nyuuzyou/wikis huggingface dataset:
Paths Crossing Rectangle Necessarily Meet

== Theorem ==
Let $R = \closedint a b \times \closedint c d$ be a closed rectangle.
Let $p, q : \closedint {-1} 1 \to R$ be paths in $R$.
Suppose:
* $\map {p_1} {-1} = a$
* $\map {p_1} 1 = b$
* $\map {q_2} {-1} = c$
* $\map {q_2} 1 = d$
where:
:$\map p t = \tuple {\map {p_1} t, \map {p_2} t}$
:$\map q t = \tuple {\map {q_1} t, \map {q_2} t}$
Intuitively, that is to say:
:The path $p$ begins on the left edge of $R$, and ends on the right edge.
:The path $q$ begins on the bottom edge of $R$, and ends on the top edge.
Then, there exist $t_p, t_q \in \closedint {-1} 1$ such that:
:$\map p {t_p} = \map q {t_q}$
which is to say, the two paths meet at some point in $R$.
== Proof ==
, suppose there are no such $t_p, t_q$.
Define $N : \closedint {-1} 1^2 \to \R_{\ge 0}$ as:
:$\map N {s, t} = \map \max {\size {\map {p_1} s - \map {q_1} t}, \size {\map {p_2} s - \map {q_2} t}}$
By our assumption:
:$\map p s \ne \map q t$
for all $s, t \in \closedint {-1} 1$.
Therefore, for each $s, t$, at least one of $\map {p_1} s \ne \map {q_1} t$ or $\map {p_2} s \ne \map {q_2} t$ holds.
Thus, $\map N {s, t} > 0$ for all $s, t \in \closedint {-1} 1$.
Additionally, by Combination Theorem for Continuous Mappings on Metric Space:
:$N$ is continuous on $\closedint {-1} 1^2$
Now, we define $F : \closedint {-1} 1^2 \to \closedint {-1} 1^2$ as:
:$\map F {s, t} = \tuple {\dfrac {\map {q_1} t - \map {p_1} s} {\map N {s, t}}, \dfrac {\map {p_2} s - \map {q_2} t} {\map N {s, t}}}$
First, we must show that $F$ is well-defined.
Since:
:$\map N {s, t} \ge \size {\map {q_1} t - \map {p_1} s}$
:$\map N {s, t} \ge \size {\map {p_2} s - \map {q_2} t}$
it follows that the image of $F$ is indeed contained in $\closedint {-1} 1^2$.
Next, we want to prove that:
:$\Img F \subseteq \closedint {-1} 1 \times \set {-1, 1} \cup \set {-1, 1} \times \closedint {-1} 1$
or in other words, for all $s, t \in \closedint {-1} 1$:
:$1 \in \set {\size {\map {F_1} {s, t}}, \size {\map {F_2} {s, t}}}$
Observe that, by definition of $N$:
:$\map N {s, t} = \size {\map {p_i} s - \map {q_i} t}$
for at least one of $i \in \set {1, 2}$.
Then:
By:
* Brouwer's Fixed Point Theorem
* Compact Convex Sets with Nonempty Interior are Homeomorphic
* Fixed-Point Property is Topological
it follows that $F$ must have a fixed point $\tuple {s_0, t_0}$.
That is:
:$\map F {s_0, t_0} = \tuple {s_0, t_0}$
By the above argument, at least one of the following holds:
* $s_0 = -1$
* $s_0 = 1$
* $t_0 = -1$
* $t_0 = 1$
Suppose, for instance, $s_0 = -1$.
Then, , $\map {p_1} {s_0} = a$.
Since $\map {q_1} {t_0} \in \closedint a b$, it follows that:
:$\map {q_1} {t_0} \ge \map {p_1} {s_0}$
Therefore:
which contradicts our supposition that $\map F {s_0, t_0} = \tuple {s_0, t_0}$
In the same manner, we can show:
* $s_0 = 1 \implies \map {F_1} {s_0, t_0} \le 0$
* $t_0 = -1 \implies \map {F_2} {s_0, t_0} \ge 0$
* $t_0 = 1 \implies \map {F_2} {s_0, t_0} \le 0$
all of which are clearly contradictory.
Since every case leads to a contradiction, we conclude that there must exist $t_p, t_q$ such that:
:$\map p {t_p} = \map q {t_q}$
== Sources ==
* 
Category:Paths (Topology)
Paths of Minimal Length from Vertex form Tree

