nyuuzyou/wikis huggingface dataset:
Similarity of Polygons is Equivalence Relation

== Theorem ==
Let $A, B, C$ be polygons.
If $A$ and $B$ are both similar to $C$, then $A$ is similar to $B$.
It is also worth noting that:
:$A$ is similar to $A$, and so similarity between polygons is reflexive.
:If $A$ is similar to $B$, then $B$ is similar to $A$, and so similarity between polygons is symmetric.
Hence the relation of similarity between polygons is an equivalence relation.
== Proof ==
:
We have that $A$ is similar to $C$.
From , it is equiangular with it and the sides about the equal angles are proportional.
We also have that $B$ is similar to $C$.
Again, from , it is equiangular with it and the sides about the equal angles are proportional.
So by definition $A$ is similar to $B$.
The statements of reflexivity and symmetry are shown similarly.
It follows that if $A$ is similar to $B$, and $B$ is similar to $C$, then $A$ is similar to $C$.
Thus similarity between polygons is transitive.
Hence the result, by definition of equivalence relation.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Polygons
Category:Examples of Equivalence Relations
Simple Algebraic Field Extension consists of Polynomials in Algebraic Number

== Theorem ==
Let $F$ be a field.
Let $\theta \in \C$ be algebraic over $F$.
Let $\map F \theta$ be the simple field extension of $F$ by $\theta$.
Then $\map F \theta$ consists of polynomials that can be written in the form $\map f \theta$, where $\map f x$ is a polynomial over $F$.
== Proof ==
Let $H$ be the set of all numbers which can be written in the form $\map f \theta$.
We have that:
:$H$ is closed under addition and multiplication.
:$H$ contains $0$ and $1$
:For every element of $H$, $H$ also contains its negative.
Let $\map f \theta \ne 0$.
Then $\theta$ is not a root of $\map f x$.
Hence from Polynomial with Algebraic Number as Root is Multiple of Minimal Polynomial:
:the minimal polynomial $\map m x$ in $\theta$ does not divide $\map f x$.
From Minimal Polynomial is Irreducible, the GCD of $\map m x$ and $\map f x$ is $1$.
Therefore:
:$\exists \map s x, \map t x: \map s x \map m x + \map t x \map f x = 1$
Substituting for $\theta$:
:$\map s \theta \, \map m \theta + \map t \theta \, \map f \theta = 1$
Because $\map m \theta = 0$ it follows that:
:$\map t \theta \, \map f \theta = 1$
We have that $\map t \theta \in H$.
Thus $\map t \theta$ is the product inverse of $\map f x$ in $H$.
Thus $H$ is a field.
A field containing $F$ and $\theta$ must contain $1$ and all the powers of $\theta$ for positive integer index.
Hence such a field also contains all linear combinations of these, with coefficients in $F$.
So a field containing $F$ and $\theta$ contains all the elements of $H$:
:$H \subseteq \map F \theta$
But by definition, $\map F \theta$ is the smallest field containing $F$ and $\theta$.
That is:
:$\map F \theta \subseteq H$
Thus:
:$\map F \theta = H$
and the result follows.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Theorem $72$
Category:Field Extensions
Simple Connectedness is Preserved under Homeomorphism

== Theorem ==
Let $\struct {T_1, \tau_1}, \struct {T_2, \tau_2}$ be topological spaces.
Let $\phi: T_1 \to T_2$ be a homeomorphism.
Let $S \subseteq T_1$ be a subset of $T_1$.
Let $S$ be simply connected in $\struct {T_1, \tau_1}$.
Then $\phi \sqbrk S$ is simply connected in $\struct {T_2, \tau_2}$.
That is, simple connectedness is a topological property.
== Proof ==
By , $S$ is path-connected.
Path-Connectedness is Preserved under Homeomorphism shows that $\phi \sqbrk S$ is path-connected.
Let $x \in S$.
Let:
:$\gamma_1 : \closedint 0 1 \to \phi \sqbrk S$
:$\gamma_2 : \closedint 0 1 \to \phi \sqbrk S$
be two loops in $\phi \sqbrk S$.
Let $\gamma_1, \gamma_2$ both have base point $x$.
Composite of Continuous Mappings is Continuous shows that $\phi^{-1} \circ \gamma_1$ and $\phi^{-1} \circ \gamma_2$ are continuous mappings.
So $\phi^{-1} \circ \gamma_1 , \phi^{-1} \circ \gamma_2$ are loops in $S$ with base point $\map {\phi^{-1} } x$.
By , there exists a path homotopy $H: \closedint 0 1 \times \closedint 0 1 \to S$ such that:
Composite of Continuous Mappings is Continuous shows that $ \phi \circ H : \closedint 0 1 \times \closedint 0 1 \to \phi \sqbrk S$ is a continuous mapping.
Then:
It follows that $\phi \circ H$ is a path homotopy between $\gamma_1$ and $\gamma_2$.
By , $\phi \sqbrk S$ is simply connected.
== Sources ==
* : $\S 52$
Category:Simply Connected Spaces
Category:Topological Properties
Simple Events are Mutually Exclusive

