nyuuzyou/wikis huggingface dataset:
Lower Closure of Meet of Lower Closures

== Theorem ==
Let $\struct {S, \preceq}$ be a meet semilattice.
Let $D_1, D_2$ be subsets of $S$.
Then:
:$\set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}^\preceq = \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}^\preceq$
where
:$D_1^\preceq$ denotes the lower closure of $D_1$.
== Proof ==
By Lower Closure is Closure Operator:
:$D_1 \subseteq D_1^\preceq$ and $D_2 \subseteq D_2^\preceq$
Then
:$\set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2} \subseteq \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}$
By Lower Closure is Closure Operator:
:$\set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}^\preceq \subseteq \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}^\preceq$
By definition of set equality it remains to prove that
:$\set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}^\preceq \subseteq \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}^\preceq$
Let $x \in \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}^\preceq$.
By definition of lower closure of set:
:$\exists y \in \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq} \land x \preceq y$
Then:
:$\exists i_1 \in D_1^\preceq, i_2 \in D_2^\preceq: y = i_1 \wedge i_2$
By definition of lower closure of set:
:$\exists d_1 \in D_1: i_1 \preceq d_1$
and
:$\exists d_2 \in D_2: i_2 \preceq d_2$
By Meet Semilattice is Ordered Structure:
:$y \preceq d_1 \wedge d_2$
By definition of transitivity:
:$x \preceq d_1 \wedge d_2$
:$d_1 \wedge d_2 \in \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}$
Thus by definition of lower closure of set:
:$x \in \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}^\preceq$
Thus the result.
== Sources ==
* 
* 
Category:Join and Meet Semilattices
Lower Closure of Singleton

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $s$ be an element of $S$.
Then:
:$\set s^\preceq = s^\preceq$
where:
:$\set s^\preceq$ denotes the lower closure of $\set s$
:$s^\preceq$ denotes the lower closure of $s$.
== Proof ==
== Sources ==
* 
Category:Lower Closures
Lower Closure of Subset is Subset of Lower Closure

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $X, Y$ be subsets of $S$.
Then
:$X \subseteq Y \implies X^\preceq \subseteq Y^\preceq$
where $X^\preceq$ is the lower closure of $X$.
== Proof ==
Let $X \subseteq Y$.
Let $x \in X^\preceq$.
By definition of lower closure of subset:
:$\exists y \in X: x \preceq y$
By definition of subset:
:$y \in Y$
Thus by definition of lower closure of subset:
:$x \in Y^\preceq$
== Sources ==
* 
Category:Order Theory
Category:Lower Closures
Lower Closures are Equal implies Elements are Equal

== Theorem ==
Let $L = \struct {S, \preceq}$ be an ordered set.
Let $x, y \in S$ such that
:$x^\preceq = y^\preceq$
where $x^\preceq$ denotes the lower closure of $x$.
Then $x = y$
== Proof ==
By definitions of lower closure of element and reflexivity:
:$x \in x^\preceq$ and $y \in y^\preceq$
By definition of lower closure of element:
:$x \preceq y$ and $y \preceq x$
Thus by definition of antisymmetry:
:$x = y$
== Sources ==
* 
Category:Lower Closures
Lower Section is Convex

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ be a lower section.
Then $T$ is convex in $S$.
== Proof ==
Let $a, c \in T$.
Let $b \in S$.
Let $a \preceq b \preceq c$.
Since:
:$c \in T$
:$b \preceq c$
:$T$ is a lower section
it follows that:
:$b \in T$
This holds for all such $a$, $b$, and $c$.
Hence, by definition, $T$ is convex in $S$.
== Also see ==
* Upper Section is Convex
Category:Lower Sections
Category:Convex Sets (Order Theory)
Lower Section is Dual to Upper Section

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
The following are dual statements:
:$T$ is a lower section in $S$
:$T$ is an upper section in $S$
== Proof ==
By definition, $T$ is a lower section in $S$ :
:$\forall t \in T: \forall s \in S: s \preceq t \implies s \in T$
The dual of this statement is:
:$\forall t \in T: \forall s \in S: t \preceq s \implies s \in T$
by Dual Pairs (Order Theory).
 
