## And That Is The Rest Of The Story

Ask me anything   throw idea's my way!   Hello everyone, my name is Aidan Patterson, and I am the owner of my own thoughts and actions. My hobbies include Mathematics, Karate, Physics and anything music related. Never be afraid to tell me I am wrong, but be prepared to prove it, and may I say; good luck proving it to me!

Solutions to the general equations

ax+b=0

ax^2+bx+c=0

ax^3+bx^2+cx+d=0

ax^4+bx^3+cx^2+dx+e=0

There does not exist a general solution by radicals for equations of degree 5 or more - a fact that can be proved using Galois Theory and the fact that Sn is not solvable for n greater than or equal to 5.

— 18 hours ago with 305 notes

Honey, I don’t think you gotta worry about that anytime soon.

A few things.

1) Your TAGS are DISGUSTING. You are a misogynistic, fat-shaming, rape apologist. The idea that a woman, regardless of how she looks, would be OPTIMISTIC to be raped is just stomach churning.

2) The image you have posted is CLEARLY PHOTOSHOPPED. Here is this woman’s original picture with her ACTUAL message:

Thanks for proving her point.

You gross ass sexist crumblefuck.

I didn’t even make the picture, I fucking found it on the internet. Thanks for not being able to take a fucking joke.

Oh wow here we fucking go again, but let’s go in categories this time to make this easy for you.

Rape

There is so much more than can be said but fuck you, go look that shit up yourself.

Fat Shaming

• Fat shaming is damaging, bullying and wrong period. It is fucked up to make fun of someone for any reason. The only reason why someone should be spoken to in a negative fashion is if they have actually done something wrong and harmful. (Like you did!)
• The BMI scale is a crock of shit. It doesn’t actually determine what you as an individual need to be functioning properly. It goes by a shitty, rigid “weight by height” requirement. It doesn’t take into account personal family genetics, for instance.
• Some people are just fat. That’s it. Genetics, disorders, pregnancy, diseases, illness and body composition just fucking happen to some people. Body shaming does not make fat go away.
• White culture leads to rigid beauty standards: thin, white, able-bodied and minded, symmetrical face with fat in all the “right places”. Only about 10% of women actually fit this “standard” and are ultimately starving for it.
• Fat people suffer from not only peer abuse and media ridicule, but eating disorders and other heinous things to try and match their thinner counterparts.
• Fat people are systematically oppressed. IE they don’t get jobs like their thin counterparts based on people thinking they are “lazy” and other things.
• A case in Australia about how a couple could not adopt a child because the woman was too fat. (Also, England.)
• None of this shit actually has to do with these people as individuals, their skill sets, etc. This is discrimination against them as being fat only, regardless if they can “help it” or not. (IE genetics versus simply lifestyle.)

So fuck you.

THANKS FOR THE SOURCES! I AM USING THEM IN AN ENGLISH ESSAY!

— 1 week ago with 3138 notes

"Wear pink in support of anti-bullying"

I did my best :)

— 1 week ago with 26 notes
#me  #gay  #pink  #bullying  #school
Fractional Calculus

Hey everyone, sorry I haven’t been posting things for a while, I’ve been swamped with homework.

Anyway, I was in Advanced Functions the other day, and I was reading through a book on applications of differential equations. At one point, the author makes note of the fact that you can do ordinary algebra on derivative function if you are careful and write them in operator form. As a comment on this he said: “it is possible to extend these cases to those where n is not an integer.” Meaning he wished to take a non-integer number of derivatives.

This was kind of strange to think about at first, but made a lot of sense afterwards. When you want to take a derivative, you can use the limit definition to figure it out, but you could also just use a general form. For instance, the general form for the nth derivative of x^k is:

$D^nx^k=\frac{k!}{(k-n)!}x^{k-n}$

If I wanted to extend this to the field of rational numbers, I would have to find a way to extend the factorial function to the rational numbers. Thankfully, this was already done for me by Euler, and improved by Gauss, Weierstrass and Legendre, who used analytic continuation to prove that

$n!=\Gamma(n+1)=\int_{0}^{\infty}u^ne^{-u}ds$

This isn’t as random as it seems, the integral follows naturally from something called Laplace transformations, but that’s for another post.

