Let's prove that K5,9 is not planar: First, how many vertices and how many edges does K5,9 have? v= and e = Suppose, for the sake of contradiction, that K5,9 is planar. Then how many faces would it have? f = 4. However, since every face is bounded by at least possible based on this line of reasoning? f≤ This is a contradiction, so K5,9 is not planar. QED. edges, and every edge borders exactly faces, we can get a bound on the number of faces. What is the largest number of faces

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Let's prove that K5,9 is not planar:
First, how many vertices and how many edges does K5,9 have?
and e =
V=
Suppose, for the sake of contradiction, that K5,9 is planar. Then how many faces would it have?
f =
However, since every face is bounded by at least
possible based on this line of reasoning?
f<
This is a contradiction, so K5,9 is not planar. QED.
edges, and every edge borders exactly
faces, we can get a bound on the number of faces. What is the largest number of faces
Transcribed Image Text:Let's prove that K5,9 is not planar: First, how many vertices and how many edges does K5,9 have? and e = V= Suppose, for the sake of contradiction, that K5,9 is planar. Then how many faces would it have? f = However, since every face is bounded by at least possible based on this line of reasoning? f< This is a contradiction, so K5,9 is not planar. QED. edges, and every edge borders exactly faces, we can get a bound on the number of faces. What is the largest number of faces
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