== Theorem ==
Let $G = \struct {V, E}$ be a simple graph.
Let $r \in V$ be a vertex in $G$.
Let $P$ be the set of minimal length paths beginning at $r$.
Let $p, q \in P$.
Let $E'$ be defined as follows:
$\set {p, q} \in E'$ either:
:$q$ is formed by extending $p$ with one edge of $E$ and one vertex of $V$
or:
:$p$ is formed by extending $q$ with one edge of $E$ and one vertex of $V$.
Then $T = \struct {P, E'}$ is a tree.
== Proof ==
Let $\tuple r$ be the $0$-length $G$-path whose only vertex is $r$. 
=== Connected ===
We will show that there is a $T$-path from $\tuple r$ to each element $p$ of $P$.
We proceed by induction on the length of $p$.
Let $p$ have length $0$.
Then by definition:
:$p = \tuple r$
Thus the $0$-length $T$-path $\tuple {\tuple r}$ connects $\tuple r$ to $p$.
Suppose that there is a $T$-path from $\tuple r$ to each element of $P$ of length $n$.
Let $p \in P$ have length $n + 1$ and final vertex $z$.
Let $p^*$ be the $G$-path obtained from $p$ by removing the last vertex and the last edge.
Then $p^*$ is a $G$-path of length $n$.
We must show that $p^* \in P$.
Let $w$ be the last vertex of $p^*$.
 $p^* \notin P$.
Then by definition of $P$, there is a $G$-path $m$ from $r$ to $w$ which is shorter than $p^*$.
But then appending the last edge and vertex of $p$ to $m$, one obtains a $G$-path from $r$ to $z$ which is shorter than $p$.
This contradicts the supposition that $p \in P$.
Thus we conclude that $p^* \in P$.
By the inductive hypothesis, there is a $T$-path $b$ from $\tuple r$ to $p^*$.
Then by definition of $E'$, there is an edge from $p^*$ to $p$.
Appending that edge and $p$ to $b$ yields a $T$-path from $\tuple r$ to $p$.
As such a $T$-path exists for each $p \in P$, $T$ is connected.
=== Acyclic ===
Category:Paths (Graph Theory)
Category:Tree Theory
Paving Lemma