By definition, this means $T$ is an upper section in $S$.
The converse follows from Dual of Dual Statement (Order Theory).
== Also see ==
* Duality Principle (Order Theory)
Category:Upper Sections
Category:Lower Sections
Category:Dual Pairs (Order Theory)
Lower Section of Natural Number is Provable

== Theorem ==
Let $x \in \N$ be a natural number.
Then the following WFF:
:$\forall y: y = 0 \lor y = \map s 0 \lor \dotso \lor y = \sqbrk {x - 1} \lor \neg \paren {y < \sqbrk x}$
is a theorem of minimal arithmetic.
== Proof ==
Proceed by induction on $x$.
=== Basis for the Induction ===
Let $x = 0$.
Then the result follows immediately from Axiom $\text M 7$.
=== Induction Hypothesis ===
Suppose that there is a formal proof of:
:$\forall y: y = 0 \lor y = \map s 0 \lor \dotso \lor y = \sqbrk {x - 1} \lor \neg \paren {y < \sqbrk x}$
=== Induction Step ===
The following is a formal proof:
:By Universal Instantiation, let $y_0$ be arbitrary.
:Suppose $y_0 = 0 \lor y_0 = \map s 0 \lor \dotso \lor y_0 = \sqbrk {x - 1}$.
::Then, the result follows from Rule of Addition.
:Otherwise, $\neg {\paren {y_0 < \sqbrk x} }$.
::By Law of Excluded Middle, $y_0 = \sqbrk x \lor y_0 \ne \sqbrk x$.
::Suppose $y_0 = \sqbrk x$.
:::Then, the result again follows from Rule of Addition.
::Finally, $y_0 \ne \sqbrk x$.
:::By Rule of Conjunction and Conjunction of Negations, $\neg \paren {y_0 < \sqbrk x \lor y_0 = \sqbrk x}$.
:::By Axiom $\text M 8$ and Modus Tollendo Tollens: $\neg \paren {y_0 < \map s {\sqbrk x} }$.
:::The result follows from Rule of Addition.
The result in question is:
:$y_0 = 0 \lor y_0 = \map s 0 \lor \dotso \lor y_0 = \sqbrk {x - 1} \lor y_0 = \sqbrk x \lor \neg \paren {y_0 < \map s {\sqbrk x} }$
But as $\map s {\sqbrk x}$ is the same as $\sqbrk {\map s x}$, the induction step is satisfied by Universal Generalisation.
The result follows from Principle of Mathematical Induction.
Category:Natural Numbers
Category:Proofs by Induction
Lower Section with no Greatest Element is Open in GO-Space

== Theorem ==
Let $\struct {S, \preceq, \tau}$ be a generalized ordered space.
Let $L$ be a lower section in $S$ with no greatest element.
Then $L$ is open in $\struct {S, \preceq, \tau}$.
== Proof ==
By Maximal Element in Toset is Unique and Greatest, $L$ has no maximal element.
By Lower Section with no Maximal Element:
:$\ds L = \bigcup \set {l^\prec: l \in L}$
where $l^\prec$ is the strict lower closure of $l$.
By Open Ray is Open in GO-Space and the fact that a union of open sets is open, $L$ is open.
== Also see ==
* Upper Section with no Smallest Element is Open in GO-Space
Category:Generalized Ordered Spaces
Lower Section with no Maximal Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $L \subseteq S$.
Then:
:$L$ is a lower section in $S$ with no maximal element
:
:$\ds L = \bigcup \set {l^\prec: l \in L}$
where $l^\prec$ is the strict lower closure of $l$.
== Proof ==
By Dual Pairs (Order Theory):
* Lower section is dual to upper section.
* Maximal element is dual to minimal element.
* Strict lower closure is dual to strict upper closure.
Thus the theorem holds by the duality principle applied to Upper Section with no Minimal Element.
== Also see ==
* Upper Section with no Minimal Element
Category:Lower Sections
Lower Sections in Totally Ordered Set form Chain

