avewright/proofwiki-math huggingface dataset:
Theorem: Set of Subgroups of Abelian Group form Subsemigroup of Power Structure

Let $\struct {G, \circ}$ be an abelian group.
Let $\struct {\powerset G, \circ_\PP}$ denote the power structure of $\struct {G, \circ}$.
Let $\SS$ be the set of all subgroups of $G$.
Then $\struct {\SS, \circ_\PP}$ is a subsemigroup of $\struct {\powerset G, \circ_\PP}$.

Proof:
From Power Structure of Semigroup is Semigroup:
:$\struct {\powerset S, \circ_\PP}$ is a semigroup.
Let $A$ and $B$ be arbitrary subgroups of $G$.
We need to show that $A \circ_\PP B$ is also a subgroup of $G$.
Let $x$ and $y$ be arbitrary elements of $A \circ_\PP B$.
Then:
{{begin-eqn}}
{{eqn | q = \exists a_x \in A, b_x \in B
      | l = x
      | r = a_x \circ b_x
      | c = 
}}
{{eqn | q = \exists a_y \in A, b_y \in B
      | l = y
      | r = a_y \circ b_y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x \circ y^{-1}
      | r = \paren {a_x \circ b_x} \circ \paren {a_y \circ b_y}^{-1}
      | c = 
}}
{{eqn | r = \paren {a_x \circ b_x} \circ \paren { {b_y}^{-1} \circ {a_y}^{-1} }
      | c = Inverse of Group Product
}}
{{eqn | r = \paren {a_x \circ {a_y}^{-1} } \circ \paren {b_x \circ {b_y}^{-1} }
      | c = {{Defof|Abelian Group}}: $\circ$ is commutative
}}
{{eqn | ll= \leadsto
      | q = \exists a \in A, b \in B
      | l = x \circ y^{-1}
      | r = a \circ b
      | c = where $a = a_x \circ {a_y}^{-1}$ and $b = b_x \circ {b_y}^{-1}$
}}
{{eqn | ll= \leadsto
      | l = x \circ y^{-1}
      | o = \in
      | r = A \circ_\PP B
}}
{{end-eqn}}
Hence the result from the One-Step Subgroup Test.
{{qed}}
Theorem: Lamé's Theorem/Lemma

Suppose it takes $n$ cycles around the Euclidean Algorithm to find $\gcd \set {a, b}$.
Then $\min \set {a, b} \ge F_{n + 2}$, where $F_n$ denotes the $n$th Fibonacci number.

Proof:
{{WLOG}}, suppose $a \ge b$.
Let $q_i, r_i$ be the quotients and remainders of each step of the Euclidean Algorithm, that is:
{{begin-eqn}}
{{eqn | l = a
      | r = q_1 b + r_1
      | c = 
}}
{{eqn | l = b
      | r = q_2 r_1 + r_2
      | c = 
}}
{{eqn | l = r_1
      | r = q_3 r_2 + r_3
      | c = 
}}
{{eqn | l = \cdots
      | o = 
      | c = 
}}
{{eqn | l = r_{n - 2}
      | r = q_n r_{n - 1} + r_n
      | c = 
}}
{{eqn | l = r_{n - 1}
      | r = q_{n + 1} r_n + 0
      | c = 
}}
{{end-eqn}}
so $r_n = \gcd \set {a, b}$.
We prove that $r_{n - m} \ge F_{m + 2}$ for $0 \le m < n$ by the Principle of Mathematical Induction:
Theorem: Product of Subgroup with Inverse

Let $\struct {G, \circ}$ be a group.
Then:
:$\forall H \le \struct {G, \circ}:$
::$H^{-1} \circ H = H$
::$H \circ H^{-1} = H$
where $H \le G$ denotes that $H$ is a subgroup of $G$.

