nyuuzyou/wikis huggingface dataset:
Slice Category is Category

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $C \in \mathbf C_0$ be a object of $\mathbf C$.
Let $\mathbf C / C$ be the slice category of $\mathbf C$ over $C$.
Then $\mathbf C / C$ is a metacategory.
== Proof ==
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
Suppose that $a: f \to g$ and $b: g \to h$ are morphisms in $\mathbf C / C$.
To show that $b \circ a: f \to h$ is a morphism as well, compute:
Hence $b \circ a: f \to h$ is a morphism.
For $(C2)$, observe that for $a: f \to g$, with objects $f: X \to C$ and $g: Y \to C$, we have:
Hence $(C2)$ is shown to hold.
Since the composition in $\mathbf C / C$ is inherited from $\mathbf C$, satisfaction of the associative property $(C3)$ is also inherited.
Hence $\mathbf C / C$ is a metacategory.
Category:Slice Categories
Slice Category of Order Category

== Theorem ==
Let $\mathbf P$ be a order category, and denote its ordering by $\preceq$.
Let $p \in \mathbf P_0$ be an object of $\mathbf P$.
Then:
:$\mathbf P \mathbin / p \cong p^\preceq$
where:
:$\mathbf P \mathbin / p$ is the slice of $\mathbf P$ over $p$
:$p^\preceq$ is the order category defined by the weak lower closure of $p$.
== Proof ==
The objects of $\mathbf P \mathbin / p$ are morphisms $q \to p$ of $\mathbf P$.
The morphisms are $q \to r$ fitting into a commutative diagram:
$\quad\quad \begin{xy}\xymatrix@C=1em{
q
\ar[rr]
\ar[rd]
& &
r
\ar[ld]
\\ &
p
}\end{xy}$
Define a functor $U: \mathbf P \mathbin / p \to \mathbf P$ by:
:$U \left({q \to p}\right) := q$
:$U \left({q \to r}\right) := q \to r$
Because there is at most one morphism $q \to p$ for each $q$, $U$ is injective on objects.
That $U$ is faithful is trivial.
Hence by Functor is Embedding iff Faithful and Injective on Objects, $U$ is an embedding.
That is, $\mathbf P \mathbin / p$ is isomorphic to the image of $U$.
Now the objects of the image of $U$ in $\mathbf P$ are those $q$ such that $q \preceq p$.
That is, the image of $U$ has as objects $p^\preceq$.
By Commutative Diagram in Preorder Category, every morphism $q \to r$ in $p^\preceq$ fits into a commutative triangle as above.
Thus all morphisms of $p^\preceq$ are in the image of $U$, so that $p^\preceq$ is precisely the image of $U$.
Hence the result.
== Sources ==
* : $\S 1.6.4$
Category:Slice Categories
Category:Order Categories
Slope/Examples/Arbitrary Example 1

== Examples of Slope ==
A slope or gradient of $1$ in $4$ indicates a vertical distance of $1$ unit for every $4$ units travelled along the road.
This is also indicated as a ratio $1 : 4$ or a fraction $1 / 4$, or can also be expressed as the angle of slope, that is, $\arcsin \dfrac 1 4$ or a percentage, that is, a $25 \%$ gradient.
== Sources ==
* 
* 
Category:Examples of Slope
Slope of Lemniscate of Bernoulli at Origin

== Theorem ==
Consider the lemniscate of Bernoulli $M$ embedded in a Cartesian plane such that its foci are at $\tuple {a, 0}$ and $\tuple {-a, 0}$ respectively.
Let $O$ denote the origin.
The tangents to $M$ at $O$ are at an angle of $45 \degrees = \dfrac \pi 4$ to the $x$-axis.
:
== Proof ==
Recall the parametric definition of lemniscate of Bernoulli:
:$\begin {cases} x = \dfrac {a \sqrt 2 \cos t} {\sin^2 t + 1} \\ \\ y = \dfrac {a \sqrt 2 \cos t \sin t} {\sin^2 t + 1} \end {cases}$
Note that $\tuple {x, y} = \tuple {0, 0}$ $\cos t = 0$.
At these points we have $\sin t = \pm \sqrt {1 - \cos^2 t} = \pm 1$.
We have:
By substituting $\cos t = 0$ and $\sin t = \pm 1$, we have:
As $\map \tan {\pm 45 \degrees} = \pm 1$, these tangents at $O$ are at an angle of $45 \degrees$ to the $x$-axis.
== Sources ==
* : $\S 11$: Special Plane Curves: Lemniscate: $11.3$
* : $\S 9$: Special Plane Curves: Lemniscate: $9.3.$
Category:Lemniscate of Bernoulli
Slope of Normal is Minus Reciprocal of Tangent

