nyuuzyou/wikis huggingface dataset:
Largest Known Lead by 4n+1 in Prime Number Race

== Theorem ==
In the prime number race between prime numbers of the form $4 n - 1$ and $4 n + 1$, the highest known stretch of integers where $4 n + 1$ is not less than $4 n - 1$ is between $18 \, 465 \, 126 \, 293$ and $19 \, 033 \, 524 \, 538$.
== Proof ==
First note that:
where it can be seen that prime numbers of the form $4 n + 1$ are locally increasing, and:
where it can be seen at that point prime numbers of the form $4 n - 1$ may now be in front, but this is inconclusive.
== Sources ==
* : $18,465,126,293$
* : $18,465,126,293$
Category:Prime Number Races
Largest Mutually Coprime Subset of Initial Segment of Natural Numbers

== Theorem ==
Let $n \in \N$ be a natural number.
Consider the set $\N_n$ defined as:
:$\N_n = \closedint 1 n = \set {1, 2, \ldots n}$
Let $Q_n$ be the largest subset of $\N_n$ such that no element of $Q_n$ is the divisor of another element of $Q_n$.
Let $\map f n$ be the cardinality of $Q_n$.
Then for sufficiently large $n$:
:$0 \cdotp 6725 \ldots \le \dfrac {\map f n} n \le 0 \cdotp 6736 \ldots$
== Proof ==
== Sources ==
* 
* : $0,6725 \ldots \leqslant d \leqslant 0,6736 \ldots$
Category:Number Theory
Largest Number Not Expressible as Sum of Fewer than 8 Cubes

== Theorem ==
$8042$ is (probably) the largest positive integer that cannot be expressed as the sum of fewer than $8$ cubes.
== Proof ==
It is believed that this entry is a mistake.
$8042 = 1^3 + 4^3 + 4^3 + 10^3 + 10^3 + 10^3 + 17^3$, among many other expressions.
However:
$8042$ is conjectured to be the largest positive integer that cannot be expressed as the sum of fewer than $\bf 7$ cubes.
$\bf {454}$ is proven to be the largest positive integer that cannot be expressed as the sum of fewer than $8$ cubes.
== Historical Note ==
== Sources ==
* : $8042$
Category:Sums of Cubes
Category:8042
Largest Number Not Expressible as Sum of Fewer than 8 Cubes/Mistake

== Source Work ==
:
:The Dictionary
::$8042$
== Mistake ==
:This is probably the largest integer that cannot be represented as the sum of fewer than $8$ cubes.
== Correction ==
:$(1) \quad$ This should probably read:
::::This is probably the largest integer that cannot be represented as the sum of fewer than $7$ cubes
:::as we have:
::::$8042 = 1^3 + 4^3 + 4^3 + 10^3 + 10^3 + 10^3 + 17^3$
:::among many other expressions.
:$(2) \quad$ Also note that:
::::$8042 = 1340^3 + 1338^3 + \paren {-1339}^3 + \paren {-1339}^3 + 2^3$
:::so what it really ought to say is:
::::This is probably the largest integer that cannot be represented as the sum of fewer than $7$ positive cubes.
This entry has been excluded from of $1997$, so perhaps it was not even true in the first place.
== Sources ==
* : $8042$
Category:Mistakes/Curious and Interesting Numbers
Largest Number not Expressible as Sum of Less than 32 Positive Fifth Powers

== Theorem ==
The largest positive integer which cannot be expressed as the sum of less than $32$ positive fifth powers is $466$:
:$466 = 18 \times 1^5 + 14 \times 2^5$
== Proof ==
== Also see ==
* Hilbert-Waring Theorem
== Sources ==
* : $466$
Category:Fifth Powers
Category:Hilbert-Waring Theorem
Category:466
Largest Number not Expressible as Sum of Less than 37 Positive Fifth Powers

