nyuuzyou/wikis huggingface dataset:
Division Theorem for Polynomial Forms over Field

== Theorem ==
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental over $F$.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Let $d$ be an element of $F \sqbrk X$ of degree $n \ge 1$.
Then $\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ such that either:
:$(1): \quad r = 0_F$
or:
:$(2): \quad r \ne 0_F$ and $r$ has degree that is less than $n$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Definition:Polynomial Division
Category:Field Theory
Category:Polynomial Theory
Category:Division Theorem for Polynomial Forms over Field
Category:Division Theorem
Division Theorem for Polynomial Forms over Field/Proof 1

== Theorem ==
== Proof ==
From the equation $0_F = 0_F \circ d + 0_F$, the theorem is true for the trivial case $f = 0_F$.
So, if there is a counterexample to be found, it will have a degree.
 there exists at least one counterexample.
By a version of the Well-Ordering Principle, we can assign a number $m$ to the lowest degree possessed by any counterexample.
So, let $f$ denote a counterexample which has that minimum degree $m$.
If $m < n$, the equation $f = 0_F \circ d + f$ would show that $f$ was not a counterexample.
Therefore $m \ge n$.
Suppose $d \divides f$ in $F \sqbrk X$.
Then:
:$\exists q \in F \sqbrk X: f = q \circ d + 0_F$
and $f$ would not be a counterexample.
So $d \nmid f$ in $F \sqbrk X$.
So, suppose that:
We can create the polynomial $\paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d$ which has the same degree and leading coefficient as $f$.
Thus $f_1 = f - \paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d$ is a polynomial of degree less than $m$.
Since $d \nmid f$, $f_1$ is a non-zero polynomial.
There is no counterexample of degree less than $m$.
Therefore:
:$f_1 = q_1 \circ d + r$
for some $q_1, r \in F \sqbrk X$, where either:
:$r = 0_F$
or:
:$r$ is non-zero with degree strictly less than $n$.
Hence:
Thus $f$ is not a counterexample.
From this contradiction follows the result.
== Sources ==
* : $\S 65.1$ Some properties of $F \sqbrk X$, where $F$ is a field
Category:Division Theorem for Polynomial Forms over Field
Division Theorem for Polynomial Forms over Field/Proof 2

== Theorem ==
== Proof ==
Suppose $\map \deg f < \map \deg d$.
Then we take $\map q X = 0$ and $\map r X = \map a X$ and the result holds.
Otherwise, $\map \deg f \ge \map \deg d$.
Let:
:$\map f X = a_0 + a_1 X + a_2 x^2 + \cdots + a_m X^m$
:$\map d X = b_0 + b_1 X + b_2 x^2 + \cdots + b_n X^n$
We can subtract from $f$ a suitable multiple of $d$ so as to eliminate the highest term in $f$:
:$\map f X - \map d X \cdot \dfrac {a_m} {b_n} x^{m - n} = \map p X$
where $\map p X$ is some polynomial whose degree is less than that of $f$.
If $\map p X$ still has degree higher than that of $d$, we do the same thing again.
Eventually we reach:
:$\map f X - \map d X \cdot \paren {\dfrac {a_m} {b_n} x^{m - n} + \dotsb} = \map r X$
where either $r = 0_F$ or $r$ has degree that is less than $n$.
This approach can be formalised using the Principle of Complete Induction.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 27$. Euclidean Rings: Theorem $52$
Category:Division Theorem for Polynomial Forms over Field
Division Theorem for Polynomial Forms over Field/Proof 3

