Mathematics - 2014-01-19 (page 1 of 5)

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12:04 AM

@PedroTamaroff That should have been $\mu(E) = \inf \{\mu(O) : O \supseteq E, O \text{ open}\}$, shouldn't it? I have met the terms "outer regular" and "inner regular" for the respective properties, and "regular" can mean "both, outer and inner regular".

@DanielFischer Yes, I copy pasted and missed it.

I know ;)

@DanielFischer Naïve question: why doesn't "closed" work?

@PedroTamaroff For inner regularity?

@DanielFischer Aha.

12:06 AM

That was meant to be a question "Why doesn't closed work?" For inner regularity?

Right.

Don't know, actually.

Why is it so hard to find questions on m.se

@Nick What are you looking for?

@Nick here

Lots of questions ;)

12:11 AM

XD

I'm looking for something that proves $\infty - \infty$ is indeterminite

$n-n\to 0$

$2n-n\to \infty$.

Done =)

.... how does $n\to \infty$ prove that you can't determine $\infty - \infty$?

he just showed you two different things that both go to infinity whose difference goes to different things

... Oh!

I can't determine what I can determine!

12:18 AM

@Pedro: that's actually pretty good.

@TedShifrin Captain!

similarly you can make that limit anything you want

hey guysssssssss

@Pedro!

anyone wanna help me? This question is confusing

12:19 AM

I hate not being able to determine something.

Whar? @usukidoll

If I can't use a truth table... what do I do?
http://assets.openstudy.com/updates/attachments/52da4347e4b0b71c4f8dd63d-usukidoll-1390035831646-scan1401170001.jpg
it's 1.4.15 I need an example. All I read is that if I put a negation sign on one side of the expression, I would get a contradiction

which is nothing but F's in the column

The quality is too damn high.

I've drawn the truth tables, I do have tautology all around but I can't even use them...so do I use...demorgan's?

better to have high quality than low @Pedro

@Pedro: that's what Mcdonalds said

12:22 AM

Huh? @Pedro

@TedShifrin Cf. usukidoll's book scan.

@TedShifrin he was talking about my scan I think

yeah

what previous results do they mean? @usukidoll

Is there another way for this question

that's the thing I don't know what they're talking about... should I scan the previous page and have you read it?

12:23 AM

Trial and error is so last century.

No... You should figure it out! I assume you understand that $P\implies Q$ fails precisely when $P$ holds and $Q$ fails.

yeah that's when P is true and Q is false

$P \rightarrow Q$
P is true...Q is false... $P \rightarrow Q$ is false
P is false...Q is True... $P \rightarrow Q$ is true
P is false...Q is false... $P \rightarrow Q$ is true

forgot that when P is true and Q is true $P \rightarrow Q$ is true

If I can't draw a truth table, what should....I do ? x.x

@Pedro: Is there some symbol that I can use instead of equality while working with non-finite numbers? $\infty + \infty = \infty$ doesn't feel right.

@Nick It should because it is right.

I have drawn them, but the problem says don't use it which I don't get it because I do have a bunch of T's in one column

12:28 AM

Right. But we're doing truth tables. Maybe there's a result earlier that gives an equivalent to $P\implies Q$?

earlier???

@Pedro: logically, based on the definition of infinity maybe but not on the basis of equality. If you think about it $\infty = \infty$ is something that really doesn't make sense.

hmm the question says that verify that the following are tautologies by citing the appropriate previous result... guess I have to find the appropriate previous result for those sets...

Earlier. Something they've stated or proved. Like $P\implies Q$ is the same as not $P$ or $Q$?

@Nick It does make sense. Something is always equal to itself.

12:31 AM

$\lnot P$ has F F T T
$\lnot Q$ has F T F T

@Pedro: Maybe your infinity and my infinity aren't the same guys.

@Nick $\rm potato = potato$

my infinity is bigger than yours, @Nick!

@DanielFischer I have a prawblem.

@Ted: Yeah, that's a big problem.

12:33 AM

@PedroTamaroff What sort of p0rbelm?

so.. $ P \rightarrow Q$ is T F T T
but $ \lnot P \rightarrow \lnot Q$ is hmmm I should use a table one sec

Hi @Ted, by the way.

Hi @Daniel

@Ted: is the number of numbers from 0 to 0.5 lesser than the number of numbers from 0 to 1?

Nope, @Nick. Same.

