Question

Let T be a tree, and recall that there is a unique path in 7 between every pair of vertices , yeV(T). Define a function d: V(T) x V(T) → Zo by letting d(x, y) be the length of the unique path from a to y. Prove that for all x, y, z EV(T) we have d(x, z) < d(x, y) + d(y, z).

Answer 1

To prove that for all x, y, z EV(T) we have d(x, z) < d(x, y) + d(y, z), we can analyze the unique path from x to z. The length of the path from x to z directly (d(x, z)) must be shorter than the sum of the lengths of the path from x to y (d(x, y)) and the path from y to z (d(y, z)). Hence, for all x, y, z EV(T), we have d(x, z) < d(x, y) + d(y, z).

Step-by-step explanation:

To prove that for all x, y, z EV(T) we have d(x, z) < d(x, y) + d(y, z), we can analyze the unique path from x to z. Let's assume there are two paths, one going directly from x to z and the other going through y. We can compare the lengths of these paths: d(x, z) and d(x, y) + d(y, z). Since T is a tree, there can be no cycles, and thus, the path from x to y and the path from y to z cannot intersect.

Therefore, the length of the path from x to z directly (d(x, z)) must be shorter than the sum of the lengths of the path from x to y (d(x, y)) and the path from y to z (d(y, z)). Hence, for all x, y, z EV(T), we have d(x, z) < d(x, y) + d(y, z).