How do we prove all of those people right? Well,

$\Gamma(n+1)=\int_{0}^{\infty}u^ne^{-u}ds$

If we integrate by parts, we get:

$\Gamma(n+1)=(u^n)(-e^-u)|^{\infty}_0-\int_{0}^{\infty}(nu^{n-1})-(e^{-u})du$

$\Gamma(n+1)=n\int_{0}^{\infty}u^{n-1}(e^{-u})du$

$\Gamma(n+1)=n\Gamma(n)$

This is called the recursive form of the gamma function, and from this definition, it is easy to see that it will act like the factorial function for all n values:

$\Gamma(1)=1!,\, \Gamma(2)=2\cdot \Gamma(1)=2!,\, \Gamma{3}=3\cdot \Gamma(2)=3!...$

With this in mind, we can rewrite the nth derivative of x^k as:

$D^nx^k=\frac{\Gamma(k+1)}{\Gamma(k-n+1)}x^{k-n}$

Since the gamma function can handle non-integer numbers, the derivative can be extended to non-integer derivatives.

This  can be extended to other functions as well, such as e^(ax)

$D^ne^{ax}=a^ne^{ax}$

no work needs to be done on this one, unless you’re really ambitious and want to take an irrational derivative of this function when a is negative, but I’ll leave that one up to you to figure out.

There is more too! You could extend this idea so that you can take imaginary derivatives and complex derivatives as well!

That’s about all I have time for today, but let me know if you if you thought this was cool. This area of math is called fractional calculus, and it encompasses the taking of derivatives and integrals of non-integer order. If you would like a cool resource for this kind of mathematics, I found an online pdf that explains the stuff I left out.

Happy mathing :)

— 1 week ago with 26 notes
#mathematics  #maths  #math  #physics  #mathema  #calculus  #derivative  #integral  #gamma  #function  #Euler

# Computers are providing solutions to math problems that we can’t check

Good news! A computer has solved the longstanding Erdős discrepancy problem! Trouble is, we have no idea what it’s talking about — because the solution, which is as long as all of Wikipedia’s pages combined, is far too voluminous for us puny humans to confirm.

A few years ago, the mathematician Steven Strogatz predicted that it wouldn’t be too much longer before computer-assisted solutions to math problems will be beyond human comprehension. Well, we’re pretty much there. In this case, it’s an answer produced by a computer that was hammering away at the Erdős discrepancy problem.

Full Story: Io9

(via dieboldt)

— 1 week ago with 723 notes
An old mathematical puzzle soon to be unraveled?.

It is one the oldest mathematical problems in the world. Several centuries ago, the twin primes conjecture was formulated. As its name indicates, this hypothesis, which many science historians have attributed to the Greek mathematician Euclid, deals with prime numbers, those divisible only by…

— 2 weeks ago with 26 notes
Why $$0!=1$$? Idea of factorials →

This is another arithmetic rule that is difficult to understand intuitively. But firstly, what does the symbol “$$!$$” mean?

A factorial of a natural number $$n$$ is denoted as $$n!$$ and its value is given by the recurrence function $$f(n)=n\cdot (n-1)!$$ where $$f(0)=0!=1$$. Note that…

Alternate proof!

I once asked my math teacher this question, as I was having a hard time deriving the Taylor series of e to the x, which is written as the sum from n=0 to infinity of (x^n)/(n!). I wanted to start at n=1 and add (1)/(2^n)because the concept of 0! didn’t make sense to me. His reply was: "We use 0! a lot in stats class, and we say that it is one, simply because it fits in with what we need it too."

I HATED THIS SO MUCH!

So I went and investigate, and I found out that:

$n!>\sqrt{2\pi n}\left(\frac{n}{e} \right )^n$

meaning 0! has to be greater than 0, This alone should convince you that 0! isn’t 0, but why would it be one? this answer is a lot more complicated.

