nyuuzyou/wikis huggingface dataset:
Content of Rational Polynomial is Multiplicative

== Theorem ==
Let $h \in \Q \sqbrk X$ be a polynomial with rational coefficients.
Let $\cont h$ denote the content of $h$.
Then for any polynomials $f, g \in \Q \sqbrk X$ with rational coefficients:
:$\cont {f g} = \cont f \cont g$
== Proof 1 ==
== Proof 2 ==
Category:Gauss's Lemma (Polynomial Theory)
Category:Content of Polynomial
Category:Content of Rational Polynomial is Multiplicative
Content of Rational Polynomial is Multiplicative/Proof 1

== Theorem ==
== Proof ==
Let $f^* = \dfrac 1 {\cont f} f$, $g^* = \dfrac 1 {\cont g} g$
By Content of Scalar Multiple:
:$\cont {f^*} = \cont {g^*} = 1$
That is, $f^*$ and $g^*$ are primitive.
By Gauss's Lemma on Primitive Rational Polynomials, it follows that $f^* g^*$ is primitive.
Then:
Category:Content of Rational Polynomial is Multiplicative
Content of Rational Polynomial is Multiplicative/Proof 2

== Theorem ==
== Proof ==
From Rational Polynomial is Content Times Primitive Polynomial, let $\map f X$ and $\map g X$ be expressed as:
:$\map f X = \cont f \cdot \map {f^*} X$
:$\map g X = \cont g \cdot \map {g^*} X$
where:
:$\cont f, \cont g$ are the content of $f$ and $g$ respectively
:$f^*, g^*$ are primitive.
We have, by applications of Rational Polynomial is Content Times Primitive Polynomial:
:$\cont h \cdot \map {h^*} X = \cont f \cont g \cdot \map {f^*} X \, \map {g^*} X$
By Gauss's Lemma on Primitive Rational Polynomials we have that $\map {f^*} X \, \map {g^*} X$ is primitive.
As $\cont f > 0$ and $\cont g > 0$, then so is $\cont f \cont g > 0$.
By the uniqueness clause in Rational Polynomial is Content Times Primitive Polynomial, the result follows.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Theorem $62$
Category:Content of Rational Polynomial is Multiplicative
Content of Scalar Multiple

== Theorem ==
Let $f$ be a polynomial with rational coefficients.
Let $\cont f$ denote the content of $f$.
Let $q \in \Q$ be a rational number.
Then:
:$\cont {q f} = q \cont f$
== Proof ==
Let $q = \dfrac a b$ with $a, b \in \Z$.
Let $\Z \sqbrk X$ denote the ring of polynomials over $\Z$.
Let $n \in \Z$ such that $n f \in \Z \sqbrk X$. 
Then we have:
:$b n \paren {q f} = a n f \in \Z \sqbrk X$
By the definition of content, and using that $a \in \Z$:
:$\cont {b n q f} = a \cont {n f}$
By definition of content:
:$\cont {q f} = \dfrac 1 {b n} \cont {b n q f}$
Combining the above with the definition of $\cont f$:
:$\cont {q f} = \dfrac a b \dfrac 1 n \cont {n f} = q \cont f$
Category:Polynomial Theory
Category:Content of Polynomial
Content of Subdivided Rectangle

