@SohamChowdhury Oh you giving IISER, thats cool... though the curriculum for IISER mohali is better

For mathematics

Intrinsincally the collatz conjecture is about how base 2 transformations act on base 3 expansion and vice verse

@LeakyNun hm

I think you misunderstand

I meant that a number such as 6 has a ld of 2.

2/2 = 1

what is ld?

lowest divisor

(other than 1)

I don't see how that has anything to do with the conjecture

So $$T^k(n) \approx \left(\frac{1}{2}\right)^{\frac{k}{2}}\left(\frac{3}{2}\right)^{\frac{k}{2}} n = \left(\frac{\sqrt{3}}{2}\right)^k n$$

Sorry, I forgot that you were so much smarter than me.

Which holds for large $k$ and $n$ - so you expect the function to decrease eventually.

And yes, @LeakyNun, of course I know this isn't a proof and maths isn't based on chances.

hey guys

what's up?

But if something is still a conjecture, it's a good thing to have arguments why the conjecture is at least plausible

Yeah, you'd want to know if you are looking to prove or disprove it

I'll repeat

Sorry, I forgot you were so much smarter than me.

Right.

@SteamyRoot I think of a conjecture like this... it's a statement we have left unresolved. Its truthfulness is the question. We don't go into it with one impression. Rather, we seek to find evidence either way. However, I think most conjectures spring from basic intuition.

That is to say... someone notices a pattern and just asks "is it true?"

I haven't really thought about the collatz conjecture. I just think that it should be easier to prove.

To me, that's not good enough for a conjecture.

@SteamyRoot Another way to put it, I'd rather have decent question on why a conjecture matters either way.

I see many people use the word "conjecture" for everything they're like: "ooh, I think this is true because it works for these $3$ examples".

I use it for anything I have an intuition is true, but cannot prove.

So you are more of a Weil, @SteamyRoot

If your conjecture is computable for a lot of values, then compute a crapload of them.

and graph them?

If it holds under extra conditions or in special cases, prove it; and see if there's an argument why these conditions are perhaps not necessary.

@Dodsy Yeah, Collatz is kind of an embarrassment for us mathematicians

It's such a simple statement, but we have no idea how to prove it :P

I don't find it embarrassing at all, honestly :/

Mathematics is not yet prepared for such problems

What does the first person to prove it get.

international recognition?

money?

both

Mostly recognition

Satisfaction

Recognition, some. Money, probably not.

Only one way to find out

Most of it won't depend on the fact that it's proven or not.

@SteamyRoot Not even physics agrees with that mentality. I think of conjectures like physical laws: "we have enough examples to say it is true (Newtonian Physics), but we're really just waiting for some shmuck to come along and disprove it (Einsteinian Physics)". In other words, it is assumption by observance and experimentation, but we cannot prove it... yet.

It'll depend on the proof itself.

Yup, the way we prove it might lead to some astounding new techniques

If the proof only uses known techniques and offers nothing new outside the proof itself, few mathematicians will care.

What is the collatz conjecture again?

@Typhon Physics agrees that if you try something for enough values and they all work out, it's true.

Mathematics doesn't - but it can be an indication of truth.

Still holds.

@SteamyRoot that is a blatant lie. Physics says that there is no perfect mathematical model and that there is always a case where something fails to hold true.

@SteamyRoot What?!

brb

They would perhaps not care for the proof itself, but the mathematical world would at least be shocked for a day

@Typhon Physics has nothing to say over mathematical models.

You sound like you want to refer to Gödel's incompleteness theorems.

The fact that the result would be provable by known techniques would itself be shocking

Well, of course they'd be shocked. But they wouldn't care in the long run.

true.

It'd be a disappointment.

@Dodsy if even maps to 3n/2 and odd maps to (n+1)/2, then nothing would ever map to 1 except 1 itself

@Krijn It would be more embarrassing

right

I hope you've discovered that before you removed it.

I realized that

nice.

It only goes to 2.

but its the same idea.

it will decrease to 2.

@SteamyRoot "Physics agrees that if you try something for enough values and they all work out, it's true." <=== It can only be true if you try ALL the values. That's the issue. Newtonian Physics is actually false.

D:

Einstein disproved Newtonian Physics

ooo

no.

that's not exactly true.

I am planning on studying complex analysis. My background is in real analysis and linear algebra. Do I need any more prerequisites for understanding complex analysis?

he proved it fails under certain conditions

Everything in physics only holds approximately; and under certain conditions.

Einstein proved when these conditions were no longer satisfied.

And, you can't ever try all the values for a physics experiment.

