/sci/ - Science & Math

[Anonymous No.3006067]

ITT: math facts that blow your mind. I'll start.

It is unknown whether pi + e is rational or irrational.

[Anonymous No.3006079]

>>3006075
>for sufficiently large values of 2

[Anonymous No.3006075]

2+2=5

[Anonymous No.3006090]

There is no way to logically prove that math is true.

[Anonymous No.3006098]

>>3006090
That's a universal negative and hence impossible to prove.

[Anonymous No.3006105]

Mathematics=Maths=/=Math
Just like Abdominals=Abs.
L2 language Amerifags.

[Anonymous No.3006132]

I am frightened by conditional nevers, for example:
a fork in which you can't go back, esp over a timeline.
like you walk down the street and pass a free sample stand and don't stop for the snack. the next days it's gone, you will NEVER taste that snack.. EVER.

all snacks in the future will not be the snack you passed up, they are different matter.

of course I would not be so dumb to pass up free snacks.. but that is a trivial example. I'm not sure how this is expressed in math. but I find it very frightening how conditional scenarios/object encounters/presentations of information/etc.

I'll try to think of more example while the thread goes..

[Anonymous No.3006153]

how conditional scenarios/object encounters/presentations of information/etc. can be controlled by forking sequences of actions and influences (randoms, etc) over time

captcha: sequences relateun
>mfw

[Anonymous No.3006154]

Dammit /sci/, hurry up and blow my mind. I'm getting bored.

Here's one I always liked: pi = 4 - 4/3 + 4/5 - 4/7 + 4/9 - ... which is all but useless for calculating digits but eerily simple.

[Anonymous No.3006192]

>>3006090

You can't prove your own statement.

[Anonymous No.3006204]

The curvature of surfaces can be calculated by only knowing the first fundamental form.

In layman's terms: You can measure angles and distances on an arbitrary surface and can use these to calculate the curvature.

This means: There exist objects with curvature that do not have to be embedded in a surrounding space. You can use almost all the geometrical concepts you know without having to leave a surface, or even higher dimensional objects (manifolds).

http://en.wikipedia.org/wiki/Theorema_egregium

Liek this?

[Anonymous No.3006214]

>>3006090
Maybe not WITH math, but it's true practically.
Proving math is true with math is like proving the universe true with the universe. Just an aspie circle jerk.

>>3006067
Can two irrational numbers make a rational number? I didn't think so.

[t-shirt !pants1bKj2 No.3006215]

the golden ratio.

[Anonymous No.3006220]

>>3006214

i guess OP meant transcendental

[Anonymous No.3006227]

>>3006204

That's where Riemannian geometry comes from, which was more or less the starting point of general relativity.
Our 4-dimensional universe is not embedded into a higher-dimensional space, yet, it has curvature we can measure from inside this 4d-space.

[Anonymous No.3006228]

>>3006214

Pi is irrational. -pi is irrational

pi+(-pi)=0

[Anonymous No.3006233]

>>3006214
>>3006220
Nope, I meant irrational. And yes, two irrationals can add up to a rational quite easily -- sqrt(2) and -sqrt(2), for example.

[Anonymous No.3006238]

>>3006228
>>3006233

Ok, irrational numbers a,b

a + b != 0

Example?

[Anonymous No.3006240]

pi + e is only "unknown" to be rational or irrational because there's no proof either way

there's too many of these proofs and they're too difficult to do for many people to bother, pi+e, pi^2+sqroot(e) etc, is the main reason they're not done

it's almost certainly known to be irrational

[Anonymous No.3006242]

>>3006238

a = pi + 1
b = - pi

a + b = 1

QED

[Anonymous No.3006245]

>>3006242

Any examples not involving the same irrational number?

[Anonymous No.3006256]

Lie-groups are an algebraic structure, but are a special kind of surface as well.

[Anonymous No.3006264]

>>3006245

You'll find that 1 + pi and -pi are two very different numbers.

