Mathematics - 2014-12-22 (page 2 of 4)

@BalarkaSen mother of god ! I heard about another guy ! S. Naik .. he passed out from ISI Bangalore about 4 years ago .. he was just like you .. he finished Complex analysis by 9th grade !

Hippalectryon

Well, first of all, which case are you talking about ? The one with 1 sheet of paper ? The one with $n$ identical sheets ? The one with $n$ non identical sheets ? Are the sides the same length ? @BalarkaSen

Balarka Sen

2:09 PM

@r9m Bet he did excellent in S?

r9m

@BalarkaSen yes !! that too :)

Hippalectryon

What's S ?

Balarka Sen

@Hippalectryon n identical sheet is essentially equivalent to 1 sheet.

Hippalectryon

@BalarkaSen Not exactly.

@BalarkaSen If you have distinct sheets, you can rotate them around

r9m

@Hippalectryon Secondary exam .. a board exam that we have to take in 10th grade

Hippalectryon

2:10 PM

If they're all linked you don't have the same freedom in your moves

@r9m Ah ok

Balarka Sen

OK, OK, for simplicity assume you're working with a single 1 \times 1 sheet.

Hippalectryon

Ok

@r9m trololol

Balarka Sen

Hm. First we have to define what one means by "making" numbers out of folding it (I know what you mean, just trying to formalize)

Hippalectryon

Well we can assume that a number is "made" if you have a side with its length

Balarka Sen

OK, so wait.

I think what you mean is this

r9m

2:12 PM

@Hippalectryon no .. that's not him Hippa :P

Balarka Sen

You have a unit sheet of paper

Constructing $x \in \Bbb R$ means you can fold this sheet of paper to construct a polygon with a some side of length $x$.

Does that sound good?

Hippalectryon

Exactly

Balarka Sen

But this definition looks superfluous.

Since you need a single side of length $x$, the polygon looks somewhat irrelevant.

Hippalectryon

Well it's not needed, indeed

The shape can be anything, as long as there is a straight side of length $x$

Balarka Sen

That's why I am saying that your problem is equivalent to consider a huge sheet of paper with a single straightline of length 1 drawn in it

and your asked to fold the paper so that you get a segment of length x from that line.

Hippalectryon

2:17 PM

Well it's not 100% the same. Here, you start with a square of side 1.

With just one line, it would be way harder, for instance, to produce $\sqrt{2}$

Balarka Sen

You're somewhat right.

Hippalectryon

~~I believe I'm left, but whatever~~

Balarka Sen

You're left?

Hippalectryon

Stupid pun

Balarka Sen

Oh ah

Well Sanic doesn't care about which side of the road he's taking

:P

Hippalectryon

2:20 PM

Lol

Gotta go fast, faster

Balarka Sen

it's sonic though

Hippalectryon

sAnic

Balarka Sen

:P

We'e drifting away from the problem though

Hippalectryon

True

Balarka Sen

Given that you can construct $x$, can you also construct $1/x$, if both is in $(0, \sqrt{2}]$?

Hippalectryon

2:25 PM

How would you do that ?

Balarka Sen

No idea. That's why the question mark.

Hippalectryon

Well I don't know

Balarka Sen

$1/\sqrt{2}$ you can construct.

So I wondered...

Hippalectryon

:O How do you do this one ?

Balarka Sen

$1/\sqrt{2} = \sqrt{2}/2$

Hippalectryon

2:27 PM

(Yeah i'm bad at geometry)

Oh of course

me stupid

Balarka Sen

OK, a better formulation :

You have a huge sheet of paper and a unit square drawn on it.

So your problem now boils down to origami on that square ;)

Hippalectryon

Kind of

Balarka Sen

Yeah so now you can do Huzita-Hatori but still some work involved I guess

Hippalectryon

Well I'm not so sure about H-H

Balarka Sen

Maybe the way to go would be to formalize as follows : Folding is equivalent to choosing an axis and flipping by mirror reflection.

Yeah I am not sure either

Hippalectryon

2:31 PM

See my comment to Achille : I'm not sure that those axioms can be applied here. For instance, axiom 1 states that for any points p1,p2 there exists a unique fold that passes through both, and gives a parametric equation for the line. However, I hardly see how by just folding your sheet (you have no other geometric tool available) you will be able to create that fold for random points. Or am I missing something ?

You must clarify the axioms you can use to fold, like "can fold along diagonals" or "can fold in half". Without some preassumed axioms, you cannot fold anything rigorously, or in other words, mathematically. — cjackal 55 secs ago

^ let's address that first

Well a fold can be defined by two points. Two edges. Does that seem ok ? Am I missing cases ?

Balarka Sen

You're choosing an axis first.

