ADDITIONAL PROPERTIES OF BETWEENNESSNOTE. The following should be inserted before the subheading Separation axioms on page47 of the file geometrynotes2a.pdf.Here are some additional facts about betweenness with vector proofs (however, manycomputational steps have been omitted).Example 1. Suppose that a 6= b and for i = 1, 2, 3 we have xi= a + ti(b − a) whereti∈ R. If t1< t2< t3, then we have the ordering relation x1∗ x2∗ x3.SOLUTION. The distance between xiand xjis equal to |xi− xj|, and if we substitute theexpressions for the vectors in question we see that|xi− xj| = |ti− tj| · |b − a| .Therefore we have|x1− x3| = (t3− t1) · |b − a| = (t3− t2) · |b − a| + (t2− t1) · |b − a| =|x2− x3| + |x1− x2|and therefore by the definition of betweenness we have x1∗ x2∗ x3.Example 2. Suppose that a∗b∗c. Then we have the inclusion of closed rays [bc ⊂ [ac.SOLUTION. By hypothesis we have b = a+t(c−a) for some scalar t satisfying 0 < t < 1.If x ∈ [bc, then x = b + u(c − b) for some scalar u satisfying u ≥ 0. Substituting theexpression for b in the preceding equation, we find thatx = a + (t + u − tu) · (c − a) .To show that x ∈ [ac, we need to check that the coefficient of c − a is positive. However,we have t + u − tu = u(1 − t) + t; the first term on the right hand side is a product ofnonnegative factors and hence is nonnegative, and since t > 0 it follows that the entireexpression is positive.Example 3. Suppose that a ∗ b ∗ c and b ∗ x ∗ c. Then we have a ∗ x ∗ c.SOLUTION. This is similar to the preceding example. By assumption we haveb = a + t(c − a) x = b + u(c − b)where 0 < t, u < 1. It follows thatx = a + (t + u − tu) · (c − a) .1The desired conclusion will follow if we can verify that the coefficient of c − a lies strictlybetween 0 and 1. Now(1 − t) (1 − u) = 1 − u − t + tuand since the left hand side lies between 0 and 1 it follows that the same is true for theright hand side. But this means thatu + t − tu = 1 − (1 − u − t + tu)must also lie strictly between 0 and 1.Here is a final example along the lines of the first one.Example 4. Suppose that x and y are distinct points such that a ∗ x ∗ c and a ∗ y ∗ care both true. Then either a ∗ x ∗ y or a ∗ y ∗ x is true.SOLUTION. We may writex = a + t(c − a) y = a + u(c − a)where 0 < u, t < 1. Clearly we also havea = a + 0(c − a) .Either 0 < t < u or 0 < u < t is true. In the first case we have a ∗ x ∗ y, and in the secondcase we have a ∗ y ∗
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