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < n$
=== Basis for the Induction ===
$\map P 0$ is the statement that $q$ and $r$ exist when $f = 0$.
This is shown trivially to be true by taking $q = r = 0$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < k$
Then we need to show:
:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < k + 1$
=== Induction Step ===
This is our induction step:
Let $f$ be such that $\map \deg f = n$.
Let:
:$f = a_0 + a_1 \circ x + a_2 \circ x^2 + \cdots + a_n \circ x^n$ where $a_n \ne 0$
Let:
:$d = b_0 + b_1 \circ x + b_2 \circ x^2 + \cdots + b_j \circ x^j$ where $b_j \ne 0$
If $n < l$ then take $q = 0, r = f$.
If $n \ge l$, consider:
:$c := f - a_n b_j^{-1} x^{n - j} d$
This has been carefully arranged so that the coefficient of $x^n$ in $c$ is zero.
Thus $\map \deg c < n$.
Therefore, by the induction hypothesis:
:$c = d q_0 + r$
where $\map \deg r < \map \deg d$.
Therefore:
Thus the existence of $q$ and $r$ have been established.
As for uniqueness, assume:
:$d q + r = d q' + r'$
with $\map \deg r < \map \deg d, \map \deg {r'} < \map \deg d$
Then:
:$d \paren {q - q'} = r' - r$
By Degree of Sum of Polynomials:
:$\map \deg {r' - r} \le \max \set {\map \deg {r'}, \map \deg r} < \map \deg d$
and by Degree of Product of Polynomials over Integral Domain:
:$\map \deg {d \paren {q - q'} } = \map \deg d + \map \deg {q - q'}$
That is:
:$\map \deg d < \map \deg d + \map \deg {q - q'}$
and the only way for that to happen is for:
:$\map \deg {q - q'} = -\infty$
that is, for $q - q'$ to be the null polynomial.
That is, $q - q' = 0_F$ and by a similar argument $r' - r = 0_F$, demonstrating the uniqueness of $q$ and $r$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ such that either:
::$(1): \quad r = 0_F$
:or:
::$(2): \quad r \ne 0_F$ and $r$ has degree that is less than $n$.
== Sources ==
* : $\S 3.2$: Polynomial rings: Lemma $3.8$ 
Category:Division Theorem for Polynomial Forms over Field
Division by Zero

== Mistake ==
Let $x$ and $y$ be numbers such that $y = 0$.
The quantity $\dfrac x y$ is undefined.
It is a common mistake to forget this fact when evaluating formulas.
== Also see ==
* L'Hôpital's Rule: while $\dfrac 0 0$ is undefined, the limit of a rational function whose denominator approaches $0$ is not necessarily undefined.
== Historical Note ==
== Sources ==
* : Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers: $(\text P 8)$
* : Chapter $1$: Equations
* : $\S 1$: Real Numbers: $\S 1.3$: Arithmetic
* 
Category:Fallacies and Mistakes
Category:Real Division
Division by Zero/Historical Note

== Historical Note on Division by Zero ==
 stated that:
:positive or negative divided by cipher is a fraction with that for denominator.
This was then referred to as quantity with zero as denominator.
 writes in Ganita Sara Samgraha of c. $850$ CE:
:A number multiplied by zero is zero and that number remains unchanged which is divided by, added to or diminished by zero.
which is of course seriously questionable.
== Sources ==
* : $0$ Zero
* : $0$ Zero
Category:Historical Notes
Division of Circle into 4 Equal Parts by 3 Equal Length Lines

== Problem ==
:Divide a circle into $4$ equal parts by $3$ lines of equal length.
By equal here, we mean of equal area.
== Solution ==
:
Each of the curved lines is a semicircle whose radius is $\dfrac {n r} 4$, where:
:$n$ goes from $1$ to $3$
:$r$ is the radius of the main circle.
== Proof ==
Let the main circle have radius $4 a$, to make the numbers convenient.
Let the area of the main circle be $\AA$.
From Area of Circle:
:$\AA = \pi \paren {4 a}^2 = 16 \pi a^2$
The area $\AA_1$ of one of the outside shapes is:
:half the area of the main circle
:less half the area of the circle whose radius is $3 a$
:plus half the area of the circle whose radius is $a$
That is:
The area $\AA_2$ of one of the inside shapes is:
:half the area of the circle whose radius is $3 a$
:less half the area of the circle whose radius is $2 a$
:plus half the area of the circle whose radius is $2 a$
:less half the area of the circle whose radius is $a$
That is:
== Sources ==
* 
* : Rational Amusements for Winter Evenings: $156$
Category:Circles
Category:Dissections
Category:Division of Circle into 4 Equal Parts by 3 Equal Length Lines
Division of Complex Numbers

