So it holds for that, but still, it's over-complicating it I am afraid

if A is an adjacency matrix of a graph G and m>1 an integer do you know what the entries of A^m represent?

the (i,j) entry of A^m counts the number of m-step paths from i to j

one can prove this by induction on m

say we have a cycle of length $m$. take a multiple of $m$ greater than $n$, say $jm$. what do we know about $A^{jm}$?

Yes, I know that

I proved it, too

number of walks of length jm, and it means the diagonal elements have walks of length m

cycles*

so if i go around the cycle $j$ times that's a walk of length $jm$ from vertex $i$ to $i$, for some vertex $i$

What does it mean to "write the polynomial as the product of factors that are irreducible over the rationals". My first inclination is to find the zeros but for this specific problem there are no zeros, is that possible? Im using synthetic division to find the zeros of a polynomial with a leading root higher than 2

@Brittany "irreducible over the rationals" means that it can't be factored to any polynomials of smaller degree that have rational coefficients.

Yes, @Alex, I see where this is going

But that proves one direction only, doesn't it?

@Brittany every polynomial factors as a product of irreducible polynomials, just as every natural number factors as a product of irreducible numbers ("primes")

That if A^n=0 there are no cycles

matrices blow

The other direction is that if there are no cycles, A^n=0, and with that I struggle

@Studentmath what if $A^n\ne 0$?

is there a cycle?

There's a walk of length n

I need to prove that if there is a walk of length n,

it oughta be a cycle

@seaturtles thank you. So is it fair to assume they all mean the same thing, they are just saying "find the zeros" in a hard to understand way?

since there are only n vertices

@Brittany nope. why do you think finding irreducible factors and finding zeros are the same over the rationals?

@Studentmath ah so $n$ is the order of the graph?

Correct

@Brittany what they're saying is actually a bit deep

That probably made it much easier and I sound extremely stupid, doesn't it? :P

is it true that the length of a cyclic of a digraph must divide the order of the graph?

@seaturtles nah, you can just add vertices arbitrarily and it won't affect anything

@seaturtles I dont know, I am just trying to understand what they are asking =P I like exact precise understandings so I tend to group them together to understand a concept. In this case though its not true and I'm confused =$

@Brittany do you know about the conjugate pairs theorem, for example?

@Brittany which part of the problem don't you understand? you know what a factor is, you know what irreducible means, and so therefore you should know what it means to find all irreducible factors (even if you don't know how). no?

Lots of thanks for the patience and help @Alex!

@Studentmath did ya get it?

for example, over the real numbers, $x^4+1$ doesn't have any zeros, but it has two irreducible factors: $$x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1).$$ so obviously finding irreducible factors doesn't just mean finding zeros.

@AlexanderGruber If (a+bi) is a root of a polynomial then (a-bi) is also a root

I hopefully did, will try to prove that if there is a walk of n length it oughta be a cycle

But then again I may be off :P

@Brittany correct. so here's an example: $x^4-1=(x^2-1)(x^2+1)=(x+1)(x-1)(x^2+1)$

one can think of it this way: over a field, given a polynomial there is a finest factorization - a factorization into a product of polynomials which cannot be reduced further (by factoring those factors further).

@seaturtles yes you're right, this is helping me clear this up thank you, let me look at the problem again

If anon was here he could explain this to you :P

heh heh

but the roots of $x^2+1$, $i$ and $-i$, aren't rational numbers. so we can't factor it anymore if we're only working over the rationals.

also, $x^8-1=(x^4+1)(x^2+1)(x+1)(x-1)$ over $\Bbb Q$. we can't factor $x^4+1$ any further over the rationals.

turns out there's a similar theorem, for example, for polynomials that have roots involving $\sqrt{2}$. if $a+b\sqrt{2}$ is a root, so is $a-b\sqrt{2}$.

wooo galois conjugates

you'd say in this case that $\left(x-(a+b\sqrt{2})\right)\left(x-(a-b\sqrt{2})\right)$ was irreducible over the rationals (when multiplying out)

*after

@Studentmath think about it this way, if you start with a vertex $i$, try to make a walk of length $n$ that doesn't contain a cycle. how many vertices can you visit?

So simple. n-1.

@studentmath: What I said was garbage. I was thinking of the boundary matrix, not the adjacency matrix. Sorry.

@Studentmath boom!

