Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I do not fully understand the proof in Wikipedia, the first paragraph of this.

$$ \begin{cases} v − e + f = 2 \\ 2e \ge 3f \end{cases} $$

Firstly $2e \ge 3f$ means that a facet has at least 3 edges, at the same time one edge exists besides exactly 2 facets - I guess that's how it is, right?

The conclusion $v \le 2f − 4$ however is still unclear. They simply took a legit $f \le 2v − 4$ and exchanged the variables. How is that allowed?

share|improve this question
See en.wikipedia.org/wiki/Dual_graph –  user7530 Feb 19 '13 at 17:35
add comment

1 Answer 1

up vote 2 down vote accepted

You left out the justification given for the resulting inequality: "by duality."

If you need help understanding what "by duality" means and why it gives the resulting inequality, see this link to dual graph. In particular, read the properties, and the summary:

"Because of the duality, any result involving counting faces and vertices can be dualized by exchanging them." [http://en.wikipedia.org/wiki/Dual_graph]

share|improve this answer
I'd love to come up with the same property for $\mathbb{R}^4$. Do you think that's possible? Given that the above applies only to vertices-facets... –  Ranas Feb 19 '13 at 17:58
Over spaces, over 300 today! Nealy 30k :-)) –  B. S. Feb 19 '13 at 18:14
@Babak: I've "capped" on upvote points (until 6 hours from now). I meant "accept" as in an accepted answer...there is no limit on points obtained from accepted answers. –  amWhy Feb 19 '13 at 18:18
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.