Plane Geometry - Bruce E. Shapiro

64 SECTION 14. BETWEENNESSTheorem 14.21 Distance is a metric.Proof. Let P and Q be points. Then we need to show that each of thefollowing hold:(a) P Q = QP(b) P Q ≥ 0(c) P Q = 0 ⇐⇒ P = QBy lemma 14.20 there is a line l that contains both P and Q.By the ruler postulate (axiom 14.11), there is a one to one function f : l ↦→R, with let x = f(P ) and y = f(Q) such that the distance is given byTo see (a),To see (b),P Q = |f(P ) − f(Q)| = |x − y|P Q = |x − y| = |y − x| = QPP Q = |x − y| ≥ 0To see (c), first suppose that P Q = 0. Then0 = P Q = |x − y|⇒ x = y⇒ P = Qwhere the last line follows because f is one-to-one. To verify the converseof (c) suppose that P = Q. Then x = f(P ) = f(Q) = y so thatwhich verifies the converse of (c).P Q = |x − y| = 0Example 14.1 (Euclidean Metric) Let P = (x 1 , y 1 ), Q = (x 2 , y 2 ). Then»d(P, Q) = (x 2 − x 1 ) 2 + (y 2 − y 1 ) 2To show that this is a metric, calculate»d(P, Q) = (x 2 − x 1 ) 2 + (y 2 − y 1 ) 2 ]»= (x 1 − x 2 ) 2 + (y 1 − y 2 ) 2 ]= d(Q, P )« CC BY-NC-ND 3.0. Revised: 18 Nov 2012