Prerequisites:
When working with numbers, a geometric representation is often helpful in trying to highlight trends or properties that are less likely to emerge from a purely numerical analysis; for this reason, we are looking for a way to represent Goldbach pairs on a Cartesian plane.
For example, by combining the Hypothesis H.1.T and the Propositions L.C.5 and L.C.6, we can obtain a representation on the Cartesian plane of the Goldbach pairs of even numbers 2n having as validity dashed line T_3 = (2, 3, 5), as follows:
- Given a Goldbach pair (p, q) where p and q are spaces of the dashed line T_3, recalling that the first component of the dashed line is 2 and therefore the values of the dashes in the first row are all even numbers from 0 onwards, we’ll have that p - 1 and q + 1, which are even numbers, are the values of two dashes in the first row.
- Denoting respectively with x and y the ordinals of the hyphens thus identified, we’ll have that p = \mathrm{t\_value}(x) + 1, q = \mathrm{t\_value}(y) - 1, that is, \mathrm{t\_value}(x) immediately precedes the space p and \mathrm{t\_value}(y) immediately follows the space q. Conversely, we can say that p is the space following the dash number x and q is the space following the dash number y, where both dashes have component n_1 (which in our case is 2, but we’ll keep the variable because the discussion can be extended to other dashed lines).
- Applying Proposition L.C.6, we’ll have that x is such that (n_2 + n_3) \cdot n_1 x \mathrm{\ mod\} (n_1 n_2 + n_1 n_3 + n_2 n_3) \in S_T(1) with S_T(1) = \{n_2 a + n_3 b \mid a \in \{1, \ldots, n_3 - 2, n_3\}, b \in \{1, \ldots, n_2 - 2, n_2\}\};
- The value of p is known, so we can calculate that of x, and also n_1, n_2 and n_3 are known, so we can substitute all the values in the previous relations: we’ll thus obtain the Diophantine equation n_2 a + n_3 b = (n_2 + n_3) \cdot n_1 x \mathrm{\ mod\} (n_1 n_2 + n_1 n_3 + n_2 n_3) in the variables a and b. The equation is generally solvable because n_2 and n_3 are coprime, but in this case the existence of a solution such that a \in \{1, \ldots, n_3 - 2, n_3\} and b \in \{1, \ldots, n_2 - 2, n_2\} is a direct consequence of Proposition L.C.6. With a little trial and error, we can then derive the values of a and b.
- The pair (a, b) can be seen as a pair of coordinates in a Cartesian plane;
- We apply a similar procedure for q, starting from the fact that y follows q, so q is the space that precedes y. We can then apply Proposition L.C.5, from which we’ll obtain another pair of coordinates (a', b');
- In total, therefore, from the Goldbach pair (p, q) we’ll obtain two pairs of coordinates (a, b) and (a', b'), which we can join with an oriented segment;
- We associate the midpoint of the segment we have obtained with the Goldbach pair (p, q);
- We apply the procedure described to all Goldbach pairs formed by spaces, also with inverted terms, for the even number 2n that we have fixed.
Let’s start for example from the even number 2n = 26. We chose this value because it falls within the validity interval of the dashed line T_3 = (2, 3, 5), which goes from p_3 + 1 = 5 + 1 = 6 to (p_4)^2 - 1 = 7^2 - 1 = 48 inclusive. In fact, in this interval all the spaces of the dashed line are prime numbers, so if p and q are two spaces the sum of which is 2n, we can be sure that they are prime, except in the case where p = 1 or q = 1, which however, being an isolated case, can be easily excluded.
The table representing the dashed line, with the dashes numbered and the spaces highlighted, is the following:
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 2 | — | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 16 | 17 | 19 | 22 | 23 | 26 | |||||||||||||
| 3 | — | 2 | 6 | 8 | 12 | 14 | 18 | 21 | 24 | ||||||||||||||||||
| 5 | — | 4 | 10 | 15 | 20 | 25 |
The Goldbach pairs present are (7, 19), (13, 13), (19, 7) (for example, we do not have the pair (3, 23) because it is not formed by spaces; in fact 3 is the value of a hyphen).
Let’s apply the procedure to the first pair (7, 19), starting from its first element p = 7:
- With the help of the table, we see that the space p follows the dash number x = 5 of the first row;
- By Proposition L.C.6, x is such that (n_2 + n_3) \cdot n_1 x \mathrm{\ mod\} (n_1 n_2 + n_1 n_3 + n_2 n_3) \in S_T(1) with S_T(1) = \{n_2 a + n_3 b \mid a \in \{1, \ldots, n_3 - 2, n_3\}, b \in \{1, \ldots, n_2 - 2, n_2\}\};
- Let’s make the substitutions: we have n_1 = 2, n_2 = 3, n_3 = 5, x = 5, so we’ll have that (3 + 5) \cdot 2 \cdot 5 \mathrm{\ mod\} (2 \cdot 3 + 2 \cdot 5 + 3 \cdot 5) \in S_T(1) with S_T(1) = \{3 a + 5 b \mid a \in \{1, \ldots, 5 - 2, 5\}, b \in \{1, \ldots, 3 - 2, 3\}\};
- Doing the calculations, we’ll get 18 \in S_T(1) with S_T(1) = \{3 a + 5 b \mid a \in \{1, 2, 3, 5\}, b \in \{1, 3\}\};
- We must then find the values of a and b that satisfy the Diophantine equation 3 a + 5 b = 18 such that 3 a + 5 b \in S_T(1);
- Now we should solve the equation, considering that a and b must belong to \{1, 2, 3, 5\} and \{1, 3\} respectively; to find them, let’s make some attempts:
- For a = 1 and b = 1 we’ll have the element 3 \cdot 1 + 5 \cdot 1 = 8;
- For a = 1 and b = 3 we’ll have the element 3 \cdot 1 + 5 \cdot 3 = 18; now we have the value we were looking for, so we can stop.
