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Perspectives in Digital Literacy/More Than a Number

Introduction
In the episode “Nosedive” of the dystopian science fiction television series Black Mirror, Lacie, a woman obsessed with climbing up the social ladder but unable to secure the apartment of her dreams because of her low social media score, follows the advice of a social media consultant by attempting to impress “High Fours,” people whose score is closest to her society’s standard of social achievement, the five-star rating. Conceived by Black Mirror creator Charlie Brooker as “a satire of acceptance and the image we like to portray”, “Nosedive” speaks to present social media anxieties and our excessive adoration of social media influencers. But to actually lose one’s job because of a low social rating, as the episode portrays, is far-fetched, right? Not exactly. In fact, this is exactly what may be happening in China today due to its social crediting system.

By the Numbers
China has the strongest presence on the Internet in the world, with nearly one billion users. China also has some of the strictest media censorship, and most of the websites and Web 2.0 services used in the Western world are blocked completely and subsequently replaced by “copycats” of their own: while the Western world is used to Google, the Chinese have Baidu. Instead of Twitter, the Chinese have Weibo. In place of Facebook, the Chinese have created Renren. Instead of viewing videos on YouTube, Chinese people view them on Youku and Tudou.

By keeping Chinese people on their own websites that usually record and store all their personal data within buildings in the country’s capital, Beijing, the government can constantly monitor what is done online and by whom. And now that China is implementing a national social crediting system, this government surveillance coupled with the Chinese people’s embrace of their censored Internet could lead to a disastrous result. History has shown that people in power -- especially those with access to innovative technology -- tend to favor the wealthy and elite.

Social Credit
So, what is the new Chinese social credit system and how does it work? Formally introduced in 2011 by Chinese Premier Wen Jiabao, the social credit system is the Chinese government’s project to create a unified record system to track and evaluate its citizens, businesses, and regional administrations for trustworthiness. Some regional administrations rate trustworthiness using a numerical value, while the central government uses mostly blacklisting and whitelisting. It was officially set to begin in 2020, but the deadline was pushed back because of the COVID-19 pandemic. Some aspects of it are similar to the credit checks that banks run--using credit rating companies such as Experian in the United States--before providing someone with a loan. It also resembles the way that social media value is measured in the West, which usually depends on an Internet user’s follower count or the amount of likes they receive on online posts. The Chinese system, however, will apply these ratings to all aspects of Chinese life. A citizen or business can be given a high ranking for donating to charity, or have value deducted for grievances such as playing music loudly.

Once the national system is fully implemented, Chinese citizens, businesses, and local governments will be assigned a number based on the data gathered from Chinese “big tech,” that is, the innovative technology across the country that collects personal information as well as other data gathered from websites that people and small businesses may access.

One such technology key to information gathering is Artificial Intelligence (AI) that uses facial recognition technology to monitor citizens and their movements. Since the transgressions that the Chinese government will look for can range from jaywalking and unsound business practices to evidence of sharing anti-government rhetoric, once the system is in place, AI and the cameras will be able to do most of the work in tracking the citizens and alerting the government of violations.

The negative impact of a bad score for businesses and people will also varies. As mentioned, the central government already has a running blacklist of businesses and people who they feel have acted poorly. By being blacklisted, the government publicly shames the transgressors while banning some personal rights and potentially limiting their ability to continue to do business. Some local governments have offered positive incentives for having a good score to try to increase trust in the system, such as prioritized healthcare and deposit-free housing.

China and the Internet
As stated earlier, the Chinese government blocks Western social networks such as Facebook, Twitter, and YouTube, replacing them with their own. By allowing only these websites to exist and monitoring the content their citizens are allowed to post, the Chinese government can control the narrative and opinions around the country.

Say something about a censorship law or something damning about your official? One can be expected to be taken offline, prosecuted, and even in some cases permanently blacklisted from basic human rights such as access to healthcare and transportation. Say something good or release a story about an official the government may already want ousted? Expect to gain some notoriety when the government allows your opinion to be circulated for days before they finally take it down. This type of selective censorship may raise alarms to a Chinese citizen who now faces the threat of the government applying scores that influence opinion on who is worthy of access to basic human liberties.

With China being the most populated country in the world, holding about 20% of the world’s population according to their last census, the public’s desire for a watchful eye in the form of social crediting is expected, but may also be skewed because China is an authoritarian regime:  "Overall, [the Chinese] report a high degree of approval of SCSs [social crediting systems], with 80% of [participants] either approving or strongly approving SCSs. Only 19% of [participants] perceive the SCS in value neutral terms (neither disapprove nor approve) while just 1% reported either strong or somewhat disapproval. To some extent, the high degree of approval of SCSs and the almost non-existent disapproval we found might reflect the nature of conducting a survey in an authoritarian setting—while were clearly informed that that the data were anonymized and to be used for research purposes only, some more cautious [participants] may have falsified their preferences to a degree due to concerns about expressions of disapproval resulting in reprisals from the state." This same ideology may contribute to the public’s general acceptance of their censored Internet. As Chinese people do not get the option to speak out against the government, one must hope that the new implementation of technology will cause more good than harm, especially now that the Chinese government is pushing the surveillance of children, as illustrated in the Wall Street Journal’s special report “How China Is Using Artificial Intelligence in Classrooms." At one point, the report focuses on an electroencephalography (EEG) headband that young students must wear to gauge their attention during lessons, whose output is monitored by the teacher and shared with all the students' parents. When the reporters show the footage to neuroscientist Theodore Santo of the University of California, San Francisco, he is surprised, noting that the technology is susceptible to false readings, so he is not sure how useful it would be for measuring attention. The Chinese teachers, however, explain that the use of the headbands themselves have "forced to make students more disciplined." In other words, the technology is molding the behavior of students at a very early age.

Xinjiang
Furthermore, China has long used technology to monitor and suppress their ethnic minority populations. In Xinjang, a region on the border of China and home to the country’s largest Muslim population, adherence is instilled in the forms of virtual checkpoints and the swiping of ID cards. The Chinese government has completely digitized policing and is creating a state of constant surveillance to keep the population in line. Some people are allowed to bypass these checkpoints and are not subject to digitized policing. Chris Buckley and Paul Mozur of the New York Times report that “at many checkpoints, privileged groups — Han Chinese, Uighur officials with passes, and foreign visitors — are waved through “green channels.” Although the Chinese may seek more trust and policing, this is an obvious form of targeted surveillance that can have the possibility of trickling into more minority groups.

Conclusion
With China already implementing its social crediting system and the fear of being blacklisted at the forefront, I understand that resisting the government is a daunting task. I also understand that this is a great moral issue in which an increased number of Chinese people will be stripped of their basic human rights. Technological innovation is growing at a faster pace than anyone can keep up with and blindly following the government’s application of said technology will prove disastrous for all parties, but as history has shown, the minority will always be left with the biggest burden.

I implore the Chinese people to resist the use of such a credit system. The first step would be choosing not to opt-in to the private aggregators of data. Before the mass implementation of the social crediting system, private companies have begun to gather information on an opt-in basis, I implore you to resist. The government wins every time one of their allotted big corporations is favored. Small businesses and anyone with hiring power should also disregard consideration of the crediting scores when making hiring decisions. By doing this, the people could nullify the crediting system and leave it to the government to use it solely. Though Chinese citizens might think that the benefits of a surveillance state (including safety and security) outweigh the harms outlined in this paper, I ask you to reconsider. Further, I argue that the risk of loss of human rights and liberty for minor transgressions such as jaywalking should outweigh any benefits.

At the end of “Nosedive,” Lacie is finally overwhelmed by her lifelong desire to secure a high score. She realizes that her desire to fit into the pastel-perfect life of her disingenuous friend is made up of a societal standard that will never allow her to truly live a fulfilling life. Within a day, she was heaved to the bottom of the social ladder, along with the very people she ignored and chastised her entire life. By drinking out of a red flask of whiskey that finally released her from her flawed perception,* she was able to finally live again and see the people for who they really are, past their assigned scores. I ask Chinese citizens to do the same: take the red flask and see past whatever social crediting system may be implemented among them. Your children, your neighbor, and yourself will thank you in the long run. We are more than a number.

Notes

* This is reference to a scene from the 1999 sci-fi film The Matrix, where the protagonist is given the choice between learning the hard truth by taking a red pill or living in a comfortable illusion by taking a blue pill.

