposition/II/1}}
:400px
Let $A$ and $BC$ be two straight lines.
Let $BC$ be cut at random at points $D$ and $E$.
Then the rectangle contained by $A$ and $BC$ is equal to the sum of the rectangles contained by $A$ and $BD$, by $A$ and $DE$, and by $A$ and $EC$, as follows:
Construct $BF$ perpendicular to $BC$.
Construct $BG$ on $BF$ equal to $A$.
Construct $GH$ through $G$ parallel to $BC$.
Construct $DK, EL, CH$ through $D, E, C$ parallel to $BG$.
Then we have that:
: $\Box BCHG = \Box BDKG + \Box DELK + \Box ECHL$
Now $\Box BCHG$ is the rectangle contained by $A$ and $BC$, because it is contained by $BC$ and $BG$, and $BG = A$.
Similarly, from Opposite Sides and Angles of Parallelogram are Equal:
: $\Box BDKG$ is the rectangle contained by $A$ and $BD$, because it is contained by $BD$ and $BG = A$
: $\Box DEKL$ is the rectangle contained by $A$ and $DE$, because it is contained by $DE$ and $DK$, and $DK = A$
: $\Box ECHL$ is the rectangle contained by $A$ and $EC$, because it is contained by $EC$ and $EL$, and $EL = A$.
Hence the result.
{{qed}}
Theorem: Pointwise Inverse in Induced Structure

Let $\struct {G, \oplus}$ be a group whose identity is $e_G$.
Let $S$ be a set.
Let $\struct {G^S, \oplus}$ be the structure on $G^S$ induced by $\oplus$.
Let $f \in G^S$.
Let $f^* \in G^S$ be the pointwise inverse of $f$:
:$\forall x \in S: \map {f^*} x = \paren {\map f x}^{-1}$
Then $f^*$ is the inverse of $f$ for the pointwise operation induced on $G^S$ by $\oplus$.

Proof:
Let $f \in G^S$.
{{begin-eqn}}
{{eqn | l = \map {\paren {f \oplus f^*} } x
      | r = \map f x \oplus \map {f^*} x
      | c = 
}}
{{eqn | r = \map f x \oplus \paren {\map f x}^{-1}
      | c = 
}}
{{eqn | r = e_G
      | c = 
}}
{{end-eqn}}
Similarly for $\map {\paren {f^* \oplus f} } x$.
{{qed}}
Theorem: Quaternion Group is Hamiltonian

The quaternion group $Q$ is Hamiltonian.

Proof:
For clarity the Cayley table of $Q$ is presented below:
{{:Quaternion Group/Cayley Table}}
By definition $Q$ is Hamiltonian {{iff}}:
:$Q$ is non-abelian
and:
:every subgroup of $Q$ is normal.
$Q$ is non-abelian as demonstrated by the counter-example:
:$a b \ne b a$
From Subgroups of Quaternion Group:
{{:Subgroups of Quaternion Group}}
From Trivial Subgroup and Group Itself are Normal:
:$Q$ and $\set e$ are normal subgroups of $Q$.
From Center of Quaternion Group, $\gen {a^2} = \set {e, a^2}$ is the center of $Q$.
From Center of Group is Normal Subgroup, $\set {e, a^2}$ is normal in $Q$.
The remaining subgroups of $Q$ are of order $4$, and so have index $2$.
From Subgroup of Index 2 is Normal it follows that all of these order $4$ subgroups of $Q$ are normal.
That accounts for all subgroups of $Q$.
Hence the result.
{{qed}}
Theorem: Finite Subset of Metric Space has no Limit Points

Let $M = \struct {A, d}$ be a metric space.
Let $X \subseteq A$ such that $X$ is finite.
Then $X$ has no limit points.

Proof:
Let $x \in X$.
From Point in Finite Metric Space is Isolated, $x$ is an isolated point.
The result follows by definition of isolated point:
:$x$ is an isolated point {{iff}} $x$ is not a limit point.
{{qed}}
Theorem: Point in Finite Metric Space is Isolated

Let $M = \struct {A, d}$ be a metric space.
Let $X \subseteq A$ such that $X$ is finite.
Let $x \in X$.
Then $x$ is isolated in $X$.

