nyuuzyou/wikis huggingface dataset:
Number of Parameters of Autoregressive Model

== Theorem ==
Let $S$ be a stochastic process based on an equispaced time series.
Let the values of $S$ at timestamps $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be deviations from a constant mean level $\mu$:
:$\tilde z_t = z_t - \mu$
Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a sequence of independent shocks at timestamps $t, t - 1, t - 2, \dotsc$
Let $M$ be an autoregessive model on $S$ of order $p$:
:$\tilde z_t = \phi_1 \tilde z_{t - 1} + \phi_2 \tilde z_{t - 2} + \dotsb + \phi_p \tilde z_{t - p} + a_t$
Then $M$ has $p + 2$ parameters.
</onlyinclude>
== Proof ==
By definition of the parameters of $M$:
Thus:
:there are $p$ parameters of the form $\phi_j$
:$1$ parameter $\mu$
:$1$ parameter $\sigma_a^2$.
That is: $p + 1 + 1 = p + 2$ parameters.
== Sources ==
* :
::$1$: Introduction:
:::$1.2$ Stochastic and Deterministic Dynamic Mathematical Models
::::$1.2.1$ Stationary and Nonstationary Stochastic Models for Forecasting and Control: Autoregressive models
Category:Autoregressive Models
Number of Parameters of Moving Average Model

== Theorem ==
Let $S$ be a stochastic process based on an equispaced time series.
Let the values of $S$ at timestamps $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
Let $\tilde z_t$ be the deviation from a constant mean level $\mu$:
:$\tilde z_t = z_t - \mu$
Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a sequence of independent shocks at timestamps $t, t - 1, t - 2, \dotsc$
Let $M$ be an moving average model on $S$ of order $q$:
:$\tilde z_t = a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
Then $M$ has $q + 2$ parameters.
</onlyinclude>
== Proof ==
By definition of the parameters of $M$:
Thus:
:there are $q$ parameters of the form $\theta_j$
:$1$ parameter $\mu$
:$1$ parameter $\sigma_a^2$.
That is: $q + 1 + 1 = q + 2$ parameters.
== Sources ==
* :
::$1$: Introduction:
:::$1.2$ Stochastic and Deterministic Dynamic Mathematical Models
::::$1.2.1$ Stationary and Nonstationary Stochastic Models for Forecasting and Control: Moving average models
Category:Moving Average Models
Number of Partial Derivatives of Order n

== Theorem ==
Let $u = \map f {x_1, x_2, \ldots, x_m}$ be a function of the $m$ independent variables $x_1, x_2, \ldots, x_m$.
There are $m^n$ partial derivatives of $u$ of order $n$.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:There are $m^n$ partial derivatives of $u$ of order $n$.
$\map P 0$ is the degenerate case where $f$ is not partially differentiated at all:
:$\map f {x_1, x_2, \ldots, x_m}$
and it is apparent that there is only $1 = m^0$ such.
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
:There are $m^1 = m$ partial derivatives of $u$ of order $1$.
These can be instanced as:
and it is apparent that there are $m$ such.
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:There are $m^k$ partial derivatives of $u$ of order $k$.
from which it is to be shown that:
:There are $m^{k + 1}$ partial derivatives of $u$ of order $k + 1$.
=== Induction Step ===
This is the induction step:
Let $g$ be one of the partial derivatives of $u$ of order $k$.
Then by the basis for the induction, there are $m$ partial derivatives of $g$.
By the induction hypothesis, there are $m^k$ partial derivatives of $u$ of order $k$.
Thus by the Product Rule for Counting, there are $m \times m^n$ partial derivatives of $u$ of order $k + 1$.
That is, a total of $m^{k + 1}$ partial derivatives of $u$ of order $k + 1$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}:$ there are $m^n$ partial derivatives of $u$ of order $n$.
== Examples ==
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.3$ Higher Order Derivatives
Category:Partial Differentiation
Category:Number of Partial Derivatives of Order n
Number of Partial Derivatives of Order n/Examples

== Examples of Use of Number of Partial Derivatives of Order $n$ ==
=== Partial Derivatives of Order $1$ of Function of $2$ Variables ===
=== Partial Derivatives of Order $2$ of Function of $2$ Variables ===
=== Partial Derivatives of Order $n$ of Function of $2$ Variables ===
Category:Number of Partial Derivatives of Order n
Number of Partial Derivatives of Order n/Examples/Order 1 of 2 Variables