Proof:
From Inverse of Subgroup:
:$H = H^{-1}$
From Product of Subgroup with Itself:
:$H \circ H = H$
The result follows.
{{qed}}
Theorem: Equivalence Relation inducing Closed Quotient Set of Magma is Congruence Relation

Let $\struct {S, \circ}$ be a magma.
Let $\circ_\PP$ be the operation induced by $\circ$ on $\powerset S$, the power set of $S$.
Let $\RR$ be an equivalence relation on $S$.
Let $S / \RR$ denote the quotient set of $S$ induced by $\RR$.
Let the algebraic structure $\struct {S / \RR, \circ_\PP}$ be closed.
Then:
:$\RR$ is a congruence relation for $\circ$
and:
:the operation $\circ_\RR$ induced on $S / \RR$ by $\circ$ is the operation induced on $S / \RR$ by $\circ_\PP$.

Proof:
Let $x_1, y_1, x_2, y_2 \in S$ be arbitrary, such that:
{{begin-eqn}}
{{eqn | l = x_1
      | o = \RR
      | r = x_2
}}
{{eqn | l = y_1
      | o = \RR
      | r = y_2
}}
{{end-eqn}}
To demonstrate that $\RR$ is a congruence relation for $\circ$, we need to show that:
:$\paren {x_1 \circ y_1} \mathrel \RR \paren {x_2 \circ y_2}$
We have:
{{begin-eqn}}
{{eqn | l = x_1, x_2
      | o = \in
      | m = \eqclass {x_1} \RR
      | mo= \in
      | r =  S / \RR
}}
{{eqn | l = y_1, y_2
      | o = \in
      | m = \eqclass {x_1} \RR
      | mo= \in
      | r =  S / \RR
}}
{{end-eqn}}
Since $\struct {S / \RR, \circ_\PP}$ is closed:
:$\eqclass {x_1} \RR \circ_\PP \eqclass {y_1} \RR \in S / \RR$
From the definition of a quotient set:
:$\eqclass {x_1} \RR \circ_\PP \eqclass {y_1} \RR = \eqclass z \RR$ for some $z \in S$
From the definition of an operation induced on $\powerset S$:
:$\eqclass z \RR = \set {a \circ b: a \in \eqclass {x_1} \RR, b \in \eqclass {y_1} \RR}$
hence:
:$x_1 \circ y_1, x_2 \circ y_2 \in \eqclass z \RR$
From the definition of an equivalence class:
:$\paren {x_1 \circ y_1} \mathrel \RR \paren {x_2 \circ y_2}$
This shows that $\RR$ is a congruence relation for $\circ$.
We also have, by the equivalence of statements $(2)$ and $(4)$ in Equivalence Class Equivalent Statements:
:$\eqclass {x_1 \circ y_1} \RR = \eqclass z \RR = \eqclass {x_1} \RR \circ_\PP \eqclass {y_1} \RR$
This shows that the operation induced on $S / \RR$ by $\circ_\PP$ is the operation $\circ_\RR$ induced on $S / \RR$ by $\circ$.
{{qed}}
Theorem: Set of Normal Subgroups of Group is Subsemigroup of Power Set Semigroup

Let $\struct {G, \circ}$ be a group.
Let $\circ_\PP$ be the operation induced by $\circ$ on $\powerset G$, the power set of $G$.
Let $\HH$ be the set of all normal subgroups of $\struct {G, \circ}$.
Then the algebraic structure $\struct {\HH, \circ_\PP}$ is a subsemigroup of the algebraic structure $\struct {\powerset G, \circ_\PP}$.

Proof:
From Power Structure of Group is Semigroup, we have that $\struct {\powerset G, \circ_\PP}$ is a semigroup.
Note that $\HH \subseteq \powerset G$.
From Subset Product of Normal Subgroups is Normal, $\struct {\HH, \circ_\PP}$ is closed.
By Subsemigroup Closure Test, $\struct {\HH, \circ_\PP}$ is a subsemigroup of $\struct {\powerset G, \circ_\PP}$.
{{qed}}
Theorem: Set of Normal Subgroups of Group is Subsemigroup of Power Set under Intersection

Let $\struct {G, \circ}$ be a group.
Let $\HH$ be the set of all normal subgroups of $\struct {G, \circ}$.
Then the algebraic structure $\struct {\HH, \cap}$ is a subsemigroup of the algebraic structure $\struct {\powerset G, \cap}$.

