nyuuzyou/wikis huggingface dataset:
Push Theorem

== Theorem ==
Let $f$ be a real function which is continuous on the open interval $\openint a \to$, $a \in \R$, such that:
:$\ds \lim_{x \mathop \to +\infty} \map f x = +\infty$
Let $g$ be a real function defined on some open interval $\openint b \to$ such that, for sufficiently large $x$:
:$\map g x > \map f x$
Then:
:$\ds \lim_{x \mathop \to +\infty} \map g x = +\infty$
== Proof ==
Let $\ds \lim_{x \mathop \to +\infty} \map f x = +\infty$.
By the definition of infinite limit at infinity, this means:
:$\forall M_1 \in \R_{>0}: \exists N_1 \in \R_{>0}: x > N_1 \implies \map f x > M_1$
Now, the assertion that $\map g x \to +\infty$ is:
:$\forall M_1 \in \R_{>0}: \exists N_2 \in \R_{>0}: x > N_2 \implies \map g x > M_1$
By the premise that $\map g x > \map f x$ for sufficiently large $x$, there is an $N$ such that:
:$x > N \implies \map g x > \map f x$
Now, given $M_1$, let $N_2$ be greater than both $N_1$ and $N$.
Then since $N_2 > N$ and $N_2 > N_1$ respectively:
:$\map g x > \map f x > M_1$
whence:
:$\ds \lim_{x \mathop \to +\infty} \map g x = +\infty$
== Note on Terminology ==
The author of this page has not found a name for this theorem in any English source.
Push Theorem is the translation of the Dutch name of this theorem: Duwstelling.
== Also see ==
* Comparison Test for Divergence
Category:Unbounded Mappings
Pushforward Measure is Measure

== Theorem ==
Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $f: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping.
Then the pushforward measure $f_* \mu: \Sigma' \to \overline \R$ is a measure.
== Proof ==
To show that $f_* \mu$ is a measure, it will suffice to check the axioms $(1)$, $(2)$ and $(3')$ for a measure.
=== Axiom $(1)$ ===
The statement of axiom $(1)$ for $f_* \mu$ is:
:$\forall E' \in \Sigma': \map {f_* \mu} {E'} \ge 0$
Now observe:
=== Axiom $(2)$ ===
Let $\sequence {E'_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.
The statement of axiom $(2)$ for $f_* \mu$ is:
:$\ds \map {f_* \mu} {\bigcup_{n \mathop \in \N} E'_n} = \sum_{n \mathop \in \N} \map {f_* \mu} {E'_n}$
Now compute:
Note that the second equality uses Preimage of Intersection under Mapping and Preimage of Empty Set is Empty to confirm that $\sequence { f^{-1} \sqbrk {E'_n} }_{n \mathop \in \N}$ is pairwise disjoint:
:$f^{-1} \sqbrk {E'_n} \cap f^{-1} \sqbrk {E'_m} = f^{-1} \sqbrk {E'_n \cap E'_m} = f^{-1} \sqbrk \O = \O$
=== Axiom $(3')$ ===
The statement of axiom $(3')$ for $f_* \mu$ is:
:$\map {f_* \mu} \O = 0$
Now compute:
Thus $f_* \mu$, satisfying the axioms, is seen to be a measure.
== Sources ==
* : $7.6$
Category:Measures
Category:Pushforward Measures
Pushforward of Lebesgue Measure under General Linear Group