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $\LL$ be a set of lower sections in $S$.
Then $\LL$ is a chain.
That is, $\LL$ is totally ordered by $\subseteq$.
== Proof ==
Let $L, M \in \LL$.
Suppose that $M \nsubseteq L$.
Then:
:$\exists x \in M: x \notin L$
Let $y \in L$.
Then since $\preceq$ is a total ordering, $x \preceq y$ or $y \preceq x$.
If $x \preceq y$, then since $L$ is a lower section: $x \in L$, a contradiction.
Thus $y \preceq x$.
Since $M$ is a lower section, $y \in M$.
Since this holds for all $y \in L$, $L \subseteq M$.
Hence, for all $L, M \in \LL$:
:$M \subseteq L$ or $L \subseteq M$
That is, $\LL$ is a chain.
Category:Lower Sections
Category:Total Orderings
Category:Chains (Order Theory)
Lower Sections of Well-Ordered Classes are Order Isomorphic at most Uniquely

== Theorem ==
Let $\struct {A, \preccurlyeq_A}$ and $\struct {B, \preccurlyeq_B}$ be well-ordered classes.
Let $L_A$ be a lower section of $\struct {A, \preccurlyeq_A}$.
Then:
:$(1): \quad L_A$ is order isomorphic to at most one lower section $L_B$ of $\struct {B, \preccurlyeq_B}$
:$(2): \quad$ If such an $L_B$ exists, there exists exactly one order isomorphism from $L_A$ to $L_B$.
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 2$ Isomorphisms of well orderings: Corollary $2.7$
Category:Well-Orderings
Category:Lower Sections
Category:Order Isomorphisms
Lower Sum of Refinement

== Theorem ==
Let $\closedint a b$ be a closed interval.
Let $P$ be a finite subdivision of $\closedint a b$. 
Let $Q$ be a refinement of $P$. 
Then: 
:$\map L {f, P} \le \map L {f, Q}$
where $\map L {f, P}$ and $\map L {f, Q}$ denotes the lower Darboux sum of $f$ with respect to $P$ and $Q$.
== Proof ==
Write: 
:$P = \set {x_0, x_1, \ldots, x_k}$
and:
:$Q = \set {y_0, y_1, \ldots, y_l}$
where:
:$a = x_0 < x_1 < \ldots < x_k = b$
and:
:$a = y_0 < y_1 < \ldots < y_l = b$
Since $P \subseteq Q$, we have $k \le l$ from Cardinality of Subset of Finite Set.
Set: 
:$m_i = \inf \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$
for each $1 \le i \le k$.
Also set: 
:${\tilde m}_j = \inf \set {\map f x : x \in \closedint {y_{j - 1} } {y_j} }$
for each $1 \le j \le l$. 
Consider a pair of elements $\tuple {x_{i - 1}, x_i}$ in $P$. 
Since $P \subseteq Q$, there exists $a_i, b_i$ such that: 
:$\tuple {x_{i - 1}, x_i} = \tuple {y_{a_i}, y_{b_i} }$
We can see that: 
:$a_1 = 0$
and:
:$b_k = l$ 
We also clearly have: 
:$b_{i - 1} = a_i$ for each $1 \le i \le k$. 
Note that: 
:$\closedint {y_{j - 1} } {y_j} \subseteq \closedint {x_{i - 1} } {x_i}$
for all $a_i + 1 \le j \le b_i$. 
So:
:$\set {\map f x : x \in \closedint {y_{j - 1} } {y_j} } \subseteq \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$
for all $a_i + 1 \le j \le b_i$. 
So, from Infimum of Subset, we have:
:$\inf \set {\map f x : x \in \closedint {y_{j - 1} } {y_j} } \ge \inf \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$
for all $a_i + 1 \le j \le b_i$.
That is: 
:$m_i \le {\tilde m}_j$
for each $\tuple {i, j}$ with $a_i + 1 \le j \le b_i$
We can then write: 
:$\ds x_i - x_{i - 1} = \sum_{j \mathop = a_i + 1}^{b_i} \paren {y_j - y_{j - 1} }$
for each $1 \le i \le k$, giving:
== Sources ==
* : $2.5$: The Riemann Integral
Category:Real Analysis
Lower Topology is Unique