== Theorem ==
The largest positive integer which cannot be expressed as the sum of less than $37$ positive fifth powers is $223$:
:$223 = 31 \times 1^5 + 6 \times 2^5$
== Proof ==
== Also see ==
* Hilbert-Waring Theorem
== Sources ==
* : $223$
Category:Fifth Powers
Category:Hilbert-Waring Theorem
Category:223
Largest Number not Expressible as Sum of Less than 37 Positive Fifth Powers/Mistake

== Source Work ==
:
:The Dictionary
::$466$
== Mistake ==
:The largest number which cannot be represented with less than $32$ $5$th powers.
It needs to be specified that they ought to be positive $5$th powers, otherwise, for example:
:$466 = 2 \times 3^5 + 20 \times \paren {-1}^5$
... a mere $22$ $5$th powers.
== Sources ==
* : $466$
Category:Mistakes/Curious and Interesting Numbers
Largest Number not Expressible as Sum of Multiples of 23 and 28

== Theorem ==
The largest integer $n$ that cannot be expressed in the form:
:$n = 23 x + 28 y$
for $x, y \in \Z_{>0}$ is $593$.
== Proof ==
By Largest Number not Expressible as Sum of Multiples of Coprime Integers, the largest such number is:
:$\paren {23 - 1} \times \paren {28 - 1} - 1 = 593$
== Historical Note ==
== Sources ==
* : $593$
* : $593$
Category:Integer Combinations
Largest Number not Expressible as Sum of Multiples of 23 and 28/Historical Note

== Historical Note on Largest Number not Expressible as Sum of Multiples of 23 and 28 ==
Wilhelm Fliess thought he had stumbled upon a remarkable fact about the numbers $23$ and $28$: that combinations of them could be used to explain practically every earthly phenomenon.
From this "discovery" be built the pseudoscience of biorhythms.
However, it needs to be noted that all but a finite set of positive integers can be represented in the form $23 x + 28 y$ for positive integers $x$ and $y$.
The number $593$ is the largest which cannot.
Every pair of coprime integers has the same property as $23$ and $28$.
== Sources ==
* : $593$
* : $593$
Category:Historical Notes
Largest Number not Expressible as Sum of Multiples of Coprime Integers

== Theorem ==
Let $a, b$ be coprime integers, each greater than $1$.
Then the largest number not expressible as a sum of multiples of $a$ and $b$ (possibly zero) is the number:
:$a b - a - b = \paren {a - 1} \paren {b - 1} - 1$
=== Generalization ===
== Proof ==
First we show that $a b - a - b$ is not expressible as a sum of multiples of $a$ and $b$.
 $a b - a - b = s a + t b$ for some $s, t \in \N$.
Note that $t b \le s a + t b < a b - b = \paren {a - 1} b$.
This gives $t < a - 1$.
We also have $\paren {a - t - 1} b = \paren {s + 1} a$.
Hence $a \divides \paren {a - t - 1} b$.
Since $a$ and $b$ are coprime, by Euclid's Lemma:
:$a \divides a - t - 1$
As $a - t - 1 > 0$, by Absolute Value of Integer is not less than Divisors:
:$a \le a - t - 1$
which is a contradiction.
Hence $a b - a - b$ is not expressible as a sum of multiples of $a$ and $b$.
Next we need to show that all numbers greater than $a b - a - b$ can be so expressed.
 assume that $a > b$ and we split the numbers into two cases:
=== Case $1$: $x = a b - a - b + k$ for $1 \le k \le b$ ===
For $k = b$, we have $a b - a = a \paren {b - 1}$.
Notice that for $1 \le k < b$ and $0 \le s \le b - 2$:
and we see that, by Absolute Value of Integer is not less than Divisors:
:$a b - a - b + k - \paren {b - 1} a = k - b$
cannot be a multiple of $b$.
We claim that one of $a b - a - b + k - s a$ is divisible by $b$, where $0 \le s \le b - 1$.
Suppose not. Then we consider each of the remainders when dividing by $b$.
There are $b - 1$ remainders excluding $0$.
However we have a set of $b$ integers.
By Pigeonhole Principle, two integers must share the same remainder.
Suppose we have $a b - a - b + k - s_1 a$ and $a b - a - b + k - s_2 a$ both having remainder $r$ with $s_1 \ne s_2$:
:$\exists p, q \in \Z: a b - a - b + k - s_1 a = p b + r, a b - a - b + k - s_2 a = q b + r$
Then:
Hence $b \divides \size {s_1 - s_2} a$.
By Euclid's Lemma, $b \divides \size {s_1 - s_2}$.
But $0 < \size {s_1 - s_2} \le b - 1$.
This contradicts Absolute Value of Integer is not less than Divisors.
Therefore one of $a b - a - b + k - s a$ must be divisible by $b$, where $0 \le s \le b - 1$.
We have shown that $s \ne b - 1$.
Hence for some $s$ with $0 \le s \le b - 2$, $a b - a - b + k - s a$ is a positive multiple of $b$.
This gives:
:$a b - a - b + k = s a + t b$ for some $s, t \in \N$
as required.
=== Case $2$: $x > a b - a$ ===
By Division Theorem:
:$\exists q, r \in \Z: 0 \le r < b: x - \paren {a b - a - b + 1} = q b + r$
We have:
hence $q > 0$.
Moreover:
:$x - q b = a b - a - b + \paren {r + 1}$
which falls into Case $1$.
We have shown that:
:$\exists s, t \in \N: a b - a - b + \paren {r + 1} = s a + t b$
Hence:
:$x = s a + t b + q b = s a + \paren {t + q} b$
as required.
Category:Largest Number not Expressible as Sum of Multiples of Coprime Integers
Category:Integer Combinations
Largest Number not Expressible as Sum of Multiples of Coprime Integers/Generalization

