{ "format": "recallweave-deck/1", "title": "Mathematical induction: a base, a bridge, every integer", "attribution": "Original questions, options, explanations, transfer prompts and examples authored for Jacob’s RecallWeave project with AI assistance. Background: MIT OpenCourseWare, Mathematics for Computer Science, Chapter 3: Induction (Fall 2010), https://ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2010/resources/mit6_042jf10_chap03/. Standard induction and arithmetic identities are used; no source exercises, passages or figures are copied.", "license": "Original course content authored for Jacob’s RecallWeave project; no additional reuse license is granted here. Reference material retains its published terms.", "concepts": [ "Statement and domain", "Successor algebra", "Finding proof gaps", "Coverage and exact certificates" ], "items": [ { "id": "ind-domain", "concept": "Statement and domain", "prerequisites": [], "prompt": "A statement P(n) is intended for every integer n ≥ 3. Which induction plan reaches exactly that whole intended domain?", "options": [ "Prove P(0), then prove P(n) implies P(n + 1) only for n ≥ 3.", "Prove P(3), then prove P(n) implies P(n + 1) for every integer n ≥ 3.", "Check P(3), P(4) and P(5), then stop.", "Assume P(n + 1), then use it to assert P(n)." ], "answer": 1, "explanation": "The base starts at the smallest intended integer, 3. The successor argument applies at every reachable n: 3 reaches 4, 4 reaches 5, and so on. A base at 0 cannot cross the missing 0-to-3 interval when the given step starts at 3. A finite list alone is not this universal argument.", "transfer": "Suppose the intended domain is n ≥ 7. State a sufficient base and the precise range of the ordinary successor implication." }, { "id": "ind-empty-base", "concept": "Statement and domain", "prerequisites": [], "prompt": "Let S(n) = sum of (2k − 1) for k = 1 through n, with S(0) defined as the empty sum 0. For the claim S(n) = n² on integers n ≥ 0, what is its base check?", "options": [ "S(0) = −1 because substituting k = 0 creates one term.", "S(0) = 1 because the first odd number is 1.", "S(1) = 1 is the required check for a domain starting at 0.", "S(0) = 0 and 0² = 0, so the stated base is true." ], "answer": 3, "explanation": "At n = 0 the index range 1 through 0 has no terms, so its sum is the stated empty sum 0. There is no k = 0 term. The right side is 0² = 0. Checking n = 1 is useful but does not replace the specified base at 0.", "transfer": "For T(n) = sum from k = 1 through n of (4k + 1), state T(0) and T(1). Explain which is the base if the claim starts at 0." }, { "id": "ind-hypothesis", "concept": "Statement and domain", "prerequisites": [], "prompt": "You want to prove 1 + 2 + ⋯ + n = n(n + 1)/2 for every integer n ≥ 0, with the empty sum at 0. Which is a legitimate hypothesis inside the successor step?", "options": [ "For an arbitrary integer n ≥ 0, assume the formula holds at n and use that assumption to derive it at n + 1.", "Assume the formula holds for every integer, which already gives the desired conclusion.", "Assume the formula at n + 1 and present that same expression as the proof.", "Check the formula at n = 10 and treat n as fixed at 10." ], "answer": 0, "explanation": "The hypothesis is conditional at an arbitrary allowed n. You derive the next case using that condition, producing P(n) implies P(n + 1). Separately proving the base gives the starting point for repeated use. Assuming the final universal claim would skip the proof.", "transfer": "Write the hypothesis and target for proving the odd-number sum S(n) = n² by induction starting at 0. Keep the assumed and to-be-proved statements distinct." }, { "id": "ind-odd-step", "concept": "Successor algebra", "prerequisites": [ "Statement and domain" ], "prompt": "Assume sum from k = 1 through n of (2k − 1) equals n². Which calculation justifies adding the next term?", "options": [ "n² + (2n − 1) = (n + 1)².", "n² + 2(n + 1) = (n + 1)².", "n² + [2(n + 1) − 1] = n² + 2n + 1 = (n + 