A long time ago, Gauss proved that,

$n!=\Gamma (n+1)$

Where the gamma function (the half “t" looking thing) is expanded to, (with the second part being Euler’s expansion of the gamma function)

$\Gamma(z)=\int_{0}^{\infty}t^{z-1}e^{-t}dt=\lim_{n \rightarrow \infty}\frac{n^zn!}{\left ( \prod_{k=0}^{n}(z+k) \right )}$

Long story short, and all the math aside, these are just functions of z, and it turns out, gamma(1) is 1, which means that 0! is 1.

you can check it for yourself, evaluate the integral:

$\Gamma(1)=\int_{0}^{\infty}\frac{1}{e^t}dt$

Happy proofing :)

— 1 month ago with 26 notes
#math  #mathematics  #maths  #physica  #mathema

Reports emerged this weekend that Kazakh mathematician Mukhtarbay Otelbaev has published a proposed solution to the Navier-Stokes existence and smoothness problem, one of the seven Millennium Prize problems offered by the Clay Mathematics Institute. Today I want to explain some of the background of this problem, what is known about Otelbaev’s proposed solution, and what a solution would mean for fluid dynamics.

The Navier Stokes Equation

The Navier-Stokes equation is one of the governing equations of fluid dynamics and is an expression of conservation of momentum in a fluid. With the exception of a few very specific and simplified cases, there is no known general solution to equation. Instead, the equation, or a simplified model, is solved numerically using supercomputers as part of direct numerical simulation (DNS) or other forms of computational fluid dynamics (CFD). These methods allow scientists and engineers to solve the equations of fluid motion for practical problems from flow through a pipe to flow around a re-entering spacecraft.

Existence and Smoothness

Although the Navier-Stokes equation has been known for more than 150 years and can be solved numerically for many situations, some basic mathematical aspects of the equation have not yet been proven. For example, no one has proven that a general solution always exists in three-dimensions and that the energy of such a solution is bounded at all points. Colloquially, this is known as the Navier-Stokes existence and smoothness problem. The Clay Mathematics Institute has a very specific problem statement (PDF) asking for a proof (or counter-proof) of the existence and smoothness of the Navier-Stokes equation for an incompressible fluid in three-dimensions. Otelbaev contends that he has provided such a proof.

Otelbaev’s Proposed Solution

Mukhtarbay Otelbaev is an experienced mathematician with numerous papers addressing related mathematical problems. His latest paper, entitled “Existence of a strong solution to the Navier-Stokes equation,” is freely available online (PDF, in Russian, with an English abstract at the end). There is an ongoing project to translate the paper into English, and mathematicians are already evaluating the validity of this proposed solution. From what I can gather of the paper, it specifically address the Millennium Prize problem and presents Otelbaev’s proposed solution for the existence and smoothness of an incompressible fluid in three dimensions with periodic boundary conditions.

What It Means

As with any announcement of a major technical breakthrough, skepticism is warranted while experts evaluate the proposal. If the mathematical community upholds the validity of Otelbaev’s proof, he may be offered the Millennium Prize and other honors. More importantly, his solution could lead to a better understanding of the nature of the equation and the flows it describes. It is not, in itself, a general solution to the Navier-Stokes equation, but it may be a stepping stone in the path toward one. In the meantime, scientists and engineers will continue to rely on a combination of theory, experiment, and computation to progress our understanding of fluid dynamics.

For More

The story of Otelbaev’s proof and the community’s evaluation of its validity is on-going. You can follow @fyfluiddynamics and the #NavierStokes hashtag on Twitter for updates and commentary. I’d like to specially thank Catriona Stokes, Praveen C, David Sarma, and Glenn Carlson for their helpful links and observations as this story develops.

(via stochastastic)

— 1 month ago with 787 notes
#math  #mathematics  #maths  #physics  #mathema
Fun Sequence Question

Okay, so I was in math today, and I decided to be a bit of a troll and see if I could stump ,my teacher using infinite series (because were are learning about discrete functions), and I gave him a funny one.

So, say you have the series

f(n)= (-1)(n-1)

if we wanted to sum this series, we would say that:

Sn=1-1+1-1+1-1…

Now, there are a few ways we could do this. One way would be to group terms in two’s beginning at the first term.

