nyuuzyou/wikis huggingface dataset:
Perpendicular Distance from Straight Line in Plane to Point/General Form/Proof 1

== Theorem ==
== Proof ==
We have that $\LL$ has the equation:
:$(1): \quad a x + b y + c = 0$
Let a perpendicular be dropped from $P$ to $\LL$ at $Q$.
The perpendicular distance $d$ that we are to find is then $PQ$.
In order to simplify the algebra that will inevitably follow, we are to make a transformation as follows.
Let $\MM$ be constructed parallel to $\LL$.
Construct a perpendicular from $\MM$ to pass through the origin.
Let this perpendicular intersect $\MM$ at $R$ and $\LL$ at $S$.
We have that $PQSR$ is a rectangle, and so $RS = PQ$.
It remains to establish the length of $RS$.
We can manipulate $(1)$ into slope-intercept form as:
:$y = -\dfrac a b x - \dfrac c b$
Thus the slope of $\LL$ is $-\dfrac a b$.
From Condition for Straight Lines in Plane to be Perpendicular, the slope of $RS$ is then $\dfrac b a$.
The next step is to find the coordinates of $R$ and $S$.
From Equation of Straight Line in Plane: Point-Slope Form, the equation of $\MM$ can be given as:
:$y - y_0 = -\dfrac a b \paren {x - x_0}$
or:
:$(2): \quad y = \dfrac {-a x + a x_0 + b y_0} b$
From Equation of Straight Line in Plane: Slope-Intercept Form, the equation of $RS$ can be given as:
:$(3): \quad y = \dfrac b a x$
$\MM$ and $RS$ intersect where these are equal:
:$\dfrac b a x = \dfrac {-a x + a x_0 + b y_0} b$
which gives us:
:$x = \dfrac {a \paren {a x_0 + b y_0} } {a^2 + b^2}$
Substituting back for $y$ in $3$, we find that:
:$R = \tuple {\dfrac {a \paren {a x_0 + b y_0} } {a^2 + b^2}, \dfrac {b \paren {a x_0 + b y_0} } {a^2 + b^2} }$
Now to find the coordinates of $S$, which is the intersection of $\LL$ and $RS$.
We can express $\LL$ as:
:$y = -\dfrac {a x + c} b$
and so:
:$\dfrac b a x = -\dfrac {a x + c} b$
which leads to:
:$x = -\dfrac {a c} {a^2 + b^2}$
Substituting back for $y$ in $3$, we get (after algebra):
:$S = \tuple {\dfrac {-a c} {a^2 + b^2}, \dfrac {-b c} {a^2 + b^2} }$
It remains to find the length $d$ of $RS$.
From the Distance Formula:
Category:Perpendicular Distance from Straight Line in Plane to Point
Perpendicular Distance from Straight Line in Plane to Point/General Form/Proof 2

== Theorem ==
== Proof ==
Let a perpendicular be dropped from $P$ to $\LL$ at $Q$.
Let $PQ$ make an angle $\alpha$ with the $x$-axis.
Let $p$ be the length of $PQ$.
Then the coordinates of $Q$ are given by:
:$Q = \tuple {x_0 + p \cos \alpha, y_0 + p \sin \alpha}$
$Q$ lies on $a x + b y + c$, and so:
:$a \paren {x_0 + p \cos \alpha} + b \paren {y_0 + p \sin \alpha} + c = 0$
But from Condition for Straight Lines in Plane to be Perpendicular:
The minus sign has no immediate significance, and the result follows.
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line: The length of the perpendicular
Category:Perpendicular Distance from Straight Line in Plane to Point
Perpendicular Distance from Straight Line in Plane to Point/Normal Form