Further learning

 * Antonia Hmaidi. "'The' Social Credit System in China." (Video). 35C3. 27 December 2018. A lecture for the 2018 Chaos Computer Club Congress that seeks to provide a more nuanced view on the Chinese Social Credit System than what the Western media offers.
 * Victoria Adelmant. "Social Credit in China: Looking Beyond the 'Black Mirror' Nightmare." Center for Human Rights and Global Justice, New York University School of Law. April 20, 2021. A discussion of the human rights implications of the Chinese Social Credit system. With many additional resources and explanations of the organizations involved.
 * Eunsun Cho. "The Social Credit System: Not Just Another Chinese Idiosyncrasy." Journal of Public and International Affairs. 1 May 2020.

Discussion questions

 * What are forms of financial, work performance, and social credit scoring in the United States? How are they similar to the Chinese system? How are they different? In your opinion, could the institutions that run our financial and our social credit scores be turned into a loosely unified system similar to that of China?
 * What kind of citizen is the Chinese government attempting to obtain through its system of credit rewards and punishments? What could be the reasons for making every citizens' score public, accessible, and displayable? What could be the reasons for deploying monitoring technologies in school?
 * What could be some unintended consequences of manipulating a population's social behaviors?


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Solutions To Mathematics Textbooks/Algebra (9780132413770)/Chapter 1

Exercise 1.7
We show that

$$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & n & n(n+1)/2 \\ 0 & 1 & n \\ 0 & 0 & 1 \end{pmatrix}$$.

When $$n = 1$$, the equation holds. Assume it holds for $$n = k$$ and consider $$n=k+1$$. Then we have

$$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & k & k(k+1)/2 \\ 0 & 1 & k \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & k+1 & (k+1)(k+2)/2 \\ 0 & 1 & k+1 \\ 0 & 0 & 1 \end{pmatrix}$$.

Exercise 3.4
Using row operations we can simplify a matrix to the row-echelon form. If also column operations are allowed, starting from the first row, we can scale and subtract the pivot elements from all the non-zero elements on a row. Hence, we can bring the matrix to a form where only the pivot elements are ones, and the rest are 0. If the matrix has full rank, only row operations are enough.

Exercise 4.6
Let $$ A \in \mathbb{R}^{n \times n}, D \in \mathbb{R}^{k \times k}, B \in \mathbb{R}^{n \times k} $$ and $$M = \begin{pmatrix} A & B \\ 0 & D \end{pmatrix} \in \mathbb{R}^{n+k}$$. We want to show that $$\det M = \det A \cdot \det D$$.

To show this, we use the formula (1.6.4) for the determinant: $$\det M = \sum_{p \in S_{n+k}} \textrm{sign}(p) m_{1, p(1)}\cdots m_{n, p(n)}\cdots m_{n+1, p(n+1)}\cdots m_{n+k, p(n+k)} $$. Since $$m_{n+i, p(n+i)} = 0 $$ whenever $$1 \leq i \leq k, p(n+i) \leq n $$, the products in the sum are nonzero only when $$p $$ is a permutation such that $$p(\{1,2,\ldots,n\}) = \{1,2, \ldots,n\} $$and $$p(\{n+1,n+2,\ldots,n+k\}) = \{n+1,n+2, \ldots,n+k\} $$. Therefore we can write

$$\begin{alignat}{1} \det M  & = \sum_{\pi \in S_n, \sigma \in S_k} \textrm{sign}(\pi)\textrm{sign}(\sigma) m_{1, \pi(1)}\cdots m_{n, \pi(n)}\cdots m_{n+1, \sigma(1)}\cdots m_{n+k, \sigma(n+k)} \\ & = \sum_{\pi \in S_n} \textrm{sign}(\pi)m_{1, \pi(1)} \cdots m_{n, \pi(n)} \sum_{\sigma \in S_k} \textrm{sign}(\sigma)m_{1, \sigma(1)} \cdots m_{n, \sigma(k)} \\ & = \det A \cdot \det D

\end{alignat} $$.

Exercise 5.1
$$(12)(13)(14)(15) = (15432) $$

$$(123)(234)(345) = (12)(54) $$

$$(1234)(2345) = (12453) $$

$$(12)(23)(34)(45)(51) = (2345) $$

Exercise 5.2
Consider the permutation $$p = (1324) $$.

 * a) The permutation matrix associated to p is $$P = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\end{pmatrix} $$.
 * b) $$p = (1324) = (14)(12)(13) $$
 * c) sign(p) = -1.

Exercise 6.2
Let $$A \in \mathbb{Z}^{n \times n} $$. Claim: $$A $$ is invertible and $$A^{-1} $$ has integer entries if and only if $$\det A = \pm 1 $$.

Assume '''$$A^{-1} $$''' exists and has integer entries. First notice that from the determinant formula

$$\det A = \sum_{p \in S_{n}} \textrm{sign}(p) a_{1, p(1)}\cdots a_{n, p(n)} $$

We immediately see that if the entries $$a_{i,j} $$ are integral, then the determinant must be integral as a sum of products of integers. Conversely, if the determinant is not an integer, at least one of the entries has to be non-integral.

Next we observe that since '''$$\det(A) \neq 0 $$, $$\det(AA^{-1}) = \det(A)\det(A^{-1}) = 1 \Longrightarrow \det(A^{-1}) = \frac{1}{\det(A)} $$'''. So unless $$\det A = \pm 1 $$, $$\det (A^{-1}) $$ is not an integer and hence $$A^{-1} \notin \mathbb{Z}^{n \times n} $$ contradicting the assumption.

Then assume that $$\det A = \pm 1 $$. Then $$A^{-1} $$ exists and the cofactor matrix formula (Theorem 1.6.9) tells us that $$A^{-1} = \frac{1}{\det A} \textrm{cof}(A) $$. The entries of the cofactor matrix are given by $$\textrm{cof}(A)_{i,j} = (-1)^{i+j} \det A_{ji} $$, where $$A_{ji} $$ is the matrix $$A $$ with row $$i $$ and column $$j $$ removed. Clearly if $$A \in \mathbb{Z}^{n \times n} $$, the cofactor matrix then has integral entries and if $$\det A = \pm 1 $$, also the inverse has integral entries.

Exercise M.8
a) The only problem here is assuming that $$L $$ would be the right inverse of $$A $$. Necessarily, if $$A \in \mathbb {R}^{m \times n} $$, we have to have $$L \in \mathbb {R}^{n \times m} $$, in which case the product $$AL $$ is not even defined.

b) The sequence is correct, and shows that the equality $$AX = ALB $$ holds. If $$L $$ was also the right inverse of $$A $$, we would necessarily have $$m = n $$ in order to have the product $$AL $$ defined.

Exercise M.11
a) We have the variables $$x_{0,1},x_{-1,0},x_{0,0},x_{1,0},x_{0,-1}$$and the boundary conditions $$\beta_{2,0}=\beta_{1,-1}=\beta_{-2,0}=\beta_{-1,-1}=\beta_{0,-2}=0$$ and $$\beta_{1,1}=\beta_{0,2}=\beta_{-1,1}=1$$. Using the discrete Laplace equation these conditions translate to the equations

$$\begin{array}{llcr} (0,1): & \beta_{0,2} + x_{0,0} + \beta_{1,1} + \beta_{-1,1} -4 x_{0,1} & = & 0 \\ (-1,0): & \beta_{-1,1} + x_{0,0} + \beta_{-2,0} + \beta_{-1,-1} -4 x_{-1,0} & = & 0 \\ (0,0): & x_{1,0} + x_{-1,0} + x_{0,1} + x_{0,-1} -4 x_{0,0} & = & 0 \\ (1,0): & \beta_{2,0} + x_{0,0} + \beta_{1,1} + \beta_{1,-1} -4 x_{1,0} & = & 0 \\ (0,-1): & \beta_{0,-2} + x_{0,0} + \beta_{-1,-1} + \beta_{1,-1} -4 x_{0,-1} & = & 0 \\

\end{array}$$,

which simplifies to the linear system

$$\begin{pmatrix} -4 & 0 & 1 & 0 & 0 \\ 0 & -4 & 1 & 0 & 0 \\ 1 & 1 & -4 & 1 & 1 \\ 0 & 0 & 1 & -4 & 0 \\ 0 & 0 & 1 & 0 & -4 \\ \end{pmatrix} \begin{pmatrix} x_{0,1} \\ x_{-1,0} \\ x_{0,0} \\ x_{1,0} \\ x_{0,-1} \\ \end{pmatrix} = \begin{pmatrix} -3 \\ -1 \\ 0 \\ -1 \\ 0 \\ \end{pmatrix}$$.

