Mathematics - 2017-09-14 (page 2 of 6)

Leaky Nun

04:00

@Dodsy 1. the square root of -1 is not > 0; 2. the square root of 0 is not > 0

Dodsy

what's the square root of zero?

what's the square root of -1?

Leaky Nun

the square root of zero is zero

the square root of -1 is undefined in the reals

Dodsy

so, what's your point?

I was talking about the reals.

as you can see above.

Leaky Nun

you said the square root of any real number must be > 0

i'm just pointing out that the square root of any real number doesn't have to be define; and even if it's defined, it can be = 0

Dodsy

if I forgot a line under my > sign, then you are being quite pedantic to suggest that my entire statement was wrong

and no, you aren't just saying

Leaky Nun

04:04

you don't have to interpret every statement maliciously

Semiclassical

oy vey.

Dodsy

04:23

@Semiclassical best place to sketch a graph

Semiclassical

if you can't use desmos/wolfram alpha/etc

then by hand is pretty much what's left to you

Leaky Nun

G={5,15,25,35} is a group under multiplication modulo 40 :O

@Semiclassical geogebra!!

Kasmir Khaan

finally

I found leaky nun

Semiclassical

meh. divide everything by five and you get {1,3,5,7} mod 8 i.e. (Z/8)*

Kasmir Khaan

:D

Leaky Nun

04:26

@KasmirKhaan my timezone is UTC+8.

Kasmir Khaan

let me see what is that in my time ><

Leaky Nun

@Semiclassical but 5 isn't the identity; it's 25

Semiclassical

huh.

Kasmir Khaan

7 HOURS difference ><

Leaky Nun

@Semiclassical 25*25=25; 5*5=25

Semiclassical

04:27

sure, I see it

i'm just a bit puzzled

Leaky Nun

@KasmirKhaan I sleep quite late, so the difference may not be that huge

Kasmir Khaan

Anyway leaky if you are free ,i kinna need some help putting things on texstudio

Leaky Nun

@KasmirKhaan ok

Secret

[Chemistry]
To be submitted:
file=025 stage=1 bash masterscript
file=026 stage=1 bash masterscript
file=027 stage=1 bash masterscript
file=028 stage=1 bash masterscript
file=029 stage=1 bash masterscript
file=030 stage=1 bash masterscript
file=031 stage=1 bash masterscript
file=032 stage=1 bash masterscript
file=033 stage=1 bash masterscript
file=034 stage=1 bash masterscript
file=035 stage=1 bash masterscript
file=036 stage=1 bash masterscript
file=037 stage=1 bash masterscript
file=038 stage=1 bash masterscript

and this is why I spent 4 weeks to wrote that python bash stuff previously

Kasmir Khaan

@LeakyNun Ill send you the examples one by one =p they are a mess btw, gave up at the end on using corect latex , but just let me send it :D

Semiclassical

04:28

Youtube: Stop trying to get me to play League of Legends. So tired of these ads that go into breathless detail about it

Leaky Nun

@KasmirKhaan ok

@Semiclassical so what's your explanation?

Kasmir Khaan

@LeakyNun Sent =P Ill walk you thru what I did so you can correct me =p

Semiclassical

well, the elements of G are 5m where m=1,3,5,7

so if 5m is the identity in G then it acts as (5m)(5n)=40a+(5n) -> 5mn=8a+n

Leaky Nun

@KasmirKhaan what is the question? still the isomorphism between S3 and GL(2,Z/2)?

Kasmir Khaan

Yes

I found a good way on how to solve it =p

using generators

Semiclassical

04:32

so the identity in (Z/8)* will be 5m

and since the identity in (Z/8)* is 1, we need 5m=1 mod 8 and that's m=5

Leaky Nun

I see

so the identity in U(10) is 1 = 9x9 = 3x7

Semiclassical

I think I should've gone the opposite direction, mind

Leaky Nun

if we consider 3U(10) = {3,9,21,27}

Semiclassical

start from (Z/8)* and go to G

Leaky Nun

under multiplication mod 30

then m=7 is the identity, i.e. 21?