== Theorem ==
Let $\EE$ be an experiment.
Let $e_1$ and $e_2$ be distinct simple events in $\EE$.
Then $e_1$ and $e_2$ are mutually exclusive.
== Proof ==
By definition of simple event:
for some elementary events $s_1$ and $s_2$ of $\EE$ such that $s_1 \ne s_2$.
It follows that:
The result follows by definition of mutually exclusive events.
== Sources ==
* : Chapter $2$: Probability and Discrete Sample Spaces: $2.2$ Sample spaces and events
Category:Events
Simple Finite Continued Fraction has Rational Value

== Theorem ==
Let $n \ge 0$ be a natural number.
Let $\tuple {a_0, \ldots, a_n}$ be a simple finite continued fraction‎ of length $n$.
Then its value $\sqbrk {a_0, \ldots, a_n}$ is a rational number.
== Proof ==
This will be proved by induction on the number of partial denominators.
For all $n \in \N$, let $\map P n$ be the proposition that the continued fraction $\sqbrk {a_0, a_1, \ldots, a_n}$ has a rational value.
=== Basis for the Induction ===
$\map P 0$ is true, as $\sqbrk {a_0}$ is an integer and therefore rational.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
The continued fraction $\sqbrk {a_0, a_1, \ldots, a_k}$ has a rational value.
Then we need to show that the continued fraction $\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} }$ also has a rational value.
=== Induction Step ===
Consider the continued fraction $\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} }$.
By definition of value of a finite continued fraction:
:$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = a_0 + \dfrac 1 {\sqbrk {a_1, a_2, \ldots, a_k, a_{k + 1} } }$
$a_0$ is an integer, as we have seen.
By the induction hypothesis, $\sqbrk {a_1, a_2, \ldots, a_k, a_{k + 1} }$ is a rational number, and so is its reciprocal.
Hence as $a_0 + \dfrac 1 {\sqbrk {a_1, a_2, \ldots, a_k, a_{k + 1} } }$ is the sum of two rational numbers, it is itself rational.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore for all $n \in \N$, the continued fraction $\sqbrk {a_0, a_1, \ldots, a_n}$ has a rational value.
== Also see ==
* Correspondence between Rational Numbers and Simple Finite Continued Fractions
Category:Simple Continued Fractions
Category:Rational Numbers
Simple Finite Continued Fraction is Almost Determined by Value

