Other Problems

D.1

Coordinatization of the Affine plane of 9 points:

Let A be an affine plane of 9 points. Let the points be denoted by O, 1, A, B, C, D, E, F, G. Let the lines of A be

Each set of three lines is a parallel class. Denote the lines OFG and ODA as the x-axis and y-axis respectively. Let OIC be called the diagonal. Let be the set . Fix a bijection from the points of the diagonal ( ) to elements of such that , , and .

Points on the Diagonal: Let the coordinates of O be (0,0), the coordinates of I be (1,1), and the coordinates of C be (2,2).

Points not on Diagonal: Let P be a point not on the diagonal. The coodinates of P are given in the same manner as was described in the coordinatization of the general affine plane above. For example, consider E. E is on the line CEG which is parallel to the x-axis, and CEG intersects the diagonal at point C(2,2). E is also on the line DIE parallel to the y-axis. DIE intersects the diagonal at point I(1,1). Let E have coordinates (2,1).

Slope: This is assigned in the same manner as described in section one. Let the line BIF be the line of slopes. Since ADO, BIF, and CEG are in the parallel class of the y-axis, let them have undefined slope. Since ABC, DIE, and OFG are in the parallel class of the x-axis (which intersects the line of slopes BIF at F(1,0)), they have slope 0. The lines DBG, FEA, and OIC are in the same parallel class and have slope 1 since the line OIC contains the point (0,0) and crosses the lines of slopes at (1,1). Similarily, the lines FDC, BEO, and AIG have slope 2 since BEO intersects the line of slopes at (1,2).

Ternary operation: Define the ternary operation as in section 1. Then is the second coordinate of the intersection of the line parallel to the y-axis through the point (2,0) (ie, line CEG) and the line with slope 2 through A(0,2) (ie, AIG). The intersection point is G(2,0), thus the second coordinate is 0, and . Similarly, is the second coordinate of the intersection point of the line BIF and the line FDC (with slope 2, through (0,1)). The intersection point is F(1,0), thus, . Also, since the intersection of CEG and DBG is G(2,0).

Equation of lines: We may write equations of lines using slope and y-intercept. For example, consider line AEF. It has slope 1, and contains point A(0,2). Thus, its equation is . Also, line AIG has slope 2 and contains point A(0,2), thus its equation is . For lines with undefined slope, such as CEG, we have cooresponding to the first coordinate of every point on the line. This particular example is linear, that is, . This can be checked with the equations of each line, and each point on that line. For example, is ths equation of AEF, and if we let , we see that point A(0,2) gives us , a true statement, as well as E(2,1) : , and F(1,0) : .

D.9

Coordinatization of the general projective plane of order n:

Let X, Y, O, I be four points, no three collinear in a projective plane of order n. This is called the coordinatizing quadrangle. Every set of four points in a projective plane that has the property that no three are collinear can be used to coordinatize a projective plane, and they will determine the ternary operation (this will not be the same for each set of four points). Let be a set of order n including the elements 0, and 1. Let the line OX be the x-axis, OY be the y-axis, OI be called l, and XY be called . Now fix a bijection between the points on l besides such that and .

Points on l (excluding those on ): If A is a point on l such that , let its coordinates be (a,a).

Points not on l (excluding those on ): Suppose that P is not on l. Then the line PY does not contain the point since Y is already on . Thus, PY crosses l somewhere, say point B(b,b). Similarily, the line PX crosses l somewhere, say point C(c,c). Let the coordinates of P be (b,c).

 
Figure 7: The Point P(b,c)

For the points on the line , let the line through (0,0) and (1,m) intersect at the point (m). Let Y be labeled , and note that X is 0.

 
Figure 8: Points (m)