== Theorem ==
Let $z_1 := a_1 + i b_1$ and $z_2 := a_2 + i b_2$ be complex numbers such that $z_2 \ne 0$.
The operation of division is performed on $z_1$ by $z_2$ as follows:
:$\dfrac {z_1} {z_2} = \dfrac {a_1 a_2 + b_1 b_2} {a_2^2 + b_2^2} + i \dfrac {a_2 b_1 - a_1 b_2} {a_2^2 + b_2^2}$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 1$. Complex Numbers: $\text {(iv)}$ The definition of division
* : $\S 1.2$. The Algebraic Theory: $(1.13)$
* : $\S 6$: Complex Numbers: Division of Complex Numbers: $6.5$
* : $\S 4$: Complex Numbers: Arithmetic of Complex Numbers: $4.5.$
Category:Division of Complex Numbers
Category:Complex Division
Division of Complex Numbers/Also presented as

== Division of Complex Numbers: Also presented as ==
The operation of complex division on $z_1$ by $z_2$ can also be presented as:
:$\dfrac {z_1} {z_2} = \dfrac {a_1 a_2 + b_1 b_2 + i \paren {a_2 b_1 - a_1 b_2} } { {a_2}^2 + {b_2}^2}$
== Sources ==
* : $\S 1.2$. The Algebraic Theory: $(1.14)$
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Multiplication and Division: $3.7.13$
* : $1$: Complex Numbers: Fundamental Operations with Complex Numbers: $4$. Division
Category:Division of Complex Numbers
Division of Complex Numbers in Exponential Form

== Theorem ==
Let $z_1 := r_1 e^{i \theta_1}$ and $z_2 := r_2 e^{i \theta_2}$ be complex numbers expressed in exponential form.
Then:
:$\dfrac {z_1} {z_2} = \dfrac {r_1} {r_2} e^{i \paren {\theta_1 - \theta_2} }$
== Proof ==
== Sources ==
* : $\S 7$: Operations with Complex Numbers in Polar Form: $7.26$
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $19$
Category:Complex Division
Division of Complex Numbers in Polar Form

== Theorem ==
Let $z_1 := \polar {r_1, \theta_1}$ and $z_2 := \polar {r_2, \theta_2}$ be complex numbers expressed in polar form, such that $z_2 \ne 0$.
Then:
:$\dfrac {z_1} {z_2} = \dfrac {r_1} {r_2} \paren {\map \cos {\theta_1 - \theta_2} + i \map \sin {\theta_1 - \theta_2} }$
or:
:$\dfrac {z_1} {z_2} = \dfrac {r_1} {r_2} \map \cis {\theta_1 - \theta_2}$
== Proof ==
== Examples ==
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 3$. Geometric Representation of Complex Numbers
* : $\S 2$. Geometrical Representations
* : $\S 6$: Complex Numbers: Multiplication and Division of Complex Numbers in Polar Form: $6.8$
* : $1$: Complex Numbers: De Moivre's Theorem
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $19 \ \text{(b)}$
* : $\S 4$: Complex Numbers: Multiplication and Division of Complex Numbers in Polar Form: $4.9.$
Category:Complex Division
Category:Polar Form of Complex Number
Category:Division of Complex Numbers in Polar Form
Division of Complex Numbers in Polar Form/Examples

== Examples of Use of Division of Complex Numbers in Polar Form ==
=== Example: $\dfrac {\paren {2 \cis 15 \degrees}^7} {\paren {4 \cis 45 \degrees}^3}$ ===
=== Example: $\dfrac {\paren {8 \cis 40 \degrees}^3} {\paren {2 \cis 60 \degrees}^4}$ ===
=== Example: $\dfrac {\paren {3 \cis \dfrac \pi 6} \paren {2 \cis \dfrac {-5 \pi} 4} \paren {6 \cis \dfrac {5 \pi} 3} } {\paren {4 \cis \dfrac {2 \pi} 3}^2}$ ===
Category:Examples of Use of Division of Complex Numbers in Polar Form
Division of Complex Numbers in Polar Form/Examples/(2 cis 15)^7 (4 cis 45)^-3

== Example of Use of Division of Complex Numbers in Polar Form ==
:$\dfrac {\paren {2 \cis 15 \degrees}^7} {\paren {4 \cis 45 \degrees}^3} = \sqrt 3 - i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $26 \ \text {(b)}$
Category:Examples of Use of Division of Complex Numbers in Polar Form
Division of Complex Numbers in Polar Form/Examples/(3 cis pi by 6) (2 cis -5 pi by 4) (6 cis 5 pi by 3) (4 cis 2 pi by 3)^-2