== Examples of Use of Number of Partial Derivatives of Order $n$ ==
Let $u = \map f {x, y}$ be a real function of $2$ variables.
There are $2$ partial derivatives of $u$ of order $1$.
These are:
== Proof ==
From Number of Partial Derivatives of Order $n$, there are $m^n$ partial derivatives of order $n$ of a function of $m$ independent variables.
In this case, $m = 2$ and $n = 1$.
Thus there are $2^1 = 2$ partial derivatives of $u$ of order $1$.
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.3$ Higher Order Derivatives
Category:Number of Partial Derivatives of Order n
Number of Partial Derivatives of Order n/Examples/Order 2 of 2 Variables

== Examples of Use of Number of Partial Derivatives of Order $n$ ==
Let $u = \map f {x, y}$ be a real function of $2$ variables.
There are $4$ partial derivatives of $u$ of order $2$.
These are:
== Proof ==
From Number of Partial Derivatives of Order $n$, there are $m^n$ partial derivatives of order $n$ of a function of $m$ independent variables.
In this case, $m = 2$ and $n = 2$.
Thus there are $2^2 = 4$ partial derivatives of $u$ of order $2$.
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.3$ Higher Order Derivatives
Category:Number of Partial Derivatives of Order n
Number of Partial Derivatives of Order n/Examples/Order n of 2 Variables

== Examples of Use of Number of Partial Derivatives of Order $n$ ==
Let $u = \map f {x, y}$ be a real function of $2$ variables.
There are $2^n$ partial derivatives of $u$ of order $n$.
== Proof ==
From Number of Partial Derivatives of Order $n$, there are $m^n$ partial derivatives of order $n$ of a function of $m$ independent variables.
In this case, $m = 2$.
Thus there are $2^n$ partial derivatives of $u$ of order $n$.
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.3$ Higher Order Derivatives
Category:Number of Partial Derivatives of Order n
Number of Partitions as Coefficient of Power Series

== Theorem ==
The number of partitions $\map p n$ of a (strictly) positive integer $n$ is equal to the coefficient of $x^n$ when the expression:
:$\map f n = \dfrac 1 {\paren {1 - x} \paren {1 - x^2} \paren {1 - x^3} \cdots}$
is expanded into a power series.
That is:
:$\map f n = 1 + \map p 1 x + \map p 2 x^2 + \map p 3 x^3 + \cdots$
or:
:$\ds \sum_{n \mathop \in \Z_{\ge 0} } \map p n q^n = \prod_{j \mathop \in \Z_{>0} } \dfrac 1 {1 - q^j}$
where $\map p 0 := 1$.
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $6$
* : Chapter $\text {A}.21$: Euler ()
* : $\S 1.2.9$: Generating Functions: Exercise $8$
* 
Category:Partition Theory
Category:Number Theory
Number of Partitions with no Multiple of 3 equals Number of Partitions where Parts appear No More than Twice

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map t n$ denote the number of ways $n$ can be partitioned into parts which are specifically not multiples of $3$.
Let $\map v n$ denote the number of ways $n$ can be partitioned such that no part appears twice.
Then:
:$\forall n \in \Z_{>0}: \map t n = \map v n$
== Proof ==
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $14$
Category:Examples of Integer Partitions
Number of Paths on Graph along X-axis using Diagonal Steps

== Theorem ==
The number of different paths that can be taken on a Cartesian plane from the origin to $\tuple {2 n + 2, 0}$, using only diagonal steps, and never touching the $x$-axis except at the beginning and the end of the path, is the Catalan number $C_n$:
:
== Proof ==
== Sources ==
* : $42$
* : $42$
Category:Catalan Numbers
Number of Permutations

== Theorem ==
Let $S$ be a set of $n$ elements.
Let $r \in \N: r \le n$.
The number of $r$-permutations of $S$ is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the falling factorial symbol, this can also be expressed:
:${}^n P_r = n^{\underline r}$
=== Corollary ===
== Proof 1 (Informal) ==
== Proof 2 (Formal) ==
== Examples ==
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations
* 
* 
* 
Category:Number of Permutations
Category:Permutations (Ordered Selections)
Category:Combinatorics
Number of Permutations/Examples/4 from 52