== Theorem ==
Let $S$ be an open subset of the Euclidean space $\R^m$ or the set of complex numbers $\C$.
Let $\gamma: \closedint a b \to S$ be a path in $S$.
Then there exists $K \in \R_{>0}$ such that:
:For all $\epsilon \in \openint 0 K$, there exists a normal subdivision $\set {x_0, x_1, \ldots, x_{n - 1}, x_n}$ of the closed interval $\closedint a b$ such that:
::$\ds \bigcup_{i \mathop = 0}^n \map {B_\epsilon} {\map \gamma {x_i} } \subseteq S$
:and for all $i \in \set {0, 1, \ldots, n - 1}$:
::$\map \gamma {\closedint {x_i} {x_{i + 1} } } \subseteq \map {B_\epsilon} {\map \gamma {x_i} }$
Here, $\map {B_\epsilon} {\map \gamma {x_i} }$ denotes the open ball of $\map \gamma {x_i}$ with radius $\epsilon$.
== Proof ==
=== Finding the constant ===
First, suppose that $S$ is a subset of $\R^m$.
From Closed Real Interval is Compact in Metric Space, it follows that $\closedint a b$ is compact.
Then Continuous Image of Compact Space is Compact shows that $\map \gamma {\closedint {x_i} {x_{i + 1} } }$ is compact.
From the Heine-Borel Theorem, it follows that $\map \gamma {\closedint {x_i} {x_{i + 1} } }$ is bounded and closed.
Suppose that $S \ne \R^m$.
As $\R^m \setminus S$ is closed, it follows from Distance between Closed Sets in Euclidean Space that:
:$\map d {\map \gamma {\closedint a b}, \R^m \setminus S} > 0$
Put $K = \map d {\map \gamma {\closedint a b}, \R^m \setminus S}$, the distance between $\map \gamma {\closedint a b}$ and $\R^m \setminus S$.
If instead $S = \R^m$, then $K$ may be any strictly positive real number.
=== Finding the subdivision ===
Let $\epsilon \in \openint 0 K$.
From the Heine-Cantor Theorem, it follows that $\gamma$ is uniformly continuous.
Then there exists $\delta \in \R_{>0}$ such that:
:$\forall y, z \in \closedint a b : \size {y - z} < \delta$ implies $\map d {\map \gamma y, \map \gamma z } < \epsilon$
where $d$ denotes the Euclidean metric.
Choose $n \in \N$ such that $\dfrac {b - a} n < \delta$, and put $x_i = a + i \dfrac {b - a} n$ for all $i \in \set {0, 1, \ldots, n}$.
Then $\set {x_0, x_1, \ldots, x_n}$ is a normal subdivision of $\closedint a b$.
It follows that $\map {B_\epsilon} {\map \gamma {x_i} }$ and $\R^m \setminus S$ are disjoint, as either:
:$\epsilon < \map d {\map \gamma {\closedint a b }, \R^m \setminus S}$
or:
:$\R \setminus S = \O$
Then: 
:$\ds \bigcup_{i \mathop = 0}^n \map {B_\epsilon} {\map \gamma {x_i} } \subseteq S$
Let $x \in \closedint {x_i} {x_{i + 1} }$.
Then:
:$\size {x - x_i} \le \dfrac{b - a} n < \delta$
As the uniorm continuity condition applies, it follows that:
:$\map \gamma {\closedint {x_i} {x_{i + 1} } } \subseteq \map {B_\epsilon} {\map \gamma {x_i} }$
=== Complex plane case ===
Suppose that $S$ is a subset of $\C$.
This theorem only uses the properties of $\C$ that depends on the metric.
From Complex Plane is Metric Space, it follows that $\C$ and $\R^2$ are homeomorphic.
It follows that the proof that applies for $\R^2$ also applies for $\C$.
== Sources ==
* $\S 5.1$
Category:Complex Analysis
Category:Euclidean Space
Category:Named Theorems
Payoff Table/Examples/Arbitrary Example 1

== Examples of Payoff Tables ==
Consider a two-person game $G$ with players are $\text A$ and $\text B$ such that:
:player $\text A$ make make any of $3$ moves
:player $\text B$ make make any of $4$ moves.
The payoff table for player $\text A$ will be in the form:
Here, $a_{i j}$ is the amount $\text A$ wins if $\text A$ makes move $i$ and $\text B$ makes move $j$.
If $G$ is a zero-sum game, then player $\text B$'s payoff table will be the same as for player $\text A$, but with $a_{i j}$ replaced with $-a_{i j}$.
== Sources ==
* 
* 
Category:Examples of Payoff Tables
Peak Point Lemma

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$ which is infinite.
Then $\sequence {x_n}$ has an infinite subsequence which is monotone.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
This result is also known as the Spanish hotel theorem, as is apparent from proof 2.
Category:Limits of Sequences
Category:Named Theorems
Category:Peak Point Lemma
Peak Point Lemma/Proof 1

== Theorem ==
== Proof ==
There are $2$ cases to consider.
First, suppose that every set $\set {x_n: n > N}$ has a maximum.
If this is the case, we can find a sequence $n_r \in \N$ such that:
:$\ds x_{n_1} = \max \set {x_n: n > 1}$
:$\ds x_{n_2} = \max \set {x_n: n > n_1}$
:$\ds x_{n_3} = \max \set {x_n: n > n_2}$
and so on.
From the method of construction, $n_1 < n_2 < n_3 < \cdots$, so at each stage we are taking the maximum of a subset of the previous set.
At each stage, the previous maximum has already been taken as the previous term in the sequence.
Thus, $\sequence {x_{n_r} }$ is a decreasing subsequence of $\sequence {x_n}$.
Second, suppose it is not true that every set $\set {x_n: n > N}$ has a maximum.
Then there will exist some $N_1$ such that $\set {x_n: n > N_1}$ has no maximum.
So it follows that, given some $x_m$ with $m > N_1$, we can find an $x_n$ following that $x_m$ such that $x_n > x_m$.
(Otherwise, the biggest of $x_{N_1 + 1}, \ldots, x_m$ would be a maximum for $\set {x_n: n > N_1}$.)
So, we define $x_{n_1} = x_{N_1 + 1}$.
Then $x_{n_2}$ can be the first term after $x_{n_1}$ such that $x_{n_2} > x_{n_1}$.
Then $x_{n_3}$ can be the first term after $x_{n_2}$ such that $x_{n_3} > x_{n_2}$.
And so on.
Thus we get an increasing subsequence of $\sequence {x_n}$.
There exist only these two possibilities.
From each one we get a subsequence that is monotone:
:one is decreasing
:one is increasing.
We can of course choose instead to investigate whether every set $\set {x_n: n > N}$ has a minimum.
The same conclusion will be reached.
== Sources ==
* : $\S 5$: Subsequences: Bolzano-Weierstrass Theorem: $\S 5.9$
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Peak Point Lemma
Peak Point Lemma/Proof 2