== Theorem ==
Let $R$ be a closed $n$-rectangle.
Let $P$ be a subdivision of $R$.
Then:
:$\ds \map V R = \sum_{r \mathop \in P^*} \map V r$
where:
:$\map V R$ denotes the content of the rectangle $R$
:$P^*$ denotes the set of subrectangles of $P$.
== Proof ==
We will proceed by induction on $n$.
=== Basis for the Induction ===
First, consider the case $n = 1$.
Let $R = \closedint a b$.
Let us write:
:$P = \tuple {\set {x_0, x_1, \dotsc, x_m}}$
where:
:$a = x_0 < x_1 < \dotso < x_m = b$
Then, we have:
Therefore, the theorem holds for the case $n = 1$, which is the basis for the induction.
=== Induction Hypothesis ===
Let us now assume as the induction hypothesis that the theorem holds for some fixed $n = q \ge 1$.
We will then deduce the theorem for $n = q + 1$.
=== Induction Step ===
Let $R = R' \times \closedint a b$, where $R'$ is a closed $q$-rectangle.
Write:
:$P = \tuple {P_1, P_2, \dotsc, P_q, P_{q + 1}}$
and let:
:$P' = \tuple {P_1, P_2, \dotsc, P_q}$
so that $P'$ is a subdivision of $R'$.
Let:
:$P_{q + 1} = \set {x_0, x_1, \dotsc, x_m}$
where:
:$a = x_0 < \dotso < x_m = b$
For each $k \in \set {1, 2, \dotsc, m}$, let $S_k$ be the set of all subrectangles $r$ of $P$ with the form:
:$r = r' \times \closedint {x_{k - 1}} {x_k}$
where $r'$ is a subrectangle of $P'$.
By definition, each subrectangle of $P$, is an element of exactly one $S_k$.
Thus:
Therefore, we may apply the Principle of Mathematical Induction to deduce the theorem in its full generality.
Category:Jordan Content
Contingency Table/Examples/Arbitrary Example 1

== Example of Contingency Table ==
The following is an example of a contingency table:
:$\begin{array}{r|ccc|c}
& \text {Superior} & \text {Average} & \text {Poor} & \text {Factory totals} \\
\hline
\text {Factory A} & 12 & 20 & 8 & 40 \\
\text {Factory B} & 14 & 39 & 7 & 60 \\
\hline
\text {Category totals} & 26 & 59 & 15
\end{array}$
== Sources ==
* 
* 
Category:Examples of Contingency Tables
Continued Fraction Algorithm

== Algorithm ==
The Continued Fraction Algorithm is a method for finding the continued fraction expansion for any irrational number to as many partial denominators as desired.
Let $x_0$ be the irrational number in question.
The steps are:
Then $x_0 = \sqbrk {a_0, a_1, a_2, \ldots}$ is the required continued fraction expansion.
== Examples ==
== Proof 1 ==
Category:Continued Fraction Algorithm
Category:Continued Fractions
Continued Fraction Algorithm/Proof 1

== Theorem ==
== Proof == 
Let $x_0$ be an irrational number.
We seek $a_0, a_1, \ldots \in \Z$ such that $x_0 = \sqbrk {a_0, a_1, \ldots}$.
From Bound for Difference of Limit of Simple Infinite Continued Fraction with Convergent:
:$x_0$ lies strictly between any successive pair of its convergents.
So, for a start, it has to lie between $C_0 = a_0$ and $C_1 = a_0 + \dfrac 1 {a_1}$.
In particular, as $a_1 \ge 1$, we know that $a_0 < x_0 < a_0 + 1$.
So:
:$a_0 = \floor {x_0}$
where $\floor {x_0}$ is the floor function of $x_0$.
We now write:
:$x_0 = \floor {x_0} + \fractpart {x_0}$
where $\fractpart {x_0}$ is the fractional part of $x_0$.
Then:
:$x_0 = a_0 + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \cfrac 1 \ddots} } = \floor {x_0} + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \cfrac 1 \ddots} } = \floor {x_0} + \dfrac 1 {\sqbrk {a_1, a_2, a_3, \ldots} }$
From Real Number minus Floor:
:$0 \le \fractpart {x_0} < 1$
But because $x_0$ is irrational:
:$\fractpart {x_0} \ne 0$
So:
:$0 < \fractpart {x_0} < 1$
So:
:$\sqbrk {a_1, a_2, a_3, \ldots} = \dfrac 1 {x_0 - \floor {x_0} } = \dfrac 1 {\fractpart {x_0} }$
Now we write:
:$x_1 = \dfrac 1 {\fractpart {x_0} }$
Then:
:$x_1 = \sqbrk {a_1, a_2, a_3, \ldots}$
As $\fractpart {x_0} < 1$ we have that:
:$x_1 = \dfrac 1 {\fractpart {x_0} } > 1$
So $x_1$ is an irrational number greater than $a_1$ which is positive.
Repeating the argument leads to $a_1 = \floor {x_1}$ and so $a_1$ is determined uniquely from $x_1$ and hence from $x_0$.
We continue on in the same manner, :
:$x_{k + 1} = \dfrac 1 {\fractpart {x_k} }$ and so $x_{k + 1} = \sqbrk {a_{k + 1}, a_{k + 2}, a_{k + 3}, \ldots}$
Hence the result.
Category:Continued Fraction Algorithm
Continued Fraction Expansion of Euler's Number/Convergents