== Theorem ==
Let $\LL$ be a straight line in the Cartesian plane.
Let $\LL$ be expressed in normal form:
:$x \cos \alpha + y \sin \alpha = p$
Let $P$ be a point in the cartesian plane whose coordinates are given by:
:$P = \tuple {x_0, y_0}$
Then the perpendicular distance $d$ from $P$ to $\LL$ is given by:
:$\pm d = x_0 \cos \alpha = y_0 \sin \alpha - p$
where $\pm$ depends on whether $P$ is on the same side of $\LL$ as the origin $O$.
== Proof ==
First suppose that $P$ is on the opposite side of $\LL$ from the origin $O$.
Let $MP$ be the ordinate of $P$.
Let $N$ be the point of intersection between $\LL$ and the perpendicular through $O$.
Let $ON$ be produced to $N'$ where $PN'$ is the straight line through $P$ parallel to $\LL$.
:
We have that:
:$d = NN'$
and so:
:$x_0 \cos \alpha + y_0 \sin \alpha = ON' = p + d$
That is:
:$d = x_0 \cos \alpha + y_0 \sin \alpha - p$
By a similar construction, if $P$ is on the same side of $\LL$ as the origin $O$:
:$-d = x_0 \cos \alpha + y_0 \sin \alpha - p$
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $6$. Distance of a point from a line
Category:Perpendicular Distance from Straight Line in Plane to Point
Perpendicular from Center of Circle to Side of Inscribed Pentagon

== Theorem ==
== Proof ==
:
Let $ABC$ be a circle.
Let $BC$ be the side of a regular pentagon which has been inscribed within $ABC$.
Let $D$ be the center of the circle $ABC$.
Let $DE$ be drawn from $D$ perpendicular to $BC$, and produce it both ways to meet $ABC$ in $A$ and $F$.
It is to be demonstrated that $DE$ is half the sum of:
:the side of a regular hexagon which has been inscribed within $ABC$
and:
:the side of a regular decagon which has been inscribed within $ABC$
Let $DC$ and $CF$ be joined.
Let $G$ be the point on $AE$ such that $GE = EF$.
Let $GC$ be joined.
We have that:
:the circumference of $ABC$ is $5$ times $\smile BFC$
and:
:half the circumference of $ABC$ is $\smile ACF$
where $\smile BFC$ and $\smile ACF$ are used to denote the arc $BFC$ and arc $ACF$.
while:
:$\smile FC$ is half of $\smile BFC$.
Thus:
Also:
But from :
:$DE$ is the side of a regular hexagon which has been inscribed within $ABC$
and from the construction:
:$FC$ is the side of a regular decagon which has been inscribed within $ABC$.
Hence the result.
== Sources ==
* : The So-Called Book $\text{XIV}$, by Hypsicles
Category:Circles
Category:Pentagons
Category:Hexagons
Category:Decagons
Perpendicular from Point to Straight Line in Plane is Unique

== Theorem ==
Let $BC$ be a straight line in the plane.
Let $AM$ be perpendicular to $BC$ at $M$.
Let $P$ be an arbitrary point on the same side of $BC$ as $A$ is, but not on $AM$.
Then $PM$ cannot be perpendicular to $BC$.
== Proof ==
:
 let $M$ bisect $BC$.
 $PM \perp BC$.
By definition $PM$ is a perpendicular bisector of $BC$ at $M$.
Then by definition of perpendicular bisector:
:$PC = PB$
Find $P'$ on $AM$ such that $P'C = PC$.
Since $AM$ is also a perpendicular bisector of $BC$:
:$P'B = P'C$
:$\leadsto PB = PC = P'C = P'B$
This is a contradiction.
Therefore, no such $P$ exists.
All points equidistant from $B$ and $C$, on the same side of $BC$ as $A$, lie on $AM$.
== Also see ==
* Perpendicular Bisector is Locus of Points Equidistant from Endpoints
* Definition:Perpendicular Bisector
Category:Perpendicular from Point to Straight Line in Plane is Unique
Category:Perpendiculars
Category:Planes
Perpendicular in Right-Angled Triangle makes two Similar Triangles

