nyuuzyou/wikis huggingface dataset:
Quotient of Group by Center Cyclic implies Abelian

== Theorem ==
Let $G$ be a group.
Let $\map Z G$ be the center of $G$.
Let $G / \map Z G$ be the quotient group of $G$ by $\map Z G$.
Let $G / \map Z G$ be cyclic.
Then $G$ is abelian, so $G = \map Z G$.
That is, the group $G / \map Z G$ cannot be a cyclic group which is non-trivial.
== Proof ==
Suppose $G / \map Z G$ is cyclic.
Then by definition:
:$\exists \tau \in G / \map Z G: G / \map Z G = \gen \tau$
Since $\tau$ is a coset by $\map Z G$:
:$\exists t \in G: \tau = t \map Z G$
Thus each coset of $\map Z G$ in $G$ is equal to $\paren {t \map Z G}^i = t^i \map Z G$ for some $i \in \Z$.
Now let $x, y \in G$.
Suppose $x \in t^m \map Z G, y \in t^n \map Z G$.
Then $x = t^m z_1, y = t^n z_2$ for some $z_1, z_2 \in \map Z G$.
Thus:
This holds for all $x, y \in G$, and thus $G$ is abelian.
Thus by definition of abelian group:
:$\map Z G = G$
Therefore Quotient of Group by Itself it follows that $G / \map Z G$ is the trivial group.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \epsilon$
* : $\S 51.1$ The quotient group $G / Z$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.21$
Category:Quotient Groups
Category:Cyclic Groups
Category:Abelian Groups
Category:Centers of Groups
Quotient of Group by Itself

== Theorem ==
Let $G$ be a group.
Let $G / G$ be the quotient group of $G$ by itself.
Then:
:$G / G \cong \set e$
That is, the quotient of a group by itself is isomorphic to the trivial group.
== Proof ==
Let the homomorphism $\phi: G \to \set e$ be defined as:
:$\forall g \in G: \map \phi g = e$
Then:
:$\map \ker \phi = G$
and:
:$\Img \phi = \set e$
By the First Isomorphism Theorem:
:$G / \map \ker \phi \cong \Img \phi$
Hence the result:
:$G / G \cong \set e$
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.7$
Category:Examples of Quotient Groups
Quotient of Homogeneous Functions

== Theorem ==
Let $\map M {x, y}$ and $\map N {x, y}$ be homogeneous functions of the same degree.
Then:
:$\dfrac {\map M {x, y} } {\map N {x, y} }$
is homogeneous of zero degree.
== Proof ==
Let:
:$\map Q {x, y} = \dfrac {\map M {x, y} } {\map N {x, y} }$
where $M$ and $N$ are homogeneous functions of degree $n$.
Let $t \in \R$.
Then:
The result follows from the definition.
== Sources ==
* : $\S 2.7$: Homogeneous Equations
Category:Homogeneous Functions
Quotient of Independent Random Variables with Chi-Squared Distribution Divided by Degrees of Freedom has Snedecor's F-Distribution

== Theorem ==
Let $n$ and $m$ be strictly positive integers.
Let $X$ and $Y$ be independent random variables. 
Let $X \sim \chi_n^2$ where $\chi_n^2$ is the chi-squared distribution with $n$ degrees of freedom. 
Let $Y \sim \chi_m^2$ where $\chi_m^2$ is the chi-squared distribution with $m$ degrees of freedom.
Then: 
:$\dfrac {X / n} {Y / m} \sim F_{n, m}$
where $F_{n, m}$ is Snedecor's $F$-distribution with $\tuple {n, m}$ degrees of freedom.
 
