/sci/ - Science & Math

[Anonymous No.3639375]

>Atheist vs religion threads
>hw threads
>no string theory threads
>no actual math threads

You guys suck

[Anonymous No.3639404]

>>3639375
okay sucker
>string theory
>math

ask me anything.

[Anonymous No.3639405]

>>3639375
>thinking string theory is scientific

ISHYGDDT

[Anonymous No.3639415]

0.9999... < 1

[Anonymous No.3639427]

>implying string theory and math are the only things you can talk about in a math and science board

[Anonymous No.3639455]

Let's talk about maths, then.

[Anonymous No.3639460]

>>3639404
>>3639404
You're a red headed girl, what could you possibly know?

[Anonymous No.3639474]

>>3639404
how can a photon "interfere" with itself? Is there quantum superposition of many EM fields?

[Anonymous No.3639488]

>>3639455
<span class="math">2730 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 13[/spoiler]

Obviously <span class="math">2|n(n^12-1)[/spoiler] since it's a product of an even and an odd number. Can't be bothered doing 3,5,7 and 13 but it's an easy uninteresting problem.

[Anonymous No.3639489]

>>3639455

First of all, that problem is bullshit.
Since 1 is a natural number, it implies 12 is divisible by 2730.

>>3639427
>>3639405

You guys do realize that any theoretical physicist who's to be taken seriously researches string theory, right? Theories relating to string theory (well M-theory, now) represent the only model we have that accurately explains all the natural phenomena around us.

You guys must all be pseudo-intellectual undergrads..

[Anonymous No.3639490]

String theory isn't really a scientific theory. It's a shitty hypothesis. So yeah, you suck.

[Josef crossed the Atlantic !!nUf2NflSAyw No.3639492]

>>3639455
As a physicist, the first <span class="math">10^7[/spoiler] naturals are enough for all interesting applications

[Anonymous No.3639505]

>>3639455
http://www.wolframalpha.com/input/?i=Table[Mod[n^13-n%29%2C2730]%2C{n%2C1%2C10000}]

It works for the first 10000 n's. So as a physicist, I define it to be true.

>>3639474
A single photon wouldn't interfere with anything. However a single free photon is a weird concept anyway, since there are charged particles around.

[Anonymous No.3639510]

>>3639455
<span class="math">1^{13} - 1 = 0[/spoiler]

[Anonymous No.3639513]

>>3639505
You're a fucking dumb shit.

You only have to check the first 2730 numbers. learn modular arithmetic if you want to stop sounding like a dumb shit. Splitting the congruence into primes let you check even fewer cases.

It's disgusting how idiotic statements like yours are used to demonize case checking in mathematics.

[Anonymous No.3639520]

>>3639510
> He doesn't know that everything divides 0.

[Anonymous No.3639533]

>>3639520
Of course it does. I was explaining why it holds for n=1.

>>3639505
There have been a few important problems where counterexamples don't start showing up for quite a while.

[Anonymous No.3639539]

>>3639513
Someone's mad.
(however, I have to admit, that atm I don't see why you could stop checking at 2730)

>>3639492
lol, physicist hivemind

[Anonymous No.3639559]

proving <span class="math">2730 | n^{13}-n[/spoiler] is equivalent to proving <span class="math">n^{13}-n \equiv 0 \pmod 2730[/spoiler]. This means you only need to check <span class="math">n\equiv 0,1,2,\ldots,2729[/spoiler]. Even faster would be to check <span class="math">n^{13}-n \equiv 0 \pmod {2,3,5,7,13}[/spoiler], this decomposition is valid due to Sun Zis theorem.

[Anonymous No.3639582]

>>3639559
why does this imply you only need to check n=1,...2370?
you take the mod of f(n):=n^13-n, why does the bound translate to the argument n?

[Anonymous No.3639578]

> my face when people studying shit like string theory don't know basic math.

what the fuck? You physics niggers use lie groups, get your shit together.

[Josef crossed the Atlantic !!nUf2NflSAyw No.3639588]

>>3639582
<div class="math">n^{13}-n=n(n^{12}-1)</div>

[Anonymous No.3639595]

Half figured it out.

The factorization of <span class="math">n^13 - 1[/spoiler] is:
<div class="math"> n (n+1) (n-1) (n^2 + n + 1) (n^2 - n + 1) (n^2 + 1) (n^4 - n^2 + 1) </div>
Now, the first 3 terms <span class="math"> (n, n+1, n-1) [/spoiler] force the resulting number to be divisible by 2 and 3, since they are a set of three consecutive numbers. Presumably, the addition of the other 4 terms force the resultant number to be divisible by 5, 7, and 13, but it's a lot of trouble to prove that...

[Anonymous No.3639596]

>>3639588
still don't see it.

[Anonymous No.3639599]

>>3639404
That is one ugly bitch.

[Anonymous No.3639601]

>>3639595
Typo: I meant to write:
"The factorization of <span class="math"> n^{13} - n <span class="math"> is..."[/spoiler][/spoiler]

[Anonymous No.3639602]

>>3639595
Double Typo: I meant to write:
"The factorization of <span class="math"> n^{13} - n [/spoiler] is..."

Christ, is my typing going to hell today?

[Anonymous No.3639608]

>>3639582
"mod m" is a number system where there are exactly "m" numbers. For example mod 3 we have number A B and C which satisfy:

A+A=A
A+B=B
A+C=C
B+B=C
B+C=A
C+C=B

A*A=A
A*B=A
A*C=A
B*B=B
B*C=C
C*C=B

we tend to write 0,1,2 instead of A,B,C though. The reason is because there's a inclusion from the integers into each of these modular arithmetics. Based on this inclusion some properties of the integers can be discovered/proved/etc just by studying modular arithmetics (especially prime powers, due to the Sun Zi theorem again).

