Explain Mathematical induction and principle of mathematical induction with examples

 

Understanding Mathematical Induction

Mathematical induction is a powerful method of mathematical proof used to establish that a given statement is true for all natural numbers. It is particularly useful for proving statements about sequences, series, and other properties that are indexed by natural numbers.

Principle of Mathematical Induction

The principle of mathematical induction consists of two main steps:

1. Base Case: Show that the statement is true for the initial value (usually $n = 1$).
2. Inductive Step: Assume that the statement is true for some arbitrary natural number $k$ (this is called the inductive hypothesis). Then, show that the statement must also be true for $k + 1$.

If both steps are successfully completed, the statement is proven to be true for all natural numbers.

Example 1: Sum of the First $n$ Natural Numbers

Let's use mathematical induction to prove that the sum of the first $n$ natural numbers is given by the formula:

$S(n) = \frac{n(n + 1)}{2}$

Step 1: Base Case

For $n = 1$:

$S(1) = \frac{1(1 + 1)}{2} = \frac{2}{2} = 1$

The formula holds for $n = 1$.

Step 2: Inductive Step

Assume the formula is true for some arbitrary natural number $k$, i.e.,

$S(k) = \frac{k(k + 1)}{2}$

We need to show that the formula holds for $k + 1$:

$S(k + 1) = S(k) + (k + 1)$

Using the inductive hypothesis:

$S(k + 1) = \frac{k(k + 1)}{2} + (k + 1)$

Factor out $(k + 1)$:

$S(k + 1) = \frac{k(k + 1) + 2(k + 1)}{2}$

$S(k + 1) = \frac{(k + 2)(k + 1)}{2}$

Thus, the formula holds for $k + 1$, completing the induction.

Example 2: Proving an Inequality

Let's prove by induction that for all natural numbers $n \geq 1$, $2^n \geq n + 1$.

Step 1: Base Case

For $n = 1$:

$2^1 = 2$ $1 + 1 = 2$

The inequality holds for $n = 1$.

Step 2: Inductive Step

Assume the inequality holds for some arbitrary natural number $k$, i.e.,

$2^k \geq k + 1$

We need to show that the inequality holds for $k + 1$:

$2^{k + 1} = 2 \cdot 2^k$

Using the inductive hypothesis:

$2^{k + 1} = 2 \cdot 2^k \geq 2(k + 1)$

Since $2(k + 1) > (k + 1) + 1$ for $k \geq 1$:

$2^{k + 1} \geq k + 2$

Thus, the inequality holds for $k + 1$, completing the induction.

Summary

Mathematical induction is a method used to prove statements about natural numbers by confirming the base case and proving the inductive step. This technique ensures that if the statement holds for an arbitrary natural number $k$, it also holds for $k + 1$, thereby proving the statement for all natural numbers.