12:34 AM

@DanielFischer $\mu$ is a measure induced by a monotone right continuous function $f:\bf R\to\bf R$, (a Lebesgue Stieltjes measure) and we also call $\mu$ the completion of $\mu$ induced by the outer measure induced by $\mu$ yadda yadda...

@Ted: Then my infinity is as big as yours.

Now.

Nah, @Nick, you don't know which mine is!

@DanielFischer Beacuse of slightly obvious reasons (right continuity), we work with half open intervals $(a,b]$ as elementary sets, extend to the Borel sigma algebra and induce the otuer measure... so we know then $\mu(E)=\inf\{\sum \mu(a_i,b_i]:E\subseteq \bigcup (a_i,b_i]\}$

@usukidoll: We're not supposed to be making more truth tables.

12:36 AM

appropriate previous result... x.x

@TedShifrin Reminds me of the child's game "biggest number". "Ultimo", "Infinity", "Ben Gurion".

if I can figure out what those three words are ...the problem would be a bit easier

@Ted: Then I define "Trinity" to be a variable greater than infinity! Then my non-finite number is bigger than your non-finite number.

any examples?

@Nick No way you can beat Ben Gurion.

12:37 AM

@DanielFischer Now, the guy proves that $\mu(E)$ equals to the infumum when $E$ is covered by open intervals.

Look at propositions they've proved, @usukidoll

Not half open.

And now he wants to prove inner and outer regularity.

do I have to modify the problems like add a negation or something?

@PedroTamaroff Outer regularity is already done, innit?

-_- the other problems in that text are easier than this... hmmm :/

12:38 AM

@DanielFischer Well, I would say so.

Take any open, express it as countable disjoint union of intervals

Look at what I wrote earlier and see if it's there somewhere

@DanielFischer: How is Ben-Gurion, non-finite? He's completely expressible in finite space.

user image

@DanielFischer Then it is clear that $\mu(E)\leqslant \nu(E)$ where $\nu$ is the measure over the inf of open sets.

i.e. we want to prove $\mu=\nu$.

@Nick, Nothing is greater than him.

OK, nevermind.

12:40 AM

@Daniel: You contradict yourself. Are you saying he's a negative number?

@Nick I didn't say "zero" was greater.

:/

@Daniel: How many sheep do you have?

@PedroTamaroff Did the "nevermind" mean you have figured it out, or did you think I was too occupied talking nonsense with Nick?

@DanielFischer I think I have figured out.

12:44 AM

@Nick Twice as many as my brother.

@Daniel: ... how many constellations are under your possession?

@Nick this sort?

use demorgan's law? X/

@Daniel: No, this sort.

@Daniel: Is that information classified?

Then, how many sheep does your brother have?

@Nick I believe in the freedom of stars, I have none under my possession. And you'd have to ask that my brother, I don't know if he'd approve me telling you.

12:52 AM

@Daniel: ...dang, then is 0 lesser than or greater than nothing?

@Nick Neither. $0$ is less than something ($1$, for example), and it is greater than something.

@Daniel: But something is better than nothing, am I not correct?

Incomparable.

Dang, I thought you'd throw up an error and malfunction by now.

...I think I have one more trick up my sleeve

@Nick I haven't been coded in any of C++, Perl, PHP, VB. Why would I throw up an error?

12:57 AM

This sentence is false.

Is it?

error

error

@Nick That should be #error to have an effect.

user image

The only portable way to ensure that a conforming C compiler fails to compile a programme is an #error directive that is not excluded by other preprocessing directives.

1:03 AM

Just because I'm a Carbon-based life form, doesn't mean I'm written in C.

@Daniel: You've enlightened me, the cardinality of nothing is zero.

I preferred the first version ;)

$n(\lbrace \rbrace) = 0$

I've actually never thought of it that way.

:S

@Nick No, the cardinality of the empty set is zero, so the cardinality of something is indeed zero.

@Mike: But I've already proved something is infinite.

1:09 AM

@DanielFischer OK, I don't buy this guy.

Sounds like a contradiction.

bangs head this question is driving me nuts

@PedroTamaroff Slave trading is illegal, it's good that you don't.

@DanielFischer Oh, Daniel. You so jolly.

hmmm maybe I should eat lunch and go back to this.

1:10 AM

What's the problem? Errors in a proof?

@Mike: Why did you have to use the 'c' word. You've ruined the whole charade.

@DanielFischer Well, let me share.