Proof:
As $X$ is finite, its elements can be placed in one-to-one correspondence with the elements of $\N^*_n$ for some $n \in \N$.
So let $X = \left\{{x_1, x_2, \ldots, x_n}\right\}$.
Now let $\epsilon := \min \left\{{\forall i, j \in \N^*_n: i \ne j: d \left({x_i, x_j}\right)}\right\}$.
That is, $\epsilon$ is the minimum distance between any two elements of $X$.
We have that $\epsilon > 0$ from the definition of metric.
Let $N_\epsilon \left({x_k}\right)$ be the open $\epsilon$-ball of any $x_k \in X$.
By the method of construction of $\epsilon$, it is clear that $N_\epsilon \left({x_k}\right) = \left\{{x_k}\right\}$.
Hence $x_k$ is an isolated point.
{{qed}}
Theorem: Point in Finite Metric Space is Isolated

Let $M = \struct {A, d}$ be a metric space.
Let $X \subseteq A$ such that $X$ is finite.
Let $x \in X$.
Then $x$ is isolated in $X$.

Proof:
A Metric Space is Hausdorff.
Therefore Point in Finite Hausdorff Space is Isolated can be applied.
Hence the result.
{{qed}}
Theorem: Point in Finite T2 Space is Isolated

Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Let $X \subseteq S$ such that $X$ is finite.
Let $x \in X$.
Then $x$ is isolated in $X$.

Proof:
As $X$ is finite, its elements can be placed in one-to-one correspondence with the elements of $\set {1, 2, \ldots, n}$ for some $n \in \N$.
So let $X = \set {x_1, x_2, \ldots, x_n}$.
From the definition of $T_2$ space:
: $\forall x_i, x_j \in X: x_i \ne x_j: \exists U, V \in \tau: x_i \in U, x_j \in V: U \cap V = \O$
Let $x_k \in X$.
Then by definition:
:$\forall x_i \in X, x_i \ne x_k: \exists U_i, V \in \tau: x_k \in U_i, x_i \in V: U_i \cap V = \O$
Note that $U_i$ may be different for each $X_i$.
That is:
:$\forall x_i \in X: \exists U_i \in \tau: x_k \in U_i, x_i \notin U_i$
Now consider:
:$\ds U = \bigcap_{i \mathop \ne k} U_i$
We have by definition of a topology that $U \in \tau$, as the intersection $U$ is of a finite number of sets $U_i$.
We have that:
: $\forall i: x_k \in U_i$
: $\forall i: x_i \notin U_i$
So:
: $U \cap X = \set {x_k}$
and so $U$ is an open set of $T$ which contains no points of $X$ other than $x_k$.
So by definition, $x_k$ is an isolated point of $X$.
Hence the result.
{{qed}}
Theorem: Point in Finite T2 Space is Isolated

Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Let $X \subseteq S$ such that $X$ is finite.
Let $x \in X$.
Then $x$ is isolated in $X$.

Proof:
Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Let $X \subseteq T$ be finite.
From Separation Properties Preserved in Subspace, it follows that $\struct {X, \tau_X}$ is also a $T_2$ space.
From $T_2$ Space is $T_1$ Space it follows that $\struct {X, \tau_X}$ is a $T_1$ space.
From Finite $T_1$ Space is Discrete, it follows that $\struct {X, \tau_X}$ is a discrete space.
The result follows from Topological Space is Discrete iff All Points are Isolated.
{{qed}}
Theorem: Subsequence of Sequence in Metric Space with Limit

Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a sequence in $M$.
Let $x$ be a limit point of $S = \set {x_n: n \in \N}$, the set of members of $\sequence {x_n}$.
Then $\sequence {x_n}$ has a subsequence which converges to $x$.