I think Einstein is a little bit overhyped

he even proved that time itself isn't just one dimensional. There isn't one universal frame of reference.

like how they looked at his brain

@Dodsy ffs. It was an analogy...

XD

At least physicists use five or six sigma, unlike some not-so-exact sciences...

@SteamyRoot True. The best example of this would be thermodynamic processes. We always build systems that resemble the actual system but the actual system can never be that efficient.

I was just saying that I think of conjectures like people attempt to prove physical laws. We try as many values as we can and then we say "it is likely true". However, we cannot ever prove that it is true under all conditions.

In other words, I cannot go and say that Newton's Laws are true 100% of the time.

@Typhon I don't see what "under all conditions" has to do with a mathematical conjecture at all.

Okay I have a Dodsy Conjecture take any integer, if the integer is even, multiply it by 3 and divide by 2, if the integer is odd, add one and divide by 2. The number will always eventually reach 2.

bam

If you pose a mathematical conjecture, you specify when, for what, ... you expect something to hold.

world famous

@SteamyRoot Newton claimed his physics always worked. In the same way, people test several hundred values and presume their conjecture is true for all numbers of a certain type.

hence I think of conjectures like physical laws. They're not bulletproof but heavily tested.

granted, I'm referring to true conjectures

writing a conjecture in a dinky little proof in a paper to prove something already well known for practice is really not what we mean here. We are referring to unsolved or important conjectures.

i.e. Conjecture - a statement well believed to be true but not provable at this time by the people involved.

@Dodsy I would rather call that a proposition.

it doesn't necessarily need to believed to be true.

Einstein is overrated, half of his discoveries were already found by Poincare

Ummm

You only call something a proposition once it's proven.

@Zee not relevant...

@SteamyRoot a proposition is something you're proposing?

like as a sub-proof in a larger proof.

By general relativity, I think it's pretty relevant

@Dodsy true, but generally the notable ones have been tested to see if any immediate computable counter-examples exist.

A proposition is just a statement that is either true or false.

Usually, in textbooks or papers, you use it to name something that's more than a lemma but not worth calling a theorem.

@Zee I meant my statement was just that someone disproved Newtonian Physic's claim that it holds under all conditions. It really doesn't matter who did it.

hm

@SteamyRoot everything is a statement that is either true or false except for "this statement is false."

There's plenty more paradoxical and/or unprovable statements...

No one proved newton laws don't hold in all cases, we just take that on faith since it suits our tastes better

@Zee what?

My conjecture actually does work

take any integer, if the integer is even, multiply it by 3 and divide by 2, if the integer is odd, add one and divide by 2. The number will always eventually reach 2.

That not a conjecture but a theorem, a really bad one

why is it bad

It's too specific

It's like the collatz conjecture but better

@Zee so you've already proved it?

Yes

@Dodsy can you prove it?

@Zee how so?

I don't think so.

1 will not reach 2

@Zee show us your proof

@Typhon except 1

oh true

except 1

I should put that in there

@LeakyNun that's what I just said...

(1 + 1) / 2 = 1

bam

True

never mind

but I should put that in there

"except 1"

any integer "except 1"

so you just really need to find one exception other than 1

to disprove my conjecture

Assume n is even then (n3)/2 = p3 where p is even and smaller than n, you can do that till p=2 in which case you end up with 6 and then 3 and then 4 and then 2 . The odd case is easier

Wow zee

you are a genius in a bottle

off to dinner lads.

and lasses

proving it myself

different method

its not hard just put it all into symbols

i know...

@Dodsy 0. It cannot be 0.

they have to be > 1

@Zee you can't do that till p=2, but you can do that till p is odd

just take n=28

@LeakyNun the first step maps all integers of the form $2^k(2n+1)$ to $3^k(2n+1)$

interestingly enough....

28 -> 42 -> 63 -> 32 -> ... -> 243 -> ...

ALL integers fall into that form

@Typhon yes, I agree.

Seriously, I don't think 28 terminates.

@LeakyNun so it is a matter of showing that the odd numbers map down to a significantly small number. I think it is just induction...

i know...

Sorry, brain fart.

I meant that if we prove it for all numbers 2-10

and show that everything comes down to 2-10

then it is proved

(example set)

I'm wrong: 28 does terminate.

28
42
63
32
48
72
108
162
243
122
183
92
138
207
104
156
234
351
176
264
396
594
891
446
669
335
168
252
378
567
284
426
639
320
480
720
1080
1620
2430
3645
1823
912
1368
2052
3078
4617
2309
1155
578
867
434
651
326
489
245
123
62
93
47
24
36
54
81
41
21
11
6
9
5
3
2

After quite a few steps.

