avewright/proofwiki-math huggingface dataset:
Theorem: Simple Infinite Continued Fraction Converges to Irrational Number

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.
{{qed}}
Category:Simple Continued Fractions
Category:Irrational Numbers
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Theorem: Uniqueness of Analytic Continuation

Let $U \subset V \subset \C$ be open subsets of the complex plane.
Let $V$ be connected.
Suppose:
:$(1): \quad F_1, F_2$ are functions defined on $V$
:$(2): \quad f$ is a function defined on $U$.
Let $F_1$ and $F_2$ be analytic continuations of $f$ to $V$.
Then $F_1 = F_2$.

Proof:
We have {{hypothesis}} that $F_1$ and $F_2$ are analytic continuations of $f$.
Therefore they are {{afortiori}} analytic functions.
Let $\map g z = \map {F_1} z - \map {F_2} z$.
By various combination laws, $g$ is also an analytic function.
By definition of agreement:
:$\forall z \in U: \map g z = 0$
That is, every element of $U$ is a zero of $g$.
$U$ is {{hypothesis}} open.
Hence the zeroes of $g$ are not isolated.
From Zeroes of Non-Constant Analytic Function are Isolated, it follows by transposition that $g$ is constant everywhere in its domain $V$.
Hence we have:
{{begin-eqn}}
{{eqn | q = \forall z \in U
      | l = \map g z
      | r = 0
      | c = {{hypothesis}}
}}
{{eqn | q = \exists x \in V
      | l = \map g x
      | r = 0
      | c = Existential Instantiation
}}
{{eqn | q = \forall z \in V
      | l = \map g z
      | r = 0
      | c = {{Defof|Constant Function}}
}}
{{eqn | ll= \leadsto
      | l = \map {F_1} z - \map {F_2} z
      | r = 0
      | c = Definition of $g$
}}
{{eqn | ll= \leadsto
      | l = F_1
      | r = F_2
      | c = {{Defof|Equality of Mappings}}
}}
{{end-eqn}}
Hence:
:$\forall z \in V: \map {F_1} z - \map {F_2} z = 0$
and so:
:$F_1 = F_2$
{{qed}}
Theorem: Zeroes of Non-Constant Analytic Function are Isolated

Let $U \subset \C$ be some open set.
Let $f$ be an analytic function defined on $U$.
Then either $f$ is a constant function, or the set $\set {z \in U: \map f z = 0}$ is totally disconnected.

Proof:
Suppose $f$ has no zeroes in $U$.
Then the set described in the theorem is the empty set, and we're done.
So we suppose $\exists z_0 \in U$ such that $\map f {z_0} = 0$.
Since $f$ is analytic, there is a Taylor series for $f$ at $z_0$ which converges for $\cmod {z - z_0} < R$.
Now, since $\map f {z_0} = 0,$ we know $a_0 =0$.
Other $a_j$ may be $0$ as well.
So let $k$ be the least number such that $a_j = 0$ for $0 \le j < k$, and $a_k \ne 0$.
Then we can write the Taylor series for $f$ about $z_0$ as:
:$\ds \sum_{n \mathop = k}^\infty a_n \paren {z - z_0}^n = \paren {z - z_0}^k \sum_{n \mathop = 0}^\infty a_{n + k} \paren {z - z_0}^n$ 
where $a_k \ne 0$ (otherwise, we'd just start at $k + 1$).
Now we define a new function $\map g z$, as the sum on the right hand side, which is clearly analytic in $\cmod {z - z_0} < R$.  
Since it is analytic here, it is also continuous here.
Since $\map g {z_0} = a_k \ne 0, \exists \epsilon >0$ so that $\forall z$ such that $\cmod {z - z_0} < \epsilon, \cmod {\map g z - a_k} < \dfrac {\cmod {a_k} } 2$.
But then $\map g z $ cannot possibly be $0$ in that disk.
Hence the result.
{{explain|It is shown that each zero is isolated, but how to show the set of all zeros is totally disconnected?}}
{{qed}}
Category:Complex Analysis
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Theorem: Irrational Number is Limit of Unique Simple Infinite Continued Fraction

Let $x$ be an irrational number.
Then the continued fraction expansion of $x$ is the unique simple infinite continued fraction that converges to $x$.