The solution to this system is given by multiplying the equation from the left with the inverse of the coefficient matrix.

b) Assume the maximum value is achieved on some point $$x_{u,v}$$ inside the region $$R$$. Since $$x_{u,v}$$ is the average of its four neighbors, one of the neighbors must have a larger value than $$x_{u,v}$$, contradicting the assumption.

c) Let $$A $$ be the matrix obtained from writing the linear system of the discrete Dirichlet problem with entries $$a_{ij}$$. We have $$a_{ii} = -4$$ and $$a_{ij} \in \{0, 1\}$$ for $$i \neq j$$. Using the diagonal element of each row, we can eliminate the corresponding column for every other row. There are at most 4 non-zero, non-diagonal elements on each column, each having the value at most 1. Hence, when conducting the row eliminations, we will never eliminate a diagonal element. Therefore, we can reduce the matrix $$A$$ to the row-echelon form with no rows of only 0 elements. Hence, the matrix is invertible and so the linear system has a unique solution.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text: User:Olena.wiki.ua

title: User:Olena.wiki.ua


text:
Wikijunior:How Things Work/Telescope

Who invented it?
Hans Lippershey was a German-Dutch eyeglass maker who is believed to the inventor of the telescope as he tried to patent it in 1608, in his patent he claimed it was "for seeing things far away as if they were nearby", but some believe that he was not the first person to make a telescope, he also did not get the patent, since other eyeglass makers also discovered how to make a telescope, this kind of telescope uses lenses and is called a refracting telescope or refractor for short.

It's not really known how Lipperhey actually discovered how to make a telescope, one legend says that he did not discover it himself but some children in his shop who were playing with lenses, finding out that a distant weather vane appeared closer if viewed through two lenses, others believe that Lipperhey copied the design.

The reason why early telescopes were discovered by eyeglass makers is because they had experience working with lenses and optics.

Lipperhey did not call it the telescope however, he called it the "Dutch Perspective Glass" the word telescope wasn't coined until three years later by a scientist named Giovanni Demisiani.

These early telescopes also did not have a lot of power, the power of a telescope is how near it can make far away things look, early telescopes only had a power of about 3x, meaning that a far away object will only look 3x bigger, that's not a lot, due to how they worked, the image was also upside down.

Galileo Galilei was the first scientist to improve and work on telescopes, his first attempts weren't very powerful, but he eventually made a telescope with 30x power, he also found out how to make the image the right way up, by using a concave eyepiece and not a convex one, it was good enough for him to observe the moon with, and he also sold them, they were popular among sailors.

How dangerous is it?
Most things are safe to look at through them, but NEVER look at the sun or another bright object through a telescope, doing so can damage your eyes.

Using a telescope to look inside someones house and spy on them is against the law.

Bigger telescopes are really heavy so don't drop them.

Always put caps on your telescope and keep it away from windows when your not using it, as they can focus the sun and start a fire.

The lenses or mirrors are made of glass and glass can be really sharp when broken.

What does it do?
It makes things that are far away look closer, a good telescope may also make it brighter.

How does it vary?
Some use mirrors, these are called reflecting telescopes, they are popular with amateur astronomers, some telescopes use lenses, these are called refracting telescopes, the kind covered in this article, some telescopes use both lenses and mirrors, these are called catadioptric (cat-a-die-op-trick) telescopes.

A small collapsible refractor for looking at distant objects on land is called a spyglass or terrestrial telescope, it's what pirates are said to use.

A really small refracting telescope with a low power like 8x that can fit in your hand is called a monocular.

Two refracting telescopes connected to each other are called binoculars, unlike a monocular telescope, it allows you to see depth as it uses both eyes.

Telescopes for looking at the night sky are usually on a tripod stand and something called a mount, which lets the telescope move or "look around" this is because astronomical telescopes are so powerful that you cannot hold one because your natural shakiness will make the image look really shaky.

The most common kind of telescope mount is the altitude-azimuth, it lets the telescope pan left and right, and tilt up and down.

How has it changed the world?
The telescope changed in the world in the fact that there's nothing else like it when you need one, a telescope or binoculars is the first thing you look for when you see something strange in the distance, they were used on sea before binoculars were invented to see distant objects.

Telescopes let people watch dangerous things from a safe distance as if they were next to it, this is very useful for scientific research of dangerous things like explosions.

Some camera lenses are like telescopes, they are called telephoto lenses, they are used by sports and wildlife photographers to make subjects look very close in pictures.

Telescopes are sometimes attached to guns to make shooting distant objects easier, telescopes are also used by surveyors, with something called a crosshair, a telescope can be precisely pointed to a small object in the distance, the telescopes used by surveyors are called theodolites (thee-odd-o-lights) they measure angles on a precisely graduated mount, the telescope makes them very accurate and easy to use to sight distant targets.

The field which telescopes changed the most is astronomy, early astronomers just looked at the sky with their own eyes, but powerful telescopes let us see distant planets and other celestrial objects, these telescopes can be very big, some huge telescopes are even sent into space, such as the Hubble space telescope, there are even radio telescopes, which make an image out of radio waves.

How does it work?
The principle behind the refracting telescope is actually so simple you can make one yourself with a few pieces, early refracting telescopes usually had no less then 3 or 4 pieces.

Refracting telescopes work by bending light, called refraction, with glass lenses, the light is bent or focused to a single point.

In a simple refracting telescope, there are two lenses:

The lens at the front, or the objective, is a big lens whose purpose is to gather light and focus it to a point, it's optical power and diameter can determine how zoomed in the image is going to be, the image from this lens is focused to a point but is not viewable by the human eye, it needs something else to focus onto and that's what the eyepiece is for.

So named because it's the lens you look through to use the telescope, it's smaller then the objective lens, this is because both lenses have to be different sizes, or the image will not be magnified, the properties of the eyepiece also affect the image, this is why high end telescopes allow the user to swap in different eyepieces, to make the image bigger, smaller, brighter or darker depending on application.

The objective lens is mounted at the end of a tube matching the diameter of the lens, in the back of this tube is inserted another tube containing the eyepiece, so the eyepiece tube can be pulled in and out, this is because the telescope needs to be focused for a clear image of what is to be viewed through it, that is the eyepiece needs to be on or around the focal point of the objective for a clear image to form.

The view through this kind of simple telescope is always upside down, this is because the light coming through the top of the objective gets bent downwards and reaches the bottom of the eyepiece, and the light from the bottom of the objective gets bent upward and reaches the top of the eyepiece, this is usually corrected by putting a special prism in the telescope tube, or using an eyepiece which has a negative optical power, these are usually concave, as opposed to standard lenses which are convex (outward).

Refracting telescopes can be very complex and expensive, using more then one lens to make the image look clearer and less-prone to a rainbow effect seen on the edges of bright objects.

Refracting telescopes reached their peak in the 1890s but now are rarely used for astronomical purposes in favor of reflecting telescopes.

What idea(s) and/or inventions had to be developed before it could be created?
Optics and lenses, Optics is the study of how light acts and behaves, lenses are pieces of glass that focus light to a single point.

title: Wikijunior:How Things Work/Telescope


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The Legend of Zelda: Link's Awakening (2019 video game)/Storming the Castle

World Updates
Leaving Bottle Groto, head east to just north of Dampé's shack. Lift the boulder there to unlock a fast travel point.

Return Bow Wow to Madam MeowMeow to be rewarded with a kiss.

Visit the Town Tool Shop to see they are now carrying Bombs! The price is 10 bombs for 10 rupees. Link can carry up to 30, and that's how many should be purchased now.

Link can now enter the Dream Shrine in the north end of town, however he is ill prepared to tackle it. Hold off on visiting it for now.

With the power bracelets, it should now be possible to reach Crazy Tracy's. Crazy Tracy's is just north of the Witches hut. Crazy Tracy offers secret medicine for 28 rupees. Buy it to get an item which, upon Link running out of hearts, is consumed to revive Link automatically! Link can only ever hold one dose of secret medicine at a time since Crazy Tracy will not allow him to buy more while he is holding one.

Exploring Ukuku Prairie
Go to Ukuku Prairie from the village and enter the cave to the north. Push things out of the way to get the piece of heart in the bottom right corner. Exit and repeat to make a path to the chest, which contains 50 rupees.

Stand on the stone grey pedestal to create a very useful warp point.

At the ridge below the cemetery, bomb the cracked wall to open a cave with a great fairy. A regular fairy also spawns here, which is useful for refilling bottles.

