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2019-04-15, 13:55   #23
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

169E16 Posts

Quote:
 Originally Posted by MathDoggy It is impossible because we constructed an arbitrary number Q which is the product of the finite list of the twin primes and adding 1 to the product
2×3×5×7×11×13+1 = 59×509

And if 2 isn't a twin prime then 3×5+1 = 2×2×2×2

Either way there are always possible divisors that aren't in the original set of primes you multiplied together.

Last fiddled with by retina on 2019-04-15 at 13:55

2019-04-15, 14:05   #24
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

438710 Posts

Quote:
 Originally Posted by MathDoggy It is impossible because we constructed an arbitrary number Q which is the product of the finite list of the twin primes and adding 1 to the product
This is not a reason. You have constructed a number Q which is not divisible by your list of twin primes, and then you assume it must be divisible by a twin prime. You have no justification for the belief that Q must be divisible by a twin prime; in fact, you have constructed Q specifically such that it is NOT divisible by a twin prime, and then you claim a contradiction by just saying Q is divisible by a twin prime.

Your Q is divisible by 2 (do you see why?). All other claims about properties of Q's factors must be proven. Prove your claim.

 2019-04-15, 14:46 #25 MathDoggy   Mar 2019 5·11 Posts I will try to fix the proof once again Assume that there exist a finite amount of twin prime numbers. Then we can construct a list, which in this case will be S, S= A1,A2,A3,A4...An Let P be the product of all twin prime numbers in S, P= A1*A2*A3*A4..An Let Q=P+1 If Q is a twin prime number then S is not complete If Q is composite then some prime factor p divides Q, if this factor p were in our list S then it would divide P, but p divides P+1=Q. If p divides P and Q then p would have to divide the difference of the two numbers, which is (P+1)-P or just 1. Since no twin prime number divides 1, p can not be on the list. This means that at least one twin prime number exists beyond those in the list. Last fiddled with by MathDoggy on 2019-04-15 at 15:30
2019-04-15, 14:55   #26
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

2×3×5×193 Posts

Quote:
 Originally Posted by MathDoggy I will try to fix the proof once again Assume that there exist a finite amount of twin prime numbers. Then we can construct a list, which in this case will be S, S= A1,A2,A3,A4...An Let P be the product of all twin prime numbers in S, P= A1*A2*A3*A4..An Let Q=P+1 If Q is a twin prime number then S is not complete If Q is composite then some prime factor p divides Q, if this factor p were in our list S then it would divide P, but p divides P+1=Q. If p divides P and Q then p would have to divide the difference of the two numbers which, which is (P+1)-P or just 1. Since no twin prime number divides 1, p can not be on the list. This means that at least one twin prime number exists beyond those in the list.
Q is divisible by two. See my previous post.

Also, p doesn't have to be a twin prime, it can be some other prime, like 2 for example.

2019-04-15, 15:03   #27
wblipp

"William"
May 2003
New Haven

22×32×5×13 Posts

Quote:
 Originally Posted by MathDoggy I will try to fix the proof once again ... Since no twin prime number divides 1, p can not be on the list.
Looking good to here.

Quote:
 Originally Posted by MathDoggy This means that at least one twin prime number exists beyond those in the list.
Ooops. This means at least one prime (p) exists beyond those in the list. But why must it be a twin prime?

 2019-04-15, 15:03 #28 MathDoggy   Mar 2019 3716 Posts What do I have to fix, retina
2019-04-15, 15:16   #29
MathDoggy

Mar 2019

5·11 Posts

Quote:
 Originally Posted by wblipp Looking good to here. Ooops. This means at least one prime (p) exists beyond those in the list. But why must it be a twin prime?
I am thinking

 2019-04-15, 15:43 #30 MathDoggy   Mar 2019 5510 Posts Ooops. This means at least one prime (p) exists beyond those in the list. But why must it be a twin prime?[/QUOTE] There is no such implication
 2019-04-15, 16:02 #31 MathDoggy   Mar 2019 5·11 Posts Another attempt with an alternate method Just ignore this Last fiddled with by MathDoggy on 2019-04-15 at 16:13
2019-04-15, 16:11   #32
CRGreathouse

Aug 2006

23×3×13×19 Posts

Quote:
 Originally Posted by MathDoggy A factorial number x! of a positive integer is divisible by every integer from 2 to x, inclusive. Hence, x!+1 is either a twin prime number or divisible by a prime larger than x. In either case, for every positive integer x, there is a least one twin prime bigger than x. The conclusion is that there exist infinitely many twin prime numbers.
How does "divisible by a prime larger than x" show that there is a twin prime bigger than x?

Last fiddled with by CRGreathouse on 2019-04-15 at 16:18

 2019-04-15, 16:22 #33 MathDoggy   Mar 2019 5·11 Posts Should I quit to try to fix the proof? It is because nothing is working Last fiddled with by MathDoggy on 2019-04-15 at 16:23

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