Mathematics - 2017-02-19 (page 4 of 7)

user400188

08:00

I assume that theu are almost the same but for one the order matters and it restricted to a pair.

is it possible to write something like (a,b,c ... n)

DHMO

yes

you have ordered triplets

user400188

still ordered but not in a pair?

ah ok

Ill give that second question a go now

Paul Plummer

There is a way to translate ordered tuples to sets, but it is somewhat awkward...

user400188

sounds like something I will want to look up later

DHMO

@PaulPlummer translate it to function and then to set of ordered pairs

Paul Plummer

08:04

No I mean pairs too

user400188

Is there a thing similar to boolean algebra but for sets? (union and conjection)?

DHMO

I like the function encoding more

Paul Plummer

Well how do you define a function in terms of sets? You can but it is similarly awkward

DHMO

@user400188 A union B = {x|x in A or x in B}

user400188

@DHMO thanks I can probaly work my way from there

DHMO

08:05

@PaulPlummer functions are relations. Relations are sets of ordered pairs.

Paul Plummer

Yah and what about ordered pairs?

I just thought I would mention that you could translate ordered pairs into sets since you seem to be doing this sort of thing, but it isn't really worth dwelling on @user400188

DHMO

the usual encoding is (a,b) = {{a},{a,b}}

user400188

@DHMO A conjunction B' is just {x|x in A & x not in B} correct?

DHMO

not or

user400188

@PaulPlummer I would really like to be able to reduce everything to first order logic or predicate logic if I could. Looking up the way to do that later might help along those lines.

*Forgot the & when I copied

DHMO

08:09

correct

user400188

If I have a set in set builder notation; and I want to put another set in the same notation inside the first; do I drop the {} around the outside or do I not?

DHMO

what do you mean?

user400188

say I have {x| x in A$\cap$B or x in A$\cap$B'}

when I write A$\cap$B fully; will I be allowed to drop the { } that go on the outside of its notation?

DHMO

"x in {y|P(y)}" is equivalent to P(x)

user400188

Ok I think I have it now. Just give me some time to write it and some of the steps down.

jublikon

08:16

Good morning

user400188

Morning Jubl

Daminark

Yo

user400188

@DHMO
Ok here is is. There will probaly be some typos so expect a few edits once its up.

As $A\cup B=\{x|[x\in A]\lor [x\in B]\}$

and as $A\cap B=\{x|[x\in A]\land [x\in B]\}$

$\therefore$ we have $(A\cap B)\cup (A\cap B')=\{x|[x\in A]\land [x\in B]\}\cup \{x|[x\in A]\land [x\in B']\}$

which is $\{x|[x\in A\cap B]\land [x\in A\cap B']\}$

which in turn is $\{x|[(x\in A)\land (x\in B)]\lor [(x\in A)\land(x\in B')]\}$

which is finally: $\{x|[(x\in A)\land[(x\in B)\land(x\in B')]\}$

giving $\{x|[(x\in A)\land[T]\}$

Therefore $(A\cap B)\cup (A\cap B')=\{x|x\in A\}$

jublikon

I have a function two images $f:X \rightarrow Y$ and $g:Y \rightarrow X$

The task is to *proove* or *disproove* the following statement:

If $f$ is invertible and $g=f^{-1}$ then $g \circ f = id_{X}$

My guess is that this is correct. **Question**: Is that true?

solution so far:

If f is invertible, then f has to be bijective.
If f is bijective, then f is left- and rightinverse.

So that means that:
$f^{-1} \circ f = id_{X}$ and $f \circ f^{-1}= id_X$

$\equiv f^{-1} \circ f = id_X \equiv g \circ f = id_X$

user400188

whats Idx?

jublikon

08:30

öhm, 1?

user400188

I remember doing a proof like that once. But I cant remember the details (it was in 2012 when I did it)

jublikon

cause $x \cdot x^{-1}=1$

what I did: does it seem like a proof to you or does it miss something?

user400188

When I did it all I needed to do was look at the domain and codomains. I also needed to know that f was 1 to 1 before it was inverted.

We didnt need bijective and surjective becuase my course back then hadn't taught the meaning to us.