[Anonymous No.3006265]

>>3006245
Not that I know of.

[Anonymous No.3006273]

>>3006264

both these numbers involve Pi.

[Anonymous No.3006284]

if you know a komplex function on a closed curve, you know it at all points inside the curve

> http://en.wikipedia.org/wiki/Cauchy's_integral_formula

[Anonymous No.3006289]

>>3006256
what's mind blowing about that? it's just definition...similar to elliptic curves.

[Anonymous No.3006300]

>>3006273
If you can prove that the sum of two irrational numbers is rational, I think you're sort of saying by definition that the two irrationals "involve" one another. For irrationals a,b and integers c,d,
a + b = c/d
a = c/d - b
so one can be written in terms of the other with some boring rational number added in.

I get what you're asking but I'm not sure the question ultimately makes sense.

[Anonymous No.3006315]

>Can two irrational numbers make a rational
>number? I didn't think so.
( (sqrt2)^(sqrt2) ) ^ sqrt2 = 2

[Anonymous No.3006334]

>>3006289

You can't impose an algebraic structure/geometrical structure on everything, so for me it's pretty mindblowing to see something that has both.

And yeah, elliptic curves are also fucking neat.

[Anonymous No.3006341]

>>3006300

It makes sense if you're pondering whether pi+e is irrational. If irrational numbers only add up to a rational number if they "involve" the same basic irrational number, the answer is obvious.

[Anonymous No.3006343]

>>3006315
He was clearly referring to the sum of irrationals. Plus your example is ridiculously overcomplicated, why not just say pi/pi = 1 and be done with it.

[Anonymous No.3006352]

>>3006341
But if pi + e is rational, then pi = (some rational number) - e, so there's really only one "basic" irrational number involved.

[Anonymous No.3006376]

>>3006352

Yeah, in that case

a = (some rational number) - e

b = e

Ah wait, i see. Thanks. I am kind of slowpoke today.

[Anonymous No.3006394]

roots of negative real numbers are not defined.

logs of negative real numbers are not defined.

[Anonymous No.3006421]

You can use any base you want to give names to numbers.

irrational number bases for example are possible.

In those bases, there are other prime numbers and other theorems for number theory, which i did not have found any paper for and will probably study in the future.

[Anonymous No.3006430]

One can also define a 1/2 derivative or any other value. The -1 derivative would be an integral etc..

In this framework, sveral (physics and maths) problems and theorems can be written in a compact way.

[Anonymous No.3006438]

>>3006421

Hm, if i get that right, for OP's theorem you'd just have to test wether e has a finite representation in base pi, right?

[Josef !!nUf2NflSAyw No.3006470]

>>3006394
Yes they are. Selecting a branch in the complex case is exactly what you've been doing in the real root case anyway, defining <span class="math">\sqrt{a}\geq0[/spoiler].

[Anonymous No.3006487]

1=2.

[Anonymous No.3006515]

>>3006470

The limits do not converge from both sides, so they are actually not defined, at least if you use log and root as functions.

[Anonymous No.3006511]

>1/7=0,142857(142857)
>142857*2=285714
>142857*3=428571
>142857*4=571428
>....
>142857*18=2571426
>2571426 --> {2}{571426} -->{57142}{(6+2)} --> 571428

[Josef !!nUf2NflSAyw No.3006539]

>>3006515
Let <span class="math">\sqrt[n]z\equiv \sqrt[n]{|z|}\;e^{i\frac{\mathrm{arg}(z)}n}[/spoiler]. Tadaa, complex root defined. You're free to alter your definition by adding <span class="math">\frac{2i\pi k}{n}\,;\;k\in\{1\ldots n\}[/spoiler] to the exponent of course.
Same thing's possible for the log.

[Anonymous No.3006558]

>>3006539
that was sum fail.

[Anonymous No.3006593]

The total sum of all integer positive numbers is a negative fraction.