== Theorem ==
Let $M \in \GL {n, \R}$ be a nonsingular matrix.
Let $\lambda^n$ be $n$-dimensional Lebesgue measure.
Then the pushforward measure $M_* \lambda^n$ satisfies:
:$M_* \lambda^n = \size {\det M^{-1} } \cdot \lambda^n$
== Proof ==
From Linear Transformation on Euclidean Space is Continuous, $M^{-1}$ is a continuous mapping.
Thus from Continuous Mapping is Measurable, it is measurable, and so $M_* \lambda^n$ is defined.
Now let $B \in \map \BB {\R^n}$ be a Borel measurable set, and let $\mathbf x \in \R^n$.
Then:
Thus $M_* \lambda^n$ is a translation invariant measure.
From Translation Invariant Measure on Euclidean Space is Multiple of Lebesgue Measure, it follows that:
:$M_* \lambda^n = \map {M_* \lambda^n} {\openint 0 1^n} \cdot \lambda^n$
Lastly, using Determinant as Volume of Parallelotope it follows that:
:$\map {M_* \lambda^n} {\openint 0 1^n} = \map {\lambda^n} {\map {M^{-1} } {\openint 0 1^n} } = \size {\det M^{-1} }$
Hence the result.
== Sources ==
* : $4.9 \ \text{(iii)}$
* : $7.10$
Category:Lebesgue Measure
Putzer Algorithm

== Algorithm ==
The Putzer Algorithm is a method for analytically evaluating Matrix Exponentials using only eigenvalues and components in the solution of a relatively simple linear system.
It is particularly useful for matrices that cannot be diagonalized because it avoids the use of Jordan-Canonical Form.
== Method ==
Let $\lambda_1, \lambda_2, \dotsc, \lambda_n$ be the (not necessarily distinct) eigenvalues of the matrix $A$.
Then:
:$\ds e^{\mathbf A t} = \sum_{k \mathop = 0}^{n - 1} \map {p_{k + 1} } t M_k$
This formula is constructed by setting $M_0 = \mathbf I$:
:$\s M_k = \prod_{i \mathop = 1}^k \paren {\mathbf A - \lambda_i I}$
where:
:$\mathbf I$ is the identity matrix
and:
:the functions $\map {p_1} t, \map {p_2} t, \dotsc, \map {p_n} t$ are taken to be components of the vector function solution to the IVP:
:$p' = \begin{pmatrix}
\lambda_1 & 0 & 0 & \cdots & 0 \\ 
1 & \lambda_2 & 0 & \cdots & 0 \\ 
0 & 1 & \lambda_3 & \cdots & 0 \\ 
\vdots & \vdots & \vdots & \ddots & \vdots \\ 
0 & \cdots & 0 & 1 & \lambda_n
\end{pmatrix} p$
such that:
:$\map p 0 = \begin{pmatrix}
1 \\ 
0 \\ 
0 \\ 
\vdots\\ 
0 
\end{pmatrix}$
== Proof of Validity ==
Let $\map \Phi t$ represent the finite matrix sum, derived above, as the candidate for $e^{\mathbf A t}$. 
By the uniqueness theorem, it suffices to show that $\Phi$ satisfies the IVP:
:$X' = \mathbf A X: \map X 0 = \mathbf I$
By definition:
:$\map {p_1'} t = \lambda_1 \map {p_1} t$
:$\map {p_i'} t = \map {p_{i - 1} } t + \lambda_i \, \map {p_i} t: i > 1$
and:
:$M_0 = \mathbf I$
:$M_k = \paren {\mathbf A - \lambda_k \mathbf I} M_{k - 1}$
Note also that $M_n = 0$ by the Cayley-Hamilton Theorem.
Then:
== Sources ==
* 
Category:Matrix Exponential
Pythagoras's Theorem

== Theorem ==
Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
== Proof 6 ==
== Proof 7 ==
== Proof 8 ==
== Algebraic Proof ==
== The Classic Proof ==
== Short Algebraic Proof ==
== Also known as ==
== Also see ==
* Pythagoras's Theorem (Inner Product Space), a generalisation to the context of inner product spaces.
* Definition:Pythagorean Triangle
* Definition:Pythagorean Triple
* Sum of Squares of Sine and Cosine
* Square equals Sum of Squares implies Right Triangle: its inverse
== Historical Note ==
== ... there are even jokes about it ... ==
== Sources ==
* Pythagorean Theorem and its many proofs - $96$ proofs of Pythagoras's Theorem
* : Chapter $\text{III}$: Metric Spaces: Pythagoras' Theorem
* : $2$
* : $5$
* 
* : Chapter $1$: Hippocrates' Quadrature of the Lune ( )
* : Chapter $\text {B}.1$: The Pythagorean Theorem
* : The Nine Chapters
* : $2$
* : $5$
* 
* 
* : Chapter $2$: The Logic of Shape: Pythagoras
* 
Category:Pythagoras's Theorem
Category:Right Triangles
Pythagoras's Theorem/Algebraic Proof