Proof:
From Power Set with Intersection is Commutative Monoid, we have that $\struct {\powerset G, \cap}$ is ''a fortiori'' a semigroup.
Note that $\HH \subseteq \powerset G$.
Let $H_1$ and $H_2$ be normal subgroups of $\struct {G, \circ}$.
From Intersection of Normal Subgroups is Normal, $H_1 \cap H_2$ is also a normal subgroup of $\struct {G, \circ}$.
Hence $\struct {\HH, \cap}$ is closed.
By Subsemigroup Closure Test, $\struct {\HH, \cap}$ is a subsemigroup of $\struct {\powerset G, \cap}$.
{{qed}}
Theorem: Condition for Subgroup of Power Set of Group to be Quotient Group

Let $\struct {G, \circ}$ be a group.
Let $\circ_\PP$ be the operation induced by $\circ$ on $\powerset G$, the power set of $G$.
Let $\struct {\LL, \circ_\PP}$ be a subgroup of the algebraic structure $\struct {\powerset G, \circ_\PP}$.
Then:
:there exists a subgroup $H$ of $G$
:and a normal subgroup $K$ of $H$
:such that $\struct {\LL, \circ_\PP}$ is the quotient group $H / K$
{{iff}}:
:the identity element of $\struct {\LL, \circ_\PP}$ is a subgroup of $\struct {G, \circ}$.

Proof:
From Power Structure of Group is Semigroup, we have that $\struct {\powerset G, \circ_\PP}$ is a semigroup.
Theorem: Operation on Set for which Every Equivalence Relation is Congruence

Let $S$ be a set with at least $3$ elements.
Let $\circ$ be an operation on $S$ such that every equivalence relation on $S$ is a congruence relation for $\circ$.
Then $\circ$ is one of the following:
:the right operation $\to$
:the left operation $\gets$
:the constant operation $\sqbrk c$ for some $c \in S$.

Proof:
First we note that from:
:Equivalence Relation is Congruence for Constant Operation
:Equivalence Relation is Congruence for Left Operation
:Equivalence Relation is Congruence for Right Operation
every equivalence relation on $S$ is a congruence relation for the right operation, the left operation and the constant operation.
Let $\circ$ be an operation on $S$ which is ''not'' one of those three.
{{questionable|It is not a given that these three conditions hold at the same time. It also seems overly strong for the conclusion. But an immediate fix eludes me}}
Then there exists $x_1, x_2, y_1, y_2 \in S$ such that:
:$x_1 \circ y_1 = z_1$
:$x_2 \circ y_2 = z_2$
where:
:$z_1 \ne z_2$
:$\map \lnot {z_1 = x_1 \land z_2 = x_2}$
:$\map \lnot {z_1 = y_1 \land z_2 = y_2}$
The above are possible only if there are more than $3$ elements in $S$.
Let $\RR$ be an equivalence relation on $S$ such that:
:$x_1 \mathrel \RR x_2$
:$y_1 \mathrel \RR y_2$
but that:
:$\map \lnot {z_1 \mathrel \RR z_2}$
Then $\RR$ is not a congruence relation.
The result follows from the Rule of Transposition.
{{qed}}
Theorem: Condition for Cosets of Subgroup of Monoid to be Partition

Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.
Let $\struct {H, \circ}$ be a subgroup of $\struct {S, \circ}$.
Let the identity element of $\struct {H, \circ}$ be $e$.
Then:
:the set of left cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
and:
:the set of right cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$.

