nyuuzyou/wikis huggingface dataset:
Polynomial Time/Examples/Traveling Salesman Problem

== Example of Polynomial Time ==
Consider the traveling salesman problem.
This can be expressed:
:Is there a route shorter in length than some number $C$?
An algorithm which, given a particular randomly-selected route, determines whether it is shorter than $C$ would run in polynomial time.
== Sources ==
* 
* 
Category:Examples of Polynomial Time
Category:Traveling Salesman Problem
Polynomial X^2 + 1 is Irreducible in Ring of Real Polynomials

== Theorem ==
Let $\R \sqbrk X$ be the ring of polynomials in $X$ over the real numbers $\R$.
Then the polynomial $X^2 + 1$ is an irreducible element of $\R \sqbrk X$.
== Proof ==
 $x^2 + 1$ has a non-trivial factorization in $\R \sqbrk X$.
Then:
:$\exists \alpha, \beta \in \R: \paren {X - \alpha} \paren {X - \beta}$
and from the Polynomial Factor Theorem:
:$\alpha^2 + 1 = 0$
But that means:
:$\alpha^2 = -1$
and such an $\alpha$ does not exist in $\R$.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 29$. Irreducible elements: Example $58$
* : Chapter $9$: Rings: Exercise $24$
Category:Irreducible Elements of Rings
Category:Polynomial Theory
Polynomial has Integer Coefficients iff Content is Integer

== Theorem ==
Let $f$ be a polynomial with rational coefficients.
Let $\cont f$ be the content of $f$.
$f$ has integer coefficients $\cont f$ is an integer.
== Proof ==
If $f \in \Z \sqbrk X$ then $\cont f \in \Z$ by definition of content.
Conversely, suppose that:
:$f = a_d X^d + \cdots + a_1 X + a_0 \notin \Z \sqbrk X$
Let $m = \min \set {n \in \N : n f \in \Z \sqbrk X}$.
Then, by definition of content:
:$\cont f = \dfrac 1 m \gcd \set {m a_d, \ldots, m a_0}$
So $\cont f \in \Z$ would mean that this GCD is a multiple of $m$.
This, however, means that for each $i$, $\dfrac {m a_i} m = a_i$ is an integer, which contradicts our assumption that $f \notin \Z \sqbrk X$.
Category:Content of Polynomial
Polynomial is Linear Combination of Monomials

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $R \sqbrk X$ be a polynomial ring over $R$ in the variable $X$.
Let $P \in R \sqbrk X$.
Then $P$ is a linear combination of the monomials of $R \sqbrk X$, with coefficients in $R$.
== Proof using specific model ==
== Proof using universal property ==
=== Outline of Proof ===
We show that the subset $S$ of linear combinations of monomials forms a subring.
We use the universal property to exhibit a ring homomorphism $R \sqbrk X \to S$.
We use uniqueness to show that it is necessarily a bijection.
=== Proof ===
Let $S \subset R \sqbrk X$ be the subset of all elements that are linear combinations of monomials.
Let $\iota: S \to R \sqbrk X$ denote the inclusion mapping.
Suppose for the moment that $S$ is a commutative ring with unity.
Then by Universal Property of Polynomial Ring, there exists a ring homomorphism $g: R \sqbrk X \to S$ with $\map g X = X$.
By Inclusion Mapping on Subring is Homomorphism, $\iota: S \to R \sqbrk X$ is a ring homomorphism.
By Composition of Ring Homomorphisms is Ring Homomorphism, $\iota \circ g: R \sqbrk X \to R \sqbrk X$ is a ring homomorphism.
By construction, $\map {\paren {\iota \circ g} } X = X$.
By Universal Property of Polynomial Ring, there exists a unique ring homomorphism $h : R \sqbrk X \to R \sqbrk X$ with $\map h X = X$.
We have that $\iota \circ g$ is such a ring homomorphism.
By Identity Mapping is Ring Automorphism, the identity mapping $I$ on $R \sqbrk X$ is one too.
By uniqueness, $\iota \circ g = I$.
By Identity Mapping is Surjection and Surjection if Composite is Surjection, $\iota$ is a surjection.
By Inclusion Mapping is Surjection iff Identity, $S = R \sqbrk X$.
It remains to show that $S$ is a commutative ring with unity.
== Also see ==
* Monomials form Basis of Polynomial Ring
Category:Polynomial Theory
Category:Monomials
Polynomial is of Exponential Order Epsilon

