OpenLLM-France/wikimedia huggingface dataset:
# Writing/variety of phrases

Here are various alternatives to "he said" that you can use to add variety to your dialogue attributions: Choose the appropriate pronoun in each case.
* He mentioned
* He remarked
* He commented
* He stated
* He declared
* He noted
* He added
* He reported
* He affirmed
* He conveyed
* He replied
* He responded
* He acknowledged
* He answered
* He uttered
* He voiced
* He articulated
* He disclosed
* He revealed
* He expressed
* He asserted
* He explained
* He narrated
* He recited
* He divulged
* He recounted
* He summarized
* He clarified
* He whispered
* He mumbled
* He shouted
* He exclaimed
* He muttered
* He interjected
* He suggested
Using a variety of these attributions can make your dialogue more engaging and prevent repetition.
# Hans Moravec

This original article describes Hans Moravec, an Austrian-born American roboticist, researcher of artificial intelligence and a book author. The article takes advantage of Wikiversity being less strict about content inclusion than Wikipedia.
He is noted for his book Mind Children from 1988.
His publication list page has a lot of online articles capturing some of his thought and philosophy.
# ISO 8601 and computing differences between dates

This brief article describes how to compute differences between dates manually using ISO 8601 and suggests that it may be easier to remember dates in ISO 8601 format.

## ISO 8601

ISO 8601 is an international standard for dates and times. For example, 01/02/03: Is that January 2, 2003, or 1 February 2003 or 2001 - February 3? You can't tell by looking, and each of these three alternative interpretations is used someplace in the world. The first two are both used in the US, so it's not easy to tell which interpretation is intended unless one of the three numbers is greater than 12.
ISO 8601 solves this problem by insisting we write YYYY-MM-DD. Thus, Christmas Day 2003 for most people in the West is 2003-12-25 in ISO 8601 format. Similarly, Valentine's Day 2001 is 2001-02-14.
What's the time difference between Valentine's Day 2001 and Christmas 2003? Easy:
    (2003-12-25) - (2001-02-14) = (2-10-11) = 2 years, 10 months, and 11 days.
Computing the difference between Christmas 2001 and Valentines 2003 is harder but not too hard:
    (2003-02-14) - (2001-12-25)
We can't subtract 25 from 14 and get a positive number, but we can think of February 14 as the 45th day of January, so 2003-02-14 is the same as 2003-01-45. With this equivalence, we can subtract 25 from 45 and get a difference of 20 days.
Similarly, we can't subtract 12 from 1 and get a positive number. However, again, we can think of January 2003 as the 13th month of 2002. With this, we rewrite 2003-02-14 as 2003-01-45 and then as 2002-13-45. Now we can subtract as follows:
    (2002-13-45) - (2001-12-25) = (1-01-20) = one year plus 1 month and 20 days.
Oh, that was almost hard ;-)

## Roman numerals

Have you ever tried to do arithmetic with Roman numerals? It's much easier with so-called Arabic numerals, which seem to have been invented in India and brought to Europe by Arab traders.
At least for the present author it's much easier to do arithmetic with dates in ISO 8601 format than in other formats. Moreover, it seems easier to remember dates in ISO 8601 format, perhaps because it's easier for me to remember the number for each month than the name; it's certainly more obvious that June comes after March, etc.
# Wikimedia Foundation

Learn about the Wikimedia Foundation and its sister projects by clicking on the icons above.
The mission of the Wikimedia Foundation is to help projects like Wikiversity collect and develop free and openly licensed educational content.
# Aircraft

An aircraft is vehicle capable of flight. There are many different kinds of aircraft.
Some aircraft keep in the sky by moving air over their wings, including aeroplanes (airplanes), helicopters, and gliders. Others, like balloons and airships, float.
Most aircraft use engine power while others use no power, such as gliders and balloons. Few aircraft use muscle power.
Big aeroplanes that transport people are called airliners. Airliners can fly at more than 850 km/h (530 mph). Less than 100 years ago, the quickest ships needed more than seven days to travel across the Atlantic Ocean because it is 4,800 km (3,000 miles) wide. Now airliners need less than seven hours to travel across the Atlantic Ocean. Airliners are the quickest way to travel. Airliners can fly over mountains and bad weather. Airliners have complex technology to make them fly quickly, safely, and for less money.
Fighting aircraft fly at 3,200 km/h (2,000 mph).