[Anonymous No.3006594]

>>3006539

Let's say <span class="math">z = re^(it)[/spoiler]
To make the log one-to one, we have to take <span class="math">t \elementof (-\pi,\pi)[/spoiler] right? So let's see what happens, if we see what z -> -1 yields. Set r = 1.

<span class="math">z_1 = e^(i(\pi-\epsilon))[/spoiler]
<span class="math">z_2 = e^(i(-\pi+\epsilon))[/spoiler]

with <span class="math">\epsilon > 0[/spoiler] sufficiently small. In both of these cases, the argument is an element of the function's domain.

Now let's see what we get if we take the Ln:

<span class="math">lim_{\epsilon->0} Ln(e^(i(t-\epsilon))) = lim_{\epsilon->0} i(\pi-\epsilon) = i\pi[/spoiler]

and


<span class="math">lim_{\epsilon->0} Ln(e^(i(-\pi+\epsilon))) = lim_{\epsilon->0} i(-\pi+\epsilon) = -i\pi[/spoiler]

So those are not the same number, therefore, the limit does not converge, therefore, Ln(-1) does not have a solution if you apply mathematics rigorously.

[Anonymous No.3006598]

>>3006594

Shit, Latex failure.

I hope you still get what i mean.

[Anonymous No.3006609]

>>3006539

Cleaned up a bit:

Let's say <span class="math">z = re^{it}[/spoiler]
To make the log one-to one, we have to take <span class="math">t \element (-\pi,\pi)[/spoiler] right? So let's see what happens, if we see what z -> -1 yields. Set r = 1.

<span class="math">z_1 = e^{i(\pi-\epsilon)}[/spoiler]
<span class="math">z_2 = e^{i(-\pi+\epsilon)}[/spoiler]

with <span class="math">\epsilon > 0[/spoiler] sufficiently small. In both of these cases, the argument is an element of the function's domain.

Now let's see what we get if we take the Ln:

<span class="math">lim_{\epsilon->0} Ln(e^{i(t-\epsilon)}) = lim_{\epsilon->0} i(\pi-\epsilon) = i\pi[/spoiler]

and


<span class="math">lim_{\epsilon->0} Ln(e^{i(-\pi+\epsilon)}) = lim_{\epsilon->0} i(-\pi+\epsilon) = -i\pi[/spoiler]

So those are not the same number, therefore, the limit does not converge, therefore, Ln(-1) does not have a solution if you apply mathematics rigorously.

[Anonymous No.3006615]

>>3006609

Sigh, we have to restrict t to -Pi < t < Pi.

[Anonymous No.3006632]

>>3006609

Yes. Asking "Ln(-1) = ?" Is not the same as asking "What numbers x can i put into e^x so that e^x = -1?".

Same with roots. x² = 4 has 2 solutions, -2 and 2, whereas root(4) only has one solution.

[Josef !!nUf2NflSAyw No.3006634]

>>3006609
That's an inconsistent definition of the logarithm, you'll have to limit your result to a certain complex phase.
Define <span class="math">\log(z)=\log(|z|)+i\,\mathrm{arg}(z)[/spoiler] and you've got a perfectly well-behaved function everywhere but at the branch point.

[Anonymous No.3006650]

>>3006634

It is not inconsistent, it's the usual convention of choosing where to slice the complex plane so the logarithm becomes a function.

You can also choose another direction in which you cut the complex plane, but why would one do that? This way, the logarithm is compatible to most mathematical applications and all fields in which it is used. No exceptions as far as i know.

Same goes for the root of course.

[Josef !!nUf2NflSAyw No.3006658]

>>3006650
The phase of your result is <span class="math">\pm\pi[/spoiler], but you have to limit yourself an interval like <span class="math">[0,2\pi)[/spoiler] or you'll jump over the branch cut.

[Anonymous No.3006665]

>>3006090
actually there is, you should see the proof for 1+1=2. shit is incredible, and once you can prove that logically... math generally starts with that assumption

[Anonymous No.3006719]

>>3006658

My branch cut was at \pi. I did not cross it as far as i see it.

