Mathematics - 2014-02-02 (page 2 of 5)

Pedro Tamaroff

04:00

@TedShifrin Not sure.

Ted Shifrin

But the bigger problem, @Mike, was people who did fine on quals but had no one wanting to be their adviser ... or students too shy to seek out potential advisers and do reading courses with them.

Mike

@Ted How would the former occur?

Ted Shifrin

@Pedro: About what? My original true/false?

Pedro Tamaroff

@TedShifrin No, I am sure that is false.

I'm saying the second one.

Mike

Also, when does one start trying to find an advisor? Or if one has advisor(s) in mind, when does one approach them?

@Pedro I'm fairly sure it's true, but you certainly couldn't do it without the structure theorem.

Ted Shifrin

04:01

It occurred, @Mike. Berkeley's grad courses are typically hands-off on the part of the faculty. Very few assign homework, and so they don't get to know the students very well unless the student is aggressive and pro-active. One problem with a big department.

Oh, the abelian case, @Pedro.

It's truly a counting argument, once one knows the structure theorem.

My advice, @Mike, is not to be shy. Be polite but aggressive. Take advanced courses from the person if possible and make sure he/she gets to know you. And make your intentions known.

Pedro Tamaroff

But I told you I don't know that effing theorem =)

Ted Shifrin

Well, @Pedro, learn that effing theorem. :D

usukidoll

lol

Ted Shifrin

It's easy to state.

Mike

@Pedro He told you when he posed the question that "assuming the structure theorem..."

usukidoll

04:02

anyway 3 down 4 more to go on this c**p

Ted Shifrin

Did he @Mike?

Mike

Yep

Pedro Tamaroff

@Mike No, he didn't.

Mike

Ok, I'll be more explicit in my statement. The strucure theorem says that "Finitely generated abelian groups are as nice as the Chinese Remainder Theorem can make them."

Pedro Tamaroff

15 mins ago, by Ted Shifrin

Another cute group theory question, @Pedro, of which I'm reminded by a question on main: If order $a$ is $k$ and order $b$ is $\ell$, is order $ab$ equal to $\text{lcm}(k,\ell)$ if $a$ and $b$ commute?

Mike

04:04

We're talking about different problems.

How do I quote?

Ted Shifrin

Yup @Mike. I was wondering.

usukidoll

$A+(B+C) = (A+B) +C$ associative law in a nutshell.

so they must be equivalent to each other.

$A + (B \cup C) \backslash (B \cap C)$

= $(A \cup B) \backslash (A \cap B) + C$

Ted Shifrin

@Pedro: For the problem you just cited. Prove it's true when gcd = 1. Then play around.

ZealotSveta

test

Pedro Tamaroff

@TedShifrin I know it is true when GCD=1

@TedShifrin My comment here.

Ted Shifrin

04:06

Ah, ok, @Pedro. So most of us then think the lcm statement is true in general.

Pedro Tamaroff

@ZealotSveta test failed: retry (y/n)?

Ted Shifrin

Ah, @Pedro: that was the question I noticed on main that triggered my challenge to you.

usukidoll

@PedroTamaroff bunnehhhh can you look at my stuff for a bit x)

Ted Shifrin

I have tennis at 8 am, so I'm departing. Take care, all.

Mike

What's $+$ for you?

See ya @Ted

usukidoll

04:07

k throws tennis ball

Pedro Tamaroff

@Mike $\triangle$

@usukidoll Start by writing $A\triangle B$ like normal people.

usukidoll

D:

Mike

@Pedro I guess some people write it that to be indicative of a boolean algebra?

Pedro Tamaroff

@Mike Sure.

You know I like to be an ass though.

usukidoll

$A \triangle (B \triangle C) = (A \triangle B) \triangle C$

$A \triangle (B \cup C) \backslash (B \cap C)$

= $(A \cup B) \backslash (A \cap B) \triangle C$

satisfied bugs bunny?

Mike

04:22

@Pedro I know and approv.e

Pedro Tamaroff

@Mike Also, I got your mail but I was too lazy to answer.

So sorry about that.

Mike

I am always lazy.

I forgive, and already forgot

Pedro Tamaroff

Now I sleep. Byes.

Mike

04:44

@usukidoll Look carefully at the hint in the comments and Asaf's answer

The two of them together should give you the idea you want

agent154

Just want to make sure I'm not missing something...

On a 10-question test, how many ways are there to answer exactly eight questions correctly?

Is this just 10 choose 8? I can't see any reason why not. Seems too simple.

usukidoll

which is the C point

like there may not be any elements in A and B but there are some in C

Mike

what does that mean?