We're trying to prove inner regularity.

well I don't understand what they mean by citing the appropriate previous result on this http://assets.openstudy.com/updates/attachments/52da4347e4b0b71c4f8dd63d-usukidoll-1390035831646-scan1401170001.jpg
I have the truth tables, but it says not to draw them.
maybe I should approach the problems the same way as this question
http://math.stackexchange.com/questions/377439/prove-that-statements-forms-are-tautologies

@Nick There's no charade here; we've proven now that $1=0$, something mathematicians have been working on for centuries!

@DanielFischer First, assume $E$ is bounded.

1:12 AM

Easy.

If it is closed, there is nothing to prove, for it is comapct.

So assume it is not closed.

Consider thus the set $\overline E\smallsetminus E$.

Given $\varepsilon> 0$, by outer regularity, we can find an open set $O\supset \overline E\smallsetminus E$ such that $\mu(O)\leqslant \mu(\overline E\smallsetminus E)+\varepsilon$.

@Mike: No, we proved that $\text{something} = 0$. You're the one blatantly saying that $\text{something} = 1$

@Nick Well, we agree that $0 = \infty$

So now we simply need to divide both sides by $\infty$.

@DanielFischer Now, the set $\bar E\setminus O$ is closed and bounded hence compact, and $K\subseteq E$. But how do we know it is not empty?

And we have the world's first indeterminate equation!

Might as well divide "not defined" by zero to redefine it again. I wonder if we'll get HD picture quality.

1:16 AM

@PedroTamaroff We don't. At least, not yet.

@DanielFischer Well, then the author proves $\mu(K)\geqslant \mu(E)-\varepsilon$.

Hence proving the claim for bounded $E$s.

From the argument, it is seen $K$ empty implies $E$ has measure zero.

@PedroTamaroff $$\mu(K) \leqslant \mu(E) \leqslant \mu(K\cup O),$$ right?

@Mike: Too bad the monkey who supplies infinities died during our conversation.

@DanielFischer Note $O$ is an open set containing $\overline E\smallsetminus E$.

@PedroTamaroff Yes, and $K = \overline{E}\setminus O$.

1:20 AM

Aha.

So $\overline{E}\subset K\cup O$.

Yes.

@Mike: Cutting the chitchat, are you good with basic algebra?

Depends on what you mean by that, but probably

@Mike "Basic Algebra II" OK kids, let's study Crystallographic Groups.

1:22 AM

@PedroTamaroff So we have $\mu(K) \geqslant \mu(\overline{E}) - \mu(O)$.

I hear crickets no one is responding to me D:

@DanielFischer Why not equal?

@PedroTamaroff Because in general $\overline{E} \subsetneqq K\cup O$.

@DanielFischer OK.

stares

1:25 AM

@usukidoll at anything in particular?

no just trying to figure how to approach this... it says "by citing the previous appropriate result...do not make truth tables"

@PedroTamaroff And so $\mu(K) \geqslant \mu(\overline{E}) - \mu(O) \geqslant \mu(\overline{E}) - \mu(\overline{E}\setminus E) - \varepsilon$.

If I can't make a truth table to verify that it's a tautology, What am I supposed to do? Write a simple proof about $ \lnot (\lnot P) \leftrightarrow P$? @DanielFischer

:/

@usukidoll Can you cite previous results to use $A \leftrightarrow B \equiv (A \to B \land B \to A)$, and $A\to B \equiv B\vee \lnot A$?

the thing is...what the heck is the previous result? There's ... http://assets.openstudy.com/updates/attachments/52da4347e4b0b71c4f8dd63d-usukidoll-1390035831646-scan1401170001.jpg
I can understand the last line after exercise 1.4.15. If I put a $\lnot$ on one side, it will be a contradiction truth table of all F's
@DanielFischer

but it's not asking all F's it's asking verify for a tautology without using a truth table which suxz

hmmm let A ... be $\lnot (\lnot P)$ and B be $P$?

1.4.14 was easy to do... why isn't 1.4.15 straight forward?

@PedroTamaroff @Mike any ideas?

should I do it similarly to this? math.stackexchange.com/questions/358750/…

1:37 AM

Hmm. It says $X$ is a tautology iff $\lnot X$ is a contradiction. $\lnot (A \iff B) \equiv (A \land \lnot B) \vee (\lnot A \land B)$. So $\lnot (\lnot(\lnot P) \iff P) \equiv (\lnot(\lnot P) \land \lnot P)\vee (\lnot(\lnot(\lnot P))\land P)$.

that's a lot of $\lnot$!