== Theorem ==
== Proof ==
We start with the algebraic definitions for sine and cosine:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
:$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$
From these, we derive the proof that $\cos^2 x + \sin^2 x = 1$.
Then from the Equivalence of Definitions of Trigonometric Functions, we can use the geometric interpretation of sine and cosine:
:
:$\sin \theta = \dfrac {\text{Opposite}} {\text{Hypotenuse}}$
:$\cos \theta = \dfrac {\text{Adjacent}} {\text{Hypotenuse}}$
Let $\text{Adjacent} = a, \text{Opposite} = b, \text{Hypotenuse} = c$, as in the diagram at the top of the page.
Thus:
Category:Pythagoras's Theorem
Pythagoras's Theorem/Also known as

== Pythagoras's Theorem: Also known as ==
Pythagoras's Theorem was known to the Pythagoreans as the Theorem of the Bride, from its numerological significance.
To the ancient Chinese it was called the Gougu theorem.
== Sources ==
* : $5$
* : The Nine Chapters
* : $5$
Category:Right Triangles
Category:Pythagoras's Theorem
Pythagoras's Theorem/Classic Proof

== Theorem ==
== Proof ==
:
Let $ABC$ be a right triangle whose angle $BAC$ is a right angle.
Construct squares $BDEC$ on $BC$, $ABFG$ on $AB$ and $ACKH$ on $AC$.
Construct $AL$ parallel to $BD$ (or $CE$).
Since $\angle BAC$ and $\angle BAG$ are both right angles, from Two Angles making Two Right Angles make Straight Line it follows that $CA$ is in a straight line with $AG$.
For the same reason $BA$ is in a straight line with $AH$.
We have that $\angle DBC = \angle FBA$, because both are right angles.
We add $\angle ABC$ to each one to make $\angle FBC$ and $\angle DBA$.
By common notion 2, $\angle FBC = \angle DBA$.
By Triangle Side-Angle-Side Congruence, $\triangle ABD = \triangle FBC$.
We have that the parallelogram $BDLM$ is on the same base $BD$ and between the same parallels $BD$ and $AL$ as the triangle $\triangle ABD$.
So, by Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $BDLM$ is twice the area of $\triangle ABD$.
Similarly, we have that the parallelogram $ABFG$ (which happens also to be a square) is on the same base $FB$ and between the same parallels $FB$ and $GC$ as the triangle $\triangle FBC$.
So, by Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $ABFG$ is twice the area of $\triangle FBC$.
So $BDLM = 2 \triangle ABD = 2 \triangle FBC = ABFG$.
By the same construction, we have that $CELM = 2 \triangle ACE = 2 \triangle KBC = ACKH$.
But $BDLM + CELM$ is the whole of the square $BDEC$.
Therefore the area of the square $BDEC$ is equal to the combined area of the squares $ABFG$ and $ACKH$.
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
* : Chapter $2$: The Logic of Shape: Euclid
Category:Pythagoras's Theorem
Pythagoras's Theorem/Historical Note

== Historical Note on Pythagoras's Theorem ==
Pythagoras's Theorem, or at least certain specific instances of it, was known to the ancient Egyptians, and it may date back further than that.
Little is actually known about himself, and it is uncertain whether he actually proved it or not.
Some traditional sources suggest that when he discovered this theorem, he was so delighted he sacrificed an ox to the gods in thanksgiving. However, given his philosophical outlook, this is vanishingly unlikely.
== Sources ==
* : Chapter $\text {A}.2$: Pythagoras (ca. B.C.)
* : Chapter $2$: The Logic of Shape: Pythagoras
Category:Historical Notes
Pythagoras's Theorem/Proof 1