== Theorem ==
== Proof ==
A coastal town in Spain has an infinite row of hotels along a road leading down to the beach.
The tourist bureau which rates these hotels has a special designation for any hotel having a view of the sea.
Any hotel which is at least as tall as the rest of the hotels on the road to the sea receives this view-rating.
Starting with the hotel farthest from the sea, let $x_1, x_2, \ldots$ denote the heights of the hotels as we travel along the road to the beach.
One possibility is that an infinite number of hotels are view-rated.
In this case the heights of these hotels form a decreasing (i.e. nonincreasing) subsequence of the original sequence of heights.
The other possibility is that only a finite number of hotels are view-rated.
In this case we can obtain an increasing subsequence as follows.
Walk along the road to the beach past all the view-rated hotels.
Let $n_1$ be the index of the next hotel.
Since it does not have a view of the sea, there is a taller hotel nearer the shore which blocks the view.
Let $n_2$ be the index of that taller hotel.
Since it is not view-rated, it does not have a view of the sea.
Thus there must be an even taller hotel nearer the shore which blocks the view.
In this way we generate a list of indices $n_1 < n_2 < n_3 \cdots$ of taller and taller hotels.
This gives an increasing subsequence $x_{n_1}, x_{n_2}, x_{n_3}, \ldots$ of the original sequence.
== Also known as ==
This result is also known as the Spanish hotel theorem.
== Sources ==
* : $\S 5.9$
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Peak Point Lemma
Peano Structure is Unique

== Theorem ==
Let $\struct {P, s, 0}$ and $\struct {P', s', 0'}$ be Peano structures.
Then there is a unique bijection $f: P \to P'$ such that:
== Proof ==
First to establish uniqueness of $f$.
Suppose that $f, g: P \to P'$ both satisfy the conditions.
Define $A \subseteq P$ as:
:$A := \set {n \in P: \map f n = \map g n}$
Then the first condition implies that $0 \in A$.
Now suppose that $n \in A$. Then:
Hence $\map s n \in A$.
Since $P$ is a Peano structure, it follows that $A = P$.
Hence by Equality of Mappings, $f = g$.
For the existence of such an $f$, we apply the Principle of Recursive Definition with $0' \in P'$ and $s': P' \to P'$.
Lastly, it is to be established that the $f$ so obtained is a bijection.
Let $A' \subseteq P$ be defined as:
:$A' := \set {n \in P: \forall m \in P: \map f n = \map f m \implies n = m}$
It follows from Axiom $(P4)$ that $\map f n = 0'$ implies $n = 0$.
Hence $0 \in A'$.
Suppose now that $n \in A'$, but $\map s n \notin A'$.
Then, since $\map f {\map s n} \ne 0'$, there exists some $m \in P, m \ne n$ such that:
:$\map f {\map s n} = \map f {\map s m}$
That is, such that:
:$\map {s'} {\map f n} = \map {s'} {\map f m}$
But by Axiom $(P3)$, $s'$ is injective.
Hence $\map f n = \map f m$, contradicting the assumption that $n \in A'$.
Therefore $\map s n \in A'$, so that $A' = P$ by Axiom $(P5)$.
That is, $f$ is an injection.
Next, define $A \subseteq P'$ as:
:$A := \set {n' \in P': \exists n \in P: \map f n = n'}$
Since $\map f 0 = 0'$, we have $0' \in A$.
Suppose now that $n' \in A$, and let $\map f n = n'$.
Then:
meaning $\map {s'} {n'} \in A$.
Hence by Axiom $(P5)$, $A = P'$.
That is, $f$ is surjective.
It follows that $f$ is the sought bijection.
Category:Abstract Algebra
Pedal Circle of Excenter is Excircle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $H$ be an excenter of $\triangle ABC$.
Then the pedal circle of $H$ is the excircle of $\triangle ABC$ whose center is $H$.
== Proof ==
Let $\EE$ denote the excircle of $\triangle ABC$ whose center is $H$.
 $\EE$ is not the pedal circle of $H$.
By definition, the sides of $\triangle ABC$ are tangent to $\EE$.
Hence from , the line through $H$ to those point of tangency to $\EE$ are perpendicular to the sides.
Hence $\EE$ is the pedal circle of $H$ by definition.
Hence by Proof by Contradiction, the pedal circle of $H$ is the excircle of $\triangle ABC$ whose center is $H$.
== Also see ==
* Pedal Circle of Incenter is Incircle
Category:Pedal Circles
Category:Excircles of Triangles
Pedal Circle of Incenter is Incircle

