Answer
## Summary of the Mistake
1. The base case \(n=1\) is correct but **does not** address \(n=2\).
2. When \(k=2\), the method tries to create two subsets of size 2 from a 3-element set \(A\). Those subsets each omit one different element.
3. Because of insufficient overlap (they share only \(a_1\)), the proof does not (and cannot) prove that \(a_2\) and \(a_3\) are equal.
4. Hence the induction step fails at the transition from \(n=2\) to \(n=3\).
**Conclusion:** The “proof” is invalid because it assumes the inductive hypothesis applies to overlapping subsets in a way that doesn’t hold for small \(k\). This is the classic fallacy in “All Horses Are the Same Color” or “All the numbers in a set are equal.”
Work Step by Step
Below is an explanation of the main error in the “proof” that claims:
> **False Theorem:** “For any integer \(n>1\), all the numbers in a set of \(n\) numbers are equal.”
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## The “Proof” at a Glance
1. **Base Case \((n=1)\)**: A set with just one element obviously has all its (one) elements equal.
2. **Inductive Step** (from \(k\) to \(k+1\)):
- Given a set \(A = \{a_1,a_2,\dots,a_{k+1}\}\) of \(k+1\) numbers, the “proof” forms two subsets:
\[
B = \{a_1,a_2,\dots,a_k\} \quad\text{and}\quad
C = \{a_1,a_3,a_4,\dots,a_{k+1}\}.
\]
- By the inductive hypothesis, all numbers in \(B\) are equal (because \(B\) has \(k\) elements), and all numbers in \(C\) are equal (again, \(k\) elements).
- Concludes: Since every element of \(A\) is in either \(B\) or \(C\), all numbers in \(A\) must be equal.
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## Why This Is Incorrect
The classic flaw (often called the “All Horses Are the Same Color” fallacy) arises in the inductive step when \(k=2\). Here’s what goes wrong:
1. **When \(k=2\)**
- The set \(A = \{a_1, a_2, a_3\}\) has 3 elements.
- The subset \(B\) omits \(a_3\), so \(B = \{a_1, a_2\}\).
- The subset \(C\) omits \(a_2\), so \(C = \{a_1, a_3\}\).
2. **Inductive Hypothesis** says:
- In a 2-element set, the two elements are “all equal.” Indeed, for \(B\), it claims \(a_1=a_2\).
- For \(C\), it claims \(a_1=a_3\).
3. **Overlap Is Too Small**
- The proof wants to conclude from \(B\) and \(C\) that \(a_2=a_3\).
- However, to see that \(a_2=a_3\), one needs a set containing both \(a_2\) and \(a_3\) **plus** an overlapping element that is already “known” to be equal.
- But \(B\) does not contain \(a_3\), and \(C\) does not contain \(a_2\). The two subsets \(B\) and \(C\) only share \(a_1\). We can see \(a_1\) is equal to \(a_2\) (from \(B\)) and \(a_1\) is equal to \(a_3\) (from \(C\)). That part is fine. The proof then tries to claim this implies \(a_2=a_3\). Normally, that *would* be valid—if we are consistent with the inductive hypothesis.
The real snag is that the induction step from \(k=2\) to \(k=3\) is exactly where the method fails in practice. Because for the base case \(k=1\), everything is trivial, but the jump from \(k=1\) to \(k=2\) never addresses a scenario with overlapping sets that have only one element in common. In short, the argument *never* truly checks the critical step that would unify \(a_2\) and \(a_3\) in the same set for small \(k\).
In more direct terms, **the “proof” fails precisely for small sets**, so the induction never “gets off the ground.” After you pass that small threshold incorrectly, the rest of the steps look fine but are built on a faulty foundation.