== Theorem ==
Let $a, b$ be integers greater than $1$.
Let $d = \gcd \set {a, b}$.
Then the largest multiple of $d$ not expressible as a sum of multiples of $a$ and $b$ (possibly zero) is the number:
:$\dfrac {a b} d - a - b$
== Proof ==
By Integers Divided by GCD are Coprime:
:$\dfrac a d \perp \dfrac b d$
By Largest Number not Expressible as Sum of Multiples of Coprime Integers, the largest number not expressible as a sum of multiples of $\dfrac a d$ and $\dfrac b d$ is the number:
:$\dfrac {a b} {d^2} - \dfrac a d - \dfrac b d$
Let $k d$ be a multiple of $d$ expressible as a sum of multiples of $a$ and $b$:
:$\exists s, t \in \N: s a + t b = k d$
Then:
:$s \dfrac a d + t \dfrac b d = k$
showing that $k$ is a sum of multiples of $\dfrac a d$ and $\dfrac b d$.
This argument reverses.
Hence the largest multiple of $d$ not expressible as a sum of multiples of $a$ and $b$ is the number:
:$d \paren {\dfrac {a b} {d^2} - \dfrac a d - \dfrac b d} = \dfrac {a b} d - a - b$
Category:Largest Number not Expressible as Sum of Multiples of Coprime Integers
Category:Integer Combinations
Largest Pandigital Square

== Theorem ==
The largest pandigital square (in the sense where pandigital includes the zero) is $9 \, 814 \, 072 \, 356$:
:$9 \, 814 \, 072 \, 356 = 99 \, 066^2$
== Proof ==
We check all the squares of numbers from $99 \, 067$ up to $\floor {\sqrt {9 \, 876 \, 543 \, 210} } = 99 \, 380$, with the following constraints:
Since all these squares has $9$ as its leftmost digit, the number cannot end with $3$ or $7$.
The number cannot end with $0$ since its square will end in $00$.
A pandigital number is divisible by $9$, so our number must be divisible by $3$.
These constraints leaves us with the following $74$ candidates:
By inspection, none of these numbers are pandigital.
== Sources ==
* : $9,814,072,356$
* : $9,814,072,356$
Category:Square Numbers
Category:Pandigital Integers
Category:9,814,072,356
Largest Parallelogram Contained in Triangle

