2.1.11. Theorem. If G is a simple graph, then diam G ≥ 3 ⇒ diam & ≤3. Proof: When diam G > 2, there exist nonadjacent vertices u, v = V(G) with no common neighbor. Hence every x = V (G) - {u, v} has at least one of (u, v) as a nonneighbor. This makes x adjacent in G to at least one of {u, v} in G. Since also uv E(G), for every pair x, y there is an x, y-path of length at most 3 in G through (u, v). Hence diam G≤ 3.
2.1.11. Theorem. If G is a simple graph, then diam G ≥ 3 ⇒ diam & ≤3. Proof: When diam G > 2, there exist nonadjacent vertices u, v = V(G) with no common neighbor. Hence every x = V (G) - {u, v} has at least one of (u, v) as a nonneighbor. This makes x adjacent in G to at least one of {u, v} in G. Since also uv E(G), for every pair x, y there is an x, y-path of length at most 3 in G through (u, v). Hence diam G≤ 3.
Principles of Microeconomics
7th Edition
ISBN:9781305156050
Author:N. Gregory Mankiw
Publisher:N. Gregory Mankiw
Chapter22: Frontiers Of Microeconomics
Section: Chapter Questions
Problem 6PA
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
Transcribed Image Text:2.1.11. Theorem. If G is a simple graph, then diam G ≥ 3 ⇒ diam & ≤3.
Proof: When diam G > 2, there exist nonadjacent vertices u, v = V(G) with no
common neighbor. Hence every x = V (G) - {u, v} has at least one of (u, v) as a
nonneighbor. This makes x adjacent in G to at least one of {u, v} in G. Since
also uv E(G), for every pair x, y there is an x, y-path of length at most 3 in G
through (u, v). Hence diam G≤ 3.
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