== Example of Number of Permutations ==
The number of ways of choosing $4$ objects in order from a set of $52$ (for example, cards from a deck) is:
:${}^{52} P_4 = 52 \times 51 \times 50 \times 49 = \dfrac {52!} {48!} = 6 \, 497 \, 400$
== Proof ==
From Number of Permutations:
:${}^{52} P_4 = \dfrac {52!} {\paren {52 - 4}!}$
== Sources ==
* : $24$
* : $24$
Category:Number of Permutations
Number of Permutations/Examples/Even Integers from 1, 2, 3, 4

== Example of Number of Permutations ==
Let $N$ be the number of even integers which can be made using all the digits $1$, $2$, $3$ and $4$.
Then:
:$N = 12$
== Proof ==
An integer formed using the digits $1$, $2$, $3$ and $4$ is even it ends in $2$ or $4$.
Those $4$ digit integers ending in $2$ consist of the $3$ digits integers that can be made with $1$, $3$ and $4$
Those $4$ digit integers ending in $4$ consist of the $3$ digits integers that can be made with $1$, $2$ and $3$
From Number of Permutations, the total number of integers which can be made using $3$ different digits is $3!$.
:$N = 2 \times 3! = 12$
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Exercises $\text I$: $6$
Category:Number of Permutations
Number of Permutations/Examples/Numbers Greater than 200 from 1, 2, 3, 4

== Example of Use of Number of Permutations ==
Let $N$ be the number of ways you can make a number greater than $200$ using the digits $1$, $2$, $3$ and $4$ no more than once each.
Then:
:$N = 42$
== Proof ==
From Number of Permutations, you can make $24$ numbers using all $4$ digits once each, and all will be over $200$.
You cannot make a number greater than $200$ using $2$ digits or fewer.
From Number of Permutations, you can make $\dfrac {4!} {\paren {4 - 3}!} = \dfrac {24} 1 = 24$ numbers using only $3$ digits.
$6$ of those begin with $1$, and so are not greater than $200$.
Hence:
:$N = 24 + 24 - 6$
Hence the result.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Exercises $\text I$: $5$
Category:Number of Permutations
Number of Permutations/Proof 1

== Theorem ==
== Proof ==
We pick the elements of $S$ in any arbitrary order.
There are $n$ elements of $S$, so there are $n$ options for the first element.
Then there are $n - 1$ elements left in $S$ that we haven't picked, so there are $n-1$ options for the second element.
Then there are $n - 2$ elements left, so there are $n - 2$ options for the third element.
And so on, to the $r$th element of our selection: we now have $n - \paren {r - 1}$ possible choices.
Each mapping is independent of the choices made for all the other mappings, so by the Product Rule for Counting, the total number of ordered selections from $S$:
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: The value of ${}_n P_r$
* : Problems
* : $\text {3-1}$ Permutations and Combinations: Theorem $\text {3-1}$
* : $\S 1.2.5$: Permutations and Factorials
Category:Number of Permutations
Number of Permutations/Proof 2

== Theorem ==
== Proof ==
By definition, an $r$-permutation of $S$ is an ordered selection of $r$ elements of $S$.
Also, an $r$-permutation of $S$ is an injection from a subset of $S$ into $S$.
The result is immediate from Cardinality of Set of Injections‎.
Category:Number of Permutations
Number of Permutations of All Elements

== Theorem ==
Let $S$ be a set of $n$ elements.
The number of permutations of $S$ is $n!$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $24$
* : $24$
* 
* 
* 
Category:Number of Permutations
Number of Permutations of All Elements/Proof 1

== Theorem ==
== Proof ==
We are seeking to calculate the number of $r$-permutations of $S$, that is ${}^n P_r$, where $r = n$.
Hence:
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: The number of ways of arranging $n$ things in line
* : $\text {3-1}$ Permutations and Combinations: Theorem $\text {3-1}$
* : $\S 1.2.5$: Permutations and Factorials
Category:Number of Permutations
Number of Permutations of All Elements/Proof 2