## Resources

Special:Prefixindex/Aircraft/
# Vector space/K/Inner product/Orthogonal basis/Definition

Let $V$ be a ${\mathbb {K} }$-vector space, endowed with an inner product. A basis $v_{i}$, $i\in I$, of $V$ is called an orthogonal basis if
    $\left\langle v_{i},v_{j}\right\rangle =0{\text{ for }}i\neq j$
holds.
# Vector space/K/Inner product/Orthonormal basis/Definition

Let $V$ be a ${\mathbb {K} }$-vector space, endowed with an inner product $\left\langle -,-\right\rangle$. A basis $v_{i}$, $i\in I$, of $V$ is called an orthonormal basis if
    $\left\langle v_{i},v_{i}\right\rangle =1{\text{ for all }}i\in I{\text{ and }}\left\langle v_{i},v_{j}\right\rangle =0{\text{ for }}i\neq j$
holds.
# Vector space/K/Inner product/Orthonormal basis/Coefficients/Fact

Let $V$ be a ${\mathbb {K} }$-vector space, endowed with an inner product, and let $u_{i}$, $i\in I$, be an orthonormal basis of $V$.
$v$
    $v=\sum _{i\in I}\left\langle v,u_{i}\right\rangle u_{i}\,.$
# Vector space/K/Inner product/Orthonormal basis/Coefficients/Fact/Proof

Since we have a basis, there exists a unique representation
    $v=\sum _{j\in I}a_{j}u_{j}\,$
(where all $a_{j}$ are $0$ up to finitely many). Therefore, the claim follows from
    $\left\langle v,u_{i}\right\rangle =\left\langle \sum _{j\in I}a_{j}u_{j},u_{i}\right\rangle =\sum _{j\in I}a_{j}\left\langle u_{j},u_{i}\right\rangle =a_{i}\,.$
# Vector space/K/Inner product/Finite-dimensional/Orthonormalization/Fact

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product, and let $v_{1},v_{2},\ldots ,v_{n}$ be a basis of $V$.
orthonormal basis $u_{1},u_{2},\ldots ,u_{n}$ of $V$ with
    $\langle v_{1},v_{2},\ldots ,v_{i}\rangle =\langle u_{1},u_{2},\ldots ,u_{i}\rangle \,$
for alle
$i=1,\ldots ,n$
# Vector space/K/Inner product/Finite-dimensional/Orthonormalization/Fact/Proof

We prove the statement by induction over $i$, that is, we construct successively a family of orthonormal vectors spanning the same linear subspaces. For $i=1$, we just have to normalize $v_{1}$, that is, we replace it by $u_{1}={\frac {v_{1}}{\Vert {v_{1}}\Vert }}$. Now suppose that the statement is already proven for $i$. Let a family of orthonormal vectors $u_{1},\ldots ,u_{i}$ fulfilling $\langle u_{1},\ldots ,u_{i}\rangle =\langle v_{1},\ldots ,v_{i}\rangle$ be already constructed. We set
    $w_{i+1}=v_{i+1}-\left\langle v_{i+1},u_{1}\right\rangle u_{1}-\cdots -\left\langle v_{i+1},u_{i}\right\rangle u_{i}\,.$
Due to
    ${\begin{aligned}\left\langle w_{i+1},u_{j}\right\rangle &=\left\langle v_{i+1}-\left\langle v_{i+1},u_{1}\right\rangle u_{1}-\cdots -\left\langle v_{i+1},u_{i}\right\rangle u_{i},u_{j}\right\rangle \\&=\left\langle v_{i+1},u_{j}\right\rangle -\sum _{k\leq i,k\neq j}\left\langle v_{i+1},u_{k}\right\rangle \left\langle u_{k},u_{j}\right\rangle -\left\langle v_{i+1},u_{j}\right\rangle \left\langle u_{j},u_{j}\right\rangle \\&=\left\langle v_{i+1},u_{j}\right\rangle -\left\langle v_{i+1},u_{j}\right\rangle \\&=0,\end{aligned}}$
this vector is orthgonal to all $u_{1},\ldots ,u_{i}$, and also
    $\langle u_{1},\ldots ,u_{i},w_{i+1}\rangle =\langle v_{1},\ldots ,v_{i},v_{i+1}\rangle \,$
holds. By normalizing $w_{i+1}$, we obtain $u_{i+1}$.
# R^3/Kernel of 2x+3y-z/Orthonormal basis/Gram-Schmidt/Example

