9_9

September 9, 2009

Today, we worked on the proofs for Propositions 1.2, 1.4, 1.5, and 1.6

Restating the incidence axioms, we have

I-1: For any two distinct points there exists a unique line that passes through both of them. I-2: For any line there exist at least two distinct points incident with it. I-3: There exist three distinct points with the property that no line is incident with all three of them.

We started with **Proposition 1.2:** If //l// and //m// are two distinct lines that are not parallel, then //l// and //m// have a unique point in common.

On September 2, 2009, we defined parallel as follows: Two lines are //parallel// if they have no points in common (if no point lies on both of them).

Given that //l// and //m// are //not// parallel, we conclude that there must be a point P that is incident with both //l// and //m//. We know that P is the only point incident with both //l// and //m// since otherwise, there would be a point Q, distinct from P, that is also incident with both //l// and //m.// But Axiom I-1 implies that since P and Q are distinct, there is a unique line that passes through both points, and //l// and //m// are the same line. Because we were given that //l// and //m// are distinct lines, P can be the only point that is incident with both //l// and //m.//


 * Definition 1.3:** A collection of lines is //concurrent// if there is a point incident with all them.


 * Proposition 1.4:** There exist three distinct lines that are not concurrent.

We first sought to find three distinct lines. Our starting point is Axiom I-3, which states there exist three distinct points with the property that no line is incident with all of them. Name these three points A, B, and C. Then Axiom I-1 implies that there is a unique line //l// passing through A and B, a unique line //m// passing through B and C, and a unique line //n// passing through A and C.

We know that //l, m,// and //n// are distinct. If they were not, and //l// and //m// were the same line, for example, then we would have one line passing through A, B, and C. Then, A, B, and C would not be the points with the property described in Axiom I-3.

We struggled a bit at this point in our proof. We initially thought the fact that none of //l, m,// or //n// passed through all of A, B, and C was sufficient to prove that the lines were not concurrent. We did not consider the possibility of another distinct point D, through which //l, m,// and //n// all do pass.

Fortunately, Proposition 1.2 came to the rescue. Since //l// and //m// are both incident with point B; //m// and //n// are both incident with point C; and //n// and //l// are both incident with point A, no pair of these lines is parallel. Since A, B, and C are the points of incidence, or the points at which the lines intersect, Proposition 1.2 assures us that there are no other points at which the lines can intersect. So the possibility of some fourth distinct point D at which all the lines intersect has been eliminated.

Therefore, //l, m,// and //n// are distinct and nonconcurrent.


 * Proposition 1.5:** For every line there is at least one point not lying on it.

This statement seems intuitive, doesn't it? At first glance, it seems as if the proof will be a snap. But we sure ran into some walls here!

Our proof begins innocently enough: Let //l// be a line. Then the question arose, "How can we just say there is a line?" We came up with a couple of answers to that question. First, we have Proposition 1.1, which states that there is a line. But we also talked about how it was okay logically to assume the existence of certain objects, and then prove an idea based on the assumption. Here, for example, we assume the existence of a line, and now we want to show that there //must// be a point not lying on the line.

At any rate, let //l// be a line. Axiom I-2 states that two distinct points lie on this line. Call them A and B. Here, we invoked Axiom I-3, saying that there must be a third distinct point C not lying on the line, and we thought we were done! Then our bubble burst.

Axiom I-3 states that there exist three distinct points with the property that no line passes through all of them. This axiom does not guarantee that A and B are two of those points, however. Now, our proof breaks down into two cases.

Case 1: A and B //are// two of the three points provided by Axiom I-3. Since //l// passes through A and B, it does not pass through the third distinct point C. Point C does not lie on //l//.

Case 2: One or both of A and B are not the points provided by Axiom I-3. That's okay though, because Axiom I-3 still provides three distinct points such that no line passes through all of them, so line //l// does not pass through all of them, either.

Case 2 brought up the question, "Do we now know that //five// distinct points must exist?" The answer was that we can neither prove nor disprove that five distinct points exist, at least not with the three axioms we have. We only knew that A and B, through which line //l// passes, may or may not be two of the three points provided by Axiom I-3.

Emina briefly mentioned that the proof of Proposition 1.5 could be presented more simply. Let //l// be a line. Axiom I-3 states that there exist three distinct points with the property that no line passes through all of them. Line //l// is included in the lines that do not pass through all three points. Therefore, for any line there is a point not lying on it.


 * Proposition 1.6:** For every point, there is at least one line not passing through it.

We began with a point P, and then immediately got into trouble. We invoked Axiom I-3, and claimed the existence of two other points, Q and R, such that no line passes through all three points P, Q, and R. The line passing through Q and R does not pass through P, and we thought we were done! Once again, we assumed that P was one of the points guaranteed in Axiom I-3. Our proof really breaks down into two cases.

Case 1: P is one of the points guaranteed by Axiom I-3. In this case, the line passing through Q and R really does not pass through P, and we are done.

Case 2: P is not one of the points guaranteed by Axiom I-3. In this case, there exist distinct points A, B, and C such that no line passes through all of them. It is from the points A, B, and C that we will find a line not passing through P.

According to Axiom I-1, there is a unique line //l// passing through P and A. Then line //l// does not pass through at least one of the other points B or C. Suppose line //l// does not pass through B. Invoke Axiom I-1 once again to find the unique line //m// that passes through A and B. Line //m// does not pass through P since line //m// already intersects line //l// at A, and A is the unique intersection of lines //l// and //m,// by Proposition 1.2.