Hippalectryon

Choosing two points gives you an axis

user21820

Folding by reflection alone can't do a lot of things..

Hippalectryon

What other folds do you see ?

user21820

one can also fold in such a way as to mimic a compass construction

in the standard way to get an equilateral triangle for example

Balarka Sen

2:41 PM

I think one can construct a regular octagon @Hippa

user21820

reflection stays within the base field, which will be Q for a square

Balarka Sen

In fact regular hexagon is also possible.

@user21820 Not really

user21820

wait how do you construct a hexagon with only reflection?

Balarka Sen

You can extend to Q(2^(1/2))

Just fold through the diagonal of the square

user21820

i don't think that's the right way to think of the problem

because that doesn't prove that you can get a regular octagon for example

Balarka Sen

2:42 PM

I can construct a regular octagon

Hippalectryon

@user21820 Could you give me an example of other folds ?

Balarka Sen

fold the corners of the squares appropriately.

user21820

@Hippalectryon: Wait a moment.. I want to see Balarka's construction of an octagon.. maybe we don't have the same idea of what is allowed

Balarka Sen

You can prove that it's regular, if you're thinking the other way

user21820

@BalarkaSen: What is "appropriately"?

given just a square piece of paper, how do you get an octagon?

Balarka Sen

2:44 PM

@user21820 OK, dis : You have the square

Fold so you get a 1 \times 1/2 rectangle

user21820

how?

sorry

Balarka Sen

?

user21820

i saw sqrt(2)..

lol

Balarka Sen

Oh OK

user21820

please carry on

Balarka Sen

2:46 PM

OK, now fold the a corner of your rectangle such that if you halve the hypotenuse of the right triangle you get after the fold, it is of the same length as the rest of side of length 1/2

user21820

do you want to name the points so that it's easier to describe?

Balarka Sen

Sure.

user21820

and do you at any point do an angle bisection?

that isn't a reflection

Balarka Sen

Angle bisection? No.

user21820

ok good

Hippalectryon

2:48 PM

You guys could use twiddla.com/1915075 to visualize the shapes

@user21820 @BalarkaSen Wouldn't this edit allow what you tried to do ? Here, given two points, __we'll define a fold by the action of reuniting those two edges__ (e.g putting two opposite corners together, that would makes a diagonal fold) __and we'll also consider folding along an axis defined by two points__

__Is defined as a point what is, or has been in previous folds, an edge__

Or, would that allow too easily the construction of any real ?

user21820

ok first we start with a piece of paper with some marked points

Balarka Sen

ok, sure

user21820

in general that will just be the 4 corners

assume that those are {0,1}*{0,1}

so only coordinates are 0 and 1

Hippalectryon

Note that with my last edit, more "marked points" can appear without being currently edges

user21820

yes

just a little moment

any 2 points determine a line.. we can fold to crease the paper along that line..

we can also use any intersection of creases as a new point

Balarka Sen

3:00 PM

mmhmm

user21820

but each time we unfold the whole paper

and look at the points with respect to the original coordinate system

note that by folding we can reflect a point across a crease

and can use that point in subsequent folds

Balarka Sen

sure

how does that differ from my definition?

user21820

so if we unfold everything.. we can see that the only points we can construct are closed under intersections of lines between 2 pairs of points and reflections across lines

wait wait

now define a real number to be reflection-constructible if it is a coordinate of some point that can be constructed as above.. by reflections alone

Balarka Sen

ok

ok i see

user21820

now we can check that any reflection-constructible point has rational coordinates

we need a compass like mechanism to construct more

Balarka Sen

3:04 PM

but @user21820 you're defining stuff by coordinates

i was defining them by length

user21820

i know.. but by length it is difficult to say much about what you can and cannot do with folds

with coordinates we can have precise control

in particular this shows that you cannot create the regular hexagon or regular octagon by reflection folds alone

Balarka Sen

but the problem @Hippa intended was about lengths, i think.

user21820

wait a minute.. we can get lengths once we add in compass-like folds

Hippalectryon

@TedShifrin Haha math.stackexchange.com/review/low-quality-posts/324047

Balarka Sen

what do you mean by compus-like folds

HAHAH @Hippa

user21820

3:06 PM

low quality lol!

actually i would put that in a comment rather than an answer

hahaha

Jasper Loy

@Hippalectryon My answers often appear there because they are one line.

Hippalectryon

Plz I wanted a full answer because I don't want to work, don't give me a hint. FLAG

user21820

let me call it a compass-fold which is as follows: for points A,B,C, we are allowed to construct D on AB such that AD = AC

Hippalectryon

At worst, flag it as "not an answer", not a "low qual"

user21820

there are two solutions, and we are allowed to construct both of course

Balarka Sen

3:08 PM

i'm confused. draw me it, @user21820

user21820

sure

John Doe

I once was lost but now I'm found...