== Theorem ==
The convergents of the continued fraction expansion to Euler's number $e$ are:
:$2, 3, \dfrac 8 3, \dfrac {11} 4, \dfrac {19} 7, \dfrac {87} {32}, \dfrac {106} {39}, \dfrac {193} {71}, \dfrac {1264} {465}, \dfrac {1457} {536}, \dfrac {2721} {1001}, \ldots$
These best rational approximations are accurate to $0, 0, 1, 1, 2, 3, 3, 4, 5, 5, \ldots$ decimals.
The fraction $\dfrac {878} {323}$ is exceptionally easy to remember:
:$\dfrac {878} {323} = 2 \cdotp 71826 \, 625 \ldots$
although this does not occur in the above continued fraction expansion.
== Proof ==
== Sources ==
* : $2 \cdotp 718 \, 281 \, 828 \, 459 \, 045 \, 235 \, 360 \, 287 \, 471 \, 352 \, 662 \, 497 \, 757 \, 247 \, 093 \, 699 \ldots$
* : $2 \cdotp 71828 \, 18284 \, 59045 \, 23536 \, 02874 \, 71352 \, 66249 \, 77572 \, 47093 \, 69995 \ldots$
* 
Category:Continued Fraction Expansion of Euler's Number
Continued Fraction Expansion of Euler's Number/Proof 1

== Theorem ==
== Proof == 
From the recursive definition of continued fractions, we have:
Let:
In other words:
:$a_{3 n + 1} = 2 n$
and:
:$a_{3 n + 0} = a_{3 n + 2} = 1$
Then $p_i$ and $q_i$ are as follows:
:$\begin{array}{r|cccccccccc}
\ds i & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
p_i & 1 & 1 & 2 & 3 & 8 & 11 & 19 & 87 & 106 & 193 \\
q_i & 1 & 0 & 1 & 1 & 3 & 4 & 7 & 32 & 39 & 71 \\
\hline
\end{array}$
Furthermore, $p_i$ and $q_i$ satisfy the following $6$ recurrence relations:
Our ultimate aim is to prove that:
:$\ds \lim_{n \mathop \to \infty} \frac {p_n} {q_n} = e$
In the pursuit of that aim, let us define the integrals:
=== Lemma ===
We assert that $A_n$, $B_n$ and $C_n$ all converge to $0$ as $n \to \infty$:
We now have:
from which we conclude:
== Sources ==
* 
Category:Continued Fraction Expansion of Euler's Number
Continued Fraction Expansion of Euler's Number/Proof 1/Lemma