Proof:
Because of the fact that it is not necessarily the case that $x \in S$ has an inverse, we cannot invoke the result Left Congruence Modulo Subgroup is Equivalence Relation.
Instead we prove partitionhood directly.
First we recall the definition of left coset and right coset:
:The '''left coset of $H$ by $x$''', is:
::$x \circ H = \set {y \in S: \exists h \in H: y = x \circ h}$
:The '''right coset of $H$ by $x$''', is:
::$H \circ x = \set {y \in S: \exists h \in H: y = h \circ x}$
Next we recall the definition of partition:
:A '''partition''' of $S$ is a set of non-empty subsets $\Bbb S$ of $S$ such that each element of $S$ lies in exactly one element of $\Bbb S$.
Testing each of the criteria for a partition as follows:
Theorem: Condition for Subgroup of Monoid to be Normal

Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.
Let $\struct {H, \circ}$ be a subgroup of $\struct {S, \circ}$.
Then:
::the set of left cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
:and:
::the set of right cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
:such that the equivalence relations induced by those partitions are congruence relations for $\circ$
{{iff}}:
:$\struct {H, \circ}$ is a normal subgroup of $\struct {S, \circ}$.

Proof:
=== Necessary Condition ===
Let $\struct {H, \circ}$ be a normal subgroup of $\struct {S, \circ}$.
Then by definition:
:$e \in H$
Hence from Condition for Cosets of Subgroup of Monoid to be Partition, the set of left cosets and the set of right cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$.
{{ProofWanted}}
Theorem: Existence of One-Sided Inverses on Natural Numbers whose Composition is Identity Mapping

Consider the set of natural numbers $\N$.
There exist mappings $f: \N \to \N$ and $g: \N \to \N$ such that:
:$g \circ f = I_\N$
where:
:$\circ$ denotes composition of mappings
:$I_\N$ denotes the identity mapping on $\N$
such that neither $f$ nor $g$ are permutations on $\N$.

Proof:
Let $f: \N \to \N$ be the mapping defined as:
:$\forall x \in \N: \map f x = x + 1$
Let $g: \N \to \N$ be the mapping defined as:
:$\forall x \in \N: \map g x = \begin {cases} x - 1 & : x > 0 \\ 0 & : x = 0 \end {cases}$
It is apparent by inspection that:
:$f$ is injective
but
:$f$ is not surjective, as there exists no $x \in \N$ such that $x + 1 = 0$
It is also apparent by inspection that:
:$g$ is surjective
but
:$g$ is not injective, as $\map g 0 = \map g 1 = 0$
Thus neither $f$ nor $g$ are bijective, and hence not permutations on $\N$.
However, it is also apparent by inspection that:
:$g \circ f = I_\N$
{{qed}}
Theorem: Condition for Partition between Invertible and Non-Invertible Elements to induce Congruence Relation on Monoid/Counterexample

Let $\circ$ be such that it is neither a cancellable operation nor a commutative operation.
Then it is not necessarily the case that either:
:$\RR$ is a congruence relation on $\circ$
or:
:the quotient structure $\struct {S / \RR, \circ_\RR}$ is isomorphic to $\struct {\Z_2, \times_2}$, the multiplicative monoid of integers modulo $2$.

Proof:
Consider the set of natural numbers $\N$.
Let $\N^\N$ denote the set of all mappings from $\N$ to $\N$.
Now consider the algebraic structure $\struct {\N^\N, \circ}$, where $\circ$ in this context denotes composition of mappings.
From Set of all Self-Maps under Composition forms Monoid, $\struct {\N^\N, \circ}$ is a monoid.
From Composition of Mappings is not Commutative, $\circ$ is not a commutative operation.
Let $g: \N \to \N$ be the mapping defined as:
:$\forall x \in \N: \map g x = \begin {cases} x - 1 & : x > 0 \\ 0 & : x = 0 \end {cases}$
$g$ is not injective, as:
:$\map g 0 = \map g 1 = 0$
From Injection iff Left Cancellable, $g$ is not cancellable.
Thus we have that $\circ$ is not a cancellable operation.
So we have a monoid $\struct {\N^\N, \circ}$ whose operation is neither cancellable nor commutative.
From Existence of One-Sided Inverses on Natural Numbers whose Composition is Identity Mapping, there exist mappings $f: \N \to \N$ and $g: \N \to \N$ such that:
:$g \circ f = I_\N$
such that:
:$g$ is surjective but not injective
:$f$ is injective but not surjective.
By definition of a bijection, neither $f$ nor $g$ are invertible elements of $\struct {\N^\N, \circ}$ since they are not bijective.
Hence we have $f, g \in N$ while $g \circ f = I_\N \in H$.
Also note that $f \circ f$ is not surjective, so $f \circ f \in N$.
By definition of $\RR$:
:$x \mathrel \RR y \iff \set {x, y} \subseteq N \lor \set {x, y} \subseteq H$
We now have:
:$g \mathrel \RR f \land f \mathrel \RR f$
:$\neg \paren {\paren {g \circ f} \mathrel \RR \paren {f \circ f}}$
thus $\RR$ is not a congruence relation, and the quotient structure $\struct {S / \RR, \circ_\RR}$ is not well-defined.
{{qed}}
Theorem: Condition on Congruence Relations for Cancellable Monoid to be Group