Kasmir Khaan

04:34

@LeakyNun matrix X = (1 0 ) first row and second row is (1 1 )

Semiclassical

lemme see. the identity in U(10) acts as (1)(m)=10a+m

Leaky Nun

3 x 3 = 9
9 x 9 = 1
21 x 21 = 21
27 x 27 = 9

wait what

Kasmir Khaan

@LeakyNun matrix Y is : (0 1 ) first row (1 0 ) second row

Semiclassical

I guess a=0 is sorta superfluous there.

Kasmir Khaan

@Semiclassical what are you guys trying to solve? =p

Semiclassical

04:35

(1)(m)=m, (1)(3m)=(3m)

Leaky Nun

why doesn't 3U(10) form a group ;_;

10 mins ago, by Leaky Nun

G={5,15,25,35} is a group under multiplication modulo 40 :O

Semiclassical

so one needs 3n=1 mod 30

Leaky Nun

@KasmirKhaan we're trying to understand and generalize that result

Semiclassical

...which is not going to happen :/

Leaky Nun

@Semiclassical interesting

Semiclassical

04:36

lemme go back to the original example in this approach

Leaky Nun

but 5n=1 mod 40 also doesn't happen

Kasmir Khaan

But there is no formula for Z/n *

Semiclassical

the identity in U(7) acts as (1)(m)=m

Kasmir Khaan

as far as I know

Leaky Nun

@KasmirKhaan so what

Semiclassical

04:37

so (1)(5m)=5m...hrm. this doesn't work, no

Leaky Nun

@Semiclassical I don't understand what you mean by "act as"

Semiclassical

I mean that the defining feature of the identity in a group is that eg=g

Leaky Nun

@Semiclassical oh, you can instead use e^2=e

Semiclassical

hmm

Kasmir Khaan

that group is isomorphic to Z/8*

what else you want to do

Semiclassical

04:39

We know.

But the identity element in {5,15,25,35} isn't 5. it's 25.

Leaky Nun

@KasmirKhaan we want to know why you can form another group by multiplying each element in U(8) by 5

Kasmir Khaan

oh

Hmm

Semiclassical

@LeakyNun I think what we really want to say is that while x->5x is a bijection with G, it's not an isomorphism because it doesn't map the identity to the identity

Leaky Nun

@Semiclassical I suspect it isn't multiplying by 5 so much as multiplying by 25, amirite?

Semiclassical

so the actual isomorphism between (Z/8)* and G is different.

I think so, yeah.

Leaky Nun

04:41

back to U(10) x 3

mod 30, but multiply each element by 9

Semiclassical

{1,3,5,7}->{25,35,5,15}

Leaky Nun

{1,3,7,9} -> {9,27,3,21} wtf am I doing here

we do have a group

this is very interesting

Semiclassical

yeah. 3^2, 5^2

(1)(n)=10a+n -> (25m)(5n)=5(40a)+5n

that don't look right

Leaky Nun

$$\begin{array}{r|r}
\cdot&9&27&3&21\\\hline
9&21&3&27&9\\\hline
27&3&9&21&27\\\hline
3&27&21&9&3\\\hline
21&9&27&3&21
\end{array}$$

You're right, the identity is 3m=1 -> m=7 -> 3x7=21

@KasmirKhaan are you sending me an email?

Kasmir Khaan

@LeakyNun I did , just did not tell you becauseyou were busy

Semiclassical

04:48

i feel like there's a better description than this, but eh

Leaky Nun

@KasmirKhaan just because I'm talking to Semi doesn't mean I'm busy

Kasmir Khaan

@Semiclassical Semi I got a deadline on 18th and another on 21th , let leaky help me :D

Semiclassical

lol

Kasmir Khaan

@LeakyNun oh okay =p its a mess btw , warned you again :D

Leaky Nun

@Kasmir, I can respond to two people at the same time

Kasmir Khaan

04:49

this teacher want things in pdf form

I used to do them handwritten

But at some point I need to learn how to use tex so =P

@LeakyNun that is nice, sometimes I got trouble responding to 1 person =p

Leaky Nun

@Semiclassical conjectures:
1. let a be an element of U(n). Then, a^2U(n)/(an) forms a group.
2. a^2U(n)/(an) = aU(n)/(an) as sets

@Kasmir, cut the opening

Semiclassical

{1,3,5,7}->5{1,3,5,7}={5,7,1,3}

Kasmir Khaan

@LeakyNun what do you mean ?

