Question

- opinions please

thank you very much !

Answer 1

About??

Answer 2

1.



One Possible Proof of Goldbach’s Conjecture

Andrew John Gábor
Gizella út 6/b
1143 Budapest - Hungary
gbrndrs1968@yahoo.com




,,every even integer greater than 2 can be expressed as the sum of two primes “

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. proof:- 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











2.



- normally induction would start with n=1 but the theorem concerns n>=3 so that's our first point. Is the theorem true for n=3 ? - this is trivial ...
- prove. 3=1+1+1
- so to prove that the theorem is true by induction, we prove that it's true for the lowest value (3) and then prove that if it's true for n, it is also true for n+1.
- for n=3, we see that n=1+1+1 so it works for a=1 and b=1.
- now we assume that the theorem is true for every value up to n, and we
test whether it's true for n+1. So assuming we can write n = a+b+1, so we write n+1 as the sum of two numbers plus 1
- so n+1=a+b+1+1
- and the right hand side can be grouped so that it is two numbers plus 1
- like n+1=a+(b+1)+1
- so then n+1 = a+b+1+1 = a+(b+1)+1 and we have our two numbers, a and b+1 satisfying the equation.
- since the set of natural numbers is generated by starting with 1 and adding 1 repeatedly, this process covers every possible natural number. QED by induction.
*** - step 2.
- ,,every even integer greater than 2 can be expressed as the sum of two primes”
m=p+k
- prove by ,,reductio ad absurdum"
- so then m is not equal p+k
- so 2n is not equal 2a+1+2b+1
- so 2n is not equal 2a+2b+2 / divide both sides by 2
- so then 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 m=p+k is proved that is true so ,,every even integer greater than 2 can be expressed as the sum of two primes” -

- q.e.d. -

Answer 3

What is written above??

By seeing this, I am thinking that I should log out now..

Ha ha ha..

Answer 4

May be I liked the style to wrote it!