Mathematics Homework Solutions
Problem
#6813

Proving connectedness.

Prove that if G is a disconnected graph, the complement graph G^G is connected, and in fact, diam(G^)<=2.


Solution Summary

A proof is offered for the following statement: If G is a disconnected graph, the complement graph G^G is connected, and in fact, diam(G^)<=2.

Solution
What is this?
By OTA - Overall OTA Rating
Changping Wang, MA - 4.9/5
Purchase Cost Now
$2.19 CAD
Included in Download
  • Plain text response
$2.19 Instant Download
Add to Cart
Why you can trust BrainMass.com
  • Your Information is Secure
  • Best Online Academic Help Service
  • Students find real academic Success
Related Solutions
  • Prove Connectedness - Prove that G with at least (n-1)(n-2)/2+1 edges is connected, where n is the order of G.
  • Connectedness - Let G be a graph of order n such that deg(v)>=(n-1)/2. Prove that G is connected.
  • Circles that cannot be homeomorphic - 23. Using the intuitive notion of connectedness, argue that a circle and a circle with a spike attached cannot be homeomorphic. (Question is also included in attachment)
  • General and Differential Topology - This is one of the basic courses for students beginning study towards the Ph.D. degree in mathematics. Content: Topological and metric spaces, continuity, subspaces, products and quotient topology, ...
  • General and Differential Topology - This is one of the basic courses for students beginning study towards the Ph.D. degree in mathematics. Content: Topological and metric spaces, continuity, subspaces, products and quotient topology, ...
Browse