Question

true or false ? opinions please !

Answer 1

what is the question?

Answer 2

I'm so excited.

Answer 3

Both.

Answer 4

Neither.

Answer 5

1. subst. - let p and k , two prime numbers greater or equal 2,from the set of prime numbers, P, in the form : p=2a + 1 and k=2b + 1 , such that a and b are natural numbers,from the set of natural numbers N, - let m=2n ,m greater or equal 4,even number,from the set of natural numbers N and n grater or equal 2,natural number from set of natural numbers N,
2. concl. - every even integer greater than 2 can be expressed as the sum of two primes
3. prove:- by ,,reductio ad absurdum” * - step 0. for n=2 --- m=4 --- 4=2+2 ** - step 1. for - if n is greater or equal 3 so always will be a number a and b such that n=a + b + 1 - prove. 3=1+1+1 4=2+1+1 5=2 +2+1 ................ n=a+b+ 1 - so for n=k than k=a+b+1 - suppose that is true - for k+1=(a+b+1)+1=(k)+1=k+1 - so for k+1 is true - for n grater than 2 always will be a number a and b such that n =a+b+ 1 *** - step 2. - every even integer greater than 4 can be expressed as the sum of two primes m=p+k - prove by ,,reductio ad absurdum" - so than m is not equal p+k - so than 2n is not equal 2a+1+2b+1 - so than 2n is not equal 2a+2b+2 / divide both sides by 2 - so than n is not equal a+b+1 - so what is in contradiction with the proof from step 1. where n=a+b+1 was proved that is true - so than m=p+k is proved that is true so,, every even integer greater than 4 can be expressed as the sum of two primes” --- q.e.d.

Answer 6

Woah, We are not going to do your homework.

Answer 7

true

Answer 8

This math is just too hard for you guys, stop making excuses and shut up!

Answer 9

then you do it for him

Answer 10

And he asked three questions simultaneously...

Answer 11

but its a true or false question and that made it really confusing for me

Answer 12

He actually didn't, you just can't read. He introduced the problem in 1, made a statement in 2 and he's trying to prove it in 3.

Answer 13

ok

Answer 14

Don't give up so easily guys. Give it a try.

Answer 15

Ok, you solve it for him, then. I don't know math theories thingy.

Answer 16

how much is understandably ?

Answer 17

yes sure just for people who know from what problem is the theme ,what conjecture ...

Answer 18

Phew. I got through it and it makes a lot of sense, it was just hard to understand, try to make your work more clear. Basically you prove that n can always be expressed as n=a+b+1. Then you use "reductio ad absurdum" to prove your point. Let's say that 2n≠p+k≠2a+1+2b+1. Then, 2n≠2a+2b+2, n≠a+b+1, which is impossible because you proved that is always true. Good work!

Answer 19

thank you @BluFoot ! so but you know what conjecture is proven here than in this case ?

Answer 20

so and this your words,reply mean that you accept this proof for seriosly true ?

Answer 21

http://en.wikipedia.org/wiki/Goldbach's_conjecture

Answer 22

yes wann being the Goldbach's conjecture

Answer 23

great

thank you

Answer 24

Hmm, wikipedia says that there isn't actually a proof. Your proof looked pretty good but I guess it's not? not sure.

Answer 25

why ?

in your first above reply you have told me that you have understood it very well and you say that is correct and acceptably true ?

Answer 26

so because this is an easy ,simple proof of by reductio ad absurdum

Answer 27

if the conjecture has been around since the 1700s, and it doesn't have a proof yet, i dont think there is going to be a "simple" proof for it.

Also, if you want to verify if your proof is correct, you need to send it to an academic journal, so a referee can look at it.

Answer 28

ok thank you but do you can help me with this journal address ,email where i can sending it for deciding is true or false ,acceptably or not ?

Answer 29

im not too sure exactly who to talk to, but the Mathematics Association of America is a good place to start:

http://www.maa.org/

Or the American Mathematics Society:

http://www.ams.org/home/page

Answer 30

ok

thank you very very much

good luck

bye

Answer 31

A book featuring the Goldbach's Conjecture
http://www.amazon.com/Uncle-Petros-Goldbachs-Conjecture-Mathematical/dp/1582341281

Answer 32

well question is good
but this would really involve much more time also....
i' ll get back if i find something useful

Answer 33

thank you

Answer 34

http://plus.maths.org/content/mathematical-mysteries-goldbach-revisited

http://plus.maths.org/content/os/issue5/turing/index