What you guys are doing is cosets btw

Leaky Nun

What am I doing, just combine 1 and 2 to get aU(n)/(an) forms a group @Semi

@KasmirKhaan it isn't coset

it's poor notation on my part

Kasmir Khaan

Okay then I dont know=p

@LeakyNun Did you get the email ?

Leaky Nun

04:52

@Semiclassical refined conjecture: let $a$ be an element of $U(n)$ and let $ab=1$ under $U(n)$. Then, $x \mapsto abx \pmod {an}$ gives you a group under multiplication mod $an$.

Semiclassical

so b=a^{-1} in U(n)

Leaky Nun

@KasmirKhaan I got it but you sent it to the wrong address

Kasmir Khaan

@LeakyNun what do you mean ? you only gave me that one :D

Leaky Nun

@KasmirKhaan email me the address you sent to

(I know it sounds paradoxical and stupid; just do it)

Kasmir Khaan

Okay ><

Done @LeakyNun

Leaky Nun

04:56

interesting, I received it under the proper address now

am I supposed to fill in the blank where it says "X = ..."?

zhk

https://i.sstatic.net/Pxje6.png

wha type of function is E here?'

Leaky Nun

latex rule 1: never use \\ outside math context :p

zhk

wha type of function is E here?

Kasmir Khaan

@LeakyNun haha could not type matrices there

matrix Y is : (0 1 ) first row (1 0 ) second row

Semiclassical

Evidently it's part of what you get when you take the inverse Laplace transform of $\dfrac{1}{s^\alpha+\nu k^2}$

which...ew

Kasmir Khaan

04:59

@LeakyNun matrix X = (1 0 ) first row and second row is (1 1 )

@LeakyNun matrix Y is : (0 1 ) first row (1 0 ) second row

Semiclassical

I mean, the case of $\alpha=2$ isn't too bad.

But for the general case I dunno what function would produce that

Leaky Nun

@KasmirKhaan do you want $\begin{pmatrix}1&0\\1&1\end{pmatrix}$ or $\begin{bmatrix}1&0\\1&1\end{bmatrix}$?

Semiclassical

Finding a good table of Laplace transforms is surprisingly annoying, alas. The usual ones are pretty short and spare.

zhk

@Semiclassical I agree. But here I am unable to identify the function E_alpha,alpha.

Kasmir Khaan

@LeakyNun first one is correct

@LeakyNun zero only on the row1, colum2 postion

Leaky Nun

05:05

@KasmirKhaan they're the same

the difference being () vs []

Kasmir Khaan

@LeakyNun well the second could not read it thats y >< ehm does not matter really =p parenteses are better

they look better just that

zhk

Kasmir Khaan

the matrix Y is like the identity matrix with 0's and 1's swtiched

Daminark

Analysis or Model Theory?

Secret

Matrix of ones

In mathematics, a matrix of ones or all-ones matrix is a matrix over the real numbers where every element is equal to one. Examples of standard notation are given below: J 2 = ( 1 1 1 1 ...

lol, this guy has standard notation

Leaky Nun

05:09

@KasmirKhaan you can't write S3 = <a,b|a^3=b^2=1,ab=ba^-1> on one line and S3 = <a,b> on the other

although I know what you mean

Secret

Exchange matrix

In mathematics, especially linear algebra, the exchange matrix (also called the reversal matrix, backward identity, or standard involutory permutation) is a special case of a permutation matrix, where the 1 elements reside on the counterdiagonal and all other elements are zero. In other words, it is a 'row-reversed' or 'column-reversed' version of the identity matrix. J 2 = ( 0 ...

uh, what?

Both are J

Kasmir Khaan

@LeakyNun Oh thanks, can you please correct it? =p am really bad at this tex thing :D

Semiclassical

@zhk Source?

Leaky Nun

@KasmirKhaan this isn't about tex.