Sn=(1-1)+(1-1)+(1-1)…

=0+0+0…

=0

But, we could also group them in two’s starting with the second term

Sn=1+(-1+1)+(-1+1)…

Sn=1+0+0…

Sn=1

Alright, we have two answers so far, but why not go for one more, this time more algebraic?

Sn=1-1+1-1+1-1…

(-1)+Sn=-1+1-1+1-1+1…

[(-1)+Sn]+Sn=0

2Sn-1=0

Sn=1/2

This is pretty cool, but which one is it? The answer may or may not even exist, but why not give it a try!?

— 1 month ago with 38 notes
#Math  #Mathematics  #Maths  #Mathema  #Physics
alwayssearchingforthings asked: I JUST HAD MY FIRST GOOD (live) MATH LECTURE SINCE AT LEAST A YEAR AND A HALF AGO sorry I'm kind of excited

I AM EXTREMELY PROUD OF YOU!! :D

— 1 month ago with 6 notes

Oh! Sorry, I forget sometimes that not everyone knows all of the terms. A McDojo is a school that cares more about monetary gain than anything else; at these schools, guys will change belts extraordinarily quickly, or you’ll have kids as young as 11 with black belts that they claim are on par with the adult belts. They would rather make money than be a credible school, because the more students they get the more money they make. It’s like this:
Kid A wants to do karate. Kid A’s parents take him to a McDojo, where he instantly learns flashy moves and techniques. Kid A’s parents pay maybe $50/month. Kid A’s friends see how quickly he moved up in rank. So ten of Kid A’s friends go to the same school, each paying$50/month. See what I mean now?

While Kid B also wants to learn karate. He finds a  different school, a school where he actually learns good technique with solid teaching. Kid B earns belt slower, but learns better stuff.

Kid A and Kid B meet up. Kid A has black belt, Kid B still has colored belt. They spar,  Kid B easily dominates. Kid A is confused.

Kid A: “but I have a black belt, and can do spinning kicks? What?”

Kid B: *shrugs, and leaves*

That’s what the rest of us have dealt with. I was Kid B. Taking five years to get my first black belt, it definitely did look odd to some. They assumed that I wasn’t as smart or talented or hard-working as Kid A.
Not true. I don’t regret those five years, or the three and a half it took me to get my second black belt. I did very well against McDojo black belts as a colored belt. And it was because my training was tough, structured, and had a clear purpose.

i think it took me 3 to earn my belt.

but i went every day at home and at the dojo, and took my black belt test in front of the masters, even training in okinawa for two weeks

93 signs that a dojo is a Mc’dojo

$50? HA! MORE LIKE$1000 a month in aus for the trophy catching mcdojo’s

Whoa, wait, did you just add an extra cero to exaggerate, or are you serious a \$1000? xO

gosh darn it, i meant 100, and that is if you pay for the whole year up front, it gets worse the smaller the the amount of time you are paying for

because everyone wants to get payed, and rent in a good place in aus is crazy expensive

I can understand setting a price to pay rent and win some on the side for yourself for teaching the knowledge. If you want to live off of martial arts, and you have the rank to do it, it’s understandable. But there are way to many that do it as a business, a gym where you train to win competitions, instead of a place for knowledge and learning the way of martial arts as a way of life.

Is it wrong then for a dojo to advertise themselves for fitness and competitions, so that people come in and learn the “way of martial arts” and knowledge while they are there? I joined karate for a hobby. Once I joined I got interested in the teachings.

I’ve never gone to a place that advertised as such. I’ve gone to tournaments and I’ve gotten more fit but I’ve never gone to a place that would directly say that those were two things to gain.

Hey! I thought I would share a few things about my training in Karate.

I started when I was three in Barry Ontario learning Shotokan Karate. In this school, I practiced for 4 years, and left with my green belt ( the grading system put this as the fourth belt, where as black was the ninth).

I moved after my fourth year to Stratford Ontario, where I took up Okinawan Goju Ryu karate at a local dojo run by Shihan Chuck Hassin (Facebook [ you can see me on the left of the cover photo] and website)

I spent 9 years relearning everything I did at the old dojo, as well as going through equivalency training and gradings. My Shihan took this process very seriously, as he is well known for having one of the most reputable schools in Ontario, and possibly Canada. My Shihan was not easy on me, and put me through a four hour blue belt grading (5th belt) when I was 10 (with people in their late teens), which I walked away from crying. I passed, but it was not fun.

