Answer: P(n) is simply false for n=2 onward, so no valid inductive step can exist (any apparent proof has an algebraic error).
- A The base case itself is actually wrong here, since one squared does not truly equal one in this context
- B P(n) is simply false for n=2 onward, so no valid inductive step can exist (any apparent proof has an algebraic error)
- C Induction as a method rarely applies cleanly to quadratic statements like n<sup>2</sup> = n in general cases like this
- D The flaw lies mainly in how the final concluding sentence of the proof happens to be phrased, not in the underlying logic
Correct answer: B. P(n) is simply false for n=2 onward, so no valid inductive step can exist (any apparent proof has an algebraic error)
Explanation: Since P(2): 4 = 2 is false, the statement is false in general. Any inductive step that appears to prove P(k+1) from P(k) here must contain a hidden algebraic error, since a true base case plus a genuinely valid inductive step cannot produce a false conclusion.
Mathematical induction works like a row of dominoes: proving the base case P(1) tips the first domino, and proving the inductive step (P(k) ⟹ P(k+1)) guarantees each domino knocks over the next - together these two facts guarantee ALL dominoes fall, without checking each one individually.
Concept context
A proof technique used to establish that a statement is true for every natural number, using a base case and an inductive step.