- From p = 7 we then obtain the point (1, 3).
Let’s move on to the second element q = 19:
- It precedes the dash number y = 19 of the first row;
- By Proposition L.C.5, y is such that (n_2 + n_3) \cdot n_1 y \mathrm{\ mod\} (n_1 n_2 + n_1 n_3 + n_2 n_3) \in P_T(1) with P_T(1) = \{n_2 a' + n_3 b' \mid a' \in \{2, \ldots, n_3\}, b' \in \{2, \ldots, n_2\}\};
- We apply the substitutions: we have n_1 = 2, n_2 = 3, n_3 = 5, y = 19, so we’ll have that (3 + 5) \cdot 2 \cdot 19 \mathrm{\ mod\} (2 \cdot 3 + 2 \cdot 5 + 3 \cdot 5) \in P_T(1) with S_T(1) = \{3 a' + 5 b' \mid a' \in \{2, \ldots, 5\}, b' \in \{2, \ldots, 3\}\};
- Doing the calculations, we’ll get 25 \in P_T(1) with P_T(1) = \{3 a' + 5 b' \mid a' \in \{2, 3, 4, 5\}, b' \in \{2, 3\}\};
- As can be easily verified by calculating the elements of P_T(1), the values we are looking for are a' = 5 and b' = 2;
- From q = 19 we’ll then obtain the point (5, 2).
Starting from the Goldbach pair (7, 19) we thus obtained an oriented segment that goes from (1, 3) to (5, 2), the midpoint of which, as it can be easily verified through a graphical representation, is (3, \frac{5}{2}).
Applying the same procedure to the other two pairs, we’ll obtain:
- For the pair (13, 13), the segment from (2, 3) to (4, 2), the midpoint of which is (3, \frac{5}{2});
- For the pair (19, 7), the segment from (3, 3) to (3, 2), the midpoint of which is (3, \frac{5}{2}).
Let’s graphically represent the segments we have obtained, this way:
- Each segment is represented as an arrow going from the point corresponding to p to the one corresponding to q;
- At each extreme, we indicate the ordinal of the dash to which it refers; if there is more than one, the ordinals are indicated as a list, in the order in which they are found in the dashed line;
- Let’s draw a graph for each even number 2n such that 26 \le 2n \le 48; in fact, these even numbers are all and only those having as validity dashed line the dashed line (2, 3, 5) that we considered initially.
By analyzing all these graphs we can deduce that, for all even numbers 2n that have a third-order validity dashed line:
- When segments intersect, the intersections coincide with their midpoints;
- By grouping them appropriately in pairs, the midpoints share a coordinate; at a graphic level, for example, some are aligned horizontally, others vertically.
- If several arrows representing Goldbach pairs have the same midpoint, the sum of the ordinals of the dashes to which each arrow corresponds is always the same throughout the group.
- If, given two arrows, the sum of the corresponding ordinals is different, then they have different midpoints; however, if the sum of the ordinals is equal, the midpoints are not necessarily equal.
- By joining any pair of points referred to spaces the sum of which is different from the starting even number, we’ll obtain a segment the midpoint of which is always different from the ones of the arrows representing Goldbach pairs.
About the third and fourth points, for example, for 2n = 48 we can observe that:
- In the top left group, formed by two overlapping arrows with the same direction, referring to different pairs, the sum of the ordinals is 36 + 11 = 5 + 42 = 47;
- In the top right group, formed by two overlapping arrows with opposite directions, the sum of the ordinals is 29 + 17 = 17 + 29 = 36;
- In the lower right group, formed by three arrows, two of which are overlapping, the sum of the ordinals is 7 + 40 = 38 + 9 = 19 + 28 = 47;
- The upper left and lower right arrows have the same sum of ordinals, but have different midpoints.
The last point is easily verifiable: for example, for 2n = 26, joining the points (1, 3) and (3, 2), which refer to spaces the sum of which is different from 2n, we obtain a segment with a midpoint different from the ones of the arrows in the graph.
This geometric representation of Goldbach pairs can be the starting point of a larger study, which, through graphs like these, could let us see properties that do not emerge so easily from a study based only on the theory of dashed line or on numerical analyses. The fact that the arrows are grouped, for example, suggests that the set of Goldbach pairs can be partitioned into subsets that could be studied individually.
The characterization of the spaces that allowed us to associate coordinates to the spaces of a dashed line has been developed and proved up to the third order, but we are confident that the formulas can be extended without difficulty to orders higher than the third. The problem in this case would be the number of coordinates, since in general starting from a dashed line of order k we would obtain k - 1 coordinates for each space, therefore each space would be a point of a space of dimension k - 1. The case we have addressed in this article, with k = 3, is the simplest to visualize, because the points are two-dimensional; in spaces of higher dimension the visualization would be more complicated, but this would not prevent us from making geometric considerations.