Question

The following is a (wrong) proof by induction that each nonnegative integer equals twice itself, i.e. n = 2n for all integers n > 0. n+1 Base case n = 0: P(n) holds for n -0. Inductive step: let us assume that n 2n holds for some nonnegative integer n. We multiply both sides of this equation by the quantity This yields n +1 = 2(n + 1), which is P(n + 1). This completes the proof by induction. n = = Explain what is wrong with this proof. As usual, we will use the notation P(n) (n = 2n). The base case is not n = 0, it's n =1 , for which the equation n - 2n is false. 1. The base case is n = 0 but was incorrectly verified. P(0) is false. 2. The conditional P(n) + P(n + 1) is false for all n, and the algebra in the inductive step that verifies this conditional is mistaken in all cases. 3. The conditional P(n) + P(n + 1) is false (only) for n=0, and the algebra in the inductive step that verifies this conditional is mistaken (only) in that case. 04. The conditional P(n) + P(n + 1) is false (only) for n=1, and the algebra in the inductive step that verifies this conditional is mistaken (only) in that case. 05. The conditional P(n) → P(n + 1) is false (only) for n=2, and the algebra in the inductive step that verifies this conditional is mistaken (only) in that case. 6. There is nothing wrong with either the base case or the inductive step. The principle of induction just breaks down in this situation.

Answer 1

The base case of the proof is incorrectly verified, leading to an incorrect conclusion.

Step-by-step explanation:

The wrong part of this proof by induction is that the base case is incorrectly verified. The proof attempts to show that for any nonnegative integer n, n = 2n holds true. However, the base case should be n = 1, not n = 0. The equation n - 2n is false for the base case n = 1.

Therefore, option 1 is the correct answer: The base case is n = 0 but was incorrectly verified. P(0) is false.

Answer 2

The error in the proof is the false assumption that each nonnegative integer equals twice itself, which the inductive step erroneously tries to validate. The base case for n=0 is true, but the conditional P(n) → P(n + 1) is false for all n > 0.

Step-by-step explanation:

The error in this proof by induction lies in the incorrect assumption made during the inductive step. Inductive reasoning requires a correct base case, and then for each step n to n+1, a correct logical argument must demonstrate that if proposition P(n) is true, then proposition P(n + 1) must also be true. In this case, the base case for P(0) is indeed true since 0 = 2*0, but stating that P(n) implies P(n + 1) by simply multiplying both sides of P(n) by (n + 1) results in an inherently faulty conclusion when n ≠ 0. The proposition that any number equals twice itself (n = 2n) is false for any n > 0.

Therefore, the correct answer is that the conditional P(n) → P(n + 1) is false for all n > 0, but the explanation in the question incorrectly assumes it becomes true when multiplying both sides by (n + 1), which is a mistaken application of the inductive step.