Problem 4 - A Gentle Introduction to the Art of Mathematics

Question

Suppose that the rules of the game for PMI were changed so that one did the following: \- Basis. Prove that \(P(0)\) is true. \- Inductive step. Prove that for all \(k, P_{k}\) implies \(P_{k+2}\) Explain why this would not constitute a valid proof that \(P_{n}\) holds for all natural numbers \(n\). How could we change the basis in this outline to obtain a valid proof?

Step-by-Step Solution

Answer

The current proof skips some values (like P(1)). Fix it by proving both P(0) and P(1) in the base case.

Step 1 - Understand the Current Proof Outline

The outline given uses two steps: proving the base case for n=0 and the inductive step where proving for a generic k implies the statement for k+2. This leaves out proving for k+1.

Step 2 - Identify the Issue

In standard PMI (mathematical induction), the inductive step usually proves that if the statement is true for k, it must also hold for k+1. Here, the step skips k+1 and directly goes to k+2, potentially missing intermediate values.

Step 3 - Consider Specific Values

If the rule only proves the statement for n=0 and n=2, values like n=1 and n=3 are unchecked. The base case alone does not ensure those values if the inductive step skips them.

Step 4 - Adjust the Base Case

To validate the proof, change the base case to prove that the statement holds for both P(0) and P(1). This ensures all subsequent statements can be reached through the inductive step. This updated base strengthens the induction process, covering every natural number by either P(k) or P(k+1) going to P(k+2).

Key Concepts

Base Case

The base case is the first step in a proof by mathematical induction. In this step, we prove that our statement holds true for the initial value. Usually, this initial value is the smallest natural number, commonly 0 or 1. Proving the base case creates a foundation for the rest of the proof.
For example, if we're proving the statement for all natural numbers starting from 0, we need to show that it holds for \(P(0)\). This gives us a starting point.

• If we have the base case as \(P(0)\), we know the statement is true for n = 0.

When adjusting an induction proof to cover gaps, like in the provided exercise, we might need to add another base case, such as \(P(1)\). This adjustment helps to cover all possible values without skipping any number.

Inductive Step

The inductive step is the second part of proof by induction. Here, we prove that if the statement holds for an arbitrary natural number \(k\), it also holds for \(k+1\).
In the modified rules of the game for PMI (provided in the exercise), they applied the inductive step to prove \(P(k)\) implies \(P(k+2)\). This step means that knowing \(P(k)\) is true, we need to show that \(P(k+2)\) is also true.

• This method may skip some natural numbers, leaving them unchecked.
• For example, proving \(P(0)\) and then proving \(P(2)\) from \(P(0)\) does not prove that \(P(1)\) is true.

To ensure the proof covers every natural number, it's vital to prove that the statement works for the next consecutive number, typically from \(k\) to \(k+1\). By adjusting the inductive step carefully, we ensure the statement holds for every subsequent natural number.

Natural Numbers

Natural numbers are the set of positive integers starting from 0 or 1, depending on the context. They are usually denoted by \( \mathbb{N} \) and include numbers like 0, 1, 2, 3, and so on.
In the context of mathematical induction, natural numbers provide the domain for which the statement needs to be proven. We aim to show that a claim holds for every element in this set.
Let's break down important aspects:

• Natural numbers are sequential and continuous.
• Each number in a sequence is followed by another natural number, with no gaps in between.

When using induction, we leverage this continuity.
The exercise mentioned involves proving statements for these numbers. However, when skipping from \(k\) to \(k+2\), intermediate natural numbers like \(k+1\) might be skipped, leading to an incomplete proof for the entire set.