Let $V$ be the kernel of the linear mapping
    $\mathbb {R} ^{3}\longrightarrow \mathbb {R} ,(x,y,z)\longmapsto 2x+3y-z.$
As a linear subspace of $\mathbb {R} ^{3}$, $V$ carries the induced inner product. We want to determine an orthonormal basis of $V$. For this, we consider the basis consisting of the vectors
    $v_{1}={\begin{pmatrix}1\\0\\2\end{pmatrix}}\,\,{\text{ and }}\,\,v_{2}={\begin{pmatrix}0\\1\\3\end{pmatrix}}.$
We have $\Vert {v_{1}}\Vert ={\sqrt {5}}$; therefore,
    $u_{1}={\frac {v_{1}}{\Vert {v_{1}}\Vert }}={\begin{pmatrix}{\frac {1}{\sqrt {5}}}\\0\\{\frac {2}{\sqrt {5}}}\end{pmatrix}}\,$
is the corresponding normed vector. According to orthonormalization process, we set
    ${\begin{aligned}w_{2}&=v_{2}-\left\langle v_{2},u_{1}\right\rangle u_{1}\\&={\begin{pmatrix}0\\1\\3\end{pmatrix}}-\left\langle {\begin{pmatrix}0\\1\\3\end{pmatrix}},{\begin{pmatrix}{\frac {1}{\sqrt {5}}}\\0\\{\frac {2}{\sqrt {5}}}\end{pmatrix}}\right\rangle {\begin{pmatrix}{\frac {1}{\sqrt {5}}}\\0\\{\frac {2}{\sqrt {5}}}\end{pmatrix}}\\&={\begin{pmatrix}0\\1\\3\end{pmatrix}}-{\frac {6}{\sqrt {5}}}{\begin{pmatrix}{\frac {1}{\sqrt {5}}}\\0\\{\frac {2}{\sqrt {5}}}\end{pmatrix}}\\&={\begin{pmatrix}0\\1\\3\end{pmatrix}}-{\begin{pmatrix}{\frac {6}{5}}\\0\\{\frac {12}{5}}\end{pmatrix}}\\&={\begin{pmatrix}-{\frac {6}{5}}\\1\\{\frac {3}{5}}\end{pmatrix}}.\end{aligned}}$
We have
    $\Vert {w_{2}}\Vert =\Vert {\begin{pmatrix}-{\frac {6}{5}}\\1\\{\frac {3}{5}}\end{pmatrix}}\Vert ={\sqrt {{\frac {36}{25}}+1+{\frac {9}{25}}}}={\sqrt {\frac {70}{25}}}={\frac {\sqrt {14}}{\sqrt {5}}}\,.$
Therefore,
    $u_{2}={\frac {\sqrt {5}}{\sqrt {14}}}{\begin{pmatrix}-{\frac {6}{5}}\\1\\{\frac {3}{5}}\end{pmatrix}}={\begin{pmatrix}-{\frac {6}{\sqrt {70}}}\\{\frac {\sqrt {5}}{\sqrt {14}}}\\{\frac {3}{\sqrt {70}}}\end{pmatrix}}\,$
is the second vector of the orthonormal basis.
# Vector space/K/Inner product/Finite-dimensional/Orthogonal complement/Direct sum/Fact

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product, and let $U\subseteq V$ denote a linear subspace.
    $V=U\oplus U^{\perp }\,,$
that is, $V$ is the direct sum of $U$ and its
# Vector space/K/Inner product/Finite-dimensional/Orthonormal basis/Existence/Fact

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product.
orthonormal basis
$V$
# Vector space/K/Inner product/Finite-dimensional/Orthonormal basis/Existence/Fact/Proof

This follows directly from fact.
# Vector space/K/Inner product/Finite-dimensional/Orthogonal complement/Direct sum/Fact/Proof

