Question

Extended Principle of Mathematical Induction The Extended Principle of Mathematical Induction states that if Conditions I and II hold, that is, (I) A statement is true for a natural number \(j\). (II) If the statement is true for some natural number \(k \geq j\), then it is also true for the next natural number \(k+1\). then the statement is true for all natural numbers \(\geq j\). Use the Extended Principle of Mathematical Induction to show that the number of diagonals in a convex polygon of \(n\) sides is \(\frac{1}{2} n(n-3)\) [Hint: Begin by showing that the result is true when \(n=4\) (Condition I).]

Short answer

The formula \[ \frac{1}{2} n(n-3) \]holds for all n ≥ 4 by the Extended Principle of Mathematical Induction.

Step-by-step solution

Establish the base case (Condition I)

Verify that the given formula is true for the polygon with the initial number of sides, in this case, a polygon with 4 sides (a quadrilateral). For a quadrilateral (n=4), the formula should be verified to give the correct number of diagonals.

Verify the formula for n=4

In a quadrilateral (n=4), we can calculate the number of diagonals. A quadrilateral has 2 diagonals. Check the formula with n=4: \[ \frac{1}{2} n(n-3) = \frac{1}{2} 4(4-3) = \frac{1}{2} 4 \times 1 = 2 \] The formula holds as it gives 2, which is the correct number of diagonals for a quadrilateral.

Assume formula holds for n=k (Inductive Hypothesis)

Assume that the formula is true for a polygon with k sides, i.e., \[ \frac{1}{2} k(k-3) \] represents the number of diagonals in a polygon with k sides.

Prove the formula for n=k+1

Now, demonstrate that if the formula holds for a polygon with k sides, then it should also hold for a polygon with (k+1) sides. A polygon with (k+1) sides can be seen as a polygon with k sides plus one additional side. Adding an extra side to a k-sided polygon creates k new diagonals (one for each of the k vertices).

Number of diagonals in (k+1)-sided polygon

The new number of diagonals when we add one extra side is given by adding the k new diagonals to the existing number of diagonals for k sides:\[ \text{Number of diagonals in (k+1)-sided polygon} = \frac{1}{2} k(k-3) + k \]Simplify the expression:\[ = \frac{1}{2} k^2 - \frac{3}{2} k + k \]\[ = \frac{1}{2} k^2 - \frac{1}{2} k \]\[ = \frac{1}{2} k(k-1) \text{ diagonals} \]

Justify the modified formula

Notice that k(k-3) + 1 is the correct number of diagonals for a polygon with k sides plus the new vertex added: \[ = \frac{1}{2} (k+1)((k+1)-3) \]\[ = \frac{1}{2} (k+1)(k-2) \]

By the principle of induction

Since the formula has been verified for n=4 and the induction step holds (k to k+1), by the Extended Principle of Mathematical Induction, the formula \[ \frac{1}{2} n(n-3) \]holds for all natural numbers n ≥ 4.

Key concepts

mathematical induction

Mathematical induction is a powerful proof technique used in mathematics. It allows us to establish the truth of an infinite number of cases by proving a base case and an inductive step. This method is particularly useful because it breaks down what might seem like an impossible task into manageable steps. Mathematical induction comprises two main parts:

• The Base Case: We show that a statement holds for a particular initial value, often the smallest value in the domain.
• The Inductive Step: We assume the statement holds for some arbitrary case, say n = k, and then prove it holds for the next case, n = k + 1. This step relies on the assumption being true for n = k.

When both steps are verified, we can conclude that the statement is true for all natural numbers starting from the base case.

polygon diagonals

A diagonal in a polygon is a line segment connecting two non-adjacent vertices. In a convex polygon, the diagonals do not curve outside the polygon, making them easy to count. The number of diagonals in a polygon with n sides can be determined using the formula:\[ \frac{1}{2} n(n-3) \]This formula arises because each vertex connects to (n-3) other vertices (excluding itself and its two neighbors). Multiplying n by (n-3) counts all possible diagonal connections, but since each diagonal is counted twice, we divide by 2. For the initial case, consider a quadrilateral (4 sides). Using the formula, the number of diagonals is:\[ \frac{1}{2} 4(4-3) = 2 \]which matches the actual count of diagonals in a quadrilateral.

proof by induction

Proof by induction is essential when dealing with statements involving natural numbers. Here, we'll use it to show that a given formula for the number of diagonals in a convex polygon is valid for all polygons with at least four sides.
First, we verify the base case (n = 4). Then, we assume that the formula holds for a polygon with k sides. This assumption is our inductive hypothesis. Next, we prove that if the formula holds for a polygon with k sides, it must also hold for a polygon with (k+1) sides.
We add a side to the k-sided polygon, which introduces k new diagonals. Therefore, the new total for a (k+1)-sided polygon becomes:
\[ \frac{1}{2} k(k-3) + k \]Simplifying this expression confirms our formula for the polygon with (k+1) sides.

base case in induction

The base case is the first critical step in a proof by induction. It verifies that the formula or statement works for the smallest value in the context, giving a concrete starting point for the induction. In our problem, we set the base case with n = 4 (a quadrilateral). We calculate the number of diagonals using our given formula:
\[ \frac{1}{2} 4(4-3) = 2 \]We count the actual diagonals in a quadrilateral and confirm that our formula works. Establishing the correctness at this base step is essential because it sets the stage for the subsequent inductive step. Without a verified base case, the entire inductive proof would collapse.