The number of diagonals in any polygon is whole number. but on the other side the formula of the number of diagonals in any convex polygon is in a fraction form   \frac{n(n-3)}{2[tex]  . how can you explain this?

I think the formula for the number of diagonals of a convex polygon is actually

This will always give a result of a whole number because 
if n is odd, then n-1 is even, or
if n is even, n-1 must be odd.
So the division by 2 will always give a whole number because there is always an even number in the numerator.

By the way, this number is also solution to the "hand-shake" problem, which is "how many possible handshakes in a group of n people, if everyone shakes hand with everyone else exactly once?"  Therefore, again, it is a whole number.

Hey dear, hoping i'll be