Inductive Proof Quiz: Practice Mathematical Induction Basics

Rewriting 3^(k+1) = 3·3^k allows substitution of the inductive hypothesis. Then 3^(k+1) - 1 = 3(2m+1) - 1 = 6m+2 = 2(3m+1), which is clearly even. Option B assumes what must be proved. Options A and C skip essential reasoning.

The base case guarantees the first domino falls and the implication P(k)→P(k+1) ensures every subsequent domino falls in turn. Together they give P(n) for every n at or beyond the base. Option D is incorrect because the chain is unbroken — every value is reached, not just infinitely many scattered values.

The answer is True. Strong induction assumes P(1) through P(k) all hold, which includes the weaker assumption that just P(k) holds. Any argument using only P(k) to derive P(k+1) can be replicated under strong induction since P(k) is a subset of the stronger hypothesis. The converse is not generally true.

The Induction Axiom and the Well-Ordering Principle — every nonempty subset of ℕ has a least element — are logically equivalent and each can be derived from the other. This equivalence grounds induction in the foundational structure of the natural numbers rather than in empirical observation or geometric axioms.

Despite the name, mathematical induction is fully deductive. From the base case and the proven implication P(k)→P(k+1) we logically conclude P(n) holds for all n, with no uncertainty or generalization from examples. Statistical induction generalizes from observed samples and carries probability rather than certainty, making it fundamentally different.

The base case must correspond to the smallest n for which the claim is asserted. If the theorem states P(n) for all n≥4, the base case is n=4. If it states n≥0, the base case is n=0. Neither 0 nor 1 is universally required. Option D is false — a missing base case invalidates any induction proof regardless of the inductive step.

From P(1) and the step k→k+2 the chain gives P(1), P(3), P(5), P(7), … covering all odd integers ≥ 1. Even integers are unreachable from an odd starting point with a step of 2. The proof is meaningful but limited in scope to odd values.

Since the inductive step reduces k+1 by subtracting 4 or 5, the proof needs four consecutive verified cases so every reduction lands on an already-proven value. Verifying 12, 13, 14, 15 provides this foundation. Verifying only P(12) is insufficient because the step might reduce to 13, 14, or 15 which have not been checked.

Strong induction assumes every case from the base through k is true simultaneously. This collection of facts gives more leverage when proving P(k+1), particularly in arguments where the value at k+1 depends on earlier values other than just k. Weak induction is the special case where only P(k) is needed.

In strong induction the hypothesis assumes P(1), P(2), …, P(k) all hold, not just P(k) alone. This richer assumption is useful when proving P(k+1) requires more than just the immediately preceding case, such as in number theory or recursive sequences. The conclusion is equally strong — P(n) holds for all n.

Mathematical induction uses a base case and a step from n=k to n=k+1, directly mirroring how the natural numbers are constructed. This sequential structure makes induction ideal for propositions indexed by n∈ℕ. Real numbers, complex numbers, and rationals lack the discrete successor structure that induction requires.

The claim starts at n=1 so the base case verifies n=1. Computing 3^1 - 1 = 2 which is divisible by 2 confirms the property holds at the anchor. Starting at n=2 skips the first required value. Starting at n=0 gives 0 which works but is outside the stated range. Option C is the inductive hypothesis.

The goal of the inductive step is to prove P(k+1): the sum of the first k+1 integers equals (k+1)(k+2)/2. Starting from P(k) and adding k+1 to both sides gives k(k+1)/2 + (k+1) = (k+1)(k/2+1) = (k+1)(k+2)/2, completing the step. Option A is just the inductive hypothesis restated.

P(k) is the formula instantiated at n=k, asserting the sum of the first k integers equals k(k+1)/2. This is the inductive hypothesis from which P(k+1) is derived by adding k+1 to both sides and simplifying. Option A is the general statement. Option B is P(k+1). Option C is P(1).

P(k+1) is obtained by substituting k+1 for n in the general statement. The inductive step assumes P(k) — a k-element set has 2^k subsets — and derives P(k+1) by considering what happens when one new element is added to the set, doubling the number of subsets.

The claim begins at n=4, so the base case must verify the statement at the lowest claimed value. Checking 4! = 24 > 16 = 2^4 confirms it. Starting at n=1 or n=0 would check values where the statement is actually false, which cannot serve as the anchor. Option C is the inductive hypothesis, not the base case.

The base case establishes the first true instance and the inductive step shows truth propagates forward. Both are strictly necessary — a base case without an inductive step proves only one value, and an inductive step without a base case has no anchor and proves nothing. Options A, B, and D do not correctly name the two standard components.

The inductive step proves that truth at step k guarantees truth at step k+1. Combined with the base case, this domino implication covers every natural number beyond the starting point. Option A reverses the direction incorrectly. Option B only handles one transition. Option C always returns to the first case rather than advancing.

The inductive hypothesis assumes P(k) holds for an arbitrary but fixed integer k. From this assumption you derive that P(k+1) must also be true, creating the forward chain. Assuming P(1) alone gives you nothing for general k. Assuming P(k+1) is circular. Assuming the statement is false leads to contradiction, a different technique.

The base case anchors the entire chain by verifying the statement holds for the smallest value, typically n=0 or n=1. Without it, even a valid inductive step would have no starting point and the proof would establish nothing. Option A comes after the base case. Option D is part of the inductive step, not a separate component before it.