# Mathematical Induction

## Trigonometry # Mathematical Induction

Mathematical Induction is introduced to prove certain things and can be explained with this simple example. Garima goes to a garden which has different varieties of flowers. The colour of all the flowers in that garden is yellow. She picks a flower and brings it home. Now if she picks up a rose then what colour is it? Is it too difficult to answer? No, obviously the colour of the rose is yellow as every flower in that garden is yellow. What did we just do here? We used the concept of logical reasoning to deduce the colour of the rose. Similar to this analysis, in Mathematics the use of the concept of mathematical thinking can be applied to reach conclusions. Consider a mathematical example:

• All the numbers lying on the real number line are known as real numbers
• All the real numbers greater than zero are positive real numbers
• 25 is a real number

From the above statements, we can say that if the first two statements are true then the third one is definitely true. Let us learn more here.

## What is Mathematical Induction?

It is the art of proving any statement, theorem or formula which is thought to be true for each and every natural number n.

In mathematics, we come across many statements that are generalized in the form of n. To check whether that statement is true for all natural numbers we use the concept of mathematical induction.

This concept of induction is generally based on the fall of dominoes concept.

It’s just like all the dominoes will fall one by one if the first one arranged in the queue is pushed. Similar to this in induction we prove that if a statement is true for the first number (n = 1) and then show that it is true for n = kth number then it can be generalized that the given statement is true for every n.

It is important to mention here that a set of N natural numbers is the smallest subset of the set of real numbers R with the given property:

 A set S is said to be inductive if 1 is an element of S and x + 1 is also an element of S when it is given that x is an element of S.

Now as N is a subset of the inductive set R then it can be concluded that any subset of R that is inductive must consist of N.

Suppose to find out the sum of positive natural numbers we use the formula:

 1 + 2 + 3 … … … n = [n(n+1)/2]

But is the formula valid? To check the validity of such formula, we use mathematical induction. We check the validity for the smallest possible and then continue for higher values and then if it is true for higher we accept the validity for entire n.

## Mathematical Induction Steps

Below are the steps that help in proving the mathematical statements easily.

Step (i): Let us assume an initial value of n for which the statement is true. Here, we need to prove that the statement is true for the initial value of n.

Step (ii): Now, assume that the statement is true for any value of n say n = k. Then, prove the given statement is true for n = k + 1 also.

Step (iii): Finally, we have to split n = k + 1 into two parts; one part is n = k (already proved in the second step), and we have to prove the other part.

In the above procedure, proving the given statement for the initial value is considered as the base step of mathematical induction and the remaining procedure is known as the inductive step.

## Mathematical Induction Examples

Q.1: Show that, 1 + 2 + 3 … … … n = [n(n+1)/2] is true for n = 5.

Solution: Given, n = 5

First, let us find the L.H.S = 1 +2+3+4+5 = 15

Now, R.H.S = [5(5+1)]/2 =  (5 x 6)/2 = 30/2 = 15

Since, L.H.S = R.H.S.

Hence, 1 + 2 + 3 … … … n = [n(n+1)/2] is true for n = 5.

Q.2: Show that 1 + 3 +…+(2n-1) = n2 for n = 3.

Solution: Given, n =3

2n – 1 = (2 x 3) – 1 = 6 -1 = 5

So, LHS = 1 + 3 + 5 = 9

RHS = 32 = 9

Since, L.H.S = R.H.S.

Hence, 1 + 3 +….+(2n-1) = n2 for n = 3.

### Mathematical Induction Problems

Practice the mathematical induction questions given below for the better understanding of the concept.

1. Using mathematical induction to prove that 1⋅2⋅3 + 2⋅3⋅4 + … + n(n + 1)(n + 2) = [n(n + 1)(n + 2)(n + 3)]/4  for n ∈ N.
2. Prove that 2n > n for all positive integers n.

3. For every positive integer n, prove that 7n – 3n is divisible by 4.