== Theorem ==
Let $n,m \geq 0$ be natural number.
Let $\sequence {a_k}_{0 \mathop \le k \mathop \le m}$ and $\sequence {b_k}_{0 \mathop \le k \mathop \le n}$ be simple finite continued fractions in $\R$.
Let $\sequence {a_k}_{0 \mathop \le k \mathop \le m}$ and $\sequence {b_k}_{0 \mathop \le k \mathop \le n}$ have the same value.
Then either:
*$n = m$, and the sequences are equal.
*$n = m + 1$, $a_k = b_k$ for $k < m$, $a_m = b_m-1$ and $b_{m + 1} = 1$
*$m = n + 1$, $a_k = b_k$ for $k < n$, $b_n = a_n-1$ and $a_{n + 1} = 1$
== Proof ==
== Also see ==
* Simple Infinite Continued Fraction is Uniquely Determined by Limit
Category:Simple Continued Fractions
Simple Function is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a simple function.
Then $f$ is $\Sigma$-measurable.
== Proof ==
Let $f$ be written in the following form:
:$f = \ds \sum_{i \mathop = 1}^n a_i \chi_{S_i}$
where $a_i \in \R$ and the $S_i$ are $\Sigma$-measurable.
Next, for each ordered $n$-tuple $b$ of zeroes and ones define:
:$\map {T_b} i := \begin{cases}
S_i & : \text {if $\map b i = 0$}\\
X \setminus S_i & : \text {if $\map b i = 1$}
\end{cases}$
and subsequently:
:$T_b := \ds \bigcap_{i \mathop = 1}^n \map {T_b} i$
From Sigma-Algebra Closed under Intersection, $T_b \in \Sigma$ for all $b$.
Also, the $T_b$ are pairwise disjoint, and furthermore:
:$f = \ds \sum_b a_b \chi_{T_b}$
where:
:$a_b := \ds \sum_{i \mathop = 1}^n \map b i a_i$
Now we have, for all $\lambda \in \R$:
:$\set {x \in X: \map f x > \lambda} = \ds \bigcup \set {T_b: a_b > \lambda}$
which by Sigma-Algebra Closed under Union is a $\Sigma$-measurable set.
From Characterization of Measurable Functions: $(5)$ it follows that $f$ is measurable.
== Sources ==
* : $8.5 \ \text{(ii)}$
Category:Measurable Functions
Category:Simple Functions
Simple Graph/Examples/Arbitrary Order 4

== Example of Simple Graph ==
Let $V = \set {v_1, v_2, v_3, v_4}$.
Let $\RR = \set {\tuple {v_1, v_2}, \tuple {v_1, v_3}, \tuple {v_2, v_1}, \tuple {v_2, v_3}, \tuple {v_3, v_1}, \tuple {v_3, v_2}, \tuple {v_3, v_4}, \tuple {v_4, v_3} }$.
Then:
:$E = \set {\set {\tuple {v_1, v_2}, \tuple {v_2, v_1} }, \set {\tuple {v_1, v_3}, \tuple {v_3, v_1} }, \set {\tuple {v_2, v_3}, \tuple {v_3, v_2} }, \set {\tuple {v_3, v_4}, \tuple {v_4, v_3} } }$
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs
Category:Examples of Simple Graphs
Simple Graph/Examples/Arbitrary Order 5

== Example of Simple Graph ==
Let $G = \struct {V, E}$ be a simple graph such that:
:$V = \set {v_1, v_2, v_3, v_4, v_5}$
:$E = \set {v_1 v_2, v_1 v_4, v_1 v_5, v_2 v_3, v_3 v_5, v_4 v_5}$
Then $G$ can be presented in diagram form as:
:
The underlying relation $\RR$ on $V$ which defines the edge set of $G$ is:
:$\RR = \set {\tuple {v_1, v_2}, \tuple {v_2, v_1}, \tuple {v_1, v_4}, \tuple {v_4, v_1}, \tuple {v_1, v_5}, \tuple {v_5, v_1}, \tuple {v_2, v_3}, \tuple {v_3, v_2}, \tuple {v_3, v_5}, \tuple {v_5, v_3}, \tuple {v_4, v_5}, \tuple {v_5, v_4} }$
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $15$
Category:Examples of Simple Graphs
Simple Graph of Maximum Size is Complete Graph

== Theorem ==
Let $G$ be a simple graph of order $n$ such that $n \ge 1$.
Let $G$ have the largest size of all simple graphs of order $n$.
Then:
:$G$ is the complete graph $K_n$
:its size is $\dfrac {n \paren {n - 1} } 2$.
== Proof ==
By definition, $K_n$ is the simple graph of order $n$ such that every vertex of $K_n$ is adjacent to all other vertices.
So, let $G$ have the largest size of all simple graphs of order $n$.
Then by definition of largest size, it is not possible for another edge to be added to $G$.
The only way that could be is if all the vertices of $G$ are already adjacent to all other vertices.
Hence $G = K_n$ by definition.
The size of $G$ then follows from Size of Complete Graph.
== Examples ==
== Also see ==
* Size of Complete Graph
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $18 \ \text {(d)}$
Category:Complete Graphs
Category:Simple Graphs
Category:Simple Graph of Maximum Size is Complete Graph
Simple Graph of Maximum Size is Complete Graph/Examples/Order 3

