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Graph Theory – Proofs that K5 and K3,3 are not planar Copyright © R F Barrow 2009, all rights reserved

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K5K5 K 3,3 The Proofs that K 5 and K 3,3 are not planar Complete graph with 5 nodes Complete bipartite graph with nodes

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Proof that K 5 is not planar Proof by contradiction Assume K 5 is planar So K 5 is not planar There’s a contradiction

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Proof that K 5 is not planar Proof by contradiction Assume K 5 is planar Then it obeys Euler’s Relationship So K 5 doesn’t obey Euler’s Relationship. So K 5 is not planar R + N = A + 2 N = 5A = 10so R = 7 Each region is bounded by at least three arcs So there are at least 3R region boundaries So there are at least 3R ÷ 2 arcs, since each arc is 2 region boundaries 3 × 7 ÷ 2 = 10.5, so at least 11 arcs But there are 10 arcs There’s a contradiction QED

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Proof that K 3,3 is not planar Proof by contradiction Assume K 3,3 is planar Then it obeys Euler’s Relationship So K 3,3 doesn’t obey Euler’s Relationship. So K 3,3 is not planar R + N = A + 2 N = 6A = 9so R = 5 Each region is bounded by at least four arcs So there are at least 4R region boundaries So there are at least 4R ÷ 2 or 2R arcs, since each arc is 2 region boundaries 2 × 5 = 10, so at least 10 arcs But there are 9 arcs There’s a contradiction QED

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A Waldomaths clip this, along with other videos and more than 100 applets, all on mathematical themes, can be found on the free website Copyright © R F Barrow 2009, all rights reserved visit Waldomaths today

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