== Theorem ==
Let $P: \R \to \mathbb F$ be a polynomial, where $\mathbb F \in \set {\R, \C}$.
Then $P$ is of exponential order $\epsilon$ for any $\epsilon > 0$ arbitrarily small in magnitude.
== Proof ==
If $P = 0$, the theorem holds trivially. 
Let $P_n$ be a polynomial of degree $n$, where $n \ge 0$.
The proof proceeds by induction on $n$, where $n$ is the degree of the polynomial.
=== Basis for the Induction ===
Let $P_0$ be of degree zero.
Then $P_0$ is a constant polynomial.
By Constant Function is of Exponential Order Zero, $P_0 \in \EE_0$.
Therefore, by Raising Exponential Order, it is of exponential order $\epsilon$ as well.
This is the basis for the induction.
=== Induction Hypothesis ===
Fix $n \in \N$ with $n \ge 0$.
Assume:
:$P_n \in \EE_\epsilon$
That is:
:$\size {\map {P_n} t} < K e^{\epsilon t}$
for some $K > 0$, for $\epsilon > 0$ arbitrarily small.
This is our induction hypothesis.
=== Induction Step ===
Let $P_{n + 1}$ be of degree $n + 1$.
By the definition of polynomial,
:$P_{n + 1} = P_n + a_{n + 1} x^{n + 1}$
for some polynomials of degree $n$.
$P_n$ is of exponential order $\epsilon$ by the induction hypothesis.
Thus, by:
:Sum of Functions of Exponential Order
:Scalar Multiple of Function of Exponential Order
:Natural Number Power is of Exponential Order Epsilon
We have that $P_{n + 1}$ is of degree $\epsilon$.
The result follows by the Principle of Mathematical Induction.
Category:Exponential Order
Category:Proofs by Induction
Polynomial over Field has Finitely Many Roots

== Theorem ==
Let $F$ be a field.
Let $F \left[{x}\right]$ be the ring of polynomial functions in the indeterminate $x$.
If $p \in F \left[{x}\right]$ be non-null, then $p$ has finitely many roots in $F$.
== Proof ==
Let $n \ge 1$ be the degree of $p$.
We argue that $p$ has at most $n$ roots in $F$.
Let $A$ be the set of roots of $p$.
Let $a \in A$.
By the Polynomial Factor Theorem:
:$p \left({x}\right) = q_1 \left({x}\right) \cdot \left({x - a}\right)$
:where $\deg q_1 = n - 1$.
Let $a' \in A$ such that $a' \ne a$.
Then since:
:$p \left({a'}\right) = 0$
but:
:$\left({a' - a}\right) \ne 0$
it follows that:
:$q_1 \left({a'}\right) = 0$
Again by the Polynomial Factor Theorem:
:$q_1 \left({x}\right) = q_2 \left({x}\right) \cdot \left({x - a'}\right)$
Therefore:
:$p \left({x}\right) = q_2 \left({x}\right) \cdot \left({x - a'}\right) \cdot \left({x - a}\right)$
where $\deg q_2 = n-2$.
We can repeat this procedure as long as there are still distinct roots in $A$.
After the $i$th iteration we obtain a polynomial $q_i$ of degree $n-i$.
Now the degree of $q_i$ decreases each time we factor out a root of $p$, so $q_n$ is necessarily a constant term.
Therefore $q_n$ can share no roots with $p$.
So this procedure must stop by at most the $n$th step.
That is, we can pick at most $n$ distinct elements from the set $A$, and:
:$\left\vert{A}\right\vert \le n$
Category:Polynomial Theory
Polynomial over Field is Reducible iff Scalar Multiple is Reducible