== Theorem ==
Let $T$ be a triangle.
Let $P$ be a parallelogram contained within $T$.
Let $P$ have the largest area possible for the conditions given.
Then:
:$(1): \quad$ One side of $P$ is coincident with part of one side of $T$, and hence two vertices lie on that side of $T$
:$(2): \quad$ The other two vertices of $P$ bisect the other two sides of $T$
:$(3): \quad$ The area of $P$ is equal to half the area of $T$.
== Proof ==
We will first find the maximum area of $P$ when $(1)$ is satisfied, that is, when $P$ is inscribed in $T$.
=== Proof of $(2)$ ===
Consider the diagram below.
:
Here $DEGF$ is our inscribed parallelogram $P$.
Since $FG \parallel BC$, by Equiangular Triangles are Similar:
:$\triangle AFG \sim \triangle ABC$
Let $FG : BC = 1 : r = \text {Height of } \triangle AFG : \text {Height of } \triangle ABC$.
The area of $T$, which is fixed, is given by:
:$\dfrac {BC \times \text {Height of } \triangle ABC} 2$
The area of $P$ is given by:
which is $\dfrac {2 \paren {r - 1} } {r^2}$ of the area of $T$.
Notice that:
Equality holds $r = 2$ for the first inequality.
Therefore the maximum of $\dfrac {2 \paren {r - 1} } {r^2}$ occurs at $r = 2$.
This implies $AF : AB = AG : AC = 1 : 2$.
Hence both sides $AB$ and $AC$ are both bisected by vertices of $P$.
=== Proof of $(3)$ ===
At $r = 2$:
:$\dfrac {2 \paren {r - 1} } {r^2} = \dfrac {2 \paren {2 - 1} } {2^2} = \dfrac 1 2$
therefore the maximum area of $P$ is equal to half the area of $T$.
=== Proof of $(1)$ ===
Now we consider the cases where $P$ is not inscribed in $T$.
Suppose less than $3$ vertices of $P$ lie on the sides of $T$.
By constructing parallel lines to the sides of $T$, we can find a smaller triangle $T'$ that is similar to $T$, and the sides of $T'$ touches at least $3$ vertices of $P$.
:
In the above figure, $\triangle A'B'C'$ is our $T'$.
The case where all $4$ vertices of $P$ lie on the sides of $T'$ has been covered above.
Suppose, then, that only $3$ vertices of $P$ lie on the sides of $T'$.
We can split $T'$ using a line parallel to one of the sides of $P$ that passes through a vertex of $T'$.
We connect the base of that line to the vertex of $P$ that does not lie on the sides of $T'$.
:
In the figure above, we can see that $P$ has also been split into two parallelograms, both are inscribed in the two smaller triangles $\triangle XB'C'$ and $\triangle XYC'$.
By our results above:
so the area of $P$ cannot exceed half the area of $T$.
Hence we see that half the area of $T$ is indeed the maximum area of $P$, and occurs when all three conditions above are satisfied.
== Sources ==
* : $3.4.$ Miscellaneous Results in Geometry: Problem $3$
Category:Triangles
Category:Parallelograms
Largest Penholodigital Square

== Theorem ==
The largest penholodigital square is $923 \, 187 \, 456$:
:$923 \, 187 \, 456 = 30 \, 384^2$
== Proof ==
== Sources ==
* : $923,187,456$
* : $923,187,456$
Category:Square Numbers
Category:Penholodigital Integers
Category:923,187,456
Largest Prime Factor of Euler Phi Function