== Theorem ==
== Proof ==
Consider the triangle shown below.
:
We can take $4$ copies of this triangle and form them into a square using isometries, specifically rotations and translations.
This new figure is shown below.
:
This figure is clearly a square, since all the angles are right angles, and the lines connecting the corners are easily seen to be straight.
Now to calculate the area of this figure.
On the one hand, we can add up the area of the component parts of the square.
Specifically, we can add up the $4$ triangles and the inner square.
Thus we calculate the area of the large square to be:
:$4 \paren {\dfrac 1 2 a b} + c^2 = 2 a b + c^2$
On the other hand, we can similarly calculate the area of the large square to be:
:$\paren {a + b}^2 = a^2 + 2 a b + b^2$
Now these two expressions have to be equal, since they both represent the area of the square.
Thus:
:$a^2 + 2 a b + b^2 = 2 a b + c^2 \iff a^2 + b^2 = c^2$
== Sources ==
* : Chapter $\text {B}.1$: The Pythagorean Theorem
Category:Pythagoras's Theorem
Pythagoras's Theorem/Proof 2

== Theorem ==
== Proof ==
:
We have:
:$\dfrac b c = \dfrac d b$
and:
:$\dfrac a c = \dfrac e a$
using the fact that all the triangles involved are similar.
That is:
:$b^2 = c d$
:$a^2 = c e$
Adding, we now get:
:$a^2 + b^2 = c d + c e = c \paren {d + e} = c^2$
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.1$: The Pythagorean Theorem
* 
Category:Pythagoras's Theorem
Pythagoras's Theorem/Proof 3

== Theorem ==
== Proof ==
:
The area of the big square is $c^2$.
It is also equal to $4 \dfrac {a b} 2 + \paren {a - b}^2$.
So:
== Sources ==
* : Chapter $\text {B}.1$: The Pythagorean Theorem
Category:Pythagoras's Theorem
Pythagoras's Theorem/Proof 4

== Theorem ==
== Proof ==
:
Dissect the square on the left (which has area $c^2$) as shown.
Rearrange the pieces to make the two squares on the right, with areas $a^2$ and $b^2$.
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.1$: The Pythagorean Theorem
* 
Category:Pythagoras's Theorem
Pythagoras's Theorem/Proof 5

== Theorem ==
== Proof ==
:
The two squares both have the same area, that is, $\paren {a + b}^2$.
The one on the left has four triangles of area $\dfrac {a b} 2$ and a square of area $c^2$.
The one on the right has four triangles of area $\dfrac {a b} 2$ and two squares: one of area $a^2$ and one of area $b^2$.
Take away the triangles from both of the big squares and you are left with $c^2 = a^2 + b^2$.
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.1$: The Pythagorean Theorem
* 
* 
Category:Pythagoras's Theorem
Pythagoras's Theorem/Proof 6

== Theorem ==
== Proof ==
:
We have that $CH = BS = AB = AJ$.
Hence the result follows directly from Pythagoras's Theorem for Parallelograms.
== Sources ==
* : Chapter $\text {A}.8$: Pappus (fourth century A.D.)
* : Chapter $\text {B}.1$: The Pythagorean Theorem
Category:Pythagoras's Theorem
Pythagoras's Theorem/Proof 7

