Twin prime conjecture proof error I am absolutely sure this is wrong but I can't find why. 
For every integer $n$ there exist a finite number of primes less than $n$. Take the set containing those primes and multiply them together to get $x$. Aren't $x+1$ and $x-1$ prime, implying there is an infinite number of twin primes? 
Follow up question is there guaranteed to be a prime between n and $x^{.5}$? What about for large n? this prime wouldn't have to devide x just exist in the given range
 A: Your proof most likely stems off of a proof that there are an infinite number of primes. It says that if $\mathbb{P}$ is the set of all primes, and $|\mathbb{P}|<\aleph_0$, then intuitively $$\left(\pm1+p_{|\mathbb{P}|}\#\right)\notin\mathbb{P} \:\text{and}\:\forall p\in\mathbb{P},p\nmid\left(\pm1+p_{|\mathbb{P}|}\#\right)$$  (where $p_n\#$ is the $n$th primorial) implying that $\left(\pm1+p_{|\mathbb{P}|}\#\right)$ is prime, proving that there are an infinite number of primes by reductio ad absurdum. However, since $|\mathbb{P}|=\aleph_0$, $\pm1+p_n\#$ is prime is not necessarily true for arbitrary $n$.
The type of primes that are in the form of $\pm1+p_n\#$ are called primorial primes and you can read more about them here.
A: Let $n = 8$. Then all primes less than $8$ are $7, 5, 3, 2$. The product of these is $x = 210$. 
$x + 1 = 211$ which is prime,
$x - 1 = 209 = 11\times19.$
A: I had the same question once I heard about the twin prime conjecture. The confusion stems from the following.
Take the first n primes: $p_1, p_2, .... p_n$, and define: 
$$A_n = p_1 p_2 .... p_n +1$$
$$B_n = p_1 p_2 .... p_n -1$$
Now we know that neither $A_n$ nor $B_n$ is divisible by any of the primes $p_1, p_2, .... p_n$. 
However this does not rule out the possibility that there may exist other prime numbers greater than $p_n$ that divide $A_n$ or $B_n$.
