# Exist at least $21$ lines and $21$ points

Given $6$ axioms :

$(1)$ For every point $P$ and for every point $Q$ not equalto $P$ there exists a unique line $l$ incident with $P$ and $Q$ .

$(2)$ For every line $l$ there exist at least two distinct points incident with $l$ .

$(3)$ There exist three distinct points with the property that no line is incident with all three of them.

$(4)$ If $A * B * C$ ( it's mean $A,B,C$ are three distinct points all lying on the same line and $B$ is between $A,C$) then $C*B*A$

$(5)$ Given any two distinct points $B,D$ there exist points $A,C,E$ lying on the line through $B,C$ such that $A*B*D,B*C*D,B*D*E$

$(6)$ If $A,B,C$ are three distinct points lying on the same line , then one and only one of the points is between the other two .

Prove any model of axioms $(1) \to (6)$ must have at least $21$ points and $21$ lines . And give a model that have exactly $21$ points and $21$ lines

• What does "between" mean in this context? Commented Feb 1, 2017 at 6:31
• You can think it likes $AB<AC$ , Commented Feb 1, 2017 at 6:34
• $(4)$ means " between " $A,C$ is same as between $C,A$ Commented Feb 1, 2017 at 6:36
• Hm. No, I doubt one can think that. The problem probably means that there is a set of "points", a set of "lines", a incidence relation between points and lines, and a ternary relation of betweenness on the set of points incident to each line… If you intend something like distances between points, so as to compare them, or something, you should be considerably more precise. Commented Feb 1, 2017 at 6:36
• I am still trying to understand what you are asking! Commented Feb 1, 2017 at 6:51

Let's consider a model with 21 points and 21 lines. There are 210 pairs of different points and each such pair determines a line. Suppose that

all lines have the same number of points, $N$.

Then each line contains $\binom N2$ pairs of lines, and we must have $N=5$. This suggest that we look at the projective plane $PG(2,4)$ over the field with four elements, which gives us a configuration of 21 lines and 21 points, with each line containing 5 points. It satisfies your conditions 1, 2 and 3.

To see whether this can be completed to an example, we have to see whether we can define a ternary relation on each line, which is a set of 5 elements, satisfying the conditions 4, 5, and 6. Indeed, as the betweenness relation is essentially independent of the incidence relation, we can do this line by line.

Consider the latin square

1   4   2   5   3

4   2   5   3   1

2   5   3   1   4

5   3   1   4   2

3   1   4   2   5


And view it as an operation $\star$ on the set $X=\{1,2,3,4,5\}$. If $a$ and $b$ are two distinct points in $X$, we declare that the unique point in $X$ which is between $a$ and $b$ is $a\star b$. You can check (somewhat laboriously) that this satisfies the conditions. Condition 4 is a consequence of the fact that the latin square is symmetric. The following Mathematica code checks condition 5

tt = {
{1, 4, 2, 5, 3},
{4, 2, 5, 3, 1},
{2, 5, 3, 1, 4},
{5, 3, 1, 4, 2},
{3, 1, 4, 2, 5}
};
qq = Flatten[Table[{i, tt[[i, j]], j}, {i, 5}, {j, 5}], 1];
Length@Union@qq[[All, #]] & /@ Subsets[Range[3], {2}]


The result expected in {25,25,25}.

• The latin squarde I got by googling "symmetric idempotent latin square of order 5". I have no idea if it is special in some way or if any one works... Commented Feb 1, 2017 at 8:12

To show that any model of those six axioms must have at least 21 points and 21 lines, you can construct any model and slowly show that no matter what, there must exist at least 21 points and 21 lines.

By (3), we have three distinct non-collinear points. By (1), we draw lines AB, AC, and BC connecting all three points. You can easily show that all these lines are distinct. For example, if line AB = AC, then points A, B, and C are collinear. Contradiction.

By (5), points A and B have to have a point C in between. Now you have to show that C cannot be any other point. The reason you need to show that is because we are proving no matter how you construct any model, you have to have at least that number of points. Since you can add a point in between A and B, B and C, and A and C, you will have shown that so far, you will need at least 6 points. You will continue to do this until you get to 21 points. Similarly, you will do the same to show that you need at least 21 lines.

To show that a model with 21 points and 21 lines exist, consider the following. A projective plane of order n has (n^2) + n + 1 number of points and lines. If you have a field, you can construct a projective plane over that field. So we need a finite field with 4 elements.

Since you can construct a finite Galois Field for any prime number p and natural number n such that p^n is the number of elements of the GF field. We have a GF field with 2^2 elements.

Since projective planes satisfy the incidence axioms, you only have to show that the order axioms (4), (5), and (6) are satisfied. To show that, consider any line in the projective plane of order 4 we defined above. That line will have 5 points. You can put those 5 points together in a way that satisfies all the order axioms.

To visualize how that is done, consider a pentagon with vertices A, B, C, D, and E. Any two points have a point in between that corresponds to the opposite point. For example, point D is between points A and B, point E is between points B and C, etc. Also, any two points have a point before and after them that corresponds to their order. For example, point E is before A and B, point C is after points A and B, etc.

You will end up with the following relations: A∗B∗C, B∗C∗D, C∗D∗E, D∗E∗A, E∗A∗B, A∗D∗B, B∗E∗C, C∗A∗D, D∗B∗E, A∗C∗E,

That is actually why the plane separation axiom was introduced. You can check all the relations and see that they satisfy all the order axioms.

Thus, we've constructed a projective plane that satisfies all the 6 axioms.