== Theorem ==
== Proof ==
:
Let $\triangle ABC$ be a right triangle and $h_c$ the altitude from $c$.
We have:
:$\angle CAB \cong \angle DCB$
:$\angle ABC \cong \angle ACD$
Then we have:
:$\triangle ADC \sim \triangle ACB \sim \triangle CDB$
Use the fact that if $\triangle XYZ \sim \triangle X'Y'Z'$ then by Ratio of Areas of Similar Triangles:
:$\dfrac {\paren {XYZ} } {\paren {X' Y' Z'} } = \dfrac {XY^2} {X'Y'^2}$
where $\paren {XYZ}$ denotes the area of $\triangle XYZ$.
This gives us:
:$\dfrac {\paren {ADC} } {\paren {ACB} } = \dfrac {AC^2} {AB^2}$
and
:$\dfrac {\paren {CDB} } {\paren {ACB} } = \dfrac {BC^2} {AB^2}$
Taking the sum of these two equalities we obtain:
:$\dfrac {\paren {ADC} } {\paren {ACB} } + \dfrac {\paren {CDB} } {\paren {ACB} } = \dfrac {BC^2} {AB^2} + \dfrac {AC^2} {AB^2}$
Thus:
:$\dfrac {\overbrace {\paren {ADC} + \paren {CDB} }^{\paren {ACB} } } {\paren {ACB} } = \dfrac {BC^2 + AC^2} {AB^2}$
Thus:
:$AB^2 = BC^2 + AC^2$
as desired.
Category:Pythagoras's Theorem
Pythagoras's Theorem/Proof 8

== Theorem ==
== Proof ==
Let $\Box ABCD$ be an arbitrary rectangle with opposing sides $AB = CD$ and $AD = BC$.
Let $O$ be the point where the diameters of $\Box ABCD$ meet.
By Diagonals of Rectangle are Equal:
:$AC = BD$
By Diameters of Parallelogram Bisect each other:
:$OA = OB = OC = OD$
Let a circle be constructed on center $O$ with radius $OA$.
Then $\Box ABCD$ is inscribed in the circle.
This also follows from the converse to Thales' Theorem.
By Ptolemy's Theorem:
:$AD \cdot BC + AB \cdot CD = AC \cdot BD$
But $AD = BC$ and $AB = CD$ and $AC = BD$ so:
:$AD \cdot AD + AB \cdot AB = AC \cdot AC$
Hence the result follows.
Category:Pythagoras's Theorem
Pythagoras's Theorem/Short Algebraic Proof

== Theorem ==
== Proof ==
Looking at the hypotenuse and height of the three similar triangles, we can write the following products and ratios relationships, then multiply them:
:$ {a} {a′} +  {b} {b′}   =   {c} {c′}$   (products = 2 x areas)
:$ \dfrac {a} {a′} = \dfrac {b} {b′} = \dfrac {c} {c′}$     (ratios)
⇒     $ a^2 +  b^2   =   c^2$
Note, dividing them naturally also gives us: $ {a′}^2 +  {b′}^2   =   {c′}^2$
Category:Pythagoras's Theorem
Pythagoras's Theorem (Inner Product Space)

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $\norm \cdot$ be the inner product norm on $\struct {V, \innerprod \cdot \cdot}$. 
Let $f_1, \ldots, f_n \in V$ form an orthogonal set.
Then:
:$\ds \norm {\sum_{i \mathop = 1}^n f_i}^2 = \sum_{i \mathop = 1}^n \norm {f_i}^2$
== Proof ==
== Also see ==
* Pythagoras's Theorem, the well-known instance of this theorem with $V = \R^2$ and the usual inner product.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: $2.2$ The Pythagorean Theorem
Category:Inner Product Spaces
Category:Pythagoras's Theorem
Pythagoras's Theorem for Parallelograms

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $ACDE$ and $BCFG$ be parallelograms constructed on the sides $AC$ and $BC$ of $\triangle ABC$.
Let $DE$ and $FG$ be produced to intersect at $H$.
Let $AJ$ and $BI$ be constructed on $A$ and $B$ parallel to and equal to $HC$.
Then the area of the parallelogram $ABIJ$ equals the sum of the areas of the parallelograms $ACDE$ and $BCFG$.
== Proof ==
:
From Parallelograms with Same Base and Same Height have Equal Area:
:$ACDE = ACHR = ATUJ$
and:
:$BCFG = BCHS = BIUT$
Hence the result.
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.8$: Pappus (fourth century A.D.)
Category:Triangles
Category:Parallelograms
Pythagorean Triangle/Examples

