Proof:

Let $T = (V, E)$ be a tree, and let $u, v \in V$ be any two nodes. We aim to prove that the height of the rooted tree $(T, u)$, denoted $h_u$, is at most twice the height of the rooted tree $(T, v)$, denoted $h_v$.

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Key Definitions:

• Height of a tree rooted at $u$: The length of the longest path from $u$ to any leaf in the tree.
• Distance between two nodes $u$ and $v$: The number of edges in the shortest path between $u$ and $v$, denoted $d(u, v)$.

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Step 1: Define the farthest node

Let $w$ be the farthest node from $u$. Thus, the path $P(u, w)$ defines $h_u$, the height of $(T, u)$, because $w$ maximizes the distance from $u$: $h_u = d(u, w).$

Step 2: Analyze the role of $v$

• The path $P(u, w)$ must pass through some intermediate node, potentially $v$, or it avoids $v$ entirely.
• If $v$ lies on the path $P(u, w)$, then: $d(u, w) = d(u, v) + d(v, w).$
• If $v$ does not lie on $P(u, w)$, then $d(v, w)$ could be shorter than $d(u, w)$, but we analyze the worst-case scenario below.

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Step 3: Height of the tree rooted at $v$

The height $h_v$ corresponds to the farthest distance from $v$ to any node in the tree: $h_v = \max_{x \in V} d(v, x).$

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Step 4: Bounding $h_u$ by $h_v$

Now, consider the distance from $u$ to any leaf in the tree. If $v$ lies between $u$ and $w$, then: $d(u, w) = d(u, v) + d(v, w).$ Since $d(v, w) \leq h_v$ and $d(u, v) \leq h_v$ (because all nodes in the tree are at most $h_v$ distance from $v$), we have: $h_u = d(u, w) \leq d(u, v) + d(v, w) \leq h_v + h_v = 2h_v.$

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Step 5: Case where $v$ is not between $u$ and $w$

If $v$ is not on the path $P(u, w)$, let $x$ be the farthest node from $v$. The triangle inequality ensures: $d(u, x) \leq d(u, v) + d(v, x).$ Here, $d(v, x) \leq h_v$ and $d(u, v) \leq h_v$, so: $d(u, x) \leq h_v + h_v = 2h_v.$ Therefore, the worst-case scenario still satisfies: $h_u \leq 2h_v.$

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Discussion:

When $v$ is not between $u$ and its farthest node $w$:

• The contribution of $v$ to the distances from $u$ becomes indirect.
• Even in this scenario, the triangle inequality ensures that the height of the tree rooted at $u$ cannot exceed twice the height of the tree rooted at $v$.

Thus, regardless of $v$'s position, the height $h_u$ is at most $2h_v$.

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Follow-Up Questions:

1. How does the proof change if the graph is not a tree but a general connected graph?
2. Can we construct a counterexample where $h_u > 2h_v$ for some graph structure?
3. What happens when $T$ is a directed acyclic graph (DAG) instead of an undirected tree?
4. How would the result be affected if the tree $T$ is weighted?
5. How do distances change if we consider multiple intermediate nodes between $u$ and $w$?

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Tip:

When working with rooted trees, visualize paths using shortest-path metrics to better understand distance relationships