Proof:
Follows from:
* Continued Fraction Expansion of Irrational Number Converges to Number Itself
* Simple Infinite Continued Fraction is Uniquely Determined by Limit.
{{Qed}}
Theorem: Analytic Continuation of Riemann Zeta Function

The Riemann zeta function is meromorphic on $\C$.

Proof:
The (yet to be confirmed) meromorphic continuation of the Riemann zeta function to the half-plane $\set {s: \map \Re s > 0}$ is given by Integral Representation of Riemann Zeta Function in terms of Fractional Part:
:$(1): \quad \ds \map \zeta s = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$
where $\fractpart x$ is the fractional part of $x$.
Let $\map \Re s \le 0$.
Then the value of $\map \zeta s$ can be computed from the relation:
:$\map \Gamma {\dfrac s 2} \pi^{-s/2} \map \zeta s = \map \Gamma {\dfrac {s - 1} 2} \pi^{\frac {1 - s} 2} \map \zeta {1 - s}$
That is:
:$\map \xi s = \map \xi {1 - s}$
where $\xi$ is the completed zeta function.
First we show that $(1)$ is analytic for $\map \Re s > 0$.
For $n \ge 1$, let:
{{begin-eqn}}
{{eqn | l = a_n
      | r = s \int_n^{n + 1} \fractpart x x^{-s - 1}\rd x
      | c = 
}}
{{eqn | o = \ll
      | r = s \int_n^{n + 1} x^{-s - 1} \rd x
      | c = 
}}
{{eqn | r = \bigintlimits {-x^{-s} } n {n + 1}
      | c = 
}}
{{eqn | r = \frac 1 {n^s} - \frac 1 {\paren {n + 1}^s}
      | c =
}}
{{eqn | r = \frac {\paren {n + 1}^s - n^s} {n^s \paren {n + 1}^s}
      | c =
}}
{{end-eqn}}
Here $\ll$ is the order notation.
By the Mean Value Theorem, for some $n \le \theta \le n + 1$:
:$\paren {n + 1}^s - n^s = s \theta^{s - 1} \le s \paren {n + 1}^{s-1}$
Thus if $s = \sigma + i t$:
:$\size {a_n} \le \size {\dfrac s {n^{s + 1} } } = \dfrac {\sigma^2 + t^2} {n^{\sigma + 1} }$
Since:
:$\ds \map \zeta s = \frac s {s - 1} - \sum_{n \mathop \ge 1} a_n$
it follows that this representation converges absolutely uniformly on $\map \Re s > 0$.
Thus by Uniform Limit of Analytic Functions is Analytic $\map \zeta s$ is analytic for $\map \Re s > 0$ and $s \ne 1$.
For all $s$ with $\map \Re s < \dfrac 1 2$, $\map \zeta s$ is simply the reflection of $\zeta$ in the upper half-plane.
Therefore, $\zeta$ is also analytic for all $s$ with $\map \Re s < 0$.
{{qed}}
Category:Riemann Zeta Function
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Theorem: Accuracy of Convergents of Continued Fraction Expansion of Irrational Number

Let $x$ be an irrational number.
Let $(a_0, a_1, \ldots)$ be its continued fraction expansion.
Let $\left \langle {C_n}\right \rangle_{n \geq 0}$ be its sequence of convergents.
Let $p_0, p_1, p_2, \ldots$ and $q_0, q_1, q_2, \ldots$ be its numerators and denominators.
Then:
:$\forall k \ge 1: \left\vert{x - \dfrac {p_{k + 1} } {q_{k + 1} } }\right\vert < \dfrac 1 {q_{k + 1} q_{k + 2} } \le \dfrac 1 {2 q_k q_{k + 1} } < \left\vert{x - \dfrac {p_k} {q_k} }\right\vert$
Thus:
:The {{LHS}} of the inequality gives an indication of how close each convergent gets to its true value.
:The {{RHS}} gives a bound that limits its accuracy.