== Theorem ==
Let $\triangle ABC$ be a triangle whose incenter is $H$.
Then the pedal circle of $H$ is the incircle of $\triangle ABC$.
== Proof ==
Let $\CC$ denote the incircle of $\triangle ABC$.
 $\CC$ is not the pedal circle of $H$.
By definition, the sides of $\triangle ABC$ are tangent to $\CC$.
Hence from , the line through $H$ to those point of tangency to $\CC$ are perpendicular to the sides.
Hence $\CC$ is the pedal circle of $H$ by definition.
Hence by Proof by Contradiction, the pedal circle of $H$ is the incircle.
== Also see ==
* Pedal Circle of Excenter is Excircle
Category:Pedal Circles
Category:Incircles of Triangles
Pedal Triangle of Point on Circumcircle is Straight Line

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $P$ be an arbitrary point on the circumcircle of $\triangle ABC$.
The pedal triangle of $\triangle ABC$ $P$ degenerates to a straight line segment.
== Proof ==
== Sources ==
* 
Category:Pedal Triangles
Category:Circumcircles of Triangles
Peirce's Law/Formulation 2

== Theorem ==
:$\vdash \paren {\paren {p \implies q} \implies p} \implies p$
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : Chapter $2$: The Propositional Calculus $2$: $2$: Theorems and Derived Rules: Exercise $5 \ \text{(c)}$
* : $\text{I}$: 'NOT' and 'IF': $\S 3$: Exercises, Group $\text{I}: \ 14$
* : $\text{I}$: 'NOT' and 'IF': $\S 5$: Theorem $\text{T23}$
* : $\S 1.14$: Exercise $12 \ (11)$
Category:Peirce's Law
Peirce's Law/Formulation 2/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connective are $\T$ for all boolean interpretations.
$\begin{array}{|ccccc|c|c|}\hline
((p & \implies & q) & \implies & p) & \implies & p \\
\hline
\F & \T & \F & \F & \F & \T & \F \\
\F & \T & \T & \F & \F & \T & \F \\
\T & \F & \F & \T & \T & \T & \T \\
\T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : $\S 1.4.1$: The meaning of logical connectives: Exercise $1.8: \ 2$
Category:Peirce's Law
Category:Truth Table Proofs
Peirce's Law/Historical Note