1)².", "n² + (n + 1)² = 2n² + 1." ], "answer": 2, "explanation": "The next index is n + 1, so its odd term is 2(n + 1) − 1 = 2n + 1. Under the hypothesis the new sum is n² + 2n + 1, which factors as (n + 1)². Using the old term 2n − 1 would advance the index incorrectly.", "transfer": "Use the odd-number induction step at n = 4. State the old sum, new term and next sum." }, { "id": "ind-triangular-step", "concept": "Successor algebra", "prerequisites": [ "Statement and domain" ], "prompt": "Assume the sum 1 + 2 + ⋯ + n is n(n + 1)/2. What does adding the next term n + 1 produce?", "options": [ "n(n + 1)/2 + 1 = (n + 1)(n + 2)/2.", "n(n + 1)/2 + n = (n + 1)(n + 2)/2.", "n(n + 1)/2 + (n + 1) = n(n + 2)/2.", "n(n + 1)/2 + (n + 1) = (n + 1)(n + 2)/2." ], "answer": 3, "explanation": "Factor out n + 1: n(n + 1)/2 + (n + 1) = (n + 1)(n/2 + 1) = (n + 1)(n + 2)/2. This is the same proposed formula with n replaced by n + 1. The next term is n + 1, not 1 or n.", "transfer": "Using the triangular formula at n = 4, calculate the fifth partial sum by adding the next term and by substituting into the formula." }, { "id": "ind-base-essential", "concept": "Finding proof gaps", "prerequisites": [ "Successor algebra" ], "prompt": "For S(n) = sum from k = 1 through n of (2k − 1), someone proposes S(n) = n² + 5 for every integer n ≥ 0. The proposed right side increases by exactly 2n + 1 when n becomes n + 1. What does that establish?", "options": [ "The whole claim is true because matching successive differences makes the base unnecessary.", "The successor difference matches, but the base fails: S(0) = 0 whereas 0² + 5 = 5.", "The base is true, but the successor difference is 2n + 5.", "The claim becomes true for all n ≥ 1, because the first case can be skipped." ], "answer": 1, "explanation": "Adding a constant does not change successive differences: (n + 1)² + 5 − (n² + 5) = 2n + 1. However, S(0) is 0 and the proposed base is 5. Without the true base this step cannot establish the claim. In fact the proposed expression stays 5 above the actual sum at every n.", "transfer": "Replace the constant 5 by any integer c in the proposed odd-number formula n² + c. Which c passes the base at 0, and what happens to the successor difference?" }, { "id": "ind-step-essential", "concept": "Finding proof gaps", "prerequisites": [ "Successor algebra" ], "prompt": "A proposed formula for 1 + 2 + ⋯ + n is n² for integers n ≥ 1. Its base at n = 1 is true. What happens at the first successor?", "options": [ "The next sum is 4 because a true base forces the formula to remain true.", "The next sum is 2 because only the new term counts.", "The next sum is 1 + 2 = 3, while the proposed value 2² is 4; the base alone is insufficient.", "The discrepancy is harmless because induction allows one exception after its base." ], "answer": 2, "explanation": "At n = 1 the formula gives the right value 1, but at n = 2 the actual sum is 3 and the proposal is 4. A successful base is one required part, not evidence that every needed implication has been justified. This is a direct counterexample in the claimed domain.", "transfer": "For the same false triangular proposal n², compare the true and proposed totals at n = 3. State the numerical difference." }, { "id": "ind-finite", "concept": "Finding proof gaps", "prerequisites": [ "Successor algebra" ], "prompt": "For S(n) = 1 + 2 + ⋯ + n, consider Q(n) = (3n² − 5n + 4)/2. It matches S(n) at n = 1 and n = 2. Which observation correctly challenges a claim that these two matches prove equality for every n ≥ 1?", "options": [ "At n = 3, S(3) = 6 but Q(3) = 8, so two matching cases did not prove the universal claim.", "At n = 3, both values are 6; matching twice determines every possible formula.", "At n = 3, S(3) = 8 but Q(3) = 6.", "Two matches are sufficient because Q is a polynomial." ], "answer": 0, "explanation": "Q(1) = (3 − 5 + 4)/2 = 1 and Q(2) = (12 − 10 + 4)/2 = 3. But Q(3) = (27 − 15 + 4)/2 = 8, while 1 + 2 + 3 = 6. Two numerical matches leave this false proposal possible; they