@usukidoll Are we talking about classical logic or intuitionistic?

I guess it's logic?!

i assume it's for a discrete math class, so the former

yeah

1:39 AM

There are different brands of logic.

more of a intro to advanced mathematics class...which I like to call proofwriting for beginners or something

what happens if there's a problem with an R?

@DanielFischer Sorry, was away.

What are you trying to get at with those calculations?

@PedroTamaroff Just seeing that indeed $\mu(K) \geqslant \mu(E) - \varepsilon$.

1:55 AM

@DanielFischer Right, thought so.

@PedroTamaroff Okay, so we have inner regularity for bounded sets.

@DanielFischer Yiss.

And for unbounded, you probably intersect with $(-n,n]$, and approximate that.

They guy did $(j,j+1]$. Potato, pohtato.

That's the other common way.

Well, night, everybody, it's bedtime.

2:03 AM

@DanielFischer Night,

Cheers.

I'm off to watch "The Network":

back

well I'm doomed

hey it's grumpy cat!

2:29 AM

anyone here?

@anon wanna help me to the right direction? x.x

2:58 AM

@Waffle'sCrazyPeanut u there?

Waffle's Crazy Peanut

@usukidoll Yep!

I need help with something I can't seem to figure out what the heck is going on

http://assets.openstudy.com/updates/attachments/52da4347e4b0b71c4f8dd63d-usukidoll-1390035831646-scan1401170001.jpg
If I can't use a truth table for 1.4.15 do I have to write a simple proof for each of the following problems?

Waffle's Crazy Peanut

@usukidoll I'm afraid I won't be helpful if that's too mathey

:/

Waffle's Crazy Peanut

@usukidoll Yeah, just prove them :)

3:01 AM

oh! So is it something similar to this problem?
http://math.stackexchange.com/questions/358750/prove-the-following-is-a-tautology

the first one is a double negative law and I see De Morgan's on b and c

Waffle's Crazy Peanut

@usukidoll Yep :D

so I need to prove that these are tautologies without drawing a table

Waffle's Crazy Peanut

yeah, that's what it says (unless you know any other method that I'm unaware of) :P

math.stackexchange.com/questions/358750/… could be similar to this one too

Waffle's Crazy Peanut

@usukidoll That's what you gave me first -_-

3:04 AM

oh my bad x.x

I could do some substitution ..I'm viewing a video right now...youtube.com/watch?v=iPbLzl2kMHA

A HA! Logical identities...THat's what these problems were X)

what's up @IanMateus

3:23 AM

@usukidoll hello, nothing at all by now

I'm just trying another attempt at this problem .x.x

sighhhhhhhhhhhhhhh D:

@IanMateus I'm s o lost x.x

@usukidoll what's the problem?

http://assets.openstudy.com/updates/attachments/52da4347e4b0b71c4f8dd63d-usukidoll-1390035831646-scan1401170001.jpg
1.4.15 I just realized that most of the problems are logical identities...but the problem is how can I verify that it's a tautology if I can't draw a truth table? @IanMateus

by using various tautologies you already know [from truth tables]

what does it mean by cite an appropriate previous result? XX

3:32 AM

You can use axioms, right?

use exercise .14

or actually just do the first thing i said

use de morgans law, your various axioms, etc

I have done .14 it's somewhere....

how do I apply de morgan's law on this X_X

i looked and .14 probably won't help, at least for the first few

i'm giving examples of things you can use

this is what they mean by previous results

meaning demorgans on the first problem/

but that's just a double negation

this is 1.4.14 btw
https://www.dropbox.com/s/jus7mufhutqh0ns/Scan-140118-0002.jpg

at least I'm trying....to figure this out....
https://www.dropbox.com/s/oyem1eojjz61o2e/Scan-140118-0003.jpg
@IanMateus @Mike

I have drawn the truth tables for those problems...but the directions are like NO!!!!!!

truth tables
https://www.dropbox.com/s/4al9m6kvue6vh7t/Scan-140117-0002.jpg
https://www.dropbox.com/s/v5p5hhel8vecipc/Scan-140117-0003.jpg

$\begin{align*}\neg (\neg P)&\iff P\\
\neg\neg (\neg P)&\iff \neg P\\
(\neg P)&\iff \neg P\\\\&\text {True}
\end{align*}$

hmmmmmmm let the X be $\lnot (\lnot P)$ and Y $P$?

ughhhhhhhhhh everything was fine until that italics thing came along

@PedroTamaroff save me D:

:(

I need an example or something

1 hour later…

5:04 AM

@usukidoll: If you have only 2 truth values ( true and false) then isn't ¬(¬P) = P ?

but argh the question had it at $\lnot (\lnot)P \leftrightarrow P$

isn't the $\leftrightarrow$ a biconditional arrow?'