No problem @TedShifrin, got me studying a bit about nullspaces. @alexander that's just amazing, thank you!

@Studentmath didn't we prove this earlier?

@Brittany think of it as a quadratic polynomial in the variable x^2

@Mike: You still live!

@Ted Tragic, I know

@Mike we showed something far more interesting, I think, but not relevant to the question. Because it is only in effect if G is a simple digraph

@Studentmath pigeonhole principle is OP.

Otherwise it doesn't hold

@Mike: I'm going to a concert, but you owe me a topology update.

yay

@Ted We went over problems; by the end of the quarter we're to do the appendix of ch2 and universal coefficient theorem

No, I meant the problem we were working on ...

Which one was that?

@seaturtles can you rephrase that? I dont know what you mean, I promise Im not a 3rd grader! This is just a new set of ideas for me =)

The $A$ contractible in $X$ one ... But I have to leave ...

@Brittany can you factor $u^2+6u-27$? then notice that this is exactly the problem you have, with $u=x^2$.

Oh, that was ultimately geometric and not very alge raic

That's what I suggested ...

You can show me later :)

I think I got it via a different geometric thing than you did. But yes, it was. :P

I have bigger fish to fry than topology, anyway

@seaturtles no, okay so since its not factor-able...

Enjoy the concert!

I shall, @Mike, tanx.

@TedShifrin who're you seeing?

Ted doesn't kiss and tell

@Brittany you can factor $u^2+6u-27$ ! did you try?

find two numbers that multiply to -27 and add to +6

@Brittany When you talk about "factor-able" you have to talk about factorable over a given set of numbers

@seaturtles holy shit you can, oops!!

@skullpatrol not relevant at the moment

@Danny I'm sorry, I have not reallly worked with $\chi^2$ distributions...

@seaturtles 3, -9

@Brittany add to +6

@robjohn just square root and get $\chi$, no big deal.

@Brittany as skull says, it's the other way around: -3 and +9. so $x^4+6x^2-27=(x^2+9)(x^2-3)$. can either of $x^2+9$ or $x^2-3$ be factored over $\Bbb Q$?

@skullpatrol -3, 9

@AlexanderGruber Thanks, I'll unpot a square plant and use the root...

:)

@robjohn you ever heard that joke, what do you get when you drink root beer out of a square cup?

@AlexanderGruber nope...

straight beer?

@robjohn tipsy

I like it

huh?

@skullpatrol Can you explain the Q?

@AlexanderGruber here I was trying to find the punch line... :-)

@skullpatrol because $\sqrt{\text{beer}}^2=\text{beer}$

icic

@Brittany let sea turtle...

@skullpatrol oops i meant @seaturtles sorry

np

:)

@Brittany Q is the set of rational numbers

" expressible as a ratio of integers."

@seaturtles P= (+,_) 1, 3, 9, 27

@seaturtles Q is just 1

huh? x20

@seaturtles story of my life

I asked you if x^2+9 or x^2-3 could be factored further over the rationals

@seaturtles i know i dont know what that means

meaning, do there exist rational-coefficient polynomials a(x), b(x) such that x^2+9=a(x)b(x), or x^2-3=a(x)b(x)

@seaturtles no

correct

any idea why?

@seaturtles this is completely factored

@seaturtles because math

@seaturtles haha no idea

I'll do it for x^2+9. Suppose x^2+9=a(x)b(x) for two nonconstant polynomials a,b. Then both a(x) and b(x) would be degree 1, which means they would have rational roots, which would mean x^2+9=0 has a rational root. but there is no square root of -9 in the rationals.

the same argument works for x^2-3.

challenge: which polynomials satisfy $fg \mid f\circ g-g\circ f$?

@seaturtles here's another argument

by the quadratic formula, the roots are $\pm i3$, which aren't rational

@seaturtles I see, okay great thank you! So now I am asked to write it irreducible over the reals, does this mean with imaginary numbers?

@AlexanderGruber the roots of x^4+1 aren't real, doesn't mean it doesn't factor over R

@seaturtles yup, works for quadratics though.

@AlexanderGruber indeed, but someone new to this should know why.

don't bring out a sledge hammer to kill a fly :)

@seaturtles absolutely, i agree. but for heuristics, this is good to know.

