Difference between revisions of "Twin primes"

Revision as of 08:03, 25 September 2018

Definition

Let $p,q\in$$\mathbb{P}$ be given; here $\mathbb{P}$ denotes the set of all primes.

We say $p$ and $q$ are twin primes if(f)^{[Note 1]}:

• $\vert p-q\vert\eq 2$ where $\vert\cdot\vert$ refers to the absolute value
  • This is to say that {{M|p}] and $q$ differ by $2$ or:
    • either we have:
      1. $p+2\eq q\ \iff\ p\eq q-2$, or
      2. $p-2\eq q\ \iff\ p\eq q+2$

Motivation

The idea is to be able to talk about primes that come as close together as possible, and any such primes are called twin primes. The number $2$ is prime on a technicality, as it has exactly 2 divisors, $1$ and $2$ itself all the other primes are odd numbers.

For any prime, $p$, that isn't 2, we have that $p$ is odd, then $p+1$ would be even, but even numbers are divisible by 2, thus $p+1$ is divisible by 1, itself ($p+1$) and 2 - that's 3 divisors! So $p+1$ cannot be prime (if we ignore $p\eq 2$)

As such given any prime $p$ that isn't 2, we can be sure that $p+1$ is not prime.

However:

If $p$ is a prime that isn't 2, then as stated, $p+1$ is even, so $p+2$ is odd - and all primes except 2 are odd. Thus $p+2$ might be prime. We can't rule it out (without more information)

Note that if we consider $p$ to be any prime (so allowing 2) and using $p+1$ as the maybe next prime, the only "twins" we could have are $2$ and $3$ - these are the only primes to come $1$ apart for the reasons above.

• That is to say if we have $p,q\in\mathbb{P}$ and called them twins if $\vert p-q\vert\eq 1$ (that they differ by 1) - then the only twins that exist are $2$ and $3$.
  • This isn't very interesting, so we don't waste "twins" on these only 2 exhibits.

See the proof ruling out $p+1$ as the next prime below for a rigorous proof of the above.

Proofs

Ruling out (any prime)+1 as the next prime

Recall that:

• All primes greater than or equal to 3 are odd (note this is the same as: all primes except 2 are odd)

Then let $p\in\mathbb{P}_{\ge 3}$ be given. That is to say "let $p$ be an arbitrary prime number greater than or equal to $3$"

• From the above theorem, we have that $\forall p\in\mathbb{P}[p\ge 3\implies (p\text{ is odd})]$^{[Note 2]}
  • As $p\in\mathbb{P}_{\ge 3}$ by definition of $p$, we see $p$ must be odd.

By this point the reader should understand that we have deduced our $p$ is an odd number (and by definition is a prime greater than or equal to $3$)

If we wish to consider the immediate next prime number, we could try $p+1$, however note that:

• adding 1 to an odd number yields an even number
  • Explicitly we now know that $p+1$ is an even number

Now:

• $p\ge 3$, this means $p+1\ge 4$
  • For a number, say $n$, to be a prime number, $n$ must have exactly two distinct numbers that divide it, as we are only considering the numbers $1,2,3,\ldots$ we can state that "for any n we can always divide it by 1" and "we can always divide $n$ by $n$ itself" - note that if $n\eq 1$ then dividing it by $n$ is the same as dividing it by $1$ - so only one thing divides $1$, not two. This is why 1 is not prime. Prime numbers always have two distinct divisors, for example $n$-n\eq 2 has 1 and 2 as divisors (hence 2 is a prime number), 3 has 1 and 3 for divisors so is also prime, 4 has 1,2 and 4 as divisors (which is 3 not 2 divisors) so 4 is not prime, so on.
    • However we know that $p+1$ is an even number - which means we can divide it by $2$
      • The only way $p+1$ could be a prime number is if $1$ (which we can divide everything by) and $2$ (which we just showed we can divide $p+1$ by) are its only divisors
        • We also know that $p+1\ge 4$, that is $p+1$ is greater than or equal to $4$, it is at least $4$.
          • This means that there is no way $p+1\eq 2$, because if this were so then:
            • $2\eq p+1\ge 4$ which would give $2\ge 4$ - and 2 is greater than or equal to 4 is absurd, so we can't have $p+1\eq 2$ (as if we did, we get the nonsense: $2\ge 4$)
          • Thus we know $p+1\neq 2$ (the symbol: $\neq$ means "not equal")
      • Now we have shown that $1$ divides $p+1$, along with $2$ divides $p+1$ (as $p+1$ is even) and also that $p+1\ge 4$ (which let us show that $p+1\neq 2$)
        • As $p+1\neq 2$, when we say "we can divide it by 2" we know this is not dividing it by itself.
          • We can always divide a number (we are working with $1,2,3,\ldots$ - so 0 can't play any role here) by itself: $p+1$ divided by $p+1$ is of course 1.
            • Thus we claim $1,2,p+1$ are all divisors of $p+1$
              • We have shown $p+1\neq 2$ - so 2 and $p+1$ are different, it is obvious that $1\neq 2$, and if we have $p+1\neq 1$ then we have shown all 3 of these are different numbers
                • But a prime only has exactly 2 divisors (this is why 1 isn't prime, it can only be divided by 1, we need to be able to divide it by exactly two things for it to be prime)
                  • So if we have 3 divisors, it can't be prime
      • Let us now show that dividing by $p+1$ is not the same as dividing by $1$ (this is almost identical to the proof that dividing by 2 is not the same as dividing by $p+1$)
        • Suppose that $p+1\eq 1$, then we'd have $1\eq p+1\ge 4$ which would give $1\ge 4$ - which is absurd
          • Thus we can't have $p+1\eq 1$, as that would lead to absurd conclusions, so we must have $p+1\neq 1$
    • We have now shown that $1$, $2$ and $p+1$ are 3 distinct divisors of $p+1$ - thus $p+1$ cannot possibly be prime.

Notes

1. Jump up ↑ See: Definitions and iff
2. Jump up ↑ Read $\forall p\in\mathbb{P}[p\ge 3\implies (p\text{ is odd})]$ as:
   • forall $p$ in the set of primes we have [ if we have $p\ge 3$ then we have ($p$ is an odd number) ]