I discovered an obvious thing some days back, but is there a solid proof for it? Prove that there are \(\rm x + 1\) faces if the base of a pyramid has \(\rm x\) sides. I know the concept of it, but just need to know if there's a proof.

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I discovered an obvious thing some days back, but is there a solid proof for it? Prove that there are \(\rm x + 1\) faces if the base of a pyramid has \(\rm x\) sides. I know the concept of it, but just need to know if there's a proof.

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there is.
Please.
You mean for pyramids?

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The base of a cube has 4 sides but it has 4+2 faces.. I didn't get it!
Oops, yes, pyramids.
It seems pretty obvious. There is one face attached to each of the base polygon's sides, plus the base itself.
Yeah... but I am asking for a solid proof.
@ParthKohli needs a proof involving serious 'mathematics'
For a pentagonal prism: |dw:1350485379807:dw|
Pfft, drawing pictures is serious enough for me.
Oh Lord, a solid proof.
LOL, I get it.
Induction?
Well, @CliffSedge , your diagram says it all...
I need serious mathematics, sire, serious mathematics.
Euler's formula might come into play if you want to get all fancy about it. V-E+F=2
to prove something, you need an equation
And that formula is valid also for a cube!!
Euler's formula is valid for all polyhedra, so it will be applicable to the special case of pyramids. Here F=x+1
The number of vertices, V, also equals x+1 (x for the vertices of the base polygon, plus the top vertex of the pyramid).
And the number of edges, E = 2x (x for each side of the base polygon, plus the x edges coming up from its x vertices). These are all based on definitions of pyramids, so V-E+F=2 F=x+1 V=x+1 E=2x (x+1)-2x+(x+1)=2 Simplify and identity is verified.
Or since your task is to show that F=x+1, start with (x+1)-2x+F=2, and solve for F. This all seems quite circular, since it is merely restating definitions of pyramids.

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