== Theorem ==
Let $C$ be a curve defined by a real function which is differentiable.
Let $P$ be a point on $C$.
Let the curvature of $C$ at $P$ be non-zero.
Let $r$ be the slope of the tangent to $C$ at $P$.
Let $s$ be the slope of the normal to $C$ at $P$. 
Then:
:$r = -\dfrac 1 s$
== Proof ==
By definition, the normal to $C$ at $P$ is defined as being perpendicular to the tangent at $P$ and in the same plane as $P$.
The result follows from Condition for Straight Lines in Plane to be Perpendicular.
Category:Normals to Curves
Category:Tangents
Category:Analytic Geometry
Slope of Orthogonal Curves

== Theorem ==
Let $C_1$ and $C_2$ be curves in a cartesian plane.
Let $C_1$ and $C_2$ intersect each other at $P$.
Let the slope of $C_1$ and $C_2$ at $P$ be $m_1$ and $m_2$.
Then $C_1$ and $C_2$ are orthogonal :
:$m_1 = -\dfrac 1 {m_2}$
== Proof ==
Let the slopes of $C_1$ and $C_2$ at $P$ be defined by the vectors $\mathbf v_1$ and $\mathbf v_2$ represented as column matrices:
:$\mathbf v_1 = \begin{bmatrix} x_1 \\ y_1 \end{bmatrix} , \mathbf v_2 = \begin{bmatrix} x_2 \\ y_2 \end{bmatrix}$
By Non-Zero Vectors are Orthogonal iff Perpendicular:
:$\mathbf v_1 \cdot \mathbf v_2 = 0$ $C_1$ is orthogonal to $C_2$
where $\mathbf v_1 \cdot \mathbf v_2$ denotes the dot product of $C_1$ and $C_2$.
Thus:
Category:Analytic Geometry
Slope of Parabola at Point

== Theorem ==
Let $a$ be a constant.
Let $T$ be the parabola which is the locus of points $\tuple {x, y}$ satisfying $y = a x^2$.
The slope of the tangent to $y = a x^2$ at $x = c$ is $2 a c$.
== Proof ==
By Derivative of Power of Function the derivative of $y = ax^2$ is:
:$\dfrac {\d y} {\d x} = 2 a x$
So the derivative of $y = a x^2$ at $x = c$ is $2 a c$.
The result follows from Slope of Tangent to Curve at Point equals Value of Derivative.
Category:Parabolas
Category:Slope
Slope of Secant

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let the graph of $f$ be depicted on a Cartesian plane.
:
Let $AB$ be a secant of $f$ where:
:$A = \tuple {x, \map f x}$
:$A = \tuple {x + h, \map f {x + h} }$
Then the slope of $AB$ is given by:
:$\dfrac {\map f {x + h} - \map f x} h$
== Proof ==
The slope of $AB$ is defined as the change in $y$ divided by the change in $x$.
Between $A$ and $B$:
:the change in $x$ is $\paren {x + h} - x = h$
:the change in $y$ is $\map f {x + h} - \map f x$.
Hence the result.
== Sources ==
* : Chapter $8$: The System of the World: Calculus
Category:Analytic Geometry
Slope of Straight Line joining Points in Cartesian Plane

