6/24/2009 Theorems, Proofs, Logic And Incidence Geometry Summer 2009 MthEd/Math 362 Chapter 3 1 Theorems, Proofs, and Logic • You’ve all had 190 (mostly). If it worked for you, you should know all you need to about proof for this class. If not, talking about “how to prove things” for a day or two probably won’t help. Instead, we’ll work on some simple theorems of Incidence Geometry together. • There. That takes care of Sections 3.1 –3.5. Summer 2009 MthEd/Math 362 Chapter 32 Incidence Geometry • Undefined terms: point, line, lie on. • Axioms: 1. For every pair of distinct points P and Q there exists exactly one line l such that both P and Q lie on l. 2. For every line l there exist two distinct points P and Q such that both P and Q lie on l. 3. There exist three points that do not all lie on any one line. Summer 2009 MthEd/Math 362 Chapter 33 1 6/24/2009 Incidence Geometry • Undefined terms: point, line, lie on. • Axioms: 1. For every pair of distinct points P and Q there exists exactly one line l such that both P and Q lie on l. 2. For every line l there exist (at least) two distinct points P and Q such that both P and Q lie on l. 3. There exist (at least) three points that do not all lie on any one line. Summer 2009 MthEd/Math 362 Chapter 34 Theorem 3.6.2 Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Summer 2009 MthEd/Math 362 Chapter 35 Theorem 3.6.2 Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Summer 2009 MthEd/Math 362 Chapter 36 2 6/24/2009 Theorem 3.6.2 Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. or, Let l be an arbitrary line. or, Choose any line l. or, Suppose l is an arbitrary line. Summer 2009 MthEd/Math 362 Chapter 37 Theorem 3.6.2 Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. Summer 2009 MthEd/Math 362 Chapter 38 Theorem 3.6.2 Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. (This means we need to stop and think about how to find that point. Which axiom guarantees the existence of a point?) • That’s right, Axiom 2 or Axiom 3. Why is 3 better? Summer 2009 MthEd/Math 362 Chapter 39 3 6/24/2009 Theorem 3.6.2 Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. That’s right. Because Axiom 3 gives us 3 points that DO NOT LIE ON the same LINE. Summer 2009 MthEd/Math 362 Chapter 310 Theorem 3.6.2 Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. By Axiom 3, there are three points that do not lie on the same line. These three points cannot all lie on l, so one of them must not lie on l. Summer 2009 MthEd/Math 362 Chapter 311 Theorem 3.6.2 Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. By Axiom 3, there are three points that do not lie on the same line. These three points cannot all lie on l, so one of them must not lie on l. Summer 2009 MthEd/Math 362 Chapter 312 4 6/24/2009 Theorem 3.6.2 Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. By Axiom 3, there are three points that do not lie on the same line. These three points cannot all lie on l, so one of them must not lie on l. T.I.J. (That’s It, Jack!) Summer 2009 MthEd/Math 362 Chapter 313 Theorem 3.6.2 Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. By Axiom 3, there are three points that do not lie on the same line. These three points cannot all lie on l, so one of them must not lie on l. Q.E.D. Summer 2009 MthEd/Math 362 Chapter 314 Theorem 3.6.2 (Alternate Path) Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. By Axiom 3, there are three points that do not lie on the same line. Summer 2009 MthEd/Math 362 Chapter 315 5 6/24/2009 Theorem 3.6.2 (Alternate Path) Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. Suppose for contradiction that there were no point P that did not lie on l. Summer 2009 MthEd/Math 362 Chapter 316 Theorem 3.6.2 (Alternate Path) Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. Suppose for contradiction that there were no point P that did not lie on l. Now What? Summer 2009 MthEd/Math 362 Chapter 317 Theorem 3.6.2 (Alternate Path) Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. Suppose for contradiction that there were no point P that did not lie on l. Then every point would lie on the line l, which contradicts Axiom 3. Summer 2009 MthEd/Math 362 Chapter 318 6 6/24/2009 Theorem 3.6.2 (Alternate Path) Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. Suppose for contradiction that there were no point P that did not lie on l. Then every point would lie on the line l, which contradicts Axiom 3. Thus some such P not on l must exist. Summer 2009 MthEd/Math 362 Chapter 319 Theorem 3.6.2 (Alternate Path) Theorem 3.6.2: If l is any line, then there exists at least one point P such that P does not lie on l. Proof: Let l be any line. Suppose for contradiction that there were no point P that did not lie on l. Then every point would lie on the line l, which contradicts Axiom 3. Thus some such P not on l must exist. Q.E.D Summer 2009 MthEd/Math 362 Chapter 320 Theorem 3.6.3 Theorem: If P is any point, then there are at least two distinct lines l and m such that P lies on both l and m. Proof: Let P be any point. By Axiom 3, there exist three points not all on the same line. Call them A, B, and C. We need to be careful because one of these three points could be the point P we already chose. So. Summer 2009 MthEd/Math 362 Chapter 321 7 6/24/2009 Theorem 3.6.3 Theorem: If P is any point, then there are at least two distinct lines l and m such that P lies on both l and m. Proof: Let P be any point. By Axiom 3, there exist three points not all on the same line. Call them A, B, and C. Summer 2009 MthEd/Math 362 Chapter 322 Theorem 3.6.3 Theorem: If P is any point, then there are at least two distinct lines l and m such that P lies on both l and m. Proof: Let P be any point. By Axiom 3, there exist three points not all on the same line. Summer 2009 MthEd/Math 362 Chapter 323 Theorem 3.6.3 Theorem: If P is any point, then there are at least two distinct lines l and m such that P lies on both l and m. Proof: Let P be any point. By Axiom 3, there exist three points not all on the same line. We consider two cases. Case 1: P is one of these three points; call the other two A and B. By Axiom 1, there is exactly one line l such that P and A lie on l. Summer 2009 MthEd/Math 362 Chapter 324 8 6/24/2009 Theorem 3.6.3 Proof (continued) By Axiom 1, there is exactly one line m such that P and B lie on m.