Concept on Euler's formula Is there a much better way to proof and derive Euler's formula in geometrical figures? In that,F+V-2=E. For example an enclosed cube with  8 vertices, 6 faces and 12 edges. It is true that the edges, E=14-2
E=12
The idea is, where integer 2 comes in place in  the equation as an abstract value.
I will appreciate anyone's contribution.
Thanks
 A: A typical proof is by induction (best done for planar graphs). Imagine you have a connected graph drawn in the plane with no edge crossings and you are redrawing the graph. 
You start by drawing a single vertex. Thus, in your new drawing you've got $V = 1$, $F = 1$, and $E = 0$, so $F-E+V = 2$. So the 2 is right there from the start. 
That it stays 2 comes from the following observation. You can use two moves to draw the rest of your graph (this requires a little argument, of course). Either draw an edge you haven't drawn yet and end it at a vertex you haven't drawn yet. This does not divide a face and thus if we had thus far drawn $V$ vertices, $E$ edges and $F$ faces, we have now drawn $V+1$ vertices, $E+1$ edges, and $F$ faces and therefore $$F - (E+1) + (F+1) = F - E + F$$ and thus we haven't changed the value of $V - E + F$. 
Similarly, the second move is to add a single edge between two already existing drawn vertices. Because of the connectedness this always subdivides a face into two. In this case, then, we go from V, E, F in our drawing to V, E+1, F+1, but again: 
$$(F+1) - (E+1) + V = F - E + V.$$
So, if you accept that all planar connected graphs can be drawn using those two moves starting from a single vertex (which I didn't really prove but is fairly intuitive if you play with a few examples), and see that both moves preserve the value of $V - E + F$, then the fact that $V - E + F = 2$ just comes from the base case of the induction. 