== Theorem ==
Let $p_1 := \tuple {x_1, y_1}$ and $p_2 := \tuple {x_2, y_2}$ be points in a cartesian plane.
Let $\LL$ be the straight line passing through $p_1$ and $p_2$.
Then the slope of $\LL$ is given by:
:$\tan \theta = \dfrac {y_2 - y_1} {x_2 - x_1}$
where $\theta$ is the angle made by $\LL$ with the $x$-axis.
== Proof ==
:
The slope of a straight line is defined as the change in $y$ divided by the change in $x$.
The change in $y$ from $p_1$ to $p_2$ is $y_2 - y_1$.
The change in $x$ from $p_1$ to $p_2$ is $x_2 - x_1$.
By definition of tangent of $\theta$:
:$\tan \theta = \dfrac {y_2 - y_1} {x_2 - x_1}$
Hence the result.
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line
* : $\S 10$: Formulas from Plane Analytic Geometry: Slope $m$ of Line joining Two Points $\map {P_1} {x_1, y_1}$ and $\map {P_2} {x_2, y_2}$: $10.2$
* : $\S 8$: Formulas from Plane Analytic Geometry: Slope $m$ of Line joining Two Points $\map {P_1} {x_1, y_1}$ and $\map {P_2} {x_2, y_2}$: $8.2.$
Category:Straight Lines
Category:Slope
Slope of Tangent to Curve at Point equals Value of Derivative

== Theorem ==
Let $\CC$ be a curve embedded in the Cartesian plane described using the equation:
:$y = \map f x$
where $f$ is a real function.
Let there exist a unique tangent $\TT$ to $\CC$ at a point $P = \tuple {x_0, y_0}$ on $\CC$.
Then the slope of $\CC$ at $P$ is equal to the derivative of $f$ at $P$.
== Proof ==
We have been given that there exists a unique tangent $\TT$ to $\CC$ at $P$.
By definition of tangent, $\TT$ has a slope $M$ given by:
:$m = \ds \lim_{h \mathop \to 0} \frac {\map f {x_0 + h} - \map f {x_0} } h$
This is the definition of the derivative of $f$ at $P$.
== Examples ==
== Sources ==
* : Chapter $\text {I}$: Equations of the First Order and Degree: $1$. Definitions
* 
* 
* 
* 
Category:Slope of Tangent to Curve at Point equals Value of Derivative
Category:Tangents
Category:Differential Calculus
Category:Analytic Geometry
Slope of Tangent to Curve at Point equals Value of Derivative/Examples/Arbitrary Example 1

== Example of Use of Slope of Tangent to Curve at Point equals Value of Derivative ==
Let $\CC$ be the curve defined by the equation $y = 2 x^2$.
Let $\LL$ be the tangent line to $\CC$ at the point $\tuple {3, 18}$.
The equation of $\LL$ is:
:$y = 12 x - 18$
== Proof ==
Let the equation of $\LL$ be stated in slope-intercept form as:
:$y = m x + c$
where:
:$m$ is the slope of $\LL$
:$c$ is the $y$-intercept of $\LL$.
From Power Rule for Derivatives, the derivative of $\CC$ is:
:$\dfrac {\d y} {\d x} = 4 x$
At the point of contact where $x = 3$ the slope of $\LL$ is therefore $12$.
Hence $\LL$ is given by:
:$(1): \quad y = 12 x + c$
and it remains to find $c$.
We substitute $x = 3$ and $y = 18$ into $(1)$ and get:
:$c = -18$
Hence the result.
== Sources ==
* 
* 
Category:Examples of Use of Slope of Tangent to Curve at Point equals Value of Derivative
Slope of Tangent to Cycloid

== Theorem ==
Let $C$ be a cycloid generated by the equations:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
The slope of the tangent to $C$ at the point $\tuple {x, y}$ is given by:
:$\dfrac {\d y} {\d x} = \cot \dfrac \theta 2$
== Proof 1 ==
== Proof 2 ==
== Historical Note ==
Category:Slope of Tangent to Cycloid
Category:Cycloids
Slope of Tangent to Cycloid/Historical Note

== Historical Note on Slope of Tangent to Cycloid ==
Some sources suggest that the first to solve the problem of constructing a tangent to the cycloid was , along with his pupil at the time .
 also discovered this result, publishing it in $1643$.
Other sources cite the geometric proof of , which dates to $1638$, although yet other sources suggest it was in the same year.
It has been documented that solved the problem immediately once it had been proposed to him.
== Sources ==
* : Chapter $\text{V}$: "Greatness and Misery of Man"
* : Chapter $\text {A}.15$: Torricelli ()
* : Chapter $\text {B}.21$: The Cycloid
* 
* 
* 
* 
Category:Historical Notes
Slope of Tangent to Cycloid/Proof 1