== Theorem ==
:For $n \in \Z , n \ge 0$:
== Proof == 
To prove the assertion, we begin by demonstrating the relationships hold for the initial conditions at $n = 0$:
The final step needed to validate the assertion, we must demonstrate that the following three recurrence relations hold:
To prove the first relation, we note that the derivative of the integrand of $A_n$ is equal to the sum of the integrand of $A_n$ with the integrand of $B_{n - 1}$ and the integrand of $C_{n - 1}$.
By integrating both sides of the equation, we verify the first recurrence relation:
To prove the second relation, we note that the derivative of the integrand of $C_n$ is equal to the sum of the integrand of $B_n$ with two times n times the integrand of $A_{n}$ minus the integrand of $C_{n - 1}$.
By integrating both sides of the equation, we verify the second recurrence relation:
To prove the third relation, we have:
From the first relation, combined with the initial condition at $n = 0$ being satisfied, we have:
From the second relation, combined with the initial condition at $n = 0$ being satisfied, we have:
From the third relation, combined with the initial condition at $n = 0$ being satisfied, we have:
== Sources ==
* 
Category:Continued Fraction Expansion of Euler's Number
Continued Fraction Expansion of Fourth Power of Pi

== Theorem ==
The continued fraction expansion of the $4$th power of $\pi$ begins:
:$\pi^4 \approx 97 + \cfrac 1 {2 + \cfrac 1 {2 + \cfrac 1 {3 + \cfrac 1 {1 + \cfrac 1 {16 \, 539} } } } }$
The continued fraction expansion shown above evaluates to $\dfrac {2143} {22}$ or $\paren {97 \cdotp 4 \dot 0 \dot 9}$ which is $\approx 1.2 \times 10^{-7}$ lower than the exact value.
== Proof ==
== Sources ==
* : $97 \cdotp 40909 \, 10340 \, 0 \ldots$
Category:Pi
Category:Fourth Powers
Category:Examples of Continued Fractions
Continued Fraction Expansion of Golden Mean

== Theorem ==
The golden mean has the simplest possible continued fraction expansion, namely $\sqbrk {1, 1, 1, 1, \ldots}$:
:$\phi = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$
=== Successive Convergents ===
=== Rate of Convergence ===
== Proof ==
Let:
:$x = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$
Then:
The result follows from Golden Mean as Root of Quadratic.
== Sources ==
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* 
* 
Category:Continued Fraction Expansion of Golden Mean
Category:Golden Mean
Category:Fibonacci Numbers
Category:Examples of Continued Fractions
Continued Fraction Expansion of Golden Mean/Rate of Convergence

== Theorem ==
Consider the continued fraction expansion to the golden mean:
:$\phi = \sqbrk {1, 1, 1, 1, \ldots} = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$
This continued fraction expansion has the slowest rate of convergence of all simple infinite continued fractions.
== Proof ==
== Sources ==
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
Category:Continued Fraction Expansion of Golden Mean
Continued Fraction Expansion of Golden Mean/Successive Convergents

== Theorem ==
Consider the continued fraction expansion to the golden mean:
:$\phi = \sqbrk {1, 1, 1, 1, \ldots} = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$
The $n$th convergent is given by:
:$C_n = \dfrac {F_{n + 1} } {F_n}$
where $F_n$ denotes the $n$th Fibonacci number.
== Proof ==
The proof proceeds by induction.
Listing the first few convergents, which can be calculated:
:$C_1 = \dfrac 1 1$
:$C_2 = \dfrac 2 1$
:$C_3 = \dfrac 3 2$
:$C_4 = \dfrac 5 3$
and so on.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$C_n = \dfrac {F_{n + 1} } {F_n}$
$\map P 1$ is the case:
:$C_1 = \dfrac {F_{n + 1} } {F_n}$
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
$\map P 2$ is the case:
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$C_k = \dfrac {F_{k + 1} } {F_k}$
from which it is to be shown that:
:$C_{k + 1} = \dfrac {F_{k + 2} } {F_{k + 1} }$
=== Induction Step ===
This is the induction step:
Let $C_n$ be expressed as $\dfrac {p_n} {q_n}$ for any given $n$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{>0}: C_n = \dfrac {F_{n + 1} } {F_n}$
== Sources ==
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
Category:Continued Fraction Expansion of Golden Mean
Continued Fraction Expansion of Irrational Number Converges to Number Itself