Proof:
The Continued Fraction Algorithm gives the following system of equations:
{{begin-eqn}}
{{eqn | l = x
      | r = \left[{a_0, x_2}\right]
      | c = 
}}
{{eqn | r = \left[{a_0, a_1, x_3}\right]
      | c = 
}}
{{eqn | r = \ldots
      | c = 
}}
{{eqn | r = \left[{a_0, a_1, \ldots, a_n, x_{n + 1} }\right]
      | c = 
}}
{{eqn | r = \ldots
      | c = 
}}
{{end-eqn}}
and
{{begin-eqn}}
{{eqn | l = \left\vert{x - \frac {p_n} {q_n} }\right\vert
      | r = \left\vert{\left[{a_0, a_1, \ldots, a_n, x_{n + 1} }\right] - \frac {p_n} {q_n} }\right\vert
      | c = 
}}
{{eqn | r = \left\vert{\frac {x_{n + 1} p_n + p_{n - 1} } {x_{n + 1} q_n + q_{n - 1} } - \frac {p_n} {q_n} }\right\vert
      | c = Value of Finite Continued Fraction equals Numerator Divided by Denominator
}}
{{eqn | r = \left\vert{\frac {p_{n - 1} q_n - p_n q_{n - 1} } {q_n \left({x_{n + 1} q_n + q_{n - 1} }\right)} }\right\vert
      | c = 
}}
{{eqn | r = \frac 1 {q_n \left({x_{n + 1} q_n + q_{n - 1} }\right)}
      | c = Difference between Adjacent Convergents of Simple Continued Fraction
}}
{{end-eqn}}
Now from the Continued Fraction Algorithm:
:$x_{n + 1} = \left[{a_{n + 1}, a_{n + 2}, a_{n + 3}, \ldots}\right]$
So:
: $a_{n + 1} < x_{n + 1} < a_{n + 1} + 1$
Therefore:
:$\left\vert{x - \dfrac {p_n} {q_n} }\right\vert < \dfrac 1 {q_n \left({a_{n + 1} q_n + q_{n - 1} }\right)} = \dfrac 1 {q_n q_{n + 1} }$
This gives the {{LHS}} of the inequality when $n = k + 1$.
We also have:
:$\left\vert{x - \dfrac {p_n} {q_n} }\right\vert > \dfrac 1 {q_n \left({\left({a_{n + 1} + 1}\right) q_n + q_{n - 1} }\right)}$
{{begin-eqn}}
{{eqn | l = \left\vert{x - \frac {p_n} {q_n} }\right\vert
      | o = >
      | r = \frac 1 {q_n \left({\left({a_{n + 1} + 1}\right) q_n + q_{n - 1} }\right)}
      | c = 
}}
{{eqn | r = \frac 1 {q_n \left({a_{n + 1} q_n + q_{n - 1} }\right) + q_n^2}
      | c = 
}}
{{eqn | r = \frac 1 {q_n \left({q_{n + 1} + q_n}\right)}
      | c = 
}}
{{eqn | o = >
      | r = \frac 1 {q_n \left({q_{n + 1} + q_{n + 1} }\right)}
      | c = 
}}
{{eqn | r = \frac 1 {2 q_n q_{n + 1} }
      | c = 
}}
{{end-eqn}}
This gives the {{RHS}} of the inequality when $n = k$.
For the middle inequality, note that:
:$q_{k + 2} = a_{k + 2} q_{k + 1} + q_k > q_k + q_k = 2 q_k$
So:
:$\dfrac 1 {q_{k + 1} q_{k + 2} } \le \dfrac 1 {2 q_k q_{k + 1} }$
{{qed}}
Theorem: Trivial Zeroes of Riemann Zeta Function are Even Negative Integers

Let $s = \sigma + i t$ be a zero of the Riemann zeta function not contained in the critical strip:
:$0 \le \map \Re s \le 1$
Then:
:$s \in \set {-2, -4, -6, \ldots}$
These are called the '''trivial zeros''' of $\zeta$.