== Example of Use of Division of Complex Numbers in Polar Form ==
:$\dfrac {\paren {3 \cis \dfrac \pi 6} \paren {2 \cis \dfrac {-5 \pi} 4} \paren {6 \cis \dfrac {5 \pi} 3} } {\paren {4 \cis \dfrac {2 \pi} 3}^2} = -\dfrac {9 \sqrt 2} 8 - \dfrac {9 \sqrt 2} 8 i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $89 \ \text{(d)}$
Category:Examples of Use of Division of Complex Numbers in Polar Form
Division of Complex Numbers in Polar Form/Examples/(3 cis pi by 6) (2 cis -5 pi by 4) (6 cis 5 pi by 3) (4 cis 2 pi by 3)^-2/Mistake

== Source Work ==
:Chapter $1$: Complex Numbers
::Supplementary Problems: De Moivre's Theorem: $89 \ \text {(d)}$
== Mistake ==
:Evaluate each of the following:
::... $\dfrac {\paren {3 e^{\pi i / 6} } \paren {2 e^{- 5 \pi i / 4} } \paren {6 e^{5 \pi i / 3} } } {\paren {4 e^{2 \pi i / 3} }^2}$
:Ans. $3 \sqrt 2 / 2 - \paren {3 \sqrt 3 / 2} i$
== Correction ==
The correct solution is:
:$-\dfrac {9 \sqrt 2} 8 - \dfrac {9 \sqrt 2} 8 i$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $89 \ \text{(d)}$
Category:Mistakes/Theory and Problems of Complex Variables
Division of Complex Numbers in Polar Form/Examples/(8 cis 40)^3 (2 cis 60)^-4

== Example of Use of Division of Complex Numbers in Polar Form ==
:$\dfrac {\paren {8 \cis 40 \degrees}^3} {\paren {2 \cis 60 \degrees}^4} = -16 - 16 \sqrt 3 i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $89 \ \text{(c)}$
Category:Examples of Use of Division of Complex Numbers in Polar Form
Division of Fractions

== Theorem ==
Let $a, b, c, d \in \Z$ such that $b c d \ne 0$.
Then:
:$\dfrac a b \div \dfrac c d = \dfrac {a d} {b c}$
== Proof ==
== Examples ==
== Sources ==
* 
* 
Category:Division of Fractions
Category:Fractions
Category:Division
Division of Straight Line into Equal Parts using Rusty Compass

== Theorem ==
Let $AB$ be a line segment.
Using a straightedge and rusty compass, it is possible to divide $AB$ into as many equal parts as required.
== Construction ==
:
Let $n + 1$ divisions be required.
Using Construction of Perpendicular using Rusty Compass, construct straight line at right angles to $AB$ from the endpoints $A$ and $B$, but in opposite directions.
Mark off $n$ points along each of those perpendiculars, equally spaced by the length determined by the opening of your rusty compass.
Let the points from $A$ be $A_1, A_2, \dotsc, A_n$.
Let the points from $B$ be $B_1, B_2, \dotsc, B_n$.
Join the points:
:$A_1$ to $B_n$
:$A_2$ to $B_{n - 1}$
:$\cdots$
:$A_{n - 1}$ to $B_2$
:$A_n$ to $B_1$
The divisions are seen where these lines intersect $AB$.
== Historical Note ==
== Sources ==
* 
* : Abul Wafa (): $44$
Category:Lines
Category:Rusty Compass Constructions
Division on Numbers is Not Associative

== Theorem ==
The operation of division on the numbers is not associative.
That is, in general:
:$a \div \paren {b \div c} \ne \paren {a \div b} \div c$
== Proof ==
By definition of division:
So we see that:
:$a \div \paren {b \div c} = \paren {a \div b} \div c \iff c = 1$
and so in general:
:$a \div \paren {b \div c} \ne \paren {a \div b} \div c$
== Sources ==
* 
* 
Category:Numbers
Category:Division
Category:Examples of Associative Operations‎
Divisions of Numbers in Unit Interval with Numbers in Different Intervals