Proof:
By Finite Subset of Metric Space has no Limit Points, $S$ is infinite (or it has no limit points).
We may assume that $x_n$ never equals $x$.
Otherwise we delete all instances of $x$ from $\sequence {x_n}$ and create a new sequence $x_m$ which is a subsequence of $x_n$ which ''does'' never equal $x$.
Then $S \setminus \set x$ is still infinite, and it still has $x$ as a limit point.
So, since $x$ is a limit point of $S$, there is an integer $\map n 1$, say, such that $x_{\map n 1} \in \map {B_1} x$, where $\map {B_1} x$ is the open $1$-ball of $x$.
Suppose we choose the integers $\map n 1 < \map n 2 < \cdots < \map n k$ so that $x_{\map n i} \in \map {B_{1/i} } x$ for $i = 1, 2, \ldots, k$.
Now we put $\epsilon = \min \set {\dfrac 1 {k+1}, \map d {x_1, x}, \map d {x_2, x}, \map d {x_{\map n k}, x} }$.
Since $x_n \ne x$ for any $n$, it follows that $\epsilon > 0$.
Since $x$ is a limit point of $S$, there exists an integer $\map n {k+1}$ such that $x_{\map n {k+1} } \in \map {B_\epsilon} x$.
Now $\epsilon$ has been chosen so as to force $\map n {k+1} > \map n k$, since $\map d {x_i, x} \ge \epsilon$ for all $i \le \map n k$.
Also, note that $x_{\map n {k+1} } \in \map {B_{\frac 1 {k+1}} } x$.
This completes the inductive step in constructing a subsequence.
This converges to $x$, because $\forall k_0 \in \N_{>0}$ and $\forall k \ge k_0$ we have $\map d {x_{\map n k}, x} < \dfrac 1 k \ge \dfrac 1 {k_0}$.
Hence the result.
{{qed}}
Theorem: Primes of form Power Less One

Let $m, n \in \N_{>0}$ be natural numbers.
Let $m^n - 1$ be prime.
Then $m = 2$ and $n$ is prime.

Proof:
First we note that by Integer Less One divides Power Less One:
:$\paren {m - 1} \divides \paren {m^n - 1}$
where $\divides$ denotes divisibility.
Thus $m^n - 1$ is composite for all $m \in \Z: m > 2$.
Let $m = 2$, and consider $2^n - 1$.
Suppose $n$ is composite.
Then $n = r s$ where $r, s \in \Z_{> 1}$.
Then by {{Corollary|Integer Less One divides Power Less One}}:
:$\paren {2^r - 1} \divides \paren {2^{r s} - 1}$
Thus if $n$ is composite, then so is $2^n - 1$.
So $2^n - 1$ can be prime only when $n$ is prime.
{{qed}}
Theorem: Lebesgue's Number Lemma/Sequentially Compact Space

Let $M = \struct {A, d}$ be a metric space.
Let $M$ be sequentially compact.
Then there exists a Lebesgue number for every open cover of $M$.

Proof:
{{AimForCont}} $\UU$ is an open cover of $M$ for which no Lebesgue number exists.
Then for any $n \in \N$, there exists some $x_n \in M$ such that $\map {B_{1 / n} } {x_n} \subseteq U$ is false for each $U \in \UU$. (Otherwise $1/n$ would be a Lebesgue number for $\UU$.)
As $M$ is sequentially compact, $\sequence {x_n}$ has a subsequence, say $\sequence {x_{\map n r} }$ which converges to some $x \in M$.
Since $\UU$ covers $M$, $x \in U_0$ for some $U_0 \in \UU$.
Since $U_0$ is open, $\exists m \in \N: \map {B_{2 / m} } x \subseteq U_0$.
Now $\map {B_{1 / m} } x$ contains $x_{\map n r}$ for all $r \ge R$, say.
Choose $r \ge R$ such that $\map n r \ge m$ and write $s = \map n r$.
Then $\map {B_{1 / s} } {x_s} \subseteq \map {B_{2 / m} } x$ since:
{{begin-eqn}}
{{eqn | l = \map d {x_s, y}
      | o = <
      | r = \frac 1 s
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map d {x, y}
      | o = \le
      | r = \map d {x, x_s} + \map d {x_s, y}
      | c = 
}}
{{eqn | o = <
      | r = \frac 1 m + \frac 1 s
      | c = 
}}
{{eqn | o = \le
      | r = \frac 2 m
      | c = 
}}
{{end-eqn}}
So $\map {B_{1 / s} } {x_s} \subseteq U_0$ which contradicts the choice of $x_s$.
So, by Proof by Contradiction, there has to be a Lebesgue number for $\UU$.
{{qed}}
Theorem: First Subsequence Rule

Let $T = \struct {A, \tau}$ be a Hausdorff space.
Let $\sequence {x_n}$ be a sequence in $T$.
Suppose $\sequence {x_n}$ has two convergent subsequences with different limits.
Then $\sequence {x_n}$ is divergent.

