Goldbach's Conjecture

[D.D.A. !L.mWDGnT2A No.3012035]

Few months ago I had an idea about Goldbach's conjecture, but since there were no math savvy people around me, I'm gonna post this here.

So basically, Goldbach's conjecture says that every integer bigger than 5 is actually a sum of 3 primes. So let's take an array P of prime numbers, p1,p2,p3,..p(n-1),pn,p(n+1)... where p1=2,p2=3,p4=5 etc. It is obvious that pn > pn-1 > pn-2 > ... > p2 > p1. All clear and basic by now.
Now, largest number between any two primes is larger prime -1: pn > pn-1 > p(n-1)
Sum of any three numbers is smaller than largest term times three:
3p(n-1) > p(n-1) + p(n-2) + p(n-3)
So, since
pn - 1 ≤ p(n-1) + p(n-2) + p(n-3)
we can conclude that
pn - 1 < 3p(n-1)
(will continue in next post)

[D.D.A. !L.mWDGnT2A No.3012059]

Here comes the handy thing I accidentally found - Bertrand's postulate:
If n > 3 is an integer, then there always exists at least one prime number p with n < p < 2n − 2. A weaker but more elegant formulation is: for every n > 1 there is always at least one prime p such that n < p < 2n.

So, from pn - 1 < 3p(n-1) we conclude that Goldbach's conjecture is wrong iff difference between two consecutive prime numbers is greater then smaller prime multiplied by three. However Bertrand's postulate tells us that there is always at least one prime number between some n and 2n-2. If n=p(n-1), then
2p(n-1)>pn>p(n-1).

Q.E.D.

Now tell me where I'm wrong, because sure as fuck it can't be this simple.

[D.D.A. !L.mWDGnT2A No.3012090]

Bumping for people who actually know shit.

[Anonymous No.3012109]

i don't see where you got this:
pn - 1 ≤ p(n-1) + p(n-2) + p(n-3)

[D.D.A. !L.mWDGnT2A No.3012141]

>>3012109
pn-1 (aka the largest number below prime pn) is either equal to the sum of three preceding primes, or some other three primes which are smaller than those, out of which you can conclude that the sum of three preceding primes was bigger than pn-1.

[Anonymous No.3012139]

>So, since
>pn - 1 ≤ p(n-1) + p(n-2) + p(n-3)

This is demonstrably false. Use pn = 17. 16 is not less then or equal to 13 + 11 + 7

[Anonymous No.3012144]

>>3012139
17-1<13+11+7
16<31

What are you talking about ?

[D.D.A. !L.mWDGnT2A No.3012149]

>>3012141
After reading this, I conclude that I really need to work on my English writing skills. Sorry anon, I hope you got the point.

[Anonymous No.3012155]

>>3012035
1st Observation:

>that every integer bigger than 5 is actually a sum of 3 primes

That is the weak conjecture. When mathematicans usually refer to goldbach conjetcure (without implicit specification), they are talking about the strong conjecture (not the weak). Although this is not really too important a point, you may want to note this, if you ever plan to do any serious work in this field.

[D.D.A. !L.mWDGnT2A No.3012183]

>>3012155
I'm aware of that, however I wanted to keep it simple for start. Any other observations ?

[Anonymous No.3012194]

>>3012059
You are wrong. and you logic is incoherent as well. Please, use an actual formal "proof" if you want us to be able to follow and find the flaws in your shit.

Being a good mathematican, means knowing how to write "proofs"

[D.D.A. !L.mWDGnT2A No.3012204]

Bumping once again.

[D.D.A. !L.mWDGnT2A No.3012220]

>>3012194
>Being a good mathematican
Never said I was a mathematician, especially a good one. And if you expected a formal, precise and correct proof, I believe that you are on a wrong website. But thanks on the bump.

[Anonymous No.3012236]

so does

>pn - 1 ≤ p(n-1) + p(n-2) + p(n-3)

follow from assuming GBC is correct?
just trying to grasp the proof here

[D.D.A. !L.mWDGnT2A No.3012256]

>>3012236
Yes.
It's either
pn-1 = p(n-1)+p(n-2)+p(n-3)
or
pn-1 = p(n-x)+p(n-y)+p(n-z)
where x,y,z>1

[Anonymous No.3012262]

>>3012059
What exactly do you attempt to prove? It is very unclear.

Do you actually prove anything? What is the point of this?

[Anonymous No.3012284]

alright so;

"So, from pn - 1 < 3p(n-1) we conclude that Goldbach's conjecture is wrong iff difference between two consecutive prime numbers is greater then smaller prime multiplied by three."

you have only proved one way, namely IF GBC is true THEN pn - 1 < 3p(n-1). not the other way around.
so you'll have to be very careful with that, now checking the last part.

[Anonymous No.3012300]

>>3012035
Can you try re-writing this? They way you present it is pretty shitty, and not easy to follow.

[Anonymous No.3012309]

>>3012284
using that you go to:

GBH => for all n: pn - 1 < 3p(n-1), which you use to show

for all n pn - 1 >= 3p(n-1) => not GBH,

which is not logically correct since it should be;

there exist an n such that pn - 1 >= 3p(n-1) => not GHB.