From $u\in U\cap U^{\perp }$ we get directly
    $\left\langle u,u\right\rangle =0\,,$
thus $u=0$. This means that the sum direct. Let $u_{1},\ldots ,u_{k}$ be an orthonormal basis of $U$. We extend it to an orthonormal basis $u_{1},\ldots ,u_{n}$ of $V$. Then we have
    $U^{\perp }=\langle u_{k+1},\ldots ,u_{n}\rangle \,;$
Therefore, $V$ is the sum of the linear subspaces.
# Vector space/K/Inner product/Finite-dimensional/Orthogonal complement/Structural properties/Fact

Let $V$ be a ${\mathbb {K} }$-vector space, endowed with an inner product $\left\langle -,-\right\rangle$. Then the following statements hold.
* For a linear subspace $U\subseteq U'\subseteq V$, we have
    $U^{\perp }\supseteq U'^{\perp }\,.$
* We have $0^{\perp }=V$ and $V^{\perp }=0$.
* Let $V$ be finite-dimensional. Then we have
    ${\left(U^{\perp }\right)}^{\perp }=U\,.$
* Let $V$ be finite-dimensional. Then we have
    $\dim _{K}{\left(U^{\perp }\right)}=\dim _{K}{\left(V\right)}-\dim _{K}{\left(U\right)}\,.$
# President of the United States/Joe Biden

| 44th President | 45th President | 46th President | 47th President | 48th President |
| Barack Obama 2009 - 2017 | Donald Trump 2017 - 2021 | Joe Biden 2021 - ? | Who will be the next President? | Who will be the next President? |
# Vector space/K/Inner product/Finite dimension/Linear subspace/Orthogonal projection/Orthonormal basis/Fact/Proof

We extend the basis to an orthonormal basis $u_{1},\ldots ,u_{n}$ of $V$. The orthogonal complement of $U$ is
    $U^{\perp }=\langle u_{m+1},\ldots ,u_{n}\rangle \,.$
Due to fact, we have
    $v=\sum _{i=1}^{n}\left\langle v,u_{i}\right\rangle u_{i}={\left(\sum _{i=1}^{m}\left\langle v,u_{i}\right\rangle u_{i}\right)}+{\left(\sum _{i=m+1}^{n}\left\langle v,u_{i}\right\rangle u_{i}\right)}\,.$
Therefore, $\sum _{i=1}^{m}\left\langle v,u_{i}\right\rangle u_{i}$ is the projection onto $U$ along $U^{\perp }$.
# Cross product/K^3/Introduction/Section

A special feature of $\mathbb {R} ^{3}$ is the so-called cross product. This assigns, to two given vectors, a vector that is orthogonal to them.

## Definition

K^3/Cross product/Direct/Definition
The cross product is also called the vector product. To remember this formula, one might think
    $x\times y=\det {\begin{pmatrix}e_{1}&x_{1}&y_{1}\\e_{2}&x_{2}&y_{2}\\e_{3}&x_{3}&y_{3}\end{pmatrix}}\,,$
where $e_{1},e_{2},e_{3}$ are the standard vectors, and where we expand formally with respect to the first column. In this way, the cross product is defined with respect to the standard basis.

## Lemma
### Proof

$\Box$
The expression $\left\langle x\times y,z\right\rangle$ from (5), that is, the determinant of the three vectors, considered as a column vector, is also called triple product.

## Lemma
### Proof

$\Box$
# K^3/Cross product/Direct/Definition

Let $K$ be a field. The operation on $K^{3}$, defined by
    $x\times y={\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}\times {\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}:={\begin{pmatrix}x_{2}y_{3}-x_{3}y_{2}\\-x_{1}y_{3}+x_{3}y_{1}\\x_{1}y_{2}-x_{2}y_{1}\end{pmatrix}}\,,$
is called the cross product.
# R^3/Cross product/Computation/1/Example

The cross product of the vectors ${\begin{pmatrix}5\\8\\-2\end{pmatrix}},{\begin{pmatrix}3\\4\\1\end{pmatrix}}\in \mathbb {R} ^{3}$ is
    ${\begin{pmatrix}5\\8\\-2\end{pmatrix}}\times {\begin{pmatrix}3\\4\\1\end{pmatrix}}={\begin{pmatrix}8\cdot 1-4\cdot (-2)\\-5\cdot 1-2\cdot 3\\5\cdot 4-8\cdot 3\end{pmatrix}}={\begin{pmatrix}16\\-11\\-4\end{pmatrix}}\,.$
# K^3/Cross product/Properties/Fact

