/sci/ - Science & Math

>>12736801
I’m new, what is this?

>>12736801
>The (((Goldbach))) Cojecture

>>12736845
I hadn't realized but I should give a quick introduction. It's a very easy problem to state but no one has ever solved it.
>It states that every even whole number greater than 2 is the sum of two prime numbers.
For example
>4
2+2
>6
3+3
>8
5+3
>10
5+5, 7+3
>12
5+7
and on and on and on. It's true for a gorillion numbers but no one has ever proved that it's true for every even integer yet.
If you want to keep your sanity you probably shouldn't seriously spend your time on this problem but this thread is just meant to dick around in.
https://www.youtube.com/watch?v=MxiTG96QOxw
https://en.wikipedia.org/wiki/Goldbach's_conjecture

Here's my approach. I'm going to sound like a retard but whatever.
Break the even number, 2n, into two sets opposite n so that they sum vertically. This only works for numbers greater than 4 but we know 4 is true so this is ok.
For example
>6
A: 1, 2
B: 5, 4
>8
A: 1, 2, 3
B: 7, 6, 5
>10
A: 1, 2, 3, 4
B: 9, 8, 7, 6
[math]\vdots[/math]
We can take out all the columns with even numbers because we know 2 is the only even prime. So it becomes:
>6
A: 1
B: 5
>8
A: 1, 3
B: 7, 5
>10
A: 1, 3
B: 9, 7
[math]\vdots[/math]
Similarly, we know 1 isn't a prime number so we can take that column out as well. (Notice 6 is gone, we know it's true too so that's also fine.)
>8
A: 3
B: 5
>10
A: 3
B: 7
>12
A: 3, 5
B: 9, 7
>14
A: 3, 5
B: 11, 9
>16
A: 3, 5
B: 9, 7
>18
A: 3, 5
B: 9, 7
[math]\vdots[/math]
The Goldbach conjecture is true if we can prove that there will always be a prime p in set A such that p = (n-k), and a prime q in set B such that q = (n+k).

>>12736993
Bertrand's postulate means there will always be at least one prime in set B, and our construction means there will always be at least on prime in set A (namely 3).
I think Goldbach is true if we can prove that
>For every even number greater than 6, there are more primes in set A than there are in set B.
or something like that.

>>12736868
>Hehe, this will keep the smart Goys busy, hehehe.

>>12736801
If I solve it do I win a prize?

>>12737037
This its solution has no groundbreaking implications

>>12736868
Goldbach was actually a german dude tho... not a kike
the original golds

What happens when you look at sums of primes in other ordered rings?

>>12737042
A whole lot a' math hoes would be hitting you up trying to suck yo dick.
You would probably win the Abel prize and Fields Medal as well.

>>12737086
if I solve it, how do I prevent someone else from stealing my solution?
Mail it to a math professor directly?

>>12737091
Upload it to a free online journal like Academia.edu or vixra to claim primacy then try to advertise it here or somewhere else.
Most people wouldn't bother reading it though because they would think you're some kind of crank, and you probably would be, but this place is probably the only place where even educated people would actually take time to read your non-name non-peer-reviewed paper.
pic rel is a meme here and he did the same thing.

>>12737055
Spoken like a true mathlet.

>>12736993
>>12737025
Can someone explain this shit to me? What is this anon doing here...
I'm too retarded to tell if he's right or not.

My attempt. Find the average numerical reduction from each iteration. You'd need some recursion.
>If odd, do 3n+1 then do n/2 repeatedly until ending on odd
>If even, do n/2 repeatedly until ending on odd
The key would be determining the average reduction rate if the number is even. I suspect it's less than 33%, meaning the average iteration chops a number into a third of its original value.

>>12737091
Use an alias on viXra.

>>12736801
Every non-prime integer greater than 1 is exactly in between two primes (the proof is left as exercise for the reader)

If n>1 is prime, 2n = n+n which is a sum of two primes.

If n is not prime then n = (p1+p2)/2 -> 2n = p1+p2

Therefore for any n>1, 2n is a sum of two primes. QED.

>>12737114
QRD on this guy?

>>12738871
Bogdanovs fear him

2 x 3 x 5 = 30

30 - 1 = 29
30 - 7 = 23
30 - 11 = 19
30 - 13 = 17
30 - 17 = 13
30 - 19 = 11
30 - 23 = 7
30 - 1 = 29

That is the gist of it, I have it formalized somewhere, but never bothered to publish. You can probably use this to find a proof though, if you have half a brain.

>>12738952
last one came out backwards :
30 - 29 = 1

>>12737086
>abel prize
no way, that’s only for a lifetime of work. The only way you’d have a chance of getting abel would be if you waited until you were 41 to publish so you’d be disqualified from fields

>>12738952
Nice try.

2 x 3 x 5 x 7= 210

210 - 1 = 209 (not prime)
210 - 3 = 207 (not prime)
etc.

>>12738977
You still did not spot the pattern... try doing it without the primes in the product.

>>12740282
20 = 5 x 2 x 2
20 - 11 = 9
k.

>>12737086
Math hoes are a thing?

>>12737080
What are some examples of other ordered rings?
This does seem like a good approach.

>>12737080
Nice try, but probably not.

>>12737646
>Every non-prime integer greater than 1 is exactly in between two primes
8 isn't.

>>12740312
yes, fortunately.

>>12736801
1. find a formula for primes
2. solve goldbach

If you tards want to actually make some sort of progress on this problem you MUST learn p-adic analysis

>>12741896
8 is exactly in between 5 and 11, retard.

>>12741917
Therefore 6n+-1 is probably the answer, implying that there are infinitely many twin primes, thus solving the goldbach by the n/2 approach
>>12743275
t. retard