Question

Challenge , but not easy
(again)

Answer 1

\(\Large \Large \text {let P be the set of first nth odd primes } \\\Large P= \left \{ p_1,p2,....p_n \right\} \\ \Large \text{ and } \Large S= \left \{ p_i+p_j :\forall i,j\leq n \right \} \\ \Large \text{ then , show that there exist x(n) } \\ \Large \left \{ 6,8,10,.....,2p_{x(n)} \right \}\subseteq S \text { note that }\\ \Large
x(n) \text{is a function of n } \\ \Large \text{with Range
and domain of N and } x(n)<n
\\ \large \text{ PS :- } \left \{ 6,8,10,.....,2p_{x(n)} \right \} =\left \{ 2\times 3 ,2\times 4,..., 2(p_{x(n ) }-1) ,2p_{x(n)} \right \} \)

Answer 2

This looks like the goldbach conjecture haha.

Answer 3

awww missed the fun so quick :P
ok try to show it :D

Answer 4

It looks confusing :/

Answer 5

eh why so haley :O

Answer 6

I haven't learned this yet :P

Answer 7

I thought about what I was talking earlier and realized that there is this weird sort of infinite structure for numbers that's the same no matter how much you zoom in or out but it's kind of hard to describe uhhh it's like :

...
0, 1/4, 1/2, 3/4, 1
0, 1/2, 1, 3/2, 2
0, 1, 2, 3, 4
0, 2, 4, 6, 8
0, 4, 8, 12, 16
0, 8, 16, 24, 32
0, 16, 32, 48, 64
...

So you can always view an interval and look at it cut in half and each half interval is the same as the entire interval kind of, sort of like a fractal I guess?

Answer 8

or you can cut an interval into 3rds or 7ths and each 7th interval can be cut up into 7 intervals and you can keep going infinitely smaller or zoom outward and see that your interval of 7 is part of 6 others you know? The point is that I'm trying to find out how to see the holes in the composite numbers which are just cycles that start with primes uhhh... It's related to the goldbach conjecture I swear I'm not hijacking your question which seems impossible to me unless its an easier version of the goldbach conjecture :X

Answer 9

Oh goodness this is too confusing.....I have to go before my brain explodes XD

Answer 10

ur 100% on the right track @Kainui

Answer 11

@ganeshie8 ermm

Answer 12

it looks same as your previous problem ?

Answer 13

>.< its the same eh , still confused about its phrase thingy ?

Answer 14

tbh im still not able to comprehend the problem fully :/ wil try again in the morning

Answer 15

hmm if u dont get it yet then its not phrasing problem :|

Answer 16

i give up, this is not my level

Answer 17

@ganeshie8 Nahhhh you just need to rest. She's just trying to say the goldbach conjecture in a confusing set theory way I think.

Answer 18

i made proving Goldbatch more easy to prove :P
i noted if chose x(n) sickly less than n then its true . ( like very small x(n) comparing to n
how ever i just wanted someone to say its just an equvilent to goldbatch :P
so by proving this , we would prove Goldbatch itself

im trying to make x(n) something with log and floor

Answer 19

so unlike u said @Kainui its more easy now
jusr mapping functions , moving from space to another

Answer 20

Hmmm interesting. You might be able to use a prime counting function to do it or maybe use euler's totient function? I actually came up with a real proof of the goldbach conjecture, unfortunately it is only valid for small numbers, if you want to increase the accuracy then you have to continually add a correction factor like the sieve of eratosthenes kind of.

Answer 21

i know what u mean thats why i need very small x(n) here
i wanna very slow function like this

Answer 22

if im lucky enough , i'll found something in topology else i would fail
i tried testing in abstract playing with order not good enough ....

also me myself not feeling well to continue :|
im just spreading idea , mostly i would be wrong

Answer 23

I think you should change it to this instead of 2px(n), only 1px(n) {6,8,10,.....,px(n)}⊆S The reasoning is you might not be able to get all the numbers otherwise.

Answer 24

Bingo !

Answer 25

but note what i have is the equivalence :P

Answer 26

Oh maybe I'm confused, let me make an example: P={3,5,7,11} then S should have all the even numbers from 6 to 22 right? But to get 20 we need 13+7 or 17+3 which aren't in P.

Answer 27

@Marki

Answer 28

bleh I see I'm dumb

Answer 29

x(n)<n ok now I understand I was confusing myself but now that I explained it now I understand ok... XD

Answer 30

Yeah this is awesome I see what you mean now! As we get larger it should sort of get larger with the prime counting function as well, like 1/2 the prime counting function about I think?

Answer 31

ٍS={6,8,10,12,14,16,18,22}
so i can chose
{6,8,10,12,14,16,18} sub of S
so last multiple let last one be 14=2*7
7 is the third prime
x(4)=3
so
x(n)=n-1 ( for this case )

Answer 32

Hmm I think the problem is we still have to show that all the even numbers can be reached by adding primes. There might be a point where when you get higher up like 20 except no matter how many more primes you add, you can never get a combination to add to it, so I don't know how we could show that at all.

Answer 33

yes exactly , when i had small lest of primes i took linear equations then it got more harder i took sqrt , then it get more tough started with log
maybe next time i'll see how about 1/x

Answer 34

i agree to u !
but at least i found strong equivalence :P
so not sad at all for time i've wasted on it

Answer 35

Yeah this is interesting hmmm this is giving me ideas. What can we prove if we assume that the goldbach conjecture is true?

Answer 36

we can prove this and other stuff