I thought gradings before the back belt grading were taken seriously, but my black belt grading in june of last year (after 9 years of training with my current Shihan, and 4 years of training with my previous Sensi) was absolutely grueling. To even be allowed the option of participating, I was required to write a 90 page essay on how karate has effected my life (I still have it, I wouldn’t recommend reading it though XD). The actual grading was on the 21st of June, and it was a balmy 35 degrees outside, while just eight of us were being graded in a tiny wooden dojo in Toronto (2.5 hour drive). The grading lasted four hours as well, but we were being graded by four 6th degree or higher black belts, one of which was Shihan Ron Yamanka, a ninth degree black belt who founded the YKKF (Youdansha Kobujutsu Karate-Doh) in Canada. I also passed this grading as well, and learned more amazing things than I care to write down right now.

My Shihan did push me to get my black belt at 16 because I knew what I was doing, but does not do so for others as often. You must be 16 to learn the kata’s which are required for the grading (again, I got a bit of an exception for this, and started learning them at 15), and so no one under 16 is allowed to go to the gradings.

This is the kind of push a good dojo will give. The other one in Stratford is a mixed karate kickboxing sort of thing (absolutely a Mcdojo), and I constantly get students coming up to me and saying that we should spar because we have the same belt. What I find weird is that they not only just started, but that they have the need to prove themselves which is obviously not a virtuous quality taught in the martial arts.

I have to say that I did get lucky twice though, so I’m very grateful for that. I just wish Karate was taught at the same caliber everywhere.

— 1 month ago with 53 notes
#karate  #martial arts  #jujitsu  #black belt
Anonymous asked: post a selfie of yourself pls thx

— 2 months ago with 15 notes
#me

legitimate criticisms of education

• most teaching methods only compliment a few ways of learning
• large workloads that require students to compromise their physical or emotional health
• Cs are supposed to be average but are treated by teachers and parents as failure
• tenure

non-legitimate criticisms of education

• "if teachers only teach one subject how do they expect students to learn 5 subjects?"

Legitimate criticisms of education

• Emphasis is on getting good grades, but not on understanding the material
• English-like subjects grade work on its qualitative properties, and can be very subjective
• People who like certain subjects are grouped with those who hate those same subjects, meaning those who want to learn are held back, and those who don’t care feel like they are stupid
• Those who do well are held to different standards than those who don’t
• Material taught in core courses is not representative of the subject at all
• Women are given a lesser chance of achieving anything academically due to nothing other than their sex.
• Just as above: Anyone who is not a Cishet white male is given a lesser chance of achieving anything academic
• Teachers are predominantly white, showing less representation for those who aren’t. This makes it hard for them to find a role model
• Curriculum is set in stone with little room for negotiation on the teachers part. You could have an amazing teacher with a mastery of the subject they teach, but they would still be stuck teaching what the system tells them too.

Illegitimate criticisms of education

• Art students are forced to take years of Math and Science, why aren’t Math and Science students forced to take years of Art-intensive subjects?

Quick explanation: this is not about torturing those who enjoy art, contrary to popular belief! Mathematics and Science courses teach people how to reason and think in new ways, they can provide solutions to problems you’ll face in very day life (I solved a system of linear equations, which is a grade 10 technique, earlier today for someone who wanted a discount on a weighted item at work… I pushed up the scale to the right amount). Art on the other hand; while good for confidence, is inessential, as other subjects like English work to achieve the same result through different means. English is essential because we use it every day, and teaches us to be confident with spoken and written word, while Art is about expressing yourself through your chosen means; a different process and end result.

Science and Math are useful, Art-like subjects are “fun” and expressive.

• I’m smarter than my teachers, there is nothing left for them to teach me, why am I still here? I should be in university, let me leave!

Okay, I am guilty, guilty, guilty of this one; not anymore, but I was at one point.

