@JayeshBadwaik My argument was incomplete. In a monoid $M$, the maps $x\mapsto mx$ and $x\mapsto nx$ can be identical even if $n\ne m$. As a concrete example, take $M=\{n,m\}$, where the multiplication of two elements returns whichever element is on the right. (Exercise: verify associativity.) Then both induced maps are the identity. The fix is if $M$ doesn't have an identity element, you can by fiat give it one: set $M^\circ=M\cup\{1\}$ and extend the multiplication table in the obvious way...

Then $M$ is isomorphic to a submonoid of $(M^\circ)^{M^\circ}$, in particular the submonoid of left multiplication maps $\varphi_m:M^\circ\to M^\circ:x\mapsto mx$ for the $m\in M$. (Exercise: check any two maps $\varphi_m,\varphi_n$ are distinct if and only if $m,n$ are distinct.)

@MeAndMath Marylin is a nice name.

@MarianoSuárez-Alvarez Hola, Mariano. Estoy tratando de ver como demostrar el teorema delos intervalos encajados (de Cantor)

qué enunciado?

@MarianoSuárez-Alvarez TEOREMA Considerese la sucesion de intervalos cerrados $I_1=[a_1,b_1]$,$I_2=[a_2,b_2]$,$\dots$,$I_n=[a_n,b_n]$ tal que $a_n\leq a_{n+1}$ y $b_{n+1}\leq b_n$ para todo $n$. Entonces existe un $x$ tal que $x\in I_n$ para todo $n$, es decir $$\bigcap_{n=1}^\infty I_n\neq \varnothing$$

ok

y qué intentaste?

@MarianoSuárez-Alvarez Estoy pensando en considerar $\sup\{a_n\}$ y e $\inf \{b_n\}$.

basta considerar el primero

ya que solo querés ver que la intersección no es vacía

@MarianoSuárez-Alvarez Esta claro que $a_n\leq b_n$ para todo $n$.

Por lo tanto $\sup\{a_n\}\leq \inf\{b_n\}$

no es que está claro; está dicho en el enunciado :-)

@MarianoSuárez-Alvarez jajaja bueno.

Por hipotesis "bla bla"

olvidate del inf de los b's

ocupate de las aes nomás

@MarianoSuárez-Alvarez OK. Me pongo a pensar.

llamá alpha al sup de las aes

y mostrá que ese número está en todos los intervalos

@MarianoSuárez-Alvarez Por curiosidad, no serviria tambien usar $\inf b_n$? Quiero pensar que a la larga debe ser $\inf b_n=\sup a_n$, con lo que existe uno y solo un $x$.

no es cierto que el inf y el sup sean iguales

@MarianoSuárez-Alvarez OK.

podria ser que a_n=0 y b_n=1 para todo n, por ejemplo!

o a_n = -1/n y b_n = 1+1/n, para un ejemplo menos tonto

@MarianoSuárez-Alvarez Cierto. No son estrictamente crecientes.

aunque lo sean: cf. mi segundo ejemplo

@MarianoSuárez-Alvarez OK.

Moving on...

@MarianoSuárez-Alvarez No parece ser tan complicado. Si $\alpha=\sup A$, entonces $\sup A\leq b_n$ para cada $n$, y es $a_n\leq \sup A$ para cada $a_n$ =P ????

con lo que $\alpha \in I_n$ para cada $n$.

nadie dijo que fuera complicado :-)

@MarianoSuárez-Alvarez ok, ok

porque es sup A \leq b_n para todo n?

cada afirmación que hagas tiene que estar probada

@MarianoSuárez-Alvarez Porque cada $b_n$ es cota superior de $A$.

ok

pero la prueba es mejor organizarla fijando el n antes que nada

@MarianoSuárez-Alvarez Como fijando $n$?

«sea $n\geq1$. Como $b_n$ es una cota superior de $A$, es $\alpha\leq b_n$; como $a_n\in A$, es $a_n\leq \alpha$. Entonces $\alpha\in[a_n,b_n]$.»

@MarianoSuárez-Alvarez Ah, bien.

vemos entonces que $\alpha\in\bigcap_{n\geq1}[a_n,b_n]$.

