Mathematical Induction: Principle, Steps and Applications

Mathematical induction aids in the proof of mathematical conclusions and theorems for all natural numbers. 

• One method for proving certain algebraic propositions that are expressed in terms of n, a natural number, is the principle of mathematical induction. 
• Any mathematical assertion or expression is established as true for n = 1, n = k, and n = k + 1 before being established for n = k + 2.

Key Terms: Mathematical Induction, Natural Numbers, Theorem, Subset, Set, Real Number

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What is Mathematical Induction? 

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It is the art of proving any statement, theorem, or formula that is thought to be true for all natural numbers n.

• Numerous statements in mathematics are generalized in the form of n. 
• The concept of mathematical induction is used to determine if that assertion is true for all natural numbers.
• This induction notion is often based on the domino effect.
• If the first one arranged in the queue is pushed, all the dominoes will fall one by one.
• Induction works similarly in that if a proposition is true for the first number (n = 1) and then shown to be true for the n = kth number, it may be generalised to be true for every n.

It is worth noting that a set of N natural numbers is the smallest subset of the set of real numbers R that has the following property:

• A set S is considered inductive if it contains the number 1 and, whenever it contains a number x, it also contains the number x + 1.
• Because N is a subset of the inductive set R, any inductive subset of R must contain N.
• Assume use the following formula to calculate the sum of positive natural numbers:

1 + 2 + 3 … … … n = [n(n+1)/2]

Read More: Ordered Pairs

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Principle of Mathematical Induction Statement

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Let us now establish the mathematical induction principle and explain how it is used to prove claims step by step:

Consider the following natural number n proposition P(n):

• The statement is true for n = 1, i.e., P(1) is true.
• If the proposition is true for n = k, where k is a natural integer, it is also true for n = k+1, implying that the truth of P(k) implies the truth of P(k+1).

As a result, P(n) is true for all natural numbers n.

Step 1- Prove P(1) is true

Step 2- Assume P(k) is true for some n=k

Step 3- Prove P(k+1) is true

Before getting started answering problems utilising the Principle of Mathematical Induction, let's go over some key points.

• The first step is to state a fact. 
• Some mathematical statements are correct for n ≥ 5. 
• In this scenario, step 1 will begin with n = 5 to demonstrate the outcome using the Principle of Mathematical Induction.
• The second step is a conditional attribute. 
• The Principle of Mathematical Induction does not claim that proposition P(n) is true for n=k. 
• Instead, it asserts that if P(k) is true for some natural number k, then P(k+1) must also be true. 
• In other words, the principle requires you to assume that P(k) is true and then demonstrate that P(k+1) is also true.

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Mathematical Induction- Steps to Solve

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Each step required to prove the theorem or proposition using mathematical induction now has a name. Each stage is labelled as follows:

• Base step- To prove P(1) is true.
• Assumption step- Assume that P(k) is true for some k in N
• Induction step- Prove that P(k+1) is true.

After proving these three stages, may state that "P(n) is true for all n in N according to the principle of mathematical induction." The Induction hypothesis refers to the assumption made in the second step that P(n) holds for some natural integer n = k.

Read More: Types of Sets

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Application of Mathematical Induction 

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Now that students have a better understanding of the notion of mathematical induction, let's look at an example to see how it might be used.

Example 1: Using the idea of mathematical induction, demonstrate that the formula for the sum of n natural numbers holds for all natural numbers, that is, 1 + 2 + 3 + 4 + 5 +.... + n = n(n+1)/2.

Solution: Suppose P(n): 1 + 2 + 3 + 4 + 5 + .... + n = n(n+1)/2

Utilize the principle of mathematical induction to demonstrate this in three steps.

Base Step: To prove P(1) is true.

For n = 1, LHS = 1

RHS = 1(1+1)/2 = 2/2 = 1

Hence LHS = RHS ⇒ P(1) is true.

Assumption Step: Assume P(n) holds for n = k, i.e. P(k) is correct.

⇒ 1 + 2 + 3 + 4 + 5 + .... + k = k(k+1)/2 --- (1)

Induction Step: Shall now demonstrate that P(k+1) is correct.

To prove: 1 + 2 + 3 + 4 + ... + (k+1) = (k+1)(k+2)/2

Consider LHS = 1 + 2 + 3 + 4 + ... + (k+1)

= 1 + 2 + 3 + 4 + ... k + (k+1)

= (1 + 2 + 3 + 4 + ... + k) + k+1

= k(k+1)/2 + k+1 [Using (1)]

= [k(k+1) + 2(k+1)]/2

= (k+1)(k+2)/2

= RHS

⇒ P(n) is true for n = k+1

As a result, P(n) is true for all natural numbers n, according to the concept of mathematical induction.

Example 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, LHS = RHS

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

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Things to Remember 

• For a natural number n, each mathematical statement is assumed to be P(n).
• Start with n = 1, then suppose n = k, and ultimately establish n = k+1.
• After writing the kth phrase (before the (k+1)th word), the result of the "assumption step" is used.
• If deriving the RHS from the LHS appears challenging, separate the LHS and the RHS and demonstrate that they are equal.
• The mathematical Induction is used to determine if a statement is true for all natural numbers.
• It's a technique used to prove mathematical statements for natural numbers.

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