== Historical Note ==
's own statement and proof of the Peirce's Law:
:A fifth icon is required for the principle of excluded middle and other propositions connected with it. One of the simplest formulae of this kind is:
::$\paren {\paren {x \mathop {-\!\!\!<} y} \mathop {-\!\!\!<} x} \mathop {-\!\!\!<} x$
:This is hardly axiomatical. That it is true appears as follows. It can only be false by the final consequent $x$ being false while its antecedent $\paren {x \mathop {-\!\!\!<} y} \mathop {-\!\!\!<} x$ is true. If this is true, either its consequent $x$ is true, when the whole formula would be true, or its antecedent $x \mathop{-\!\!\!<} y$ is false. But in the last case the antecedent of $x \mathop{-\!\!\!<} y$, that is $x$, must be true.
Peirce goes on to point out an immediate application of the law:
:From the formula just given, we at once get:
::$\paren {\paren {x \mathop {-\!\!\!<} y} \mathop {-\!\!\!<} a} \mathop {-\!\!\!<} x$
:where the $a$ is used in such a sense that $\paren {x \mathop {-\!\!\!<} y} \mathop {-\!\!\!<} a$ means that from $\paren {x \mathop {-\!\!\!<} y}$ every proposition follows. With that understanding, the formula states the principle of excluded middle, that from the falsity of the denial of $x$ follows the truth of $x$.
Note the use by of the sign of illation $-\!\!\!<$ for implication.
== Sources ==
* 
Category:Historical Notes
Peirce's Law/Strong Form/Formulation 1

== Theorem ==
:$\paren {\paren {p \implies q} \implies p} \dashv \vdash p$
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||c|}\hline
((p & \implies & q) & \implies & p) & p \\
\hline
\F & \T & \F & \F & \F & \F \\
\F & \T & \T & \F & \F & \F \\
\T & \F & \F & \T & \T & \T \\
\T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Category:Peirce's Law
Category:Truth Table Proofs
Peirce's Law is Equivalent to Law of Excluded Middle

== Theorem ==
Peirce's Law:
:$\paren {p \implies q} \implies p \vdash p$
is logically equivalent to the Law of Excluded Middle:
:$\vdash p \lor \neg p$
That is, Peirce's Law holds the Law of Excluded Middle holds.
== Proof ==
=== Law of Excluded Middle implies Peirce's Law ===
Let the truth of the Law of Excluded Middle be assumed.
Then:
:$\paren {p \lor \neg p} \vdash \paren {\paren {p \implies q} \implies p} \implies p$
is demonstrated, as follows.
The result follows by an application of Modus Ponendo Ponens:
:$\paren {\paren {p \implies q} \implies p} \implies p, \paren {p \implies q} \implies p \vdash p$
=== Peirce's Law implies Law of Excluded Middle ===
Category:Peirce's Law
Category:Law of Excluded Middle
Pell's Equation/Examples

== Examples of Solutions of Pell's Equation ==
=== Pell's equation: $x^2 - 2 y^2 = 1$ ===
=== Pell's equation: $x^2 - 2 y^2 = -1$ ===
=== Pell's equation: $x^2 - 8 y^2 = 1$ ===
=== Pell's equation: $x^2 - 13 y^2 = 1$ ===
=== Pell's equation: $x^2 - 29 y^2 = 1$ ===
=== Pell's equation: $x^2 - 61 y^2 = 1$ ===
Category:Pell's Equation
Pell's Equation/Examples/13

== Theorem ==
Pell's equation:
:$x^2 - 13 y^2 = 1$
has the smallest positive integral solution:
:$x = 649$
:$y = 180$
== Proof ==
From Continued Fraction Expansion of $\sqrt {13}$:
:$\sqrt {13} = \sqbrk {3, \sequence {1, 1, 1, 1, 6} }$
The cycle is of length $5$.
By Solution of Pell's Equation, the only solutions of $x^2 - 13 y^2 = 1$ are:
:${p_{5 r} }^2 - 13 {q_{5 r} }^2 = \paren {-1}^{5 r}$
for $r = 1, 2, 3, \ldots$
When $r = 1$ this gives:
:${p_5}^2 - 13 {q_5}^2 = -1$
which is not the solution required.
When $r = 2$ this gives:
:${p_{10} }^2 - 13 {q_{10} }^2 = 1$
From Convergents of Continued Fraction Expansion of $\sqrt {13}$:
:$p_{10} = 649$
:$q_{10} = 180$
although on that page the numbering goes from $p_0$ to $p_9$, and $q_0$ to $q_9$.
== Sources ==
* : $13$
Category:Pell's Equation
Category:13
Pell's Equation/Examples/2

