What is it that you're proving? Postulates are axioms; that is, they're assumed true.
Consider the system [S, L, P], where S contains exactly four points A, B, C, and D, the lines are the sets with exactly two points, and the planes are sets with exactly three points. This "space" is illustrates by the following figure:
Here it should be remembered that A, B, C, and D are the only points that count. Show that there is at least one plane.
Here are the relevant Incidence Postulates in the chapter:
I-0)All lines and planes are sets of points.
I-1) Given any two different points, there is exactly one line containing them.
I-2) Given any three different noncollinear points, there is exactly one plane containing them.
I-3) If two points lie in a plane, then the line containing them lies in the plane.
I-4) If two planes interesect, then their intersection is a line.
I-5) Every line contains at least two points. S contains at least three noncollinear points. Every plane contains at least three noncollinear points. And S contains at least four noncoplanar points.
Here is my proof, but I have no idea whether or not it is correct. A push in the right direction would be greatly appreciated!
Let points A, B, C, and D in S be given. If there isn’t at least one plane, this means that any set of three points is collinear.
A, B, and C lie on a line.
A, B, and D lie on a line.
A, C, and D lie on a line.
B, C, and D lie on a line.
However, the four statements above contradict I-1, which states that given any two points, there is exactly one line containing them because each possible pair of points lies on two lines (i.e. A and B lie on line ABC and line ABD). In conclusion, there is at least one plane in the figure.
*Random question - Is there also a way to prove that there exist at least two planes?
ojones, he wants to prove there exist at least one plane- which is NOT a postulate.
Deveno, we are given that a plane is a set of three points. It does not follow from that alone that every set of three points is a "plane".
Theorem010, you assert that "i.e. A and B lie on line ABC and line ABD" contradicts the fact that two points, A and B here, can lie on only one line. To complete that you will need to show that ABC and ABD are not the same line. If they were the same line, of course, A, B, C, and D would lie on the same line. To show that is not true, use the fact that "S contains at least four noncoplanar points."
ok, so we have the following candidates for planes:
{A,B,C}, {A,B,D}, {A,C,D} and {B,C,D}.
and what you're saying is, that what needs to be shown is that one of these sets of three points is noncollinear.
again, perhaps i'm missing something, but isn't that what I-5 asserts (in which case I-2 asserts the existence of the plane)?
it seems to me that I-5 implies that we also have for our plane {P1,P2,P3}, the fact that P4 does not lie on it.
suppose {P1,P2,P4} were collinear. then the line P1P2 equals the line P1P4 (appeal to I-1)
similarly if {P1,P3,P4} are collinear, then the line P1P3 equals the line P1P4.
and therefore {P1,P2,P3} is collinear, a contradiction.
thus one of the two sets {P1,P2,P4}, or {P1,P3,P4} must be a noncollinear set, thus a plane, by I-2.
if this second plane is actually the same as {P1,P2,P3}, then all 4 points are coplanar, contradicting I-5.
hence S contains at least 2 planes.