Let b be the statement, \"Given any throe distinct lines l.m.n. if l || m and m

Question

Let b be the statement, "Given any throe distinct lines l.m.n. if l || m and m || n, then l|| This is a true statement about the Cartesian plane model of incidence geometry. (a) Consider the axiomatic system that consists of the axioms of incidence geometry plus the Euclidean parallel Property. Prove the statement P from these axioms. the desired model cannot satisfy the Elliptic parallel Property because such models have no parallel lines, and it cannot satisfy the Euclidean parallel Property due to what you proved in part (a).

Explanation / Answer

3. let p be the statement "given any three distinct lines l m n. if l || m and m|| n then l || n. this is a tue statement about the cartesian plane model o fincidence geometry

a) consider the axiomatic system that consists of the axiom of incidence geometry plus the euclidean parallel property. prove the statement p from the axioms

Solution: This statement is independent. As justification we give two models: one that satisfies the statement and another that does not

Model 1: The standard three-point model satisfies the statement. This is because there are no parallel lines in the three-point model, so the “if” part of the statement is always false, making the “if-then” statement always true

Model 1 (alternative): . We already know it is a model, so we won’t verify the axioms again. It remains to show that the statement given in the problem is satisfied. Suppose we have three (distinct) lines l, m, and n in our model. To be more specific, let us say l = {P, Q} and m = {R, S} where P, Q, R, S {A, B, C, D} (we used new names P, Q, R, S because we do not know which pair of points is on l and which pair is on m; these new names allow forthe possibility that P = A or P = B or P = C or P = D, for example). Assume further that l || m and n || l. We want to show that n m. Since l || m, these two lines have no point in common, so P, Q, R, S must be four distinct points. Since n l, n has a point in common in l by definition of parallel. We also know by Proposition 2.1 that this point is unique. Without loss of generality, say that this common point is P. (We actually do not know that this point is P; it could possibly be Q. By saying “without loss of generality”, we are indicating that the argument we are about to give will apply just as well to Q.) Now that we know P is one of the points on n, the other point on n must be either R or S (it cannot be Q because then by the uniqueness part of I-1, n would equal l, contrary to our assumption that the three lines are distinct). In either case, n will have a point in common with the line m = {R, S}, so n is not parallel to m.

Model 2: There are exactly five points in our model: A, B, C, D, E. The lines are the sets consisting of exactly two points (there are 10 lines total). “Lie on” will mean set membership. We wish to show that our model does not satisfy the statement in the problem; that is, our model satisfies the negation of the statement: There exist three lines l, m, and n such that l || m, l n, and n || m. It suffices to find three lines that have the properties specified in the statement. Here are three such lines: l = {A, B}, m = {C, D}, n = {A, E}.

Let l, m and n be three distinct lines. Suppose that l || m and n l. Since n l there is a point P that lies on both n and l. Note that P is not on m since l and m have no points in common. By the Euclidean parallel postulate there is only one line through P that is parallel to m. Since n and l are two different lines passing through P and l is parallel to m, n must not be parallel to m.

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