== Theorem ==
Let $K$ be a field.
Let $K \sqbrk X$ be the ring of polynomial forms over $K$.
Let $P \in K \sqbrk X$.
Let $\lambda \in K \setminus \set 0$.
Then $P$ is irreducible in $K \sqbrk X$ $\lambda P$ is also irreducible in $K \sqbrk X$.
== Proof ==
=== Necessary Condition ===
Let $P$ be irreducible.
Suppose further that $ \lambda P$ has a non-trivial factorization:
:$\lambda P = Q_1 Q_2$
that is, such that $Q_1$ and $Q_2$ are not units of $K \sqbrk X$.
By Units of Ring of Polynomial Forms over Field it follows that $\deg Q_1 \ge 1$ and $\deg Q_2 \ge 1$.
Let $Q_1' = \lambda^{-1} Q_1$.
This implies that:
:$P = Q_1' Q_2$
with $\deg Q_1' = \deg Q_1 \ge 1$.
But this is a non-trivial factorization of $P$ in $K \sqbrk X$.
This contradicts our supposition that $P$ is irreducible.
Therefore $\lambda P$ has no non-trivial factorization, that is, $\lambda P$ is irreducible.
=== Sufficient Condition ===
Let $\lambda P$ be irreducible.
Let $Q = \lambda P$.
From the necessary condition, we know that any scalar multiple of $Q$ is irreducible.
In particular:
:$\lambda^{-1} Q = \lambda^{-1} \lambda P = P$
is irreducible, the required result.
Category:Polynomial Theory
Polynomial over Real Numbers/Examples/Arbitrary Example 1

== Example of Polynomial over Real Numbers ==
The polynomial:
:$x^3 - 2 x + 6$
is of degree $3$.
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(d)}$ Polynomials
Category:Examples of Polynomials over Real Numbers
Polynomial which is Irreducible over Integers is Irreducible over Rationals

== Theorem ==
Let $\Z \sqbrk X$ be the ring of polynomial forms over the integers in the indeterminate $X$.
Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X \in \Z \sqbrk X$ be irreducible in $\Z \sqbrk X$.
Then $\map f X$ is also irreducible in $\Q \sqbrk X$.
== Proof ==
 $\map f X = \map g X \, \map h X$ for some $\map g X, \map h X \in \Q \sqbrk X$.
By hypothesis:
:$\map f X \in \Z \sqbrk X$
and so by definition has coefficients all of which are integers.
But from Factors of Polynomial with Integer Coefficients have Integer Coefficients it follows that $\map f X$ can be expressed as:
:$\map f X = \map {g'} X \, \map {h'} X$
where both $\map {g'} X$ and $\map {h'} X$ are elements of $\Q \sqbrk X$ which have coefficients all of which are integers.
That is:
:$\map {g'} X \in \Z \sqbrk X$
and:
:$\map {h'} X \in \Z \sqbrk X$
This contradicts the statement that $\map f X$ is irreducible in $\Z \sqbrk X$.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients
Category:Polynomial Theory
Polynomial with Algebraic Number as Root is Multiple of Minimal Polynomial

== Theorem ==
Let $F$ be a field.
Let $\map P x$ be a polynomial in $F$.
Let $z$ be a root of $\map P x$.
Then $\map P x$ is a multiple of the minimal polynomial $\map m x$ in $z$ over $F$.
== Proof ==
For $z$ to be a root of $F$, $z$ must be algebraic over $F$.
Let us write:
:$\map P x = \map m x \, \map q x + \map r x$
where $\map q x$ and $\map r x$ are polynomials in $F$.
Then either $\map r x = 0$ or $\map \deg {\map r x} < \map \deg {\map m x}$.
Then:
:$\map P z = \map m z \, \map q z + \map r z$
But as $z$ is a root of both $\map P x$ and $\map m x$, we have that:
:$\map P z = \map m z = 0$
and so:
:$\map r z = 0$
So if $\map r x \ne 0$ we have that $\map r x$ is a polynomial of smaller degree than $\map m x$.
This contradicts the minimality of $\map m x$.
Thus $\map r x = 0$ and so $\map P x$ is a multiple of $\map m x$.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Theorem $71 \ \text{(ii)}$
Category:Minimal Polynomials
Polynomials Closed under Addition/Polynomial Forms