== Examples of Use of Simple Graph of Maximum Size is Complete Graph ==
Let $G$ be the simple graph of order $3$ whose edge set $E$ is as large as possible.
Then the size of $G$ is given by:
:$\size E = 3$
== Proof ==
By Simple Graph of Maximum Size is Complete Graph:
:$G = K_3$
From Size of Complete Graph:
:$\size E = \dfrac {3 \times \paren {3 - 1} } 2 = \dfrac 6 2 = 3$
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $18 \ \text {(a)}$
Category:Simple Graph of Maximum Size is Complete Graph
Simple Graph of Maximum Size is Complete Graph/Examples/Order 4

== Examples of Use of Simple Graph of Maximum Size is Complete Graph ==
Let $G$ be the simple graph of order $4$ whose edge set $E$ is as large as possible.
Then the size of $G$ is given by:
:$\size E = 6$
== Proof ==
By Simple Graph of Maximum Size is Complete Graph:
:$G = K_4$
From Size of Complete Graph:
:$\size E = \dfrac {4 \times \paren {4 - 1} } 2 = \dfrac {12} 2 = 6$
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $18 \ \text {(b)}$
Category:Simple Graph of Maximum Size is Complete Graph
Simple Graph of Maximum Size is Complete Graph/Examples/Order 5

== Examples of Use of Simple Graph of Maximum Size is Complete Graph ==
Let $G$ be the simple graph of maximum size whose vertex set is:
:$V = \set {v_1, v_2, v_3, v_4, v_5}$
Then the edge set $E$ such that $\size E$ is as large as possible is:
:$E = \set {v_1 v_2, v_1 v_3, v_1 v_4, v_1 v_5, v_2 v_3, v_2 v_4, v_2 v_5, v_3 v_4, v_3 v_5, v_4 v_5}$
Thus:
:$\size E = 10$
and it is seen by inspection that $G = K_5$, that is, the complete graph of order $5$.
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $18 \ \text {(c)}$
Category:Simple Graph of Maximum Size is Complete Graph
Simple Graph where All Vertices and All Edges are Adjacent

== Theorem ==
Let $G$ be a simple graph in which:
:every vertex is adjacent to every other vertex
and:
:every edge is adjacent to every other edge.
Then $G$ is of order no greater than $3$.
== Proof ==
It is seen that examples exist of simple graphs which fulfil the criteria where the order of $G$ is no greater than $3$:
:
The cases where the order of $G$ is $1$ or $2$ are trivial.
When the order of $G$ is $3$, the criteria can be verified by inspection.
Let the order of $G = \struct {V, E}$ be $4$ or more.
Let $v_1, v_2, v_3, v_4 \in V$.
Suppose every vertex is adjacent to every other vertex.
As $v_1$ is adjacent to $v_2$, there exists the edge $v_1 v_2$.
As $v_3$ is adjacent to $v_4$, there exists the edge $v_3 v_4$.
But $v_1 v_2$ and $v_3 v_4$ both join two distinct pairs of vertices.
Thus $v_1 v_2$ and $v_3 v_4$ are not adjacent, by definition.
So when there are $4$ or more vertices in $G$, it cannot fulfil the criteria.
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $19$
Category:Simple Graphs
Simple Graph whose Vertices Incident to All Edges

== Theorem ==
Let $G = \struct {V, E}$ be a simple graph whose vertices are incident to all its edges.
Then $G$ is either:
:the star graph $S_2$, which is also the complete graph $K_2$
:an edgeless graph of any order.
== Proof ==
If $G$ has no edges, then all the vertices are incident to all the edges vacuously.
So any of the edgeless graphs $N_n$ for order $n \in \Z_{\ge 0}$ fulfils the criterion.
Suppose $G$ has more than $2$ vertices $v_1, v_2, v_3$ and at least one edge.
, let one edge be $v_1 v_2$.
But $v_3$ cannot be incident to edge $v_1 v_2$.
So $G$ can have no more than $2$ vertices.
Furthermore, there can be only one edge joining those two vertices.
The result follows from inspection of $K_2$ and $S_2$.
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $22$
Category:Graph Theory
Simple Graph whose Vertices all Incident but Edges not Adjacent