Let $\struct {S, \circ}$ be a cancellable monoid whose identity element is $e$.
Then:
:$\struct {S, \circ}$ is a group
{{iff}}:
:every non-trivial congruence relation on $\struct {S, \circ}$ is induced by a normal subgroup of $\struct {S, \circ}$.

Proof:
=== Necessary Condition ===
Let $\struct {S, \circ}$ be such that every non-trivial congruence relation on $\struct {S, \circ}$ is induced by a normal subgroup of $\struct {S, \circ}$.
Hence, let $\RR$ be an arbitrary non-trivial congruence relation.
From Condition for Subgroup of Monoid to be Normal, there exists a normal subgroup $\struct {H, \circ}$ such that:
:the set of left cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
and:
:the set of right cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
such that the equivalence relations induced by those partitions is $\RR$.
By the definition of normal subgroup, the set of left cosets is the same as the set of right cosets.
It remains to be shown that $\struct {S, \circ}$ is a group.
We already have that:
:{{Group-axiom|0}}
:{{Group-axiom|1}}
:{{Group-axiom|2}}
are {{apriori}} satisfied by the fact that $\struct {S, \circ}$ is a monoid.
Hence it remains to prove {{Group-axiom|3}}.
{{ProofWanted|The suggestion is to use Condition for Partition between Invertible and Non-Invertible Elements to induce Congruence Relation on Monoid}}
Theorem: Condition on Congruence Relations for Cancellable Monoid to be Group/Counterexample

Let $\struct {S, \circ}$ be a monoid which is not cancellable.
Let every non-trivial congruence relation on $\struct {S, \circ}$ be induced by a normal subgroup of $\struct {S, \circ}$.
Then it is not necessarily the case that $\struct {S, \circ}$ is a group.

Proof:
{{Improve|Is all this necessary, considering we have Group is Cancellable Monoid?}}
Consider the Multiplicative Monoid of Integers Modulo $3$ $\struct {\Z_3, \times_3}$, defined by its Cayley table:
{{:Modulo Multiplication/Cayley Table/Modulo 3}}
It is noted that $\struct {\Z_3, \times_3}$ is not a cancellable monoid.
Indeed:
:$0 \times_3 1 = 0$
:$0 \times_3 2 = 0$
and so $\times_3$ is not cancellable for $0$.
Hence from Group is Cancellable Monoid:
:$\struct {\Z_3, \times_3}$ is not a group.
We note that $\struct {\Z_3, \times_3}$ has an identity element $\eqclass 1 3$.
We list all the equivalence relations on $\struct {\Z_3, \times_3}$ by the partitions they induce:
:$\set {\set 0, \set 1, \set 2}$ is induced by the diagonal relation
:$\set {\set {0, 1, 2}}$ is induced by the trivial relation
:$\set {\set {0, 1}, \set 2}$ is not a congruence relation because we have $0 \mathrel \RR 1$ and $2 \mathrel \RR 2$ but not $\paren {0 \times 2} \mathrel \RR \paren {1 \times 2}$
:$\set {\set {0, 2}, \set 1}$ is not a congruence relation because we have $0 \mathrel \RR 2$ and $2 \mathrel \RR 2$ but not $\paren {0 \times 2} \mathrel \RR \paren {2 \times 2}$
:$\set {\set 0, \set {1, 2}}$ is induced by the equivalence relation induced by the normal subgroup $\set 0$.
{{explain|Can a normal subgroup induce an equivalence relation when $S$ is not cancellable, since all cosets of $\set 0$ is $\set 0$?}}
We see that the first four relations are not non-trivial congruence relation on $\struct {S, \circ}$.
Therefore every non-trivial congruence relation on $\struct {S, \circ}$ is induced by a normal subgroup of $\struct {S, \circ}$.
But, {{apriori}}, $\struct {\Z_3, \times_3}$ is not a group.
{{qed}}
Theorem: Riesz-Markov-Kakutani Representation Theorem