South of that cave, bomb the skull rock to clear a path.

Climb up and down the plateau, then go south around pothole field. This eventually leads to Richard's Villa. Richard wants five golden leaves from the castle. From here until the end of the quest, collected golden leaves will appear under the hearts in the upper left corner of the screen.

Follow the signs to reach the castle. Logically, the front door of the castle is locked tight. Take a detour just to the east to visit Seashell mansion, and dig in the glowing dirt to the right of the mansion to establish a warp point.

Go northeast to find Kiki the Monkey. Trade the bananas as outlined in the Trading Sequence. This causes the monkeys to build a bridge.

Old Man Ulrira will mention where the golden leaves are on the phone.

Cut the bush west of the telephone booth to unveil stairs. Do a short platforming section to get inside the castle. The knights here are much like moblins in tactics. Let them strike your shield to stun them, and then strike back to beat them.

Golden Leaf Locations
On the castle perimeter there is a pentagon of holes with a center hole. An enemy will pop out of the holes and toss a bomb at Link. This is basically whack a mole, so try to predict where it comes up and strike it. Eventually it is beat, and it gives up a golden leaf.

On the western wall there is a crow relaxing on a tree. Provoke it into attacking Link by placing a bomb at the tree, and then strike the bird down when it attacks Link in response. As a result, the crow yields a golden leaf.

Go inside the castle. In the first room, follow the blue carpet to fight two knights and a red slime. When they are defeated Link receives a golden leaf.

Head north to immediately face two knights. Continue right to face more enemies. Press the button to open the front gate to the castle. This makes coming and going to the castle easier.

Head up the stairs to enter a room with a patrolling energy ball and two cracked walls. The right wall contains an enemy knight. The left wall also contains an enemy knight, but this one is holding a golden leaf. Defeat it to claim the leaf.

Head to the right to see a preview of the castle boss. Use the balcony to renter the castle through the right section. To open the door in this room, throw a pot at it. To beat the boss, hold your shield up and then strike quickly when possible. When it goes down the final golden leaf is yielded.

The Reward
Return the golden leaves to Richard and he grants Link a reward, stepping back to allow access. Push the statue to the right to reveal a hidden staircase. Descend into the tunnels to begin the process of claiming it.

Mind the green slimes and head down the left fork. Push the boulder down the pit, then jump over the pit to reach the treasure chest. This chest contains a Secret Shell.

Head back to the fork and this time go northeast. Head up the stairs. Exit the cavern and enter pothole field. This treacherous expanse of grass and shrub is littered with pits. Slash forward with Links sword, and move slowly to safely progress. In a pinch, bombs can clear shrubs from a distance. Head to the outer perimeter, then shimmy along it to proceed through the field. The dungeon key is hidden in the dirt just south of the owl statue. Dig there uncover the key, then press A to grab the Slime Key.

Continue left to make it to the dungeon key way. How convenient. Use the key to open the bars to the dungeon. However there is no direct path to the dungeon from the keyway. Head northeast to see Tarin waiting by a bee hive on a tree. He notices the stick Link has and wants it. Give it to him and watch him knock down the hive. After Tarin is chased away by the angry bees, Link gains the now abandoned honeycomb.

Continue south to reach a pond dotted with small islands. From the southeastern shore, leap from island to island to the west, until Link reaches the western shore.

Before entering the dungeon, climb up to the area just above the entrance and go to the patchy grass area. Dig one diagonal square from the northwest corner to find a secret seashell hidden there.

Now enter Key Cavern, the next dungeon.


text: The Legend of Zelda: Link's Awakening (2019 video game)/Pieces of Heart


text:
Wikijunior:Dinosaurs/Nigersaurus

Nigersaurus taqueti, was a dinosaur that lived on earth 115 to 110 million years ago, during the Cretaceous period.

What was their body shape?

Nigersaurus taqueti was small for a dinosaur, and was only 9 meters (29 and a half feet) long.

What did they eat?
This dinosaur was a herbivore, and consumed vast quantities of vegetation. Most of the vegetarian this dinosaur ate was close to the ground. This dinosaur had more than 500 teeth, which it regularly replaced every 14 days.

When did they live?
This dinosaur lived 115 to 110 million years ago, during the Cretaceous period.

What was their habitat?
This dinosaur lived in a Riparian zone, a floodplain near water.

How were they discovered?
This dinosaur was found in the Gadoufaoua region of the country of Niger, a desert region now famous for it's dinosaur fossils. It was first discovered in 1976.

title: Wikijunior:Dinosaurs/Nigersaurus


text:
0 A.D./Musicians

On June 8th, 2018 the group Materia Collective published a soundtrack for the game.

title: 0 A.D./Musicians


text:
User talk:193.114.113.173

--ЗAНИA talk 05:34, 18 June 2021 (UTC)

title: User talk:193.114.113.173


text:
Solutions To Mathematics Textbooks/Algebra (9780132413770)/Chapter 2

Exercise 1.1
$$a(bc)=ab=a$$ and $$(ab)c=ac=a$$. If $$1,a \in S$$ and $$a \neq 1$$, we have $$1 \cdot a = 1$$, so the only set $$S$$ with this composition law and identity is the set $$S = \{1\}$$.

Exercise 2.3
a) $$y = x^{-1}w^{-1}z$$

b) $$yzx = x^{-1}xyzx = x^{-1}x=1 $$, but $$yxz \neq 1 $$ in general. Consider for example the permutations $$x = (12), y = (13), z = (123) $$, so $$xyz = (12)(13)(123) = 1 $$, but $$yxz = (13)(12)(123) = (132) $$.

Exercise 2.4
a) $$GL_n(\mathbb{R}) $$ is a subgroup of $$GL_n(\mathbb{C}) $$

b) $$\{-1,1\} $$ is a subgroup of $$\mathbb{R}^\times $$.

c) Positive integers are not a subgroup of $$\mathbb{Z}^+ $$, as they do not contain inverse elements of even 0.

d) The positive reals form a subgroup of $$\mathbb{R}^\times $$.

e) The set $$H $$ is not a subgroup of $$GL_2(\mathbb{R}) $$, as it does not contain the identity matrix.

Exercise 2.5
Let $$H $$ be a subgroup of $$G $$. If $$H $$ has an identity element (possibly different from the identity of $$G $$), then we have $$1_H h = h = 1_G h $$, and multiplying both sides by $$h^{-1} $$ shows the identity elements are the same. Similarly if $$h $$ has an inverse in $$H $$, we necessarily have $$hh^{-1}_H = 1 = hh^{-1}_G $$ implying the inverses are the same.

Exercise 3.1
Using the standard Euclidean algorithm we get $$\gcd(321, 123) = 3 $$ and $$-18 \cdot 321 + 47 \cdot 123 = 3 $$.

Exercise 3.2
Let $$a, b \in \{1,2,3,...\} $$ be positive integers such that $$a+b $$ is a prime number. Then, if $$\gcd(a,b) = d $$, we have $$a = a_0 d, b = b_0d $$, so $$a + b = d(a_0 + b_0) $$. Because we want $$a + b $$ to be prime, the only possibility is $$d = 1 $$.

Exercise 4.3
Let $$a,b \in G$$ and $$k$$ such that $$(ab)^k = 1$$. Then $$(ab)^{k-1} = (ab)^{k-1} abb^{-1}a^{-1} = (ab)^k (ab)^{-1} = (ab)^{-1}$$ so $$(ba)^k = baba\ldots baba = b (ab)^{k-1} a = b (ab)^{-1} a = b b^{-1}a^{-1}a = 1$$.

Exercise 4.4
Assume that a group $$G $$ doesn't have a proper subgroup. Then

 * $$G $$ must be finite. If $$G $$ is infinite, take any $$g\in G $$ such that $$g $$ doesn't generate $$G $$, and consider the set $$H = \{g^k | k \in \mathbb{Z}\} $$. Since $$g $$ does not generate $$G $$, $$H $$ is a proper subgroup of $$G $$, as is easy to show.
 * $$G $$ has to be cyclic. Indeed, assume $$G $$ is not cyclic, so no single element generates $$G $$. Then, for any element $$g \in G $$, $$H = \{g^k | k \in \mathbb{N}\} $$ is a subgroup of $$G $$, since $$G $$ is finite.
 * $$G $$ has to have a prime number order. For contradiction, assume the order of $$G $$ is $$pq $$. Then, since $$G $$ is a finite cyclic group, generated by an element $$g $$, we have that $$H = \{g^{pk} | k \in \mathbb{N}\} $$ is a proper subgroup.
 * Finally, if $$G $$ has a prime order, no element of $$G $$ generates a proper subgroup of $$G $$. This is because any element is of the form $$g^k $$ for some element $$g \in G $$ and integer $$k $$. From Proposition 2.4.3 we know that the order of $$g^k $$ is $$n/d $$, where $$n $$ is the order of the group and $$d = \gcd(n, k) $$. In this case $$n $$ is a prime number, so the order of $$g^k $$ is either $$n $$ or 1 (in the case that $$k = n $$).