Proof:
{{finish|First it needs to be established that those points are in fact zeroes. This follows directly and trivially from Riemann Zeta Function at Non-Positive Integers, but needs to be made explicit here.}}
First we note that by Zeroes of Gamma Function, $\Gamma$ has no zeroes on $\C$.
Therefore, the completed Riemann zeta function:
:$\map \xi s = \dfrac 1 2 s \paren {s - 1} \pi^{-s/2} \map \Gamma {\dfrac s 2} \, \map \zeta s$
has the same zeroes as $\zeta$.
Additionally by Functional Equation for Riemann Zeta Function, we have:
:$\map \xi s = \map \xi {1 - s}$
for all $s \in \C$.
Therefore if:
:$\map \zeta s \ne 0$
for all $s$ with $\map \Re s > 1$, then also:
:$\map \zeta s \ne 0$
for all $s$ with $\map \Re s < 0$.
Let us consider $\map \Re s > 1$.
We have:
:$\ds \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
where here and in the following $p$ ranges over the primes.
Therefore, we have:
:$\ds \map \zeta s \prod_p \paren {1 - p^{-s} } = 1$
All of the factors of this infinite product can be found in the product:
:$\ds \prod_{n \mathop = 2}^\infty \paren {1 - n^{-s} }$
which converges absolutely.
This follows because from Convergence of P-Series:
:$\ds \sum_{k \mathop = 1}^\infty k^{-s}$ converges absolutely.
Hence:
:$\ds \prod_p \paren {1 - p^{-s} }$ converges absolutely.
So by the fact that:
:$\ds \map \zeta s \prod_p \paren {1 - p^{-s} } = 1$
we know $\map \zeta s$ cannot possibly be zero for any point in the region in question.
{{qed}}
Theorem: Poles of Gamma Function

The gamma function $\Gamma: \C \to \C$ is analytic throughout the complex plane except at $\set {0, -1, -2, -3, \dotsc}$ where it has simple poles.

Proof:
First we examine the location of the poles.
We examine the Weierstrass form of the Gamma function:
:$\dfrac 1 {\map \Gamma z} = z e^{\gamma z} \ds \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z / n} }$
The factors of the continued product clearly do not tend to zero.
So the continued product is zero only if one of the factors is zero.
This occurs when $1 + \dfrac z n = 0$.
This occurs at $z \in \set {-1, -2, \dotsc}$.
We also have the expression outside the continued product to consider.
From Exponential Tends to Zero and Infinity, the exponential function is never $0$.
So this expression is zero whenever $z = 0$.
Furthermore, if some factor in the continued product is zero, this excludes the possibility that any other factor is zero.
So the zeros are simple.
Hence:
:$\dfrac 1 {\map \Gamma z} = 0 \iff z \in \set {0, -1, -2, -3, \ldots}$
Therefore $\Gamma$ has simple poles at these points.
Next we show that $\Gamma$ is analytic on $\map \Re z > 0$. 
By Gamma Difference Equation this proves that $\Gamma$ is analytic on $\C \setminus \set {0, -1, -2, -3, \dotsc}$.
Let the functions $\Gamma_n$ be defined by
:$\ds \map {\Gamma_n} z = \int_0^n t^{z - 1} e^{-t} \rd t$
Clearly each $\Gamma_n$ is analytic.
So by Uniform Limit of Analytic Functions is Analytic it is sufficient to show that $\Gamma_n \to \Gamma$ locally uniformly.
By Modulus of Complex Integral:
:$\ds \cmod {\map \Gamma z - \map {\Gamma_n} Z } \le \int_n^\infty t^{x - 1} e^{-t} \rd t$
where $x = \map \Re z$. 
Let $a \in \C$ with $\map \Re a > 0$.
Let $D$ be an open disk of radius $r$ about $a$.
By definition of exponential function, we have the expansion:
:$e^z = 1 + z + \dfrac {z^2} 2 + \dotsb$
from which we see that:
:$\forall \alpha \in \R_{>0}: z^\alpha = \map \oo {e^z}$
where $\oo$ denotes little-$\oo$ notation.
In particular:
:$\exists c \in \R_{>0}: \forall t \ge 1: e^{-t} \le c t^{-\map \Re a - r - 1}$
Then:
{{begin-eqn}}
{{eqn | l = \cmod {\map \Gamma z - \map {\Gamma_n} z}
      | o = \le
      | r = \int_n^\infty t^{x - 1} c t^{-\map \Re a -r - 1} \rd t
}}
{{eqn | o = \le
      | r = c\int_n^\infty t^{-2} \rd t
}}
{{eqn | o = \le
      | r = \frac c n
}}
{{end-eqn}}
So $\Gamma_n \to \Gamma$ uniformly in $D$, and the proof is complete.
{{qed}}
Theorem: Zeroes of Gamma Function