Let $\struct {X, \tau}$ be a locally compact Hausdorff space.
Let $\map {C_c} X$ be the space of continuous complex functions with compact support on $X$.
Let $\Lambda$ be a positive linear functional on $\map {C_c} X$.
There exists a $\sigma$-algebra $\MM$ over $X$ which contains the Borel $\sigma$-algebra of $\struct {X, \tau}$.
There exists a unique complete Radon measure $\mu$ on $\MM$ such that:
:$\ds \forall f \in \map {C_c} X: \Lambda f = \int_X f \rd \mu$

Proof:
=== Lemma $1$ ===
{{:Riesz-Markov-Kakutani Representation Theorem/Lemma 1}}{{qed|lemma}}
Theorem: Even Integers not Sum of Two Abundant Numbers

The even integers which are not the sum of $2$ abundant numbers are:
:All even integers less than $24$;
:$26, 28, 34, 46$

Proof:
From Sequence of Abundant Numbers, the first few abundant numbers are:
:$12, 18, 20, 24, 30, 36, 40, 42, 48$
Immediately we see that any number less than $2 \times 12 = 24$ cannot be expressed as a sum of $2$ abundant numbers.
The sum of the first $2$ abundant numbers is $12 + 18 = 30$, so $26$ and $28$ are not sums of $2$ abundant numbers.
{{begin-eqn}}
{{eqn | l = 34 - 12
      | r = 22
}}
{{eqn | l = 34 - 18
      | o = <
      | r = \frac {34} 2
}}
{{eqn | l = 46 - 12
      | r = 34
}}
{{eqn | l = 46 - 18
      | r = 28
}}
{{eqn | l = 46 - 20
      | r = 26
}}
{{eqn | l = 46 - 24
      | o = <
      | r = \frac {46} 2
}}   
{{end-eqn}}
Since none of the differences above are abundant numbers, $34$ and $46$ are not sums of $2$ abundant numbers.
We demonstrate that $32$ and all even numbers from $36$ to $66$ except $46$ are sums of $2$ abundant numbers:
{{begin-eqn}}
{{eqn | l = 32
      | r = 12 + 20
}}
{{eqn | l = 38
      | r = 18 + 20
}}
{{eqn | l = 40
      | r = 20 + 20
}}
{{eqn | l = 44
      | r = 20 + 24
}}
{{eqn | l = 50
      | r = 20 + 30
}}
{{eqn | l = 52
      | r = 12 + 40
}}
{{eqn | l = 56
      | r = 20 + 36
}}
{{eqn | l = 58
      | r = 18 + 40
}}
{{eqn | l = 62
      | r = 20 + 42
}}
{{eqn | l = 64
      | r = 24 + 40
}}
{{end-eqn}}
The numbers $36, 42, 48, 54, 60$ and $66$ are multiples of $6$.
By Multiple of Perfect Number is Abundant, any multiple of $6$ greater than $6$ is abundant.
Hence these numbers can be expressed as:
:$12 + \paren {n - 12}$
which are sums of $2$ multiples of $6$ greater than $6$.
Now that we show that all even numbers greater than $66$ are sums of $2$ abundant numbers.
By Multiple of Abundant Number is Abundant, any multiple of $20$ is abundant.
By Largest Number not Expressible as Sum of Multiples of Coprime Integers/Generalization, any even number greater than:
:$\dfrac {6 \times 20} {\gcd \set {6, 20}} - 6 - 20 = 34$
is a sum of (possibly zero) multiples of $6$ and $20$.
Hence any even number greater than:
:$34 + 6 \times 2 + 20 = 66$
is a sum of a multiple of $6$ greater than $6$ and a non-zero multiple of $20$, which by above is a sum of $2$ abundant numbers.
This shows that the list above is complete.
{{qed}}
Theorem: Sum of Reciprocals of Sequence of Pairs of Odd Index Consecutive Fibonacci Numbers is Reciprocal of Golden Mean