Hippalectryon

@JohnDoe hello, Found

Jasper Loy

One can get lost again after being found.

John Doe

True dat @JasperLoy

Jasper Loy

3:11 PM

@JohnDoe What made you utter that utterance anyway?

Balarka Sen

ok, how do you propose to construct \sqrt{2} this way @user21820?

user21820

hmm you have (1,1)

just compass it down to the x-axis

centred at (0,0)

Balarka Sen

oh right

user21820

then you have (sqrt(2),0)

Hippalectryon

But we're not allowed to use a compass... or am I missing something ?

Balarka Sen

3:13 PM

yeah that looks like a way interesting aspect of doing it

user21820

no this is equivalent to using the compass

so if we have the compass-fold, defining constructability of real numbers by either coordinates or lengths don't matter anymore.. both are equivalent..

Hippalectryon

Ok

user21820

and then we can get the hexagon and regular octagon by mimicking the compass constructions using the appropriate folds

John Doe

somewhere in my head there is a collection of religious songs...not something I'm proud of...but I was whistling it...and then I thought what's that...turned out to be 'amazing grace'...anyway. @JasperLoy

Balarka Sen

yes

user21820

3:14 PM

but we still are stuck in quadratic extensions

Balarka Sen

your field is quadratic then

user21820

are you familiar with a bit of field theory?

any 2-power extension is obtainable

Balarka Sen

ahem @user21820 yes.

my interest is in galois theory, in fact

user21820

there are more types of folds

that allow us to construct even more

i'm not an origami expert but i saw those somewhere online before

you may want to search them out

Balarka Sen

well we're just closing in at the huzita-hatori extension

John Doe

3:16 PM

Just spending your life doing maths is that anyway to live?

any way

Balarka Sen

Q adjoined with 2^n-degrees adjoined with 3^m-degress adjoined with fermat prime degrees

Hippalectryon

@JohnDoe What's wrong with it ? (btw, up arrow to edit posts)

Balarka Sen

such a boring extension

user21820

ah yes that's what i saw before

hahaha

it's axiom 6

that is special

anyway it's just a fun excursion into field theory

John Doe

@Hippalectryon I guess nothing is wrong with it, it just seems like a very confined way to live sometimes.

user21820

3:18 PM

better than just talking about polynomials all day.. especially if students are bored

Balarka Sen

haha

user21820

hey since you know field theory.. maybe i should be asking you some questions

Balarka Sen

ok

John Doe

@Hippalectryon I want to go to France.

@Hippalectryon ce que je peux vous rendre visite

Balarka Sen

French algebraists are ugh

user21820

3:20 PM

@BalarkaSen: if you have [F(x):F] and [F(y):F] coprime.. what can we say about [F(x+y):F]?

where x,y are algebraic over F

Jasper Loy

@BalarkaSen What do you mean?

@user130018 Hi!

Hippalectryon

@JohnDoe Why France ?

Balarka Sen

@user21820 F(x + y) = F(x, y)

user21820

why?

i don't see why

Balarka Sen

well one inclusion is obvious

John Doe

3:22 PM

@Hippalectryon Je ai un endroit pour rester

Balarka Sen

now for the other inclusion, consider the basis of F(x, y) over F

this is {1, x, ..., y, y^2, ..., x^i y^j, ..., x^n y^m}

user21820

yup

Balarka Sen

now what's the basis of F(x + y) over F?

user21820

powers of (x+y) also but i don't know the degree

Jasper Loy

@user130018 What happened?

Balarka Sen

3:24 PM

degree is \leq mn, @user21820

user21820

yup i know that

it's a factor too

Hippalectryon

@JohnDoe Well I doubt you could see me lol, I am just a student

John Doe

la vie est dure!

Balarka Sen

actually we have to show that x, y \in F(x + y)

user21820

yea how?

John Doe

3:26 PM

I don't want to see you I just want somewhere to stay lol @Hippalectryon

user130018

@JasperLoy Just wanted to drop in and say hi to you before school

user21820

haha in fact i wrote something one year ago that implied what you said but i couldn't prove it again a few days ago and now doubt that it was correct

Jasper Loy

@user130018 OK, good luck for your exams! We'll talk again after they are over!

Hippalectryon

@JohnDoe I'm afraid I can't help with that :/

Balarka Sen

actually @user21820 you need some assumptions

like F(x) \cap F(y) = F

John Doe

3:28 PM

pas de problème @Hippalectryon

user21820

ok what do you get with that assumption?

wait that assumption is true

John Doe

la vie est dure!!!