The cross product on $K^{3}$ fulfills the following properties (where $x,y,z\in K^{3}$ and $a,b\in K$).
* We have
    $x\times y=-(y\times x)\,.$
* We have and
    $(ax+by)\times z=a(x\times z)+b(y\times z)\,$
    $z\times (ax+by)=a(z\times x)+b(z\times y)\,.$
* We have if and only if $x$ and $y$ are linearly dependent.
    $x\times y=0\,$
* We have
    $x\times (y\times z)+y\times (z\times x)+z\times (x\times y)=0\,.$
* We have where $\left\langle -,-\right\rangle$ denotes the formal evaluation in the sense of the standard inner product.
    $\left\langle x\times y,z\right\rangle =\det(x\,,y\,,z)\,,$
* We have where $\left\langle -,-\right\rangle$ denotes the formal evaluation in the sense of the standard inner product.
    $\left\langle x,x\times y\right\rangle =0=\left\langle y,x\times y\right\rangle \,,$
# K^3/Cross product/Properties/Fact/Proof

(1) is clear from the definition.
(2). We have
    ${\begin{aligned}{\left(a{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}+b{\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}\right)}\times {\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}}&={\begin{pmatrix}ax_{1}+by_{1}\\ax_{2}+by_{2}\\ax_{3}+by_{3}\end{pmatrix}}\times {\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}}\\&={\begin{pmatrix}(ax_{2}+by_{2})z_{3}-(ax_{3}+by_{3})z_{2}\\-(ax_{1}+by_{1})z_{3}+(ax_{3}+by_{3})z_{1}\\(ax_{1}+by_{1})z_{2}-(ax_{2}+by_{2})z_{1}\end{pmatrix}}\\&={\begin{pmatrix}ax_{2}z_{3}-ax_{3}z_{2}\\-ax_{1}z_{3}+ax_{3}z_{1}\\ax_{1}z_{2}-ax_{2}z_{1}\end{pmatrix}}+{\begin{pmatrix}by_{2}z_{3}-by_{3}z_{2}\\-by_{1}z_{3}+by_{1}z_{3}\\by_{1}z_{2}-by_{2}z_{1}\end{pmatrix}}\\&=a{\begin{pmatrix}x_{2}z_{3}-x_{3}z_{2}\\-x_{1}z_{3}+x_{3}z_{1}\\x_{1}z_{2}-x_{2}z_{1}\end{pmatrix}}+b{\begin{pmatrix}y_{2}z_{3}-y_{3}z_{2}\\-y_{1}z_{3}+y_{1}z_{3}\\y_{1}z_{2}-y_{2}z_{1}\end{pmatrix}}\\&=a(x\times z)+b(y\times z).\end{aligned}}$
The second equation follows from this and from (1).
(3). If $x$ and $y$ are linearly dependent, then we can write $x=cy$ (or the other way round). In this case,
    ${\begin{pmatrix}cy_{1}\\cy_{2}\\cy_{3}\end{pmatrix}}\times {\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}={\begin{pmatrix}cy_{2}y_{3}-cy_{2}y_{3}\\-cy_{1}y_{3}+cy_{3}y_{1}\\cy_{1}y_{2}-cy_{2}y_{1}\end{pmatrix}}=0\,.$
If the cross product is $0$, then all entries of the vectors ${\begin{pmatrix}x_{2}y_{3}-x_{3}y_{2}\\-x_{1}y_{3}+x_{3}y_{1}\\x_{1}y_{2}-x_{2}y_{1}\end{pmatrix}}$ equal $0$. For example, let $y_{1}\neq 0$. From $x_{1}=0$, we can deduce directly
    $x_{2}=x_{3}=0\,,$
and $x$ is the zero vector. So suppose that $x_{1}\neq 0$. Then $y_{2}={\frac {y_{1}}{x_{1}}}x_{2}$ and $y_{3}={\frac {y_{1}}{x_{1}}}x_{3}$; therefore, we get
    $y={\frac {y_{1}}{x_{1}}}x\,.$
(4). See exercise.
(5). We have
    ${\begin{aligned}\left\langle x\times y,z\right\rangle &=\left\langle {\begin{pmatrix}x_{2}y_{3}-x_{3}y_{2}\\-x_{1}y_{3}+x_{3}y_{1}\\x_{1}y_{2}-x_{2}y_{1}\end{pmatrix}},{\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}}\right\rangle \\&=z_{1}x_{2}y_{3}-z_{1}x_{3}y_{2}-z_{2}x_{1}y_{3}+z_{2}x_{3}y_{1}+z_{3}x_{1}y_{2}-z_{3}x_{2}y_{1}.\end{aligned}}$
This coincides with the determinant, due to Sarrus.
(6) follows from (5).
# R^3/Cross product/Oriented orthonormal basis/Fact