BAYMAX

@arctictern @PaulPlummer I got it for <g> $\neq$ Z as it has many subgroups like 2Z which is normal and hence cant be simple ,but what about the factor group $Z/mZ$, m is composite is isomorphic to $Z_{m}$ and i forgot the normal subgroups of $Z_{m}$,any help ?

arctic tern

@BAYMAX if you forgot the subgroups of Z_m, find them again

DHMO

08:40

@user400188 did you put line 4 to test if i read?

line 6 also

your proof is good

BAYMAX

i took an example like $Z_{4}$ and i found that ${0,1,3}$ is anormal subgroup of $Z_{4}$ b which violates that $G$ is simple, is iit correct?

DHMO

@user400188 A \ B = ?

BAYMAX

@arctictern

arctic tern

@BAYMAX no, {0,1,3} is not a subgroup of Z/4Z

BAYMAX

nono

i said for $Z_{4}$

arctic tern

08:45

yes, Z/4Z

BAYMAX

Now ?

arctic tern

Z_4 and Z/4Z both stand for integers mod 4 under addition

{0,1,3} is not a subgroup

e.g. 1+1=2 which is not in {0,1,3}

BAYMAX

ohh if i add that 2 , then it becomes the whole group?

so trivial choice is the group itself

@arctictern

arctic tern

what?

BAYMAX

Normal subgroup of $Z_{4}$ is the same group itself

user400188

08:48

Sorry I was gone for a while there.

arctic tern

@BAYMAX Z_4 is a normal subgroup of itself. any group G is a normal subgroup of itself. but that doesn't contradict being simple.

simple groups have no other normal subgroups besides the trivial subgroup {id} and the whole group.

BAYMAX

yes,

yes

user400188

@DHMO
A\B would just be a the set A, with all members it has in common with B missing.

arctic tern

if m is composite, Z/mZ will have subgroups besides just the trivial subgroup {0} and the whole group

including m=4

DHMO

@user400188 in set comprehension notation?

BAYMAX

08:52

subgroups of Z/mZ are like nZ/mZ where n $\geq$ 1 , but what can be the normal subgroups of thatose?

@arctictern

arctic tern

every subgroup of an abelian group is normal

BAYMAX

ojhhh .... Z/mZ is cyclic

Hence Abelian

Hence any of its subgroups will be normal , thus Contradictiing that G is simple

@arctictern

arctic tern

well, you have to find out that it has proper nontrivial subgroups

user400188

I suspect that it would in fact just be $A\cap B'$.

If it was not that; then it would at least be $A\cap (A\cap B)'$

BAYMAX

Ok , divisors of m

arctic tern

08:56

if p is prime then Z/pZ will have no proper nontrivial subgroup. if m is composite then Z/mZ will have proper nontrivial subgroups.

@BAYMAX it will have subgroups dZ/mZ where is a divisor of m, right

user400188

I'm atempting set notation now but I am having trouble with the first guess. So I'm going back and doing it again with the second to see if it works

BAYMAX

was my anwer correct ther??

answer

@arctictern

arctic tern

what answer?

DHMO

@user400188 the first is right and the second is redundant

BAYMAX

Hence the proof , right@arctictern

arctic tern

09:01

what?

user400188

I'm having trouble writing B'. lets say I had sets A={1,2} and B={1,3}. What would $A\cap B'$ be? I can't put my head around writing B' here. I would get {1,2}$\cap${.....}

would it be everying imaginable excpet whats in b?

DHMO

@user400188 you just did it in your proof

yes

user400188

While I did it in the proof I did not try at that step to see what an example would look like. Now I have it appears that the B' set would be vast.

BAYMAX

$<g>$ is isomorphic to $Z_{p}$ p is a prime, which is prime cyclic and hence the proof follows,right @arctictern

user400188

Which made me think I had made a mistake somehow.

DHMO

09:02

@user400188 try

user400188

anyway the answer would be:
A\B=$\{x|[x\in A]\land[x\in B']\}$

DHMO

I don't like B'

user400188

me nether

Im currently trying to figure out a way to remove it

BAYMAX

@arctictern but no where the fact of $G$ to be Abelian was taken???

user400188

I wonder if (not in B) is the same as B'?