== Proof ==
Category:Chi-Squared Distribution
Category:Snedecor's F-Distribution
Quotient of Integers is Primitive Recursive

== Theorem ==
Let $\operatorname{quot}_\Z : \N^2 \to \N$ be defined as:
:$\map {\operatorname{quot}_\Z} {a_\Z, b_\Z} = \begin{cases}
q_\Z & : b \ne 0 \\
0 & : b = 0
\end{cases}$
where:
:$k_\Z$ denotes the code number for the integer $k$
:$q$ is the quotient of $a$ on division by $b$
Then $\operatorname{quot}_\Z$ is primitive recursive.
== Proof ==
By definition:
:$a = q b + r$
where $0 \le r < \size b$.
We will first show that $\operatorname{quot}_{\Z,\N} : \N^2 \to \N$ defined as:
:$\map {\operatorname{quot}_{\Z,\N}} {a_\Z, b} = \begin{cases}
q_\Z & : b \ne 0 \\
0 & : b = 0
\end{cases}$
with the difference being that $b \in \N$ instead of $\Z$.
If $a \ge 0$, we have $\operatorname{quot} : \N^2 \to \N$ from Quotient is Primitive Recursive.
If $a < 0$, we have:
As $0 \le r < b$, we have $0 \le b - r - 1 < b$.
Thus, $-\paren {q + 1}$ is the quotient of $-\paren {a + 1}$ on division by $b$, by definition.
Additionally, since $a < 0$, we have $-\paren {a + 1} = -a - 1 \ge 0$.
Therefore, we can use the Quotient is Primitive Recursive form again.
If $q' = - \paren {q + 1}$, then we have:
:$q = -q' + -1$
Putting it all together:
:$\map {\operatorname{quot}_{\Z,\N}} {a_\Z, b} = \begin{cases}
\paren {\map {\operatorname{quot}} {\size a, b}}_\Z & : b \ne 0 \land \neg \paren {a < 0} \\
\paren {-1 + -\map {\operatorname{quot}} {\map {\operatorname{pred}} {\size a}, b}}_\Z & : b \ne 0 \land a < 0 \\
0 & : b = 0
\end{cases}$
which is primitive recursive by:
* Definition by Cases is Primitive Recursive
* Quotient is Primitive Recursive
* Predecessor Function is Primitive Recursive
* Addition of Integers is Primitive Recursive
* Absolute Value of Integer is Primitive Recursive
* Code Number for Non-Negative Integer is Primitive Recursive
* Code Number for Non-Positive Integer is Primitive Recursive
* Constant Function is Primitive Recursive
* Set Operations on Primitive Recursive Relations
* Ordering Relations are Primitive Recursive
* Equality Relation is Primitive Recursive
* Set of Strictly Negative Integers is Primitive Recursive
Finally, we can define $\operatorname{quot}_\Z$.
Observe that:
:$a = q b + r \iff a = \paren {-q} \paren {-b} + r$
Thus:
:$q$ is the quotient of $a$ on division by $b$ $-q$ is the quotient of $a$ on division by $-b$
Therefore, we can reduce to the case above:
:$\map {\operatorname{quot}_\Z} {a_\Z, b_\Z} = \map {\sgn_\Z} b \times_\Z \map {\operatorname{quot}_{\Z,\N}} {a_\Z, \size b}$
which is primitive recursive by:
* Multiplication of Integers is Primitive Recursive
* Signum Function on Integers is Primitive Recursive
* Absolute Value of Integer is Primitive Recursive
Category:Primitive Recursive Functions
Quotient of Modulo Operation with Modulus

== Theorem ==
Let $x, y \in \R$ be real numbers.
Let $x \bmod y$ denote the modulo operation:
:$x \bmod y := \begin{cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end{cases}$
where $\floor {\dfrac x y}$ denotes the floor of $\dfrac x y$.
Let $y \ne 0$.
Then:
:$0 \le \dfrac x y - \floor {\dfrac x y} = \dfrac {x \bmod y} y < 1$
== Proof ==
From Real Number minus Floor:
:$\dfrac x y - \floor {\dfrac x y} \in \hointr 0 1$
Thus by definition of half-open real interval:
:$0 \le \dfrac x y - \floor {\dfrac x y} < 1$
Then:
Hence the result.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: $(2)$
Category:Modulo Arithmetic
Quotient of Normal Distributions has Cauchy Distribution

