# Property of non-regular polyhedral graph the bound for number of faces

With just considering the number of edges that meet at a vertex, prove that f >= 2 + 1/3(e)

So what I have is that a polyhedron is 3D so at least 3 edges will meet at each vertex, that's equivalent to its degree. Since the sum of all vertex totals = 2e then 3v =< 2e

Then since a polyhedron is a plane graph, we can use V - E + F = 2 => F = 2 - V + E

But using all of the above I got f >= 2 - (-2/3)v + e = 2 + 5/3(e), which is not what it should be.

Where did I make an incorrect assumption? Thanks!

I'm thinking my error is in assuming v =< 2/3 (e) is equal to v >= -2/3(e)...

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It's $V-E+F=2$. – Gerry Myerson Feb 13 '13 at 3:44
I typed it wrong in question but have it right on paper whoops – DJ_ Feb 13 '13 at 3:48
Good. But now you've replaced $v$ with $-(2/3)v$, and that can't be right. – Gerry Myerson Feb 13 '13 at 3:55
The real problem is that $v\le(2/3)e$ wants to give you $-v\ge-(2/3)e$. – Gerry Myerson Feb 13 '13 at 3:57
So v >= 2/3 not v >= -2/3? – DJ_ Feb 13 '13 at 4:03

## 1 Answer

Everything you have done so far is right, just be careful with your inequality. Like what Gerry said above, we have v <= (2/3)e, so that -v >= -(2/3)e and so, by Euler's formula, we have:

f = 2 + e - v >= 2 + e -(2/3)e = 2 + (1/3)e

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