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Using Geometric Postulates for Theorems in 3 Dimensions

Asked by Tim O'Brien, teacher, Bremen High School on September 20, 1997:
I am trying to prove that any four noncoplanar points of a three space determine that three space, using the following postulates and theorems:

P1: If a and b are distinct points, there is at least on line on both a and b.

P2: If a and b are distinct point, there is not more than one line on both a and b.

P3: If a, b and c are points not all on the same line, and d and e are distinct points such that b, c, and d are on a line and c, a, and e are on a line, there is a point f such that a, b, and f are on a line and also d, e, and f are on a line.

P4: There exists one line.

P5: There are at least 3 distinct points on every line.

P6: Not all points are on the same line.

P7: Not all points are on the same plane.

Thm. 2-1 If two points of a line are on a given plane, then every point of the line is on that plane.

Thm. 2-2 Any two distinct coplanar lines intersect in a unique point.

Any help would be appreciated.

Tim

An axiomatic system like this is not the usual method for studying geometry in three and higher dimensions. One would normally employ the language of vector spaces, linear independence, bases, and so on. That gives you a much cleaner theory that is dimension-independent.

To do it using postulates and theorems such as the ones you describe requires that you first of all give a definition of what a "three space" is! You could either do this axiomatically by giving postulates involving planes and three-spaces (similar to your existing postulates for points and lines), or else you could adopt some sort of definition (for example, saying that "the three-space generated by non-coplanar points A, B, C, and D is defined to be the set of all points P for which the (unique) line through P and D intersects the plane described by A, B, and C".

(This definition would not work for ordinary geometry since it would omit points that lie in the plane through D parallel to the plane through A, B, and C; however, your postulates do not describe ordinary geometry because of postulate P3. Your postulates are for a form of projective geometry, and the definition I gave above would work for that).

Of course, you would also need to define what you mean by a plane. In fact, you need to do that even before you can prove theorems 2-1 and 2-2; they do not follow from the postulates, since the postulates make no mention of planes except for P7 which is not enough to capture what "plane" means.

Then, the way you would prove your desired statement would depend on the particular definitions you chose for "plane" and "three-space". For instance, if you adopted the definition I suggested, your task would then be to prove

If E, F, G, and H are in three-space(A,B,C,D) then three-space(E,F,G,H) = three-space(A,B,C,D).
You could start by proving basic facts about planes, such as the fact that if two points are in a plane then the entire line containing them is in the plane, and that any three non-collinear points in a plane determine that plane. Then you could try to prove these facts about three-spaces.

For example, to prove that if P and Q each belong to the three-space S = three-space(A,B,C,D) and R is on the line through P and Q then R is also in S, you could start by saying that the line PD intersects plane ABC in some point P' and the line QD intersects the plane ABC in some point Q' (by the definition of three-space). Having already proven that when two points are in a plane the entire line joining them lies in the plane, it follows that the entire lines PD, QD, and PQ lie in the plane PQD, so all six points P, P', Q, Q', D, and R lie in the plane PQD. Thm 2-2 guarantees that the lines P'Q' and RD intersect in a point X. Since P' and Q' are in plane ABC and X is on the line P'Q', it follows that X is on plane ABC, so line RD intersects plane ABC, showing that R is in S.

Next you could show that if P, Q, and R are in S, then the entire plane PQR is in S. You could do this by letting T be some point in plane PQR, and observing that lines TP and QR intersect in some point X (Thm 2-2). By the previous result and the fact that X is on the line QR, we know X is in S; since T is on the line PX, it follows that T is in S also.

Now you're in a position to prove that every point P in three-space(E,F,G,H) is also in S = three-space(A,B,C,D): line PH intersects plane EFG in a point X. Since E, F, and G are in S, so is X; since P is on line HX and both H and X are known to be in S, so is P.

Finally, you must prove the converse: that every point in S is also in three-space(E,F,G,H). You would use similar sorts of arguments to do this.

Please bear in mind, though, that if the particular definitions of "plane" and "three-space" you're working with are different from the ones I suggested, you would need different proofs, appropriate for whatever definitions or postulates you use for these concepts.

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