== Theorem ==
Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
:$\ds g = \sum_{k \mathop \in Z} b_k \mathbf X^k$
be polynomials in the indeterminates $\set {X_j: j \in J}$ over the ring $R$.
Then the operation of polynomial addition on $f$ and $g$:
Define the sum:
:$\ds f \oplus g = \sum_{k \mathop \in Z} \paren {a_k + b_k} \mathbf X^k$
Then $f \oplus g$ is a polynomial.
That is, the operation of polynomial addition is closed on the set of all polynomials on a given set of indeterminates $\set {X_j: j \in J}$.
== Proof ==
It is immediate that $f \oplus g$ is a map from the free commutative monoid to $R$, so we need only prove that $f \oplus g$ is nonzero on finitely many $\mathbf X^k$, $k \in Z$.
Suppose that for some $k \in Z$, $a_k + b_k \ne 0$
This forces at least one of $a_k$ and $b_k$ to be non-zero.
This can only be true for a finite number of terms because $f$ and $g$ are polynomials.
The result follows.
Category:Polynomial Theory
Polynomials Closed under Addition/Polynomials over Integral Domain

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Then $\forall x \in R$, the set $D \sqbrk x$ of polynomials in $x$ over $D$ is closed under the operation $+$.
== Proof 1 ==
== Proof 2 ==
Category:Polynomial Theory
Polynomials Closed under Addition/Polynomials over Integral Domain/Proof 1

== Theorem ==
== Proof ==
Let $p, q$ be polynomials in $x$ over $D$.
We can express them as:
:$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k$
:$\ds q = \sum_{k \mathop = 0}^n b_k \circ x^k$
where:
:$(1): \quad a_k, b_k \in D$ for all $k$
:$(2): \quad m, n \in \Z_{\ge 0}$, that is, are non-negative integers.
Suppose $m = n$.
Then:
:$\ds p + q = \sum_{k \mathop = 0}^n a_k \circ x^k + \sum_{k \mathop = 0}^n b_k \circ x^k$
Because $\struct {R, +, \circ}$ is a commutative ring, it follows that:
:$\ds p + q = \sum_{k \mathop = 0}^n \paren {a_k + b_k} \circ x^k$
which is also a polynomials in $x$ over $D$.
, suppose $m > n$.
Then we can express $q$ as:
:$\ds \sum_{k \mathop = 0}^n b_k \circ x^k + \sum_{k \mathop = n \mathop + 1}^m 0_D \circ x^k$
Thus:
:$\ds p + q = \sum_{k \mathop = 0}^n \paren {a_k + b_k} \circ x^k + \sum_{k \mathop = n \mathop + 1}^m a_k \circ x^k$
which is also a polynomials in $x$ over $D$.
Thus the sum of two polynomials in $x$ over $D$ is another polynomial in $x$ over $D$.
Hence the result.
Category:Polynomial Theory
Polynomials Closed under Addition/Polynomials over Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +, \circ}$ be a subring of $R$.
Then $\forall x \in R$, the set $S \sqbrk x$ of polynomials in $x$ over $S$ is closed under the operation $+$.
=== Polynomials over Integral Domain ===
The special case when $R$ is a commutative ring with unity and $D$ is an integral domain follows the same lines:
== Proof ==
Let $p, q$ be polynomials in $x$ over $S$.
We can express them as:
:$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k$
:$\ds q = \sum_{k \mathop = 0}^n b_k \circ x^k$
where:
:$(1): \quad a_k, b_k \in S$ for all $k$
:$(2): \quad m, n \in \Z_{\ge 0}$, that is, are non-negative integers.
Suppose $m = n$.
Then:
:$\ds p + q = \sum_{k \mathop = 0}^n a_k \circ x^k + \sum_{k \mathop = 0}^n b_k \circ x^k$
Because $\struct {R, +, \circ}$ is a ring, it follows that:
:$\ds p + q = \sum_{k \mathop = 0}^n \paren {a_k + b_k} \circ x^k$
which is also a polynomial in $x$ over $S$.
, suppose $m > n$.
Then we can express $q$ as:
:$\ds \sum_{k \mathop = 0}^n b_k \circ x^k + \sum_{k \mathop = n \mathop + 1}^m 0_D \circ x^k$
Thus:
:$\ds p + q = \sum_{k \mathop = 0}^n \paren {a_k + b_k} \circ x^k + \sum_{k \mathop = n \mathop + 1}^m a_k \circ x^k$
which is also a polynomial in $x$ over $S$.
Thus the sum of two polynomials in $x$ over $S$ is another polynomial in $x$ over $S$.
Hence the result.
Category:Polynomial Theory
Polynomials Closed under Ring Product