[D.D.A. !L.mWDGnT2A No.3012319]

>>3012262
>>3012300
If this anon here
>>3012284
Says this is worth anything, I'll try to write this as a formal proof, whatever that means, and I'll post it here. Just inform me on the meaning of the formal proof. Before you start throwing shit at me, this was made for fun, I'm not a mathematician.

[D.D.A. !L.mWDGnT2A No.3012334]

>>3012309
More this way :
If GBH is true: for all n: pn - 1 < 3p(n-1)

If there exists n for which the inequality is false, GBH is false. Later part with Bertrands postulate tries to prove that there is no such case.

[Anonymous No.3012337]

>>3012309
and my last point is that Bertand tells us that there exists a prime, not that it has to be pn. i.e.

11 < 19 < 22, then 19 is not the immediate successor.

all in all I'd say keep it up, you looked like you were on the way, but be very warned number theory is one of the hardest fields to make a contribution too (as there is just so much theory already done that you will need)

[Anonymous No.3012341]

>>3012334
alright my bad, sometimes it's kind of unclear what you're heading for

[Anonymous No.3012347]

>>3012334
This is not made clear, whatsoever in your writing. Try again.

[Anonymous No.3012373]

>>3012319
formal proof isn't neccesary, but it there are a few very important points (such as assuming GBC holds) and don't be afraid to put down more steps than you think are neccesary (more is usually better).

[Anonymous No.3012382]

>>3012373
number theory is actually a field that is interesting to non-mathematicians since it doesn't require a lot of context/definitions, but as I said before, really hard to do something worthwhile.

[D.D.A. !L.mWDGnT2A No.3012388]

>>3012337
>and my last point is that Bertand tells us that there exists a prime, not that it has to be pn. i.e.
>there always exists at least one prime number p

There may be other primes, but one of them sure is pn, which doesn't change my point one bit.

If you want, please continue your analysis. I'll try to write in more detail in couple of hours. Thanks for the support.

[Anonymous No.3012401]

>>3012388
well my point is, how do you know that pn is one of them? you will have to proof this without using Bertrand's postulate, as this only tells us there is one. and if you can already proof this you don't really need Bertrand's postulate. this is a very delicate detail but of vital importance, number theory makes things look a lot easier than they actually are.

[D.D.A. !L.mWDGnT2A No.3012434]

>>3012401
Because it can not not be, if I'm coherent enough.
if n=p(n-1) and n < p < 2n, pn, as a successor of p(n-1) must be one (not just one, the first of prime(s)) of primes that bertrand's postulate defines.
Your question actually made me think for a second. We are getting somewhere with this, I like it.

[Anonymous No.3012483]

>>3012434
alright, still don't think you need Bertrand's postulate for it but that's cool. give me in other words exactly what your conclusion is

[D.D.A. !L.mWDGnT2A No.3012664]

>>3012483
So I've shown that only if GC is true, inequality pn - 1 < 3p(n-1) is true, and that if there is a value of n for which inequality is false, GC is automatically false. Then I proceed to prove that pn - 1 < 3p(n-1) is always true using Bertrand's postulate. Note that the n in postulate and n in p(n) is not the same number.
So, if pn - 1 < 3p(n-1) is always true, GC is true. Bertrand's postulate shows that there is always a prime between p(n-1) and 2(p(n-1))-2. Since 2p(n-1)-2 is smaller than 3p(n-1), we conclude that pn - 1 < 3p(n-1) is true and therefore GC is true.

If I wasn't clear somewhere, just point out the place.

[Anonymous No.3012662]

>>3012483
I think I found your error;

you state GBC true => pn - 1 < 3p(n-1) for all n

there is n such that pn -1 >= 3p(n-1) => not GBC.

however then you show there is an n such that pn - 1 < 3p(n-1), and then you conclude GBC is true? there's the problem. the implication symbols once again flip around so you're back to GBC true => pn - 1 < 3p(n-1) for all n.

however if you can actually show that GBC true => pn - 1 < 3p(n-1) => GBC true or

not GBC => there is n such that pn -1 >= 3p(n-1) you are done.

[D.D.A. !L.mWDGnT2A No.3012701]

>>3012662
>however then you show there is an n such that pn - 1 < 3p(n-1), and then you conclude GBC is true

however then I show there is no such n that pn - 1 >= 3p(n-1), and then I conclude GBC is true

[Anonymous No.3012740]

>>3012090

Is it bad that picture frustrates me because I know that red is actually supposed to refract the least and all of the other colors refract more?

[Anonymous No.3012750]

>>3012701
yeah but that's not how logic implication works;

A => B is equivalent to not-B => not-A.

you know A=>B is true, and also not-B => not-A is true, but you want to use this to show B => A is also true, but that needs an entirely new proof.