DHMO

09:04

you just did it in your proof

yes

user400188

@DHMO In my proof all I did was x in B & X in B' = T
I never had to write not in

DHMO

not really

you only looked at the last line of your proof

user400188

$A\setminus B=\{x|[x\in A]\land[x\notin B]\}$

hmm whats the latex for \ ?

or mathjax

DHMO

\setminus

perfect

user400188

I've got to go now. Bye everyone

Thanks again for the questions and tutorial @DHMO
Also thank you Paul for the reference

DHMO

09:10

live long and prosper

6

Alessandro Codenotti

09:37

@Balarka I'm a few hours late, but can one prove FTA by arguing that ever real polynomial factors into a product of quadratics and linear factors using the fact that |\Bbb C:\Bbb R|=2 and then show that the second degree polynomials have roots in $\Bbb C$ because you calculate thrm explicitely?

DHMO

@AlessandroCodenotti can you solve [R:X]=2?

Alessandro Codenotti

Would throwing away a square root from a basis of $\Bbb R$ over $\Bbb Q$ work?

DHMO

I have no idea

Alessandro Codenotti

09:53

It feels like it should

DHMO

Can you solve [R:X] = |N|?

@AlessandroCodenotti ^

Alessandro Codenotti

I'd do the same, throw away countably many elements rather than just a square root

DHMO

no other approaches?

mercio

[R:X] = 2 is impossible i think

DHMO

why?

mercio

10:07

because then you would have [C:X] = 4 and by the artin-schreier theorem this is impossible

Alessandro Codenotti

Right, that's a good point

DHMO

what is that theorem?

Alessandro Codenotti

Basically $|\bar{K}:K|=1,2$ or $\infty$

DHMO

what the hell is [H:R] then

Alessandro Codenotti

I don't remember the exact statement, but that's what mercio is using

mercio

10:12

H is not the algebraic closure of R

DHMO

oh H is not a field

mercio

it's not even a field

jublikon

Is $X^3+2X^2-3X-2$ irreducible in $\mathbb F_7[X]$?

I have checked for every $x \in [0,6]$, but the polynom never turn $0$.

Is there a possibility of factorization that the polynom will be reducible?

DHMO

@jublikon it must have a linear factor to be reducible

Alessandro Codenotti

I think nothing goes wrong if you talk about the degree of a ring extension? Anyway H is a 4 dimensional vector space over R

@jublikon A polynomial of degree 2 or 3 is irreducible iff it has no roots (prove it)

jublikon

10:22

@Allesandro But the polynomial has the root 2 ?

Alessandro Codenotti

I don't know, you said it's never 0, I didn't check

DHMO

2 is not a root

Smit

how to know what proof are we suppose to use for any given statement?

DHMO

error: question too general

Smit

for example: Odd Integers = 2n - 1 for all n belongs to integers. prove that adding power of 2 to an element of odd integers again givens an element of odd integers.

DHMO

10:30

3 + 2^0 = odd

Smit

how come 3 + 2^0 = odd? isn't 3 + 2^0 = 3 + 1 = 4 = even.

DHMO

that is a counterexample

Smit

Can you elaborate?

DHMO

look at what you asked me to prove

Smit

Oh

right..

prove that adding a positive power of 2 to an element of odd integers again givens an element of odd integers. Sorry i forgot to mention positive...

Do I need to use contraposition proof technique ?

DHMO

10:37

(2n-1) + 2^m = 2(n+2^(m-1)) - 1

that depends on the rigorosity required

Smit

it is just a 1 mark question.

DHMO

then just use what i wrote

Smit

Ok. Just to clarify. This is how you got your answer? (2n-1) + 2^m = 2n + (2.2^m)/2 - 1 = 2(n+2^(m-1)) - 1

Thanks.

DHMO

yes

Balarka Sen

@AlessandroCodenotti Well, how would you prove every poly factors into linear and quadratics?

Alessandro Codenotti

10:49

By using the algebraic closure of $\Bbb C$, I see...

user400188

Random question about defining $\neq$

Is this a good definition for $A\neq B$ ?

I define $A=B$ to be

$\forall x[(x\in A)\leftrightarrow (x\in B)]$

If ($\ref{1}$) is my definition for "$=$", then its negation is ($\ref{2}$)

$\exists x[(x\notin A)\oplus(x\notin B)]\label{2}\tag{2}$

My question: Is ($\ref{2}$) a good definition for $A\neq B$ ? If it is not, which out of ($\ref{1}$) and ($\ref{2}$) is wrong?

