BEHAVIOUR OF PRIME NUMBERS

        AN INSIGHT AT GOLDBACH’S CONJECTURE

© Angel Lauzara Formoso-2004

In 1742, a Prussian mathematician called Christian Goldbach felt interested about the relation between prime numbers (all those numbers that only can be divided by themselves and 1 –all the others are called composite numbers-) and even numbers. In a letter written to the great mathematician Leonard Euler, he proposed a problem that has been called ‘Goldbach Conjecture’, which says that any even number can be expressed as the addition of two prime numbers. This theory, which doesn’t look too difficult to probe, remains unresolved until today. In this article, starting from a graphic explanation, we’ll see that, due to how prime numbers are distributed among natural numbers, and how they combine themselves to give even numbers, we can find these combinations for any even. And not only that: we’ll also probe that the number of valid combinations for each even (we’ll call them ‘Goldbach combinations’ –GC- from now on) increase as even numbers do so. And we’ll also give the approximate number of GC’s for any even.

 In other hand, study of prime numbers provides us more questions. For example, the gap between consecutive primes. Chebyshev’s theorem says that there is always a prime between any number n and 2n: this means that any prime can’t be more that double of the immediate lower one. Starting with this theorem, we’ll find a more exact formula to establish the maximal gap between consecutive primes. Here we show:

‘GOOD’ AND ‘BAD’ COMBINATIONS

We can begin asking about how many different ways any even number e can be expressed as the addition of two odd numbers. Taking any given e, combinations are so: associating 1 with the closest odd number lower than e, 3 with the second smaller, etc. in this way:

(1, e-1), (3, e-3), (5, e-5)…………………. (e-3, 3), (e-1, 1)

If we consider the first element of the combination different than the second one, is easy to see that each e has e/2 combinations. How can be this expressed graphically? If we call x to the first element and y to the second one, we can translate them to their respective axis, as we see in figure 1.

Each intersection is a combination: for example, the black point corresponds to (23, 13) for e= 36. The problem now is that in the figure we see all the combinations, and we only need those composed for 2 prime numbers: we look for GC’s.

GOLDBACH  COMBINATIONS

What would be the next step? We are going to eliminate all invalid combinations. To do this, let’s erase those combinations in which x are composite, connecting with lines equal x (vertical lines) and then, doing the same with y (diagonal lines), as we see in figure 2.

Obviously, all those points which remain clear are all the GC’s for each e. For example, e= 42 has 10 combinations. Of course, the graphic follows as e increases; all the lines which we made continue to the infinity; every time that a prime number appears, it combines with all the others before and after it and creates 2 ‘clean lines’, it’s to say, lines where can find GC’s: 1 line vertical and 1 diagonal. In the figure 3 we see all the GC’s for every e between 2 and 500.

If we make another graphic with the same procedure to a bigger e and compare both, we see that they follow an identical pattern: ‘clean’ and ‘dirty’ lines alternate, and it seems that the number of GC’s increase as e does so. For space reasons, we only can show a graphic to e= 500, but, if we keep doing it, it’s observed that the mosaic keeps using the same pattern, whatever e we take. There is only one possibility for GC’s disappearance: if the gap between two any consecutive prime numbers p and p’ was p, as we see in figure 4. But we know that, as Chebyshev Theorem says, there is always a prime number between n and 2n, which guarantee the continuous creation of GC’s and, therefore, the veracity of Goldbach Conjecture.

More else, we even can doubt if it would be really possible reach a so critic point. Despite Chebyshev Theorem, if the structure of the mosaic doesn’t change, we must find much more primes between n and 2n that the Theorem says. Could we do this maximum distance smaller? Could we find any formula?

MAXIMUM GAP

Let’s remember now an essential theory which explains the distribution of prime numbers among natural numbers: the Theorem of Prime Numbers. This theorem says that the amount of prime numbers (called π(x)) between 1 and x is similar to x/ln x, for any x big enough. This is expressed

π(x) ~ x / ln x,                           for an x big enough

When a new prime appears, the number π(x) increases in 1 unit. Let’s call π(y) to this. Thus,

       π(x) +1 ~ π(y)

Therefore, π(y) is the function which defines the next prime following to x: if we had π(x) prime numbers between 1 and x, we now have π(y). Working on this:

π(x) +1 ~ π(y)

(x / ln x) +1 ~ y / ln y                          Let 1= ln x / ln x. Then,

(x + ln x) / ln x ~ y / ln y                     Thus,

y ~ ln y (x + ln x) / ln x

But Chebyshev Theorem specifies that y ≤ 2x. Taking the worst case y= 2x, we have

y ≤ (ln 2x).(x + ln x) / ln x

Now, with this formula we shall make a table, giving values to y relative to x –figure 6-, establishing the maximum absolute value of the increasing (g), and the proportion of increase of y relative to x (i), as we see in table 1.

