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- September 13, 2020
- By menge

(↑Exercise 3.3.2) Show that d is a metric on R, that is, show that for all x, y, z ∈ R, 1. d(x, y) = d(y, x), 2. d(x, y) = 0 iff x = y, and 3. d(x, y) + d(y, z) ≥ d(x, z). Exercise 3.3.2 Show that d : Q × Q → Q+ satisfies, for all x, y, z ∈ Q, 1. d(x, y) = d(y, x), 2. d(x, y) = 0 if and only if x = y, and 3. d(x, y) + d(y, z) ≥ d(x, z). [Hint, set q = x − y, q’ = y − z and look at the cases above.] The inequality in Exercise 3.3.2(3) is sufficiently important that it has a name, the triangle inequality. Imagine, as in Figure 3.3.2, that x, y, and z are the vertices of a triangle. Traveling along the edge from x to y covers a distance d(x, y), along the edge from y to z covers a distance d(y, z). Along the edges of a triangle, this has to be at least as long as the distance covered traveling more directly, that is, along the edge from x to z, d(x, z). Re-writing, this is d(x, y) + d(y, z) ≥ d(x, z).