Question

Let \( P(x) \) be an arbitrary predicate over the domain of natural numbers. Which ONE of the following statements is TRUE?

• A. \(\big( P(0) \land (\forall x [P(x) \Rightarrow P(x+1)]) \big) \Rightarrow (\forall x \ P(x))\)
• B. \(\big( P(0) \land (\forall x [P(x) \Rightarrow P(x-1)]) \big) \Rightarrow (\forall x \ P(x))\)
• C. \(\big( P(1000) \land (\forall x [P(x) \Rightarrow P(x-1)]) \big) \Rightarrow (\forall x \ P(x))\)
• D. \(\big( P(1000) \land (\forall x [P(x) \Rightarrow P(x+1)]) \big) \Rightarrow (\forall x \ P(x))\)

Hint:

Mathematical induction is a crucial technique in proving statements over natural numbers. Always ensure the base case holds and the inductive step properly extends the proof.

Answer 1

The Correct Option is A

Statement (A) represents the principle of mathematical induction.

Mathematical induction states that if:

1. \( P(0) \) is true (base case), and
2. \( P(x) \Rightarrow P(x+1) \) (inductive step),

then \( P(x) \) is true for all \( x \in \mathbb{N} \).

Statement (B) is incorrect because \( P(x) \Rightarrow P(x-1) \) does not ensure all values are covered in the natural number domain, which has a well-defined lower bound at \( 0 \). In other words, the principle works by proving the property holds starting from the base case (typically \( x = 0 \)) and then proving it holds for all subsequent values of \( x \). Reversing the direction (i.e., \( P(x) \Rightarrow P(x-1) \)) does not achieve this and may lead to invalid conclusions.

Statement (C) and (D) assume \( P(1000) \) instead of \( P(0) \), which does not establish \( P(x) \) for all \( x \) in \( \mathbb{N} \). Mathematical induction requires starting from the smallest element in the set (usually \( P(0) \)) to prove the property holds for all natural numbers.

Thus, the correct answer is (A).