@MarianoSuárez-Alvarez Ahora Spivak quiere que demuestre el teorema de Bolzano con esto, y lo mismo para $f$ continua en $[a,b]\Rightarrow f$ acotada en $[a,b]$

y bueno, hacelo ;)

Usando el metodo de biseccion.

@MarianoSuárez-Alvarez Jejeje, espero que no se me haga muy tarde.

siempre es la misma idea

buscás algo en un intervalo, lo partís en dos y lo buscas en las dos mitades

I think this needs closing:

Actually the match is not as close as I thought. This bucket contains four coins and the bucket in the other questions only contains three.

But it might need closed anyway.

@MarianoSuárez-Alvarez Could you please help me with this?

@WillHunting Can you help me with something?

@wj32 Clearly an infinite strictly ascending chain of submodules does not have a maximal element, so by contraposition this gives the maximal-element-$\implies$-ACC direction. For the converse, suppose $M$ has ACC and let $\cal S$ be a collection of submodules; then any ascending chain is a linear/total order and has a maximal element by ACC and hence by Zorn's lemma $\cal S$ has a maximal element.

@MeAndMath O teu nome é a versão portuguesa de Marylin? Marília?

@MeAndMath Mas eu não tava reprovando, tava achando bonitinho. =)

@WillHunting Yay! you're back!

@WillHunting I was trying to figure out in which order I should take the OCW course in order to study theoretical mathematics.

I've found the answer.

Relax, I know I have a lot of work to do on more basic stuff first.

@PeterTamaroff Outspoken Badge, Dude....

btw

I've already found the answer here:

math.mit.edu/academics/undergrad/major/course18/…

@WillHunting How do you find a way then?

@WillHunting Do you have some problem on following a university schendule?

@WillHunting If I finish the books I'm reading now, I'll have to risk it.

At least until I enter the university.

@WillHunting You mean: Finishing a chapter and going to read something else?

Are you in a hurry?

Is this a real problem, I mean...

...If someday I find something hard to understand, won't it be obvious that I don't know about the components of that problem?

You usually warn me to don't read advanced stuff - but they're kinda impossible to read.

Some time ago I tried to learn about monoids, I thought I could learn about them.

For the first three, I remeber from school, maybe I'll need only a refreshing.

Combinatorics = some probability problems?

I know things such as: "There are 26 letters, how many possible words can you write with 4 letters?" Ans: 26*26*26*26

If the letters can't be repeated, you use factorial, and things like that.

@WillHunting What would you recommend me then?

Yep.

I'm gonna start with single variable calculus.

@WillHunting Yes.

@WillHunting I'll make this experiment as soon as I go back home tomorrow. =)

Yep, I was thinking on this possibility too

For example: The only thing I know about a monoid is that it is a set and an operation.

@WillHunting I'll have to wait ~1 year to do my degree, I want to use this free time I have now.

@WillHunting Yep, the point is: "If this is the only thing I know, then there's a lot I need to learn"

What will introduce me to proofs?

I'm reading a discrete math book.

It introduce proofs in the beginning.

At the beginning of everything I study, I get afraid and cautious.

I did the same with the piano.

Now I have less fear of searching my way.

Yep. I learned the last ones by building an ALU

Yep.

I'm gonna study something now. =)

Thanks for the advice.

btw

Are you okay?

Ok

@anon Thanks man.

@anon @robjohn Close and reopen action?

Iirc it's my fault that this one is getting closed.

@JayeshBadwaik I still didn't hear the songs because my internet sucks.

Tomorrow I'll be back at home and I'll hear them.

@GustavoBandeira okay!!

Apples. And pears.

@JonasTeuwen Maybe bananas.

You're driving me bananas.

U FRUITY?!

Very much. See you guy(s).

Jonas. Why you hate me?

Does $\mathfrak{c}^\mathfrak{c}$ have a standard name?

yes, that

I prefer to write it 2^c, though

You can write it also $\beth_2$, but no one knows what that means

(maybe it is $\beth_1$: you should check...)

ah, no, it is $\beth_2$

(the character is \beth, the second letter of the hebrew alphabet, coming right after aleph)

Why?

what do you mean, why?

I guess you're thinking the same I'm thinking now.

not really

I don't know if this question should be made, I guess it's more useful to understand it as it is.

hm?

The question is: "Why the derivative is the slope of the tangent line?"

isn't this explained in every single textbook?