Exercise 4.5
Let $$G = $$ be a cyclic group. Take any subgroup $$H \subseteq G$$. We assume $$G $$ is finite (as the finite case has the main focus in the book, and in fact the definition on page 46 implicitly assumes so), so that $$H $$ is finite as well. Let $$H = \{g^{k_1}, \ldots, g^{k_m}\} $$. Then, since $$H $$ is a subgroup, for every $$a_1, \ldots a_m \in \mathbb{Z} $$ it holds $$g^{a_1k_1} \ldots g^{a_mk_m} = g^{a_1k_1 + \ldots a_mk_m} \in H $$. Hence, for the exponents $$q \in \mathbb{Z}$$ for which $$g^q \in H$$, it holds $$q \in \mathbb{Z}k_1 + \ldots + \mathbb{Z}k_m$$. Theorem 2.3.3 tells us that we can write $$\mathbb{Z}k_1 + \ldots + \mathbb{Z}k_m = \mathbb{Z}d$$ for some integer $$d$$. This shows that any element of $$H $$ is of the form $$g^{kd} $$ so $$H =  $$.

Exercise 4.8 b
Elementary matrices of the first type are matrices $$E$$ with the entries $$e_{ii} = 1$$, $$e_{ij} = a$$ for exactly one pair of indices $$i,j$$ and 0 otherwise. It is easy to see from the determinant formula that the determinant of such matrices is always 1. From the product rule for the determinant it follows that all products of such matrices have determinant 1.

For the other direction, we consider the 2 by 2 case first and let $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ with $$ad-bc = 1$$. Adding a scaled row to the other row and using the relation between the entries, we can manipulate the matrix as

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \rightarrow \begin{pmatrix} 1 & b+\frac{d(1-a)}{c} \\ c & d \end{pmatrix} \rightarrow \begin{pmatrix} 1 & \frac{d-1}{c} \\ c & d \end{pmatrix} \rightarrow \begin{pmatrix} 1 & \frac{d-1}{c} \\ 0 & 1 \end{pmatrix}\rightarrow \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$.

Since we can arrive to the identity matrix using only operations corresponding to multiplication with matrices of the first type and the condition $$ad-bc = 1$$, the original matrix $$A$$ can be produced from the identity by reversing the operations. Hence, we can generate $$A$$ with only elementary matrices of the first type.The general case follows by induction: for any $$M \in \mathbb{R}^{n \times n}$$ matrix with determinant 1, assume we can manipulate $$M$$ into the form

$$M' =\begin{pmatrix} I_{n-1} & b \\ c^T & d \end{pmatrix} $$,

where $$b,c \in \mathbb{R}^{n-1} $$ and $$d \in \mathbb{R} $$. It is easy to see that we can further manipulate $$M' $$ as follows:

$$\begin{pmatrix} I_{n-1} & b \\ c^T & d \end{pmatrix} \rightarrow \begin{pmatrix} I_{n-1} & b \\ 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} I_{n-1} & 0 \\ 0 & 1 \end{pmatrix}

$$.

Note that in the second step the element $$d$$ manipulates to 1 as we eliminate the row entries of the vector $$c$$, since otherwise the determinant of the manipulated matrix would not be equal to 1.

Exercise 4.9
Elements of order 2 necessarily swap a pair of elements, or two pairs of elements. Such permutations are $$(12), (13), (14), (23), (24), (34), (12)(34), (13)(24), (14)(23) $$ totaling 9 permutations or order 2.

Exercise 4.11
a) Any transposition that swaps elements $$a, b $$ has a permutation matrix $$T $$ that is an identity matrix where rows $$a, b $$ have been swapped. These are elementary matrices of the second type. Any permutation matrix $$P $$ corresponding to a permutation in $$S_n $$ is an identity matrix whose rows have been permuted according to the permutation. From the permutation interpretation of $$P $$ we know that $$P $$ is invertible, so by Theorem 1.2.6 it is a product of elementary matrices. It is easy to see that only elementary matrices of type 2 are essential in such a product, as scaling or adding rows together are not needed. Hence, $$P $$ can be decomposed into a product of elementary matrices of the second type, implying that the permutation is a product of transpositions.

b) First we show the following claim:

Claim: Any product of two transpositions can be written as a product of three-cycles.

Proof: Let $$(ab)(cd) $$ be a product of two transpositions, where $$a,b,c,d $$ are some integers (possible some equal). Then we can write $$(ab)(cd) = (acb)(acd) $$.

Since any permutation in $$S_n $$ is generated by transpositions, and the determinant of a transposition matrix is $$-1 $$, any permutation in $$A_n $$ is a product of an even number of transpositions. Let $$\pi \in A_n $$ be a permutation, with a decomposition into transpositions $$\pi = \tau_1\cdots\tau_{2n} $$. Then we can group the product as $$\pi = (\tau_1\tau_2)(\tau_3\tau_4)\cdots(\tau_{2n-1}\tau_{2n}) $$, where each product of two transpositions is also expressible as a product of three-cycles as shown in Claim.

Exercise 5.3
Let $$A, B \in U$$ so that $$AB = \begin{pmatrix} a_{11} & a_{12} \\ 0 & a_{22} \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \\ 0 & b_{22} \end{pmatrix} = \begin{pmatrix} a_{11}b_{11} & a_{11}b_{12} + a_{12}b_{22} \\ 0 & a_{22}b_{22} \end{pmatrix}$$. Then $$\varphi(AB) = a_{11}^2b_{11}^2 = \varphi(A)\varphi(B)$$ so $$\varphi $$ is a group homomorphism. We have $$\ker \varphi = \{A \in U : a_{11} = 0\}$$ and $$\textrm{im}~\varphi = \mathbb{R}^\times$$.

Exercise 6.2
Any homomorphism $$\varphi: \mathbb{Z}^+ \rightarrow \mathbb{Z}^+$$ satisfies $$\varphi(a + b) = \varphi(a) + \varphi(b)$$, so $$\varphi$$ must be a linear function with integer coefficient. In other words, $$\varphi(k) = m \cdot k$$ with $$m \in \mathbb{Z}$$. For $$\varphi$$ to be injective, it suffices that $$m \neq 0$$. For $$\varphi$$ to be surjective, we need $$m \in \{-1, 1\}$$, since clearly $$\textrm{im}~\varphi = \mathbb{Z}^+m$$. The surjective functions $$\varphi$$ are also bijective, so they are isomorphisms.

Exercise 6.6
Two elements $$a,b \in G$$ are conjugate, if for some $$g \in G$$, $$ga = bg$$. It is easy to see that for the given matrices, any matrix of the form $$\begin{pmatrix} 0 & s \\ s & t \end{pmatrix}$$ with $$s \neq 0$$ does the job. This matrix is also invertible, as its determinant is $$-s^2$$. Since the determinant is negative, the matrices are not conjugate in $$SL_2(\mathbb{R})$$.

Exercise 7.1
Let $$a,b \in G$$, and $$\sim$$ be a relation given by $$a \sim b$$ if and only if there exists $$g \in G$$ such that $$a = gbg^{-1}$$

Reflexivity: $$a = 1 a 1^{-1}$$, so $$a \sim a$$.

Symmetry: if $$a \sim b$$, then $$a = gbg^{-1}$$, so $$b = g^{-1}ag$$. Hence $$b \sim a$$, and vice versa.

Transitivity: if $$a \sim b$$ and $$b \sim c$$, then $$a = gbg^{-1}$$ and $$b = hch^{-1}$$, so $$a = ghch^{-1}g^{-1} = ghc(gh)^{-1}$$. Therefore $$a \sim c$$.

