Mathematics - 2014-04-24 (page 1 of 3)

Mike

00:14

@Pedro Someone is trying to convince me Euler's formula is beautiful.

@AlexanderGruber are you here?

Karl Kronenfeld

@Mike Is it the usual, "it involves all of these special numbers"?

Ted Shifrin

howdy @Karl @Mike @seaturtles @Pedro

Karl Kronenfeld

Hi @Ted.

Ted Shifrin

Anything interesting happening here?

Mike

@Karl His argument is that it depends on how you define things, and then defined $e^x$ by its power series, and $\pi$ as the ratio of a unit circle's circumference to its diameter.

because choosing specific definitions gonna convince me. :)

@Ted Never.

Ted Shifrin

00:20

Well, in that case ... I'll be back later.

Mike

Bye. :D

enjoy your dinner.

Karl Kronenfeld

@Mike That kinda makes it uglier, imo.

Mike

@Karl I think it makes it less surprising, because the power series for sine and cos jump right out.

Alexander Gruber

@Mike in and out, what's up?

Mike

@AlexanderGruber I'm going to give your cfrac bounty a shot this weekend.

Alexander Gruber

00:24

@Mike cool, what are you going to write about?

Mike

I actually know of an interesting algorithm in combinatorial commutative algebra that you might like

eh, I'm most going to point out theorems in various parts of math that use cfracs.

they show up in bizarre places.

Alexander Gruber

I'm planning to make a post about Schor's algorithm at some point over there, I did a class in quantum computing where we had to implement it in Mathematica and I still have some neat visuals somewhere. (i'm not eligible for the bounty though of course)

@Mike cool what is it?

Mike

I need to remember, but IIRC one can make a lattice ordered by set inclusion called the "intersection algebra" of a ring, and an algorithm counting the minimum number of generators depends on the continued fraction expansion of a number associated to the algebra

of a rational number associated to the algebra, I should say. which is cool. most contfrac stuf you see is about infinite continued fractions

Alexander Gruber

00:39

@Mike Yeah! that's pretty neat

Jayesh Badwaik

@AlexanderGruber Hola! How is Florida?

Alexander Gruber

@JayeshBadwaik HOT.

it's april and i have three air conditioners on in my apartment

Jayesh Badwaik

@AlexanderGruber WOW....

It's warm in here too... at night cool winds make it lovely though...

How is it going otherwise?

Alexander Gruber

@JayeshBadwaik here we have alligators at night. :)

it's not bad. today was my last class of commutative algebra, which is

very good

Jayesh Badwaik

Hahaha, hope one doesn't show up in your bathroom! :-P
I've got measure theory exam in 30 hours.... I don't think I'll be able to sleep till then... or atleast till 6 hours before it...

Alexander Gruber

00:47

@JayeshBadwaik take a yardstick with you. it'll be good luck.

Jayesh Badwaik

@AlexanderGruber Yeah.. tried that last exam, my prof was more interested in non-measurable sets then... :-(

Sawarnik

@JayeshBadwaik Which college are you in?

Alexander Gruber

as a discrete mathematician, i refuse to acknowledge the existence of non-measurable sets.

Jayesh Badwaik

@Sawarnik Tata Institute of Fundamental Research, Bangalore.

Sawarnik

Ok :)

Jayesh Badwaik

00:50

@AlexanderGruber Good for you.... sigh me has to deal with infinities all around...

@Sawarnik howz it going with you? just got up? or going to sleep?

Alexander Gruber

@JayeshBadwaik i highly recommend it. we don't need to worry about choice either.

Sawarnik

@JayeshBadwaik Just missed my school bus .. now enjoying!

Jayesh Badwaik

@AlexanderGruber Half the proofs I have to write tomorrow depend on choice.... I also have an algebra exam a fortnight later... I'll complain about different shit then... :-P

@Sawarnik cool....

@AlexanderGruber the PSA guideline appears to be new.. what changed?

Alexander Gruber

@JayeshBadwaik that PSA is just something i put up this week because people kept starring a ton of messages in the row for now reason

like this

Jayesh Badwaik

ohh, I see.....

Alexander Gruber

00:55

the location pinning one was just a fun idea i had a few weeks ago, I guess people like it because it keeps getting re-pinned/starred

I don't know how long it'll keep up there

Jayesh Badwaik

That's a cool one, I just pinned my approximate location...

Sawarnik

@JayeshBadwaik Why are you up so early?

Jayesh Badwaik

@Sawarnik I haven't slept yet... I have a class at 11... and an exam on 25th at 10 am...

So, I'm planning on staying awake till today evening and then going to sleep around 2 in the night tomorrow...

Sawarnik

@JayeshBadwaik Yes, I see you have corrected!

Jayesh Badwaik

@Sawarnik Nopes, east of Bangalore.

@Sawarnik You are in Patna it seems...

Sawarnik

01:07

@JayeshBadwaik Yes!

Jayesh Badwaik

I thought you were Parth... :-/

Sawarnik

Wat!

No, no, he is my enemy :D

@JayeshBadwaik Why did you think so? I guess because of my age and avatar?

Jayesh Badwaik

@Sawarnik I had my reasons, anyway, I cleared my confusion now...

@Sawarnik among others yes....

Pedro Tamaroff

01:59

@KarlKronenfeld

Karl Kronenfeld

@PedroTamaroff sup

Pedro Tamaroff

@KarlKronenfeld I am using Nakayama's lemma to show that if $A$ is local and $M,N$ are finitely generated $A$-modules then $M\otimes_A N=0$ implies $M=0$ or $N=0$.

If $\mathfrak m$ is our maximal ideal, and $k:=A/\mathfrak m$, we have by associativity that, denoting $M_k=k\otimes_A M\simeq M/\mathfrak mM$

$M\otimes_A N=0\implies M_k\otimes_A N_k=0$.

Now, I want to obtain that $M_k\otimes_k N_k=0$.

@KarlKronenfeld

Karl Kronenfeld

@PedroTamaroff I'd guess that it's true in general that $M_k\otimes_A N_k=M_k\otimes_kN_k$

Pedro Tamaroff

@KarlKronenfeld Yes, I buy that.

Karl Kronenfeld

Show that $M_k\otimes_AN_k$ satisfies the universal property of the other one.

Pedro Tamaroff

02:06

I mean, clearly multiplying by elements of $A$ in $M_k$ or $N_k$ obviates elements of $\mathfrak m M$.

Alexander Gruber

what is up with this question?

Pedro Tamaroff

@AlexanderGruber Oh?

What is up?

Alexander Gruber

@PedroTamaroff viewed: 23672 times

Karl Kronenfeld

holy shit

Pedro Tamaroff

@AlexanderGruber Yeah, that's ridiculous.

Alexander Gruber

02:09

and it's attracting so many weird troll posts

Karl Kronenfeld

@AlexanderGruber Sorry, I wrote a script that creates new accounts just to view that question... and write trollish answers. :P

Alexander Gruber

i just protected it so that should stave those off but i wonder why they're even coming, that is such a mundane question. :p

@KarlKronenfeld i wouldn't even be surprised if that's what was happening.

Pedro Tamaroff

@KarlKronenfeld Please let this be true.

Mike

can I see a screenshot of the deleted answers?

Anthony

Haaaaaaaaaaaaaaaaalllo

Pedro Tamaroff

02:13

"hi you can eat pinnaples or try eating your green gooey boogers!"

"L'O'L
L
'O'
L
L'O'L
L'O'L
L'O'L
L'O'L
L'O'L
L'O'L
L'O'L
L'O'L"

Mike

@pedro really?

Pedro Tamaroff

@Mike Yes.

Mike

weird.

Anthony

Anyone have advice on how to show a function is smooth?

Alexander Gruber

unfortunately we left out the best one... heh

Pedro Tamaroff

02:16

@Anthony Sit on it.

Does it hurt?

Alexander Gruber

@Anthony put it through a power sander, if it wasn't before, it will be.

Pedro Tamaroff

Then it is not smooth.

Anthony

hoho

anorton

@PedroTamaroff That's the same analogy my prof. uses. :)

Alexander Gruber

is it equal to its smoothification?

Pedro Tamaroff

02:17

@AlexanderGruber LULZ.

Anthony

I've got $e^{-1/x}$ for $x>0$ and $0$ elsewhere.

Pedro Tamaroff

le classic

you merely need to check it is smooth at $0$.

Alexander Gruber

@Anthony so you want to show the $n$th derivative exists

for starters, what's the $n$th derivative outside, at $x\ne 0$?

Anthony

So that was along the lines of what I was asking-I should find a formula for the nth derivative?

Pedro Tamaroff

No.

You should show $f^{(n)}$ exists and is continuous for every $n$.

Anthony

02:19

Indeed... Is there a better way to show existence then to explicitly give them a formula?

Not that I know there is one.

Alexander Gruber

you could find the $n$th derivative if you want though, i would just recommend grouping together whatever doesn't depend on $x$ as a constant

Pedro Tamaroff

@Anthony Here.

Spoilers ahead.

Alexander Gruber

i'm probably taking the hard way, though. as a constructivist i tend to do that.

Pedro Tamaroff

@AlexanderGruber Alex?

Constructivist.

Breathing accelerates.

Alexander Gruber

@PedroTamaroff that's where the finitism comes from my friend

:)

Pedro Tamaroff

02:22

@AlexanderGruber You said you liked finite groups. Never thought you were into finitism.

I was being fun though.

Alexander Gruber

i sort of am, i acknowledge continuums i just avoid them if i can help it

Anthony

@PedroTamaroff, should I just say the rest of the function is smooth-just because? I mean 1 is obviously the problem point, but shouldn't I say something about the rest?

Alexander Gruber

same for non-constructive proofs, they work logically, i just like constructive better

Pedro Tamaroff

@Anthony No, it is smooth because $e^x$ and $-1/x$ are.

Anthony

I don't think I've learned a composition rule

Pedro Tamaroff

02:25

Yes, you have.

Anthony

Nor have I shown that e and 1/x are smooh.

Pedro Tamaroff

$D(f\circ g)=(Df\circ g)\cdot Dg$.

Anthony

hohoho

<3

Pedro Tamaroff

Composition and product of smooth is smooth.

Inversion is smooth.

Product and composition of continuous is continuous.

So you have all there.

Done diddly doo.

Alexander Gruber

@PedroTamaroff whatchyou know bout semisimplicity?

Anthony

02:26

I still have shown that e and inversion are

But

<3

Gracias.

Fernando Martin

semisimple rings are simple, but not as simple as simple rings

Pedro Tamaroff

@AlexanderGruber Of rings?

Alexander Gruber

@PedroTamaroff yeah, should be close to where you are in CA right?

Pedro Tamaroff

Nah, not that much.

But Jacobson has a great section on them.

I scanned over it.

Will read it thoroughly sometime.

Alexander Gruber

i only have two more homeworks until i can burn my commutative algebra book and never speak of it again.

Fernando Martin

02:28

why do you dislike CA so much @AlexanderGruber?

Alexander Gruber

@FernandoMartin because i'm taking it from an awful professor.

Fernando Martin

Ahh, makes sense

Pedro Tamaroff

@AlexanderGruber Damn bad professors ruining math for everyone.

@FernandoMartin AYO.

Alexander Gruber

someday i will pick it up again once my gag reflex has become less irritated

Pedro Tamaroff

@AlexanderGruber much graphic

Alexander Gruber

02:31

i think this summer i'll be focusing on analysis actually.

there are some analytic subjects that I want to study that require reviewing a bunch of fundamental material

next year there'll be a diffgeo course here, and i'm going to fit in some type of probability if I can.

Pedro Tamaroff

@FernandoMartin Dude.

Alexander Gruber

maybe stochastic something.

Fernando Martin

sup @Pedro

Pedro Tamaroff

$$\frac{M}{\mathfrak a M}\otimes_A \frac{N}{\mathfrak a N}\simeq \frac{M}{\mathfrak a M}\otimes_{A/\mathfrak a} \frac{N}{\mathfrak a N}$$

Mike

Jesus Christ Pedro

Pedro Tamaroff

02:33

What.

Fernando Martin

Uhh, ok?

Pedro Tamaroff

Agree?

Fernando Martin

I guess, I should try to write the UP

Pedro Tamaroff

@FernandoMartin It is pretty obvious informally.

Fernando Martin

I don't see why tbh

I had overlooked that detail when I solved that since I didn't write subindices for the tensors

shame on me

Pedro Tamaroff

02:46

@FernandoMartin Maybe I am wrong then. Essentially, multiplying by elements of $A$ or of $A/\mathfrak a$ is the same in $M/\mathfrak aM$ and $N/\mathfrak aN$.

Fernando Martin

yeah but you can't make scalars jump from one side to the other

Pedro Tamaroff

Where?

Fernando Martin

it's kinda like what happens with $\Bbb C\otimes_{\Bbb C}{\Bbb C}$ vs $\Bbb C\otimes_{\Bbb R}{\Bbb C}$

Pedro Tamaroff

No, I don't think so.

Fernando Martin

maybe it works in this case

Pedro Tamaroff

02:49

My point is that if we multiply by something in $\mathfrak a$, those modules obviate it.

For $am\in \mathfrak aM$ if $a\in\mathfrak a$.

So we might as well just multiply by things lying over $\mathfrak a$.

Fernando Martin

right

it makes sense

that actually proves that the obvious map is well defined

Pedro Tamaroff

Yes.

usukidoll

04:19

hello?

skullpatrol

yo, wazzup?

usukidoll

doing homework :/

like I sort of get what's going on but something isn't clear

usukidoll

04:43

0

Q: Prove that if $y,z \in Q$ then $y^z \in A$

Question : Prove that if $y,z \in Q$ then $y^z \in A$ My attempt: Definition 2.7.8 states that a number s is an algebraic number when there exists some $p \in Z[x]$ such that $p(s) =0$. Let us denote the set $ A = [x \in C: x $ is algebraic] The set of all polynomials in x with coefficients fr...

rubito

Is this using induction?

usukidoll

I don't think so

this is rational numbers and algebraic numbers

If I was using induction I need the basis step, P(k) and P(k+1)

Mike

.

Fernando Martin

sup @Mike

rubito

yeah youre right. Idk why I said that. It's late.

Mike

04:53

@FernandoMartin Getting rid of the '21 new messages' thing

usukidoll

there's this tiny thing that's throwing me off

I was doing fine until I saw that $y^z \in A$

Fernando Martin

lel

you're popular

usukidoll

huh?

Enjoys Math

05:15

@_@

@usukidoll what's the question in gist?

usukidoll

gist?!

Enjoys Math

the gist

^_^

usukidoll

math.stackexchange.com/questions/766935/… trying to provide the answer in a comment hold up ^^

Enjoys Math

ok, lemme know if you want help

usukidoll

I'm trying to latex a huge line xD

am I on the right track @enjoysmath

Enjoys Math

05:22

idk

usukidoll

<_<

Enjoys Math

if you mean is math the right path to be on and to that I say yes

>_<

usukidoll

yeah but what about the proof part... well the recent one scroll down

Enjoys Math

you'll have to let me know specifically what's tripping you up

usukidoll

$y^z \in A$

I have to find a polynomial root of $\frac{a\b}^\frac{c}{d}$

nguh I donn't wanna edit

I mention the root in here
http://math.stackexchange.com/questions/766935/prove-that-if-y-z-in-q-then-yz-in-a/766949?noredirect=1#comment1593438_766949

Enjoys Math

05:25

you have that 'start chatJax' link?

usukidoll

yessum

usukidoll

05:52

pokes @enjoysmath

Enjoys Math

ur hot :|

usukidoll

help me D:

skullpatrol

Enjoys Math

@_@

skullpatrol

:D

Enjoys Math

05:55

@usukidoll First thing I don't understand is : "Thus each element p∈Z[x] has the form

p(x)=anxn+an−1xn−1+...+a1x+a0

where n is a non-negative integer, ai∈Z for each i∈[0,1,2,...n] and an≠0 except when n=0
"

usukidoll

what the hell? that's part of the def i the book

Enjoys Math

polynomials by def can have $a_i \neq 0$

Do you mean for $x \in A$ its associated poly is such that $a_i \neq 0$ for $0 \lt i \leq n$?

also you should use \Bbb{Q} for $\Bbb{Q}$

usukidoll

no

whatever I typed from the book is correct

anywayyyyyyyyyy

sigh view this.. I've latex an example

http://openstudy.com/study#/updates/5358994fe4b0f8d836a765a9

Enjoys Math

Question: are you trying to say $y^z \in A$ when both are rational or algebraic???

@usukidoll

usukidoll

yessssssssssssssss by that prop man

Enjoys Math

06:00

I can't read that from your post. You mean when $y, z \in A$, $y^z \in A$ or no?

I think you mean when they're rational but not sure

just a question

usukidoll

oh gawd

Enjoys Math

what?

usukidoll

can you read the question?

what does it say?

Question : Prove that if $y,z \in Q$ then $y^z \in A$

Enjoys Math

I got you

usukidoll

y and z is flipping rational numbers... I gotta prove that $y^z \in A$ which means that y to the z power is algebraic

by the prop...all rational numbers are algebraic

makes sense because y and z belong to the set of rational numbers

Enjoys Math

06:03

Okay, first show that if $z \in \Bbb{Z}$ then $z^y\in A$ when $y \in \Bbb{Q}$ and you're half-way there

usukidoll

if it was irrational, I can't prove this now can I because it break the prop

wth is that?

z is an integer y is a rational ... your point?

Enjoys Math

Show that $\sqrt[b]{z} \in A$ for all $b \in \Bbb{N}, z \in \Bbb{N}$ and you're almost there, do you see that?

usukidoll

:(

I like the other way

can we do it after we had the z and y in Q?!!?!?!

Enjoys Math

@usukidoll : $y = \frac{a}{b}$ where the $a,b$ are integers. So if you prove that integers to a certain power are algebraic then you already know that $1/b^{n}$ is algebraic since it's equal to $(b^n)^{-1}$

No, you start with smaller problems then build the bigger problem. It makes sense mostly that way

I'll write up an answer

usukidoll

noooooo I want to find a polynomial with the root $(\frac{a}{b})^{\frac{c}{d}}$ in $Z[x]$

Sawarnik

06:07

..

skullpatrol

.

Mike

@usukidoll since $x^d-\left(\frac{a}{b}\right)^c$ has that as a root, multiply thorugh by $b^c$ to get the polynomial you want

$b^cx^d - a^c$

usukidoll

there ^ that's what I want. so I multiply all of that by $b^c$

but I feel like I'm skipping steps.. there's this identical problem that multiplies, rearrange the coefficents, and then muliply the last equation by the denominator

it's something like ... $p(x) =c_nx^n+c_{n-1}x^{n-1}+...+c_1x+c_0$

such that p(y) = 0 and there exist integers a and b such that $b \neq 0$ and $ y=\frac{a}{b}$

Mike

Hello Nando

Fernando Martin

sup

usukidoll

06:15

Thus...
$c_ny^n+c_{n-1}y^{n-1}+...+c_1y+c_0=0$

Enjoys Math

@usukidoll math.stackexchange.com/questions/766935/…

Mike

@FernandoMartin its u

Fernando Martin

now I'm hungry

thanks for nothing @Mike

usukidoll

which we multiply by $y^n$ to obtain
$\frac{a^n}{b^n}y^nc_n+\frac{a^n}{b^n}c_{n-1}y^{n-1}+...+\frac{a^n}{b^n}c_1y+\frac{a^n}{b^n}c_0=0$
@Mike

then rearrange the terms and multiply by $b^n$

Enjoys Math

@usukidoll look at my post, much simpler than dealing with polys

Alex Youcis

06:19

@Mike Sup.

Mike

lol

three mikes in a minute

Alex Youcis

@Mike Mik

e's a bad name. You should get a new one.

Mike

I'll be Gaddis.

usukidoll

broken code wait gotta fix

Alex Youcis

@Mike Lame. You should be, Michaela.

usukidoll

06:22

$c_n(\frac{ay}{b})^n+c_{n_1}\frac{a}{b}$

Mike

I'll consider it.

Alex Youcis

@Mike And then, go to Coachella.

usukidoll

$(\frac{ay}{b})^{n-1}+...$

Mike

No.

usukidoll

$ +c_1 \frac{a^{n-1}}{b^{n-1})$ $(\frac{ay}{b} + \frac{a^n}{b^n}c_0=0$

ugh long code went explode

Alex Youcis

06:24

@Mike They don't sleep anymore on the beach.

Mike

I don't Do 'chella.

Fernando Martin

@AlexYoucis ;____;

usukidoll

@mike do you see the polynomial in parts ...? should I do that for the y and the z so that $y^z \in A$?

Alex Youcis

@FernandoMartin I have never understood what that emoticon meant.

Fernando Martin

a crying face

usukidoll

06:25

that means crying

Mike

@AlexYoucis Here's my favorite emoticon

.-.

Alex Youcis

@FernandoMartin What happened to :'( Did it go the way of the dinosaur?

@Mike That's terrible, and I hate you for showing it to me.

@FernandoMartin Also, why the crying face? Because that song's sad?

usukidoll

eerrrr guys what about my problem? ^^

Mike

I think he was crying because he saw you chatting, @AlexYoucis

Alex Youcis

@Mike Your half-hearted burns are like sweet rays of sun on a perfect summer's day.

Fernando Martin

06:27

Because they don't do that anymore - things changed

Alex Youcis

@FernandoMartin True, true.

Mike

I don't understand wy someone going to 'chella would sleep.

cap

i have a question. If $f(x) = \tau(n*p^x) - \tau(n)\tau(p^x)$, where $(n,p)=1$ and $\tau$ is ramanujan's function. I am trying to show $f$ is always $0$, and I was advised to show $f(x)$ is a linear combination of $f(x-1)$ and $f(x-2)$ . How does this suffice.

Alex Youcis

@Mike I don't know if you are serious. But, in case not. I references a GY!BE song, and then Fernando commiserated with me on a more global level, and then you misunderstood. That's the world as I see it.

Mike

All of this is true

Alex Youcis

06:32

@Mike A confession eh? Cuff 'im boys.

skullpatrol

takes out hand cuffs

skullpatrol

06:45

Are you going to come peacefully @Mike?

skullpatrol

06:57

@N3buchadnezzar

Enjoys Math

08:05

@usukidoll any luck yet?

usukidoll

come here and get this done for meh >:D

I also have another one... I might have some clue or not X>X

Let $a,b \in R$ where $ a < b$. Prove that there exist a rational number $c$ and an irrational number $d$ such that $ a <c<b$ and $ a<d<b$. Hint: consider decimal expansions of $a$ and $b$

Theorem 2.7.5 states that a real number is rational if and only if its decimal expansion terminates or has an infinitely repeating sequence of digits.

Set $I$ is irrational numbers

$I =[x \in R: x \notin Q]$

Set $Q$ is rational numbers

$Q = [ \frac{a}{b}:a,b, \in Z$ and $b \neq 0]$

oh geez if there is an irrational number d... the decimal expansion doesn't terminate D:

If the division process terminates, then we are done. Otherwise, since each

digit of the quotient determines in turn its duccessor, and since there are

at most $b-1$ possible remainders when dividing by $b$ (by the division

algorithim), some digit of the remainder must show up again, forcing a

sequence of digits to repeat forever. The length of the repeating cycle is

at most $b-1$

Since $d$ is irrational, the decimal expansion will terminate. Therefore,

for $r=0$, we have $a_1,a_2,...a_k$ where each $a_i \in

[0,1,2,3,4,5,6,7,8,9]$ and $a_k \neq 0$. Then

$ r = \frac{a_110^{k-1}+a_210^{k-2}+...+a_k}{10^k}$

Sawarnik

08:25

..

@usukidoll Are you a girl or a boy?

usukidoll

no

but I'm trying to get this together x.x like I sort of know how to approach it but not sure if it's going to work points up

oh no I meant to put since $c$ is rational, the decimal expansion will terminate.. since $d$ is irrational the expansion won't terminate

girl

Sawarnik

Ok :)

Chris's sis

Greetings.

usukidoll

I'm stuck D:

Chris's sis

Today I just received a short movie that was meant to impress me profoundly, but it didn't. It's related to numbers.

See this entirely (without skipping) and tell me if it works for you - youtube.com/watch?v=5Qg24O864lI

For me it didn't work at all.

usukidoll

08:34

rawr

685-252

@Chris'ssis I would like to take a Black chainsaw to that video :D

Sawarnik

@Chris'ssis Nonsense. And how the hell is it related to numbers?

Chris's sis

@Sawarnik It's probably related to the way mind usually works.

685-252

It's not related to anything, imo.

A hammer is a common tool

Sawarnik

@Chris'ssis I didn't see any connection with the calculations and the final question. Many people prefer red and hammer is a common tool could have been said in 1 line.

685-252

08:46

and red is a common color

Chris's sis

@685-252 Indeed.

@Sawarnik Yeah.

685-252

@Chris'ssis I was not impressed at all.

Chris's sis

@685-252 Neither do I.

Sawarnik

@685-252 Neither will any sensible person.

685-252

In fact, I want a refund on the 1:28 of my attention back :D

damn internet.

8 mins ago, by 685-252

@Chris'ssis I would like to take a Black chainsaw to that video :D

Chris's sis

08:52

@685-252 :-))))

usukidoll

1

Q: Let $a,b \in R$ where $ a < b$. Prove that there exist a rational number $c$ and an irrational number $d$ such that $ a <c<b$ and $ a<d<b$.

Let $a,b \in R$ where $ a < b$. Prove that there exist a rational number $c$ and an irrational number $d$ such that $ a <c<b$ and $ a<d<b$. Hint: consider decimal expansions of $a$ and $b$ Theorem 2.7.5 states that a real number is rational if and only if its decimal expansion terminates or has ...

homework proof-writing irrational-numbers rational-numbers

N3buchadnezzar

Anyone know how to access the following paper?

"An extension of integration by parts", American Mathematical Monthly, vol. 67 , 1960 , p. 372

Transcript for

$Mathematics$ Mathematics