Kasmir Khaan

@LeakyNun yes this is why I need someone good to help me write clean proofs =p

zhk

05:12

@Semiclassical Linear Partial Differential Equations for Scientists and Engineers by Tyn Myint-U Lokenath Debnath
Page 516

Leaky Nun

@KasmirKhaan I won't do your work for you

Kasmir Khaan

@LeakyNun I know I know :D just the latex thing =p

Semiclassical

Then presumably they define it somewhere in there...though I know from experience how hard it can be to find something like that in a big book @zhk

Kasmir Khaan

@LeakyNun I copied the codes you gave me, but this time its alot more new things, such as matrices and stuff thats y I was stuck

zhk

@Semiclassical thats what I thought and searched but I am unable to find.

Semiclassical

05:14

Alternatively, a very extensive table of integral transforms can be found here: authors.library.caltech.edu/43489/1/Volume%201.pdf

And by extensive I mean 400 pages long, with a good chunk of that being Laplace/ inverse Laplace transforms

zhk

@Semiclassical The only other place I found function was in the Laplace table

Semiclassical

ew.

zhk

Semiclassical

that's just unkind.

zhk

Apart from this there are quite a few typos in this book but I still like this book

Semiclassical

05:19

hmm.

zhk

@Semiclassical I got it. Its there

Sorry

Semiclassical

oh? What was it?

zhk

Mittag-Leﬄer function

Semiclassical

...yep, I have no earthly idea what that is

zhk

Oh GOD! Sorry to bother you.
me too, I have no heavenly idea about this

Semiclassical

05:21

though Wikipedia evidently does:

Mittag-Leffler function

In mathematics, the Mittag-Leffler function Eα,β is a special function, a complex function which depends on two complex parameters α and β. It may be defined by the following series when the real part of α is strictly positive: E α , β ( z ) = ∑ k = 0 ∞ z k ...

heh, no worries.

Daminark

It is a scary time when even Wikipedia is just like "uw0tm8?"

zhk

Semiclassical

yep

Leaky Nun

@KasmirKhaan skickades

Semiclassical

though, dang, the DLMF does have it: dlmf.nist.gov/10.46

Kasmir Khaan

05:22

@LeakyNun Haha tack :D

Semiclassical

Though it doesn't mention the Laplace transform.

Kasmir Khaan

@LeakyNun I sent you the next one =p ( btw I solved them all ) what did you think about the solution of the first? regor enuf?

@LeakyNun thanks alot man !you are the best :D

Semiclassical

Mathworld also has it, and does mention the Laplace transform.

Leaky Nun

@KasmirKhaan I'm surprised that the second one compiles in your TeXstudio

coz it doesn't for mine

zhk

@Semiclassical Thank you for time. Sorry to bother you with relevant stuff. I will now search more about it

Kasmir Khaan

05:24

@LeakyNun it did not open ?

Leaky Nun

@KasmirKhaan it did open

but it does not compile

Semiclassical

no problem. I'm just amazed to see a new special function

Secret

Interestingly, it seems dlmf does not have an entry for harmonic numbers and its generalisations

Daminark

What is the Laplace transform anyway?

zhk

unrelevant*. me too

Semiclassical

05:25

@Daminark $F(s)=\int_0^\infty e^{-s x}f(x)\,dx$ is the Laplace transform of $f(x)$.

zhk

@Semiclassical I know that

Secret

though the integral representation of harmonic numbers does kinda resembles that of the Mittag-Leffler function in that there's a x-stuff denominator and a stuff-more stuff numerator

Semiclassical

well, of course you do :P

Kasmir Khaan

@LeakyNun oh yeah now I also noticed some errors, did not compile it really just wrote stuff

zhk

@Semiclassical here is a new one i proposed sometime ago and then stopped working on it

N-transform

In mathematics, the Natural transform is an integral transform similar to the Laplace transform and Sumudu transform, introduced by Zafar Hayat Khan in 2008. It converges to both Laplace and Sumudu transform just by changing variables. Given the convergence to the Laplace and Sumudu transforms, the N-transform inherits all the applied aspects of the both transforms. Most recently, F. B. M. Belgacem has renamed it the natural transform and has proposed a detail theory and applications. == Formal definition == The natural transform of a function f(t), defined for all real numbers t ≥ 0, is ...