Let me brake it down; “smart” students are guilty when it comes to educational fallacies as well! The reason I say “smart” is because you probably really aren’t that “smart” if you’re making this kind of rash judgment. Just because you know quite a bit about a subject does not mean that you have it mastered. Teachers went to school for this stuff, they live it for six hours every day (plus grading), they know what they are doing, and they actually have a lot of knowledge locked up in their brains. The curriculum is mostly to blame; teachers aren’t allowed to stray too far from the curriculum, limiting what they can show you, but if you ask them, you’d realize that they have way more experienced than you.

As an example, my Mathematics teacher, who teaches grade 10 and 11 math as well as grade 12 statistics has taught me some of the coolest things I know. If you go to my blog, you’ll see that I have a lot of math stuff that reaches about second year university (if that) level in some areas. This person had loads of experience that I don’t have, and because I asked him, I am getting to learn some pretty awesome statistics! Be careful about skipping ahead unless you are a freaking genius, or you WILL miss something.

Just to prove to you again that you are not a special snowflake for knowing what seems to be a lot about one subject; or even a few, I’m going to end this with a story that a professor of Mathematics at the University of Waterloo: Brian Forest, told me when I visited him in December.

"I once had these two boys walk into my office and demand that they be exempt from their first year classes and placed in second year classes. I explained to them that they were probably making a mistake, as the workload was heavy, but they insisted. I finally decided after some time that I had had enough of them thinking that they were the smartest boys of their age, so I started asking them questions. I asked them, ‘What is So-and-so’s Theorem from Linear Algebra?' and they couldn't answer. I kept up with the questions for a while, and when I thought I had finally proven to these boys that they did not understand the curriculum, I asked them again: 'Do you still feel that you should be placed in second year classes?’ They responded with ‘Of course!' Which surprised me quite a bit. What I am trying to say is that you should never assume that you already know something; there is always room to go back and make sure that you truly understand it before you move on.”

(via amphiarus-rex)

— 2 months ago with 34870 notes
#School  #Math  #Science  #Art  #Education  #Lies  #Curriculum  #Cishet
Group Theory, Isomorphisms & Permutations

I’m not feeling social interaction today so I learned a bunch of group theory, and it just got too cool, so i had to post some of it on here!

Now, before I go on to talk about the main things which are Isomorphisms and Permutations, I should properly define groups.

Taking the axiomatic approach, groups are sets with an operation for combining elements which pass the following tests

Groups are closed under one operation, meaning if you combine any two elements using an operation of your choice (the symbols +and * are often used), you will never be able to create an element which is not in the group.

The operation is associative meaning A*(B*C)=(A*B)*C=A*B*C as long as you combine the elements as shown. This effect can be extended to any string of n elements.

There exists an element “e” such that x*e=x and e*x=x. This element is called the unit, and if we are talking about multiplication with ordinary numbers, this number would correspond to one.

For each number in a group, there will be an inverse. When we are using multiplication-like operations like we are above, we usually write the number like this a-1, and call it “a inverse” for addition we use -a and call it “the negative of a”. The inverse of the unit is the unit again, and any element combined with its inverse becomes the unit.

You can see that the commutative axiom is not listed, where A*B=B*A this is not always true, as we will see in a second. Take the group consisting of the elements

$\mathbf{I}=\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$
$\mathbf{A}=\begin{bmatrix} 0&1 \\ 1& 0 \end{bmatrix}$
$\mathbf{B}=\begin{bmatrix} 0&1 \\ -1&-1 \end{bmatrix}$
$\mathbf{C}=\begin{bmatrix} -1&-1 \\ 0&1 \end{bmatrix}$
$\mathbf{D}=\begin{bmatrix} -1&-1 \\ 1&0 \end{bmatrix}$
$\mathbf{K}=\begin{bmatrix} 1&0 \\ -1&-1 \end{bmatrix}$

under the operation of matrix multiplication which works like this.

Now, if we do the operation A*B, we get the matrix C, if we do the operation B*A, we get the matrix K, meaning that this operation is not commutative. This is not the only operation which behaves this way, so in general we have to say that the operation does not have to commute. If it does, we call the group an Abelian group.

Now that that’s out of the way, we can define an Isomorphism.