== Theorem ==
== Proof ==
Let $\triangle ABC$ be a right-angled triangle such that $\angle BAC$ is a right angle.
Let $AD$ be drawn perpendicular to $BC$.
We need to show that $\triangle ABD$ is similar to $\triangle CAD$ which is similar to $\triangle CBA$ which is in turn similar to $\triangle ABD$.
:
We have that $\angle BAC = \angle ADB$ as both are right angles.
Then $\angle ABC$ is common to both $\triangle ABC$ and $\triangle ABD$.
So from Sum of Angles of Triangle Equals Two Right Angles:
: $\angle ACB = \angle BAD$
So $\triangle ABC$ is equiangular with $\triangle ABD$.
From Equiangular Triangles are Similar it follows that $\triangle ABC$ is similar to $\triangle ABD$.
Similarly we see that $\triangle ABC$ is equiangular with $\triangle ACD$.
Therefore each of $\triangle ABD$ and $\triangle ACD$ are similar to $\triangle ABC$.
Finally, $\triangle ABD$ and $\triangle ACD$ are equiangular with each other and so similar to each other.
== Porism ==
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Right Triangles
Perpendicular is Shortest Straight Line from Point to Straight Line

== Theorem ==
Let $AB$ be a straight line.
Let $C$ be a point which is not on $AB$.
Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$.
Then the length of $CD$ is less than the length of all other line segments that can be drawn from $C$ to $AB$.
== Proof ==
Let $E$ on $AB$ such that $E$ is different from $D$.
Then $CDE$ forms a right triangle where $CE$ is the hypotenuse.
By Pythagoras's Theorem:
:$CD^2 + DE^2 = CE^2$
and so $CD < CE$.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.21 \ \text{(i)}$
Category:Lines
Category:Angles
Perpendicular through Given Point

== Theorem ==
== Construction ==
:
Let $AB$ be two points on the given infinite straight line.
Let $C$ be the given point not on it.
Let $D$ be some point not on $AB$ on the other side of it from $C$.
We construct a circle $EFG$ with center $C$ and radius $CD$.
We bisect the straight line $EG$ at the point $H$.
We draw line segments from $C$ to each of $G$, $H$ and $E$ to form the straight line segments $CG$, $CH$ and $CH$.
Then the line $CH$ is perpendicular to the given infinite straight line $AB$ through the given point $C$.
== Proof ==
As $C$ is the center of circle $BCD$, it follows from that $GC = CE$.
As $EG$ has been bisected, $GH = HE$.
Thus, as $GC = CE$ and $GH = HE$, and $CH$ is common, by Triangle Side-Side-Side Congruence, $\triangle CHG = \triangle EHG$.
Therefore $\angle CHG = \angle CHE$.
So $CH$ is a straight line set up on a straight line making the adjacent angles equal to one another.
Thus it follows from that each of $\angle CHG$ and $\angle CHE$ are right angles.
So the straight line $CH$ has been drawn at right angles to the given infinite straight line $AB$ through the given point $C$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Lines
Category:Angles
Category:Perpendiculars
Perpendicularity is Antireflexive Relation

== Theorem ==
Let $S$ be the set of straight lines in the plane.
For $l_1, l_2 \in S$, let $l_1 \perp l_2$ denote that $l_1$ is perpendicular to $l_2$.
Then $\perp$ is an antireflexive relation on $S$.
== Proof ==
By definition of perpendicular lines, for $l_1$ to be perpendicular to itself would mean it would have to meet itself in a right angle.
This it does not do.
So $l_1 \not \perp l_1$.
Thus $\perp$ is seen to be antireflexive.
== Also see ==
* Perpendicularity is Symmetric Relation
* Perpendicularity is Antitransitive Relation
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
Category:Perpendiculars
Category:Examples of Antireflexive Relations
Perpendicularity is Antitransitive Relation