== Theorem ==
Let $G = \struct {V, E}$ be a simple graph such that:
:every vertex is incident with at least one edge
:no two edges are adjacent to each other.
Then $G$ has an even number of vertices.
== Proof ==
Suppose there exists a set of $3$ vertices that are connected.
Then at least one of these vertices has at least $2$ edges.
That would mean that at least $2$ edges were incident with the same vertex.
That is, that at least $2$ edges were adjacent to each other.
So, for a simple graph to fulfil the conditions, vertices can exist only in $2$-vertex components.
So such a simple graph must have an even number of vertices.
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $20$
Category:Simple Graphs
Simple Graph with Finite Vertex Set is Finite

== Theorem ==
Let $G$ be a simple graph.
Suppose that the vertex set of $G$ is finite.
Then $G$ is a finite graph.
That is to say, its edge set is also finite.
== Proof ==
Since $G$ is simple, it can have at most one edge between each two vertices.
Therefore, the mapping:
:$v: \map E G \to \powerset {\map V G}, \map v e = \set {a_e, b_e}$
which assigns to each edge its endvertices $a_e, b_e$, is an injection.
From Power Set of Finite Set is Finite and Domain of Injection Not Larger than Codomain, it follows that $\map E G$ is finite.
Hence $G$ is finite.
Category:Graph Theory
Category:Simple Graphs
Simple Graph with no Circuits and Size One Less than Order is Connected

== Theorem ==
Let $G$ be a simple graph of order $n$ such that $G$ has no circuits.
Let the size of $G$ be $n - 1$.
Then $T$ is connected.
== Proof ==
 $G$ is not connected.
Then $G$ has at least $2$ components.
By the Pigeonhole Principle, at least one of those components has at least as many edges as vertices.
Hence by Finite Connected Simple Graph is Tree iff Size is One Less than Order, that component is not a tree.
Hence as that component is itself connected but not a tree by definition it has at least one circuit.
But $G$ has no circuits.
The result follows by Proof by Contradiction.
Category:Simple Graphs
Category:Connectedness (Graph Theory)
Simple Group of Order Less than 60 is Prime

== Theorem ==
Let $G$ be a simple group.
Let $\order G < 60$, where $\order G$ denotes the order of $G$.
Then $G$ is a prime group.
== Proof ==
First it is noted that Prime Group is Simple.
We also note from Alternating Group is Simple except on 4 Letters that the alternating group $A_5$, which is of order $60$, is simple.
Hence the motivation for the result.
It remains to be shown that all groups of composite order such that $\order G < 60$ are not simple.
Let $S$ be the set:
:$S = \set {n \in \Z: 0 < n < 60: \text { there exists a simple group of order $n$ such that $n$ is composite} }$
The aim is to show that $S$ is empty.
From Abelian Group is Simple iff Prime, all abelian groups of composite order are not simple.
Thus any simple group must be non-abelian.
Let $p$ and $q$ be prime.
From Prime Power Group has Non-Trivial Proper Normal Subgroup, no group of order $p^n$ is simple.
Thus:
:$\forall n \in \Z_{>0}: p^n \notin S$
Thus, with the primes and prime powers eliminated, we have:
:$S \subseteq \set {6, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 30, 33, 34, 35, 36, 38, 40, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58}$
From Group of Order $p q$ has Normal Sylow $p$-Subgroup:
:$p q \notin S$
The set of non-square semiprimes less than $60$ is:
:$\set {6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58}$
Thus we have so far:
:$S \subseteq \set {12, 18, 20, 24, 28, 30, 36, 40, 42, 44, 45, 48, 50, 52, 54, 56}$
From Group of Order $p^2 q$ is not Simple:
:$p^2 q \notin S$
This eliminates:
Thus we are left with:
:$S \subseteq \set {24, 30, 36, 40, 42, 48, 54, 56}$
From Normal Subgroup of Group of Order 24, a group of order $24$ has a normal subgroup either of order $4$ or order $8$.
Hence $24 \notin S$.
From Group of Order 56 has Unique Sylow 2-Subgroup or Unique Sylow 7-Subgroup, a group of order $56$ has at least one normal subgroup.
Hence $56 \notin S$.
From Group of Order 30 is not Simple:
:$30 \notin S$
Thus we are left with:
:$S \subseteq \set {36, 40, 42, 48, 54}$
$40$ is eliminated by Group of Order 40 has Normal Subgroup of Order 5‎.
$42$ is eliminated by Group of Order 42 has Normal Subgroup of Order 7‎.
$54$ is eliminated by Group of Order 54 has Normal Subgroup of Order 27‎.
$36$ is eliminated by Group of Order 36 is not Simple.
$48$ is eliminated by Group of Order 48 is not Simple.
Hence $S$ is empty.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 59 \eta$
Category:Simple Groups
Category:Groups of Order 60
Simple Harmonic Motion of Point on Tusi Couple