Kasmir Khaan

05:27

@LeakyNun that question is about the map that f : x --> x'

Semiclassical

Neat.

Leaky Nun

full question Kasmir

Kasmir Khaan

yes sir

Daminark

So is it slightly more tan cousin of Fourier?

Kasmir Khaan

ill send you the pdf file

@LeakyNun done, sent you both of the assigments for future work =p , its the first question on assigment II

Leaky Nun

05:32

ok

Semiclassical

@Daminark It's handy for some stuff, though damned if I remember what

Daminark

Semi remembers and is subsequently damned. Rip

Leaky Nun

@KasmirKhaan correct your errors so that it compiles and then send it to me.

Kasmir Khaan

@LeakyNun okay fixing it now =p

@LeakyNun \documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
\text{First implication , we assume homomorphism to get abelian}
\\$\phi(xy) =(xy^{-1})=y^{-1}x^{-1}=ba$ \text{let $a=x^{-1} and b=y^{-1}$}
\\$\phi(xy)=\phi(x)$$phi(y)=x^{-1}y^{-1}=ab$
\\$ab=ba)$ for all a,b in G
\\\text{the second implication , we assume G is abelian to get homomorphism}
\\$xy=yx$ , also $phi(xy)$ = $phi(yx)$ since G is abelian

Leaky Nun

@KasmirKhaan press Ctrl+K

Kasmir Khaan

05:41

i did that

nothing happned

af

Leaky Nun

press it when you edit the message

Kasmir Khaan

better?

Leaky Nun

Have you compiled it?

Kasmir Khaan

@LeakyNun am not making these errors on purpose , I hope you know that =p

yes

it shows fine for me

Leaky Nun

@KasmirKhaan does it produce the result you want?

no, I am not accepting it. Go fix your errors.

Kasmir Khaan

05:45

@LeakyNun its gonna ruined now :/

Leaky Nun

@KasmirKhaan ?

Kasmir Khaan

@LeakyNun it was just phi that did not go right

now its all red

Leaky Nun

then fix them

Kasmir Khaan

I typed something wrong i guess that ruined it all

:(

@LeakyNun is there a "back" function on texstudio ?

Leaky Nun

@KasmirKhaan Ctrl+Z

Kasmir Khaan

05:50

\documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
	\text{First implication , we assume homomorphism to get abelian}
	\\\(\phi(xy) =(xy^{-1})=y^{-1}x^{-1}=ba\) \text{let \(a=x^{-1} and b=y^{-1}\)}
	\\\(\phi(xy)=\phi(x)\)\(phi(y)=x^{-1}y^{-1}=ab\)
	\\\(ab=ba)\) for all a,b in G
	\\\text{the second implication , we assume G is abelian to get homomorphism}
	\\\(xy=yx\) , \text {also \(\phi(xy) = \(\phi(yx)\) since G is abelian}

:D

that was very handy =p

@LeakyNun .

Leaky Nun

1. \text is for writing in math mode; don't use it in normal mode

2. use two newlines, not \\

Kasmir Khaan

what is newlines?

Leaky Nun

what you get when you press enter

Kasmir Khaan

Just goes down 1 line @LeakyNun

Leaky Nun

@KasmirKhaan yes, it's called a newline

it's a character just like every other character

Kasmir Khaan

05:55

if I hit enter twice that will be new line ?

oh neat :D

thanks :)

Leaky Nun

now fix what i told you to

and send it to me

Kasmir Khaan

@LeakyNun dont get it man, it does not compile now

Leaky Nun

@KasmirKhaan what have you got?

Kasmir Khaan

\documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
	First implication , we assume homomorphism to get abelian


	\(\phi(xy) =(xy^{-1})=y^{-1}x^{-1}=ba\) \text{let \(a=x^{-1} and b=y^{-1}\)}
	\(\phi(xy)=\phi(x)\)\(phi(y)=x^{-1}y^{-1}=ab\)

    \(ab=ba)\) for all a,b in G
    the second implication , we assume G is abelian to get homomorphism
    \(xy=yx\) \(\phi(xy) = \(\phi(yx)\) since G is abelian

line16

shows error

that what it sais

Leaky Nun

do you think $\phi(xy) = \(\phi(yx)$ would compile?