==Theorem==
Let $\phi: \Z_{>0} \to \Z_{>0}$ be the Euler $\phi$ function.
Then the largest prime factor of $\map\phi n$ is less than the largest prime factor of $n$,
==Proof==
By Euler Phi Function of Integer, for any $n \in \Z_{>0}$, we have:
:$\map \phi n = n \paren {1 - \dfrac 1 {p_1} } \paren {1 - \dfrac 1 {p_2} } \cdots \paren {1 - \dfrac 1 {p_r} }$
where $p_1, p_2, \ldots, p_r$ are the distinct primes dividing $n$.
We express $n$ in its prime decomposition:
:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}, p_1 > p_2 > \cdots > p_r$
Then the largest prime factor of $n$ is $p_1$ and
:$\map \phi n = p_1^{k_1-1}p_2^{k_2-1} \cdots p_r^{k_r-1}\cdot(p_1-1)(p_2-1)\cdots(p_r-1)$
The largest prime factor of $(p_i-1),i=1,\dots,r$ must be less than $p_i$, hence less than $p_1$.
So the largest prime factor of $\map\phi n$ is less than $p_1$.
==Example==
For $n = 6$: 
Euler Phi Function of 6 is 2
The largest prime factor of $n$ is 3, which is greater than 2.
Category:Euler Phi Function
Largest Prime Factor of n squared plus 1

== Theorem ==
Let $n \in \Z$ be greater than $239$.
Then the largest prime factor of $n^2 + 1$ is at least $17$.
== Proof ==
We note that for $n = 239$ we have:
Thus the largest prime factor of $239^2 + 1$ is $13$.
== Also see ==
* Solution of Diophantine Equation $x^2 + 1 = 2 y^4$
== Sources ==
* 
* : $239$
Category:Number Theory
Category:239
Largest Product of Pandigital Factors

== Theorem ==
The largest integer that can be obtained by multiplying $2$ integers which between them use all the digits from $1$ to $9$ is:
:$843 \, 973 \, 902 = 9642 \times 87531$
== Proof ==
== Sources ==
* : $9642$
* : $9642$
Category:Recreational Mathematics
Largest Rectangle Contained in Triangle

== Theorem ==
Let $T$ be a triangle.
Let $R$ be a rectangle contained within $T$.
Let $R$ have the largest area possible for the conditions given.
Then:
:$(1): \quad$ One side of $R$ is coincident with part of one side of $T$, and hence two vertices lie on that side of $T$
:$(2): \quad$ The other two vertices of $R$ bisect the other two sides of $T$
:$(3): \quad$ The area of $R$ is equal to half the area of $T$.
== Proof ==
Note that a rectangle is a parallelogram.
By Largest Parallelogram Contained in Triangle, the area of $R$ cannot exceed half the area of $T$.
Hence we only need to show that when the first two conditions above are satisfied, the area of $R$ is exactly half the area of $T$.
Consider the diagram below.
:
Since $AD = DC$ and $CE = EB$:
and so the area of $R$ is equal to the area of the parts of $T$ not included in $R$.
That is, the area of $R$ is exactly half the area of $T$.
Category:Triangles
Category:Rectangles
Largest Rectangle with Given Perimeter is Square

== Theorem ==
Let $\SS$ be the set of all rectangles with a given perimeter $L$.
The element of $\SS$ with the largest area is the square with length of side $\dfrac L 4$.
== Proof ==
Consider an arbitrary element $R$ of $\SS$.
Let $B$ be half the perimeter of $R$.
Let $x$ be the length of one side of $R$.
Then the length of an adjacent side is $B - x$.
The area $\AA$ of $R$ is then given by:
:$\AA = x \paren {B - x}$
Let $\AA$ be expressed in functional notation as:
:$\map f x = x \paren {B - x}$
We have:
Thus from Derivative at Maximum or Minimum, when $B - 2 x = 0$, the area of $R$ is at a maximum or a minimum.
That is:
:$x = \dfrac B 2 = \dfrac L 4$
Geometrical analysis shows that for this length of side the area is not a minimum because that happens when $x = 0$ or $x = B$.
Hence the result.
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.13$: Fermat ()
Category:Squares
Category:Rectangles
Largest Rectangle with Given Perimeter is Square/Historical Note

== Historical Note on Largest Rectangle with Given Perimeter is Square ==
The result Largest Rectangle with Given Perimeter is Square was given by as one of the first tests to demonstrate the use of his result Derivative at Maximum or Minimum.
It is a more specific (and accessible) version of .
== Sources ==
* : Chapter $\text {A}.13$: Fermat ()
Category:Historical Notes
Largest Right-Truncatable Primes allowing 1

