Question

Find the flaw with the following proof that aⁿ=1 for all nonnegative integers n, whenever a is a nonzero real number. Basis step: a⁰=1 is true by the definition of a ⁰. Inductive step: Assume that aʲ =1 for all nonnegative integers j with j≤k. Then note that aᵏ⁺¹= aᵏ.aᵏ/aᵏ⁻¹= 1.1/1=1

Answer 1

The flaw in the proof is that it assumes that if aⁿ = 1 for all nonnegative integers j with j ≤ k, then aᵏ⁺¹ = 1.

Step-by-step explanation:

The flaw in the proof is that it assumes that if aⁿ = 1 for all nonnegative integers j with j ≤ k, then aᵏ⁺¹ = 1. To demonstrate the flaw, we can consider a counterexample. Let's take a = 2 and k = 1. According to the assumption, 2¹ = 1. However, when we calculate aᵏ⁺¹ = 2² = 4, it does not equal 1.