An Isomorphism is a one-to-one function which takes one group and converts it into another while preserving the operation. Basically, if G  is a group under the operation * denoted <G,*> and <H,+> is a group, and $a,b \in G, \, a',b' \in H$
then an Isomorphism f(x) will take f(a)=a’ and f(b)=b’ and make, f(a*b)=a’+b’ or, entirely in function notation, f(a*b)=f(a)+f(b).

To prove that two groups are isomorphic, you need to prove that the function you are using is one-to-one, that is proving that the function is both injective and surjective. To do this, you have to prove that each element of the range is the image of at most 1 element of the domain, which can be done by showing that it fits this definition $f(x_1)=f(x_2)\implies x_1=x_2$
You then have to prove that the function is surjective by showing that for some element y in he range, there exists an element who’s image is y. Together, these two things prove that the function is bijective, which means that you can turn elements from G into elements in H and vice versa.

From there, all you have to prove is that f(a*b)=f(a)+f(b) and you’re done.

Now why are isomorphisms important? Because they can be used to prove things like this Cayley’s theorem which says that every group is isomorphic to a group of permutations.  Quite a bit is known about permutations, which makes this theorem very strong in terms of its use.

We can prove this in a second, but first, we have to look at permutations.

A permutation is a function on a set S which takes the elements of S and rearranges them. Now, this must by definition be a bijective function, which takes S and maps it to its self. Taking individual permutations to be elements of a set, we can create a set of n! elements, where n is the number of numbers being used in the permutation. For instance, the set of all permutations on three numbers is:

$\sigma_0=\begin{pmatrix} 1 &2 &3 \\ 1&2 & 3 \end{pmatrix}$  $\sigma_1=\begin{pmatrix} 1 &2 &3 \\ 2&1 & 3 \end{pmatrix}$ $\sigma_2=\begin{pmatrix} 1 &2 &3 \\ 1&3 & 2 \end{pmatrix}$

$\sigma_3=\begin{pmatrix} 1 &2 &3 \\ 3&2 &1 \end{pmatrix}$ $\sigma_4=\begin{pmatrix} 1 &2 &3 \\ 3&1 &2 \end{pmatrix}$ $\sigma_5=\begin{pmatrix} 1 &2 &3 \\ 2&3 &1 \end{pmatrix}$

and has 3!=6 elements. To combine the elements, we define the permutation of a permutation to be a function which takes the elements as they are arranged by the inner permutation and then rearranges them under the outer permutation called the composition. For instance:
$\sigma_1\circ \sigma_3=\begin{pmatrix} 1& 2 &3 \\ 2& 1 &3 \end{pmatrix}\circ \begin{pmatrix} 1&2 &3 \\ 3& 2 &1 \end{pmatrix}=\begin{pmatrix} 1 & 2 &3 \\ 2&3 & 1 \end{pmatrix}=\sigma_5$
To show that this combining of permutations is closed under the operation above, we can make a chart like the one below, where the first column is multiplied on the left by the first row,

$\begin{vmatrix} \circ & \sigma_0 & \sigma_1 & \sigma_2 & \sigma_3 & \sigma_4 & \sigma_5 \\ \sigma_0 & \sigma_0 & \sigma_1 & \sigma_2 & \sigma_3 & \sigma_4 & \sigma_5 \\ \sigma_1 & \sigma_1 & \sigma_0 & \sigma_5 & \sigma_4 & \sigma_3 & \sigma_2 \\ \sigma_2 & \sigma_2 & \sigma_4 & \sigma_0 & \sigma_5 & \sigma_1 & \sigma_3 \\ \sigma_3 & \sigma_3 & \sigma_5 & \sigma_4 & \sigma_0 & \sigma_2 & \sigma_1 \\ \sigma_4 & \sigma_4 & \sigma_2 & \sigma_3 & \sigma_1 & \sigma_5 & \sigma_0 \\ \sigma_5 & \sigma_5 & \sigma_3 & \sigma_1 & \sigma_2 & \sigma_0 & \sigma_4 \end{vmatrix}$

but there are an infinite number of groups of permutations, so we must prove it. We can actually take it a step further into abstraction, and prove the composite of any two functions which are both bijective will yield another bijective function. Why are we proving the combinations of permutations are bijective? Because we started with the assumption that we had made a list of all the possible permutations for n objects, so showing that this new function is also bijective permutation means that it was on this list in the first place.