== Theorem ==
== Proof ==
By Derivative of Curve at Point, the tangent to $C$ at the point $\tuple {x, y}$ is the derivative of its equation at that point.
Thus:
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid
Category:Slope of Tangent to Cycloid
Slope of Tangent to Cycloid/Proof 2

== Theorem ==
== Proof ==
Consider a polygon $ABCD$ being rolled along a straight line in the same way as the generating circle of $C$.
Let $A', B', C', D'$ be the points around which the $ABCD$ rotates while rolling.
:
The point $A$ traces out in succession several arcs of circles with centers $B', C', D'$.
The tangent to each of these arcs is perpendicular to the line joining the point of tangency to the corresponding point of rotation.
Consider a circle as being a polygon with an infinite number of sides.
It follows that in the limit the tangent to $C$ at any given point is perpendicular to the line joining the point of tangency to the corresponding point where the generating circle touches the straight line along which the generating circle rolls.
:
Let $O$ be the center of the generating circle.
Let $P$ be the point of tangency.
Let $Q$ be the point where the generating circle touches the straight line along which the generating circle rolls.
Let $S$ be the point where the tangent to $C$ meets the straight line along which the generating circle rolls.
As $OP$ and $OQ$ are both radii of the generating circle:
:$OP = OQ = a$
and so $\triangle POQ$ is isosceles.
Let $R$ be dropped perpendicular to $PQ$ at $R$.
As $\triangle POQ$ is isosceles, $\triangle OPR = \triangle OQR$ and so $PR = RQ$.
Thus $\angle POR = \angle QOR = \dfrac \theta 2$.
As $OR \perp PR$ and $PR \perp PS$, we have that $\triangle OPR$ is similar to $\triangle QPS$.
Thus $\angle PQS = \angle POR$ and so $\angle PSQ$ is $\dfrac \pi 2 - \angle POR = \dfrac \pi 2 - \dfrac \theta 2$.
By definition, $\tan \angle PSQ$ is the slope of the tangent $\dfrac {\d y} {\d x}$ to $C$.
Thus:
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.15$: Torricelli ()
* : Chapter $\text {B}.21$: The Cycloid: Problem $5$
Category:Slope of Tangent to Cycloid
Slow g-Tower is Slowly Well-Ordered under Subset Relation

== Theorem ==
Let $M$ be a class.
Let $g: M \to M$ be a slowly progressing mapping on $M$.
Let $M$ be a slow $g$-tower.
Then $M$ is slowly well-ordered under the subset relation.
== Proof ==
Let $M$ be a slow $g$-tower.
By $g$-Tower is Well-Ordered under Subset Relation: Empty Set:
:$\O$ is the smallest element of $M$.
This is condition $(\text S_1)$ of the definition of slowly well-ordered class under subset relation.
By $g$-Tower is Well-Ordered under Subset Relation: Union of Limit Elements:
:Each limit element $x$ of $M$ is the union of its lower section $\ds \bigcup x^\subset$
This is condition $(\text S_3)$ of the definition of slowly well-ordered class under subset relation.
It remains to establish condition $(\text S_2)$ of the definition of slowly well-ordered class under subset relation.
Let $x \in M$ be arbitrary.
Let $y$ be the immediate successor of $x$.
Then $x$ is not the greatest element of $M$.
Hence by $g$-Tower is Well-Ordered under Subset Relation: Successor of Non-Greatest Element:
:$y = \map g x$
Because $g$ is slowly progressing, $y$ contains at most $1$ more element than $x$.
But we also have that $y$ is the immediate successor of $x$.
Thus $x$ is a proper subset of $y$.
So $y$ contains at least $1$ more element than $x$.
Hence $y$ contains exactly $1$ element which is not in $x$.
The result follows.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Theorem $4.3$
Category:G-Towers
Category:Well-Orderings
Category:Subset Relation
Smaller Elements of Pythagorean Triple not both Odd

