# Difference between revisions of "Twin primes"

## 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]

## Terminology

Typically for twin primes [ilmath]u,v\in\mathbb{P} [/ilmath] (which to be twins means they differ by 2) we call the smaller one [ilmath]p[/ilmath] and the larger one [ilmath]q[/ilmath] or just [ilmath]p+2[/ilmath] (as if [ilmath]p[/ilmath] is the smallest one, to be the smallest and differ by 2 from the other, the other must equal [ilmath]p+2[/ilmath], that is to say if [ilmath]q[/ilmath] is the larger one, then [ilmath]q\eq p+2[/ilmath])

And we write these as:

• "[ilmath](p,q)[/ilmath] (or [ilmath](p,p+2)[/ilmath] instead) are twin primes"

Which uses the ordered pair notation where the first is the smaller prime and the 2nd the largest.

## Examples

The first few primes are:

• [ilmath]2,3,5,7,11,13,17,19,23,29[/ilmath] and [ilmath]31[/ilmath]

A prime [ilmath]p[/ilmath] is twinned with another if [ilmath]p+2[/ilmath] is also a prime, so going through this list, adding 2 to the number and searching for this new number in the list to find its twin shows us the first few twins are (using the smallest first convention, so [ilmath](3,5)[/ilmath] is considered as 1 set of twins, where the first prime is [ilmath]3[/ilmath] and second [ilmath]5[/ilmath] is not considered distinct from [ilmath]5[/ilmath] as the first prime and [ilmath]3[/ilmath] as its twin):

• [ilmath](3,5)[/ilmath], [ilmath](5,7)[/ilmath], [ilmath](11,13)[/ilmath], [ilmath](17,19)[/ilmath], and [ilmath](29,31)[/ilmath]

We can also rule out [ilmath](31,33)[/ilmath] as being twins as [ilmath]33[/ilmath] isn't even prime, it can be divided by [ilmath]1[/ilmath] and itself ([ilmath]33[/ilmath]) yes, but also 11 and 3 (11 x 3 = 33) - which are 4 distinct divisors (there may be even more[Note 2], I only have to show there is only 1 or at least 3 distinct divisors to prove something as not prime remember)

## 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:

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 3]
• 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:

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. There isn't. As [ilmath]33\eq 3\times 11[/ilmath] and [ilmath]3[/ilmath] and {{M|11} are both prime, so we cannot divide further
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) ]