($\ref{2}$) seems to say that the elements of A are not guaranteed to be the same as B. It does not appear to say that they will never be the same no matter which element we pick. This is what confused me about the defini…

I was planning to ask it on SE but then I realised I would be slightly out of contex without refering to the fact that the definition of = is supposed to work for sets.

darn ran out of edit time'

forgot that the {} dont work in chat

(2) is meant to be ∃x[(x∉A)⊕(x∉B)]

DHMO

(2) is good

user400188

SO if (2) is good I have a somewhat weird question

lets say we have two sets {1,2} and {2,3}

and we make x=5

its not in either set so (2) is now logicaly false

DHMO

(2) does not say "for all x"

user400188

it does not say for all, but does it not permit x to be 5 ?

DHMO

11:03

exists x[P(x)]

it permits

P(5) being false does not contradict (2)

as long as P(1) is true

user400188

cuase if it did, then:
it would mean $A\neq B$=logicaly false.
which is logicaly equivilant to $\lnot (A\neq B)$

which is then A=B....

wait it would work beuase A would be 5 and B would be 5 too

unless I am mistaken

DHMO

@user400188 do you know the difference between $\forall x[P(x)]$ and $\exists x[P(x)]$?

user400188

for all is only true when they are all true. Exists is only when 1 or more is true

thats for statements not objects

DHMO

if P(5) is false

does that mean $\exists x:P(x)$ is false?

user400188

I made the mistake of assigning a domain to x again

I meant it would be false when the only choice for x was 5...

but I cant do that can I?

DHMO

11:06

correct

x can be anything

user400188

is there a way to set up an expression that allows that kind of thing?

DHMO

give me an example

user400188

lets say I want an expression that has similar properties as the exist quantifier; in that it is true when one memebr of the domain of what the we are saying exists is true.

DHMO

give me a specific example

user400188

its kind of hard to give one without figuring out how to do it

I mean: i could just give the above

DHMO

11:10

in words

user400188

ok; but you will have to take exists to mean (there exists a member for which the expression is true within the domain of the members)

DHMO

ok

$\exists x[(x \in A) \land P(x)]$

user400188

actualy I think I may have a solution..

oh wiat you might have beat me too it

DHMO

go ahead

user400188

I was going to write ∃x([(x∉A)⊕(x∉B)]$\land$(x$\in${1,2,3}))

DHMO

11:13

replace the = with $\in$

user400188

your right that would be nessesary

I think that will actualy work in restircting the domain

DHMO

yes

but that would be redundant

user400188

what do you mean?

DHMO

I mean, it is equivalent to (2)

user400188

hmm actualy; to get that to be logicaly false we would need...

∃x([(x∉A)⊕(x∉B)]∧(x∈{5}))

DHMO

11:17

yes

user400188

now it cant be in one but not the other

and we arrive at A=B which is not surprising

thats actualy really odd though. From writing A$\neq B\leftrightarrow$∃x([(x∉A)⊕(x∉B)]∧(x∈{5}))

we simplify to A=B

somehow by restricting the domain we arrive at A=B; even though our definition came out logically false

DHMO

I don't answer ill-formed questions

user400188

thats wasnt meant to be a question actualy. Now you say that tthough it makes me think I made a mistake..

Danu

Hey @TedShifrin, remember when we talked about how $\Bbb P(T\Bbb CP^n)$ can be identified with the manifolds of flags $(1)\subset (2)\subset (n)$? The identification was done by using $T\Bbb CP^n\cong \operatorname{Hom}(\gamma^1,\gamma^\perp)$ ($\gamma^1$ the taut. line bundle and $\gamma^\perp$ it's ortho. complement). Then picking a point $\ell\in \Bbb CP^n$ and a vector $v\in \ell^\perp\subset \Bbb C^{n+1}$, we got a 2-plane $P=(D_\ell \pi)^{-1}(\operatorname{span}(v))$: Then $\ell\subset P$.

