**Question Corner and Discussion Area**

I am trying to prove that any four noncoplanar points of a three space determine that three space, using the following postulates and theorems: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.P1: If

aandbare distinct points, there is at least on line on bothaandb.P2: If

aandbare distinct point, there is not more than one line on bothaandb.P3: If

a,bandcare points not all on the same line, anddandeare distinct points such thatb,c, anddare on a line andc,a, andeare on a line, there is a pointfsuch thata,b, andfare on a line and alsod,e, andfare 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

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

IfYou 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.E,F,G, andHare in three-space(A,B,C,D) then three-space(E,F,G,H) = three-space(A,B,C,D).

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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