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Fano presentation

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Fano presentation

  1. 1. Tour of Fano’s Geometry Axioms: 1. There exists at least one line. 2. Every line of the geometry has exactly three points on it. 3. Not all points of the geometry are on the same line. 4. For each two distinct points, there exists exactly one line on both of them. 5. Each two lines have at least one point on both of them.
  2. 2. A Quick Theorem Theorem (Theorem 1.7) Each two distinct lines have exactly one point in common. Proof. Suppose 1 and 2 are distinct lines in Fano’s Geometry. Axiom 5 says there is at least one point on both of them. If points P and P are on both 1 and 2 then Axiom 4 is contradicted as there are two lines on both P and P .
  3. 3. What we’ve shown Because every addition we made was strictly required to fulfill the axioms, we’ve established two facts: There are at least 7 points in Fano’s Geometry. There are at least 7 lines in Fano’s Geometry. In what seems to be the usual way, we can prove that there are no more than 7 points in Fano’s Geometry by using an argument by contradiction.
  4. 4. Part of a theorem Theorem Fano’s Geometry has exactly 7 points. Proof. We’ve established that there are at least 7 points. Suppose there is an eighth, P8. By Axiom 4, there must be a line, that contains both P1 and P8. Since must intersect the line that contains P3, P7, and P4, and that line contains no other points, it follows that contains either P3, P7, or P4. This contradicts Axiom 4.
  5. 5. The rest of the theorem Theorem Fano’s Geometry has exactly 7 lines. Proof. Suppose Fano’s geometry had a line that was not among the 7 we’ve established. By Axiom 4, each two distinct points must have exactly one line on both of them, thus can contain at most one of these 7 points. By axiom 2, though, contains 3 points. Therefore, contains at least two points that are not among the original 7. This contradicts our previous argument, that there are exactly 7 points in the geometry.

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