It's supposed to be (but I'm no expert)

@usukidoll: Wait for Pedro and if you're in no hurry, then search for a decent guide book that has the solutions for the questions in your text. Guides usually have conceptual stuff that help a lot.

I wish there was a solutions manual for passage to abstract mathematics

that's my book

5:22 AM

Downvotes were reversed earlier this morning.

@usukidoll I am no expert at prop. calc., but yes, it seems to be a bi-conditional arrow.

@Balarka: Good Morning Balarka! :D

@Nick Mornin' nick.

@Balarka: Um, do you remember how you once said that angles were weird. I didn't get what you said the last time you said it. Why do you think they can't be defined?

@Nick I do remember. I'll answer you question a Q&A style : How do you define angle, in geometry?

@Balarka: Sort of as a fraction of a turn.

In geometry, an angle is the figure formed by two rays - Wikipedia

5:36 AM

@Nick Right, so to measure that thing in radians, people draw the angle in unit circle and compare it with fractions of $2\pi$, right?

Since all circles are similar, they can all be cut up into say 360 parts (Sexagesimal system) or 100 parts (Centesimal system) or you can define an angle as a fraction of a turn.

That whole $2\pi$ thing comes along when you define an angle as arclength/radius

You're simply using the circumference as a guide to know how much you've turned from particular spot.

@Nick True. So to define sine and consine in a circle you need to know the arclength, non?

So everything comes down to integrating the circle via the arclength formula.

... um, I think the radius would suffice, don't you think?

How do you measure radius? You need arclength formula.

user image

I can whip myself a triangle using the radii

Ok, yes, I need the arclength

pouts

5:43 AM

Then when you try to find arclength, you end up with the arcsin.

i hate it how my common sense doesn't tingle when it's supposed to.

so you are defining sin via arcsin.

Yes? and how is that wrong?

Nothing wrong, but I feel particularly wimpy about it. Wait a minute, I'll formally define them via those (I have all in my notes) ...

@Balarka: Before we proceed, Is arcsin(x), the same thing as $sin^{-1}(x)$

5:48 AM

Yes.

Okay, I have the formal defs. Here they are :

Define f from [-1, 1] to R as $x\sqrt{1-x^2}/2 + \int_{x}^{1} \sqrt{1-t^2} dt$

Then define cosine in $[0, 2\in_{-1}^{1}\sqrt{1-x^2}dx]$ as the unique inverse of $f$.

such that $f(cos(x)) = x/2$

define pi as $\pi = 2\int_{-1}^{1} \sqrt{1-x^2}dx$

This is pretty neat.

Wait, it's not the end of the lecture! Define, on $[0, \pi]$, $sin$ as $sin(x) = \sqrt{1-cos(x)^2}$

Extend the domain of the function as

$\cos(x) = \cos(2\pi - x)$ and $\sin(x) = -\sin(2\pi -x)$

Extend by periodicity to the complex plane.

Note that it follows how $\cos'(x) = -\sin(x)$ and $\sin'(x) = \cos(x)$

And analytically continue it via taylor series, and we are done.

... What did we do?

@Nick Eh, what did we?

...uhm @Balarka: ... I mean what did you just prove with that?

6:00 AM

@Nick That it's not impossible to define sine with arcsine, as you stated.

Cooool

Can you prove the angle sum idenities through these?

@BalarkaSen: I can't. But probably, you can.

@Nick We know that $\cos'(x) = -\sin(x)$ and $\sin'(x) = \cos(x)$ -- just continue differentiating and then make up the taylor expansion on $\cos(x)$ and $\sin(x)$ by setting the derivatives at $x=0$. Then use them to prove Euler's formula : $e^{ix} = \cos(x) + i\sin(x)$ and finally conclude the angle-sum formulas by considering $e^{i(x + y)}$ and $e^{ix}e^{iy}$ and comparing the real and imaginary part.

oh cool!

6:12 AM

Yeah, it sure is cool. =D.

D:

6:41 AM

@Balarka: Wanna hear something weird?

@Balarka: First imagine that you are a caveman and that the wheel has not been invented. What is the most important aspect of the device that you consider?

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