== Theorem ==
Let $I = \openint 0 1$ be the open unit interval.
Let $a_1, a_2, a_3, \ldots, a_n$ be real numbers chosen in $I$ such that:
:$a_1$ and $a_2$ are in different halves of $I$
:$a_1, a_2$ and $a_3$ are in different thirds of $I$
:$a_1, a_2, a_3$ and $a_4$ are in different quarters of $I$
and so on.
Then $n \le 17$.
That is, for the conditions to be fulfilled, no more than $17$ numbers can be chosen.
== Proof ==
== Sources ==
* 
* : $17$
* : $17$
Category:17
Divisor Count Function/Examples/1

== Example of Use of Divisor Count Function ==
The value of the divisor count function for the integer $1$ is $1$.
== Proof ==
By definition, the divisor count function of an integer $n$ is the number of positive integer divisors of $n$.
There is only one positive integer which is a divisor of $1$, and that is $1$ itself.
Hence the result.
Category:Examples of Divisor Count Function
Divisor Count Function/Examples/105

== Example of Use of Divisor Count Function ==
:$\map {\sigma_0} {105} = 8$
where $\sigma_0$ denotes the divisor count function.
== Proof ==
From Divisor Count Function from Prime Decomposition:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
where:
:$r$ denotes the number of distinct prime factors in the prime decomposition of $n$
:$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$.
We have that:
:$105 = 3 \times 5 \times 7$
Thus:
The divisors of $105$ can be enumerated as:
:$1, 3, 5, 7, 15, 21, 35, 105$
Category:Examples of Divisor Count Function
Category:105
Divisor Count Function/Examples/108

== Example of Use of Divisor Count Function ==
:$\map {\sigma_0} {108} = 12$
where $\sigma_0$ denotes the divisor count function.
== Proof ==
From Divisor Count Function from Prime Decomposition:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
where:
:$r$ denotes the number of distinct prime factors in the prime decomposition of $n$
:$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$.
We have that:
:$108 = 2^2 \times 3^3$
Thus:
The divisors of $108$ can be enumerated as:
:$1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108$
Category:Examples of Divisor Count Function
Category:108
Divisor Count Function/Examples/110

== Example of Use of Divisor Count Function ==
:$\map {\sigma_0} {110} = 8$
where $\sigma_0$ denotes the divisor count function.
== Proof ==
From Divisor Count Function from Prime Decomposition:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
where:
:$r$ denotes the number of distinct prime factors in the prime decomposition of $n$
:$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$.
We have that:
:$110 = 2 \times 5 \times 11$
Thus:
The divisors of $110$ can be enumerated as:
:$1, 2, 5, 10, 11, 22, 55, 110$
Category:Examples of Divisor Count Function
Category:110
Divisor Count Function/Examples/12

== Example of Use of Divisor Count Function ==
:$\map {\sigma_0} {12} = 6$
where $\sigma_0$ denotes the divisor count function.
== Proof ==
From Divisor Count Function from Prime Decomposition:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
where:
:$r$ denotes the number of distinct prime factors in the prime decomposition of $n$
:$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$.
We have that:
:$12 = 2^2 \times 3$
Thus:
The divisors of $12$ can be enumerated as:
:$1, 2, 3, 4, 6, 12$
Category:Examples of Divisor Count Function
Category:12
Divisor Count Function/Examples/120

== Example of Use of Divisor Count Function ==
:$\map {\sigma_0} {120} = 16$
where $\sigma_0$ denotes the divisor count function.
== Proof ==
From Divisor Count Function from Prime Decomposition:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
where:
:$r$ denotes the number of distinct prime factors in the prime decomposition of $n$
:$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$.
We have that:
:$120 = 2^3 \times 3 \times 5$
Thus:
The divisors of $120$ can be enumerated as:
:$1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120$
Category:Examples of Divisor Count Function
Category:120
Divisor Count Function/Examples/3

== Example of Use of Divisor Count Function ==
:$\map {\sigma_0} 3 = 2$
where $\sigma_0$ denotes the divisor count function.
== Proof ==
We have that $3$ is a prime number.
From Divisor Count Function of Prime Number:
:$\map {\sigma_0} 3 = 2$
Category:Examples of Divisor Count Function
Category:3
Divisor Count Function/Examples/60