Proof:
As stated, let $T = \struct {A, \tau}$ be a Hausdorff space.
Let $\sequence {x_n}$ be a sequence in $T$.
Let $\sequence {y_n}$ and $\sequence {z_n}$ be convergent subsequences of $\sequence {x_n}$ with different limits.
{{AimForCont}} $\sequence {x_n}$ is convergent.
From Convergent Sequence in Hausdorff Space has Unique Limit, $\sequence {x_n}$ has exactly one limit.
From Limit of Subsequence equals Limit of Sequence, $\sequence {y_n}$ has the same limit as $\sequence {x_n}$.
This contradicts the assertion that $\sequence {y_n}$ and $\sequence {z_n}$ have different limits.
Hence by Proof by Contradiction it must be the case that $\sequence {x_n}$ is divergent.
{{qed}}
Theorem: Second Subsequence Rule

Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a sequence in $M$.
Suppose $\sequence {x_n}$ has a subsequence which is unbounded.
Then $\sequence {x_n}$ is divergent.

Proof:
Follows by the Rule of Transposition from Convergent Sequence is Bounded.
{{qed}}
Theorem: Open Set Less One Point is Open

Let $M = \struct {A, d}$ be a metric space.
Let $U \subseteq M$ be an open set of $M$.
Let $\alpha \in U$.
Then $U \setminus \set \alpha$ is open in $M$.

Proof:
Let $x \in U \setminus \set \alpha$.
Let $\delta = \map d {x, \alpha}$.
Let $\map {B_\epsilon} x \subseteq U$ be an open $\epsilon$-ball of $x$ in $U$.
Let $\zeta = \min \set {\epsilon, \delta}$.
Then:
:$\map {B_\epsilon} x \subseteq U \setminus \set \alpha$
The result follows.
{{qed}}
Category:Open Set Less One Point is Open
Category:Open Sets (Metric Spaces)
88myvnv0oep0gboir83o28jrtpy2hu3
Theorem: Divisibility by 7

An integer $X$ with $n$ digits ($X_0$ in the ones place, $X_1$ in the tens place, and so on) is divisible by $7$ {{iff}}:
:$\ds \sum_{i \mathop = 0}^{n - 1} \paren {3^i X_i}$ is divisible by $7$.

Proof:
{{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}}
Theorem: Finite Subspace of Dense-in-itself Metric Space is not Open

Let $M = \struct {A, d}$ be a metric space that is dense-in-itself.
Let $U$ be a finite subset of $A$.
Then $U$ is not an open set of $M$.

Proof:
Let $U = \set {x_0, x_1, \ldots, x_n}$.
{{AimForCont}} $U$ is open.
Let:
:$\ds D = \min_{i \mathop \ne j} \map d {x_i, x_j}$
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 $U \cap \map {B_{D/2} } {x_j}$ is also an open set.
However, we have that:
:$\dfrac D 2 < \ds \min_{i \mathop \ne j} \map d {x_i, x_j}$
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:Dense-in-itself
Category:Metric Spaces
0vr367pb4usv36ksf5ffa0vpru0t834
Theorem: Region Less One Point is Region

Let $M = \struct {A, d}$ be a dense-in-itself metric space.
Let $R \subseteq M$ be a region of $M$.
Let $x \in R$.
Then $R \setminus \set x$ is also a region of $M$.

Proof:
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 $R \setminus \set x$ 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 $R \setminus \set x$, and the proof is complete.
If $x \in \Gamma$, then we consider the open $\epsilon$-ball $\map {B_\epsilon} x$ of $x$ for some $\epsilon$ such that $\map {B_\epsilon} x \subseteq R$.
{{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
gcia0m3t7e8huwlys3p0lkfmc99xcrf
Theorem: Theorem of Even Perfect Numbers

Let $a \in \N$ be an even perfect number.
Then $a$ is in the form:
:$2^{n - 1} \paren {2^n - 1}$
where $2^n - 1$ is prime.
Similarly, if $2^n - 1$ is prime, then $2^{n - 1} \paren {2^n - 1}$ is perfect.