== Theorem ==
Let $\left({x, y, z}\right)$ be a Pythagorean triple, i.e. integers such that $x^2 + y^2 = z^2$.
Then $x$ and $y$ cannot both be odd.
== Proof ==
 $x$ and $y$ are both odd such that:
:$\exists z \in \Z: x^2 + y^2 = z^2$
Then:
:$x^2 + y^2 \equiv 1 + 1 \equiv 2 \pmod 4$
But from Square Modulo 4:
:$z^2 \equiv 0 \pmod 4$ or $z^2 \equiv 1 \pmod 4$
Thus $x^2 + y^2$ can not be square.
It follows by Proof by Contradiction that $x$ and $y$ cannot both be odd.
Category:Pythagorean Triples
Smaller Number of Wythoff Pair is Smallest Number not yet in Sequence

== Theorem ==
Consider the sequence of Wythoff pairs arranged in sequential order:
:$\tuple {0, 0}, \tuple {1, 2}, \tuple {3, 5}, \tuple {4, 7}, \tuple {6, 10}, \tuple {8, 13}, \ldots$
The first coordinate of each Wythoff pair is the smallest positive integer which has not yet appeared in the sequence.
== Proof ==
From Sequence of Wythoff Pairs contains all Positive Integers exactly Once Each, every positive integer can be found in the sequence of Wythoff pairs.
From Difference between Terms of Wythoff Pair, the first coordinate is the smaller of the coordinates of the Wythoff pair.
So consider a given Wythoff pair.
Let $p$ be the smallest positive integer which has not yet appeared in the sequence.
It has to appear somewhere.
The terms of both the lower and upper Wythoff sequences are in ascending order.
$n$ must appear in the next Wythoff pair, otherwise a larger positive integer will appear before it when it eventually does appear.
$n$ cannot appear as the second coordinate, or it will be less than whatever number appears as the first coordinate.
Hence $n$ appears as the first coordinate of the next Wythoff pair.
== 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:Beatty Sequences
Smaller of Thabit Pair is Tetrahedral

== Theorem ==
Let $\tuple {m_1, m_2}$ be a Thabit pair such that $m_1 < m_2$.
Then $m_1$ is a tetrahedral number.
== Proof ==
By Thabit's Rule:
for some $n \in \Z_{\ge 0}$.
We have that:
and so $m_1 < m_2$.
Then:
From Closed Form for Tetrahedral Numbers, this is the $k$th tetrahedral number where $k = 3 \times 2^n - 2$.
== Sources ==
* : $220$
* : $220$
Category:Amicable Pairs
Category:Tetrahedral Numbers
Smallest 10 Primes in Arithmetic Sequence

