Assume that N = {0, 1, 2, 3, . . .}.(a) Suppose that we change the induction mechanism as follows:• Base case: Prove that P (0) is true• Inductive step: Prove that for all k ≥ 0, P (k) ⇒ P (k + 2)Explain why this would not constitute a valid proof that P (n) is true for alln ∈ N. How would you change the base case to obtain a valid proof?
Question
Assume that N = {0, 1, 2, 3, . . .}.(a) Suppose that we change the induction mechanism as follows:• Base case: Prove that P (0) is true• Inductive step: Prove that for all k ≥ 0, P (k) ⇒ P (k + 2)Explain why this would not constitute a valid proof that P (n) is true for alln ∈ N. How would you change the base case to obtain a valid proof?
Solution
The induction mechanism you've described is not sufficient to prove that P(n) is true for all n in N because it only proves P(n) for even numbers. This is because you're starting with a base case of P(0) and then proving P(k) implies P(k+2), which only covers the even numbers in N.
To make this a valid proof for all n in N, you would need to add an additional base case for an odd number, say P(1). Then, you would need to prove that P(k) implies P(k+2) for all k ≥ 1. This way, you would cover both the even and odd numbers in N, thus proving that P(n) is true for all n in N.
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