The Gamma function is never equal to $0$.

Proof:
Suppose $\exists z$ such that $\map \Gamma z = 0$.
We examine the Euler form of the gamma function, which is defined for $\C \setminus \set {0, -1, -2, \ldots}$.
The Euler form, equated with zero, yields
:$\ds 0 = \frac 1 z \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac 1 n}^z \paren {1 + \frac z n}^{-1} }$
It is clear that $\dfrac 1 z \ne 0$, so we may divide this out for $z$ in the area of definition.
Now it is clear that as $n \to \infty$, each of the two halves of the term in the product will tend to $1$ for any $z$, and there is no $z$ which yields zero for any $n$ in either of the product terms.
Hence this product will not equal $0$ anywhere.
This leaves only the question of the behavior on $\set {0, -1, -2, \ldots}$, which is discussed at Poles of Gamma Function.
{{qed}}
Category:Gamma Function
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Theorem: Gamma Difference Equation

:$\map \Gamma {z + 1} = z \, \map \Gamma z$

Proof:
Let $z \in \C$, with $\map \Re z > 0$.
Then:
{{begin-eqn}}
{{eqn | l = \map \Gamma {z + 1}
      | r = \int_0^\infty t^z e^{-t} \rd t
      | c =
}}
{{eqn | r = \bigintlimits {-t^z e^{-t} } 0 \infty + z \int_0^\infty t^{z - 1} e^{-t} \rd t
      | c = Integration by Parts
}}
{{eqn | r = z \int_0^\infty t^{z - 1} e^{-t} \rd t
      | c =
}}
{{eqn | r = z \, \map \Gamma z
      | c =
}}
{{end-eqn}}
If $z \in \C \setminus \set {0, -1, -2, \ldots}$ such that $\map \Re z \le 0$, then the statement holds by the definition of $\Gamma$ in this region.
{{qed}}
Theorem: Gamma Difference Equation

:$\map \Gamma {z + 1} = z \, \map \Gamma z$

Proof:
{{begin-eqn}}
{{eqn | l = \frac {\map \Gamma {z + 1} } {\map \Gamma z}
      | r = \paren {\frac 1 {z + 1} \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \frac {\paren {1 + \frac 1 n}^{z + 1} } {1 + \frac {z + 1} n} } \div \paren {\frac 1 z \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \frac {\paren {1 + \frac 1 n}^z} {1 + \frac z n} }
      | c = {{Defof|Euler Form of Gamma Function}}
}}
{{eqn | r = \frac z {z + 1} \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \paren {\frac {\paren {1 + \frac 1 n}^{z + 1} \paren {1 + \frac z n} } {\paren {1 + \frac 1 n}^z \paren {1 + \frac {z + 1} n} } }
      | c = Product Rule for Complex Sequences
}}
{{eqn | r = \frac z {z + 1} \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \paren {\frac {n + 1} n} \paren {\frac {z + n} {z + n + 1} }
      | c = 
}}
{{eqn | r = \frac z {z + 1} \lim_{m \mathop \to \infty} \paren {\frac {m + 1} 1} \paren {\frac {z + 1} {z + m + 1} }
      | c = Telescoping Product
}}
{{eqn | r = \frac z {z + 1} \lim_{m \mathop \to \infty} \paren {\frac {m z + z + m + 1} {z + m + 1} }
      | c = expanding the product
}}
{{eqn | r = \frac z {z + 1} \lim_{m \mathop \to \infty} \paren {\frac {m z} {z + m + 1} } + 1
      | c = 
}}
{{eqn | r = \frac z {z + 1} \times \paren {z + 1}
      | c = Limit to Infinity of Complex Rational Function
}}
{{eqn | r = z
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Convergents are Best Approximations