== Theorem ==
Let $1$ be temporarily considered to be a prime number.
Under that consideration, the largest right-truncatable prime numbers are:
:$1 \, 979 \, 339 \, 333$
:$1 \, 979 \, 339 \, 339$
== Proof ==
We have that:
For both, the truncation process is the same:
== Sources ==
* : $1,979,339,339$
* : $1,979,339,339$
Category:Right-Truncatable Primes
Largest n such that 1 to n can be Partitioned for no Element to be Sum of 2 Elements in Same Set

== Theorem ==
$44$ is the largest integer $n$ such that the set of integers from $1$ to $n$ can be partitioned into $4$ subsets such that no integer in any of these subsets is the sum of $2$ other integers in the same subset:
:$\set {1, 3, 5, 15, 17, 19, 26, 28, 40, 42, 44}$
:$\set {2, 7, 8, 18, 21, 24, 27, 33, 37, 38, 43}$
:$\set {4, 6, 13, 20, 22, 23, 25, 30, 32, 39, 41}$
:$\set {9, 10, 11, 12, 14, 16, 29, 31, 34, 35, 36}$
== Proof ==
== Sources ==
* 
* : $44$
Category:Recreational Mathematics
Category:44
Largest nth Power which has n Digits

== Theorem ==
The largest $n$th power which has $n$ digits is $9^{21}$:
:$9^{21} = 109 \, 418 \, 989 \, 131 \, 512 \, 359 \, 209$
== Proof ==
The $n$th power of $10$ has $n + 1$ digits.
Hence the $n$th power of $m$ such that $m > 10$ has more than $n$ digits.
The $11$th power of $8$ has $10$ digits:
:$8^{10} = 8 \, 589 \, 934 \, 592$
and so $8^n$ where $n > 10$ has fewer than $n$ digits.
Hence the $n$th power of $m$ such that $m < 9$ and $n < 21$ has fewer than $n$ digits.
Finally we note that:
:$9^{22} = 984 \, 770 \, 902 \, 183 \, 611 \, 232 \, 881$
has $21$ digits.
Hence the $n$th power of $9$ such that $n > 21$ has fewer than $n$ digits.
Hence the result.
== Historical Note ==
== Sources ==
* : $109,418,989,131,512,359,209$
* : $109,418,989,131,512,359,209$
Category:Powers
Category:Recreational Mathematics
Category:109,418,989,131,512,359,209
Largest nth Power which has n Digits/Historical Note

== Historical Note on Largest nth Power which has n Digits ==
, in his of $1997$, attributes this result to a certain Friedlander, but no further information has so far been unearthed.
== Sources ==
* : $109,418,989,131,512,359,209$
* : $109,418,989,131,512,359,209$
Category:Historical Notes
Lasota-Yorke Inequality/One-Sided Shift Space of Finite Type

== Theorem ==
Let $\struct {X_\mathbf A ^+, \sigma}$ be a one-sided shift of finite type.
Let $F_\theta ^+$ be the space of Lipschitz functions on $X_\mathbf A ^+$.
Let $\norm \cdot_\theta$ be the Lipschitz norm on $F_\theta ^+$.
Let $\norm \cdot_\infty$ be the supremum norm on $F_\theta ^+$.
Let $f \in F_\theta ^+$.
Let $u := \map \Re f$ be the real part of $f$.
Let $\LL_f$ and $\LL_u$ denote the transfer operators.
If $\LL_u$ is normalized, then there is a $C > 0$ such that:
:$\norm {\LL_f ^n w}_\theta \le C \norm w_\infty + \theta ^n \norm w_\theta$
for all $w \in F_\theta ^+$ and $n \in \N$.
== Proof ==
Recall the basic inequality:
:There exists a $C_0 > 0$ so that we have:
:for all $w \in F_\theta ^+$ and $n \in \N$.
On the other hand, we have:
since for all $x \in X_\mathbf A ^+$:
Therefore:
Category:Ergodic Theory
Category:Functional Analysis
Last Digit of Perfect Numbers Alternates between 6 and 8