== Theorem ==
Let $x$ be an irrational number.
Then the continued fraction expansion of $x$ converges to $x$.
== Proof ==
Let $\sequence {a_0, a_1, \ldots}$ be its continued fraction expansion.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators.
Then $C_n = p_n / q_n$ is the $n$th convergent.
By Accuracy of Convergents of Continued Fraction Expansion of Irrational Number, for $n \ge 2$:
:$\size {x - \dfrac {p_n} {q_n} } < \dfrac 1 {q_n q_{n + 1} }$
By Lower Bounds for Denominators of Simple Continued Fraction:
:$q_n q_{n + 1} \ge n$ for $n \ge 5$
So from Basic Null Sequences and the Squeeze Theorem:
:$\dfrac 1 {q_n q_{n + 1} } \to 0$
as $n \to \infty$.
Thus $C_n = p_n / q_n$ converges to $x$.
That is, $(a_0, a_1, \ldots)$ converges to $x$.
== Also see ==
* Irrational Number is Limit of Unique Simple Infinite Continued Fraction
Category:Continued Fractions
Continued Fraction Expansion of Irrational Square Root

== Theorem ==
Let $n \in \Z$ such that $n$ is not a square.
Then the continued fraction expansion of $\sqrt n$ is of the form:
:$\sqbrk {a_1 \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a_1} }$
or
:$\sqbrk {a_1 \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a_1} }$
where $m \in \Z: m \ge 0$.
That is, it has the form as follows:
:It is periodic
:It starts with an integer $a_1$
:Its cycle starts with a palindromic section, either:
:::$b_1, b_2, \ldots, b_{m - 1}, b_m, b_{m - 1}, \ldots, b_2, b_1$
::or:
:::$b_1, b_2, \ldots, b_{m - 1}, b_m, b_m, b_{m - 1}, \ldots, b_2, b_1$
::which may be of length zero
:Its cycle ends with twice the first partial denominator. 
== Proof ==
We use:
:Expansion of Associated Reduced Quadratic Irrational to establish a series of reduced quadratic irrationals associated to $n$ 
and then use
:Finitely Many Reduced Associated Quadratic Irrationals to assert this series must repeat (hence be periodic) due to the finite number of such irrationals.
Let:
:$a_0 = \floor {\sqrt n}$
:$\sqrt n = a_0 + \dfrac 1 {\alpha_0}$
For all $i \in \N: i > 0$, let:
:$a_{i + 1} = \floor {\alpha_i}$
:$\alpha_i = a_{i + 1} + \dfrac 1 {\alpha_{i + 1} }$
From here, we will prove:
:$\alpha_0$ is a reduced quadratic irrational associated to $n$
:If all $\alpha_i: 0 \le i \le k$ are all reduced quadratic irrationals associated to $n$, then so is $\alpha_{k + 1}$.
Since $\dfrac 1 {\alpha_0}$ is the fractional part of the irrational $\sqrt n$, we have:
:$0 < \dfrac 1 {\alpha_0} < 1 \implies \alpha_0 > 1$
By simple algebra, we have:
:$\alpha_0 = \dfrac {a_0 + \sqrt n} {n - a_0^2}$
:$\tilde {\alpha_0} = \dfrac {a_0 - \sqrt n} {n - a_0^2}$
where $\tilde {\alpha_0}$ is the conjugate of $\alpha_0$.
Since $a_0$ is the floor of $\sqrt n$, we know that:
:$a_0 - \sqrt n < 0 \implies \tilde {\alpha_0} < 0$
Since $n \in \Z \implies \sqrt n > 1$ and $\sqrt n > a_0$, we have:
Thus $\alpha_0$ is a reduced quadratic irrational.
Since $P = a_0$ and $Q = n - a_0^2 = n - P^2$, $Q$ clearly divides $n - P^2$.
Thus $\alpha_0$ is associated to $n$ as well.
We have that each $\alpha_i$ is a reduced quadratic irrational.
So, following the recurrence defined, each $a_i \ge 1$.
Also, by Expansion of Associated Reduced Quadratic Irrational, each $\alpha_{i + 1}$ is reduced and associated to $n$ since $\alpha_0$ is.
By Finitely Many Reduced Associated Quadratic Irrationals, we only have finitely many choices for these.
Hence there must be some smallest $k$ for which $\alpha_k = \alpha_0$.
Since $\alpha_{i + 1}$ is determined completely by $\alpha_i$ we will then have:
:$\alpha_{k + j} = \alpha_j$
for all $j > 0$.
Hence the $\alpha_i$ are periodic.
Similarly, as the $a_i$ for $i > 0$ are determined completely by $\alpha_{i - 1}$, the $a_i$ must be periodic as well.
This forces the continued fraction expansion:
:$\sqrt n = a_0 + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \ddots} }$
to be periodic.
 