do not supply induction's all-n step.", "transfer": "Write Q(n) as n(n + 1)/2 plus (n − 1)(n − 2). Use that difference to explain both initial matches and the discrepancy at n = 4." }, { "id": "ind-step-size", "concept": "Coverage and exact certificates", "prerequisites": [ "Finding proof gaps" ], "prompt": "For a statement about nonnegative integers, a proof establishes P(0) and P(n) implies P(n + 2) for every integer n ≥ 0. What coverage follows from those facts alone?", "options": [ "All nonnegative integers, because +2 always increases the input.", "Every nonnegative even integer; reaching the odd integers also needs a starting odd case or another bridge.", "Only P(0), because a step of size 2 cannot be used repeatedly.", "Only the odd integers, because adding 2 changes parity." ], "answer": 1, "explanation": "Adding 2 preserves parity. Repeatedly applying the given implication from 0 reaches 2, 4, 6 and every later even integer. It never reaches 1. A true base P(1) together with the same step would also start the odd chain.", "transfer": "State two base cases that, together with the +2 successor implication, cover every integer n ≥ 0." }, { "id": "ind-missing-bridge", "concept": "Coverage and exact certificates", "prerequisites": [ "Finding proof gaps" ], "prompt": "A proof intends to establish P(n) for all integers n ≥ 4. It checks P(4), but its argument P(n) implies P(n + 1) is valid only for n ≥ 5. What is missing?", "options": [ "Nothing; the difference between 4 and 5 is too small to matter.", "An argument for P(3), because induction must always begin before the claimed domain.", "A calculation of P(100), which bypasses the first step.", "A justification reaching P(5), such as a valid 4-to-5 step or a separate true base at 5." ], "answer": 3, "explanation": "The successor argument is unavailable at n = 4, so the stated proof cannot move from its only known base to 5. Proving P(5) separately would then let the given step cover 6, 7 and every larger integer, while P(4) remains separately checked.", "transfer": "If the successor argument were valid only for n ≥ 6, which additional individual cases would bridge a checked P(4) to that range?" }, { "id": "ind-arithmetic", "concept": "Successor algebra", "prerequisites": [ "Statement and domain" ], "prompt": "Let S(n) = sum from k = 1 through n of (3k + 2), and S(0) = 0. Which proposed formula has the correct base and increases by the next term 3(n + 1) + 2 for every integer n ≥ 0?", "options": [ "S(n) = (3n² + 7n)/2.", "S(n) = (3n² + 4n)/2.", "S(n) = (3n² + 7n + 2)/2.", "S(n) = (3n² + n)/2." ], "answer": 0, "explanation": "At n = 0, (3n² + 7n)/2 is 0. Its successor difference is [3((n + 1)² − n²) + 7]/2 = (6n + 10)/2 = 3n + 5, exactly 3(n + 1) + 2. These two facts prove the stated formula by induction on all n ≥ 0.", "transfer": "Propose and verify a formula for the sum from k = 1 through n of (2k + 3), starting at n = 0." }, { "id": "ind-coefficients", "concept": "Coverage and exact certificates", "prerequisites": [ "Finding proof gaps" ], "prompt": "In an exact symbolic check, the successor residual Q(n + 1) − Q(n) − t(n + 1) is (6n − 6)/2. It is zero at n = 1. Can this residual certify the required successor identity for every integer n ≥ 1?", "options": [ "Yes; one zero residual proves an identity in n.", "Yes; the denominator 2 cancels every numerator.", "No; at n = 2 the residual is 3, so the successor identity fails within the stated domain.", "No; a symbolic expression can never be used in an induction argument." ], "answer": 2, "explanation": "Substituting n = 2 gives (12 − 6)/2 = 3. The required identity is therefore not true at every allowed n. A nonzero linear polynomial can have a particular zero without being identically zero. Exact symbolic coefficients, rather than a single successful substitution, determine that distinction.", "transfer": "A successor residual is (0n + 0)/7. What does that establish about the successor identity? What separate fact is still needed for an induction proof?" } ] }