Thread: Proof by mathematical Induction

Okay new here,
I love this site you were very helpful in my last post so I have come back with another question.

For all possible integers n >= 3, prove that the total number of interior non-crossing diagonals of a polygon with n sides, is equal to n - 3.

eg. for n = 3 sides, its a triangle which has no crossing diagonals and for n = 4 sides its a square which has 4-3=1 sides.

I have done these two as base steps to prove the proposition (P(n)=n-3).
However I am having trouble when proving that P(k)->P(k+1) as when i put in P(k+1)=k-3+k+1 it will not factorise or simplify to (k+1)-3. I'm probably doing something horribly wrong here like my proposition is wrong or my method is wrong so can someone help put me back onto the right track??

Help is much appreciated

Assume the statement is true for some k and look at a polygon with k+ 1 vertices. With A, B, and C three consecutive vertices, draw the diagonal AC. That divides the polygon into a triangle and a polygon with k vertices. Do you see that no diagonal of this new polygon crosses that original diagonal?