Kasmir Khaan

06:06

\documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
	First implication , we assume homomorphism to get abelian


	\(\phi(xy) =(xy^{-1})=y^{-1}x^{-1}=ba\) \text{let \(a=x^{-1} and b=y^{-1}\)}
	\(\phi(xy)=\phi(x)\)\(phi(y)=x^{-1}y^{-1}=ab\)

    \(ab=ba)\) for all a,b in G
    the second implication , we assume G is abelian to get homomorphism
    \(xy=yx\) \(\phi(xy) = (\phi(yx)\) since G is abelian

@LeakyNun grrrr this is really madness for me , each time i use tex

@LeakyNun how did you learn it?

Leaky Nun

You need to press enter twice if you want to get a new line

Kasmir Khaan

@LeakyNun Ill do that in future, because now when i try to do it , it gives error for some reasom

Leaky Nun

do you really think that $\phi(xy) =(xy^{-1})$?

Kasmir Khaan

yes by definition

Leaky Nun

what

Kasmir Khaan

06:09

f(xy) = (xy)^-1

or what do you mean ?

Leaky Nun

@KasmirKhaan this is right

what I quoted isn't right.

Kasmir Khaan

oh parenteses

but where do they appear

in what i sent

Leaky Nun

and check your output again.

@KasmirKhaan second line

Kasmir Khaan

oh damn it

fixed =p

Leaky Nun

fix everything and send it to me

Kasmir Khaan

06:14

\documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
	First implication , we assume homomorphism to get abelian


	\(\phi(xy) =(xy)^{-1}=y^{-1}x^{-1}=ba\) \text{let \(a=x^{-1} and b=y^{-1}\)}


	\(\phi(xy)=\phi(x)\)\(phi(y)=x^{-1}y^{-1}=ab\)

    \(ab=ba)\) for all a,b in G


    the second implication , we assume G is abelian to get homomorphism


    \(xy=yx\) \(\phi(xy) = (\phi(yx)\)

@LeakyNun done =p

Leaky Nun

press enter two times, not three times

have you seen the file I sent you for the previous assignment?

Kasmir Khaan

Yes it was perfect =p

\documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
	First implication , we assume homomorphism to get abelian


	\(\phi(xy) =(xy)^{-1}=y^{-1}x^{-1}=ba\) \text{let \(a=x^{-1} and b=y^{-1}\)}


	\(\phi(xy)=\phi(x)\)\(phi(y)=x^{-1}y^{-1}=ab\)

    \(ab=ba)\) for all a,b in G


    the second implication , we assume G is abelian to get homomorphism


    \(xy=yx\) \(\phi(xy) = (\phi(yx)\)

Leaky Nun

@KasmirKhaan it will be worthless if you don't learn from it

Kasmir Khaan

I know you are right

but this is my second time using tex

need to work more on this, ill save those you did ofc, and wont ask you for same stuff again

Leaky Nun

Check your output.

Fix it until it is exactly what you want.

ask me if you don't know how to fix something

not just give me the whole thing

Kasmir Khaan

06:19

@LeakyNun \documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
	First implication , we assume homomorphism to get abelian


	\(\phi(xy) =(xy)^{-1}=y^{-1}x^{-1}=ba\) \text{let \(a=x^{-1} and b=y^{-1}\)}


	\(\phi(xy)=\phi(x)\)\(\phi(y)=x^{-1}y^{-1}=ab\)

    \(ab=ba\) for all a,b \(\in G\)


    The second implication , we assume G is abelian to get homomorphism


    \(xy=yx\)

    \(\phi(xy) = \phi(yx)\)

@LeakyNun it is fixed now =p

@LeakyNun Ill be more specific with my problems in the future, thanks for taking time with me :)

Leaky Nun

Look at the second line in your output

did you intend the "and" to be in italics?