== Theorem ==
Let $S$ be the set of straight lines in the plane.
For $l_1, l_2 \in S$, let $l_1 \perp l_2$ denote that $l_1$ is perpendicular to $l_2$.
Then $\perp$ is an antitransitive relation on $S$.
== Proof ==
Let $l_1 \perp l_2$ and $l_2 \perp l_3$.
Then $l_1$ and $l_3$ are parallel, and not perpendicular.
Thus $\perp$ is seen to be antitransitive.
== Also see ==
* Perpendicularity is Symmetric Relation
* Perpendicularity is Antireflexive Relation
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
Category:Perpendiculars
Category:Examples of Antitransitive Relations
Perpendicularity is Symmetric Relation

== Theorem ==
Let $S$ be the set of straight lines in the plane.
For $l_1, l_2 \in S$, let $l_1 \perp l_2$ denote that $l_1$ is perpendicular to $l_2$.
Then $\perp$ is a symmetric relation on $S$.
== Proof ==
Let $l_1 \perp l_2$.
By definition of perpendicular lines, $l_1$ meets $l_2$ at a right angle.
Hence $l_2$ similarly meets $l_1$ at a right angle.
That is:
:$l_2 \perp l_1$
Thus $\parallel$ is seen to be symmetric.
== Also see ==
* Perpendicularity is Antireflexive Relation
* Perpendicularity is Antitransitive Relation
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
Category:Perpendiculars
Category:Examples of Symmetric Relations
Pfaff's Transformation

== Theorem ==
Let $a, b, c \in \C$.
Let $\size x < \dfrac 1 2$
Let $\map \Re c > \map \Re b > 0$.
Then:
:$\ds \map F {a, b; c; x} = \paren {1 - x}^{-a} \map F {a, c - b; c; \dfrac x {x - 1} }$
where:
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
== Proof ==
From Euler's Integral Representation of Hypergeometric Function, we have:
:$\ds \map F {a, b; c; x} = \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \int_0^1 t^{b - 1} \paren {1 - t}^{c - b - 1} \paren {1 - xt}^{- a} \rd t$
Letting $s = \paren {1 - t}$, we now have:
Letting $\size x < \dfrac 1 2$ and expanding the product of $\paren {1 - \dfrac {xs} {x - 1} }^{-a}$:
Therefore:
Therefore:
:$\ds \map F {a, b; c; x} = \paren {1 - x}^{-a} \map F {a, c - b; c; \dfrac x {x - 1} }$
== Also see ==
* Euler's Integral Representation of Hypergeometric Function
* Euler's Transformation
* Kummer's Quadratic Transformation
== Sources ==
* : Chapter $\text {2}$. The Hypergeometric Functions
* 
Category:Pfaff's Transformation
Category:Euler's Integral Representation of Hypergeometric Function
Category:Gaussian Hypergeometric Function
Category:Gamma Function
Pfaff-Saalschütz Theorem

== Theorem ==
Let $a, b, c \in \C$.
Let $n \in Z_{\ge 0}$.
Then:
:$\ds \map { {}_3 \operatorname F_2} { { {a, b, -n} \atop {c, 1 + a + b - c - n} } \, \middle \vert \, 1} = \dfrac {\paren {c - a}^{\overline n} \paren {c - b}^{\overline n} } { c^{\overline n} \paren {c - a - b}^{\overline n} }$
where:
:${}_3 \operatorname F_2$ is the generalized hypergeometric function
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$.
== Proof ==
By definition of the generalized hypergeometric function:
:$\ds \map { {}_3 \operatorname F_2} { { {a, b, -n} \atop {c, 1 + a + b - c - n} } \, \middle \vert \, 1} = \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} \paren {-n}^{\overline k} } { c^{\overline k} \paren {1 + a + b - c - n}^{\overline k} } \dfrac {1^k} {k!}$
From Euler's Transformation, we have:
Next, we observe that:
Equating coefficients of $x^n$ in $(1)$ above, we obtain:
== Also known as ==
Some sources refer to the Pfaff-Saalschütz Theorem as Saalschütz's Hypergeometric Theorem.
== Examples ==
== Also see ==
* Dixon's Hypergeometric Theorem
* Gauss's Hypergeometric Theorem
* Kummer's Hypergeometric Theorem
* Properties of Generalized Hypergeometric Function
== Historical Note ==
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
* Chapter $\text {2}$. The hypergeometric series
* 
Category:Pfaff-Saalschütz Theorem
Category:Hypergeometric Functions
Pfaff-Saalschütz Theorem/Examples