== Theorem ==
Let $X$ and $Y$ be independent continuous random variables each with a normal distribution with zero expectation:
Let $U$ be the continuous random variable defined as:
:$U = \dfrac X Y$
Then $U$ has the Cauchy distribution:
:$U \sim \Cauchy 0 \lambda$
where:
:$\lambda = \dfrac {\sigma_x} {\sigma_y}$
=== Corollary ===
== Proof ==
== Sources ==
* 
Category:Quotient of Normal Distributions has Cauchy Distribution
Category:Cauchy Distribution
Category:Normal Distribution
Quotient of Normal Distributions has Cauchy Distribution/Corollary

== Theorem ==
Let $X$ and $Y$ be independent continuous random variables each with a normal distribution with:
:zero expectation
:the same variance $\sigma$:
Let $U$ be the continuous random variable defined as:
:$U = \dfrac X Y$
Then $U$ has the Cauchy distribution:
:$U \sim \Cauchy 0 1$
and so does $\dfrac 1 U$:
:$\dfrac 1 U \sim \Cauchy 0 1$
== Proof ==
From Quotient of Normal Distributions has Cauchy Distribution:
:$U \sim \Cauchy 0 \lambda$
where:
:$\lambda = \dfrac {\sigma_x} {\sigma_y}$
and such that:
:$\sigma_x = \sigma_y = \sigma$
Hence $\lambda = 1$ and the result follows.
Similarly we have:
:$\dfrac 1 U = \dfrac Y X$
and again the result follows.
== Sources ==
* 
Category:Quotient of Normal Distributions has Cauchy Distribution
Quotient of Quotients of Real Numbers

== Theorem ==
:$\forall x \in \R, y, w, z \in \R_{\ne 0}: \dfrac {x / y} {w / z} = \dfrac {x \times z} {y \times w}$
== Proof ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(r)}$
Category:Real Division
Quotient of Rationally Expressible Numbers is Rational

== Theorem ==
== Proof ==
:
Let the rational area $AC$ be applied to the rational straight line $AB$ such that $BC$ is its breadth.
Let the square $AD$ be described on $AB$.
From , $AD$ is rational.
But $AC$ is also rational.
Therefore $DA$ is commensurable in length with $AC$.
From Areas of Triangles and Parallelograms Proportional to Base:
:$BD : BC = DA : AC$
Therefore from Commensurability of Elements of Proportional Magnitudes $DB$ is also commensurable in length with $BC$.
As $DB = BA$ it follows that $AB$ is also commensurable in length with $BC$.
But $AB$ is rational.
Therefore $BC$ is rational and commensurable in length with $AB$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Quotient of Rationals by Integers is Injective

== Theorem ==
Let $\struct {\Q, +}$ be the abelian group of rational numbers.
Let $\struct {\Z, +}$ be the abelian group of integers, considered as a subgroup of $\struct {\Q, +}$.
Then the quotient group $\Q / \Z$ is an injective object in the category of abelian groups.
== Proof ==
By definition, $\struct {\Q, +, \times}$ is the field of quotients of the ring of integers $\struct {\Z, +, \times}$.
By Field of Quotients is Divisible Module‎ $\Q$ is a divisible $\Z$-module.
By Quotient of Divisible Module is Divisible $\Q / \Z$ is a divisible $\Z$-module.
By Ring of Integers is Principal Ideal Domain $\struct {\Z, +, \times}$ is a principal ideal domain.
By Injective Module over Principal Ideal Domain $\Q / \Z$ is an injective $\Z$-module.
Since the category of abelian groups is equivalent to the category of $\Z$-modules, $\Q / \Z$ is an injective object in the category of abelian groups.
Category:Homological Algebra
Category:Abelian Groups
Quotient of Ring of Polynomials in Ring Element on Integral Domain by that Polynomial is that Domain

