 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-47/4/b/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

For every binary vector

x

S_{1} (x, x) = 1

, and hence

d_{2} (x, x) = \frac{1}{2} \neq = 0.

Thus $d_{2}$ is not a metric, because it fails identity of indiscernibles. It is nonnegative and symmetric. In fact it also satisfies the triangle inequality trivially: every value lies in

[1/2, 1]

, so

d_{2} (x, z) \leq 1 \leq d_{2} (x, y) + d_{2} (y, z)

. That does not repair its nonzero self-distance.

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