== Examples of Use of Pfaff-Saalschütz Theorem==
=== Example: $\map { {}_3 \operatorname F_2} {\dfrac 1 2, \dfrac 1 2, -2; \dfrac 3 2, -\dfrac 3 2; 1}$ ===
Category:Pfaff-Saalschütz Theorem
Pfaff-Saalschütz Theorem/Examples/3F2(0.5,0.5,-2;1.5,-1.5;1)

== Example of Use of Pfaff-Saalschütz Theorem ==
:$\ds \map { {}_3 \operatorname F_2} { { {\dfrac 1 2, \dfrac 1 2, -2} \atop {\dfrac 3 2, -\dfrac 3 2} } \, \middle \vert \, 1} = \dfrac {64} {45}$
== Proof ==
From Pfaff-Saalschütz Theorem:
:$\ds \map { {}_3 \operatorname F_2} { { {a, b, -n} \atop {c, 1 + a + b - c - n} } \, \middle \vert \, 1} = \dfrac {\paren {c - a}^{\overline n} \paren {c - b}^{\overline n} } { c^{\overline n} \paren {c - a - b}^{\overline n} }$
where:
:${}_3 \operatorname F_2$ is the generalized hypergeometric function
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$.
We have:
and:
Therefore:
:$\ds \map { {}_3 \operatorname F_2} { { {\dfrac 1 2, \dfrac 1 2, -2} \atop {\dfrac 3 2, -\dfrac 3 2} } \, \middle \vert \, 1} = \dfrac {64} {45}$
Category:Pfaff-Saalschütz Theorem‎
Pfaff-Saalschütz Theorem/Historical Note

== Historical Note on Pfaff-Saalschütz Theorem ==
According to by , and , the Pfaff-Saalschütz Theorem was discovered by in $1797$, and rediscovered by in $1890$.
The proof shown here is a more detailed version of a proof presented in chapter $2$ of by , as follows.
In the formula:
obtained in $\S 1.2$, equate the coefficients of $z^n$ to obtain:
Hence:
As can be seen, vaulted over several intermediate steps between steps $(2)$ and $(3)$, but his assertions were correct.
== Sources ==
* : Chapter $\text {2}$. The Hypergeometric Functions
* Chapter $\text {2}$. The hypergeometric series
Category:Historical Notes
Phase Space/Examples/Simple Harmonic Motion

== Example of Phase Space ==
Consider the differential equation:
:$\ddot x + \omega^2 x 0$
A solution is:
:$x = a \sin \omega t$
where $a$ and $\omega$ are positive constants.
Because:
:$\dot x = a \omega \cos \omega t$
the phase space is the set of all points:
:$\tuple {a \sin \omega t, a \omega \cos \omega t}$
which is the locus of an ellipse.
== Sources ==
* 
* 
Category:Examples of Phase Space
Phi in the Pentagon