== Theorem ==
Let $\struct {D, +, \times}$ be an integral domain.
Let $X \in R$ be transcencental over $D$.
Let $D \sqbrk X$ be the ring of polynomials in $X$ over $D$.
Let $D \sqbrk X / \ideal X$ denote the quotient ring of $D \sqbrk X$ by the ideal of $D$ generated by $X$.
Then:
:$D \sqbrk X / \ideal X \cong D$
== Proof ==
Let $n \in \Z_{> 0}$ be arbitrary.
Let $P = a_n X^n + a_{n - 1} X^{n - 1} + \dotsb + a_1 X + a_0$ be a polynomial over $D$ in $X$.
Consider the mapping $\phi: D \sqbrk X \to D$ defined as:
:$\forall P \in D \sqbrk X: \map \phi P = a_0$
Let:
:$P_1 = a_m X^m + a_{m - 1} X^{m - 1} + \dotsb + a_1 X + a_0$
:$P_2 = b_m X^m + b_{m - 1} X^{m - 1} + \dotsb + b_1 X + b_0$
We have that:
and:
Thus it is seen that $\phi$ is a homomorphism.
The First Ring Isomorphism Theorem can then be used.
== Sources ==
* : Chapter $9$: Rings: Exercise $20$
Category:Polynomial Rings
Category:Integral Domains
Quotient of Sylow P-Subgroup

== Theorem ==
Let $P$ be a Sylow $p$-subgroup of a finite group $G$.
Let $N$ be a normal subgroup of $G$.
Then $P N / N$ is a Sylow $p$-subgroup of $G / N$.
== Proof ==
We have that $P \le G$ and $ N \lhd G$.
So by the Second Isomorphism Theorem for Groups:
:$P N / N \cong P / \paren {P \cap N}$
We have that:
:$P N / N = \set {p N : p \in P}$
and so every element of $P N / N$ has order a power of $p$.
Hence $P N / N$ is a $p$-subgroup of $G / N$.
From Intersection of Normal Subgroup with Sylow P-Subgroup, we have that:
:$p \nmid \index G {P N}$
So $P N / N$ is a Sylow $p$-subgroup of $G / N$.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Proposition $11.14 \ \text{(ii)}$
Category:Quotient Groups
Category:Sylow p-Subgroups
Quotient of Symmetric Group by Alternating Group is Parity Group

== Theorem ==
Let $n \ge 2$ be a natural number.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $A_n$ be the alternating group on $n$ letters.
Then the quotient of $S_n$ by $A_n$ is the parity group $C_2$.
== Proof ==
From Alternating Group is Normal Subgroup of Symmetric Group, $A_n$ is a normal subgroup of $S_n$ whose index is $2$.
By definition of index, $\dfrac {S_n} {A_n}$ is a group of order $2$.
The result follows by definition of the parity group.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 81$
Category:Symmetric Groups
Category:Alternating Groups
Category:Parity Group
Quotient of Transformation Group acts Effectively

== Theorem ==
Let $G$ be a transformation group (which may or may not be effective) acting on $X$.
Then the quotient group $G / G_0$, where $G_0$ is the kernel, does act effectively on $X$.
== Proof ==
Let $g G_0 \in G / G_0$.
Let $x \in X$.
Let $G / G_0$ act on $X$ by the rule:
:$g G_0 * x := g x$
Then by definition of effective transformation group:
:$G / G_0$ acts effectively on $X$ :
::$g G_0 * x = x \implies g G_0 = G_0$
Suppose $g G_0 * x = g x = x$.
We want to show $g G_0 = G_0$.
By definition of kernel of group action of $G$ on $X$:
:$g * x = gx = x \iff g \in G_0$
So $gG_0 = G_0$ by closure of subgroups.
Hence the result.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 53 \gamma$
Category:Group Actions
Category:Quotient Groups
Quotients of 3 Unequal Numbers are Unequal