== Example of Use of Divisor Count Function ==
:$\map {\sigma_0} {60} = 12$
where $\sigma_0$ denotes the divisor count function.
== Proof ==
From Divisor Count Function from Prime Decomposition:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
where:
:$r$ denotes the number of distinct prime factors in the prime decomposition of $n$
:$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$.
We have that:
:$60 = 2^2 \times 3 \times 5$
Thus:
The divisors of $60$ can be enumerated as:
:$1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$
Category:Examples of Divisor Count Function
Category:60
Divisor Count Function from Prime Decomposition

== Theorem ==
Let $n$ be an integer such that $n \ge 2$.
Let the prime decomposition of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
Let $\map {\sigma_0} n$ be the divisor count function of $n$.
Then:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Properties of the Natural Numbers: $\S 24 \beta$
* 
* 
Category:Divisor Count Function
Category:Divisor Count Function from Prime Decomposition
Divisor Count Function from Prime Decomposition/Proof 1

== Theorem ==
== Proof ==
We have:
:$d \divides n \implies \forall i: 1 \le i \le r: d = p_1^{l_1} p_2^{l_2} \ldots p_1^{l_1}, 0 \le l_i \le k_i$
For each $i$, there are $k_i + 1$ choices for $l_i$, making $\paren {k_1 + 1} \paren {k_2 + 1} \cdots \paren {k_r + 1}$ choices in all.
By the Fundamental Theorem of Arithmetic and hence the uniqueness of prime decomposition, each of these choices results in a different number, therefore a distinct divisor.
Category:Divisor Count Function from Prime Decomposition
Divisor Count Function from Prime Decomposition/Proof 2

== Theorem ==
== Proof ==
From Divisor Count Function of Power of Prime we have:
:$\forall j \in \closedint 1 r: \map {\sigma_0} {p_j^{k_j} } = k_j + 1$
The result follows immediately from Divisor Count Function is Multiplicative‎.
Category:Divisor Count Function from Prime Decomposition
Divisor Count Function is Multiplicative

== Theorem ==
The divisor count function:
:$\ds \tau: \Z_{>0} \to \Z_{>0}: \map {\sigma_0} n = \sum_{d \mathop \divides n} 1$
is multiplicative.
== Proof ==
Let $f_1: \Z_{>0} \to \Z_{>0}$ be the constant function:
:$\forall n \in \Z_{>0}: \map {f_1} n = 1$
Thus we have:
:$\ds \map {\sigma_0} n = \sum_{d \mathop \divides n} 1 = \sum_{d \mathop \divides n} \map {f_1} d$
But from Unity Function is Completely Multiplicative, $f_1$ is multiplicative.
The result follows from Sum Over Divisors of Multiplicative Function.
== Sources ==
* 
Category:Multiplicative Functions
Category:Divisor Count Function
Divisor Count Function is Odd Iff Argument is Square

== Theorem ==
Let $\sigma_0: \Z \to \Z$ denote the divisor count function.
Then $\map {\sigma_0} n$ is odd $n$ is square.
== Proof ==
Let $n$ be an integer such that $n \ge 2$.
Let the prime decomposition of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$
Then from Divisor Count Function from Prime Decomposition we have that:
:$\ds \map {\sigma_0} n = \prod_{i \mathop = 1}^r \paren {k_i + 1}$
=== Sufficient Condition ===
Let $\map {\sigma_0} n$ be odd.
Then all factors of $\ds \prod_{i \mathop = 1}^r \paren {k_i + 1}$ are odd (and of course $\ge 3$).
Therefore all factors of $\ds \prod_{i \mathop = 1}^r \paren {k_i}$ are even.
Thus $n = p_1^{2 s_1} p_2^{2 s_2} \ldots p_r^{2 s_r}$ for $r_i = k_i / 2$ for all $i$.
Hence $n = \paren {p_1^{s_1} p_2^{s_2} \ldots p_r^{s_r} }^2$ and therefore is square.
=== Necessary Condition ===
Now suppose $n$ is square.
The above argument reverses, and we see that all factors of $\ds \prod_{i \mathop = 1}^r \paren {k_i + 1}$ are odd.
Hence $\map {\sigma_0} n$ is itself odd.
Category:Divisor Count Function
Category:Square Numbers