== Theorem ==
== Proof ==
We pick the elements of $S$ in an arbitrary order.
There are $n$ elements of $S$, so there are $n$ options for the first element.
Then there are $n - 1$ elements left in $S$ that we have not picked, so there are $n - 1$ options for the second element.
Then there are $n - 2$ elements left, so there are $n - 2$ options for the third element.
And so on, until there are $3$, then $2$, then $1$ remaining elements.
Each mapping is independent of the choices made for all the other mappings.
So, by the Product Rule for Counting, the total number of ordered selections from $S$:
== Sources ==
* : $\S 3.6$. Products of bijective mappings. Permutations: Example $54$
* : $\S 1.6$
* : Appendix: Elementary set and number theory
Category:Number of Permutations
Number of Permutations of All Elements/Proof 3

== Theorem ==
== Proof ==
From the definition, it can be seen that a bijection $f: S \to S$ is an $n$-permutation.
Hence, from Cardinality of Set of Bijections the number of $n$-permutations on a set of $n$ elements is:
:${}^n P_n = \dfrac {n!} {\paren {n - n}!} = n!$
Category:Number of Permutations
Number of Permutations of One Less

== Theorem ==
:${}^{n - 1} P_n = {}^n P_n$
where ${}^k P_n$ denotes the number of ordered selections of $k$ objects from $n$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $2$
Category:Number of Permutations of One Less
Category:Permutations (Ordered Selections)
Number of Permutations with Repetition

== Theorem ==
Set $S$ be a set of $n$ elements.
Let $\sequence T_m$ be a sequence of $m$ terms of $S$.
Then there are $n^m$ different instances of $\sequence T_m$.
== Proof ==
Let $N_m$ denote the set $\set {1, 2, \ldots, m}$.
Let $f: N_m \to S$ be the mapping defined as:
:$\forall k \in N_m: \map f t = t_m$
By definition, $f$ corresponds to one of the specific instances of $\sequence T_m$.
Hence the number of different instances of $\sequence T_m$ is found from Cardinality of Set of All Mappings:
:$\card S^{\card {N_m} }$
The result follows.
== Examples ==
== Sources ==
* : Chapter $\text I$ Permutations and Combinations: The $r$, $s$ Principle: Example $3$.
Category:Combinatorics
Category:Number of Permutations with Repetition
Number of Permutations with Repetition/Examples/7 Choices from 4

== Example of Use of Number of Permutations with Repetition ==
Residents of a boarding house are offered a choice of $1$ of $4$ main foods for breakfast:
:fish or eggs or bacon or sausages.
Over the course of a week, a resident can then order:
:$1$ of $4$ dishes on Monday
:$1$ of $4$ dishes on Tuesday
:$\ldots$
:$1$ of $4$ dishes on Sunday
The number of different breakfast arrangements over a week of $7$ days is:
:$4 \times 4 \times 4 \times 4 \times 4 \times 4 \times 4 = 16 \, 384$
== Sources ==
* : Chapter $\text I$ Permutations and Combinations: The $r$, $s$ Principle: Example $3$.
Category:Number of Permutations with Repetition
Number of Petals of Even Index Rhodonea Curve

== Theorem ==
Let $n$ be an even (strictly) positive integer.
Let $R$ be a rhodonea curve defined by one of the polar equations:
Then $R$ has $2 n$ petals.
== Proof ==
== Sources ==
* : $\S 11$: Special Plane Curves: Four-Leaved Rose: $11.17$
* 
* 
* : $\S 9$: Special Plane Curves: Four-Leaved Rose: $9.17.$
* 
* 
Category:Rhodonea Curves
Number of Petals of Odd Index Rhodonea Curve

