Question

Find the error in the following "proof" that all positive integers \(n\) are equal: Let \(p(n)\) be the statement that all numbers in an \(n\)-element set of positive integers are equal. Then \(p(1)\) is true. Now assume \(p(n-1)\) is true, and let \(N\) be the set of the first \(n\) integers. Let \(N^{\prime}\) be the set of the first \(n-1\) integers, and let \(N^{\prime \prime}\) be the set of the last \(n-1\) integers. By \(p(n-1)\), all members of \(N^{\prime}\) are equal, and all members of \(N^{\prime \prime}\) are equal. Thus, the first \(n-1\) elements of \(N\) are equal and the last \(n-1\) elements of \(N\) are equal, and so all elements of \(N\) are equal. Therefore, all positive integers are equal.

Short answer

The error is that the inductive step assumes incorrect overlap between \(N'\) and \(N''\), leading to a false conclusion.

Step-by-step solution

Identify the Inductive Base Case

The base case of the proof states that for a 1-element set of positive integers, all numbers are equal. This is trivially true, as the set has only one number, so there is no other number to compare against.

Understand the Inductive Hypothesis

The proof assumes that the statement is true for a set of size \(n-1\), meaning all numbers in any \((n-1)\)-element set of positive integers are equal.

Define the Sets Involved

Define \(N\), \(N'\), and \(N''\): \(N\) is a set of \(n\) integers, \(N'\) consists of the first \(n-1\) integers of \(N\), and \(N''\) consists of the last \(n-1\) integers of \(N\).

Apply Inductive Hypothesis

By the inductive hypothesis, the proof claims that all integers in \(N'\) are identical, and all integers in \(N''\) are identical.

Identify the Critical Assumption

The proof assumes without justification that two different, partially overlapping sets, \(N'\) and \(N''\), share common integers. This would imply that all elements in \(N\) are identical because they supposedly overlap enough to ensure equality.

Spot the Logical Flaw

Consider the smallest set \(N\) with 2 elements, \(N = \{1, 2\}\). Here, \(N' = \{1\}\) and \(N'' = \{2\}\). In this case, \(N'\) and \(N''\) do not share any elements, so the claim that their members are equal cannot hold.

Uncover the Conclusion's Fault

Since the assumed overlap of equal elements does not exist in the smallest non-trivial case, the proof's inductive step fails. Therefore, the conclusion that all positive integers are equal cannot logically follow.

Key concepts

Proof by Induction

Proof by induction is a powerful technique used in mathematics to establish the truth of an infinite number of statements. It involves two main steps: the base case and the inductive step. The base case verifies that the statement holds for the initial value, usually 1 or 0. For example, when proving a property for all positive integers, we show it is true for the smallest integer (often 1).
In the original exercise, the base case confirms that a set with just one integer contains equal elements, which is trivially true.
The inductive step assumes the statement is true for some arbitrary positive integer, let's say \(n = k\), and then demonstrates it also holds for \(n = k+1\). If both steps are successful, then by induction, the statement is true for all integers greater or equal to the base case. However, if any part of this logical chain fails, the proof is invalid.
Importantly, each step must be justified; an assumption made in the inductive step without proper validation can lead to errors, as seen in the exercise.

Logical Flaws

Logical flaws in mathematical proofs can lead to incorrect conclusions. They often occur when assumptions are made without proper justification or when logical steps are skipped or falsely established. In the context of the exercise, the logical flaw arises during the inductive step.
The proof incorrectly assumes that two overlapping sets, \(N'\) and \(N''\), are equal, which leads to the conclusion that all positive integers within the set \(N\) are equal. This assumption is flawed because overlapping sets don't necessarily contain all the same elements.

• For example, suppose \(N' = \{1\}\) and \(N'' = \{2\}\) when \(n = 2\). These sets do not overlap, and thus, the assertion that their members are equal is incorrect.

Such logical missteps show how essential it is to ensure that each assumption in a proof is fully validated before confidently extending the claim to larger sets. Failing to do so can lead to dramatically erroneous conclusions, like assuming all positive integers are the same.

Base Case

The base case is the starting point in a proof by induction. It establishes the truth of the statement for the smallest possible value. This might seem straightforward, but it is crucial for laying the foundation of the induction process.
In the exercise's "proof," the base case checks whether a single-element set of positive integers contains elements that are all the same. Since a set with one element inherently has all elements equal, this part of the proof is valid.

• The base case serves as the anchor of the inductive argument. Without it, one cannot proceed to create a chain of inductive steps supporting the broader claim.

Without a clearly stated and valid base case, the entire structure of an inductive proof lacks the foundational support needed for subsequent steps. The essence of a strong base case is its simplicity and undeniable validity, ensuring that when the induction proceeds, it does so from a solid and verified starting point.