1. Prove that if we colour the edges of K5 with 2 colours, red and green, so that the number of red edges is not equal to the number of green edges, then K5 will always have a monochromatic triangle.

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
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1. Prove that if we colour the edges of K5 with 2 colours, red and green, so that the number of
red edges is not equal to the number of green edges, then K5 will always have a
monochromatic triangle.
2. We colour the edges of the graph K₁6 using 3 colours, red, green, and blue. Prove that if the
number of red, green, and blue edges are not all equal, then we will always have a
monochromatic triangle.
Hint: First show that if r, g, and b are not all equal, then there must be a vertex of K16 with at
least 6 edges of the same colour.
Transcribed Image Text:1. Prove that if we colour the edges of K5 with 2 colours, red and green, so that the number of red edges is not equal to the number of green edges, then K5 will always have a monochromatic triangle. 2. We colour the edges of the graph K₁6 using 3 colours, red, green, and blue. Prove that if the number of red, green, and blue edges are not all equal, then we will always have a monochromatic triangle. Hint: First show that if r, g, and b are not all equal, then there must be a vertex of K16 with at least 6 edges of the same colour.
Expert Solution
Step 1: To prove this, let's proceed by

According to the question,contradiction. Suppose you color the edges of blankK subscript 5 (the complete graph on 5 vertices) with 2 colors, red and green, in such a way that the number of red edges is not equal to the number of green edges, and there is no monochromatic triangle.         

  1. Let's choose a vertex blankv in blankK subscript 5. Since blankK subscript 5 is a complete graph, blank is connected to 4 other vertices by 4 edges.

  2. Without loss of generality, assume there are more red edges than green edges in the graph. Then there are three possibilities for the edges coming out of blank:

    • 4 red and 0 green
    • 3 red and 1 green
    • 2 red and 2 green


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