== Theorem ==
The smallest $10$ primes in arithmetic sequence are:
:$199 + 210 n$
for $n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.
These are also the smallest $8$ and $9$ primes in arithmetic sequence.
== Proof ==
But note that $199 + 10 \times 210 = 2299 = 11^2 \times 19$ and so is not prime.
Now we show that this is the smallest $10$ primes in arithmetic sequence.
By Divisibility of Common Difference of Arithmetic Sequence of Primes, the common difference $d$ of any $10$ primes in arithmetic sequence must be divisible by all primes less that $10$.
That is, the common difference is a multiple of $2 \times 3 \times 5 \times 7 = 210$.
Suppose $d = 210$.
Then the first term $p_1$ of the arithmetic sequence cannot be $11$, as $11 + 210 = 221 = 13 \times 17$.
Note that $210 \equiv 1 \pmod {11}$.
Hence if the $p_1$ gives a remainder $r > 1$ upon divison by $11$, the $\paren {12 - r}^{th}$ term of the sequence is equivalent to:
:$r + \paren {12 - r - 1} \times 210 \equiv 0 \pmod {11}$
and is divisible by $11$, which is impossible as all terms are primes.
Therefore $p_1 \equiv 1 \pmod {11}$.
The first few primes of this form are:
:$23, 67, 89, 199, \dots$
and we eliminate the first three candidates because:
:$23 + 5 \times 210 = 1073 = 29 \times 37$
:$67 + 3 \times 210 = 697 = 17 \times 41$
:$89 + 1 \times 210 = 229 = 13 \times 23$
which are not primes, and we have checked that $199 + 210 n$ are indeed $10$ primes in arithmetic sequence.
Now suppose $d > 210$.
Then $d \ge 420$ and the last term is greater than $420 \times 9 = 3780$.
This shows that the one we found here is the smallest $10$ primes in arithmetic sequence.
=== Smallest $9$ primes in arithmetic sequence ===
Most of the analysis above applies, but we need to also consider $p_1 \equiv 2 \pmod {11}$.
This gives the first few primes:
:$13, 79, 101, 167, 211, \dots$
and we have:
:$13 + 6 \times 210 = 1273 = 19 \times 67$
:$79 + 1 \times 210 = 289 = 17^2$
:$101 + 3 \times 210 = 731 = 17 \times 43$
:$167 + 1 \times 210 = 377 = 13 \times 29$
:$221 > 199$
so there are no smaller $9$ primes in arithmetic sequence.
=== Smallest $8$ primes in arithmetic sequence ===
Most of the analysis above applies, but we need to also consider $p_1 \equiv 3 \pmod {11}$.
This gives the first few primes:
:$3, 47, 113, 157, 179, 223, \dots$
and we have:
:$3 + 1 \times 210 = 213 = 3 \times 71$
:$47 + 7 \times 210 = 1517 = 37 \times 41$
:$113 + 1 \times 210 = 323 = 17 * 19$
:$157 + 5 \times 210 = 1207 = 17 * 71$
:$179 + 7 \times 210 = 1649 = 17 * 97$
:$223 > 199$
so there are no smaller $8$ primes in arithmetic sequence.
== Sources ==
* : No. $72$
* : $199$
* : $199$
Category:Prime Numbers
Category:Arithmetic Sequences
Smallest 17 Primes in Arithmetic Sequence

== Theorem ==
The smallest $17$ primes in arithmetic sequence are:
:$3\,430\,751\,869 + 87\,297\,210 n$
for $n = 0, 1, \ldots, 16$.
== Proof ==
First we note that:
:$3\,430\,751\,869 - 87\,297\,210 = 3\,343\,454\,659 = 17\,203 \times 194\,353$
and so this arithmetic sequence of primes does not extend to $n < 0$.
But note that $3\,430\,751\,869 + 17 \times 87\,297\,210 = 4\,914\,804\,439 = 41 \times 97 \times 1 235807$ and so is not prime.
== Historical Note ==
== Sources ==
* 
* 
* : $3,430,751,869$
Category:Prime Numbers
Category:Arithmetic Sequences
Smallest 17 Primes in Arithmetic Sequence/Historical Note

== Historical Note on Smallest 17 Primes in Arithmetic Sequence ==
, in his of $1986$ reported that this was the $2$nd longest known arithmetic sequence of prime numbers.
It was discovered in $1977$ by .
Since that time, plenty longer have been found.
== Sources ==
* 
* 
* : $3,430,751,869$
Category:Historical Notes
Smallest 18 Primes in Arithmetic Sequence

== Theorem ==
The smallest $18$ primes in arithmetic sequence are:
:$107\,928\,278\,317 + 9\,922\,782\,870 n$
for $n = 0, 1, \ldots, 16$.
== Proof ==
First we note that:
:$107\,928\,278\,317 - 9\,922\,782\,870 = 98\,005\,495\,447 = 29 \times 149 \times 22\,681\,207$
and so this arithmetic sequence of primes does not extend to $n < 0$.
But note that $107\,928\,278\,317 + 18 \times 9\,922\,782\,870 = 286\,538\,369\,977 = 23 \times 181 \times 68\,829\,779$ and so is not prime.
== Historical Note ==
== Sources ==
* 
* : $107,928,278,317$
Category:Prime Numbers
Category:Arithmetic Sequences
Smallest 18 Primes in Arithmetic Sequence/Historical Note