Exercise 7.5

 * The set $$\{(s, s): s \in \mathbb{R}\}$$ defines an equivalence relation on $$\mathbb{R}$$ that is the same as the usual "=" relation on the reals.
 * The relation defined by the empty set satisfies symmetry and transitivity, but fails reflexivity.
 * The locus $$\{(x, y): xy + 1 = 0, x,y \in \mathbb{R}\}$$ satisfies symmetry and transitivity, but fails reflexivity (for example $$0 \sim 0$$ is not true).
 * The locus $$\{(x, y): x^2y - xy^2-x+y = 0, x,y \in \mathbb{R}\}$$ is reflexive ($$x \sim x$$ because $$x^3 - x^3 -x + x = 0$$), symmetric (since $$x^2y - xy^2-x+y = -(y^2x - yx^2-y+x)$$) and transitive (If $$(x, y)$$ and $$(y, z)$$ are solutions to the equation, then by symmetry also $$(y, x)$$ and $$(z, y)$$ are. Therefore also $$(z, x)$$ and thus $$(x, z)$$ is a solution.)

Exercise 7.6
The number of equivalence relations in a set is equal to the number of partitions of the set. If the set has 5 elements, we have the following partitions:

 * Partitions to 1 set: 1
 * Partitions to 2 sets of sizes 1, 4: 5
 * Partitions to 2 sets of sizes 2, 3: $$\binom{5}{2}\binom{3}{3} = 10$$
 * Partitions to 3 sets of sizes 1, 1, 3: $$\frac{1}{2}\binom{5}{1}\binom{4}{1}\binom{3}{3} = 10$$
 * Partitions to 3 sets of sizes 1, 2, 2: $$\frac{1}{2}\binom{5}{1}\binom{4}{2}\binom{2}{2} = 15$$
 * Partitions to 4 sets of sizes 1, 1, 1, 2: $$\frac{1}{6}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{2}{2} = 10$$
 * Partitions to 5 sets of size 1 each: 1

The total number of partitions is thus 52.

Exercise 8.3
Let $$G$$ be a group such that $$|G| = p^k$$ for some prime $$p$$. For any $$g \in G$$, $$$$ is a subgroup of $$G$$ and by Lemma 2.8.7 has order $$p^m$$ where $$m \leq k$$. If $$m = 1$$, we are done as we have found an element of order $$p$$. Otherwise, consider $$h = g^{p^{m-1}}$$ and notice that $$h^p = \left(g^{p^{m-1}}\right)^p = g^{p^m} = 1$$, so $$h$$ has order $$p$$.

Exercise 8.4
For a group $$G$$ of 35 elements we have a few cases.

 * If the group is cyclic, i.e., $$G = $$, we have that $$g^5$$ has order 7 and $$g^7$$ has order 5.
 * If the group is not cyclic, then for every $$h \in G$$ there is some integer $$k < 35$$ such that $$$$ is a subgroup of $$G$$ order $$k$$. The order of $$$$ divides the order of $$G$$, so if $$h$$ is not the identity, the order of $$h$$ is either 5 or 7. Assume $$h$$ is of order 5 and there is another element $$x$$ of order 5 that is not a power of $$h$$. Then the subgroup of $$G$$ generated by $$h$$ and $$x$$ has order 25. Indeed, all elements $$h^ix^j$$ for $$i,j \in \{0, \ldots, 4\}$$ are distinct (if not, we have for $$i_1\neq i_2, j_1 \neq j_2$$ that $$h^{i_1}x^{j_1} = h^{i_2}x^{j_2} \Rightarrow h^{i_1-i_2} =x^{j_2-j_1} $$, which implies that $$x$$ is a power of $$h$$), and there is 25 ways to choose the exponents. But 25 does not divide 35, so there cannot be another subgroup of order 5, and thus any element $$x$$ that is not a power of $$h$$ must generate a cyclic group of order 7.

Exercise 8.5
If a group $$G$$ contains an element $$g$$ of order 6 and an element $$h$$ of order 10, then $$$$ and $$$$ are subgroups of $$G$$ and thus 6 and 10 divide the order of $$G$$. Hence, we can say that $$|G|\geq 30$$.

Exercise 8.10
Let $$H\subset G$$ be a subgroup such that $$\left[G:H \right] = 2 $$. Then, since left (and right) cosets partition the group, we have $$G = H \cup gH = H \cup Hg$$ for some $$g \notin H$$. Because $$|H| = |gH| = |Hg|$$, we must have $$gH = Hg$$ and so by Proposition 2.8.17 $$H$$ is normal.

For an example that this is not true for $$H\subset G$$ such that $$\left[G:H \right] = 3 $$, consider $$G = S_3$$ and $$H = \{1, (12)\}$$. Since $$|H| = 2$$, we have indeed $$\left[G:H \right] = 3 $$. Then, $$(13)H = \{(13), (123)\}$$ and $$H(13) = \{(13), (132)\}$$, so $$H$$ is not normal.

Exercise 8.12
Claim 1: For all $$h, g \in S$$ we have $$hg \in S$$.

Proof: Assume that $$hg \notin S$$. Then $$hS \neq S$$, since $$hg \in hS$$. But since $$1 \in S$$, we have $$h \in hS$$, which contradicts the assumption that the cosets of $$S$$ partition $$G$$.

Claim 2: For all $$h \in S$$ we have $$h^{-1} \in S$$.

Proof: Assume $$h^{-1} \in aS$$ for some $$a \in G$$. Then we have $$h^{-1} = ag$$ for some $$g \in S$$. This implies $$1 = agh$$, and by Claim 1 we have $$hg \in S$$, so $$ahg \in aS$$. But then $$1 \in aS$$, so unless $$aS = S$$ we have a contradiction.

Claim 1 and Claim 2 together prove that $$S$$ is a subgroup of $$G$$.

Exercise 9.4
The numbers involved are so small that solving by brute force becomes viable. To solve $$2x \equiv_9 5$$ we note that $$2\cdot 5 = 10 \equiv_9 1$$ so $$2^{-1} \equiv_9 5$$ so $$x \equiv_9 5\cdot 5 \equiv_9 25 \equiv_9 7$$.

To investigate the congruence $$2x \equiv_6 5$$ we note by computing every possibility that there is no element $$2^{-1}$$ modulo 6. Therefore there is no solution to the congruence.

Exercise 9.5
Given equations

$$\begin{cases} 2x-y \equiv_n 1 \\ 4x+3y \equiv_n 2 \end{cases}$$

we can solve $$y \equiv_n 2x-1$$ and substitute to the other congruence equation to obtain $$10x \equiv_n 5$$. This equation has a solution exactly when there exists $$10^{-1}$$ modulo $$n$$, or in other words an integer $$m$$ such that $$10m \equiv_n 1$$. This translates to saying that the equation $$10m + an = 1$$ must have an integer solution $$a$$. But that is only possible when $$\gcd(10, n) = 1$$.

Exercise 10.1
It is not clear from the problem statement what the answer should look like. Therefore, the following candidate for answer might not be what is intended.

We show that each cycle of the form $$(a_1a_2\ldots a_n)$$ has sign $$(-1)^{n-1}$$. This is easy to see, as we can decompose $$(a_1a_2\ldots a_n) = (a_1a_n)(a_1a_{n-1})\cdots(a_1a_2)$$. Each transposition corresponds to an elementary matrix of type two, which has determinant $$-1$$. The product form of the cycle has $$n-1$$ transpositions, so the sign is as claimed. Therefore, if a permutation has a cycle decomposition to $$m$$ cycles, each having $$n_i$$ terms, the sign of the permutation is $$(-1)^{\sum n_i-1} $$.

Exercise 10.3
Let $$G = , G' = $$ and $$\varphi(x^i) = y^i$$. We have $$\ker \varphi = \{1, x^6\}$$, and $$G$$ has the subgroups $$<1>, , , , , G$$ and $$G'$$ the subgroups $$<1>, , , G'$$. The correspondences given by $$\varphi $$ are thus Notice that the subgroups $$<1>$$ and $$<x^4>$$ do not contain $$\ker \varphi$$.

Exercise 11.3
Let $$G = <x>, H = <y>$$ and assume $$K = G \times H$$ is generated by an element $$(x^n, y^m)$$. Then, for every element of $$k \in K$$ we would have $$k = (x^n, y^m)^i$$ for some $$i$$. But since $$G, H$$ are infinite, there are no $$i_1, i_2$$ such that $$(x^n, y^m)^{i_1} = (x, 1)$$ and $$(x^n, y^m)^{i_2} = (1, y)$$, since the first condition would require $$n \neq 0, m = 0$$ and the second $$n = 0, m \neq 0$$. The assumption that $$G, H$$ are infinite means that $$x^n = 0 \Leftrightarrow n = 0$$ and $$y^m = 0 \Leftrightarrow m = 0$$.

