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Principle of Mathematical Induction - Motivation

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Nature of Mathematical Induction: It is a specialized technique used to prove mathematical statements involving natural numbers n∈Nn \in \mathbb{N}. Unlike deductive reasoning which moves from general to specific, induction builds a general conclusion from a specific base case and a logical chain of inheritance.

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The Domino Analogy (Visual): Imagine an infinite row of dominoes stood on end. To ensure every domino falls, you need two things: first, the first domino must be pushed over (Base Case); second, the distance between any two adjacent dominoes must be such that if the kthk^{th} domino falls, it must necessarily knock down the (k+1)th(k+1)^{th} domino (Inductive Step).

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The Infinite Ladder Concept (Visual): Think of a ladder extending infinitely upwards. To reach every rung, you must be able to step onto the first rung, and you must have a consistent method to move from any rung kk to the next rung k+1k+1. If these two conditions are met, you can reach any height nn.

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Base Case (P(1)P(1)): This is the foundation of the proof. It involves verifying that the statement P(n)P(n) holds true for the smallest natural number, usually n=1n=1. Visually, this is like checking if the engine of a train starts successfully.

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Inductive Hypothesis (P(k)P(k)): We assume the statement P(n)P(n) is true for some arbitrary positive integer kk. This is not a proof for kk, but a temporary assumption used as a 'bridge' to reach the next value.

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Inductive Step (P(k+1)P(k+1)): This is the logical core where we prove that if P(k)P(k) is true, then P(k+1)P(k+1) must also be true. This step establishes the 'link' in the chain. Once this link is proven, the truth of P(1)P(1) triggers P(2)P(2), which triggers P(3)P(3), and so on for all nn.

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Motivation for PMI: In mathematics, verifying a formula for n=1,2,3...n=1, 2, 3... up to 100100 does not prove it is true for n=101n=101. PMI provides a rigorous way to prove a statement for an infinite set of natural numbers without having to check each one individually.

📐Formulae

Statement notation: P(n)P(n) where n∈{1,2,3,… }n \in \{1, 2, 3, \dots\}

Basis of Induction: P(1)P(1) is true

Inductive Step: P(k)  ⟹  P(k+1)P(k) \implies P(k+1) for any k∈Nk \in \mathbb{N}

Sum of first nn natural numbers: ∑i=1ni=1+2+3+⋯+n=n(n+1)2\sum_{i=1}^{n} i = 1+2+3+\dots+n = \frac{n(n+1)}{2}

Sum of first nn odd numbers: 1+3+5+⋯+(2n−1)=n21+3+5+\dots+(2n-1) = n^2

Sum of squares of first nn natural numbers: 12+22+32+⋯+n2=n(n+1)(2n+1)61^2+2^2+3^2+\dots+n^2 = \frac{n(n+1)(2n+1)}{6}

💡Examples

Problem 1:

Prove that for all n∈Nn \in \mathbb{N}, the sum of the first nn natural numbers is given by P(n):1+2+3+⋯+n=n(n+1)2P(n): 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}.

Solution:

Step 1: Base Case. For n=1n=1, LHS=1LHS = 1. RHS=1(1+1)2=1RHS = \frac{1(1+1)}{2} = 1. Since LHS=RHSLHS = RHS, P(1)P(1) is true. \nStep 2: Inductive Hypothesis. Assume P(k)P(k) is true for some k∈Nk \in \mathbb{N}, i.e., 1+2+3+⋯+k=k(k+1)21+2+3+\dots+k = \frac{k(k+1)}{2}. \nStep 3: Inductive Step. We need to prove P(k+1)P(k+1) is true. LHS=(1+2+3+⋯+k)+(k+1)LHS = (1+2+3+\dots+k) + (k+1). Using our hypothesis, we replace the sum of kk terms: LHS=k(k+1)2+(k+1)LHS = \frac{k(k+1)}{2} + (k+1). Factoring out (k+1)(k+1), we get (k+1)(k2+1)=(k+1)(k+22)=(k+1)((k+1)+1)2(k+1)(\frac{k}{2} + 1) = (k+1)(\frac{k+2}{2}) = \frac{(k+1)((k+1)+1)}{2}. This is exactly P(k+1)P(k+1). \nConclusion: By the Principle of Mathematical Induction, P(n)P(n) is true for all n∈Nn \in \mathbb{N}.

Explanation:

We first verified the formula for the smallest natural number. Then, we assumed it works for a general number kk and used that assumption to show it must work for the next number k+1k+1, completing the logical chain.

Problem 2:

Use the motivation of PMI to show that P(n):2n>nP(n): 2^n > n for all n∈Nn \in \mathbb{N}.

Solution:

Step 1: For n=1n=1, 21=22^1 = 2 and 2>12 > 1, so P(1)P(1) is true. \nStep 2: Assume 2k>k2^k > k for some k≥1k \geq 1. \nStep 3: To prove P(k+1)P(k+1), we look at 2k+12^{k+1}. 2k+1=2×2k2^{k+1} = 2 \times 2^k. From our hypothesis, 2k>k2^k > k, so 2×2k>2k2 \times 2^k > 2k. We know 2k=k+k2k = k + k. Since k≥1k \geq 1, k+k≥k+1k + k \geq k + 1. Thus, 2k+1>k+12^{k+1} > k+1. \nThis proves the inductive step.

Explanation:

This example shows induction applied to inequalities. The key is to manipulate the kthk^{th} term to look like the (k+1)th(k+1)^{th} term while maintaining the inequality sign.