Planar Ternary Rings

We may now prove that our previous coordinatizing set along with the ternary operation described geometrically is indeed a planar ternary ring by checking T1 - T5.

Proof:

T1

Note that T(0,b,c) is the second coordinate of the intersection point of the line through (0,0) parallel to the y-axis (ie, the y-axis itself) and the line with y-intercept c. Certainly this equals c. Also, T(a,0,c) is the second coordinate of the intersection point of the line through (a,0) parallel to the y-axis, and the line with slope 0 (parallel to the x-axis) with y-intercept c. This also is c.

T2

Note that T(a,1,0) is the second coordinate of the intersection point of the line through (a,0) parallel to the y-axis, and the line with slope 1 and y-intercept 0 (ie, the diagonal). The point of intersection is on the diagonal, thus has coordinates (a,a), making the second coodinate a. Also, T(1,a,0) is the second coordinate of the intersection point of the line through (1,0) parallel to the y-axis (ie, the line of slopes) and the line with slope a. This point is (1,a) by definition of the assignment of slopes, making the second coordinate a.

T3

Suppose that p is the line with slope m and y-intercept b, and q is the line with slope m' and y-intercept b'. Since , properties of an affine plane state that there is one unique point of intersection of p and q. Let this point be (a,s), and this is the unique solution to
.

T4

Let (a,b) and (a',b') be the coordinates of two points. Since , the line through these two points is unique by the affine axiom A1, and is also not parallel to the y-axis . This means that the line has a slope and a y-intercept. Let these be m and s respectively. Then m and s are the unique solution to and .

T5

Let l be a line through (a,c) parallel to the line through (0,0) and (1,m). By affine axiom A2, there is a unique line. The y-intercept of l is the unique solution to .

Proof: We can again check T1-T5:

T1

Note that and by ring properties and the additive identity.

T2

Similarily, and by multiplicative and additive identity properties.

T3

Since and , we have

by the properties of additive inverses, left distribution with addition, and is a group provided .

T4

This works in much the same way as above, and the unique solution is and .

T5

Naturally the unique solution is .