{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 1} \dfrac 1 {F_{2 k - 1} F_{2 k + 1} }
      | r = \dfrac 1 {1 \times 2} + \dfrac 1 {2 \times 5} + \dfrac 1 {5 \times 13} + \dfrac 1 {13 \times 34} + \cdots
      | c = 
}}
{{eqn | r = \phi^{-1}
      | c = 
}}
{{end-eqn}}
where:
:$F_k$ denotes the $k$th Fibonacci number
:$\phi$ denotes the golden mean.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 1} \dfrac 1 {F_{2 k - 1} F_{2 k + 1} }
      | r = \sum_{k \mathop \ge 1} \dfrac 1 {F_{2 k - 1} F_{2 k + 1} } \paren {\dfrac {F_{2 k + 1} - F_{2 k - 1} } {F_{2 k + 1} - F_{2 k - 1} } }
      | c = multiplying by $1$
}}
{{eqn | r = \sum_{k \mathop \ge 1} \paren {\dfrac 1 {F_{2 k - 1} } - \dfrac 1 {F_{2 k + 1} } } \paren {\dfrac 1 {F_{2 k + 1} - F_{2 k - 1} } }
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 1} \paren {\dfrac 1 {F_{2 k - 1} } - \dfrac 1 {F_{2 k + 1} } } \paren {\dfrac 1 {\paren {F_{2 k} + F_{2 k - 1} } - F_{2 k - 1} } }
      | c = {{Defof|Fibonacci Number}}
}}
{{eqn | r = \sum_{k \mathop \ge 1} \paren {\dfrac 1 {F_{2 k - 1} } - \dfrac 1 {F_{2 k + 1} } } \paren {\dfrac 1 {F_{2 k} } }
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 1} \paren {\dfrac 1 {F_{2 k - 1} F_{2 k} } - \dfrac 1 {F_{2 k} F_{2 k + 1} } }
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 1} \paren {-1}^{k + 1} \dfrac 1 {F_k F_{k + 1} }
      | c = 
}}
{{eqn | r = \dfrac 1 {1 \times 1} - \sum_{k \mathop \ge 2} \paren {-1}^k \dfrac 1 {F_k F_{k + 1} }
      | c = {{Defof|Fibonacci Number}}
}}
{{eqn | r = 1 - \phi^{-2}
      | c = Sum of Alternating Sign Reciprocals of Sequence of Pairs of Consecutive Fibonacci Numbers is Reciprocal of Golden Mean Squared
}}
{{eqn | r = \phi^{-1}
      | c = Power of Golden Mean as Sum of Smaller Powers: $z = 0$
}}
{{end-eqn}}
{{qed}}
Theorem: Condition for Equivalence Relation for Max Operation on Natural Numbers to be Congruence

Let $\RR$ be an equivalence relation on the set of natural numbers $\N$.
Let $\vee$ denote the max operation on $\N$:
:$\forall a, b \in \N: a \vee b := \max \set {a, b}$
Then:
:$\RR$ is a congruence relation for $\vee$ on $\N$
{{iff}}:
:every equivalence class under $\RR$ is an order-convex subset of $\N$.

Proof:
=== Necessary Condition ===
Suppose every equivalence class under $\RR$ is a order-convex subset of $\N$.
Let $x_1, x_2, y_1, y_2 \in \N$ such that $x_1 \mathrel \RR x_2$ and $y_1 \mathrel \RR y_2$.
{{WLOG}}, suppose:
:$x_1 \le x_2$
:$y_1 \le y_2$
:$x_1 \le y_1$
Then we have two cases: