Mathematics - 2012-06-28 (page 1 of 3)

user19161

12:08 AM

So @peter are you going to major in math?

Peter Tamaroff

@JasperLoy Yes. And get a PhD, too.

Hehehehe

user19161

@PeterTamaroff Why the laughter?

Peter Tamaroff

@JasperLoy It is not "hahahahaha" but "heheheheh"

Whatever.

user19161

@PeterTamaroff OK I get it, hahahaha.

Peter Tamaroff

@JasperLoy Actually, I should be studying algebra now. XD

user19161

12:11 AM

@PeterTamaroff What algebra?

Peter Tamaroff

@JasperLoy Linear Algebra

user19161

@PeterTamaroff OK then off you go!

Peter Tamaroff

@JasperLoy Actually I just finshed a Chapter, so I'm doing some excersices.

user19161

@PeterTamaroff When you win a Fields medal next time, remember to tell them I told you about the two vectors generating the plane.

Peter Tamaroff

@JasperLoy Haha, don't jinx it =P

I always have the doubt on how "excersice"

Or "practice" and "practise"

I think the former is the noun and the latter the verb, right?

Was it like that? Or the other way around?

user19161

12:14 AM

@PeterTamaroff Depends on whether you are using AmE or BrE. Any good dictionary will give you the answers.

leo

@PeterTamaroff Based on this Csonsidering $r_j$ an enumeration of the rationals of $[0,1]$. With this setting, since any integrable function can be approximated by a continuous function, you can see that $F_k=\chi_{\{r_1,\ldots,r_k\}}\in L^2[0,1]$. Actually that's the constant sequence $0$. Now let $r_{k_j}=1-1/j$, so $r_{k_j}\to 1$. Then $F_{k_j}-\chi_{\Bbb Q\cap[0,1]}$ tends to $0$ in the $L^2$ norm.

Jordan

I don't understand this polar coordinate stuff

Peter Tamaroff

@leo Great then!

@Jordan What is it?

Jordan

$2 < r < 3, 5 /pi/3 <= \theta <= 7 \pi/3$

leo

since any representat of the $L^2$ classes can be approximated by continuous functions by construction we are done

Peter Tamaroff

12:17 AM

@Jordan That means the longitude of the $r$ is to be at least $2$ and at most $3$

And the amplitude of $\theta$ is to be at least $5 \pi /3$ and at most $7 \pi /3$

That defines a piece of the polar plane.

leo

@Jordan you can use \pi and \leq next time

Jordan

I know there is a circle

Peter Tamaroff

@Jordan It is not a circle.

Jordan

but I don't know how to do the angle

leo

I must go. I have to write

Peter Tamaroff

12:18 AM

@Jordan Well, $\pi \operatorname{ rad} = 180º$ for starters

leo

see you later

user19161

@Jordan You want to draw the region right?

Jordan

yes

user19161

@Jordan For the radius think of concentric circles. For the argument, think of a pie being sliced.

Peter Tamaroff

The angles are $300º$ and $420º$

Jordan

12:20 AM

It should be 300 degrees and 120 degrees

Peter Tamaroff

@Jordan $420-360=?$

You should be getting something like this:

Jordan

I am having trouble figuring out 7pi over 3

Peter Tamaroff

@Jordan $$7\times 180/3$$ shouldn't trouble you.

Jordan

I got it now

I was trying to think of it without degrees

2pi is a full revolution, 7pi/3 makes 2 revolutions and then is left with pi/3

err 1

Peter Tamaroff

@Jordan =)

Jordan

12:34 AM

polar coordinates are really hard to imagine, it is like inception

making a graph if one step away from a formula and then polar coordinates are a step further away

Peter Tamaroff

@Jordan Hahah no man. Think it like this: I stand at the origin (center of the plane), then I rotate in place $\theta$ degrees and the move forward $\rho$ whatever (meters, cm)

Jordan

I am thinking of things like graphing snails and centroids

and roses

Peter Tamaroff

@Jordan Oh, that is cool. Just be patient, and buy some polar grids to plot in.

Peter Tamaroff

12:54 AM

@anon How's it rolling?

@anon BTW, I can't find my fancy toolbar!

anon

hey

fancy toolbar?

Peter Tamaroff

@anon Yeah, the mod powers...

anon

ah yes, that one

Peter Tamaroff

@anon That one.

anon

the word "tools" is right above the banner on the mainsite. click it.

Peter Tamaroff

1:03 AM

@anon I only have ▾Peter Tamaroff 10,03711448 review chat meta faq

Found it

anon

you then have to go to review, and there will be a "Tools | Review" selection thingie..

Peter Tamaroff

@anon yeah! got it.

I'm with algebra now.

Let $S$, $T$ be subspaces of $V$. Then $S \cup T$ is a subspace of $V \iff S \subseteq T$ or $T \subseteq S$

I got the $\Leftarrow$ part which is trivial.

Now I'm dealing with the $\Rightarrow$ part.

anon

contrapositive, show something breaks closure under addition

Peter Tamaroff

@anon Yeah, I was thinking about that.

Assume that $v,w \in S \cup T$ but $v+w \notin S\cup T$

anon

you can't assume that, you have to deduce it

Peter Tamaroff

1:11 AM

I have to assume that $S \not \subseteq T$ and show there is no closure, right?

anon

You have to assume $S\subsetneq T$ and $T\subsetneq S$ and show there is no closure (so you can't assume $v+w\not\in S\cup T$: that's your destination, not the starting point)

Peter Tamaroff

Right. Got it.

anon

anyway, pick an s in S that is not in T, and a t in T that is not in S. Is s+t in SUT?

Peter Tamaroff

@anon Why can't I deal with each containment separately?

It is $S \subseteq T$ or $T\subseteq S$ which is surely symmetric.

anon

and the negation of that is and, basic logic

$$\neg(A\vee B) \equiv (\neg A \wedge \neg B) $$

Peter Tamaroff

1:14 AM

@anon Right, De Morgan's...

anon

Crap, I mean $\neg (S\subseteq T)$, not $S\subsetneq T$.

$S\not\subseteq T$?

there we go

Eugene

the interwebs

Peter Tamaroff

@anon I got you before, =)

anon

So you start with $S\not\subseteq T$ and $T\not\subseteq S$

Peter Tamaroff

@Eugene Full of dorks, ain't them?

Eugene

1:16 AM

@PeterTamaroff sort of

Peter Tamaroff

@anon I have a writers block.

Well, mind block, so to say.

Then

$v,w \in S\cup T$

but if $v \in S$, $w \in T$, $v+w \notin S\cup T$.

anon

nope

Peter Tamaroff

No. I'm not sure about that.

robjohn

@leo On a set $K$ of finite measure, $\|f\|_1\le\sqrt{\mu(K)}\|f\|_2$

anon

nowhere did you use the inclusion hypotheses

Peter Tamaroff

1:22 AM

@anon Should it be the "exclusion" hypothesis?

anon

the lack thereof

Peter Tamaroff

@anon That means that $v \in S$ then $v \notin T$

anon

$S\not\subseteq T$ implies there is an $s\in S,\not\in T$. Symmetrically, $T\not\subseteq S$ implies there is a $t\in T,\not\in S$. Suppose $s+t\in S\cup T$. Then $s+t$ must be in $S$ or $T$ (or both), but is that possible?

robjohn

@leo and we are in $[0,1]$ so $\mu(K)=1$

This is what I get for being gone all day :-(

I've had to be away almost all day for a plumbing emergency

Peter Tamaroff

@anon Right. Because if $s+t \in S$ then $s, t \in S$ because $S$ is a subspace.

anon

1:27 AM

@PeterTamaroff I wouldn't phrase it that way. $a+b\in S$ does not generally imply both $a,b\in S$, unless it is known that one of $a,b$ are in $S$.

Peter Tamaroff

@anon But we know that $s \in S$

So $(-1)s \in S$

So $s+t+(-1) s \in S$

which is impossible!

anon

yes, it just wasn't clear if you were using valid or invalid reasoning

Peter Tamaroff

@anon Sorry! OK. I need to get my logic polished a bit...

anon

but there you have it

Peter Tamaroff

@anon Doesn't mean I don't have to work on my logic.

Eugene

1:35 AM

i haven't been very active on MSE it seems this past week

Peter Tamaroff

@Eugene Were you busy?

Eugene

@PeterTamaroff yup. also not many interesting things on MSE recently

Peter Tamaroff

@Eugene That's also true.

Mark Dominus

1:52 AM

@peter I'm pretty sure that question about subspaces showed up on the site not long ago. Would you like me to try to find it?

Peter Tamaroff

@MarkDominus Well, it's cool if you do, but I've solved it already. But I guess it will help!

Mark Dominus

I don't recall that there was any surprising solution though.

Peter Tamaroff

@MarkDominus Yeah, we just sorted out with anon.

Mark Dominus

I saw.

0

Q: How to prove that the union of two subspaces must be subsets of each other?

Possible Duplicate: Union of two vector subspaces not a subspace? $U,W\subseteq V$ are subspaces. Prove that in order for $U \cup W$ to be a subspace as well, either $U\subseteq W$ or $W\subseteq U$ Can anyone please give me a lead on this ?

Peter Tamaroff

@MarkDominus Thanks, potato!

Mark Dominus

1:59 AM

You are quite welcome.

Peter Tamaroff

I see I proved the theorem like Gerry did.

Mark Dominus

I remembered it because it is very unusual that I have anything useful to say about linear algebra questions.

But intersections of subspaces are always subspaces. That's nice. I wonder if you can make some pathological topology out of $R^n$ by taking the open sets to be complements of subspaces, or something like that.

Eugene

so many questions on the main page are analysis related these days

Mark Dominus

That's because analysis is hard.

Eugene

well at least i can get some work done

yeah number theory is really easy

Mark Dominus

2:06 AM

Sure. Everyone understands the positive integers. Nothing to it.

Eugene

@MarkDominus yeah. i keep on forgetting about that.

Mark Dominus

Go back to writing your thesis on whether there exists a positive integer between 7 and 9.

The real numbers, on the other hand, are mind-boggling, and nobody understands them. They have been a thorn in the side of mathematics since like 200 BC.

Eugene

@MarkDominus there's a number between 7 and 9??

Mark Dominus

I don't know, I haven't read the paper yet.

Eugene

yikes... this is surprising

Mark Dominus

2:08 AM

My toy octopus informs me that there is.

robjohn

@Eugene I was thinking that not enough were :-)

Eugene

@MarkDominus your toy octopus should publish that

Peter Tamaroff

@MarkDominus I'd love to study the $\gamma$ constant when I can.

That is a wild boar.

robjohn

@MarkDominus octopuses suck

or at least they have suckers :-)

Mark Dominus

True enough.

Eugene

2:10 AM

@robjohn really? because i seem to see things related to analysis everywhere. real analysis, measure theory, probability, differential geometry, calculus...

robjohn

@Eugene things get pushed off the front page, that perhaps I am missing them

Eugene

@robjohn i see. i polished off most of the elliptic curves questions and the modular form questions. it's really boring now.

@MarkDominus i can count to potato...

Mark Dominus

heh.

Rajesh D

Hello

everyone

Eugene

@MarkDominus it is an often used technique in number theory

Mark Dominus

2:24 AM

Counting to potato?

Eugene

@MarkDominus yup

@MarkDominus i'll give you an example

Mark Dominus

They should have named $\omega$ potato instead. Then even though you couldn't count to potato you could do transfinite induction up to it.

Eugene

the mordell weil group of an elliptic curve $E/\Bbb{Q}$ is finitely generated because when we count to potato, this becomes clear.

Mark Dominus

And it would explain what people mean when they say "one potato, two potato…"

Eugene

read this for the technique

Rajesh D

2:27 AM

@Eugene Hey

Mark Dominus

Nonstandard models of PA may contain a potato. But you can't count to it because it's nonstandard.

Peter Tamaroff

@MarkDominus That's not how you count to potato.

Eugene

@MarkDominus you are confusing counting to potato with the quater-tuberic number base.

Peter Tamaroff

@Eugene Hahahaha

Mark Dominus

That thing about the quater-tuberic base would have been funnier if Knuth hadn't actually invented such a thing.

Eugene

2:29 AM

@MarkDominus and that's what it is

it's mentioned in the linked article.

Mark Dominus

Yes, I know. I looked when you linked to it.

Eugene

so counting to potato is how number theorist solve all our problems

Mark Dominus

To invent something totally new and bizarre and ascribe it to Knuth is funny. To take something bizarre that Knuth actually invented and then arbitrarily insert "potato" into its name is just dumb.

Eugene

except trying to find this number between $7$ and $9$.

Mark Dominus

It just shows that Knuth is a lot cleverer than the people who edit Uncyclopedia.

Eugene

2:37 AM

@MarkDominus somehow i think we all knew this even without this.

Mark Dominus

I agree it is an overly sophisticated proof of a very weak theorem.

Probably the same theorem is also a corollary of Yoneda's lemma.

Peter Tamaroff

"third-order jerk functions"

Now those must be a pain to study,

Mark Dominus

A third-order jerk sounds like the kind of person who posts to Uncyclopedia.

user19161

Why is everyone talking about potatoes?

Peter Tamaroff

demonstrations.wolfram.com/SprottsJerkEquations

@JasperLoy We're trying to count to potato. But it seems not even Einstein or Gauss could.

Eugene

2:41 AM

it's because number theory is well understood and very easy

user19161

Yeah, just read "Basic Number Theory" by Weil.

robjohn

@Eugene Is that so...

Peter Tamaroff

I'll go eat something.

Eugene

@robjohn apparently

Peter Tamaroff

Good byes peoples.

robjohn

2:42 AM

@Eugene what are you going to eat?

Peter Tamaroff

BBS

robjohn

@PeterTamaroff laters

user19161

@robjohn Food.

Peter Tamaroff

@robjohn Was that for me?

user19161

@PeterTamaroff I think so.

Mark Dominus

2:43 AM

Remember back before food was invented and all we had to eat was clay?

Eugene

@PeterTamaroff beef bacon salami?

Mark Dominus

That was awful.

Peter Tamaroff

@Eugene LOL

Beef yes.

Not the other...

user19161

@PeterTamaroff Big Bull Shit.

Eugene

@JasperLoy watch for flags

Peter Tamaroff

2:45 AM

@JasperLoy You're disgusting, dude!

anon

pretty sure if a flag is gonna come around, it's gonna hit FUCKING BALLS first

user19161

@Eugene It's OK, I am planning to quit anyway.

anon

then again, mods did say to stop flagging old horses, or something

Mark Dominus

Nice of Ragib to draw fire away from the rest of us.

Eugene

@anon yup. plus ragib isn't around right now so i don't think he'll mind serving his sentence.

user19161

2:49 AM

@Eugene I was suspended from chat once for saying something about Dick being your neighbour and pussy being your cat.

5

Mark Dominus

That's the famous Scunthorpe problem.

user19161

You know it really depends on whether the mod knows the room well or not.

Eugene

@JasperLoy there was a flagging disaster a few days ago

user19161

Seriously there are worse things going on sometimes than using these words.

Eugene

old posts were getting flagged and people kept on getting suspended.

user19161

2:52 AM

Like people deliberately trying to hurt others with their refined words.

Mark Dominus

Like "poo".

user19161

And sometimes the most harmless things are flagged by nutcases.

Mark Dominus

Are the chat mods not the same as the se.math mods, then?

user19161

@MarkDominus I use boo as a standard greeting in another room.

user19161

@MarkDominus When there is a flag all SE mods know, regardless of the room.

user19161

2:55 AM

So really, flagging something attracts a lot of attention!

robjohn

@MarkDominus There are no "chat" mods, they are the mods from the various SE boards.

Eugene

@robjohn yup. that's right

Mark Dominus

Does being a mod of one board mean you have mod powers on all boards? I think not, correct?

user19161

Hey I already said that!

user19161

@MarkDominus No, you only have mod powers on that site, but chat is different.

Eugene

2:59 AM

well i should getting back to learning useless already solved number theory now.

bye all

robjohn

@MarkDominus Each SE board has its own mods, and they are just for that board, but chat is chat for all the boards together.

Mark Dominus

Thanks.

anon

3:10 AM

man, I've spent 20 minutes on understanding a proof just to realize there's a typo consistently throughout it.

leo

:-O

@PeterTamaroff Have you seen Mariano in here these days?

Joe L.

Can anyone give me a hint on $\lim\limits_{x \to \infty} \frac{e^{3x} - e^{-3x}}{e^{3x} + e^{-3x}}$

Mark Dominus

Divide through by $e^{3x}$.

Joe L.

I tried rewriting $e^{-3x}$ as \frac{1}{e^{3x}}$.

Gah, that was my first intention. Thanks.

Mark Dominus

You are welcome.

I wonder if there is some canned method for turning arguments of the type "$e^{-3x}$ is insignificant compared with $e^{3x}$ when $x$ is large, so we can disregard it in the numerator and denominator" into strategies of the type "Divide through by $e^{3x}$.".

The former type of argument seems to me to be the way most people handle these sorts of problems day-to-day, and much easier to learn to do, but to make it stick in Calculus class you have to come up with the second type, which seems to me to be harder to learn.

It would be a great boon to humanity if one could turn the first type into the second mechanically.

anon

3:25 AM

It's more like, if f~a and g~b, then f/g~a/b, IMO..

also, f+o(f)~f..

Mark Dominus

By f~a here you mean that lim f/a is finite?

anon

specifically that lim(f/a)=1

Mark Dominus

Then your "also" is not correct.

anon

oh?

Mark Dominus

Oh, that is a small o, not a big O.

anon

3:30 AM

indeed

Mark Dominus

Maybe the simple strategy looks something like: 1. For $lim_{x\to a} f(x)$, first calculate $f(a)$. If this is finite, you win. 2. Otherwise, if $f(x)$ has the form $g(x)/h(x)$ and you get $\infty / \infty$, then find the terms of $g$ that go to $\infty$ (there must be at least one) and divide through by the biggest one. 3. …

Part of the strategy is that $\infty + \infty = \infty$, $c\cdot\infty = \infty$, and $c/\infty = 0$, but $\infty - \infty$ requires further investigation.

I'm trying to codify how people actually calculate these things when they are not required to provide a Calculus I proof.

The seventeenth-century method, as it were.

Peter Tamaroff

@leo No. He seems to be awy.

Mark Dominus

It seems to me that a lot of this stuff could be reduced to a flowchart.

Peter Tamaroff

@anon Hhahaha link?

Joe L.

@MarkDominus Can the previous limit be computed without L'Hospitals? I did it with L'Hospitals but the textbook I'm working problems through right now (Stewart) does not introduce L'Hop 'till later on.

Mark Dominus

3:41 AM

@Joe: The one you just asked about? I would think so.

Joe L.

Yeah.

Mark Dominus

Did you divide the numerator and denominator by $e^{3x}$?

Joe L.

Yeah

Mark Dominus

Then what?

Joe L.

and then I divided through by $e^{-6x}$ then used L'Hopital to rewrite it as $\frac{6e^{6x}}{6e^{6x}}$ to conclude that $L = 1$.

Mark Dominus

3:43 AM

Do you really need L'Hospital's rule? You have $1-e^{-6x}\over 1 + e^{-6x}$, and the numerator and denominator both go to 1.

Joe L.

Ah, you're right.

Mark Dominus

Surely there is some theorem that says that if $lim_{x\to a} f(x) = p, lim_{x\to a} g(x) = q \ne 0$, then $lim_{x\to a} {f(x)\over g(x)} = \frac pq$, no?

Joe L.

I would imagine so, yes.

Mark Dominus

Disclaimer: I cheated my way all through Calculus I limits by relentlessly using L'Hospital's rule in secret.

Joe L.

Heh.

I'm just brushing up on some old intro calc topics right now.

Mark Dominus

3:46 AM

This was in 1985, so the statute of limitations has expired.

anon

@PeterTamaroff JS Milne's Group Theory notes, page 28-29, THM 1.64

I'll let you figure out what the typo is.

Actually, you probably don't know what a lot of the expressions are. Milne uses $q-1$ when he means to use $p^q-1$, and $\Bbb F_q$ when he means to use $\Bbb F_{p^q}$.

Joe L.

4:18 AM

How would I go about solving this DE? puu.sh/EdMB

anon

@JoeL you mean, find the values $\lambda$? Plug $y=e^{\lambda x}$ into the equation, then factor $e^{\lambda x}$ out of both sides when you've written the derivatives. You will have a quadratic equation in the variable $\lambda$.

Peter Tamaroff

@JoeL Substitue $e^{\lambda x}$, and you'll have to solve a quadratic in $\lambda$

Mark Dominus

Use L'Hospital's rule!

Joe L.

@anon Yes. My background in DE is limited right now only have dealt with separable DEs. I'll try your suggestion.

Peter Tamaroff

@MarkDominus That is basic algebra of limits

Mark Dominus

4:24 AM

I make joke. Laugh, comrade!

Peter Tamaroff

@MarkDominus hahhahaha I was out of context!

Darn! I find that the iteration $\frac{3-r}{r-1}$ converges to $-\sqrt 3$

I need to find one for $\sqrt 3$

anon

naively, I would guess (3+r)/(r+1)

simply by changing signs

Peter Tamaroff

@anon Worked.

Mark Dominus

@peter: I wrote a blog article about this a while back . Let me see if I can find it.

Never mind then.

Peter Tamaroff

@MarkDominus Wait!

Just link it.

Mark Dominus

4:31 AM

1m

Peter Tamaroff

@anon I'll reverse engineer that then!

Mark Dominus

blog.plover.com/math/attractors.html

Joe L.

Would $\lambda = \frac{1}{2} (1 \pm \sqrt{5})$?

Mark Dominus

blog.plover.com/math/attractors-2.html

Peter Tamaroff

@JoeL Yes!

Joe L.

4:34 AM

Cool, thanks. Sorry for the slow responses, my keyboard is acting up right now and isn't registering a lot of my keystrokes.

Mark Dominus

Many examples of iterated functions that converge to $\sqrt 2$, but the main point is to show a bunch of general methods for finding iterations that converge to whatever you need.

Eugene

@PeterTamaroff here

Peter Tamaroff

@Eugene LOL

Mark Dominus

Basic strategy: pick some function $f$ for which $\sqrt 3$ is a fixed point. if it turns out to be a repelling fixed point, perturb $f$ in one of several ways.

anon

you want a monotonic decreasing function on a large interval surrounding the desired limit (with said limit as a fixed point)

Peter Tamaroff

4:43 AM

Tell me what you think guys

Special thanks to anon for changing the sign there, hehe.

@MarkDominus Thanks!

Mark Dominus

You are quite welcome, good sir. I was only too happy to assist you.

Peter Tamaroff

@MarkDominus You're a lazy writer!

Joe L.

How would I go about solving this? http://puu.sh/Ee69

Stewart's answer: http://puu.sh/Ee5N

I computed the first three derivatives and easily saw the pattern how $f'(x) = 2 f(x), f''(x) = 4 f(x)$, etc.

Peter Tamaroff

Should I flag this يا الله اعطيني الصبر... مابدي اعصب عليه.. يا الله
as offensive?

Google gives

Mark Dominus

@JoeL That is the answer to some other problem.

Peter Tamaroff

4:48 AM

"O God, give me patience ... Mabdiy Aasb it .. O God,"

anon

@JoeL Um, the "Stewart's answer" you linked appears to be f(x)=log(x-1) or something, not f(x)=e^(ax). Just use the pattern to hypothesize a general formula, and prove the formula via induction.

Mark Dominus

I would say that a necessary (but not sufficient) condition for flagging anything as offensive is that you were actually offended by it.

Joe L.

@anon that's what I did. I supposed I copied the answer from a different section with the same problem number. Just checked again with the back of the book and there is no answer even though it is an odd number, weird.

Peter Tamaroff

Well, good byes peoples.

@anon Saw this?

anon

heh, exploting

Peter Tamaroff

4:53 AM

@MarkDominus Let me know what you think

Mark Dominus

Your step from ${1\over r+1} > {\sqrt 3 - 1\over 2}$ to ${r+3\over r+1} > \sqrt 3$ is not clear to me at 1AM.

anon

@MarkDominus double both sides then add 1

Mark Dominus

OK!

Peter Tamaroff

Darn, now Frank wants to talk about real numbers

Mark Dominus

I wonder if this is really as simple as possible. I remember an exercise in the early part of the little Rudin book that seems similar, except for $\sqrt 2$.

Frank Science

4:57 AM

@Peter The definition from Dedekind cut?

Peter Tamaroff

@FrankScience I was about to comment " If it is Dedekind Cuts I'm smelling, I know what you mean."