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$, $\ds g = \sum_{k \in Z} b_k \mathbf X^k$ be polynomials in the indeterminates $\set {X_j: j \in J}$ over $R$, where $Z$ is the set of all multiindices indexed by $\set {X_j: j \in J}$.
Define the product
:$\ds f \otimes g = \sum_{k \in Z} c_k \mathbf X^k$
where
:$\ds c_k = \sum_{\substack {p + q = k \\ p, q \in Z} } a_p b_q$
Then $f \otimes g$ is a polynomial.
== Proof ==
It is immediate that $f \otimes g$ is a map from the free commutative monoid to $R$, so we need only prove that $f \otimes g$ is nonzero on finitely many $\mathbf X^k$, $k \in Z$.
Suppose that for some $k \in Z$:
:$\ds \sum_{\substack {p + q = k \\ p, q \mathop \in Z}} a_p b_q \ne 0$
Therefore if $c_k \ne 0$ there exist $p, q \in Z$ such that $p + q = k$ and $a_p$, $b_q$ are both nonzero.
Since $f$ and $g$ are polynomials, the set $\set {p + q: a_p \ne 0, b_q \ne 0}$ is finite.
Hence the result.
Category:Polynomial Theory
Polynomials Contain Multiplicative Identity

== Theorem ==
The set of polynomials has a multiplicative identity.
== Proof ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity with multiplicative identity $1_R$ and additive identity $0_R$.
Let $\set {X_j: j \in J}$ be a set of indeterminates.
Let $Z$ be the set of all multiindices indexed by $\left\{{X_j: j \in J}\right\}$.
Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ 
be an arbitrary polynomial in the indeterminates $\set {X_j: j \in J}$ over $R$.
Let:
:$\ds N = 1_R \mathbf X^0 = \sum_{k \mathop \in Z} b_k \mathbf X^k$ 
where $b_k = 0_R$ if $k \ne 0$ and $b_0 = 1_R$.
Then:
 
 
 
Therefore:
:$f \circ N = f$
for all polynomials $f$.
Therefore $N$ is a multiplicative identity for the set of polynomials.
Category:Polynomial Theory
Polynomials in Integers is Unique Factorization Domain

== Theorem ==
Let $\Z \sqbrk X$ be the ring of polynomials in $X$ over $\Z$.
Then $\Z \sqbrk X$ is a unique factorization domain.
== Proof ==
We have that Integers form Unique Factorization Domain.
The result follows from Gauss's Lemma on Unique Factorization Domains.
== Sources ==
* : Chapter $9$: Rings: Exercise $21$
Category:Polynomial Rings
Category:Integers
Category:Unique Factorization Domains
Polynomials in Integers is not Principal Ideal Domain

== Theorem ==
Let $\Z \sqbrk X$ be the ring of polynomials in $X$ over $\Z$.
Then $\Z \sqbrk X$ is not a principal ideal domain.
== Proof ==
Let $J$ be the ideal formed from the set of polynomials over $\Z$ in $X$ which have a constant term which is even.
From Polynomials in Integers with Even Constant Term forms Ideal, $J$ is indeed an ideal.
 $J$ is a principal ideal of $\Z \sqbrk X$ such that $J = \ideal f$.
But $2 \in J$, and so $2$ is a multiple of $f$ in $\Z \sqbrk X$.
So $f = \pm 2$ or $f = \pm 1$.
But this contradicts the fact that $J = \ideal f$.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $9$: Rings: Exercise $21$
Category:Polynomial Rings
Category:Integers
Category:Principal Ideal Domains
Polynomials in Integers with Even Constant Term forms Ideal