== Theorem ==
Let $n$ be an odd positive integer.
Let $R$ be a rhodonea curve defined by one of the polar equations:
Then $R$ has $n$ petals.
== Proof ==
Let $R$ be defined by $r = a \cos n \theta$.
The tips of each of the petals of $R$ occur when $\cos n \theta = \pm 1$.
This happens whenever $n \theta \in \set {0, \pi}$.
Thus for $0 \le \theta < 2 \pi$, the tips occur at:
:$\theta \in \set {\dfrac {2 k \pi} {2 n}: k \in \set {0, 1, \ldots, 2 n - 1} }$
Let $k < n$.
Then each of $\dfrac {2 k \pi} {2 n}$ is a distinct value between $0$ and $\pi$.
Hence there are at least $n$ petals, one for each $k \in \set {0, 1, \ldots, n - 1}$.
Now consider $\theta \in \hointr \pi {2 \pi}$.
From Cosine of Angle plus Straight Angle, we have that:
:$\cos \theta = -\map \cos {\theta - \pi}$
But as $\theta$ is on the opposite side of the circle to $\theta - \pi$, the points they describe are exactly the same.
Hence the petals for $\theta \ge \pi$ are in the same place as that for $\theta < \pi$.
So for $\pi \le \theta < 2 \pi$ the pattern will be traced out again.
Hence the result.
== Sources ==
* : $\S 11$: Special Plane Curves: Three-Leaved Rose: $11.16$
* 
* 
* : $\S 9$: Special Plane Curves: Three-Leaved Rose: $9.16.$
* 
* 
Category:Rhodonea Curves
Number of Powers of Cyclic Group Element

== Theorem ==
Let $G$ be a cyclic group of order $n$, generated by $g$.
Let $d \divides n$.
Then the element $g^{n/d}$ has $d$ distinct powers.
== Proof ==
Follows directly from Order of Subgroup of Cyclic Group:
:$\order {\gen {g^{n/d} } } = \dfrac n {\gcd \set {n, n/d} } = d$
Thus from List of Elements in Finite Cyclic Group:
:$\gen {g^{n/d} } = \set {e, g^{n/d}, \paren {g^{n/d} }^2, \ldots, \paren {g^{n/d} }^{d - 1} }$
and the result follows.
== Sources ==
* : Chapter $4$: Subgroups: Proposition $4.14$ (passim)
Category:Cyclic Groups
Number of Primes is Infinite

== Theorem ==
The number of primes is infinite.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
== Proof 6 ==
== Sources ==
* : Chapter $7$: Patterns in Numbers: Euclid
Category:Euclid's Theorem
Number of Primes is Infinite/Proof 1

== Theorem ==
== Proof ==
 there exists a greatest prime number $N$.
Hence, let $S$ denote the finite set of all prime numbers.
Euclid's Theorem states that:
:For any finite set of prime numbers, there exists a prime number not in that set.
Hence there exists $N'$ not in $S$.
If $N' < N$ then $N \in S$.
Hence:
:$N' > N$
which contradicts the assumption the $N$ is the greatest prime number.
The result follows by Proof by Contradiction.
== Sources ==
* 
* 
Category:Euclid's Theorem
Number of Primes is Infinite/Proof 2

== Theorem ==
== Proof ==
Define a topology on the integers $\Z$ by declaring a subset $U \subseteq \Z$ to be an open set it is either:
:the empty set $\O$
or:
:a union of sequences $\map S {a, b}$ such that $a \ne 0$, where:
::$\map S {a, b} = \set {a n + b: n \in \Z} = a \Z + b$
In other words, $U$ is open every $x \in U$ admits some non-zero integer $a$ such that $\map S {a, x} \subseteq U$.
The open set axioms are verified as follows:
;: All unions of open sets are open:
For any set of open sets $U_i$ and $x$ in their union $U$, any of the numbers $a_i$ for which $\map S {a_i, x} \subseteq U_i$ also shows that $\map S {a_i, x} \subseteq U$.
;: The intersection of two (and hence finitely many) open sets is open:
Let $U_1$ and $U_2$ be open sets.
Let $x \in U_1 \cap U_2$ (with integers $a_1$ and $a_2$ establishing membership).
Set $a$ to be the lowest common multiple of $a_1$ and $a_2$.
Then:
:$\map S {a, x} \subseteq \map S {a_i, x} \subseteq U_1 \cap U_2$
;: By definition, $\O$ is open: $\Z$ is just the sequence $\map S {1, 0}$, and so is open as well.
The topology is quite different from the usual Euclidean one, and has two notable properties:
:$(1): \quad$ Since any non-empty open set contains an infinite sequence, no finite set can be open. Put another way, the complement of a finite set cannot be a closed set.
:$(2): \quad$ The basis sets $\map S {a, b}$ are both open and closed: they are open by definition, and we can write $\map S {a, b}$ as the complement of an open set as follows:
:$\ds \map S {a, b} = \Z \setminus \bigcup_{j \mathop = 1}^{a - 1} \map S {a, b + j}$
The only integers that are not integer multiples of prime numbers are $-1$ and $+1$, that is:
:$\ds \Z \setminus \set {-1, +1} = \bigcup_{\text {$p$ prime} } \map S {p, 0}$
By the first property, the set on the cannot be closed.
On the other hand, by the second property, the sets $\map S {p, 0}$ are closed.
So, if there were only finitely many prime numbers, then the set on the would be a finite union of closed sets, and hence closed.
Therefore by Proof by Contradiction, there must be infinitely many prime numbers.
== Also see ==
* Euclid's Theorem and its corollary
* Definition:Furstenberg Topology
== Historical Note ==
Category:Euclid's Theorem
Number of Primes is Infinite/Proof 3