Daniel Fischer

@BalarkaSen Follows from the coprimality of $[F(x) : F]$ and $[F(y):F]$ here.

Balarka Sen

ah

then it's true

user21820

yup

no i disagree

John Doe

3:29 PM

Je devrais aller travailler maintenant

Balarka Sen

looks at the galois group of the composite F(a, b)

user21820

what can you say if you have that?

Hippalectryon

@JohnDoe Bonne chance :)

Balarka Sen

Gal(F(x, y)/F) is Gal(F(x)/F) \times Gal(F(x)/F)

John Doe

merci J'en ai besoin

user21820

3:31 PM

oh okay that's assuming it's a normal and separable extension

if we don't have that?

John Doe

Je ai une nuit de Latex!!! :) eish

Balarka Sen

there are no nongalois extensions :P

what?

i am just joking

lol

3:31 PM

So I was playing Condemned 2: Bloodshot last night. That doll factory ;_;

user21820

for example if [F(x):F] = 4 and [F(y):F] = 9.. is it impossible for [F(x+y):F] to be 6?

i can prove that [F(x+y):F]^2 >= [F(x,y):F]

but nothing more

Sawarnik

@ParthKohli Yes? :)

user21820

maybe @DanielFischer can help? =)

Balarka Sen

no idea, @user21820.

Sawarnik

Hmm...Balarka's got a hat.

Balarka Sen

3:36 PM

lots of hats @Sawarnik

Sawarnik

Yup..

user21820

@DanielFischer: the problem i asked is that if F is a field and [F(x):F] and [F(y):F] are coprime, what can we say about [F(x+y):F]?

Sawarnik

@DanielFischer I voted for you just now :)

Jasper Loy

@user21820 Didn't Balarka already help you?

Balarka Sen

for galois extensions, @Jasper

not in general

user21820

3:38 PM

@JasperLoy: Yea I wanted to know for things like Q(2^(1/5)).. which seems to be an abnormal extension

Balarka Sen

it is

user21820

yea i'm joking too =P

Balarka Sen

:P

user21820

@BalarkaSen: while waiting for Daniel.. you're a high-school student?

Balarka Sen

kind of

Daniel Fischer

3:43 PM

@user21820 Obviously it's a divisor of $[F(x,y) : F]$, and it is $> 1$ (unless $x,y\in F$). Without further niceness assumptions, I don't know if we can say more, haven't done algebra for a while. There's a good chance that it is equal to $[F(x,y) : F]$, but I doubt whether it has to be.

user21820

@DanielFischer: Ah ok thanks.. i suppose the good chance is due to the primitive element theorem that a generic choice of a gives F(x+ay) = F(x,y)?

Balarka Sen

a is unlikely to be identity for most cases

user21820

oh why?

i thought it's a non-constructive proof

Balarka Sen

if F is not of char 0... things get complicated

user21820

ah

still

if F is infinite

Balarka Sen

3:46 PM

still might not be true

say, if F = Frac C[[z]]... ack.

i doubt it's true for infinite cases either.

user21820

wait do you have an example satisfying the conditions i specified but such that F(x+y) isn't F(x,y)?

i'd be curious to see one

Balarka Sen

no, but i'd have to devise one

user21820

haha

like you i suspected we may be able to get one using the finite fields

but no idea

Balarka Sen

well I think you can get one from C(z) @user21820

user21820

@BalarkaSen: Anyway I'm sleeping soon.. nice talking to you!

oh

ok i'll try another time

Balarka Sen

3:53 PM

consider C(\sqrt{z}) and C(\sqrt[3]{z+1})

C(\sqrt{z} + z\sqrt[3]{z+1}) is not C(\sqrt{z} + \sqrt[3]{z+1})

user21820

z is a constant?

Balarka Sen

z is a transcendental over C

your base field is C(z)

user21820

oh ok

Balarka Sen

so... there you go

user21820

not so fast

both have degree 2

not coprime

Balarka Sen

3:54 PM

oh you want coprimality

user21820

yup

=)

Balarka Sen

there you go

;)

it still works

user21820

hmm

wait why are they not the same?

Balarka Sen

dunno. pretty sure they aren't.

can't think about it anymore, need to go

byes

user21820

lol

pretty sure they are the same but ok see you!

Balarka Sen

3:57 PM

@user21820 hey not my fault if i handwave'

:P

user21820

no problem =)

Daniel Fischer

@user21820 Maybe they are equal, maybe not. Point is, sometimes we have $F(x+y) = F(x,y)$ and sometimes not. When I say "good chance", I don't know how good the chance is. I just think with coprime degrees, it's not too rare.

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$Mathematics$ Mathematics