== Theorem ==
Let $ABCDE$ be a regular pentagon.
The chords and sides of $ABCDE$ form three kinds of similar triangles.
* I: Isosceles, similar to $\triangle ACD$.
* II: Isosceles, similar to $\triangle ABE$.
The sides of the first type are in the ratio $\phi : 1$, where $\phi$ is the golden mean.
The sides of the second type are in the ratio $1 : \phi$.
== Proof ==
 five sides of $ABCDE$ are equal.
Draw all five chords.
From symmetry, the chords are all equal.
$\leadsto$:
:$\triangle ACD$ is an isosceles triangle.
$\leadsto$:
:$\triangle BCD$ is an isosceles triangle
Hence $A'B'C'D'E'$ has all internal angles equal and is Similar to $ABCDE$.
By Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal:
:$BCDC'$ is a parallelogram.
Also by symmetry:
:$\triangle AD'C'$ is isosceles
Let the length of $D'C' = B'C' = 1$.
Let the length of $EB' = AD' = AC' = \phi$, where $\phi$ is to be determined.
By the quadratic equation:
:$\phi = \dfrac{ 1 \pm \sqrt{5} } 2$
We take $\phi > 0$ so:
:$\phi = \dfrac{ 1 + \sqrt{5} } 2$
By definition of golden mean:
:$\phi$ is the golden mean
The length of the sides of the pentagon is $\phi + 1$.
By Equiangular Triangles are Similar:
:$\triangle ADE \sim \triangle B'ED$
$\leadsto$:
:$\dfrac {\phi + 1} { 2 \phi + 1 } = \dfrac {\phi} {\phi + 1}$
Working with the and substituting from above:
:$\dfrac {\phi} {\phi + 1} = \dfrac {\phi} {\phi^2}$
:$= \dfrac 1 \phi$
The result follows.
== Also known as ==
* Definition:Golden Mean: a classical name for $\phi$
Category:Pentagons
Category:Golden Mean
Phi is 8 has only 5 solutions

== Theorem ==
$\phi$ denotes the Euler $\phi$-function.
There are $5$ numbers $n$ with the property that $\map \phi n = 8$, and they are $15$, $16$, $20$, $24$ and $30$.
== Proof ==
Suppose $\phi(n)=8$.
By Fundamental Theorem of Arithmetic, $n$ has a prime factorization: $$n=\prod_{i=1}^k p_i^{\alpha_i}$$
where $p_i$ are all distinct primes. Since $\phi(1)=1\neq8$, we assume $k>0$. Say that $n$ has $k$ distinct prime factors, so that each $\alpha_i > 0$.
By Euler Phi Function is Multiplicative, $$\phi(n) = 8 =\prod_{i=1}^k \phi(p_i^{\alpha_i})$$
By Euler Phi Function of Prime Power, $$8 = \prod_{i=1}^k (p_i-1)p_i^{\alpha_i-1}$$
Because each $p_i-1$ is a factor of $8$, it follows that $p_i-1\leq 8$. Thus, each $p_i\leq 9$. The only primes less than $9$ are $2$, $3$, $5$, and $7$. Further, $7-1$ is not a factor of $8$, so $7$ is not a prime factor of $n$.
Thus, $n$ is of the form $2^{a}3^{b}5^{c}$, for nonnegative integers $a, b, c$.
By Euler Phi Function is Multiplicative and Euler Phi Function of Prime Power, if $b$ or $c$ are greater than $1$, then $\phi(n)$ is divisible by $3$ or $5$, respectively. This is not the case, so $3$ and $5$ each divide $n$ at most once. 
This splits the analysis into four cases. Because $\phi(n)\leq n$ for all $n\in\mathbb{N}$, $n$ cannot be $1$, $3$, or $5$. Thus, in the first three cases, $n$ must have at least one factor of $2$.
===Case 1:$b=c=0$===
Now $n=2^a$, so $\phi(n)=8=(2-1)2^{a-1}=2^{a-1}\Rightarrow a=4$. This gives one solution of $n=16$.
===Case 2: $b=1$ and $c=0$===
$n=2^a\cdot 3$, so $\phi(n)=8=(2-1)2^{a-1}\cdot(3-1)\Rightarrow 8=2^a\Rightarrow a=3$. This gives one solution of $n=24$.
===Case 3: $b=0$ and $c=1$===
$n=2^a\cdot 5$, so $\phi(n)=8=(2-1)2^{a-1}\cdot (5-1)\Rightarrow 8=2^{a-1}\cdot 4\Rightarrow a=2$.This gives one solution of $n=20$.
===Case 4: $b=c=1$===
If $n$ is divisible by $2$:
$n=2^a\cdot 3\cdot 5$, so $\phi(n)=8=(2-1)2^{a-1}\cdot(3-1)\cdot(5-1)\Rightarrow 8=2^{a-1}\cdot 8\Rightarrow a=1$. This gives one solution of $n=30$.
If $n$ is not divisible by $2$, then $n=3\cdot 5$, and so $n=15$ is the fifth and final solution.
Pi Squared is Irrational