== Theorem ==
Let $T_1 = \left({S, \preceq, \tau_1}\right)$ and $T_2 = \left({S, \preceq, \tau_2}\right)$ be relational structures with lower topologies.
Then:
: $\tau_1 = \tau_2$
== Proof ==
Define:
: $B := \left\{ {\complement_S \left({x^\succeq}\right): x \in S}\right\}$
where $x^\succeq$ denotes the upper closure of $x$.
Thus:
== Sources ==
* 
* 
Category:Topological Order Theory
Lower Triangular Matrix/Examples/Square Matrix

== Example of Lower Triangular Matrix ==
An lower triangular square matrix of order $n$:
:$\mathbf L = \begin{bmatrix}
 a_{1 1} & 0 & 0 & \cdots & 0 & 0 \\
 a_{2 1} & a_{2 2} & 0 & \cdots & 0 & 0 \\
 a_{3 1} & a_{3 2} & a_{3 3} & \cdots & 0 & 0 \\
 \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
a_{n - 1, 1} & a_{n - 1, 2} & a_{n - 1, 3} & \cdots & a_{n - 1, n - 1} & 0 \\
 a_{n 1} & a_{n 2} & a_{n 3} & \cdots & a_{n, n - 1} & a_{n n} \\
\end{bmatrix}$
Category:Examples of Triangular Matrices
Lower Triangular Matrix/Examples/m greater than n

== Example of Lower Triangular Matrix ==
A lower triangular matrix of order $m \times n$ such that $m > n$:
:$\mathbf L = \begin{bmatrix}
 a_{1 1} & 0 & 0 & \cdots & 0 & 0 \\
 a_{2 1} & a_{2 2} & 0 & \cdots & 0 & 0 \\
 a_{3 1} & a_{3 2} & a_{3 3} & \cdots & 0 & 0 \\
 \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
a_{n - 1, 1} & a_{n - 1, 2} & a_{n - 1, 3} & \cdots & a_{n - 1, n - 1} & 0 \\
 a_{n 1} & a_{n 2} & a_{n 3} & \cdots & a_{n, n - 1} & a_{n n} \\
a_{n + 1, 1} & a_{n + 1, 2} & a_{n + 1, 3} & \cdots & a_{n + 1, n - 1} & a_{n + 1, n} \\
 \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
a_{m - 1, 1} & a_{m - 1, 2} & a_{m - 1, 3} & \cdots & a_{m - 1, n - 1} & a_{m - 1, n} \\
 a_{m 1} & a_{m 2} & a_{m 3} & \cdots & a_{m, n - 1} & a_{m n} \\
\end{bmatrix}$
Category:Examples of Triangular Matrices
Lower Triangular Matrix/Examples/m less than n

== Example of Lower Triangular Matrix ==
A lower triangular matrix of order $m \times n$ such that $m < n$:
:$\mathbf L = \begin{bmatrix}
 a_{1 1} & 0 & 0 & \cdots & 0 & 0 & \cdots & 0 & 0 \\
 a_{2 1} & a_{2 2} & 0 & \cdots & 0 & 0 & \cdots & 0 & 0 \\
 a_{3 1} & a_{3 2} & a_{3 3} & \cdots & 0 & 0 & \cdots & 0 & 0 \\
 \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\
a_{m - 1, 1} & a_{m - 1, 2} & a_{m - 1, 3} & \cdots & a_{m - 1, m - 1} & 0 & \cdots & 0 & 0 \\
 a_{m 1} & a_{m 2} & a_{m 3} & \cdots & a_{m - 1, m} & a_{m m} & \cdots & 0 & 0 \\
\end{bmatrix}$
Category:Examples of Triangular Matrices
Lower and Upper Bound of Factorial