== Theorem ==
== Proof ==
 that there are only $N$ prime numbers.
Let the set of all primes be:
:$\Bbb P = \set {p_1, p_2, \ldots, p_N}$
By the Fundamental Theorem of Arithmetic, every integer $k > 1$ can be expressed in the form:
:$k = {p_1}^{a_1} {p_2}^{a_2} \dotsm {p_N}^{a_N}$
Let $n > 1$ be fixed.
Let $a$ be the largest exponent occurring in the prime decomposition of all positive integers $k \le n$.
Then:
:$\ds \sum_{k \mathop = 1}^n \frac 1 k \le \prod_{j \mathop = 1}^N \paren {\sum_{k \mathop = 0}^a \frac 1 { {p_j}^k} }$
is expressible in the form:
which can be seen by multiplying out the factors on the .
But from Sum of Infinite Geometric Sequence:
:$1 + x + x^2 + \dotsb = \dfrac 1 {1 - x}$
for all $x$ such that $\size x < 1$.
Thus the factors in $(1)$ are less than the numbers:
:$\dfrac 1 {1 - 1 / p_1}, \dfrac 1 {1 - 1 / p_2}, \dotsb, \dfrac 1 {1 - 1 / p_N}$
and so:
:$1 + \dfrac 1 2 + \dfrac 1 3 + \dotsb + \dfrac 1 n < \dfrac {p_1} {p_1 - 1} \dfrac {p_2} {p_2 - 1} \dotsm \dfrac {p_N} {p_N - 1}$
We have chosen $n > 1$ arbitrarily, so this holds for every $n > 1$.
This contradicts the result Harmonic Series is Divergent.
Hence the result, by Proof by Contradiction.
== Historical Note ==
== Sources ==
* : $\S 3$: Appendix $\text A$: Euler
Category:Euclid's Theorem
Number of Primes is Infinite/Proof 4

== Theorem ==
== Proof ==
 there exist only a finite number of primes.
From Sum of Reciprocals of Powers as Euler Product:
:$\ds \sum_{n \mathop \ge 1} \dfrac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$
When $z = 1$ this gives:
:$\ds \sum_{n \mathop \ge 1} \dfrac 1 n = \prod_p \frac 1 {1 - 1 / p}$
As there exist only a finite number of primes, the is also finite.
But from Harmonic Series is Divergent, the diverges to infinity.
The result follows by Proof by Contradiction.
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.19$: The Series $\sum 1/ p_n$ of the Reciprocals of the Primes
Category:Euclid's Theorem
Number of Primes is Infinite/Proof 5

== Theorem ==
== Proof ==
 there exist $n$ prime numbers.
Consider the Fermat number $F_n$.
From Goldbach's Theorem, $F_n$ is coprime to each of $F_0$ to $F_{n - 1}$.
Therefore there must be a prime number which is a divisor of $F_n$ which is not a divisor of any of $F_0$ to $F_n$.
But, again from Goldbach's Theorem, each of $F_0$ to $F_{n - 1}$ is coprime to every other Fermat number.
So all the $n$ prime numbers must have been exhausted being used as prime factors of $F_0$ to $F_{n - 1}$.
So there must be more prime numbers than $n$.
The result follows by Proof by Contradiction.
== Sources ==
* : $257$
Category:Euclid's Theorem