- #1

JFGariepy

- 17

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So we have a set ω containing

*n*numbers referred to as P

_{1}, P

_{2}, P

_{3}...

Each of these numbers is a

**distinct**prime number

**greater than 4**. It may or may not be relevant to the proof, but

*ω*always contains all qualifying prime numbers up to a certain number, so it would be impossible to have missing prime numbers, and the following set would be impossible (ω = P

_{1:3}= [5, 7, 13]). You can only have

*ω*that include all prime numbers up to a point, such as (ω = P

_{1:3}= [5, 7, 11]), or (ω = P

_{1:2}= [5, 7]).

Note that because all elements of ω are ordered and distinct, and that 5 is always present in the set, we can write them as prime gaps being added to 5 in the following way:

P

_{1}= P

_{1}= 5

P

_{2}= P

_{1}+ PG

_{2}= 5 + 2 = 7

P

_{3}= P

_{1}+ PG

_{2}+ PG

_{3}= 5 + 2 + 4 = 11

...

where PG

_{i}reads "the prime gap associated with P

_{i}" and where the indexing of the gaps starts at P

_{1}= 5, and PG

_{1}= 0.

So to recapitulate the conditions in a cleaner list:

__Conditions__*ω contains all distinct prime numbers between 4 and a given integer (condition 1)*

*ω has a certain size n > 1 (condition 2)*

Due to Bertrand's postulate, note that any PGDue to Bertrand's postulate, note that any PG

_{i}< P_{i}.__(this is an equivalent of the general case with m = n)__

**Specific formulation of the problem***Prove that for any given ω with number of elements n > 1, the product of all P in ω is greater than the sum of the products of all possible combinations of subsets of ω with size n - 1.*

So, for instance, for n = 2, prove that:

P

_{1}P

_{2}> P

_{1}+ P

_{2}

or, for another instance with n = 4, prove that:

P

_{1}P

_{2}P

_{3}P

_{4}> P

_{1}P

_{2}P

_{3}+ P

_{1}P

_{2}P

_{4}+ P

_{1}P

_{3}P

_{4}+ P

_{2}P

_{3}P

_{4}

__General formulation of the problem__*Prove that for any given ω with number of elements n > 1, given any chosen integer 1 > m ≤ n, the sum of the products of all possible combinations of subsets of ω with size m is greater than the sum of the products of all possible combinations of subsets of*So, for instance:

*ω with size m - 1.*for n = 4 and m = 3,

P

_{1}P

_{2}P

_{3}+ P

_{1}P

_{2}P

_{4}+ P

_{1}P

_{3}P

_{4}+ P

_{2}P

_{3}P

_{4}> P

_{1}P

_{2}+ P

_{1}P

_{3}+ P

_{1}P

_{4}+ P

_{2}P

_{3}+ P

_{2}P

_{4}+ P

_{3}P

_{4}

or for n = 5 and m = 4,

P

_{1}P

_{2}P

_{3}P

_{4}+ P

_{1}P

_{2}P

_{3}P

_{5}+ P

_{1}P

_{2}P

_{4}P

_{5}+ P

_{1}P

_{3}P

_{4}P

_{5}+ P

_{2}P

_{3}P

_{4}P

_{5}> P

_{1}P

_{2}P

_{3}+ P

_{1}P

_{2}P

_{4}+ P

_{1}P

_{2}P

_{5}+ P

_{1}P

_{3}P

_{4}+ P

_{1}P

_{3}P

_{5}+ P

_{1}P

_{4}P

_{5}+ P

_{2}P

_{3}P

_{4}+ P

_{2}P

_{3}P

_{5}+ P

_{2}P

_{4}P

_{5}+ P

_{3}P

_{4}P

_{5}

__Example proof__I've had success handling single cases by expressing all primes as being an addition of prime gaps with 5, for instance:

**Proof for m = 2 and n = 2**

Prove that P

_{1}P

_{2}> P

_{1}+ P

_{2}

We know that P

_{2}= P

_{1}+ PG

_{2}, therefore:

P

_{1}(P

_{1}+ PG

_{2}) > P

_{1}+ P

_{1}+ PG

_{2}

P

_{1}

^{2}+ P

_{1}PG

_{2}> 2P

_{1}+ PG

_{2}

We may label these 4 terms as term 1, 2, 3 and 4. It is easy to see that term 1 is greater than term 3, because P

_{1}> 2 (condition 1). It is also easy to see that term 2 is greater than term 4, because P

_{1}> 1 (condition 1).

Therefore the left side has to be greater than the right side.

**Help needed!**I'd love to see if anyone has some guidance to provide on proving the general case. It feels like this is strongly related to the Pascal triangle and multinomial theorem but I have difficulties finding where to start.