Let $x$ be an irrational number.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be the numerators and denominators of its continued fraction expansion.
Let $\dfrac {p_n} {q_n}$ be the $n$th convergent.
Let $\dfrac a b$ be any rational number such that $0 < b < q_{n + 1}$.
Then:
:$\forall n > 1: \size {q_n x - p_n} \le \size {b x - a}$
The equality holds only if $a = p_n$ and $b = q_n$.

Proof:
Let $\dfrac a b$ be a rational number in canonical form such that $b < q_{n + 1}$.
Suppose it is not true that $a = p_n$ and $b = q_n$, in which case the equality certainly holds.
Consider the system of equations:
{{begin-eqn}}
{{eqn | l = a
      | r = r p_n + s p_{n + 1}
      | c = 
}}
{{eqn | l = b
      | r = r q_n + s q_{n + 1}
      | c = 
}}
{{end-eqn}}
Multiplying the first by $q_n$, and the second by $p_n$, then subtracting, we get:
:$a q_n - b p_n = s \paren {p_{n + 1} q_n - p_n q_{n + 1} }$
After applying Difference between Adjacent Convergents of Simple Continued Fraction we get:
{{begin-eqn}}
{{eqn | l = s
      | r = \paren {-1}^{n + 1} \paren {a q_n - b p_n}
      | c = 
}}
{{eqn | l = r
      | r = \paren {-1}^{n + 1} \paren {b p_{n + 1} - a q_{n + 1} }
      | c = by a similar process
}}
{{end-eqn}}
So $r$ and $s$ are integers.
Neither of them is $0$ because:
:if $r = 0$ then $a q_{n + 1} = b p_{n + 1}$, and Euclid's Lemma means $q_{n + 1} \divides b$ as $p_{n + 1} \perp q_{n + 1}$, which contradicts $0 < b < q_{n + 1}$
:if $s = 0$ we have $\dfrac a b = \dfrac {p_n} {q_n}$ and this we have already excluded as a special case.
{{questionable|the below needs to link to a stronger result}}
From Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent, the convergents are alternately greater than and less than $x$.
Hence since $0 < b = r q_n + s q_{n + 1} < q_{n + 1}$, the integers $r$ and $s$ must have opposite sign.
It follows that $r \paren {q_n x - p_n}$ and $s \paren {q_{n + 1} x - p_{n + 1} }$ have the same sign.
This is necessary for the Triangle Inequality to hold.
So:
{{begin-eqn}}
{{eqn | l = \size {b x - a}
      | r = \size {\paren {r q_n + s q_{n + 1} } x - \paren {r p_n + s p_{n+1} } }
      | c = 
}}
{{eqn | r = \size {r \paren {q_n x - p_n} + s \paren {q_{n + 1} x - p_{n + 1} } }
      | c = 
}}
{{eqn | r = \size r \size {q_n x - p_n} + \size s \size {q_{n + 1} x - p_{n + 1} }
      | c = 
}}
{{eqn | o = >
      | r = \size r \size {q_n x - p_n}
      | c = 
}}
{{eqn | o = \ge
      | r = \size {q_n x - p_n}
      | c = 
}}
{{end-eqn}}
as we wanted to prove.
{{qed}}
Theorem: Condition for Rational to be Convergent

Let $x$ be an irrational number.
Let the rational number $\dfrac a b$ satisfy the inequality:
:$\size {x - \dfrac a b} < \dfrac 1 {2 b^2}$
Then $\dfrac a b$ is a convergent of $x$.