Let $u_{1},u_{2},u_{3}$ be an orthonormal basis of $\mathbb {R} ^{3}$ with
    $\det {\left(u_{1},\,u_{2},\,u_{3}\right)}=1\,.$
cross product $x\times y$ can be computed with the coordinates of $x$ and $y$ with respect to this basis (and the formula from
# Digital Media Concepts/David LaRochelle

David LaRochelle (born October 26, 1960) is an American author and illustrator of children's books. He can also be notable for his work on puzzles, games, and a novel for young adults.
David LaRochelle :
| Children's Literature Section in Vancouver Library ||
| Born | October 26, 1960 |
| Job | Author & illustator |
| Birth Place | Park Rapids, Minnesota |

## Early Life & Education

David LaRochelle was raised in New Brighton, Minnesota. His main sources of entertainment as a kid was watching tons of TV, reading, and drawing. He grew up with many pets and his older sister Debbie. David graduated from St. Olaf College studying art and English with the goal to become an illustrator for Hallmark Cards.

## Career

With the intention of becoming a Hallmark Card illustrator, David's career took a turn when Hallmark officials didn't think he drew well enough. Instead, David became an elementary school teacher. He continued to illustrate and draw while teaching fourth graders. David's friend called a book publisher about his work and they bought it. With success with his first book, A Christmas Guest, he began to publish many other projects. He transitioned from a teacher to a full time illustrator and author.

## Popular Books

| Year Published | Book |
| 2020 | See the Cat: Three Stories about a Dog |
| 2012 | It's a Tiger! |
| 2013 | Moo! |
| 2004 | The Best Pet of All |
| 2013 | How Martha Saved Her Parents from Green Beans |
| 2021 | See the Dog: Three Stories about a Cat |

## Awards

David LaRochelle has received many awards through his imaginative storytelling. He has won the Theodor Seuss Geisel Award for his book See the Cat: Three Stories About a Dog. His excellence in literature has won him the Minnesota Book Award numerous times. The Society of Children's Book Writers awarded David with the The Sid Fleischman Humor Award for his outstanding humor in children's books. He received the Parents' Choice Award for many of his books including It's a Tiger! and Moo!
# R^3/Cross product/Oriented orthonormal basis/Fact/Proof

Let
    $x=c_{1}u_{1}+c_{2}u_{2}+c_{3}u_{3}\,$
and
    $y=d_{1}u_{1}+d_{2}u_{2}+d_{3}u_{3}\,.$
Due to fact (2), we have
    $x\times y={\left(c_{1}u_{1}+c_{2}u_{2}+c_{3}u_{3}\right)}\times {\left(d_{1}u_{1}+d_{2}u_{2}+d_{3}u_{3}\right)}=\sum _{1\leq i,j\leq 3}c_{i}d_{j}{\left(u_{i}\times u_{j}\right)}\,.$
Due to fact (3), we have
    $u_{i}\times u_{i}=0\,,$
and, because of fact (1), we have
    $u_{i}\times u_{j}=-u_{j}\times u_{i}\,.$
According to fact (6), $u_{1}\times u_{2}$ is perpendicular to $u_{1}$ and to $u_{2}$; therefore,
    $u_{1}\times u_{2}=\lambda u_{3}\,$
with some $\lambda \in \mathbb {R}$, as this orthogonality condition defines a line. Because of fact (5) and the condition, we get
    $\lambda =\left\langle \lambda u_{3},u_{3}\right\rangle =\left\langle u_{1}\times u_{2},u_{3}\right\rangle =\det {\left(u_{1},\,u_{2},\,u_{3}\right)}=1\,;$
hence,
    $u_{1}\times u_{2}=u_{3}\,.$
Using fact (3), we obtain $u_{1}\times u_{3}=-u_{2}$ and $u_{2}\times u_{3}=u_{1}$. Altogether we get
    ${\begin{aligned}x\times y&=\sum _{1\leq i,j\leq 3}c_{i}d_{j}{\left(u_{i}\times u_{j}\right)}\\&=\sum _{i<j}(c_{i}d_{j}-c_{j}d_{i}){\left(u_{i}\times u_{j}\right)}\\&=(c_{1}d_{2}-c_{2}d_{1})u_{3}-(c_{1}d_{3}-c_{3}d_{1})u_{2}+(c_{2}d_{3}-c_{3}d_{2})u_{1},\end{aligned}}$
and this is the claim.
# Vector space/K/Inner product/Linear isometry/Introduction/Section
## Definition

