# Proof that every finite planar graph contains at least one vertex with valence less than six

`Assume, to the contrary, that`
`there is a finite planar graph with all vertices having valence >= 6. (1)`
`Let V,E,F be the number of vertices, edges, faces respectively.`
`Then 2*E = SUM(vertices v) valence of v`
` >= SUM(vertices v) 6 since each valence >= 6 by (1)`
` = 6*V`
`I.e. V <= E/3 (2)`
`Also, 2*E = SUM(faces f) valence of f`
` >= SUM(faces f) 3 (equality if the graph is triangulated)`
` = 3*F`
`I.e. F <= 2*E/3 (3)`
`Euler's formula says:`
` 2 = V - E + F`
` <= E/3 - E + 2*E/3 by (2) and (3)`
` = 0`
`Contradiction.`