Oh, I'm not with a textbook now.

the quotient $$\frac{f(x_0+h)-f(x_0)}{h}$$

I was just watching some lectures.

is the slope of the line going thourgh $(x_0,f(x_0))$ and $(x_0+h,f(x_0+h))$

that line is called a «secant» of the graph, because it goies through (at least...) two points

now, as $h$ becomes smaller and smaller, the secant approximates the tangent line at $x_0$

that is the idea

Yep, I'm also going to consult a textbook now.

@MarianoSuárez-Alvarez Thanks. I never did read much about beth numbers.

they are not very fashionable :-)

they make for pretty statements for the continuum hypotheses, tho

@MarianoSuárez-Alvarez did you read about the possible proof of abc conjecture?

It runs into 500 pages!

I read about it

that's taking intto account previous work by the guy

ohh.

@anon @MarianoSuárez-Alvarez Thanks, but I think that the other direction is wrong

(chat.stackexchange.com/transcript/message/6090517#6090517) What if I just choose two subspaces that are completely unrelated, and put them in a collection $\mathcal{S}$?

Good morning

hi

@wj32 At first glance, it looks like there is like there is a condition missing from that ACC thing you are worried about

yes, i think it's a mistake in the text - just because it satisfies the ACC doesn't mean that every collection of submodules has a maximal element

would it be true if the word "nested" were inserted somewhere?

Even if the word "nested" is not needed, a useful tactic when struggling to prove something is to add some extra conditions and then see if you can prove it - just to get a bit more insight

a bit like in number theory - if I can't prove something for all $n$, I try to prove it for the case where $n$ is a prime - or a power of a prime ...

Maybe it is not a misprint - are we talking about maximal elements in a partially-ordered set here?

it just says "collection"

If you take the partial order of set inclusion for your submodules, then look at any 2 submodules where neither is a submodule of the other, I think the are both maximal elements, are they not?

but it explicitly defines "maximal"

what is the exact definition it has for "maximal"

as an $S\in\mathcal{S}$ such that $T\in\mathcal{S} \Rightarrow T \subseteq S$

oh well, the book has some other stupid errors as well

but it's pretty damn good

Hmm - the definition I have seen for maximal says something like "$A$ is a maximal element if for any $B$ which is comparable (in the partial order), we have $B \subseteq A$"

he's not attempting to define "maximal" in general - it's just for that question

Looking in Hungerford's Algebra - he then has a theorem which looks exactly like the one you have

thanks, i'll take a look

page 373 in Hungerford, if you have access to the book

Votes for reopen please.

@Matt agreed - done

@JohnSenior ok, i see what you mean now

so the question in Advanced Linear Algebra was slightly incorrect

@wj32 I think the statement is true, if you delete the very last bit of his statement - where he explains "maximal"

- BUT - I know very little abstract algebra, so I might be talking total rubbish :)

yes, the theorem in hungerford makes sense

thanks for the help

@wj32 Yep - I think so

no problem

Yep.

There are no problems. Not even challenges.

how about this:

prove that an open, continuous mapping from $\mathbb{R}$ to $\mathbb{R}$ is monotonic

some of the messiest stuff i've ever seen

That doesn't sound too hard. A couple applications of the intermediate value theorem and the extreme value theorem should do the trick.

Or if it would not be monotonic do some divide and conquer.

@JonasTeuwen I was thinking the same. I have the proof by contradiction. Now thinking of how to do it straight, without using contradiciton.

Nah, no contradiction.

Contrapositive.

I no fancy contradiction.

The annoying thing is that monotonic is like a global statement and me no like, so contrapositive would somewhat rephrase it as a local one 8-).

But first reduce this to the fact that the function is not invertible. As if it would be then... no problem.

Etc etc. I need a walk and a brew. Bye guys.

Bye. I need some coffee too.

@WillHunting Coffee is as always heavenly.

The todd-coxeter algorithm is however not so heavenly. :P

Today many of the bounty problems are quiet elementary.

Hello

@DantheMan Hi

Substitution or addition/subtraction?

Draw a sketch first.

?

a graph?

Yes.

@JohnJunior Hi John!

@DantheMan Hi!

fooplot.com/plot/2wazgapt4n

Very nice :D

Thank you very much!

Now what?

You tell me.

substitution?

What are you looking for?

the sets of x and y

Yes, where are they on the graph?