C is unrelated to A and B

@agent154 yep

usukidoll

hold on I'm thinking... yeah I was going to say that... like $A =B$ no elements in A and B, but ... C is its own unique set/point

agent154

@Mike Thanks.

Mike

04:50

@usukidoll Think specifically about the case where $A = A \cap C$ and $B = B \cap C$

no prob

usukidoll

hmm the C is there....in both A and B

there's no elements in A and B, but there is C remaining... ... ...

Mike

huh?

you keep saying there is no elements in A and B

and I don't know why

usukidoll

because of the proposition

Mike

no, you misread the proposition

the proposition says empty sets are equal

if two sets are equal there's absolutely no reason for them to be empty

usukidoll

oh O_O

no wonder I was getting strange results

-______________- yeah x.x

they're not empty after all they were just equal

agent154

04:58

Here's a tough one that I think I have a grasp of, but I'm not certain...

What is the probability that a five-card poker hand has a straight (a set of five consecutive values)?

There are 52 choose 5 total possible hands... but I'm not quite sure how to represent the total possible ways to get a straight. I would assume you pick a card, and based on that card you have at most 8 out of 51 cards that could possibly have any chance of satisfying the requirement, but as little as 4 if the card you choose is an Ace/2 or a King. Any tips on how to approach this one?

usukidoll

I may have used the wrong definition I think

Mike

@agent154 well, for each 5-in-a-row, you can have $5^4$ possibilities (why?), and there are only so many 5-in-a-rows

@usukidoll of intersection? your def'ns fine

usukidoll

no the one about empty sets

or the $A=B$

there were two of them

Mike

you don't need empty sets here :)

$A=B$ if $A \subset B$ and $B \subset A$

usukidoll

one was ...yeah that's why I got confused as hell

Mike

05:01

@usukidoll Someone posted a more full answer on your question, but I think you're closer, so fiddle with it for a bit before reading theirs

they only gave you a hint but I bet you don't need that hint

usukidoll

it was definition 3.1.8 Let $A$ and $B$ be sets. Then A equals B, written $A=B$ when both $A \subseteq B$ and $B \subseteq A$. Thus the symbols $A$ and $B$ denote the same set

how am I closer ? O_O, lol ;P

Mike

well, now you know what to look for, and you're not going about with A and B are empty silliness

usukidoll

proposition what ever I put was on there.. what I should've done was look at def 3.1.8

both def 3.1.8 and prop 3.1.12 had $A=B$ in them

Mike

you don't need definitions, or propositions

draw some pictures!

draw a lot of pictures of various A, B, C, etc

usukidoll

?

agent154

05:04

Hmm, I'm not sure I see why. The only time I've seen something like $5^4$ was when replacements are alowed.

usukidoll

how is that a proof?

Mike

to help you figure out what your condition should be

it's not a proof of anything (here)

@agent154 Each of the five cards can be one of five suits

err

four suits

agent154

Oh... I forgot all about the suits

Mike

@usukidoll though I have a proof by pictures in a paper I wrote :P

agent154

so wouldn't it be $4^5$ then and not $5^4$?

Mike

05:08

you're right

oopsie

agent154

OK, so that satisfies the number of ways to have a straight given the four suits... but isn't there more to it? I don't know if I'm mixing up probability and combinations here

Mike

Well, first we're just counting the number of straights

agent154

That's just $4^5$ ways to have ONE straight of values. But there are many straights.

Mike

So that's the number of straights of a certain type (A-5, 2-6, etc)

But there are only a few types of these.

agent154

yes

Mike

05:13

Can you count the total number of straights of values? If we had a 13-card deck, no suits, say.

agent154

Is there a formula I can use without having to count them by hand? I suppose it should just be 13-4?

usukidoll

-_- can't believe that I used the wrong fricking definition for the problem ughhhhhhhhhhhhhhhhhhhhhhhhhhhh

that made a big difference

agent154

so a2345 up to 9,10,11,12,13

(using numbers instead of letters)

Mike

Well, remember that a string of values is unique determined by its lowest element

So pick the highest straight and its lowest element

And that's how many there are.

agent154

so there are 9 then.

Mike

05:15

Nope

Remember

Straights sort of wrap around :)

10JQKA

(If this isn't to be included, then yes, there are 9)

agent154

Well, I had no idea of that rule, and I don't know if that applies here. The full text of the question is as follows:

A straight (a set of five consecutive values)

Mike

Maybe it doesn't, but it sure does at a poker table

Anyway, it's 9 or 10

Let's call whatever it is $s$ for convenience

Then the total number o straights is $s \cdot 4^5$

And you have the total number of hands.