== Theorem ==
Pi squared ($\pi^2$) is irrational.
=== Decimal Expansion ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Pi is Irrational
== Historical Note ==
== Sources ==
* : $9 \cdotp 869 \, 604 \ldots$
* : $9 \cdotp 86960 \, 44010 \, 89358 \ldots$
Category:Pi Squared is Irrational
Category:Pi
Pi Squared is Irrational/Historical Note

== Historical Note on Pi Squared is Irrational ==
$\pi^2$ was demonstrated to be irrational in $1794$ by .
== Sources ==
* : $9 \cdotp 869 \, 604 \ldots$
* : $9 \cdotp 86960 \, 44010 \, 89358 \ldots$
Category:Historical Notes
Pi Squared is Irrational/Proof 1

== Theorem ==
== Proof ==
 $\pi^2$ is rational.
We establish a lemma:
=== Lemma ===
We will use the definition of $A_n$ from the lemma.
Then we will deduce that $A_n$ is an integer for all $n$.
First we confirm by direct integration that $A_0$ and $A_1$ are integers:
and 
Suppose $A_{k - 2}$ and $A_{k - 1}$ are integers. 
By our lemma:
:$A_k = \paren {4 n - 2} q A_{k - 1} - p q A_{k - 2}$
As $n$, $p$ and $q$ are all integers , $A_k$ is also an integer.
So $A_n$ is an integer for all $n$ by Second Principle of Mathematical Induction.
Let $x \in \closedint 0 \pi$.
From Shape of Sine Function and Real Sine Function is Bounded, we have:
:$0 \le \sin x \le 1$
From Derivative at Maximum or Minimum, we have:
Therefore, plugging $\dfrac \pi 2$ into $\map f x$, we obtain:
:$0 \le x \paren {\pi - x} \le \dfrac {\pi^2} 4$
Since $A_n$ is clearly positive and the length of the interval is $\pi$ and the integrand is bounded at $\paren {\pi^2 / 4}^n$, from Darboux's Theorem, we have:
:$0 < A_n < \pi \times \dfrac {q^n} {n!} \times \paren {\pi^2 / 4}^n$
From Power over Factorial:
:$\ds \lim_{n \mathop \to \infty} \frac {\paren {q \pi^2 / 4}^n} {n!} = 0$
It follows from the Squeeze Theorem that:
:$\ds \lim_{n \mathop \to \infty} A_n = 0$
Hence for sufficiently large $n$, $A_n$ is strictly between $0$ and $1$.
This contradicts the deduction that $A_n$ is an integer.
This depends on our supposition that $\pi^2$ is a rational number,
Hence by Proof by Contradiction it follows that $\pi^2$ is irrational.
Category:Pi Squared is Irrational
Pi Squared is Irrational/Proof 1/Lemma

== Pi Squared is Irrational: Lemma ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let it be supposed that $\pi^2$ is irrational, so that:
:$\pi^2 = \dfrac p q$
where $p$ and $q$ are integers and $q \ne 0$.
Let $A_n$ be defined as:
:$\ds A_n = \frac {q^n} {n!} \int_0^\pi \paren {x \paren {\pi - x} }^n \sin x \rd x$
Then:
:$A_n = \paren {4 n - 2} q A_{n - 1} - p q A_{n - 2}$
is a reduction formula for $A_n$.
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
Let:
Now let:
We note in passing that:
:$(1): \quad \paren {q \pi}^2 = q^2 \paren {\dfrac p q} = p q$
which, , is an integer.
Then:
Category:Pi Squared is Irrational