== Theorem ==
Pell's equation:
:$x^2 - 2 y^2 = 1$
has the positive integral solutions:
:$\begin {array} {r|r} x & y \\ \hline
3 & 2 \\
17 & 12 \\
99 & 70 \\
577 & 408 \\
3363 & 2378 \\
\end {array}$
and so on.
== Proof ==
From Continued Fraction Expansion of $\sqrt 2$:
:$\sqrt 2 = \sqbrk {1, \sequence 2}$
The cycle is of length is $1$.
By Solution of Pell's Equation, the only solutions of $x^2 - 2 y^2 = -1$ are:
:${p_r}^2 - 2 {q_r}^2 = \paren {-1}^r$
for $r = 1, 2, 3, \ldots$
From Convergents to Continued Fraction Expansion of $\sqrt 2$:
from which the solutions are obtained by taking the convergents with even indices.
Category:Pell's Equation
Category:2
Pell's Equation/Examples/2/-1

== Theorem ==
Pell's equation:
:$x^2 - 2 y^2 = -1$
has the positive integral solutions:
:$\begin {array} {r|r} x & y \\ \hline
1 & 1 \\
7 & 5 \\
41 & 29 \\
239 & 169 \\
1393 & 985 \\
\end {array}$
and so on.
== Proof ==
From Continued Fraction Expansion of $\sqrt 2$:
:$\sqrt 2 = \sqbrk {1, \sequence 2}$
The cycle is of length is $1$.
By Solution of Pell's Equation, the only solutions of $x^2 - 2 y^2 = -1$ are:
:${p_r}^2 - 2 {q_r}^2 = \paren {-1}^r$
for $r = 1, 2, 3, \ldots$
From Convergents to Continued Fraction Expansion of $\sqrt 2$:
from which the solutions are obtained by taking the convergents with odd indices.
Category:Pell's Equation
Category:2
Pell's Equation/Examples/29

== Theorem ==
Pell's equation:
:$x^2 - 29 y^2 = 1$
has the smallest positive integral solution:
:$x = 9801$
:$y = 1820$
== Proof ==
From Continued Fraction Expansion of $\sqrt {29}$:
:$\sqrt {29} = \sqbrk {5, \sequence {2, 1, 1, 2, 10} }$
The cycle is of length is $5$.
By Solution of Pell's Equation, the only solutions of $x^2 - 29 y^2 = 1$ are:
:${p_{5 r} }^2 - 29 {q_{5 r} }^2 = \paren {-1}^{5 r}$
for $r = 1, 2, 3, \ldots$
When $r = 1$ this gives:
:${p_5}^2 - 29 {q_5}^2 = -1$
which is not the solution required.
When $r = 2$ this gives:
:${p_{10} }^2 - 29 {q_{10} }^2 = 1$
From Convergents of Continued Fraction Expansion of $\sqrt {29}$:
:$p_{10} = 9801$
:$q_{10} = 1820$
although on that page the numbering goes from $p_0$ to $p_9$, and $q_0$ to $q_9$.
== Sources ==
* : $13$
Category:Pell's Equation
Category:29
Pell's Equation/Examples/61

== Theorem ==
Pell's equation:
:$x^2 - 61 y^2 = 1$
has the smallest positive integral solution:
:$x = 1 \, 766 \, 319 \, 049$
:$y = 226 \, 153 \, 980$
== Proof ==
From Continued Fraction Expansion of $\sqrt {61}$:
:$\sqrt {61} = \sqbrk {7, \sequence {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14} }$
The cycle is of length is $11$.
By Solution of Pell's Equation, the only solutions of $x^2 - 61 y^2 = 1$ are:
:${p_{11 r} }^2 - 61 {q_{11 r} }^2 = \paren {-1}^{11 r}$
for $r = 1, 2, 3, \ldots$
When $r = 1$ this gives:
:${p_{11}}^2 - 61 {q_{11}}^2 = -1$
which is not the solution required.
When $r = 2$ this gives:
:${p_{22} }^2 - 61 {q_{22} }^2 = 1$
From Convergents of Continued Fraction Expansion of $\sqrt {61}$:
:$p_{22} = 1 \, 766 \, 319 \, 049$
:$q_{22} = 226 \, 153 \, 980$
although on that page the numbering goes from $p_0$ to $p_{21}$, and $q_0$ to $q_{21}$.
== Sources ==
* : $61$
Category:Pell's Equation
Category:61