Exercise 11.4
a) $$G = R^\times$$ is isomorphic to $$H \times K = \{-1, 1\} \times \{x \in \mathbb{R} : x > 0\}$$ by the function $$\varphi: G \rightarrow H \times K$$ given by $$\varphi(g) = \left( \frac{g}{|g|}, |g| \right)$$ which has the inverse $$\varphi^{-1}((h, k)) = h \cdot k$$. Seeing that $$\varphi$$ is a homomorphism is a direct consequence of multiplication rules of real numbers and properties of absolute value.

b) Let $$G $$ be the set of invertible upper triangular matrices, $$H $$ the set of invertible diagonal matrices and $$K$$ the set of upper diagonal matrices with ones on the diagonal. In order to have a homomorphism $$\varphi: G \rightarrow H \times K$$, we must have

$$\varphi\left( \begin{pmatrix} g_1 & g_2 \\ 0 & g_3 \end{pmatrix} \right) = \left( \begin{pmatrix} h_1 & 0 \\ 0 & h_2 \end{pmatrix}, \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \right)$$,

for appropriate reals $$g_i, h_i$$ and $$k$$. Using this notation for we have

$$\varphi\left( \begin{pmatrix} g_1 & g_2 \\ 0 & g_3 \end{pmatrix} \cdot \begin{pmatrix} g'_1 & g'_2 \\ 0 & g'_3 \end{pmatrix} \right) = \varphi\left( \begin{pmatrix} g_1g_1' & g_1g_2' + g_2g_3' \\ 0 & g_3g_3' \end{pmatrix} \right) = \left( \begin{pmatrix} h_1h_1' & 0 \\ 0 & h_2h_2' \end{pmatrix}, \begin{pmatrix} 1 & h_1k' + h_2k \\ 0 & 1 \end{pmatrix} \right) $$. (1)

On the other hand,

$$\varphi\left( \begin{pmatrix} g_1 & g_2 \\ 0 & g_3 \end{pmatrix} \right) \cdot \varphi \left( \begin{pmatrix} g'_1 & g'_2 \\ 0 & g'_3 \end{pmatrix} \right) = \left( \begin{pmatrix} h_1 & 0 \\ 0 & h_2 \end{pmatrix}, \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \right) \cdot \left( \begin{pmatrix} h'_1 & 0 \\ 0 & h'_2 \end{pmatrix}, \begin{pmatrix} 1 & k' \\ 0 & 1 \end{pmatrix} \right) = \left( \begin{pmatrix} h_1h_1' & 0 \\ 0 & h_2h_2' \end{pmatrix}, \begin{pmatrix} 1 & k + k' \\ 0 & 1 \end{pmatrix} \right)$$. (2)

For $$\varphi$$ to be a homomorphism, we require these two to evaluate equal. However, the second coordinate in (1) has a dependency on $$h_1$$ and $$h_2$$, whereas (2) does not have it. Hence, (1) and (2) are not equal in general.

c) Let $$G = \mathbb{C}^\times $$, $$H = (\{\theta \in \mathbb{R} : \theta \in [0, 2\pi] \}, +) $$ (the angles of a circle with sum as the group operation) and $$H = \mathbb{R}_{> 0}^\times $$. Then $$G$$ is isomorphic to $$H \times K$$ via the homomorphism $$\varphi: G \rightarrow H \times K$$ given by $$\varphi(g) = \left( \arg(g), |g| \right)$$. That $$\varphi$$ is a bijective homomorphism follows easily from the polar form representation of complex numbers.

Exercise 12.1
Let $$H = \{1, (12)\}$$ be a subgroup of $$S_3$$. Clearly $$H $$ is not normal, as $$(13)H = \{(13), (123)\} \neq \{(13), (132)\} = H(13)$$. Then $$(13)H = \{(13), (123)\}$$ and $$(23)H = \{(23), (132)\}$$. It is a straightforward computation to check that $$(12)H(23)H = \{1, (12), (23), (132)\}$$. This set has 4 elements, but all cosets must have size 2, so it is not a coset.

Exercise 12.5
Let $$G $$ be the group of upper triangular matrices $$\begin{pmatrix} a & b \\ 0 & d \end{pmatrix} $$ with $$a, d \neq 0 $$.

a) Let $$S $$ be the subset of $$G $$ defined by $$b = 0

$$, i.e., the set of invertible diagonal matrices. It is easy to see that $$S $$ is a group, as the inverse for each such matrix is also a diagonal matrix, and the product of two diagonal matrices is a diagonal matrix, and the identity is a diagonal matrix. $$S $$ is however not a normal subgroup, since for any $$\begin{pmatrix} g_1 & g_2 \\ 0 & g_3 \end{pmatrix} \in G $$ with inverse $$\begin{pmatrix} g_1^{-1} & -g_2/(g_1g_3) \\ 0 & g_3^{-1} \end{pmatrix} \in G $$ we can compute

$$\begin{pmatrix} g_1 & g_2 \\ 0 & g_3 \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix} \begin{pmatrix} g_1^{-1} & -g_2/(g_1g_3) \\ 0 & g_3^{-1} \end{pmatrix} = \begin{pmatrix} ag_1 & dg_2 \\ 0 & dg_3 \end{pmatrix} \begin{pmatrix} g_1^{-1} & -g_2/(g_1g_3) \\ 0 & g_3^{-1} \end{pmatrix} = \begin{pmatrix} a & -ag_2/g_3+dg_2/g_3 \\ 0 & d \end{pmatrix} $$,

which is not in $$S $$ when $$a \neq d

$$.

b) Let $$S $$ be the subset of $$G $$ defined by $$d = 1

$$. Again, it is easy to see that $$S $$ is a subgroup of $$G $$. As above, we can compute

$$\begin{pmatrix} g_1 & g_2 \\ 0 & g_3 \end{pmatrix} \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} \begin{pmatrix} g_1^{-1} & -g_2/(g_1g_3) \\ 0 & g_3^{-1} \end{pmatrix} = \begin{pmatrix} ag_1 & bg_2+1 \\ 0 & g_3 \end{pmatrix} \begin{pmatrix} g_1^{-1} & -g_2/(g_1g_3) \\ 0 & g_3^{-1} \end{pmatrix} = \begin{pmatrix} a & -ag_2/g_3 + bg_2/g_3 + g_3^{-1} \\ 0 & 1 \end{pmatrix} \in S $$,

so $$S $$ is a normal subgroup. The cosets are the sets containing the matrices

$$\begin{pmatrix} g_1 & g_2 \\ 0 & g_3 \end{pmatrix} \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} ag_1 & bg_2+1 \\ 0 & g_3 \end{pmatrix}

$$,

which are distinguished by the parameter $$g_3 \neq 0 $$, as the variables $$a,b $$ can be chosen arbitrarily (as long as $$a \neq 0

$$). Hence, the quotient group is isomorphic to $$\mathbb{R}^\times $$. The homomorphism whose kernel $$S $$ is is given by $$\varphi \left( \begin{pmatrix} g_1 & g_2 \\ 0 & g_3 \end{pmatrix} \right) = g_3 $$.

c) Let $$S $$ be the subset of $$G $$ defined by $$a = d \neq 0 $$. Then again $$S $$ is clearly a subgroup of $$G $$. We have

$$\begin{pmatrix} g_1 & g_2 \\ 0 & g_3 \end{pmatrix} \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \begin{pmatrix} g_1^{-1} & -g_2/(g_1g_3) \\ 0 & g_3^{-1} \end{pmatrix} = \begin{pmatrix} ag_1 & ag_2 + bg_1 \\ 0 & ag_3 \end{pmatrix} \begin{pmatrix} g_1^{-1} & -g_2/(g_1g_3) \\ 0 & g_3^{-1} \end{pmatrix} = \begin{pmatrix} a & bg_1/g_3 \\ 0 & a \end{pmatrix} \in S $$,

so $$S $$ is a normal subgroup.To investigate the cosets of this group, we compute

$$\begin{pmatrix} g_1 & g_2 \\ 0 & g_3 \end{pmatrix} S = \left\{ \begin{pmatrix} ag_1 & ag_2 + bg_1 \\ 0 & ag_3 \end{pmatrix} : a \neq 0,b \in \mathbb{R} \right\}

$$.