@Brittany it means factor it into real-coefficient polynomials as much as possible. why would you think imaginary numbers are supposed to turn up in a real number only situation?

after all there are a lot of polynomials that are not easy to prove irreducibility for.

@AlexanderGruber also, if one knows this for quadratics (also cubics, same reason), one could solve x^2=a much easier than using the full quadratic formula

@seaturtles I dont know, you ask the best questions.

(:

why is $x^5-x+1$ irreducible over the rationals, for example?

so, you already have $(x^2+9)(x^2-3)$

@AlexanderGruber because you can't bring radicals

ba dum tss

:D

now, @Brittany. can $x^2+9$ be factored as two nonconstant real-coefficient polynomials? what about $x^2-3$?

@seaturtles i once asked a student to solve $x^2-9$ by completing the square. good times

why?

yeah, I've seen that. "but where's the $bx$ term?!?"

@skullpatrol because once they get it they understand square completin' better

@robjohn ok

dumb questions are the path to nirvana

look what happened to their lead singer

@Danny I have seen $\chi^2_k$ for $k\in\mathbb{N}$, but not $\chi^2_\alpha$

Hah, finally done. Glad I had another question afterwards to regain some confidence after that one that went so wrong.. Many thanks Alex, Ted and Mike, again!

@robjohn maybe u can help me with another thing i cannot get. have you worked with $P$ values?

other enjoyable questions include finding the horizontal asymptotes of $f(x)=1$, factoring the polynomial "x", taking the complex conjugate of $3$

@AlexanderGruber $x=\sqrt{x}\sqrt{x}$

1=1

or "where does $f(x)=1$ cross its horizontal asymptote"

that was pretty good too

y-intercept of x=0 / x-intercept of y=0

@seaturtles oh i love that one

find the point at which $f(x)=1$ attains its maximum

loaded question, using "the"

@seaturtles I would THINK that $x^2+9$ would be the one I would break down because its the bigger number. But I dont think it can be factored and two non constant real coefficient polynomials because it is positive. So $x^2-3$ im guessing would be the one to"break down" Am I looking at this right?

@robjohn compute $\int dx$

@AlexanderGruber constantly amazes me

@Brittany right. bigger has nothing to do with it. the same reasoning as before shows that if $x^2+9$ were to factor then the reals would have a square root of negative nine, which is impossible. now, try to factor $x^2-3$ over the reals.

:14059668 $\infty\ne\infty$

thanks :(

What are we discussing now?

@PedroTamaroff enjoyable precalculus questions

i have eight observations $(2,3.2,3.8,2.5,3.3,2.8,3.0,3.4)$ i need to calculate the $P$ value assuming $H_0: \mu =3.2$ against $H_1: \mu \neq 3.2$ when $\sigma=0.6$ . should i use the values that deviates at most from $3.2$ which is 2 and 3.8 ????

@robjohn that question still "haunts" me

@AlexanderGruber NAIZ.

@AlexanderGruber

My new fun activity is Riemann Stieltjes integration. UGH.

@PedroTamaroff yikes

oh wait.

nothing is fun

i actually think i like that. that's $\int \omega(x)f(x)dx$ or something like that right

some sort of weighting function thing

df, but that's the same for differentiable f (change your f to f')

@seaturtles Im left with $x\pm\sqrt{3}$

@seaturtles Sorry im responding so slow, im trying to look up meanings of some of these concepts !

indeed $x^4+6x^2-27=(x^2+9)(x+\sqrt{3})(x-\sqrt{3})$ is the full factorization over the reals

@AlexanderGruber No, Riemann Stieltjes is $$\int_a^b fd\alpha$$

Where $\alpha$ is some other function.

The sums are of the form $$\sum f(t_i)\Delta \alpha(x_i)$$

@AlexanderGruber The best thing one can do is ask $\alpha$ to be of bounded variation.

beat you @prdro

@Brittany well, I saw the $u=x^2$ trick just by looking at it, and it took another couple seconds to do the full factorizations over rationals, reals and complex numbers. this comes with lots of experience.

@Mike Your comment was too little, I didn't see it.

Irony?

Not now, on phone

I had an awfully long nap today.

Just woke up.

It's 10 p.m.

My sleep clock is fucked up now.

good

@seaturtles Yeah I liked that u trick, I remember it from awhile back.

@Brittany very good

don't worry bud, I'm sufficiently damned

@seaturtles so now to write this in "completely factored form" I would need to bring out the imaginary numbers right?

yes

@Mike You seem a little bummed today.