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
:$\dfrac {n^n} {e^{n - 1} } \le n! \le \dfrac {n^{n + 1} } {e^{n - 1} }$
== Proof ==
We have:
and similarly:
First the inequality:
Then:
Now the inequality:
Then:
Hence the result.
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $24$
Category:Factorials
Lower and Upper Bounds for Sequences

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $x_n \to l$ as $n \to \infty$.
Then:
:$(1): \quad \forall n \in \N: x_n \ge a \implies l \ge a$
:$(2): \quad \forall n \in \N: x_n \le b \implies l \le b$
=== Corollary ===
== Proof ==
$(1): \quad \forall n \in \N: x_n \ge a \implies l \ge a$:
Let $\epsilon > 0$.
Then:
:$\exists N \in \N: n > N \implies \size {x_n - l} < \epsilon$
So from Negative of Absolute Value:
:$l - \epsilon < x_n < l + \epsilon$
But $x_n \ge a$, so:
:$a \le x_n < l + \epsilon$
Thus, for any $\epsilon > 0$:
:$a < l + \epsilon$
From Real Plus Epsilon it follows that $a \le l$.
$(2): \quad \forall n \in \N: x_n \le b \implies l \le b$:
If $x_n \le b$ it follows that $-x_n \ge -b$ and the above result can be used.
== Warning ==
== Sources ==
* : $\S 4$: Convergent Sequences: Some simple properties of convergent sequences: $\S 4.23$
Category:Limits of Sequences
Lower and Upper Bounds for Sequences/Warning

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $x_n \to l$ as $n \to \infty$.
Then it is not necessarily the case that:
:$(1): \quad \forall n \in \N: x_n > a \implies l > a$
:$(2): \quad \forall n \in \N: x_n < b \implies l < b$
== Proof ==
Take the examples:
:$(1): \quad \sequence {x_n} = \dfrac 1 n$
:$(2): \quad \sequence {y_n} = -\dfrac 1 n$
Then :
:$\forall n \in \N_{>0}: \dfrac 1 n > 0, -\dfrac 1 n < 0$
From Sequence of Reciprocals is Null Sequence, we have
:$x_n \to 0$
:$y_n \to 0$
as $n \to \infty$.
However, it is clearly false that $0 > 0$ and $0 < 0$.
== Sources ==
* : $\S 4$: Convergent Sequences: Some simple properties of convergent sequences: $\S 4.24$
Category:Limits of Sequences
Lowest Common Multiple is Associative

== Theorem ==
Let $a, b, c \in \Z$.
Then:
:$\lcm \set {a, \lcm \set {b, c} } = \lcm \set {\lcm \set {a, b}, c}$
where $\lcm$ denotes the lowest common multiple.
== Proof ==
Follows directly from LCM from Prime Decomposition and Max Operation is Associative.
== Also see ==
* Greatest Common Divisor is Associative
Category:Lowest Common Multiple
Category:Examples of Associative Operations‎
Lowest Common Multiple of Integers/Examples

== Examples of Lowest Common Multiples of Integers ==
=== $6$ and $15$ ===
=== $-12$ and $30$ ===
=== $25$ and $30$ ===
=== $42$ and $49$ ===
=== $27$ and $81$ ===
=== $28$ and $29$ ===
=== $3$, $9$ and $11$ ===
=== $7$, $9$, $12$ and $14$ ===
=== $n$ and $n + 1$ ===
=== $2 n - 1$ and $2 n + 1$ ===
Category:Examples of Lowest Common Multiples of Integers
Lowest Common Multiple of Integers/Examples/-12 and 30

== Example of Lowest Common Multiple of Integers ==
The lowest common multiple of $-12$ and $30$ is:
:$\lcm \set {-12, 30} = 60$
== Proof ==
From Greatest Common Divisor: $-12$ and $30$:
:$\gcd \set {-12, 30} = 6$
Then:
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
Category:Examples of Lowest Common Multiples of Integers
Category:60
Lowest Common Multiple of Integers/Examples/25 and 30