== Examples of Pythagorean Triangles ==
=== $3-4-5$ Triangle ===
=== $6-8-10$ Triangle ===
=== $5-12-13$ Triangle ===
=== $7-24-25$ Triangle ===
=== $693-1924-2045$ Triangle ===
=== $4485-5852-7373$ Triangle ===
=== $3059-8580-9109$ Triangle ===
=== $1380-19 \, 019-19 \, 069$ Triangle ===
Category:Examples of Pythagorean Triangles
Pythagorean Triangle/Examples/1380-19,019-19,069

== Example of Primitive Pythagorean Triangle ==
The triangle whose sides are of length $1380$, $19 \, 019$ and $19 \, 069$ is a primitive Pythagorean triangle.
:
It has generator $\left({138, 5}\right)$.
== Proof ==
We have:
It follows by Pythagoras's Theorem that $1380$, $19 \, 019$ and $19 \, 069$ form a Pythagorean triple.
We have that:
It is seen that $1380$ and $19 \, 019$ share no prime factors.
That is, $1380$ and $19 \, 019$ are coprime.
Hence, by definition, $1380$, $19 \, 019$ and $19 \, 069$ form a primitive Pythagorean triple.
The result follows by definition of a primitive Pythagorean triangle.
Category:Examples of Pythagorean Triangles
Pythagorean Triangle/Examples/3-4-5

== Example of Primitive Pythagorean Triangle ==
The triangle whose sides are of length $3$, $4$ and $5$ is a primitive Pythagorean triangle.
:
== Proof ==
It follows by Pythagoras's Theorem that $3$, $4$ and $5$ form a Pythagorean triple.
Note that $3$ and $4$ are coprime.
Hence, by definition, $3$, $4$ and $5$ form a primitive Pythagorean triple.
The result follows by definition of a primitive Pythagorean triangle.
== Also see ==
* Smallest Pythagorean Triangle is 3-4-5
* Pythagorean Triangle whose Area is Half Perimeter: its area and semiperimeter are both $6$
* Pythagorean Triangle with Sides in Arithmetic Sequence
== Historical Note ==
== Sources ==
* : $5$
* : $6$
* : Chapter $1$: Hippocrates' Quadrature of the Lune ( )
* : Pythagorean Triples
* : $5$
* : $6$
Category:Examples of Pythagorean Triangles
Pythagorean Triangle/Examples/3-4-5/Historical Note

== Historical Note on $3-4-5$ Triangle ==
To the Pythagoreans, the $3-4-5$ triangle had particular significance: the sides of lengths $3$ and $4$ denoted the male and female principles, while the hypotenuse of length $5$ denoted their offspring.
== Sources ==
* : $5$
* : $5$
Category:Historical Notes
Pythagorean Triangle/Examples/3059-8580-9109

== Example of Primitive Pythagorean Triangle ==
The triangle whose sides are of length $3059$, $8580$ and $9109$ is a primitive Pythagorean triangle.
:
It has generator $\tuple {78, 55}$.
== Proof ==
We have:
It follows by Pythagoras's Theorem that $3059$, $8580$ and $9109$ form a Pythagorean triple.
We have that:
It is seen that $3059$ and $8580$ share no prime factors.
That is, $3059$ and $8580$ are coprime.
Hence, by definition, $3059$, $8580$ and $9109$ form a primitive Pythagorean triple.
The result follows by definition of a primitive Pythagorean triangle.
Category:Examples of Pythagorean Triangles
Pythagorean Triangle/Examples/4485-5852-7373