[Anonymous No.3006738]

>>3006665
Hope you don't mean the false proof which involves zero.

[Josef !!nUf2NflSAyw No.3006754]

>>3006719
The "contradicting" results of your log are <span class="math">\pm i\,\pi[/spoiler], so in one of your limits you've jumped over the cut.

[Anonymous No.3006773]

>>3006754

I don't see where i did that, as i said in
>>3006615
that i restrict t to be an element of (-\pi,\pi).

[Josef !!nUf2NflSAyw No.3006839]

I don't feel like playing error ping pong anymore. Functions may have different limits depending on approaching direction, and the log I've posted above is a well-defined function on <span class="math">\mathbb C\setminus\{0\}[/spoiler].

[Anonymous No.3006846]

>>3006839

> Functions may have different limits depending on approaching direction

No.

[EK FAN CLUB No.3006857]

>>3006846
f(x)=1/x

Approach zero from either direction.

[Anonymous No.3006859]

<span class="math">e^{\pi i}+1=0[/spoiler]

<span class="math">i^i\in\mathbb{R}[/spoiler]

[Anonymous No.3006876]

<div class="math">e^{2\pi\,i+1}=e^{2\pi\,i}e=1\cdot e=e</div><div class="math">e=e^{2\pi\,i+1}=(e^{2\pi\,i+1})^{2\pi\,i+1}=e^{(2\pi\,i+1)(2\pi\,i+1)}=e^{-4\pi^
2+4\pi\, i+1}=e^{-4\pi^2}e^{4\pi\, i}e^1=e^{-4\pi^2}e</div><div class="math">e^{-4\pi^2}=1</div>

[Anonymous No.3006895]

1-pi is irrational, pi is irrational, 1 - pi + pi = 1 which is rational.

[Anonymous No.3006903]

>>3006876

I hadn't seen that before. It's very nice.

[Anonymous No.3006924]

>>3006876

the fucks wrong with you?

[Anonymous No.3006938]

>>3006846
>has never heard of discontinuous functions

[Anonymous No.3007001]

Riemanns mapping theorem: any simply connected proper subset of the complex plane can be mapped conformally (i.e. preserving angles) to the unit sphere, is pretty mindblowing.

[Anonymous No.3007008]

I got a 15 in the last math class I took.

[Anonymous No.3008340]

the dimension of the vector space of the equivalence classes of closed forms on a manifold mod exact forms is equal to the number of "holes" in the manifold. This is FUCKING AWESOME

[Anonymous No.3008361]

every finitely presented group is the fundamental group of some symplectic 4-manifold. Fuckin awesome.

[Anonymous No.3008385]

>>3008361
I wonder if there is any correspondence between the manifolds who have a certain group as their fundamental group and the polynomials who have the group as their galois group?

[Grothendieck !Topos4qCwc No.3008422]

>>3008385
I kinda doubt it, since the fundamental group isn't giving you an enormous amount of data about the manifold, whereas the galois group is telling you quite a lot about the field extension.

You would need to layer on additional structure on the manifold to get any kind of correspondence, I imagine.

As for mind-blowing: Higher homotopy groups for spheres still haven't been calculated.

[Anonymous No.3008463]

>>3007001
>>3008340
>>3008361
>>3008385
>>3008422

Sure is nerdy in here...

ive always thought the idea that 1/x means you can get as close as you want to to infinity but you're actually getting close to zero.. if you ask me, this is proof that order exists in chaos. all and nothing combind as one.

[Anonymous No.3008478]

>>3006315


wat.

[Anonymous No.3008501]

>>3006214
Can't they?

take an irrational number 0<X<1

take a number Y such that the ith digit in the decimal expansion of Y is equal to 9 - the ith digit in the decimal expansion of X

X+Y=1

[Anonymous No.3008569]

>>3008501
x+y is not in the intermediate field extension of the hyper complex numbers though.