== Examples ==
Category:Continued Fractions
Continued Fraction Expansion of Irrational Square Root/Examples

== Examples of Continued Fraction Expansion of Irrational Square Root ==
=== Continued Fraction Expansion of $\sqrt 2$ ===
=== Continued Fraction Expansion of $\sqrt 5$ ===
=== Continued Fraction Expansion of $\sqrt 8$ ===
=== Continued Fraction Expansion of $\sqrt {13}$ ===
=== Continued Fraction Expansion of $\sqrt {29}$ ===
=== Continued Fraction Expansion of $\sqrt {61}$ ===
Category:Examples of Continued Fractions
Continued Fraction Expansion of Irrational Square Root/Examples/13

== Example of Continued Fraction Expansion of Irrational Square Root ==
The continued fraction expansion of the square root of $13$ is given by:
:$\sqrt {13} = \sqbrk {3, \sequence {1, 1, 1, 1, 6} }$
=== Convergents ===
== Proof ==
Let $\sqrt {13} = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$
From Partial Denominators of Continued Fraction Expansion of Irrational Square Root, the partial denominators of this continued fraction expansion can be calculated as:
:$a_r = \floor {\dfrac {\floor {\sqrt {13} } + P_r} {Q_r} }$
where:
: $P_r = \begin {cases} 0 & : r = 0 \\
a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\
\end {cases}$ 
: $Q_r = \begin {cases} 1 & : r = 0 \\
\dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\
\end {cases}$ 
|}
and the cycle is complete:
:$\sequence {1, 1, 1, 1, 6}$
Category:Examples of Continued Fractions
Category:13
Continued Fraction Expansion of Irrational Square Root/Examples/13/Convergents

== Convergents to Continued Fraction Expansion of $\sqrt {13}$ ==
The sequence of convergents to the continued fraction expansion of the square root of $13$ begins:
:$\dfrac 3 1, \dfrac 4 1, \dfrac 7 2, \dfrac {11} 3, \dfrac {18} 5, \dfrac {119} {33}, \dfrac {137} {38}, \dfrac {256} {71}, \dfrac {393} {109}, \dfrac {649} {180}, \ldots$
== Proof ==
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators.
Then the $n$th convergent is $\dfrac {p_n} {q_n}$.
By definition:
:$p_k = \begin {cases} a_0 & : k = 0 \\
a_0 a_1 + 1 & : k = 1 \\
a_k p_{k - 1} + p_{k - 2} & : k > 1 \end {cases}$
:$q_k = \begin {cases} 1 & : k = 0 \\
a_1 & : k = 1 \\
a_k q_{k - 1} + q_{k - 2} & : k > 1 \end {cases}$
From Continued Fraction Expansion of $\sqrt {13}$:
:$\sqrt {13} = \sqbrk {3, \sequence {1, 1, 1, 1, 6} }$
Thus the convergents are assembled:
|}
Category:Examples of Continued Fractions
Category:13
Continued Fraction Expansion of Irrational Square Root/Examples/2