== Example of Primitive Pythagorean Triangle ==
The triangle whose sides are of length $4485$, $5852$ and $7373$ is a primitive Pythagorean triangle.
:
It has generator $\tuple {77, 38}$.
== Proof ==
We have:
Hence:
It follows by Pythagoras's Theorem that $4485$, $5852$ and $7373$ form a Pythagorean triple.
We have that:
It is seen that $4485$ and $5852$ share no prime factors.
That is, $4485$ and $5852$ are coprime.
Hence, by definition, $693$, $1924$ and $2045$ form a primitive Pythagorean triple.
The result follows by definition of a primitive Pythagorean triangle.
Category:Examples of Pythagorean Triangles
Pythagorean Triangle/Examples/5-12-13

== Example of Primitive Pythagorean Triangle ==
The triangle whose sides are of length $5$, $12$ and $13$ is a primitive Pythagorean triangle.
:
== Proof ==
It follows by Pythagoras's Theorem that $5$, $12$ and $13$ form a Pythagorean triple.
Note that $5$ and $12$ are coprime.
Hence, by definition, $5$, $12$ and $13$ form a primitive Pythagorean triple.
The result follows by definition of a primitive Pythagorean triangle.
== Sources ==
* : $13$
* : $13$
Category:Examples of Pythagorean Triangles
Pythagorean Triangle/Examples/6-8-10

== Example of Pythagorean Triangle ==
The triangle whose sides are of length $6$, $8$ and $10$ is a Pythagorean triangle.
This is not a primitive Pythagorean triangle.
:
== Proof ==
It follows by Pythagoras's Theorem that $6$, $8$ and $10$ form a Pythagorean triple.
Note that $6$ and $8$ are not coprime as $\gcd \set {6, 8} = 2$.
Hence, by definition, $6$, $8$ and $10$ do not form a primitive Pythagorean triple.
The result follows by definition of a primitive Pythagorean triangle.
== Sources ==
* : $13$
* : $13$
Category:Examples of Pythagorean Triangles
Pythagorean Triangle/Examples/693-1924-2045

== Example of Primitive Pythagorean Triangle ==
The triangle whose sides are of length $693$, $1924$ and $2045$ is a primitive Pythagorean triangle.
:
== Proof ==
It follows by Pythagoras's Theorem that $693$, $1924$ and $2045$ form a Pythagorean triple.
Note that $693$ and $1924$ are coprime.
Hence, by definition, $693$, $1924$ and $2045$ form a primitive Pythagorean triple.
The result follows by definition of a primitive Pythagorean triangle.
== Sources ==
* : $13$
* : $13$
Category:Examples of Pythagorean Triangles
Pythagorean Triangle/Examples/7-24-25

== Example of Primitive Pythagorean Triangle ==
The triangle whose sides are of length $7$, $24$ and $25$ is a primitive Pythagorean triangle.
:
== Proof ==
It follows by Pythagoras's Theorem that $7$, $24$ and $25$ form a Pythagorean triple.
Note that $7$ and $24$ are coprime.
Hence, by definition, $7$, $24$ and $25$ form a primitive Pythagorean triple.
The result follows by definition of a primitive Pythagorean triangle.
== Sources ==
* : $13$
* : $13$
Category:Examples of Pythagorean Triangles
Pythagorean Triangle cannot be Isosceles

== Theorem ==
Let $P$ be a Pythagorean triangle.
Then $P$ is not isosceles.
== Theorem ==
Let $P$ be a Pythagorean triangle.
 $P$ is an isosceles.
Let the legs of $P$ be of length $a$.
Let the hypotenuse of $P$ be of length $h$.
We have from Pythagoras's Theorem that:
:$2 a^2 = h^2$
and so:
:$\dfrac h a = \sqrt 2$
By definition, $h$ and $a$ are integers.
Hence, by definition, $\sqrt 2$ is a rational number.
But that contradicts the result Square Root of 2 is Irrational.
By Proof by Contradiction, it follows that $P$ cannot be isosceles.
== Sources ==
* : $13$
* : $13$
Category:Isosceles Triangles
Category:Pythagorean Triangles