== Theorem ==
Let $\Z \sqbrk X$ be the ring of polynomials in $X$ over $\Z$.
Let $S \subseteq \Z \sqbrk X$ be the set of polynomials over $\Z$ in $X$ which have a constant term which is even.
Then $S$ is an ideal of $\Z \sqbrk X$.
== Proof ==
For example, $X + 2$ is a polynomials over $\Z$ in $X$ with an even constant term.
So $S$ is not empty.
Let $P_1 = \ds \sum_{k \mathop = 0}^n a_k X^k$ and $P_2 = \ds \sum_{k \mathop = 0}^n b_k X^k$ be elements of $S$.
We have:
The constant term of $P_1 - P_2$ is $a_0 - b_0$ which is even.
Thus $P_1 - P_2 \in S$.
Let $P_3 = \sum_{k \mathop = 0}^s c_k X^k \in \Z \sqbrk X$.
Then the constant term of $P_3 \times P_1$ is $c_0 \times a_0$.
As $a_0$ is even, so is $c_0 \times a_0$.
The result follows by Test for Ideal.
== Sources ==
* : Chapter $9$: Rings: Exercise $21$
Category:Polynomial Rings
Category:Integers
Polynomials of Congruent Integers are Congruent

== Theorem ==
Let $x, y, m \in \Z$ be integers where $m \ne 0$.
Let:
:$x \equiv y \pmod m$
where the notation indicates congruence modulo $m$.
Let $a_0, a_1, \ldots, a_r$ be integers.
Then:
:$\ds \sum_{k \mathop = 0}^r a_k x^k \equiv \sum_{k \mathop = 0}^r a_k y^k \pmod m$
== Proof ==
We have that:
:$x \equiv y \pmod m$
From Congruence of Powers:
:$x^k \equiv y^k \pmod m$
From Modulo Multiplication is Well-Defined:
:$\forall k \in \set {0, 2, \ldots, r}: a_k x^k \equiv a_k y^k \pmod m$
The result follows from Modulo Addition is Well-Defined.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $3$
Category:Polynomial Theory
Category:Modulo Arithmetic
Polynomials of Congruent Ring Elements are Congruent

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $I$ be an ideal of $R$.
Let $x, y \in R$.
Let:
:$x \equiv y \pmod I$
where the notation indicates congruence modulo $I$.
Let $\map F X \in R \sqbrk X$ be a polynomial in one variable over $R$.
Then:
:$\ds \map F x \equiv \map F y \pmod I$
== Proof ==
Let $\map F X = \ds \sum_{k \mathop = 0}^r a_k X^k$ where $X$ is the indeterminate and $a_0, a_1, \ldots, a_r \in R$.
It has to be shown:
:$\ds \sum_{k \mathop = 0}^r a_k x^k \equiv \sum_{k \mathop = 0}^r a_k y^k \pmod I$
From Left Cosets are Equal iff Product with Inverse in Subgroup:
:$\forall a, b \in R: a \equiv b \pmod I \iff a + I = b + I$
We have:
Category:Polynomial Theory
Category:Ideal Theory
Population/Examples/Smokers in London

== Example of Population ==
Consider a random sample of $300$ adult Londoners.
Suppose $75$ of them, that is, $25 \%$ of the random sample, are smokers.
Then $25 \%$ is an appropriate estimate of the proportion of the complete population of adult Londoners who smoke.
Confidence limits may be attached to this estimate.
== Sources ==
* 
* 
Category:Examples of Populations
Population/Examples/Thickness of Sheet Metal