I think this just immediately generalizes to flags $(1)\subset (k+1)\subset (n)$, obtaining $\operatorname{Gr}_k(T\Bbb CP^n)$ by picking $k$ independent vectors $v_1,\dots, v_k\in \ell^\perp$ and looking at the preimage of their span under $D_\ell \pi$

Is this story affected by identifying $T\Bbb CP^n$ with $T^{1,0}\Bbb CP^n$ as a real vector bundle?

user400188

11:34

Goodnight everyone

Astyx

Hi chat

BAYMAX

12:03

Hey guys what is the relation between $S_{n}$ and $A_{n}$ ?

groups

Danu

Have you tried to google?

BAYMAX

yeah, but due to improper math setting google gives other results

google.co.in/…

Alessandro Codenotti

How did you define $A_n$ and $S_n$?

BAYMAX

$S_{n}$is symmetric group

$A_{n}$ is alternating group

Danu

why don't you use the names

It'll be easy to find

Alessandro Codenotti

12:07

Those are not definitions (and googling the names will be more fruitful)

BAYMAX

Hey this gives me a new question

is there any software which when installed i can write similarly like as i write here

Astyx

First sentence of the "Basic Properties" tab in this wikipedia article

BAYMAX

That is simply place dollar signs , with proper LaTeX syntax

and we can generate mathematical writings

wow that would be so good

any idea guys about that one!

i think i am eating up all, sorry for that!!

Yeah thanks @Astyx

Astyx

How does one actually show that $A_n$ is the commutator subgroup of $S_n$ ? It's quite straightforward that $[G,G] \subset A_n$, but what about the other inclusion ?

Evinda

Hello!!! By solving the following problem

$\frac{dy}{dx}=\frac{1}{x-y}+1, y(1)=1$

I have found two solutions: $y(x)=x \pm \sqrt{2(1-x)}$.

Why are there two solutions, although we are given an initial condition?

Astyx

12:19

@Evinda It's not a linear equation

(You have some product between $y$ and $y'$)

s.harp

more importantly, it is not Lipschitz continuous or even everywhere defined

for example the equation $f'=\sqrt{f}$ with initial condition $f(0)=0$ has more than one globabl solution, for example $f(t)=0$ and $f(t)=\frac{1}{4}t^2$

this is because the function $x\mapsto\sqrt x$ is not locally lipschitz at $0$

Evinda

So we know that the initial value problems have an ony solution only if the differential equation is linear? @Astyx

s.harp

yes, locally Lipschitz is the condition in the Picard Lindelöf theorem that guarantees uniqueness of solutions

Astyx

Yes I made a silly confusion @s.harp

Evinda

@s.harp Is the fact that the right side of the differential equation is lipshitz continuous a necessary and sufficient condition?

Astyx

12:25

@Evinda I doubt so

s.harp

@Evinda it is a sufficient condition, but not a necessary one

Astyx

So is the lineraity of the equation

BAYMAX

Guys i am observing that you are connecting real analysis with differential equations,is there any source or book in which i can explore more @s.harp

Evinda

Can it be that the right side is not lipshitz continuous and the initial value problem has a unique solution? @s.harp

s.harp

@Evinda I think so, here is a paper describing more general settings ams.org/journals/proc/1967-018-04/S0002-9939-1967-0212240-6/…

Evinda

12:29

Ok, and if the right side is lipshitz continuous, the initial value problem has always just one solution, right? @s.harp

s.harp

@BAYMAX I haven't read it personally but Rudins book on analysis should cover ODEs, the same with Jost's book "Postmodern Analysis" (nice title I think)

@Evinda Yes

Vrouvrou

please i have $||\alpha||^2=(\alpha, \alpha)$ where $\alpha $ is $C^1$ function , what is equal to, $\frac{d}{dt} ||\alpha(t)||$

Astyx

@Vrouvrou $(\alpha', \alpha) + (\alpha, \alpha')$

BAYMAX

i will have a look 2 @s.harp

Evinda

@s.harp I see... Thank you!!!

Astyx

12:31

Unless I'm mistaken

Because $a,b \mapsto (a,b)$ is a bilinear application

Vrouvrou

@Astyx please not the derivative of the scalar pruduct , the derivative of the norm

s.harp

$(a,b)$ is the scalar product $\int_0^1 \overline{a(t)}b(t)dt$?

Astyx

My bad. Then use chain rule

Evinda

And there is no theorem for functions, that are at the right side , and are not linear and not lipshitz continous right?