Here we can choose $$b $$ as we like, so the top-right element doesn't impose any conditions on the cosets. Then, the elements of $$G $$ for which it holds $$g_1 \neq 0, g_3 = 1 $$ generates a coset of $$S $$ uniquely. This way we also obtain all cosets of $$S $$, since for every choice of $$g_1, g_3 \neq 0 $$ we can choose $$a = 1/g_3 $$ and we see that this element is in the coset generated by $$g_1' =g_1/g_3, g'_3 = 1 $$.

The quotient group is again isomorphic to $$\mathbb{R}^\times $$ with the homomorphism whose kernel $$S $$ is given by $$\varphi \left( \begin{pmatrix} g_1 & g_2 \\ 0 & g_3 \end{pmatrix} \right) = g_1/g_3 $$.

Exercise M.6 a, b
a) We observe the following:

 * Reflexivity: A point $$a \in \mathbb{R}^k$$ is connected to itself by the path $$X(t) = a$$. If $$a \in S$$, then the path is contained in $$S$$.
 * Symmetry: Let $$a,b \in \mathbb{R}^k$$ such that there exists a path $$X$$ from $$a$$ to $$b$$ contained in $$S$$. Then the path $$Y(t) = X(1-t)$$ is a path from $$b$$ to $$a$$ contained in $$S$$
 * Transitivity. If $$a,b,s \in \mathbb{R}^k$$ such that $$X$$ is a path from $$a$$ to $$b$$ contained in $$S$$ and $$Y$$ from $$b$$ to $$c$$ contained in $$S$$, then $$Z(t) = \begin{cases} X(2t), & \text{if } t < 1/2 \\ Y(2t-1), & \text{if } t \geq 1/2 \end{cases}$$ is a path from $$a$$ to $$c$$ contained in $$S$$.

b) Path connection is an equivalence relation as shown in part a). Hence, path connected subsets partition the set $$S$$. In particular, the transitivity property ensures that if two points can be path connected, any points connected to those two points can be connected with both of the points.

Exercise M.7
a) Let $$A,B,C,D \in G \subset GL_n(\mathbb{R})$$ such that $$A \sim B$$ given by $$X(t)$$ and $$C \sim D$$ given by $$Y(t)$$. Then $$Z(t) = X(t)Y(t)$$ connects $$AC$$ to $$BD$$. This path is also in $$G$$, since both paths $$X(t)$$ and $$Y(t)$$ are in $$G$$ and $$G$$ is a group.

b) We show that for any matrix $$A \sim I$$, also the matrix $$BAB^{-1} \sim I$$ for any $$B \in GL_n(\mathbb{R})$$. If $$X(t)$$ is the path joining $$A$$ and $$I$$, simply consider the path $$Y(t) = BX(t)B^{-1}$$, since $$Y(1) = BIB^{-1} = I$$. Hence, all matrices connected to identity by path form a normal subgroup.

Exercise M.8
a) Elementary matrices of the first type are matrices $$E$$ with the entries $$e_{ii} = 1$$, $$e_{ij} = a$$ for exactly one pair of indices $$i,j$$ and 0 otherwise. For each such $$E$$, there is a path to $$I$$ just by setting $$X(t)$$ as the same matrix as $$E$$, with the exception that the element $$e_{ij} = (1-t)a$$. Clearly this is a continuous path from $$E$$ to $$I$$, and since $$X(t)$$ is an elementary matrix of the first type, it also has determinant 1, and thus the path stays in $$SL_n(\mathbb{R})$$. Now, if we have the elementary matrices of the first type $$E_1, E_2 $$, we have by M.7 a) that $$E_1E_2$$ is connected to $$I$$, as $$E_1, E_2 $$ are connected to $$I$$. We have now shown that any product of elementary matrices of type 1 are path connected to $$I$$, which implies that $$SL_n(\mathbb{R})$$ is path connected, as the elementary matrices of type 1 generate $$SL_n(\mathbb{R})$$.

b) Let $$A \in GL_n(\mathbb{R})$$ such that $$\det(A) > 0$$. We can path connect $$A$$ to a matrix with determinant 1 by $$X(t) = \sqrt[n]{1 - t + \frac{t}{\det(A)}}A $$ so that $$\det(X(t)) = \left(1 - t + \frac{t}{\det(A)}\right) \det(A) $$. This path is composition of continuous functions since $$\det(A) > 0$$ and $$X(t) $$ has always positive determinant, so it is a continuous path in $$GL_n(\mathbb{R})$$. Therefore, matrices with $$\det(A) > 0$$ form a connected component of $$GL_n(\mathbb{R})$$. By the same reasoning, if we just assumed that $$\det(A) < 0$$, we can reason that the matrices with $$\det(A) < 0$$ form a connected component of $$GL_n(\mathbb{R})$$.

What remains to show is that these components are not connected. Let $$A$$ be a matrix with a positive determinant and $$X(t) $$ a path connecting $$A$$ to a matrix $$B$$ with a negative determinant. Then, $$X(t) $$ has elements $$x_{ij}(t) $$ that are continuous functions. From the determinant formula (1.6.4) we see that the determinant is then also a continuous function of the elements, as it is a sum of product of continuous functions. Then, since $$\det(X(0)) > 0 $$ and $$\det(X(1)) < 0 $$, we must have $$\det(X(t_0)) = 0 $$ for some $$t_0 \in (0, 1) $$. Therefore $$X(t) $$ is a path not contained in $$GL_n(\mathbb{R})$$, and hence $$GL_n(\mathbb{R})$$ is a disjoint union of two connected components.

Exercise M.9
Double cosets partition a group. Indeed, for every $$g \in G$$, the double coset $$HgK$$ contains $$g$$, since $$1 \in H \cap K$$. On the other hand, if $$g \in Hg_1K \cap Hg_2K$$, then there are elements $$h_1, h_2 \in H, k_1, k_2 \in K$$ such that $$h_1g_1k_1 = h_2g_2k_2$$. This implies that $$g_1 = h_1^{-1}h_2g_2k_2k_1^{-1}$$, and since $$H,K$$ are subgroups, $$h_1^{-1}h_2 \in H, k_2k_1^{-1} \in K$$, so $$g_1 \in Hg_2K$$. This implies in turn that $$Hg_1K = Hg_2K$$.

Exercise M.14
Multiplying from the left with the matrices $$E,E'$$ and their inverses corresponds to adding and subtracting rows. Multiplying from the right corresponds to adding and subtracting columns. Now, consider any $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z})$$.

Claim 1: We can bring the matrix to the form $$\begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix}$$ where $$a' = 1$$ or $$c' = 1$$ only by adding and subtracting rows.

Proof. If $$c = 0$$, we have $$d \neq 0$$ since the matrix has determinant 1, so we can add the second column to the first one to obtain another matrix with determinant 1, whose entry on the first column, second row is non-zero. So we assume from now on that $$c \neq 0$$. Using the row operations, we can perform what is essentially the Euclidean algorithm on the first column by subtracting the smaller column 1 element from the larger. Since $$\det(A) = 1$$, we have $$ad - bc = 1$$, which implies that $$\gcd(a,c) = 1$$ and so the row operation Euclidean algorithm terminates once it produces 1 on one of the rows of the 1st column.

Now, using the matrix produced in Claim 1, we can simply bring the matrix to the identity form only by applying row and column additions/subtractions. Hence, we have $$E_1\cdots E_n A F_1\cdots F_m = I$$, where $$E_i, F_j$$ are some of the matrices $$E,E'$$ and their inverses.This can be solved to show that $$A = E_1^{-1}\cdots E_n^{-1} F_1^{-1}\cdots F_m^{-1} $$.


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User talk:JJwaleng

My Topic:' Evangelism:Is The Preaching Of The Gospel Of Jesus Christ.Now Everyone attends different churches all over the world but we all speaketh about God The Almighty And Everything.But We Never Ebable In Our Minds Who Is God;How Does He Function.The Gospel Of Jesus Christ Is Preached In Many Diffferent ways across the borders of--~ t all countries but what is the message potrayed here!Watch Prophet Uebert Angel As he speaketh about God

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Brainstorming session with students on latest development in space technology

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User:Woble ItohG

I'm Woble ItohG popularly known as Blessing,i'm more curious to know what is going on in the world at large.

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User talk:95.78.48.6

Please check your edits to Children's Authors/Margaret Lapina before you save them. I have just undone one recent edit which affected the formatting of the page. There are still issues with the page now - this could be because you are using indentation when adding text which is not necessary and will make the text look strange on Wikibooks.--ЗAНИA talk 04:04, 19 June 2021 (UTC)

title: User talk:95.78.48.6