28 mins ago, by Leaky Nun

1. \text is for writing in math mode; don't use it in normal mode

Kasmir Khaan

\documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
	First implication , we assume homomorphism to get abelian


	\(\phi(xy) =(xy)^{-1}=y^{-1}x^{-1}=ba\) \text{let \(a=x^{-1} ; b=y^{-1}\)}


	\(\phi(xy)=\phi(x)\)\(\phi(y)=x^{-1}y^{-1}=ab\)

    \(ab=ba\) for all a,b \(\in G\)


    The second implication , we assume G is abelian to get homomorphism


    \(xy=yx\)

    \(\phi(xy) = \phi(yx)\)

@LeakyNun what do I use for normal text?

Leaky Nun

@KasmirKhaan nothing. It's already a text

Kasmir Khaan

Okay =p

Leaky Nun

8 mins ago, by Leaky Nun

press enter two times, not three times

Kasmir Khaan

06:24

@LeakyNun okay

am working on part b) that direct product

this what I did so far

\documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
	The first implication we assume abelian and show homomorphism


	\(\phi(x,y),(x_{1},y_{1}) =\phi(xx_{1},yy_{1})=xx_{1}yy_{1}=xyx_{1}y_{1}=\phi(x,y)\phi(x_{1},y_{1})\)

	The second implication we assume homomorphism and show that \(G\) is abelian
\end{document}

Leaky Nun

alright

@KasmirKhaan you see, you can get things done when I'm not doing everything for you

Kasmir Khaan

@LeakyNun I know you want to teach me to do this alone, and I want that, but need some help sometimes =p

really lost courage when could not put those matrices

=p

anyway am finishing in a min

the seocnd implication

@LeakyNun here?

Leaky Nun

06:41

@KasmirKhaan ?

Kasmir Khaan

I have an error

missing dollar sign

inserted

\documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
	The first implication we assume abelian and show homomorphism


	\(\phi(x,y),(x_{1},y_{1}) =\phi(xx_{1},yy_{1})=xx_{1}yy_{1}=xyx_{1}y_{1}=\phi(x,y)\phi(x_{1},y_{1})\)

	The second implication we assume homomorphism and show that \(G\) is abelian

	\(\phi(x,y),(x_{1}y_{1})=\phi(x,y)\phi(x_{1}y_{1})=xyx_{1}y_{1}\)

	\(\phi(x,y),(x_{1}y_{1})=\phi(xx_{1},yy_{1}) = xx_{1}yy_{1}

Leaky Nun

the line before that has missing \)

Kasmir Khaan

still cant spot it

( xx_{1}yy_{1})

\( xx_{1}yy_{1}\)

that line ?

GOT IT

:)

Leaky Nun

...

does it compile?

Kasmir Khaan

\documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
	The first implication we assume abelian and show homomorphism


	\(\phi(x,y),(x_{1},y_{1}) =\phi(xx_{1},yy_{1})=xx_{1}yy_{1}=xyx_{1}y_{1}=\phi(x,y)\phi(x_{1},y_{1})\)

	The second implication we assume homomorphism and show that \(G\) is abelian

	\(\phi(x,y),(x_{1}y_{1})=\phi(x,y)\phi(x_{1}y_{1})=xyx_{1}y_{1}\)

	\(\phi(x,y),(x_{1}y_{1})=\phi(xx_{1},yy_{1}\) =\( xx_{1}yy_{1}\)

Yes it does :D

Leaky Nun

06:49

why do you put the = in normal mode?

Kasmir Khaan

did not know that should be done in math mode too

ill fix it :D

Leaky Nun

well you did put the first = in math mode

Kasmir Khaan

okay ill fix those in all what I wrote

can you please tell me how to put the question on same file

\documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
	First implication , we assume homomorphism to get abelian


	\(\phi(xy) =(xy)^{-1}=y^{-1}x^{-1}=ba\) \text{let \(a=x^{-1} ; b=y^{-1}\)}


	\(\phi(xy)=\phi(x)\)\(\phi(y)=x^{-1}y^{-1}=ab\)

    \(ab=ba\) for all a,b \(\in G\)


    The second implication , we assume G is abelian to get homomorphism


    \(xy=yx\)

    \(\phi(xy) = \phi(yx)\)

@LeakyNun I tried this but there are no newlines showing up

I mean I wanted more space between them so it can be read eays

Leaky Nun

@KasmirKhaan just put them in the same file?