Proof:
=== Sufficient Condition ===
{{:Theorem of Even Perfect Numbers/Sufficient Condition}}
{{qed|lemma}}
Theorem: Composite Number has Prime Factor not Greater Than its Square Root

Let $n \in \N$ and $n = p_1 \times p_2 \times \cdots \times p_j$, $j \ge 2$, where $p_1, \ldots, p_j \in \Bbb P$ are prime factors of $n$.
Then $\exists p_i \in \Bbb P$ such that $p_i \le \sqrt n$.
That is, if $n \in \N$ is composite, then $n$ has a prime factor $p \le \sqrt n$.

Proof:
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}}
Theorem: Sum Over Divisors Equals Sum Over Quotients

Let $n$ be a positive integer.
Let $f: \Z_{>0} \to \Z_{>0}$ be a mapping on the positive integers.
Let $\ds \sum_{d \mathop \divides n} \map f d$ be the sum of $\map f d$ over the divisors of $n$.
Then:
:$\ds \sum_{d \mathop \divides n} \map f d = \sum_{d \mathop \divides n} \map f {\frac n d}$.

Proof:
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
g4on4xtf4sx8yug9tw38o0sepakax86
Theorem: Sum of Reciprocals of Divisors equals Abundancy Index

Let $n$ be a positive integer.
Let $\map {\sigma_1} n$ denote the divisor sum function of $n$.
Then:
:$\ds \sum_{d \mathop \divides n} \frac 1 d = \frac {\map {\sigma_1} n} n$
where $\dfrac {\map {\sigma_1} n} n$ is the abundancy index of $n$.

Proof:
{{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_1} n} n
      | c = {{Defof|Divisor Sum Function}}
}}
{{end-eqn}}
{{qed}}
Category:Number Theory
Category:Abundancy
l720rbauewbojjsdk61fzcuua44to8w
Theorem: Divisor Sum of Prime Number

Let $n$ be a positive integer.
Let $\map {\sigma_1} n$ be the divisor sum function of $n$.
Then $\map {\sigma_1} n = n + 1$ {{iff}} $n$ is prime.

Proof:
From Rule of Transposition, we may replace the ''only if'' statement by its contrapositive.
Therefore, the following suffices:
Theorem: Divisor Sum of Power of Prime

Let $n = p^k$ be the power of a prime number $p$.
Let $\map {\sigma_1} n$ be the divisor sum of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then:
:$\map {\sigma_1} n = \dfrac {p^{k + 1} - 1} {p - 1}$

Proof:
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:
:$\map {\sigma_1} {p^k} = 1 + p + p^2 + \cdots + p^{k - 1} + p^k = \dfrac {p^{k + 1} - 1} {p - 1}$
{{qed}}
Theorem: Divisor Count Function of Power of Prime

Let $n = p^k$ be the power of a prime number $p$.
Let $\map {\sigma_0} n$ be the divisor count function of $n$.
That is, let $\map {\sigma_0} n$ be the number of positive divisors of $n$.
Then:
:$\map {\sigma_0} n = k + 1$

Proof:
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}}
Theorem: Sum Over Divisors of Multiplicative Function

Let $f: \Z_{>0} \to \Z_{>0}$ be a multiplicative function.
Let $n \in \Z_{>0}$.
Let $\ds \sum_{d \mathop \divides n} \map f d$ be the sum over the divisors of $n$.
Then $\ds \map F n = \sum_{d \mathop \divides n} \map f d$ is also a multiplicative function.

Proof:
Let $\ds \map F n = \sum_{d \mathop \divides n} \map f d$.
Let $m, n \in \Z_{>0}: m \perp n$.
Then by definition:
:$\ds \map F {m n} = \sum_{d \mathop \divides m n} \map f d$
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:
:$\ds \map F {m n} = \sum_{r \mathop \divides m, \ s \mathop \divides n} \map f {r s}$
So, as $f$ is multiplicative:
:$\ds \map F {m n} = \sum_{r \mathop \divides m, \ s \mathop \divides n} \map f r \map f s$
But at the same time:
:$\ds \map F m \map F n = \paren {\sum_{r \mathop \divides m} \map f r} \paren {\sum_{s \mathop \divides n} \map f s}$
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}}
Theorem: Divisors of Product of Coprime Integers

Let $a \divides b c$, where $b \perp c$.
Then $\tuple {r, s}$ satisfying:
:$a = r s$, where $r \divides b$ and $s \divides c$
is unique up to absolute value with:
:$\size r = \gcd \set {a, b}$
:$\size s = \gcd \set {a, c}$

Proof:
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
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