Quod Erat Demonstrandum



Filed under: Fun — johnmayhk @ 2:03 下午

Parallel lines are usually defined as lines with no points in common. Parallelism is clearly symmetric. If line 1 has no points in common with line 2, then line 2 also has no points in common with line 1. Is parallelism reflexive? In other words, is a line parallel to itself? This appears to be a matter of convention. Since the advantage of a positive answer far outweighs the alternative, let’s modify the definition of parallel lines to be lines which do not have exactly one point in common. Finally, is parallelism transitive? Suppose line 1 is parallel to line 2 and line 2 is parallel to line 3, but line 3 is not parallel to line 1. Then line 1 and line 3 intersect at exactly one point P, which cannot be on line 2. Otherwise, being parallel to both line 1 and line 3, line 2 cannot have only one point in common with them and must coincide with both. However, line 1 and line 3, not being parallel to each other, are distinct. Through a point P not on line 2, we now have two lines parallel to line 2, which contradicts Playfair’s Axiom. So parallelism is transitive and, in turn, this implies Playfair’s Axiom. Let P be a point not on line 2. Suppose both line 1 and line 3 pass through P and are parallel to line 2. By transitivity, they are parallel to each other, and hence they cannot have exactly P in common. It follows that they are the same line, which is Playfair’s Axiom. Thus we have possibly the shortest statement of the parallel axiom. (Three words!) “Parallelism is transitive." The five-word version is: “Parallelism is an equivalence relation," and the answer to our question is: “Yes, if the geometry is Euclidean."

(Andy Liu, University of Alberta, from the College Mathematics Journal, Volume 42, Number 5, November 2011, p. 372)


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