@seaturtles How did your question do?

meh

@seaturtles Im ending with $(x\pm3i)(x\pm\sqrt{3}$

yes

@seaturtles yayayayyy

@seaturtles thank youuuuu

np

that'll be $14.99

I think she's got it :D

@seaturtles does that include sales tax?

@seaturtles haha can I pay you in trident layers

youtube.com/watch?v=YHArD6LgTAI once you get into the how its made section of youtube, there is no escape

@seaturtles I just read Wielandt's proof of Sylow I.

show-off

@DavidWheeler Ah?

lol, name-dropper

it's a very elegant argument, i will agree

It's nice because it follows the line of subsequent arguments, namely, working things by (non) divisibility by $p$.

youtube.com/watch?v=YsC0oK2Y09c This is ridiculous.

@skullpatrol well, as a point in the complex sphere, $\infty$ is well defined. what $\infty\ne\infty$ is trying to convey is that when two limits are $\infty$, the difference and the ratio of the sequences cannot be determined without more information. Sort of like when two limits are $0$, sequence 1 raised to sequence 2 cannot be determined without further information (though we can still define $0^0=1$).

@robjohn thanks :)

@PedroTamaroff primes are useful that way. Besides, counting things is easier than other kinds of stuff.

@DavidWheeler and @Mike your avatars are sooo similar :)

@PedroTamaroff Irony zone isn't working for me

@Mike Yes, seems server is down.

lame

@Mike I am doing some exercises.

If $X$ is a $G$-set, $H$ a subgroup acts transitively on $X$, for any $x\in X$, $G=H{\rm stab}\, x$

I hope you're enjoying them

@Mike Yeah, specially when I get to solve them. =)

:)

For this one, fix $x\in X$; pick $g\in G$. Then $gx\in X$, so I know there is $h\in H$, such that $hx=gx$. Then $h^{-1}g\in{\rm stab}\, x$ and $g=h(h^{-1}g)$.

A particular case of this is the Frattini Argument #namedropping

@seaturtles My question on $A_n$ is kinda meh too.

It's a simple enough argument I don't know that it deserves to be named

@Mike It's because Frattini used it in a paper when he worked on his group $\Phi(G)$.

So the name caught up.

This one is quite trivial, is it not: if $G$ has a unique $p$-Sylow for each $p\mid |G|$, then $$G=\prod_{p\mid |G|}S^{(p)}$$ Where $S^{(p)}$ is the $p$-Sylow.

@Mike Check the irony zone now.

n ot loading

@Mike Now the page is loading though, but with an "Awwwwww...." message.

Now it works.

not for me

perpetual "plugging you in"

Yeah, me too.

I was deceived.

whoa i'm in

am i?

meh.

=D

@TedShifrin what is rational homotopy theory? if i like rational numbers, will i like it?

i give up for now btw

i'll take a nap or something

@Mike Dude.

What's that song.

"X's World"?

The video is in black and whit.

A guy with a stache sings.

@PedroTamaroff if the $p$-sylows are unique, they are normal, so:

@DavidWheeler Yes.

@PedroTamaroff not only that but they're characteristic

@AlexanderGruber Yes.

Because unique order. =D

mhm :)

I am trying to prove that in the general case, if I pick a $p$-Sylow for each $p$ a divisor of $G$, they generate $G$.

Regardless of normality.

@skullpatrol Mike's is pointier

@Mike: what's up with the irony zone?

@FernandoMartin Mike played a song so lame the site crashed and burned.

@FernandoMartin Did you get your result on the test? Was it oral?

yup, 10 :)

it was oral

Cool.

@AlexanderGruber I think I got it.

@PedroTamaroff hmmm interesting one

there's some cool stuff to that, too

If $P_q$ and $P_p$ are distinct Sylows, $P_q\cap P_q=1$.

Oh, but that doens't help further.

@PedroTamaroff really?

:)

Oh, it does. I'm being stoopid.

I was thinking something else.

So yeah, that is easy too.

i'm not so sure about that.

@AlexanderGruber OK, let me explain my argument.

Say $p_1,\ldots,p_r$ are all my prime factors.

Then I claim $\langle S_1,\ldots,S_r\rangle$ has size $|G|$.

@AlexanderGruber You there?