== Theorem ==
Let $x, y, z \in \R_{\ne 0}$ be non-zero real numbers which are not all equal.
Then $\dfrac x y, \dfrac y z, \dfrac z x$ are also not all equal.
== Proof ==
 $\dfrac x y = \dfrac y z = \dfrac z x$.
This contradicts the assertion that $x, y, z$ are all unequal.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $3$
Category:Real Division
Quotients of Homeomorphic Spaces are Homeomorphic

== Theorem ==
Let $X, Y$ be topological spaces.
Let $\phi : X \to Y$ be a homeomorphism from $X$ to $Y$.
Let:
:$\RR_X \subseteq X \times X$
:$\RR_Y \subseteq Y \times Y$
be equivalence relations on $X$ and $Y$, respectively.
Suppose that, for all $x, x' \in X$:
:$\map {\RR_X} {x, x'} \iff \map {\RR_Y} {\map \phi x, \map \phi {x'}}$
Then:
:$X / \RR_X \sim Y / \RR_Y$
where $\sim$ denotes homeomorphic spaces.
== Proof ==
Let the mapping $\psi : X / \RR_X \to Y / \RR_Y$ be defined as:
:$\map \psi {\eqclass x {\RR_X}} = \eqclass {\map \phi x} {\RR_Y}$
In order for $\psi$ to be well-defined, the image needs to be independent of the choice of representative $x$.
But, for $x, x' \in X$ such that $\map {\RR_X} {x, x'}$, we have:
:$\map {\RR_Y} {\map \phi x, \map \phi {x'}}$
by hypothesis.
It follows that:
:$\eqclass {\map \phi x} {\RR_Y} = \eqclass {\map \phi {x'}} {\RR_Y}$
The inverse mapping $\psi^{-1} : Y / \RR_Y \to X / \RR_X$ can be defined as:
:$\map {\psi^{-1}} {\eqclass y {\RR_Y}} = \eqclass {\map {\phi^{-1}} y} {\RR_X}$
For the same reasons as before, $\psi^{-1}$ is well-defined.
Additionally, it is the inverse of $\psi$, since:
It remains to show that $\psi$ and $\psi^{-1}$ are continuous.
Consider an arbitrary open set $U \subseteq Y / \RR_Y$.
By definition of quotient topology, we have that:
:$\map {q_{\RR_Y}^{-1}} U$
is open in $Y$, where $q_{\RR_Y}$ is the quotient mapping induced by $\RR_Y$.
Then, since $\phi$ is continuous:
:$\map {\phi^{-1}} {\map {q_{\RR_Y}^{-1}} U}$
is open in $X$.
If we can show that:
:$\map {\psi^{-1}} U = \map {\phi^{-1}} {\map {q_{\RR_Y}^{-1}} U}$
then, since $U$ was arbitrary, $\psi$ would be continuous.
We have:
Therefore, by the above remarks, $\psi$ is a continuous mapping.
By a precisely symmetric argument, it can be shown that $\psi^{-1}$ is likewise continuous.
Therefore, since $\psi$:
* is a bijection
* is continuous
* has a continuous inverse
it is a homeomorphism by definition.
Category:Homeomorphisms (Topological Spaces)
Category:Quotient Spaces
Rabbit Problem

== Classic Problem ==
Let there be a pair of rabbits.
For the purpose of this exercise, the following assumptions are made.
Rabbits take one month to reach an age at which they may give birth.
Every month thereafter, a breeding pair of rabbits produce one further pair of rabbits.
Rabbits never die and never stop being able to produce further rabbits.
How many pairs of rabbits will be produced from this single pair of rabbits after one year?
== Solution ==
== Also presented as ==
Some sources make the additional (often unspecified) assumption that at the start of the year the rabbits are newborn (that is, are not ready to breed).
Thus they only have their first pair of offspring in the $2$nd month.
In this case, it turns out that $R \left({n}\right) = F_{n + 1}$
Hence at the end of the year there are $233$ pairs of rabbits rather than $377$.
== Also see ==
* Definition:Fibonacci Numbers
== Historical Note ==
== Sources ==
* 
* : Liber Abaci: $88$
* : $\S 1.2.8$: Fibonacci Numbers
* 
* 
Category:Rabbit Problem
Category:Fibonacci Numbers
Category:Liber Abaci
Category:Classic Problems
Rabbit Problem/Historical Note