So now finish up and find the probability.

agent154

I'll have to clarify with my prof if it's 9 or 10 then... But I see your point

Thanks.

robjohn

05:31

@AméricoTavares Sorry I didn't get back to you sooner. It is true that the antiderivative of a continuous function must be differentiable. This is because we have $$\frac{\mathrm{d}}{\mathrm{d}x}\int f(x)\,\mathrm{d}x=f(x)$$

agent154

Of a company's personnel, seven work in design, 14 in manufacturing, four in testing, five in sales, two in accounting, and three in marketing. A committee of six people is to be formed to meet with upper management. In how many ways can the committee be formed if there must be at least
two members from the manufacturing department?

Am I double counting if I have $\left.14\choose 2\right.\left.33\choose 4\right.$?

Mike

I believe so

agent154

I can't see why though.

Mike

hmm

well

let's start with the fact that that number is bigger than the total number of committees

(i.e. 35 choose 6 is smaller than that)

agent154

Indeed, that is a good point

I'll have to remember that technique when doing a test

Well, i'm off to bed. Thanks for the help @Mike

Studentmath

06:22

Got a small question if anyone can aid me with it.. I need to that every ordered countable subset of <R,<> is isomorphic to a subset of <Q,<>. But when I think about it, if said subset of <R,<> or of <Q,<> has a least and/or greatest number, then that subset is not alike to other subsets without least and/or greatest number. What am I getting wrong in here?

usukidoll

07:11

:/

Balarka Sen

07:52

@Studentmath Set isomorphic? Bijective, do you mean?

@L-L Which one? I can't see any question.

Oh, this?

How'd I know? I am not mathematically mature/young.

Ask Patel.

Or Adobe

Leave me out of these

Chris's sis

@robjohn you wanna see this one mathematica.stackexchange.com/questions/41105/…

and Greetings!!!

Balarka Sen

Man I hate real analytic functions.

Balarka Sen

08:48

@Chris'sis You there?

@Chris'ssis?

Need a little help here.

Chris's sis

@BalarkaSen I need to finish something right now. I'll be back a bit later.

(working on a nice proof)

Balarka Sen

Ok. Which proof, may I hear?

On that digamma thingy?

Chris's sis

@BalarkaSen Not really. I have a different problem right now.

brb

Balarka Sen

@Chris'ssis I see.

Chris's sis

@BalarkaSen ask me that thing and if I know I answer a bit later.

Balarka Sen

08:52

How to prove

$$\sum_{n=1}^\infty \binom{2n}{n} \frac{H_n}{4^n n} = \frac{\pi^2}{3}$$

E calls this "Bino-harmonic sum"

I am not familiar with these things though.

Chris's sis

@BalarkaSen that's very interesting.

Balarka Sen

Isn't it?

Chris's sis

@BalarkaSen does it help? $$H_n=\sum_{k=1}^{n} \int_0^1 {x^{k-1}}$$

Balarka Sen

Probably not. Or at least I cannot derive a closed form for such a double sum.

usukidoll

09:19

http://math.stackexchange.com/questions/660468/prove-a-bc-bac-c-ab-using-the-definition-of-ab

driving me nuts! >:O

Nick

09:49

@usukidoll: In set theory, what's the difference between A~B , A\B and A-B ?

usukidoll

I haven't seen A~B and A-B in my life

I have dealt with A\B
complement of B relative to A written in A\B

there are elements in A, but not in B

Nick

R~{2} is the set of real numbers without 2.

Get it?

I guess it's all the same thing

usukidoll

yeah but my book doesn't have it like that

Nick

Yeah, and that subset symbol is different in some books too

Some have it without an underline and some have it with an underline

I think the heads of mathematics around the world need to come together and clarify on these issues.

@usukidoll: There's $A\subset B$ and $A\subseteq B$

usukidoll

$\subseteq$ is what I need to use

$\subset$ no elements

Nick

10:01

... what do you mean no elements?

usukidoll

ugh this homework is forever lol... I'm determined to type a truckload of it in 2-3 hours

I've already typed out 2 of the 3 exercises yay

Nick

that's homework for you :D

congrats on moving forward

usukidoll

dude my prof gave one that's like q-h

ooops a-h

I've got most of them except g and h

and g is important otherwise I can't answer h

my first semester in pure math oy what a different realm

I think the trick to getting this right is to read like hell and understand the definitions

I was able to figure out the easy proofs

Nick

@usukidoll: :D just go steady. That's what's important

usukidoll

on top of that I have yet to do the second part of my revision omfg

ughh I am putting A S MUCH WORDS IN THIS RECENT ASSIGNMENT SO I WON'T GET A REDO OR A REJECTION NOTICE ROFL

I just submitted a revision not too long ago...yeah I kind of not understood the problem

and wham another revision..this time it was needs more words

Nick

10:07

Well, just be consoled that you're not alone in this. Millions of others have worn, are wearing and will wear similar shoes.