== Examples of Continued Fraction Expansion of Irrational Square Root ==
The continued fraction expansion of the square root of $2$ is given by:
:$\sqrt 2 = \sqbrk {1, \sequence 2}$
=== Convergents ===
== Proof ==
Thus it is possible to replace $\sqrt 2$ recursively:
The pattern repeats indefinitely, producing the continued fraction expansion:
:$\sqrt 2 = \sqbrk {1, 2, 2, 2, \ldots} = \sqbrk {1, \sequence 2}$
Category:Examples of Continued Fractions
Category:2
Continued Fraction Expansion of Irrational Square Root/Examples/2/Convergents

== Convergents to Continued Fraction Expansion of $\sqrt 2$ ==
The sequence of convergents to the continued fraction expansion of the square root of $2$ begins:
:$\dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \dfrac {1393} {985}, \dfrac {3363} {2378}, \ldots$
== Proof ==
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators.
Then the $n$th convergent is $\dfrac {p_n} {q_n}$.
By definition:
:$p_k = \begin {cases} a_0 & : k = 0 \\
a_0 a_1 + 1 & : k = 1 \\
a_k p_{k - 1} + p_{k - 2} & : k > 1 \end {cases}$
:$q_k = \begin {cases} 1 & : k = 0 \\
a_1 & : k = 1 \\
a_k q_{k - 1} + q_{k - 2} & : k > 1 \end {cases}$
From Continued Fraction Expansion of $\sqrt 2$:
:$\sqrt 2 = \sqbrk {1, \sequence 2}$
Thus the convergents are assembled:
Category:Examples of Continued Fractions
Category:Pell Numbers
Category:2
Continued Fraction Expansion of Irrational Square Root/Examples/29

== Examples of Continued Fraction Expansion of Irrational Square Root ==
The continued fraction expansion of the square root of $29$ is given by:
:$\sqrt {29} = \sqbrk {5, \sequence {2, 1, 1, 2, 10} }$
=== Convergents ===
== Proof ==
Let $\sqrt {29} = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$
From Partial Denominators of Continued Fraction Expansion of Irrational Square Root, the partial denominators of this continued fraction expansion can be calculated as:
:$a_r = \floor {\dfrac {\floor {\sqrt {29} } + P_r} {Q_r} }$
where:
:$P_r = \begin {cases} 0 & : r = 0 \\
a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\
\end {cases}$
:$Q_r = \begin {cases} 1 & : r = 0 \\
\dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\
\end {cases}$
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and the cycle is complete:
:$\sequence {2, 1, 1, 2, 10}$
Category:Examples of Continued Fractions
Category:29
Continued Fraction Expansion of Irrational Square Root/Examples/29/Convergents

== Convergents to Continued Fraction Expansion of $\sqrt {29}$ ==
The sequence of convergents to the continued fraction expansion of the square root of $29$ begins:
:$\dfrac 5 1, \dfrac {11} 2, \dfrac {16} 3, \dfrac {27} 5, \dfrac {70} {13}, \dfrac {727} {135}, \dfrac {1524} {283}, \dfrac {2251} {418}, \dfrac {3775} {701}, \dfrac {9801} {1820}, \ldots$
== Proof ==
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators.
Then the $n$th convergent is $p_n / q_n$.
By definition:
:$p_k = \begin {cases} a_0 & : k = 0 \\
a_0 a_1 + 1 & : k = 1 \\
a_k p_{k - 1} + p_{k - 2} & : k > 1\end {cases}$
:$q_k = \begin {cases} 1 & : k = 0 \\
a_1 & : k = 1 \\
a_k q_{k - 1} + q_{k - 2} & : k > 1\end {cases}$
From Continued Fraction Expansion of $\sqrt {29}$:
:$\sqrt {29} = \sqbrk {5, \sequence {2, 1, 1, 2, 10} }$
Thus the convergents are assembled:
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Category:Examples of Continued Fractions
Category:29