== Historical Note on Smallest 18 Primes in Arithmetic Sequence ==
, in his of $1986$ reported that this was the longest known arithmetic sequence of prime numbers.
As put it:
:breaking the previous record of $17$ due to Weintraub.
The Weintraub in question was , who had discovered the previous record of $17$ such primes in $1977$.
Since that time, plenty longer have been found.
== Sources ==
* 
* : $107,928,278,317$
Category:Historical Notes
Smallest 22 Primes in Arithmetic Sequence

== Theorem ==
The smallest $22$ primes in arithmetic sequence are:
:$11 \, 410 \, 337 \, 850 \, 553 + 4 \, 609 \, 098 \, 694 \, 200 n$
for $n = 0, 1, \ldots, 21$.
== Proof ==
But note that $11 \, 410 \, 337 \, 850 \, 553 + 22 \times 4 \, 609 \, 098 \, 694 \, 200 = 112 \, 810 \, 509 \, 122 \, 953 = 61 \times 107 \times 1907 \times 9063277$ and so is not prime.
== Sources ==
* 
* : $11,410,337,850,553$
Category:Prime Numbers
Category:Arithmetic Sequences
Smallest 3-Digit Permutable Prime

== Theorem ==
The smallest $3$-digit permutable prime is $113$.
== Proof ==
:$113$ is prime.
:$131$ is prime.
:$311$ is prime.
Consider the $3$-digit primes smaller than $113$:
:$101, 103, 107, 109$
They all contain a zero.
Thus, for each of these, at least one permutation ends in a zero.
Hence it is divisible by $10$ and so is not prime.
== Sources ==
* : $113$
* : $113$
Category:Permutable Primes
Category:113
Smallest 5 Consecutive Primes in Arithmetic Sequence

== Theorem ==
The smallest $5$ consecutive primes in arithmetic sequence are:
:$9 \, 843 \, 019 + 30 n$
for $n = 0, 1, 2, 3, 4$.
Note that while there are many longer arithmetic sequences of far smaller primes, those primes are not consecutive.
== Proof ==
But note that $9 \, 843 \, 019 + 5 \times 30 = 9 \, 843 \, 169 = 7^2 \times 200 \, 881$ and so is not prime.
Inspection of tables of primes (or a computer search) will reveal that this is the smallest such sequence.
== Sources ==
* 
* : $9,843,019$
Category:Prime Numbers
Category:Arithmetic Sequences
Smallest 5th Power equal to Sum of 5 other 5th Powers

== Theorem ==
The smallest positive integer whose fifth power can be expressed as the sum of $5$ other distinct positive fifth powers is $72$:
:$72^5 = 19^5 + 43^5 + 46^5 + 47^5 + 67^5$
== Proof ==
Let $\tuple {I, J, K, L, M}$ be the $5$ distinct positive fifth powers where $I < J < K < L < M$
The following program in R iterates through 97,330,464 calculations and verifies the assertion.
 for (I in 1:19) {
 	for (J in 2:43) {
 		for (K in 3:46) {
 			for (L in 4:47) {
 				for (M in 5:67) {
 					if (abs((I^5+J^5+K^5+L^5+M^5)^0.2 - round((I^5+J^5+K^5+L^5+M^5)^0.2)) < 0.00000001) {
 						print(paste((I^5+J^5+K^5+L^5+M^5)^0.2, I, J, K, L, M)) 
 					}
 				}
 			}
 		}
 	}
 }
== Sources ==
* : $72$
* : $72$
Category:Fifth Powers
Category:72
Smallest 5th Power equal to Sum of 5 other 5th Powers/Mistake

== Source Work ==
:
:The Dictionary
::$72$
:
:The Dictionary
::$72$
== Mistake ==
:$72^5 = 19^5 + 43^5 + 46^5 + 47^5 + 67^5$ is the smallest $5$th power equal to the sum of $5$ other $5$th powers.
== Correction ==
Refuted by:
What it should say is something more like:
:$72^5 = 19^5 + 43^5 + 46^5 + 47^5 + 67^5$ is the smallest positive $5$th power equal to the sum of $5$ other distinct positive $5$th powers.
== Sources ==
* : $72$
* : $72$
Category:Mistakes/Curious and Interesting Numbers