@usukidoll: just get it done with and I assure you, it will be over.

usukidoll

I know... I'm doing the last one.. I just have to regather my notes... and LATex it.. I've posted some of those super long questions on here

I know I got a right... b...was a bit nutty.... c... that commutative law... d. propositions

f associative law.. g was hell... e is somewhere

h is grrrrrrrr

errr what is the commutative law for elementary set theory? I got the boolean version in my book

Nick

10:27

@usukidoll:
A ∩ B = {x | x is an element of A and an element of B}
= {x | x is an element of B and an element of A}
= B ∩ A

usukidoll

yeah that's the intersection definition.

Nick

Same for union

union is equivalent to And

intersection is equivalent to Or

The commutativity of the logical operations serve to prove the commutativity of the corresponding set operations .

@usukidoll: Is there some other commutative law for elementary set theory independent of the logical operations?

usukidoll

oh so for commutative $A \cup B$ = $B \cup A$ ?

$A+B=(A \cup B) \backslash (A \cap B)$

$A+B=B+A$ that commutative ... so that would be
$(A \cup B) \backslash (A \cap B)$ = $(B \cup A) \backslash (B \cap A)$ right?

Balarka Sen

10:55

What's all this hubbub?

Hey, @Nick.

@usukidoll What's the problem?

usukidoll

11:28

I figured it out...

but for the associative I'm stuck

f.$A+(B+C) = (A+B) +C$\\

We need to define the symmetric difference of $A+(B+C) = (A+B) +C$\\

$A+(B+C) = A+[(B \cup C) \setminus (B \cap C)]$\\

$(A+B) +C = [(A \cup B) \setminus (A \cap B)]+ C$\\

We need to use the associative law for set theory which is $(A \cap B) \cap C = A \cap (B \cap C) $ and $(A \cup B) \cup C = A \cup (B \cup C)$\\

$$
\begin{array}{c}
A+(B+C)\\
A+[(B \cup C) \setminus (B \cap C)]\\
(A \cup [(B \cup C) \setminus (B \cap C)]) \setminus (A \cap [(B \cup C) \setminus (B \cap C)])\\

ugh oh no the array isn't loading

oh there

I see the subsitution, but I can't get rid of that garbage in the middle

@BalarkaSen

Balarka Sen

What is $+$?

usukidoll

what + ?

OH sorry about that

Balarka Sen

$A \fbox{+} B$

What is that?

usukidoll

wait I'm getting it

the first line from math.stackexchange.com/questions/660468/… that's the def of A+B

Balarka Sen

Craznuts.

Alyosha

11:32

Hello, all! I have a simple question on fields.

usukidoll

I'm trying to prove that whatever I posted above is associative

Balarka Sen

@Alyosha Fire it.

usukidoll

I can see the subsitutions but there's this stupid stuff in the middle

Alyosha

Is $C(\mathbb{R}^n)$ a ring and not a field because inverses don't necessarily exist?

usukidoll

how do I get rid of it?

D: nooo I need help moar :D

not on g... on f the assoicative one.. which I'm close if I can get rid of the middle

Alyosha

11:34

Where $C(\mathbb{R}^n)$ is the set of all continuous real-valued functions whose domain is $\mathbb{R}^n$ under the operations of multiplication and addition.

usukidoll

:(

Alyosha

For instance, $f(x)=x \in C(\mathbb{R}^n)$, but it doesn't have a continuous inverse $f(x)=x^{-1}$ isn't continuous.

Balarka Sen

Yes.

That's right.

usukidoll

. . . how do I get rid of the middle hmmm

Alyosha

Excellent, thanks.

Balarka Sen

11:36

@usukidoll I am looking at it, wait a bit.

usukidoll

k I'm taking a break from latexing

then finishing up..... almost done with this assignment

Balarka Sen

Uh, can you clarify what's the 'g' you are referring to?

usukidoll

umm that's not the problem... g is in math.stackexchange.com/questions/660468/…. I used the first line to give you what $A+B$ is

if only I could impotent the middle lol via subsitution library.thinkquest.org/C0126820/algebra.html

Balarka Sen

@Alyosha Wait, do you mean continuous or do you mean 'has a point of discontinuity'?

usukidoll

I don't know if that would work

Alyosha

11:40

@BalarkaSen I was under the impression having a point of discontinuity implied non-continuity.