== Historical Note on The Rabbit Problem ==
The Rabbit Problem was devised by and presented in his of $1202$.
It is perhaps necessary to point out that this question is not to be considered as a practical application of mathematics to the science of zoology.
As himself wrote:
:It is possible to do [the addition] in this order for an infinite number of months.
Neither is it to be confused with the ongoing endemic ecological problem of .
== Sources ==
* : Solutions: $92$. -- A Cow's Progeny
* : Answers: $175$. A Cow's Progeny
* : $\S 1.2.8$: Fibonacci Numbers
Category:Historical Notes
Rabbit Problem/Solution

== Solution to The Rabbit Problem ==
== Solution ==
At the end of the first month, the initial pair will produce $1$ new pair of rabbits, making a total of $2$ pairs.
At the end of the second month, the initial pair will produce $1$ more new pair, making a total of $3$.
At the end of the third month:
:the initial pair will produce $1$ more new pair
:the second pair will produce $1$ pair, as they are now able to breed.
:This results in $5$ pairs.
At the end of month $n$, the total count of rabbit pairs is:
:the count of the rabbits which were alive at the end of the previous month (all are still alive)
plus:
:the count of the rabbits which were alive at the end of the month before that (as all those are of an age to breed).
Let $R \left({n}\right)$ be the number of pairs of rabbits at the end of month $n$.
Thus, by the above reasoning:
:$R \left({0}\right) = 1$
:$R \left({1}\right) = 2$
:$R \left({2}\right) = 3$
:$R \left({n}\right) = R \left({n - 1}\right) + R \left({n - 2}\right)$
where $R \left({0}\right)$ is understood as being the number of pairs of rabbits at the end of month $0$, that is, the beginning of month $1$.
Thus it is seen that:
:$R \left({n}\right) = F_{n + 2}$
where $F_{n + 2}$ denotes the $n + 2$th Fibonacci number.
Hence, at the end of $12$ months, there will be $F_{14} = 377$ rabbits.
== Historical Note ==
== Sources ==
* 
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $1$
Category:Rabbit Problem
Radial Distance Function in Normal Neighborhood

== Theorem ==
Let $\struct {M, g}$ be an $n$-dimensional Riemannian manifold.
Let $U_p$ be the normal neighborhood of $p \in M$.
Let $r : U_p \to \R$ be the radial distance function.
Then $r$ is well-defined in $U_p$ independently of the choice of normal coordinates.
Furthermore, $r$ is smooth on $U_p \setminus \set p$, where $\setminus$ denotes the set difference, and $r^2$ is smooth on $U_p$.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Geodesics Are Locally Minimizing
Category:Riemannian Manifolds
Radial Geodesic Connecting Two Points in Geodesic Ball is Unique Minimizing Curve

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $U = \map {\exp_p} {\map {B_\epsilon} 0 }$ be a geodesic ball centered at $p \in M$.
Suppose $q \in U$.
Suppose $\gamma$ is a radial geodesic from $p$ to $q$.
Then, up to reparametrization, $\gamma$ is the unique minimizing curve from $p$ to $q$.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Geodesics Are Locally Minimizing
Category:Riemannian Manifolds
Radial Vector Field in Normal Neighborhood

== Theorem ==
Let $\struct {M, g}$ be an $n$-dimensional Riemannian manifold.
Let $U_p$ be the normal neighborhood of $p \in M$.
Let $\partial_r$ be the radial vector field on $U_p \setminus \set p$, where $\setminus$ denotes the set difference.
Then $\partial_r$ is well-defined in $U_p \setminus \set p$ independently of the choice of normal coordinates.
Furthermore, $\partial_r$ is smooth on $U_p \setminus \set p$.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Geodesics Are Locally Minimizing
Category:Riemannian Manifolds
Radical Ideal iff Quotient Ring is Reduced