@Typhon this is the hard part, you know

yeah...

well we could prove that any number eventually reaches a number smaller than itself?

I'm seeing a strange 11->6->9->5->3->2 pattern

@Typhon I can't.

Maybe you can.

im not saying we do

im just guessing

im saying I don't see how we can.

A miss is as good as a mile when it comes to such things

It's all well and good to feel that there should be a proof, but without an actual argument it's nothing more than a feeling.

ok...?

we're just trying to prove something dodsy posted as a challenge.

Collatz, or something weaker?

nah

just a thing dodsy posted

apparently it's "trivial"

"take any integer, if the integer is even, multiply it by 3 and divide by 2, if the integer is odd, add one and divide by 2. The number will always eventually reach 2."

...that's the Collatz conjecture, isn't it?

no...

Oh, wait

"If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1."

Yeah, I see now that the step for evens is different than Collatz

hmmm

Oh. And so is the odd. Deeeerp

hrrmm

i just realized something

the collatz conjecture could be proven by showing that the inverse operations spans all prime numbers

Why?

So if you start with an even number, Dodsy's map ultimately replace all factors of 2 in the prime factorization by factors of 3.

Doesn't the "add 1" bit mess it up? Adding 1 completely changes the prime factorization

because then repeated multiplication by two will span the other numbers...

That only gives numbers of the form $p\cdot 2^n$

@AkivaWeinberger which is all integers?

It wouldn't give 15, for example

@Typhon You're confusing "prime" with "odd"

derp

no i wasnt

i was thinking that all numbers have one prime factor other than 2

XD

Proving that the inverse map spans all odd numbers would work

however one can note that the second step might as well be (3n+1)/2

since the odd case produces an even case

immediately

Hi, considering i have a "sequence" that converges against "infinite". I shall prove that such sequence has a "minimal member". I am a bit confused how exactly i should prove this, because i thought if a sequence converges it automatically means that the sequence is "restricted" and has a minimal member ^^

gotta go

@YannikK. I'm not sure I understand. The sequence 1, 1/2, 1/3, 1/4, etc. converges to 0, but has no minimum

@AkivaWeinberger You're right. I know that now too. Bounded means only there is an upper limit and lower limit, but minimum means its included in the sequence

Yup

Incidentally, there's something called an "infimum"

which is almost the same thing as "minimum," but it doesn't need to be in the sequence

I know about those terms, but infimum and supremum are special cases of upper / lower limit

@AkivaWeinberger With the sequence you gave (1/n) it was quite obvious it has no minimum, but it has a maxium right ? since 1 is the largest element and is included

I'm pretty sure the 'proof' offered to you earlier is nonsense @dodsy

If n=6, for instance, then 3/2*6=9=3*3. This is indeed of the form 3p, but p=3 is not even

So the idea that every even case goes directly to p=2 is just nonsense

(I tend to think the statement is true, but I am highly dubious of that argument)

@AkivaWeinberger So i might be a bit confused right now, but what is then an example for a sequence that has no maximum AND minimum. Since the sequence can only rise or fall

it seems a little off for me right now that can it can have both no maximum and minimum

(-1)^n(1-1/n) @YannikK.

For even n, that's just 1-1/n which is bounded above by 1 but which never attains that maximum

For odd n, it's bounded below by -1 but again never attains it

So it has well defined inf and sup but no min or max

@Semiclassical Thanks. i played around with (-1)^n but i did not had that full approach

Np

You could also say a(2n)=b(n) but a(2n+1)=c(n), and specify these sequences appropriately

Hey, in some exercise I'm supposed to determine the open and closed sets of $Spec(\mathbb{Z}$ for the Zariski topology.

Further I shall find all generic points

@Semiclassical Well your first approach seems to be the more obvious one in my opinion

I am a bit confused: The closed sets are the $V(S)$ where $S$ is some subset of $\mathbb{Z}$. Since $V(S)=V(\langle S\rangle)$ for all $S\subset\mathbb{Z}$, it suffices to consider the ideals of $\mathbb{Z}$, right?

Now the ideals of $\mathbb{Z}$ are the sets $m\mathbb{Z}$ for any $m\in\mathbb{N}_0$

We can always write $m\in\mathbb{Z}$ in ist prime factor representation, $m=p_1^{\alpha_1}\cdots p_n^{\alpha_n}$

> With the sequence you gave (1/n) it was quite obvious it has no minimum, but it has a maxium right ?

Yes.

@YannikK.

so the elements of $Spec(\mathbb{Z})$ containing $m\mathbb{Z}$ are $(p_1),\dots,(p_n)$