Vector spaces/K/Inner product/Linear isometry/Definition
An isometry is always injective. For ${\mathbb {K} }=\mathbb {C}$, we also talk about an unitary mapping. As there are also affine isometries, we talk about a linear isometry.

## Lemma
### Proof

$\Box$
Therefore, an isometry is just a (linear) mapping that preserves distances. The set of all the vectors with norm $1$ in a Euclidean vector space is also called the sphere. Hence, an isometry is characterized by the property that it maps the sphere to the sphere.

## Lemma
### Proof

$\Box$

## Theorem
### Proof

$\Box$
# Vector spaces/K/Inner product/Linear isometry/Definition

Let $V,W$ be vector spaces over ${\mathbb {K} }$, endowed with inner products, and let
    $\varphi \colon V\longrightarrow W$
be s linear mapping. Then $\varphi$ is called an isometry if
    $\left\langle \varphi (v),\varphi (w)\right\rangle =\left\langle v,w\right\rangle \,$
holds for all $v,w\in V$.
# Vector spaces/K/Inner product/Linear isometry/Characterization/Fact

Let $V$ and $W$ be vector spaces over ${\mathbb {K} }$, both endowed with an inner product, and let $\varphi \colon V\rightarrow W$ be a linear mapping. Then the following statements are equivalent.
* $\varphi$ is an isometry.
* For all $u,v\in V$, we have $d(\varphi (u),\varphi (v))=d(u,v)$.
* For all $v\in V$, we have $\Vert {\varphi (v)}\Vert =\Vert {v}\Vert$.
* For all $v\in V$ fulfilling $\Vert {v}\Vert =1$, we have $\Vert {\varphi (v)}\Vert =1$.
# Journalism on Wikinews
## Overview

Wikinews is the news-based Wikimedia Foundation sister project to Wikiversity and Wikipedia.
Erik Möller, the current (2010) deputy CEO of the Foundation, is generally credited with being instrumental in getting the project off the ground. With a focus on news, the project is frequently overshadowed by current events on Wikipedia - which, naturally, attracts a far larger contributor base. There are stark differences between "The Free Encyclopedia you can Write", and "The Free News Source you can Write". Encyclopedic articles are continually developing - whereas news articles are a snapshot of what is known at a point in time.
Through negotiation with Google News, project bureaucrat Brian McNeil successfully arranged listing of the English-language Wikinews project in Google's index-cum-aggregator; adopting the Wikimedia extension for Flagged Revisions, and technically implementing a formal peer review process which parallels that of traditional journalism to meet Google's requirements and address a perennial contributor gripe.
This, by-and-large, makes Wikinews a unique publisher of Citizen journalism.

## Discussion questions

* What societal benefits can citizen journalism provide?
# Vector spaces/K/Inner product/Linear isometry/Characterization/Fact/Proof

The implications $(1)\Rightarrow (2)$, $(2)\Rightarrow (3)$ and $(3)\Rightarrow (4)$ are restrictions. $(4)\Rightarrow (3)$. For the zero vecto r,the statement $(3)$ is clear; so suppose that $v\neq 0$. Then, ${\frac {v}{\Vert {v}\Vert }}$ has norm $1$, and, because of
    $\varphi (v)=\varphi {\left(\Vert {v}\Vert {\frac {v}{\Vert {v}\Vert }}\right)}=\Vert {v}\Vert \varphi {\left({\frac {v}{\Vert {v}\Vert }}\right)}\,,$
we have
    $\Vert {\varphi (v)}\Vert =\Vert {v}\Vert \,.$
$(3)\Rightarrow (1)$ follows from fact.