14.1.7. Question Number 12

In question 12, I am supposed to find out the number of diagonals in an n sided polygon and I want to prove that it is n times n minus three over two, of course for all n greater than equal to three and I will prove it by induction.

In this chunk, we set up the main objective: finding the number of diagonals in an n-sided polygon. The formula we want to prove is $\frac{n(n - 3)}{2}$. Importantly, we restrict our focus to polygons with three or more sides (triangles or higher) as they are the only configurations that can have diagonals. The method of proof will use mathematical induction, where we prove the formula is true for a base case (n = 3) and then assume it holds for k sides to show it must also hold for k + 1 sides.

Imagine you have a group of friends standing in a circle, and you want to find out how many ways you can connect them with strings that don't form the circle itself (diagonals). For instance, with just three friends, you can't create any strings that skip over another friend. But with four friends, you start to see ways to connect them across the circle, and as the group grows, the number of ways increases!

Of course my base case will be three because I am making this statement to be true for n greater than equal to three onwards. So, of course the statement is true for any polygon with three sides because a polygon with three sides is nothing but a triangle and you do not have any diagonal in a triangle.

The base case is established for a triangle, which is a polygon with three sides (n = 3). Since there is no way to connect vertices in a triangle such that the connections do not form the sides of the triangle, there are zero diagonals. The formula gives $\frac{3(3-3)}{2} = 0$, confirming that it holds for the base case.

Think of a triangle made from three friends holding hands. They can only connect with each other by holding hands on the sides. There are no additional strings (diagonals) needed or possible!

Assume the statement is true for any polygon with k sides.

For our inductive step, we assume that the formula for the number of diagonals is true for a polygon with k sides. This means we assume that a k-sided polygon has $\frac{k(k - 3)}{2}$ diagonals. This assumption will be critical when we try to prove that the formula still holds when we add one more side (k + 1).

Imagine that you have managed to count all the ways your friends can connect as you added more people to the circle previously. This belief will help you examine the next step—adding yet another friend to see if your understanding holds up.

Now I want to prove that the statement is true even for the polygon with k + 1 sides. So here is a polygon with k + 1 sides and I want to count the number of diagonals here.

In this chunk, we explore how to derive the diagonal count for a polygon with k + 1 sides from the count we already know for k sides. When we add the new vertex (the new side), it will create some new diagonals. Specifically, we can create diagonals from this new vertex to all other vertices except the two adjacent ones and itself, which leads us to calculate the total number of diagonals.

Imagine you are adding another friend to the circle—now that friend can connect to everyone except their two immediate neighbors. This results in even more unique connections, just like adding edges in a growing web!

Now how many such diagonals are there, which I have not included yet namely the blue color which are k - 2. Namely, I cannot count the newly added side, or its two adjacent sides as diagonals.

When we add the new vertex, the new diagonals that can be formed connect this vertex to the others that are not adjacent to it. The count of these connections is $k - 2$ because the new vertex connects with everyone except itself and its two adjacent neighbors. This leads to the total diagonal count being the diagonals from the k-sided polygon plus these new connections.

Think about if your group of friends is in a circle, and you just added one more. They can't reach the two friends directly next to them but can connect with everyone else! You can see the new possible connections forming!

And this gives me the total number of diagonals in the overall polygon and it comes out to be what you want to show for your inductive step, that brings me to the end of tutorial number two.

Finally, when combining the numbers, we demonstrate that the total number of diagonals in a polygon with k + 1 sides is still described accurately by the formula $\frac{(k + 1)(k - 1)}{2}$. Thus, by showing both the base case and the inductive step, we conclude that the formula holds for all n >= 3.

Just picture your expanded group. Each time you add a new friend, all previous connections and new possible connections calculate how tightly knit your group truly is. By going step-by-step, you can see it's possible to keep connecting them all without missing any!

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