Frank Science

I think the definition should be come out first.

Peter Tamaroff

@FrankScience What is worrying you? It is 2 AM here, and tomorrow I need to wake up early.

@FrankScience That is a construction, and I don't think it is pertinent in this case.

Frank Science

@Peter Before definition, how can we say $\sqrt 2$?

Peter Tamaroff

@FrankScience As Spivak explains, we assume the existence of such a field. The construction is another problem.

Frank Science

5:00 AM

Well

Supposing that $x$ is a positive rational number such that $x^2<3$. You really want to show that there's another positive rational number $r$ such that $x^2<r^2<3$, but not about $\sqrt 3$.

Peter Tamaroff

@FrankScience You do realize in the proof I use that $\sqrt 3 ^2 =3$, right?

Again, we're asuming the reals are already present.

It is a delicate topic, it is true. We can talk about it some day later. Now I really need to sleep.

Frank Science

But different definitions result in different proofs.

Peter Tamaroff

@FrankScience I know! I'm sticking to whaterver the OP has, which is clearly consistent with the idae of that excercise, which is what I did.

Frank Science

For example, if in Dedekind cut, there's no doubt that there's some rational number $r$ such that $a<r<b$, where $a$, $b$ is arbitrary real number such that $a<b$.

Peter Tamaroff

@FrankScience Frank, you're not telling me anything that I don't know, really.

So, I'll find you so that we continue this talk later. I appreciate the rigor!

I'll sleep now. Bye!

Frank Science

5:06 AM

@Peter I've deleted my comment from your answer, and migrate it to the OP's question.

@Eugene I'm disappointed.

Eugene

@FrankScience why?

Frank Science

@Eugene I don't know whether I'll set a bounty of the problem cloned to MO again in MSE.

Eugene

5:21 AM

@FrankScience it's very technical in my opinion. that's why it doesn't generate much interest. sorry about that

Frank Science

@Eugene I can't get your idea clearly

Eugene

@FrankScience as in, the reason why it hasn't generated much interest on here or MO is that it is very technical in nature.

Frank Science

technical?

Eugene

as in requires a bit of technical algebra to show

Frank Science

Maybe, but somebody has told me that, the higher mathematics could be used to see through the techniques in elementary ones.

Eugene

5:28 AM

@FrankScience yeah sometimes.

user19161

@Eugene All algebra is technical dude!

Eugene

@JasperLoy that's really not true IMO

but i see you're probably being hyperbolic

user19161

@Eugene Anyway, technical is not well-defined!

Frank Science

For example, Fermat's Last Theorem is solved by higher mathematics, but not a very tricky way of elementary number theory.

Eugene

@JasperLoy i agree

@FrankScience yes but you must remember you can't always do this

Frank Science

5:30 AM

Somebody told me that it was impossible for Fermat to have proved that theorem.

Eugene

@FrankScience yes that is true

user19161

Maybe the solution is truly ingenious?

user19161

Well I'll ask Fermat when I meet him.

Frank Science

I don't exactly know the reason. Is it proved that, there's absolutely no effective elementary ways to deal with such equation?

Eugene

@JasperLoy probably not. the method he used is suspected to be infinite descent and that he proved the $n=4$ case

@FrankScience we don't know

Frank Science

5:32 AM

Yes, I know he proved $n=4$.

user19161

@frank I see you are thinking very deeply about these things which is good.

Frank Science

I will have lunch, and my computer should be suspended to save energy.

user19161

Yo @former!

user19161

@FrankScience Why not shut down?

Eugene

@FrankScience in fact the proof of FLT was conceived of completely separate of FLT

user19161

5:34 AM

@Eugene And you had a major role to play too IIRC.

Eugene

@JasperLoy in that i did nothing to help! =)

user19161

@Eugene You could at least have done the typesetting.

Eugene

@JasperLoy probably best i didn't

user19161

@Eugene Do you know a guy called Kay Jin?

Eugene

@JasperLoy nope

well i'm off to visit chow gong

bye

Frank Science

5:44 AM

@JasperLoy Suspending is much faster and easier.

anon

6:13 AM

Why is it the case that $[A\cap B:A\cap C]\le [B:C]$? It seems to be implicit in this proof.

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$Mathematics$ Mathematics