Table 1. Calculation of maximum gap relative to x.

                        x                                        y                               g                                      i

10                                                16, 0060                              7                              70 %           

100                                           120, 3498                             21                               21 %

1000                                         1107, 9442                         108                          10, 8 %

10000                                     10762, 4784                         762                          7, 62 %

100000                                  106032, 8059                        6033                           6, 03 %

1000000                              1050186, 1746                      50187                           5, 01 %

10000000                          10430059, 6621                   430060                             4, 30 %

100000000                      103762894, 0596                  3762895                            3, 76 %

1000000000                   1033447798, 7124                 33447799                            3, 34 %

Table 1. The first column indicates the number prime x. The second one gives the maximum value of y relative to x. The third column gives the maximum gap between consecutive prime numbers, and the fourth column gives the size of this gap relative to x length.

How translate this table? We see that g increases, its absolute value is bigger, but it’s also proportionally smaller respect to x, and so, as x increases, it comes more away from the critic g = x. For example, when x= 1.000.000, y ≤ 1.050.186. It’s 50.186 units bigger than x; and this amount is about 5,01 % of x, therefore, g ~ 5,01 % of x. If x= 1.000.000.000, y ≤ 1.033.447.799; the amount increases absolutely, it’s 33.447.799 units bigger; but now, proportionally it’s only a 3,34 % bigger: therefore, g ~ 3,34 % of x. And so on.

Goldbach combinations for any given e

Let’s find now a formula which gives the approximate amount of GC’s for any given e. If we express the Theorem of Prime Numbers in this way,

π(x) / x ~ 1 / ln x

we can read now that the proportion of prime numbers π (x) relative to x is similar to 1/ln x. Knowing that the probability of two different events happens together is equal to the probability of the first event multiplied by the probability of the second event, we can say that the probability P(GC) that any given combination was a GC is similar to

P(GC) ~ (1 / ln e).(1 / ln e)

Therefore, for e = 1000, the probability for any combination be a GC is

P(GC) ~ (1 / ln 1000).(1 / ln 1000) ~ 1 / 47,71 ~ 0,021

which doesn’t seem too much; but if we multiply this rate by 1000, we find that we’ll have about 21 GC’s. Also, for e = 1.000.000, we have 1 / 91 = 0.005239 %, which give us 5239 GC’s.

I have to say that this doesn’t reflect exactly the truth, because, if we remember when we make all the possible combinations (good and bad), the first combination was equal to the last one, the second was equal to the before the last, and so on. For example, let c= all possible combinations for e. Then

c(10)= (1,9), (3,7), (5,5), (7,3), (9,1)

                     c(12)= (1,11), (3,9), (5,7), (7,5), (9,3), (11,1)

This means that P(GC) has to be divided by two to get the real number of different GC’s (In fact, we must add two units before dividing to all those numbers which was double of odd numbers, because in those numbers, in the centre, the combination doesn’t repeat. But it’s a slight difference if we take very large numbers. And, despite all, GC it’s only an approximate number). Therefore, formula remains

P(GC) ~ e / 2 (ln e).(ln e)

Finally, as last probe of Goldbach Conjecture’s veracity: how many prime numbers are between any x and 2x? We know that there are π (x) primes between 1 and x. Let π (x’) = π (2x) ~ 2x /  ln (2x). Let π (x’’) = π (x’) - π (x). Thus

     π (x’’) = (2x / ln (2x)) – (x / ln x)

This means that, for example, there are about 45.131.377 primes between 10e9 and 2 x 10e9; 7699 primes between 10e5 and 2 x 10e5, and so on. This ensures enough amounts of ‘clean’ lines between x and 2x, which guarantee the veracity of Goldbach Conjecture.

Conclusions

Finally, I resume this article in the following points:

1.      Let x and y= consecutive prime numbers. The formula

y ≤ (ln (2x)).(x + ln x)/ln x

is true for any x and y

2.      Let e Є N. The formula

P(GC) (e) ~ e / 2 (ln e).(ln e)

Gives the approximate number of GC’s for any e big enough

3.      Let x= any prime number. The formula

π (x’’) (x→2x) = (2x / ln (2x)) – (x / ln x)

Gives the approximate amount of prime numbers between x and 2x.

I finish here my exposition. I would please any comments. For correspondence, please, e-mail me to angelgalicia30@hotmail.com