Balarka Sen

@Alyosha Not exactly.

Depending on what domain you are given.

Alyosha

Yes, but $C(\mathbb{R}^n)$'s domain is defined as the entirety of $\mathbb{R}^n$.

Balarka Sen

...okay, you are given $\Bbb R^n$ and thus $1/x$ is not continuous there. That's fine then.

I was confusing with codomain. =P

Alyosha

Nice harmonic sum problem, by the way.

Balarka Sen

@Alyosha Bino-harmonic =D

Clunky names E gives them

=D

Alyosha

11:45

E?

Balarka Sen

Well, he is the admin of Integrals&Series.

Alyosha

Ah

Balarka Sen

You can also call him S

usukidoll

thinks

Alyosha

That's an excellent site, but the problems are solved too quickly for me to be able to post there.

usukidoll

11:46

this is a great site

it's the only one I know that offers free tutoring for higher level math

I'm too advanced on OS... only very few know what I'm talking about

Balarka Sen

@Alyosha Haha. I only dare to touch the elliptic modular what not thingys

Studentmath

Well, yes, they are isomorphic if there is a one to one and unto function that doesn't change the order

usukidoll

now if only I can get rid of the middle part for that associative proof

ughhhhhhhh why can't I just subsitute the middle garbage by a letter and use impotent law or something

Balarka Sen

@Studentmath Yes, actually iso might mean several different things. That's why I asked.

Studentmath

@BalarkaSen

Balarka Sen

11:48

@usukidoll What's stopping you?

usukidoll

wait I could use that?

Balarka Sen

Probably.

usukidoll

because honestly the middle part of the proof is driving me nuts..like I could see that it is associative but this one middle part garbage line is preventing me from going further

Balarka Sen

I can see that.

usukidoll

so should I just subsitute with a letter?

and use impotent law?

Balarka Sen

11:49

I am not sure of a good set theorist here to look at that. All I can do is to draw you a Venn diagram, which presumably is not satisfactory to you.

usukidoll

imdeponet damn spelling

Balarka Sen

@usukidoll Godknows.

@usukidoll idempodent?

usukidoll

yes

Balarka Sen

Ok.

usukidoll

since $A \cap A = A$

Balarka Sen

11:50

You do that.

usukidoll

maybe I should use it to shut it up.. like let the middle be whatever and use imp law

and wham I got it

Balarka Sen

@usukidoll That's good.

usukidoll

I got the C's so if $A \cup B$ is subsituted with C

Studentmath

Yes, different in Cardinals than in here for example.. but still, one-to-one function that doesn't change the order. Now between sets without least and greatest I can get that, but otherwise, I mean between sets with least and greatest and sets without least and greatest, I think it falls wrong. And that's how it usually goes with such proofs, usu.

usukidoll

that would be $C \cap C$ and that's where the imp.. law applies $C \cap C = C$

would that be legal though?

Balarka Sen

11:52

So many set theorists here. I might as well run, I think.

usukidoll

NO!

I NEED YOU!

Balarka Sen

@usukidoll Sure.

usukidoll

so it is legal?

Balarka Sen

Yes. Have mercy on a child.

usukidoll

Let $A \cap B$ be $C$
Let $ A \cup B$ be $C$

then $C \cap C = C$ by imp law

Balarka Sen

11:54

Oh, no, that's not then.

usukidoll

see thought so

how the hell does it show up all of a sudden

like I'm getting there but there's the middle roadblock

any ideas?

Studentmath

Usu

Talk to me

What do you need to prove exactly?

usukidoll

Let $A$ and $B$ be sets. Define the symmetric difference of $A$ and $B$, written $A+B$ by $A+B=(A \cup B) \backslash (A \cap B)$\\

f.$A+(B+C) = (A+B) +C$\\

We need to define the symmetric difference of $A+(B+C) = (A+B) +C$\\

$A+(B+C) = A+[(B \cup C) \setminus (B \cap C)]$\\

$(A+B) +C = [(A \cup B) \setminus (A \cap B)]+ C$\\

We need to use the associative law for set theory which is $(A \cap B) \cap C = A \cap (B \cap C) $ and $(A \cup B) \cup C = A \cup (B \cup C)$\\

$$
\begin{array}{c}
A+(B+C)\\

that middle part iS BUGGING ME >:(

I'm starting from the left and going to the right.... seem to be going well until bam the middle is .. what is that?!

@Studentmath

Transcript for

$Mathematics$ Mathematics