== Conjecture ==
The last digit of the sequence of perfect numbers alternates between $6$ and $8$:
: $6$
: $28$
: $496$
: $8128$
== Refutation ==
The sequence continues:
:$33 \, 550 \, 336$
:$8 \, 589 \, 869 \, 056$
... two consecutive perfect numbers ending in $6$.
== Also see ==
* One Perfect Number for Each Number of Digits
* Perfect Number ends in 6 or 28 preceded by Odd Digit
== Historical Note ==
== Sources ==
* : $28$
* : $28$
Category:Famous False Conjectures
Category:Perfect Numbers
Last Digit of Perfect Numbers Alternates between 6 and 8/Historical Note

== Historical Note on Last Digit of Perfect Numbers Alternates between $6$ and $8$ ==
The conjecture that there is Last Digit of Perfect Numbers Alternates between $6$ and $8$ was made by in his , published some time around the $2$nd century.
It was a simple extrapolation from the knowledge of the perfect numbers at the time.
Some sources suggest that made these conjectures, but this appears to be incorrect.
== Sources ==
* : $28$
* : $28$
Category:Historical Notes
Last Element of Geometric Sequence with Coprime Extremes has no Integer Proportional as First to Second

== Theorem ==
Let $G_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers such that $a_0 \ne 1$.
Let $a_0 \perp a_n$, where $\perp$ denotes coprimality.
Then there does not exist an integer $b$ such that:
:$\dfrac {a_0} {a_1} = \dfrac {a_n} b$
== Proof ==
 there exists $b$ such that $\dfrac {a_0} {a_1} = \dfrac {a_n} b$.
Then:
:$\dfrac {a_0} {a_n} = \dfrac {a_1} b$
By Ratios of Fractions in Lowest Terms:
:$a_0 \divides a_1$
where $\divides$ denotes divisibility.
From Divisibility of Elements in Geometric Sequence of Integers:
:$a_0 \divides a_n$
But $a_0 \perp a_n$.
From this contradiction it follows that there can be no such $b$.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Geometric Sequences
Lateral Face of Prismatoid is Triangle, Trapezium or Parallelogram

== Theorem ==
Let $\PP$ be a prismatoid.
Let $\FF$ be a lateral face of $\PP$.
Then $\FF$ is one of the following:
:a triangle
:a trapezium
:a parallelogram.
== Proof ==
We have that an antiprism is a prismatoid.
The lateral faces of an antiprism are triangles.
We have that an oblique prism is a prismatoid.
The lateral faces of an oblique prism are parallelograms.
Suppose that $\FF$ were a polygon with more than $4$ sides.
Then at least $2$ of those adjacent sides, $S$ and $T$ say, are also sides of $\FF$.
But as $S$ and $T$ are not in the same plane as the lateral edges that form the other sides of $\FF$, they cannot form a polygon.
So the lateral faces can have no more than $4$ sides.
So, let $A$ and $B$ be the bases of $\PP$.
They are parallel by definition.
Let $L_1$ and $L_2$ be sides of $A$ and $B$ respectively which form the sides of one of the lateral faces $\FF$ of $\PP$.
Let $L_1$ and $L_2$ be such that they are not of the same length.
Then $\FF$ is a quadrilateral which has $2$ unequal sides parallel.
Hence by definition it is a trapezium.
== Sources ==
* 
* 
Category:Lateral Faces of Prismatoids
Category:Triangles
Category:Trapezia
Category:Parallelograms
Lateral Surface Area of Circular Cylinder

== Theorem ==
=== By Height ===
=== By Slant Height ===
== Sources ==
* : $\S 4$: Geometric Formulas: Circular Cylinder of Radius $r$ and Slant Height $l$
* : $\S 7$: Geometric Formulas: Circular Cylinder of Radius $r$ and Slant Height $l$
Category:Lateral Surface Area of Circular Cylinder
Category:Circular Cylinders
Category:Area Formulas