9_21

Today we started proving Lemma 2.4 which states that 'Points A and B are on the same side of //l// iff side(A,//l//)=side(B//,l//)'. We did this in two parts.

First, we assumed that side(A,//l//)=side(B,//l//) then we pick a point on side(A//,l)// say P and we know that P is also on side(B//,l)// by the assumption that they are equal, since P and A are on the same side (because we picked it that way) and P and B are on the same side (by the assumption that each side is equal) then by B-4 A and B are on the same side of //l.//

So we have shown that if side(A,l//)=//side(B,l) then A and B are on the same side. Now we have to prove the converse.

Next, we assumed that A and B are on the same side of //l.// Again, we pick a point P on the same side of A and also therefore in side(A,//l).// We are given that A and B are on the same side and since A and P are also on the same side then by B-4 we know that B and P are on the same side of //l//. Since P is also on the same side as B then B is also an element of side(B,//l//).

What we have just proven is that every element in side(A//,l)// is also an element in side(B,//l)// or that side(A,//l)// is a subset of side(B,//l).// In order to show that the two sets are equal we must prove that every element in side(B,//l//) is also in side(A,//l).// This argument is the same as above with one small difference. We pick a point on the same side of B and proceeding in a similar manner.

Next, we discussed that when you have an if and only if statement we have two directions to prove. Likewise, one direction might have multiple parts like the proof above. This led into finishing our discussion about Proposition 2.5 which states that every line has exactly two sides and they are disjoint. In the previous class we had proved that there are two sides and we finished by proving that the sides are disjoint in the following way:

We'll call the two sides side(A,//l)// and side(B,//l)// and every point is either in one side or the other. We know that side(A//,l)// is not equal to side(B,//l)// and by Lemma 2.4 that means that A and B must be on opposite sides. We want to show that the intersection of the two sides is the empty set which is the same as showing that the two sides are disjoint. The way we approached this was a proof by contradiction. We assumed that there was an element that is in the intersection of the two sides. This point C (that is an element of the intersection) is an element of side(A,//l//) so A and C are on the same side. Point C is also an element of side(B//,l//) so C and B are on the same side. By B-4, A, B, and C are all on the same side which is a contradiction because A and B must be on opposite sides.

Next, we moved on to Lemma 2.7 which states Lemma 2.7: If A and B are distinct points on the same side of a line //l// and lineAB intersects //l// at a point C, then A*B*C or B*A*C.

Of course, we drew a picture and it seemed like there was nothing to prove. First, we used definition 2.1 to show that if A and B are on the same side of line l then the line segment between then doesn't not intersect //l//. Then we tried to extend our segment to a line using propositon 2.3 and we discussed what a ray and a segment were. Then we used B-3 to say that since we have 3 point on a line then one point must be between the others. We already know that A and B are on the same side so C must be one of the end points. The only options left are A*B*C or B*A*C. If C were in the middle of A and B then A and B wouldn't be on the same side which is something we are given.

Next, we briefly talked about Lemma 2.8 and 2.9. Lemma 2.8- If A*B*C and l is a line passing through B that is distinct from lineAC, then A and C are on opposite sides of //l.// Lemma 2.9- If A and C are on opposite sides of a line //l,// then there exists a unique point B such that l passes through B and A*B*C. We drew a few pictures and it was mentioned that these lemmas are the converse of each other. We defined a new term which was the A,B,C are colinear if they lie on the same line. We also talked quickly about the way you would go about proving these lemmas and which previous propositions and definitions we would use to prove these.

We finished class by talking about

Proposition 2.10- If A*B*C and A*C*D, then B*C*D and A*B*D.

We drew a picture and at first glance this proposition seemed to be an easy one to prove but we ran into some problems. First, we tried proving by contradiction by writing out all possible combinations of B, C, and D and tried to find which one contradicted the assumptions. We quickly realized that we were using the proposition to try to prove the proposition. We also tried using B-2 because it seemed like that axiom told us the answer but we found that this was useless as well. A few more people tried variations of the what was already mentioned. Last, we tried using the propositions and lemmas immediately before this one (stated above). They involve a two lines which is where we left off. This argument will be continued next time.