The challenge then is to rigorously is to add ‘points at infinity’ to the real plane with two goals in mind:
1) Parallel lines meet at such a point.
2) No other new intersections are created.
Note: If Condition 2) is not satisfied, then we would no longer be able to say that two lines meet at only one point.
Here is our strategy:
1) Create equivalence classes of parallel lines. Let Pm º {All lines with slope m}
= {y = mx+b | bÎR}
To avoid confusion, let P¥ denote the equivalence class of vertical lines: P¥ º {x = b | bÎR}
2) Associate a new point, am, to each equivalence class, Pm. By associate, I mean add the point am to every line contained in Pm so that each line meets this ‘point at infinity’.
Then R2 È {am}mÎR, is the Real Projective Plane, denoted RP2 or P2(R). The ordinary doubles (x,y) are called regular points, the points am are called ideal points. Equivalently, regular lines in the Real Euclidean Plane are called ordinary lines, while the line made up of all the ideal points is called the ideal line.
We can now prove the following theorems:
Proof: Let a,bÎRP2
Case 1 – Suppose a,b are both ordinary points. Then by regular Euclidean Geometry (actually
Euclid’s 4th Postulate), only one ordinary line passes through two
distinct points.
Case 2 – Suppose a is ordinary, b is ideal. Then b = ak, for some kÎR (or k = ¥). Then look at the associated equivalence class of mutually parallel lines, Pk. The lines in this equivalence class all pass through b, and are all ordinary lines. Since a is ordinary, a = (u,v) in Cartesian coordinates. Thus if a line in Pk passes through a, the following equation is satisfied:
v = ku + c
or: c = ku - v
(or if k = ¥, i.e. vertical lines, we simply get the relationship: u = c)
So the equation of the line M passing through a, y = kx + c, is uniquely determined by the slope associated with the ideal point b, and the coordinates of a. Thus M passes through both a and b, and M is unique.
Case 3 – Suppose a,b are both ideal points. If two ordinary lines M and N, both pass through a and b, then M and N are in the equivalence class associated with a as well as the equivalence class associated with b. However, this implies that both equivalence classes are the same, which is a contradiction since a and b are assumed to be distinct. Thus the only line passing through both a and b is the ideal line.
Theorem 2 – Two distinct lines meet
in one and only one point.
Proof: Let M,N be lines in RP2
Case 1 – M, N are both ordinary, and not parallel. This implies that M and N meet at one an only one ordinary point. If M and N also met at an ideal point, say ak, then they would belong to the same equivalence class, Pk. However, this would imply that M and N were parallel, which would be a contradiction.
Case 2 – M, N are both ordinary and parallel. This implies that M and N belong to the same equivalence class, , and thus both meet at the point . Let M be defined by y = kx + b, and N by y = kx + c. If M and N met at an ordinary point as well, say (u,v) then (u,v) would satisfy:
u - u = k(v - v) + b - c
0 = b – c
b = c
Which would imply that M and N were not distinct, which would be a contradiction.
Case 3 – M is ordinary, N is ideal. Then the only point of M which meets the ideal line is the ideal point associated with M’s equivalence class. Thus M and N meet at one and only one point.
By adding these points at infinity, we have slightly modified the real plane to allow for Theorem 2, which allows one to make sense when defining objects as the intersection of lines or lines with points.
Synthetic Projective Geometry takes the above into account and proceeds to define geometry solely with a straight-edge. For Java applets which illustrate proofs of some of the important theorems (and which give a general idea of the flavor of projective geometry) see the following sites:
However, the above approach is the synthetic approach, that is defining geometry in turns of relations like parallel, incident (intersecting) etc. I could have proved the previous theorems using without resorting to equations, but mathematicians used to Algebra usually feel more comfortable describing points in terms of coordinates: that is, an analytic approach.
In order to get a bit further into Projective Geometry without having to use straight-edges (and to make things a bit easier thanks to Linear Algebra) we will now take an analytic approach. First, however we need to figure out some coordinates for RP2.
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Postulate Next: Homogeneous Coordinates