== Theorem ==
Let $\left({R, +, \circ}\right)$ be a commutative ring with unity.
Let $J$ be an ideal of $R$.
Then $J$ is a radical ideal the quotient ring $R / J$ is a reduced ring.
== Proof ==
Since $J \subset R$, it follows from:
:Quotient Ring of Commutative Ring is Commutative
and:
: Quotient Ring of Ring with Unity is Ring with Unity
that $R / J$ is a commutative ring with unity.
Let $0_{R / J}$ be the zero of $R / J$.
=== Sufficient Condition ===
Let $J$ be radical.
We need to show that:
:if $x + J \in R / J$ such that $\left({x + J}\right)^n = 0_{R / J}$ for some positive integer $n$
:then $x + J = 0_{R / J}$.
Let $\left({x + J}\right)^n = 0_{R / J}$.
Then:
:$x^n + J = 0_{R / J}$
and therefore:
:$x^n \in J$
Because $J$ is radical it follows that:
:$x \in J$
That is:
:$x + J = 0_{R / J}$
Therefore $R / J$ is reduced.
=== Necessary Condition ===
Let $R / J$ be reduced.
Let $x \in R$ be such that $x^n \in J$.
Then:
:$0_{R / J} = J = x^n + J = \left({x + J}\right)^n$
Because $A / J$ is reduced, this implies that:
:$x + J = 0_{R / J}$
That is:
:$x \in J$
This shows that $J$ is radical.
Category:Quotient Rings
Radical of Ideal Preserves Inclusion

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $\mathfrak a \subseteq \mathfrak b \subseteq A$ be an ideals.
Then we have an inclusion of their radicals:
:$\operatorname{Rad} \left({\mathfrak a}\right) \subseteq \operatorname{Rad} \left({\mathfrak b}\right)$
== Proof ==
Let $x \in \operatorname{Rad} \left({\mathfrak a}\right)$.
We show that $x \in \operatorname{Rad} \left({\mathfrak b}\right)$.
By definition of radical, there exists $n \in \N$ such that the power $x^n \in \mathfrak a$.
Because $\mathfrak a \subseteq \mathfrak b$, also $x^n \in \mathfrak b$.
Thus $x \in \operatorname{Rad} \left({\mathfrak b}\right)$.
Category:Radical of Ideals
Radical of Integer/Sequence

== Sequence of Radicals of Integers ==
The sequence of radicals of the integers beings:
|-
|align="right" | $1$
|align="right" | $1$
|align="right" | $1$
|-
|align="right" | $2$
|align="right" | $2$
|align="right" | $2$
|-
|align="right" | $3$
|align="right" | $3$
|align="right" | $3$
|-
|align="right" | $4$
|align="right" | $2^2$
|align="right" | $2$
|-
|align="right" | $5$
|align="right" | $5$
|align="right" | $5$
|-
|align="right" | $6$
|align="right" | $2 \times 3$
|align="right" | $6$
|-
|align="right" | $7$
|align="right" | $7$
|align="right" | $7$
|-
|align="right" | $8$
|align="right" | $2^3$
|align="right" | $2$
|-
|align="right" | $9$
|align="right" | $3^2$
|align="right" | $3$
|-
|align="right" | $10$
|align="right" | $2 \times 5$
|align="right" | $10$
|-
|align="right" | $11$
|align="right" | $11$
|align="right" | $11$
|-
|align="right" | $12$
|align="right" | $2^2 \times 3$
|align="right" | $6$
|-
|align="right" | $13$
|align="right" | $13$
|align="right" | $13$
|-
|align="right" | $14$
|align="right" | $2 \times 7$
|align="right" | $14$
|-
|align="right" | $15$
|align="right" | $3 \times 5$
|align="right" | $15$
|-
|align="right" | $16$
|align="right" | $2^4$
|align="right" | $2$
Category:Examples of Radicals of Integers