Assume a f(x) and g(x) are injective functions, then we have to prove that [f*g][x]=[f*g][y] implies that x=y. Well suppose  [f*g][x]=[f*g][y], then f(g(x))=f(g(y)) because f(x) is injective, g(x)=g(y), and because g(x) is injective, x=y, so [f*g][x] is injective

Assume g:A ->B and f:B ->C are surjective, then we have to prove that every element of C is f*g of some element in a. Assume z is an element of C, then because f is surjective,  f(y)=z for some y in B. Because g is surjective,f(g(x))=z so, for any element of C, there is an element for which z is its image under f*g, so [f*g][x] is surjective.

Now, assuming the functions f and g are both bijective, it follows from what was just proven that f*g is bijective as well.

By this, the group of permutations under composition is closed.

There exist inverses for each element, which return each permutation to the permutation
$\sigma_0=\begin{pmatrix} 1 &2 &3 \\ 1& 2 & 3 \end{pmatrix}$
which also happens to be identity element. Finally, combining permutations is associative, so we have a group.

Now we can prove Cayley’s theorem which again states that each group is isomorphic to a group of permutations.

Proof

Begin by assuming we have a group G, we wish to prove that it is isomorphic to a set of permutations. Well, permutations require a set, so why not use the set that G is built off of? The only set we have around is the one used to make G, so we have no other choice but to use it. We then define a function
$\pi_a:G\rightarrow G$
which is defined as:
$\pi_a(x)=ax| \, a,x\in G$
And gives all the permutations of G, changing each element a of G into ax, and then ranging X over all elements of G to give one permutation per element.

This function is injective, that is
$\pi_a(x_1)=\pi_a(x_2)\implies ax_1=ax_2\implies x_1=x_2$
and surjective, as for some y in g,
$y=a(a^{-1}y)=\pi_a(a^{-1}y)$
so y is the image of some element a^(-1)*y.

So we have a bijective function which turns an element a of G into a permutation which turns x_1 into a*x_1, x_2 into a*x_2 and so on. So we let G* represent the set of permutations of G created by pi_a(x) as a ranges through the elements of G.

Notice now that the set of permutations created is not necessarily the set of all permutations of G, but a permutation corresponding to each element. We now have to prove that it is a group, that is
$\pi_a,\pi_b\in G^*\implies \pi_a\circ\pi_b\in G^*$
and that for any permutation contained by G*
$\pi_a\circ\pi_{a^{-1}}=\pi_e$
Where pi_e is the identity permutation given by multiplying each x in G by the identity. The associative property can be assumed due to the fact that the operation is simple non-commutative multiplication.

To begin, we have to prove that
$[\pi_a\circ\pi_b][x]=\pi_{ab}(x)$
We can say that
$[\pi_a\circ\pi_b][x]=\pi_a(\pi_b(x))=abx=\pi_{ab}(x)$
Because ab is a member of G, we can say that G* is closed with respect to composition.

Finally, because G is a group with inverses,
$\pi_a\circ\pi_{a^{-1}}=\pi_{aa^{-1}}=\pi_e$
So we have proved that G* is in fact a group.

Now all we have to do is prove that G is isomorphic to G*. We already have the function ready, so we use it as the isomorphism.
$f(a)=\pi_a$
It is injective,
$f(a)=f(b)\implies a=b$
and surjective, every element pi_a of G* is the image of some f(a)

Lastly,
$f(ab)=\pi_{ab}=\pi_a\circ\pi_b=f(a)\circ f(b)$
Thus, f is an isomorphism, and
$G\cong G* \; \; \; \square$

That’s about all I have left in me for today, but I’ll try to get more up on here tomorrow! I thought Cayley’s theorem would be a good start!

— 2 months ago with 46 notes
#math  #mathematics  #maths  #physics  #mathema  #algebra  #groups  #isomorphism  #permutations
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