$(2,4),(-2,-4),(4,2),(-4,-2)$

But can I find them with substitution or addition/subtraction?

Give it a try.

I did

Don't know where to go

Look at the forms of the two equations and decide which would be a good choice to use.

Substitution

Go for it!

k

$x^2 +\frac{64}{x^2} = 20$

@DantheMan In cases you have two options and do not know where to go, sometimes, one way you can learn well is try both of them and do not worry if you take a wrong one.

ok

And my suggestion for this specific problem is to try both substitution and addition/subtraction.

$20y-y^3=8x$ with subtraction.

You can get solutions by both the methods, and I would like you to work out both of them.

Hint: Addition/Subtraction does not have to be direct, it can be of the form $(1) \pm k(2)$

where $(1)$ and $(2)$ are equations.

But now that you had chosen substitution, first complete it, and then think about subtraction, addition.

This is the substitution method

I don't know how to finish that.

So, you can solve quadratic equations right? Can we reduce this problem to a quadratic equation?

Always see what tools do you have in your toolbox, then think how can you transform the problem so that you can solve it by using one of the tools from your toolbox.

Square root each side?

What will square root give on the left hand side?

I don't know

$x + 8$?

Do not rush to conclusions without working to it.

$\sqrt{x^2 +\frac{64}{x^2}}$

This is the square root right?

yes

Can you think of any method to simplify it?

rationalize the denominator?

No, in general, can you simplify $\sqrt{x^2 + y^2}$ ?

no

right?

yes

so we cannot simplify this using square root.

So, that is not an option.

yeah...

What other option do we have?

of making this a quadratic

$x^2 +\frac{64}{x^2} = 20$

What is the degree of this equation?

2nd

isn't it already a quadratic?

Quadratic in?

?

There is a $x^2$ in the denomiator. Does it affect the degree?

I don't think so... Does it?

I just bought a book

@DantheMan What is the definition of a polynomial $p(x)$?

@JayeshBadwaik The answer is nor yes nor nor. Henceforth the question is wrong.

@JonasTeuwen Thanks.

@JayeshBadwaik It has many terms

@wj32 In a set that is given an empty partial order (i.e. no distinct elements can be compared), every element is a maximal one. Maximal simply means "has no element greater than," which is vacuously true if there are no relations. I gave an argument for both directions of the iff; is there any aspect you have a question about?

@JonasTeuwen nor nor?

Yes.

@JonasTeuwen @JohnJunior Please interrupt me whenever I am going too much off the track. My aim is to cut to the basics and construct a solution from there.

@JohnJunior Nor yes nor no.

neither yes nor no

@JohnJunior Nor.

@JonasTeuwen Neither

@JohnJunior Be careful or I'll nullify your head.

@JayeshBadwaik Why can't they be negative?

@DantheMan They are defined that way.

@JayeshBadwaik ok

@wj32 Ah, you are right in that that given definition seems to be too much stronger than the definition of maximal. I didn't even realize that! It does appear to be an error.

So, come back to our equation

$x^{2} + \frac{64}{x^2} = 20$

Suppose we rewrite this as $p(x) = x^{2} + \frac{64}{x^2} - 20$

ok

Then our job is to find the zeros of the polynomial $p(x)$

ok..?

right

continue

no

Hey, what's the fuzz about the abc-conjecture?

@JayeshBadwaik Yes. How do we do that?

@NickKidman sbseminar.wordpress.com/2012/06/12/abc-conjecture-rumor-2 ?

$x \neq 0$

@DantheMan So, since $xy \neq 0$

So now assume $q(x) = x^{4} - 20x^{2} + 64$

We have to solve $q(x) = 0$

Is $q(x)$ a polynomial?

Ah yes!

Is it quadratic in $x$ ?

no

Can we make it quadratic? possibly in some other variable which is a function of x probably?

We could foil it?

foil?

First Outer Inner Last

Factor it

Or may be we can transform it?

$y=x^2 $

@JayeshBadwaik There are some quotes now on my website!

@JayeshBadwaik Ah ok! but not $y$ because that is part of our equation.

@DantheMan yup, good. So $z$?

@JonasTeuwen Superb.

@JayeshBadwaik ok

The Old Greeks (all dead now) ROFL

$z^2-20z+64 = 0$

Can you solve it?