== Theorem ==
Let $C_1$ and $C_2$ be the epicycle and deferent respectively of a Tusi couple $H$.
Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.
Let the center of $C_1$ move at a constant angular velocity $\omega$ around the center of $C_2$.
Let $P$ be the point on the circumference of $C_1$ whose locus is $H$.
Let $C_1$ be initially positioned so that $P$ its point of tangency to $C_2$, located on the $x$-axis.
Then $P$ moves back and forward on the $x$-axis with simple harmonic motion with period $\dfrac {2 \pi} \omega$ and maximum speed $a \omega$.
== Proof ==
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $9$
Category:Tusi Couples
Category:Simple Harmonic Motion
Simple Hypothesis Test/Examples/Arbitrary Example 1

== Example of Simple Hypothesis Test ==
Let $\delta$ be a hypothesis test.
Let $H_0$ be the null hypothesis that the probability distribution $X$ of $\delta$ is:
:$X \sim \Gaussian 2 9$
where $\Gaussian 2 9$ denotes the normal distribution.
Then $\delta$ is a simple hypothesis test.
== Sources ==
* 
* 
Category:Examples of Simple Hypothesis Tests
Simple Infinite Continued Fraction Converges

== Theorem ==
Let $C = (a_0, a_1, \ldots)$ be a simple infinite continued fraction in $\R$.
Then $C$ converges.
== Proof ==
We need to show that for any SICF its sequence of convergents $\sequence {C_n}$ always tends to a limit.
Let $\epsilon > 0$.
For $m > n \ge \max \set {5, \dfrac 1 \epsilon}$:
So $\sequence {C_n}$ is indeed a Cauchy sequence.
Hence the result.
Category:Simple Continued Fractions
Simple Infinite Continued Fraction Converges to Irrational Number

== Theorem ==
The value of any simple infinite continued fraction in $\R$ is irrational.
== Proof ==
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be a simple infinite continued fraction.
Note that by Simple Infinite Continued Fraction Converges, a simple infinite continued fraction is indeed convergent, say to $x \in \R$.
Let $p_0, p_1, \ldots$ and $q_0, q_1, \ldots$ be its numerators and denominators.
Let $C_0, C_1, \ldots$ be its convergents, so that $C_n = p_n/q_n$ for $n \ge 0$.
For all $n \ge 0$, from Accuracy of Convergents of Convergent Simple Infinite Continued Fraction:
:$\size {x - \dfrac {p_n} {q_n} } < \dfrac 1 {q_n q_{n + 1} }$
Suppose $x$ is rational.
That is, let $x = \dfrac r s$ where $r, s \in \Z$ such that $s > 0$.
Then:
:$0 < \size {\dfrac r s - \dfrac {p_n} {q_n} } = \dfrac {\size {r q_n - s p_n} } {s q_n} < \dfrac 1 {q_n q_{n + 1} }$
We note that:
:$\dfrac r s \ne \dfrac {p_n} {q_n}$
because otherwise the continued fraction would be finite.
So:
:$0 < \size {r q_n - s p_n} < \dfrac s {q_{n + 1} }$
But Denominators of Simple Continued Fraction are Strictly Increasing.
That means we can choose $n$ so that $q_{n + 1} > s$.
But then $\size {r q_n - s p_n}$ would be an integer lying strictly between $0$ and $1$, which cannot happen.
So no such integers $r, s$ exist.
Thus $x$ must be irrational.
Category:Simple Continued Fractions
Category:Irrational Numbers
Simple Infinite Continued Fraction is Uniquely Determined by Limit

== Theorem ==
Let $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ be simple infinite continued fractions in $\R$.
Let $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ have the same limit.
Then they are equal.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Irrational Number is Limit of Unique Simple Infinite Continued Fraction
* Simple Finite Continued Fraction is Almost Determined by Value
Category:Simple Infinite Continued Fraction is Uniquely Determined by Limit
Category:Simple Continued Fractions