[Anonymous No.3012777]

FUCKING FIELDS MEDAL IMPEDING

[Anonymous No.3012839]

>>3012664
No.
You've got your implication wrong.
Let me illustrate it with an analogous, wrong proof:
>If there is an even number equal to 3, aliens are not real. [true statement]
>There is no number equal to 3, aliens must be real.

[Anonymous No.3012844]

>>3012750
or more accurately you're using

not-not-B => not-not-A, but that requires a different proof, what you CAN use is;

not-not-A => not-not-B

[D.D.A. !L.mWDGnT2A No.3012884]

>>3012844
>>3012839
>>3012750

if and only if A true => B true
if and only if A false => B false

I do not realize how it is wrong to prove that B is (always) true and conclude that that must mean A is (always) true as well. How can it not work backwards if it is strictly defined ?

[Anonymous No.3012915]

> if pn - 1 < 3p(n-1) is always true, GC is true

i agree that the inequality always holds by betrand's postulate, but (unless i missed something important) you've done nothing to show that <span class="math">p_n-1[/spoiler] can be expressed as the sum of three primes

[D.D.A. !L.mWDGnT2A No.3013031]

>>3012777
This is a public board, in improbable scenario it is worth shit I assume somebody will publish it under his own name and get the credits.

[Anonymous No.3013132]

How did you conclude that <span class="math">p_n-1 \leq p_{n-1} + p_{n-2} + p_{n-3}[/spoiler]?

[Anonymous No.3013188]

>>3012884
let me think of an enlightening example.
alright let's take:

n is even => n is an integer.
n is NOT an integer => n is NOT even.

now if I take n = 3 I have that n is an integer, however it is not even.

[D.D.A. !L.mWDGnT2A No.3013241]

>>3013132
It's either a sum of those 3 primes, or some other that are smaller than them.

[Anonymous No.3013249]

>>3013188
>>3013188

Not OP, but I don't get this either - he assumes a statement true (GC) and comes to a formula that is always true ONLY if the statement is true. He proves that formula is always true, so since it is always true ONLY for true statement, it should be concluded that original statement is true. Or my logic is faulty somehow ?

[Anonymous No.3013254]

It's very hard to make sense of it when it's not clear (at all), nor are there any statements that precede and conclude your proof, what are you trying to prove?

[Anonymous No.3013258]

>>3013249
that's the problem, it isn't ONLY if, he stated in his proof 'if and only if' (iff), but he has only proved one way, and the other way is basically what he wants to use.

in other words he proved that GBC is stronger than pn - 1 < 3p(n-1), but not that the other way around also holds. and that is what he wants to use

[Anonymous No.3013261]

<span class="math">p_n-1 < 3p_{n-1}[/spoiler] is a necessary condition for GBC, but not sufficient
GBC is a sufficient condition for <span class="math">p_n-1 < 3p_{n-1}[/spoiler], but not necessary

you've mistakenly assumed an 'if and only if' relation between the two

[Anonymous No.3013270]

>>3013241
Just pointing out some sloppy thinking: you are presuming that the three primes are distinct.

[D.D.A. !L.mWDGnT2A No.3013273]

So basically I have to prove that if GC is false, formula is always false ?

[Anonymous No.3013278]

>>3013273
yeah, though that would mean that the two statements are equivalent, but I think GBC is a little stronger, so I doubt it's even true.

[Anonymous No.3013279]

>>3013273

the formula is true, but knowing this does not allow you to make any conclusion about the GBC

[Anonymous No.3013283]

>>3013278
wait no, you only need to find one counter example I think

[D.D.A. !L.mWDGnT2A No.3013519]

>>3013283
Yeeeeah, because distributed computer search found a contradiction.

Lemme take some paper and see where am I right now. Everything I wrote by now is 100% correct except for the conclusion.

...

After some thinking, fuck me. GC can be expanded: every number >11 can be written as a sum of 6 primes, every number >17 can be written as a sum of 9 primes, every number >prime (6n-1) can be written as a sum of 3n primes.

Imagine if this wasn't true, if pn-1 could be written as sum of any number of primes: for 1 prime it obviously doesn't work. For 2 primes it doesn't *always* work (2n-2). For 3 primes it works. Now it get's interesting, it shouldn't be too hard to show that 4 primes are just 2+2 primes (which are shown not to work), 5 primes are 2+3 (2 not always, 3 always, therefore 5 not always), 6 works.. 3k+1 doesn't always work (3k always, 1 not always), you get my point hopefully.

I made a stupid mistake there somewhere, didn't I ?

[D.D.A. !L.mWDGnT2A No.3013539]

>For 2 primes it doesn't *always* work (2n-2)

Actually it does. I knew that here was a stupid mistake somewhere.

[Anonymous No.3013562]

> Everything I wrote by now is 100% correct except for the conclusion.

Nothing you've written demonstrates in any way that the GBC is either true or false.

[D.D.A. !L.mWDGnT2A No.3013591]

>>3013562
That would be the conclusion, thanks.

Fuck this, I'm too tired. Thanks to people who pointed out mistakes, I'm very grateful.

[Anonymous No.3013613]

>>3013591
thanks for taking the effort of posting math on this board <3