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Mathematical Induction

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Prerita Agarwal
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23 Jul, 2024 @ 01:30 PM


Mathematical Induction is a methodology for proving natural-number conclusions, a formula or existing theorems.

There are 2 steps involved in order to prove these statements using Induction:

Step 1(Base step) : 
Consider a starting point for the assertion to be true. It will always be true must prove that the statement is correct when n = starting value.


Let n0 be a fixed integer. Suppose T (n) is a statement involving the natural number n, and we wish to prove that P (n) is valid for all n ≥ n0.

T (n0) is true i.e. T (n) is true for n = n0.

Step 2(Induction Step) :
Assuming the statement to be true for any value of n = k. Then prove that the  statement is true for n = k+1.

Let n0 be a fixed integer. Suppose T (n) is a statement involving the natural number n and we wish to prove that P (n) is true for all N ≥ n0.

Assume that the P (k) is true for n = k.

Then P (K+1) should also be true.


  1. Use the Principle of Mathematical Induction to verify that, for n any positive integer, 6n-1 is divisible by 5.

For any n ≥ 1, let Pn be the statement that 6n − 1 is divisible by 5.
Base Case: The statement P1 says that 
6 1 − 1 = 6 − 1 = 5 Eq No. 1
is divisible by 5, which is true. 
Inductive Step: Fix k ≥ 1, and suppose that Pk holds, that is, 6k − 1 is divisible by 5. 

It remains to show that Pk+1 holds, that is, that 6k+1 − 1 is divisible by 5. 

6 k+1 − 1  = 6(6k ) − 1 
  = 6(6k − 1) − 1 + 6 From Eq. No-1

   = 6(6k − 1) + 5

By Pk, the first term 6(6k − 1) is divisible by 5, the second term is divisible by 5. 

Therefore the left-hand side is also divisible by 5. 

Therefore Pk+1 holds.

Thus, P(n) holds true by the principle of mathematical Induction, for all n ≥ 1.

2.  Verify the following mathematical equation using Mathematical Induction:
1 + 3 + 5 + ··· + 2n − 1 = n2 

Base Step:  We are going to prove P(1). 
LHS of P(1) = 1 
= 12
= RHS of P(1).
So P(1) is true.
Inductive Step: Now, we’ll prove that for any natural number k, “if P(k) is true then P(k +1) is true.” So assume P(k) is true, i.e.
1+3+5+ ··· + 2k − 1 = k2
Therefore for P(k + 1): 
LHS of P(k + 1) = 1 + 3 + 5 + ··· + 2k − 1 + 2(k + 1) − 1
    = (LHS of P(k)) + 2(k + 1) − 1
    = (RHS of P(k)) + 2k + 1 (by inductive assumption)
      = k+ 2k +1
    = (k + 1)2 

    = RHS of P(k + 1)

So P(k + 1) is true, if P(k) is true.

Hence, by induction P(n) is true for all natural numbers

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  1. What do mean by Strong Induction?
    Strong Induction is a different form of mathematical induction. Through this, we can prove that a propositional function, P(n) is true for all positive integers n, using the following steps −
    Step 1(Base step) − It proves that the initial proposition P(1) true.
    Step 2(Inductive step) − It proves that the conditional statement [P(1)∧P(2)∧P(3)∧⋯∧P(k)]→P(k+1)
    is true for positive integers k.
  2. What are some of the practical applications of Mathematical Induction?
    It's a great tool for recursively developing algorithms. If we can discover the answer to n using the attributes of n-1, we can construct a recursive method to compute n by n-1.



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In this article we have discussed the ‘Mathematical Induction’, Problem Solving Steps with a few examples. Check out the playlist Basic Mathematics for further topics.

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