The equivalence relevant for this situation is that if <span class="math">P(x) \equiv 0 \mod m[/spoiler] then <span class="math">m|P(x)[/spoiler] for every integer <span class="math">x[/spoiler]. Of course there much deeper relations between the modular arithmetics and the integers, but everyone should know this.

[Anonymous No.3639617]

>>3639595
you can prove it that way but..... why.. would you... almost as bad as proving it by induction (math teachers set fucking awful stupid problems like that for homework, no wonder everyone hates math and doesn't learn any of the good bits)

[Anonymous No.3639627]

>>3639617
Well, sure, you could brute force a solution by checking for every single number between and 0 and 2730, but isn't that cheating? :P

[Anonymous No.3639637]

It's shit easy using <span class="math">n^{(p-1)} \equiv 1\;mod\;p[/spoiler] on the prime factors.
<div class="math">
n^{13} - n \equiv n^{12} n - n \equiv 1n - n \equiv 0 \;mod\; 13</div>
<div class="math">n^{13} - n \equiv n^{6+7} - n \equiv 1n - n \equiv 0 \;mod\; 7</div>
<div class="math">n^{13} - n \equiv n^{4+4+4} n - n \equiv 1n - n \equiv 0 \;mod\; 5</div>
<div class="math">n^{13} - n \equiv n^{2+2} n - n \equiv 1n - n \equiv 0 \;mod\; 3</div>
<div class="math">n^{13} - n \equiv n^{1+1+1+1+1+1+1+1+1+1+1+1} n - n \equiv 1n - n \equiv 0 \;mod\; 2</div>

[Anonymous No.3639638]

you don't say what m and what P is, but I assume I know what you mean. However, I don't see why it's true.

[Josef crossed the Atlantic !!nUf2NflSAyw No.3639639]

>>3639627
Checking all possible cases is valid math.

[Anonymous No.3639641]

>>3639627
> "cheating"
> he still does things in idiotic round-about ways instead of getting shit done, because of a failed math education

[Anonymous No.3639661]

>>3639637
now you checked it for all mod p, where p are the prime factors of 2730. Why does it now follow that it's also true for 2730?

[Anonymous No.3639674]

>>3639637
What's the name of that theorem?

[Anonymous No.3639676]

>>3639661

Chinese remainder theorem

[Anonymous No.3639678]

>>3639638
P is any polynomial. It's not obviously true. You'd have to study the construction of modular arithmetic to see why. Some people define <span class="math">a \equiv b \pmod m[/spoiler] as <span class="math">m|a-b[/spoiler] but I prefer to define modular arithmetic using ring theory, so that this is a theorem instead.

The integers are a ring (in the sense that it has 0,1,+,* and a couple of axioms about how they work). The sets of multiples of a given integer are "ideals" for example 3Z={...,-3,0,3,6,9,...} is an ideal. To construct the modular arithmetic you just take the integers Z and "quotient" them by an ideal: Z/3Z. Now you can prove basic facts about it, showing that it is a ring, and some of its relations with divisibility (such as the theorem mentioned earlier).

The "quotient" thing is a very general set-theory (or category theoretic) operation and an extremely important part of mathematics necessary for understanding the development of all number systems. The majority seem to be completely ignorant about it, it's a tremendous shame. Again, shit like this is hardly anyone knows anything about math beyond how to pretend they're doing it.

[Anonymous No.3639683]

>>3639674

Fermat's little theorem

In cases where n is not coprime to p, then n^13-n being divisible by p is trivial.

[Anonymous No.3639687]

>>3639683
no

[Anonymous No.3639694]

>>3639687

yes.

p being primes of course.

[Anonymous No.3639696]

>>3639694
you are wrong. Feel free to figure out why or remain wrong.

[Anonymous No.3639703]

>>3639674
Fermat's Little Theorem

[Anonymous No.3639705]

>>3639678
ya, thx, but I know the stuff you stated here and
>>3639608
here.

Well except for the P(x)-theorem, obviously. But thanks for good intentions.

I actually like the "a-b"-definition, because it reminds me of the definitions in cohomology theory. Well, unsurprisingly I only used deRham cohomology theory, but I'm generally interested in it's generalizations.

I don't know much about number theory and only a bit about algebra, but that is of course because there are no lectures on both in physics. Only liner algebra in the first semesters. And as far as physics is concerned, you'd use modular aritmetic only as a tool in programming and to find normal subgroups of Lie groups.

[Anonymous No.3639709]

>>3639696

I am right, feel free to prove me wrong.

[Anonymous No.3639737]

>>3639696

Also, in the faint hope I am not being trolled, '13' is of course related to the above example, so p are the primes used there as well.

[Anonymous No.3640121]

thanks for bothering OP, shit needs to be said about the state of this board more often.

you are all disgraces to whatever you indentify as.

[Anonymous No.3640154]

>>3639705
Faggot

[Anonymous No.3640172]

>String theory

[Anonymous No.3642230]

>>3639375
Agreed. /sci/ mainly consists of trolls and pseudo-intellectuals

[Anonymous No.3642327]

>>3642230

And by 'mostly' you mean 'entirely'. I am both.

[Anonymous No.3642336]

>>3639375
instead let's get some "HURR DURR COLLEGE LOANS IN AMERICUH AND LET'S TALK ABOUT DEGREES AND COLLEGE AND DEGREES AND MAJORS IN MINORS MY GOD WHAT COLLEGE SHOULD I GO TO"

[Anonymous No.3642396]

>>3642336
what else do you need to know besides
>PHD MATH
>ANY JOB I WANT
>300K STARTING