Understanding Mathematical Induction with Easy Examples

Mathematical induction is a rigorous axiomatic technique for establishing the validity of statements asserted for all natural numbers $n \in \mathbb{N}$, where each such statement involves $n$ as a parameter.

Formal Principle of Mathematical Induction

The principle of mathematical induction states: If $P(n)$ is a statement about $n \in \mathbb{N}$ such that:

(i) $P(1)$ is true; and

(ii) For any $k \in \mathbb{N}$, if $P(k)$ is true, then $P(k+1)$ is also true,

then $P(n)$ is true for all $n \in \mathbb{N}$.

Mathematical Induction Overview provides more foundational context on the principle itself.

Structure of a Mathematical Induction Proof

A proof by mathematical induction consists of two essential steps:

In the base case, the statement $P(1)$ is verified directly.

In the inductive step, assume $P(k)$ holds for some $k \in \mathbb{N}$ (the induction hypothesis), and deduce $P(k+1)$.

Example 1: Induction Proof for $1 + 2 + \ldots + n = \dfrac{n(n+1)}{2}$

Let $P(n): 1 + 2 + \ldots + n = \dfrac{n(n+1)}{2}$.

For $n = 1$:

$1 = \dfrac{1 \cdot 2}{2}$

$1 = 1$

Base case holds.

Assume $P(k)$: $1 + 2 + \ldots + k = \dfrac{k(k+1)}{2}$.

We must prove $P(k+1)$: $1 + 2 + \ldots + k + (k+1) = \dfrac{(k+1)(k+2)}{2}$.

$1+2+ \ldots +k + (k+1) = [1+2+\ldots+k] + (k+1)$

Substitute the induction hypothesis:

$= \dfrac{k(k+1)}{2} + (k+1)$

Write $(k+1)$ with denominator 2:

$= \dfrac{k(k+1) + 2(k+1)}{2}$

Factor $(k+1)$:

$= \dfrac{(k+1)[k+2]}{2}$

This is precisely $P(k+1)$.

By induction, $1 + 2 + \ldots + n = \dfrac{n(n+1)}{2}$ for all $n \in \mathbb{N}$.

Example 2: Induction Proof for the Sum of First $n$ Odd Numbers

Let $P(n): 1 + 3 + 5 + \ldots + (2n - 1) = n^2$.

For $n = 1$:

$1 = 1^2$

Base case holds.

Assume $P(k): 1 + 3 + 5 + \ldots + (2k-1) = k^2$.

Consider $P(k+1):$

$1 + 3 + 5 + \ldots + (2k-1) + (2(k+1) - 1)$

$= [1 + 3 + \ldots + (2k-1)] + [2k+2-1]$

$= k^2 + (2k+1)$ (by the hypothesis)

$= k^2 + 2k + 1$

$= (k+1)^2$

Mathematical Induction Important Questions cover similar cases required for exam preparation.

Example 3: Induction Proof for $1^3 + 2^3 + \ldots + n^3 = \dfrac{n^2(n+1)^2}{4}$

Let $P(n): 1^3 + 2^3 + \ldots + n^3 = \dfrac{n^2(n+1)^2}{4}$.

Base case $n = 1$:

$1^3 = \dfrac{1^2(1+1)^2}{4} \implies 1 = \dfrac{1 \cdot 4}{4} = 1$

Base case is satisfied.

Assume $P(k): 1^3 + 2^3 + \ldots + k^3 = \dfrac{k^2(k+1)^2}{4}$.

Prove $P(k+1): 1^3 + 2^3 + \ldots + k^3 + (k+1)^3 = \dfrac{(k+1)^2(k+2)^2}{4}$.

Write:

$1^3 + 2^3 + \ldots + k^3 + (k+1)^3 = \dfrac{k^2(k+1)^2}{4} + (k+1)^3$

Bring to common denominator:

$= \dfrac{k^2(k+1)^2 + 4(k+1)^3}{4}$

Factor $(k+1)^2$:

$= \dfrac{(k+1)^2 [k^2 + 4(k+1)]}{4}$

Expand $k^2 + 4(k+1)$:

$= \dfrac{(k+1)^2 [k^2 + 4k + 4]}{4}$

$= \dfrac{(k+1)^2 (k+2)^2}{4}$

Inductive step is complete. Thus, the result holds for all $n$.

Forms of Mathematical Induction: Standard and Strong

The standard (weak) form requires assuming validity for $n=k$ to deduce for $n=k+1$.

The strong form (complete induction) allows assumption that $P(j)$ is true for all $j$ with $1 \leq j \leq k$, to prove $P(k+1)$. This is essential for statements where each case depends on several previous instances.

For distinctions relevant to discrete mathematics, refer to Mathematical Reasoning.

Induction Hypothesis and Exam Considerations

The assumption made during the inductive step is termed the induction hypothesis. It is critical not to assume $P(k+1)$ before concluding its validity; this is a standard error in induction proofs.

Some results require multiple base cases; for instance, recurrence relations of order greater than one. Verify all necessary initial cases to ensure the validity of induction.

Common Exam Problems Using Induction

Typical induction statements include: sums of arithmetic or geometric sequences, inequalities involving natural numbers, divisibility results, and summation formulas.

Use Sequences and Series for related formulaic and induction-centric results needed for JEE Main-level proofs.

Conclusion on the Mathematical Induction Principle

Mathematical induction provides a stepwise logical framework for establishing universally quantified statements concerning the set of natural numbers, ensuring both algebraic completeness and deductive rigor.