== Example of Population ==
Consider measurements of the thickness of $10$ sheets of metal selected at random from the daily production at a factory.
While the actual population which is sampled is only that for a particular day, inferences are often taken to apply to all such sheets that have been or will be produced by the factory.
The assumption made is that the standards of production do not vary appreciably from day to day.
== Sources ==
* 
* 
Category:Examples of Populations
Poset Elements Equal iff Equal Weak Lower Closure

== Theorem ==
Let $\left({S, \preccurlyeq}\right)$ be an ordered set.
Let $s, t \in S$.
Then $s = t$ :
:$s^\preccurlyeq = t^\preccurlyeq$
where $s^\preccurlyeq$ denotes weak lower closure of $s$.
That is, , for all $r \in S$:
:$r \preccurlyeq s \iff r \preccurlyeq t$
== Proof ==
=== Necessary Condition ===
If $s = t$, then trivially also:
:$s^\preccurlyeq = t^\preccurlyeq$
=== Sufficient Condition ===
Suppose that:
:$s^\preccurlyeq = t^\preccurlyeq$
By definition of weak lower closure, we have:
:$s \in s^\preccurlyeq$
:$t \in t^\preccurlyeq$
and hence:
:$s \in t^\preccurlyeq$
:$t \in s^\preccurlyeq$
which by definition of weak lower closure means:
:$s \preccurlyeq t$ and $t \preccurlyeq s$
Since $\preccurlyeq$ is antisymmetric it follows that $s = t$.
Category:Lower Closures
Poset Elements Equal iff Equal Weak Upper Closure

== Theorem ==
Let $\left({S, \preccurlyeq}\right)$ be an ordered set.
Let $s, t \in S$.
Then $s = t$ :
:$s^\succcurlyeq = t^\succcurlyeq$
where $s^\succcurlyeq$ denotes weak upper closure of $s$.
That is, , for all $r \in S$:
:$s \preccurlyeq r \iff t \preccurlyeq r$
== Proof ==
=== Necessary Condition ===
If $s = t$, then trivially also:
:$s^\succcurlyeq = t^\succcurlyeq$
=== Sufficient Condition ===
Suppose that:
:$s^\succcurlyeq = t^\succcurlyeq$
By definition of weak upper closure, we have:
:$s \in s^\succcurlyeq$
:$t \in t^\succcurlyeq$
and hence:
:$s \in s^\succcurlyeq$
:$t \in s^\succcurlyeq$
which by definition of weak upper closure means:
:$t \preccurlyeq s$ and $s \preccurlyeq t$
Since $\preccurlyeq$ is antisymmetric it follows that $s = t$.
Category:Upper Closures
Position-Ratio of Point/Examples

== Examples of Position-Ratios of Points ==
Let $P$ and $Q$ be points.
Let $R$ be a point on the straight line passing through $P$ and $Q$.
Let $k$ denote the position-ratio of $R$.
=== $R$ Approaching $P$ ===
=== $R$ Approaching $Q$ ===
=== $R$ Approaching Infinity ===
=== $R$ Midway between $P$ and $Q$ ===
:
Category:Position-Ratios
Position-Ratio of Point/Examples/R Approaching Infinity

== Example of Position-Ratio of Point ==
Let $P$ and $Q$ be points in space.
Let $R$ be a point on the straight line passing through $P$ and $Q$.
Let $k$ denote the position-ratio of $R$.
As $R$ approaches the point at infinity on the left, $k \to -1$ from above.
As $R$ approaches the point at infinity on the right, $k \to -1$ from below.
:
== Sources ==
* : Chapter $\text I$. Coordinates: $11$. Position-ratio of a point
Category:Position-Ratios
Position-Ratio of Point/Examples/R Approaching P

== Example of Position-Ratio of Point ==
Let $P$ and $Q$ be points in space.
Let $R$ be a point on the straight line passing through $P$ and $Q$.
Let $k$ denote the position-ratio of $R$.
As $R$ approaches $P$, $k \to 0$.
:
== Sources ==
* : Chapter $\text I$. Coordinates: $11$. Position-ratio of a point
Category:Position-Ratios