Vrouvrou

@Astyx you are with me ?

Astyx

12:34

@Vrouvrou Yes

You can see it as ${d\over dt}\sqrt{||\alpha||^2}$

s.harp

@Evinda There are, like Peano's theorem that gives existence of solutions if the right hand side is continuous, or Okamura which gives existence and uniqueness given other properties on the right hand side

Vrouvrou

@Astyx what we deduce then ?

Astyx

Which gives you ${d\over dt}(||\alpha||^2) \over 2\sqrt{||\alpha||^2}$

Evinda

A ok. Thank you @s.harp

Astyx

Which is $(\alpha, \alpha')\over||\alpha||$ (I'm making assumptions as to what you mean by $(\cdot, \cdot)$)

s.harp

12:40

@Astyx Do you know what a G-space is? This question makes little sense to me right now: math.stackexchange.com/questions/2151098/…

Astyx

It should be "Let $G$ be a discrete countable group $\dots$" right ?

s.harp

yes

if $X$ is discrete getting that it is locally compact doesn't need much else ;)

Astyx

True :p

Wolfram might help you more than me

DHMO

what is the difference between norm topology and metric topology?

Astyx

@DHMO metric topology is derived from a metric space, whereas norm isn't

s.harp

12:45

@DHMO whenever you have a norm, the topology it induces is the topology given by the metric that is induced by the norm.

So its the same, except that not every translation invariant metric on a vector space is induced by a norm

DHMO

what is the usual topology on Q?

Astyx

well I'm not sure about terminology, but isn't there a difference as a norm has some properties with scalar multiplication and vector addition, whereas a metric applies to other things than vector space ?

Or the other way around

s.harp

@AlessandroCodenotti thats if a norm is induced by a scalar product no?

Alessandro Codenotti

Indeed, I got confused

s.harp

@DHMO the usual topology on $\Bbb Q$ is induced by the metric $d(x,y)=|x-y|$

This has a basis of open sets $B_r(x)=\{y\in\Bbb Q\mid |x-y|<r\}$ where $r$ is real (not necessarily rational)

Alessandro Codenotti

12:50

So a norm induces a translation invariant metric, but is every translation invariant metric induced by a norm?

Astyx

Yeah that's it, a metric $d$ does not need to be compatible with operations of the space (and can in fact be defined on any set $X$). It only needs to satisfy $d(a,b) =0 \iff a=b$; $d(a,b)=d(b,a)\ge 0$; and $d(a,c) \le d(a,b) + d(b,c)$

s.harp

@AlessandroCodenotti no, consider for example $C(\Bbb R)$, where you have semi-norms for any compact $K$: $\|f\|_K:=\sup_{x\in K}|f(x)|$. You can get a translation invariant metric on $C(\Bbb R)$ via:
$$d(f,g)=\sum_{n=0}^\infty 2^{-n}\frac{\|f-g\|_{[-n,n]}}{1+\|f-g\|_{[-n,n]}}$$

this metric is not induced by any norm, indeed the topology induced by this metric also cannot be induced by any norm

Astyx

@DHMO Whereas a norm needs to be homogeneous of order 1 and satisfy the triangular innequality

s.harp

(cf Frechet spaces )

Alessandro Codenotti

Neat, thanks!

s.harp

12:55

@DHMO a sort of clarifying remark: $\Bbb Q$ has the property totally disconnected with the usual topology. This means for any open set $A\subset\Bbb Q$ you have two open sets $U,V\subset A$ that are disjoint ($U\cap B=\emptyset$) but that have $U\cup V=A$. This means every open set is disconnected

You can see this for example with $A=(0,1)\cap\Bbb Q$ by taking $U=(0,\frac1{\sqrt2})\cap\Bbb Q$ and $V=(\frac1{\sqrt2},1)\cap\Bbb Q$

DHMO

interesting!

Evinda

I want to solve the following initial value problem:

$\frac{dy}{dx}=\frac{y}{x+\sqrt{x^2+y^2}}, y(0)=1$

By making the substitution $y=xz$ I got the following differential equation:

$\frac{(1+ \sqrt{1+z^2})}{z\sqrt{1+z^2}} dz=-\frac{1}{x}dx$

How can this be solved?

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