Difference between revisions of "Twin primes"

Revision as of 08:03, 25 September 2018

Definition

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

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

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

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 [ilmath]2[/ilmath] is prime on a technicality, as it has exactly 2 divisors, [ilmath]1[/ilmath] and [ilmath]2[/ilmath] itself all the other primes are odd numbers.

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

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

However:

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

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

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

See the proof ruling out [ilmath]p+1[/ilmath] 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 [ilmath]p\in\mathbb{P}_{\ge 3} [/ilmath] be given. That is to say "let [ilmath]p[/ilmath] be an arbitrary prime number greater than or equal to [ilmath]3[/ilmath]"

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

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

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

• adding 1 to an odd number yields an even number
  • Explicitly we now know that [ilmath]p+1[/ilmath] is an even number

Now:

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

Notes

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