1st PUC Mathematics · Chapter 4

Principle of Mathematical Induction — Why the Two-Step Method Actually Proves Anything

Mathematical Induction is the chapter students find strangest — not because the method is hard, but because it’s rarely explained why it works. Most students learn it as “prove for n=1, assume for n=k, prove for n=k+1” without ever seeing why that sequence of steps constitutes a valid proof for every natural number. This guide fixes that gap.

The Domino Analogy — Why Induction Works

Picture an infinite line of dominoes. If you can guarantee two things — the first domino falls, and every domino that falls knocks over the next one — then every single domino in the line falls, no matter how long the line is. That’s the entire logic of induction. You never check every domino individually; you check that the chain reaction is guaranteed to continue forever once it starts.

The Two Steps, Precisely

Step 1 — Base Case: Prove the statement P(n) is true for n = 1 (or whatever the starting value is). This is the “first domino falls” step, and it’s usually the easiest part — just substitute n = 1 directly and verify both sides match.

Step 2 — Inductive Step: Assume P(k) is true for some arbitrary natural number k (this assumption is called the inductive hypothesis) — then use that assumption to prove P(k+1) is also true. This is the “every falling domino knocks the next one” step, and it’s where nearly all the actual algebra happens.

Once both steps are proven, the conclusion is that P(n) is true for all natural numbers n ≥ 1 — not because you checked every value, but because the base case plus the unbroken chain of implications covers every one.

Worked Example — Sum of the First n Natural Numbers

Prove: 1 + 2 + 3 + … + n = n(n+1)/2 for all n ∈ N.

Base case (n=1): LHS = 1. RHS = 1(1+1)/2 = 1. LHS = RHS, so P(1) is true.

Inductive step: Assume P(k) is true: 1 + 2 + … + k = k(k+1)/2. We need to prove P(k+1): 1 + 2 + … + k + (k+1) = (k+1)(k+2)/2.

Starting from the assumption, add (k+1) to both sides: 1 + 2 + … + k + (k+1) = k(k+1)/2 + (k+1). Factor out (k+1) on the right: = (k+1)[k/2 + 1] = (k+1)(k+2)/2. This exactly matches what P(k+1) claims, so the inductive step holds. By induction, the formula is true for all n ∈ N.

Worked Example — Divisibility

Prove: 3ⁿ − 1 is divisible by 2, for all n ∈ N.

Base case (n=1): 3¹ − 1 = 2, which is divisible by 2. True.

Inductive step: Assume 3ᵏ − 1 = 2m for some integer m (this is what “divisible by 2” means algebraically). We need: 3^(k+1) − 1 is divisible by 2. Rewrite 3^(k+1) = 3 · 3ᵏ = 3(2m + 1) = 6m + 3, using the assumption 3ᵏ = 2m + 1. So 3^(k+1) − 1 = 6m + 2 = 2(3m + 1), which is clearly divisible by 2. The inductive step holds, so the statement is true for all n by induction.

Where Students Actually Lose Marks

The single biggest mistake is skipping or rushing the base case — examiners specifically check that it’s shown explicitly, not assumed. The second is failing to actually use the inductive hypothesis inside the P(k+1) proof — if your algebra for k+1 never references the assumption from step k, you haven’t proven anything by induction, you’ve just proven the k+1 case directly (which usually isn’t possible without the assumption anyway). Always write the assumption explicitly before starting the k+1 algebra.

Frequently Asked Questions

Does the base case always have to be n = 1?

No — it starts wherever the statement is claimed to be true from. Some problems state “for all n ≥ 3,” in which case the base case is n = 3, not n = 1.

Why is it called an “assumption” if we’re proving something?

Because within the inductive step, we’re not proving P(k) is true in general — we’re proving the conditional statement “if P(k) is true, then P(k+1) is true.” Combined with the base case actually being true, the chain carries forward validly.

What kinds of statements is induction used for?

Summation formulas, divisibility statements, and inequalities involving natural numbers are the three types that appear most often in 1st PUC exams — the method above applies to all three with only minor changes in the inductive-step algebra.

Master the Logic, Not Just the Steps

Induction rewards students who understand the reasoning over those who memorise the template. Daniel Sir teaches it that way. See the PUC Mathematics Programme →