In algebra or in other discipline of mathematics, there are certain results or statements that are formulated in terms of n, where n is a positive integer. To prove such statements the well-suited principle that is used–based on the specific technique, is known as the principle of **mathematical induction**

In mathematics, we use a form of complete induction called **mathematical induction**. To understand the basic principles of mathematical induction, suppose a set of thin rectangular tiles are placed as shown in Fig 4.1.

When the first tile is pushed in the indicated direction, all the tiles will fall. To be absolutely sure that all the tiles will fall, it is sufficient to know that

- The first tile falls, and
- In the event that any tile falls its successor necessarily falls

This is the underlying principle of mathematical induction

We know, the set of natural numbers **N **is a special ordered subset of the real numbers. In fact, **N **is the smallest subset of **R **with the following property:

A set S is said to be an inductive set if **1∈ S** and **x + 1 ∈ S** whenever **x ∈ S**. Since N is the smallest subset of **R **which is an inductive set, it follows that any subset of R that is an inductive set must contain **N**

**The Principle of Mathematical Induction**

Suppose there is a given statement P(n) involving the natural number n such that

- The statement is true for n = 1, i.e., P(1) is true, and
- If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P(k) implies the truth of P (k + 1)

Then, P(n) is true for all natural numbers n

Property (i) is simply a statement of fact. There may be situations when a statement is true for all n ≥ 4. In this case, step 1 will start from n = 4 and we shall verify the result for n = 4, i.e., P(4).

Property (ii) is a conditional property. It does not assert that the given statement is true for n = k, but only that if it is true for n = k, then it is also true for n = k +1. So, to prove that the property holds , only prove that conditional proposition:

If the statement is true for n = k, then it is also true for n = k + 1.

This is sometimes referred to as the inductive step. The assumption that the given statement is true for n = k in this inductive step is called the inductive hypothesis.

**For example**, frequently in mathematics, a formula will be discovered that appears to fit a pattern like

1 = 1^{2} =1

4=2^{2 }=1+3

9=3^{2 }=1 + 3+5

9 = 3^{2}=1 +3 + 5

16 = 4^{2}== 1 + 3 + 5 + 7, etc

It is worth to be noted that the sum of the first two odd natural numbers is the square of second natural number, sum of the first three odd natural numbers is the square of third natural number and so on.Thus, from this pattern it appears that

1 + 3 + 5 + 7 + … + (2n – 1) = n^{2}

i.e, the sum of the first n odd natural numbers is the square of n.

Let us write

P(n): 1 + 3 + 5 + 7 + … + (2n – 1) = n^{2}

We wish to prove that P(n) is true for all n

The first step in a proof that uses mathematical induction is to prove that

.P (1) is true. This step is called the basic step. Obviously

1 = 1^{2} , i.e., P(1) is true.

The next step is called the inductive step. Here, we suppose that P (k) is true for some positive integer k and we need to prove that P (k + 1) is true. Since P (k) is true, we have

1 + 3 + 5 + 7 + … + (2k – 1) = k^{2} … (1)

Consider

1 + 3 + 5 + 7 + … + (2k – 1) + {2(k +1) – 1} … (2)

= k 2 + (2k + 1) = (k + 1)^{2} [Using (1)]

Therefore, P (k + 1) is true and the inductive proof is now completed.

Hence P(n) is true for all natural numbers n.