== Example of Lowest Common Multiple of Integers ==
The lowest common multiple of $25$ and $30$ is:
:$\lcm \set {25, 30} = 150$
== Proof ==
We find the greatest common divisor of $25$ and $30$ using the Euclidean Algorithm:
Thus $\gcd \set {25, 30} = 5$.
Then:
== Sources ==
* : $\text {2-2}$ Divisibility: Exercise $5 \ \text {(a)}$
Category:Examples of Lowest Common Multiples of Integers
Lowest Common Multiple of Integers/Examples/27 and 81

== Example of Lowest Common Multiple of Integers ==
The lowest common multiple of $27$ and $81$ is:
:$\lcm \set {27, 81} = 81$
== Proof ==
We find the greatest common divisor of $27$ and $81$ using the Euclidean Algorithm:
Thus $\gcd \set {27, 81} = 27$.
Then:
== Sources ==
* : $\text {2-2}$ Divisibility: Exercise $5 \ \text {(c)}$
Category:Examples of Lowest Common Multiples of Integers
Lowest Common Multiple of Integers/Examples/28 and 29

== Example of Lowest Common Multiple of Integers ==
The lowest common multiple of $28$ and $29$ is:
:$\lcm \set {28, 29} = 812$
== Proof ==
== Sources ==
* : $\text {2-2}$ Divisibility: Exercise $5 \ \text {(d)}$
Category:Examples of Lowest Common Multiples of Integers
Lowest Common Multiple of Integers/Examples/2n-1 and 2n+1

== Example of Lowest Common Multiple of Integers ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
The lowest common multiple of $2 n - 1$ and $2 n + 1$ is:
:$\lcm \set {2 n - 1, 2 n + 1} = 4 n^2 - 1$
== Proof ==
We find the greatest common divisor of $2 n - 1$ and $2 n + 1$ using the Euclidean Algorithm:
Thus $\gcd \set {2 n - 1, 2 n + 1} = 1$.
Hence by definition $n$ and $n + 1$ are coprime.
Thus:
The result follows.
== Sources ==
* : $\text {2-2}$ Divisibility: Exercise $5 \ \text {(f)}$
Category:Examples of Lowest Common Multiples of Integers
Lowest Common Multiple of Integers/Examples/3, 9, 11

== Example of Lowest Common Multiple of Integers ==
The lowest common multiple of $3$, $9$ and $11$ is:
:$\lcm \set {3, 9, 11} = 99$
== Proof ==
Hence the result.
== Sources ==
* 
* 
Category:Examples of Lowest Common Multiples of Integers
Lowest Common Multiple of Integers/Examples/3054 and 12378

== Example of Lowest Common Multiple of Integers ==
The lowest common multiple of $3054$ and $12378$ is:
:$\lcm \set {3054, 12378} = 6 \, 300\, 402$
== Proof ==
From Euclidean Algorithm: $12378$ and $3054$:
:$\gcd \set {3054, 12378} = 6$
Then:
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
Category:Examples of Lowest Common Multiples of Integers
Lowest Common Multiple of Integers/Examples/42 and 49

== Example of Lowest Common Multiple of Integers ==
The lowest common multiple of $42$ and $49$ is:
:$\lcm \set {42, 49} = 294$
== Proof ==
We find the greatest common divisor of $42$ and $49$ using the Euclidean Algorithm:
Thus $\gcd \set {42, 49} = 7$.
Then:
== Sources ==
* : $\text {2-2}$ Divisibility: Exercise $5 \ \text {(b)}$
Category:Examples of Lowest Common Multiples of Integers
Lowest Common Multiple of Integers/Examples/6 and 15