Proof:
{{AimForCont}} $\size {x - \dfrac a b} < \dfrac 1 {2 b^2}$, but that $\dfrac a b$ is ''not'' one of the convergents $\dfrac {p_n} {q_n}$ of $x$.
Let $r$ be the unique integer for which $q_r \le b \le q_{r + 1}$.
{{explain|Why does such a unique $r$ exist? Maybe related to Denominators of Simple Continued Fraction are Strictly Increasing?}}
Then:
{{begin-eqn}}
{{eqn | l = \size {q_r x - p_r}
      | o = \le
      | r = \size {b x - a}
      | c = Convergents are Best Approximations
}}
{{eqn | r = b \size {x - \frac a b}
      | c = 
}}
{{eqn | o = <
      | r = b \times \frac 1 {2 b^2}
      | c = 
}}
{{eqn | r = \frac 1 {2 b}
      | c = 
}}
{{end-eqn}}
Therefore:
:$q_r \size {x - \dfrac {p_r} {q_r} }< \dfrac 1 {2 b}$
and so:
:$\size {x - \dfrac {p_r} {q_r} } < \dfrac 1 {2 q_r b}$
Hence:
{{begin-eqn}}
{{eqn | l = \size {\frac a b - \frac {p_r} {q_r} }
      | o = \le
      | r = \size {x - \frac {p_r} {q_r} } + \size {x - \frac a b}
      | c = Triangle Inequality
}}
{{eqn | n = 1
      | o = <
      | r = \frac 1 {2 q_r b} + \frac 1 {2b^2}
      | c = 
}}
{{end-eqn}}
Now note that $q_r a - p_r b$ is a integer.
However, {{hypothesis}} $\dfrac a b$ is ''not'' one of the convergents $\dfrac {p_n} {q_n}$ of $x$.
Thus $\dfrac a b \ne \dfrac {p_r} {q_r}$.
But we have:
{{begin-eqn}}
{{eqn | l = \size {\frac a b - \frac {p_r} {q_r} }
      | r = \size {\frac {q_r a - p_r b} {q_r b} }
      | c = 
}}
{{eqn | n = 2
      | o = \ge
      | r = \frac 1 {q_r b}
      | c = 
}}
{{end-eqn}}
So, combining results $(1)$ and $(2)$, we get:
:$\dfrac 1 {q_r b} < \dfrac 1 {2 q_r b} + \dfrac 1 {2 b^2}$
This simplifies to $q_r > b$.
But {{hypothesis}} $r$ be the unique integer for which $q_r \le b \le q_{r + 1}$.
From this contradiction it follows that $\dfrac a b$ is one of the convergents of $x$
{{qed}}
Category:Continued Fractions
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Theorem: Quadratic Irrational is Root of Quadratic Equation

Let $x$ be a quadratic irrational.
Then $x$ is a solution to a quadratic equation with rational coefficients.

Proof:
Let $x = r + s \sqrt n$.
From Solution to Quadratic Equation, the solutions of $a x^2 + b x + c$ are:
:$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c}} {2 a}$
given the appropriate condition on the discriminant.
So if $x = r + s \sqrt n$ is a solution, then so is $x = r - s \sqrt n$.
Hence we have:
{{begin-eqn}}
{{eqn | l = \left({x - r + s \sqrt n}\right) \left({x - r - s \sqrt n}\right)
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \left({x-r}\right)^2 - \left({s \sqrt n}\right)^2
      | r = 0
      | c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
      | l = x^2 - 2 r x + r^2 - s^2 n
      | r = 0
      | c = 
}}
{{end-eqn}}
As $r$ and $s$ are rational and $n$ is an integer, it follows that $-2 r$ and $r^2 - s^2 n$ are also rational from Rational Numbers form Field.
Hence the result.
{{qed}}
Category:Quadratic Equations
Category:Quadratic Irrationals
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Theorem: Irrational Number has Periodic Continued Fraction iff Quadratic

Let $x$ be an irrational number.
Then $x$ is a quadratic irrational {{iff}} its continued fraction expansion is periodic.

Proof:
Follows from:
* Limit of Simple Infinite Periodic Continued Fraction is Quadratic Irrational
* Quadratic Irrational has Periodic Continued Fraction Expansion
{{qed}}
Category:Simple Continued Fractions
Category:Quadratic Irrationals
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