Kasmir Khaan

All righty =p

Leaky Nun

07:00

@KasmirKhaan \vspace{5mm}

Kasmir Khaan

@LeakyNun thanks! that was what I wanted :D

@LeakyNun Ill continue on the rest later, my eyes hurt

Leaky Nun

@KasmirKhaan ok

Kasmir Khaan

@LeakyNun thanks for all your help again ! :D ill email you once am done :)

NV-US

07:42

$G_1={(1234),(2134),(1243),(2143)}, the orbits of 1 and 2 are {1,2} and the orbits of 3 and 4 are {3,4}.$ i know the definition of orbits, but i dont see how the orbits are calculated, help please

Leaky Nun

@NV-US full question?

Leaky Nun

07:57

@NV-US did you mean (12)(34)?

NV-US

08:34

@LeakyNun i copied this from mathworld.wolfram.com/GroupOrbit.html paragraph 2

Tobias Kildetoft

@LeakyNun I think it uses one-line notation rather than cycle notation (which is a terrible idea, but then again, so is much of Mathworld)

@NV-US You just start from one element and see where you can make it go using the group elements

Leaky Nun

@TobiasKildetoft oh, ok

Tobias Kildetoft

At least, that is consistent with the orbits found

(and with those permutations forming a group in the first place)

NV-US

but then, i get, 1 can go to 2,3,2,4, wwhich is not the orbit

Tobias Kildetoft

no, $1$ can only go to $1$ or $2$

As I said, they use one-line notation rather than cycle notation

NV-US

08:46

what is 1-line notation

Tobias Kildetoft

it means that the first item is what $1$ gets sent to, then what $2$ is sent to and so on

Leaky Nun

e.g. $(2143)$ is the permutation $\sigma: \begin{cases} 1 &\mapsto& 2 \\ 2 &\mapsto& 1 \\ 3 &\mapsto& 4 \\ 4 &\mapsto& 3 \end{cases}$

Tobias Kildetoft

I don't actually think I have ever been in a situation where that notation was the best choice

NV-US

got it. ty @TobiasKildetoft @LeakyNun

Balarka Sen

09:11

back home finally

Leaky Nun

@BalarkaSen welcome back

Işık Kaplan

11:59

Hey guys, I have a question that is probably I a question because I cannot give enough attention but have to ask. I don't even know how did I notice this but on the internet there are two different sin tables. Most places like Wolfram Alpha etc. gives sin1=0.8414 but some places gives sin1=0.0174 I know the second one is calculated as sin(x*pi/180) for x=1 but I don't now what it is.

As I said this is probably something that is so simple but I can not see it and I'm about the go crazy.

Mary Star

Hello!!

Let $p$ be a prime for which it exists a $n\in \mathbb{Z}$ with $p\mid n^2+3$ and $p\mid (n+1)^2+3$. I want to show that it must hold that $p=13$ and that there are infinitely many integers $n$ so that the above relations hold.

We have that $p\mid n^2+3$ and $p\mid (n+1)^2+3 \Rightarrow p\mid n^2+2n+1+3$, so we get $p\mid (n^2+2n+1+3)-(n^2+3) \Rightarrow p\mid 2n+1$, right?

How could we continue?

Işık Kaplan

Alright, I just realized one of them is degree one of them is radian, my brain can be so slow sometimes.

Jing Weng

12:32

imgur.com/a/OqGqc For part (c) of this problem, I tried to generalize part (a) by letting E be the event that trial i comes before trial j. However, unlike part (a) where P(R intersect B intersect G)= empty_set, these trials might not be disjoint, so I couldn't just do P(E)=P(E intersect T1)+P(E intersect T2)+...+P(E intersect Tn).

Do you know any other strategies to tackle this problem?

Transcript for

$Mathematics$ Mathematics