== Example of Lowest Common Multiple of Integers ==
The lowest common multiple of $6$ and $15$ is:
:$\lcm \set {6, 15} = 30$
== Proof ==
We find the greatest common divisor of $6$ and $15$ using the Euclidean Algorithm:
Thus $\gcd \set {6, 15} = 3$.
Then:
Category:Examples of Lowest Common Multiples of Integers
Category:30
Lowest Common Multiple of Integers/Examples/7, 9, 12, 14

== Example of Lowest Common Multiple of Integers ==
The lowest common multiple of $7$, $9$, $12$ and $14$ is:
:$\lcm \set {7, 9, 12, 14} = 252$
== Proof ==
Hence the result.
== Sources ==
* 
* 
Category:Examples of Lowest Common Multiples of Integers
Lowest Common Multiple of Integers/Examples/n and n+1

== Example of Lowest Common Multiple of Integers ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
The lowest common multiple of $n$ and $n + 1$ is:
:$\lcm \set {n, n + 1} = n \paren {n + 1}$
== Proof ==
We find the greatest common divisor of $n$ and $n + 1$ using the Euclidean Algorithm:
Thus $\gcd \set {n, n + 1} = 1$.
Hence by definition $n$ and $n + 1$ are coprime.
The result follows from LCM equals Product iff Coprime.
== Sources ==
* : $\text {2-2}$ Divisibility: Exercise $5 \ \text {(e)}$
Category:Examples of Lowest Common Multiples of Integers
Lowest Common Multiple of Integers with Common Divisor

== Theorem ==
Let $b, d \in \Z_{>0}$ be (strictly) positive integers
Then:
:$\lcm \set {a b, a d} = a \lcm \set {b, d}$
where:
:$a \in \Z_{>0}$
:$\lcm \set {b, d}$ denotes the lowest common multiple of $m$ and $n$.
== Proof ==
We have that:
Suppose $n \in \Z$ such that $a b \divides n$ and $a d \divides n$.
It will be shown that $a \lcm \set {b, d} \divides n$.
So:
Thus we have:
:$a b \divides a \lcm \set {b, d} \land a d \divides a \lcm \set {b, d}$
and:
:$a b \divides n \land a d \divides n \implies a \lcm \set {b, d} \divides n$
It follows from LCM iff Divides All Common Multiples that:
:$\lcm \set {a b, a d} = a \lcm \set {b, d}$
== Sources ==
* : $\text {2-2}$ Divisibility: Exercise $6$
Category:Lowest Common Multiple
Lp Metric on Closed Real Interval is Metric

== Theorem ==
Let $S$ be the set of all real functions which are continuous on the closed interval $\closedint a b$.
Let $p \in \R_{\ge 1}$.
Let $d_p: S \times S \to \R$ be the $L^p$ metric on $\closedint a b$:
:$\ds \forall f, g \in S: \map {d_p} {f, g} := \paren {\int_a^b \size {\map f t - \map g t}^p \rd t}^{\frac 1 p}$
Then $d_p$ is a metric.
== Proof ==
=== Proof of ===
So holds for $d_p$.
=== Proof of ===
It is required to be shown:
:$\map {d_p} {f, g} + \map {d_p} {g, h} \ge \map {d_p} {f, h}$
for all $f, g, h \in S$.
Let:
:$(1): \quad t \in \closedint a b$
:$(2): \quad \size {\map f t - \map g t = \map r t}$
:$(3): \quad \size {\map g t - \map h t = \map s t}$
Thus we need to show that:
:$\ds \paren {\int_a^b \size {\map f t - \map g t}^p \rd t}^{\frac 1 p} + \paren {\int_a^b \size {\map g t - \map h t}^p \rd t}^{\frac 1 p} \ge \paren {\int_a^b \size {\map f t - \map h t}^p \rd t}^{\frac 1 p}$
We have:
So holds for $d_p$.
=== Proof of ===
So holds for $d_p$.
=== Proof of ===
From Zero Definite Integral of Nowhere Negative Function implies Zero Function we have that:
:$\map {d_p} {f, g} = 0 \leadsto f = g$
on $\closedint a b$.
So holds for $d_p$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.18$
Category:Lp Metrics