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Induction & Recursion · step 9 of 13

Spotting the flaw in a bogus induction proof

Fake induction proofs are a favourite exam device because the error is always in one identifiable place: a missing base case, a step that silently assumes n is large, or a hypothesis applied to something it does not cover. The famous "all horses are the same colour" argument fails precisely at n = 2, where the two overlapping subsets share no horse.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

  1. Test the claim on small valuesIf the conclusion is false, one concrete counterexample tells you the proof must break somewhere.
  2. Check the base case honestlyVerify it rather than accepting the author’s word. Many fakes assert a false base.
  3. Trace the step at the smallest kSubstitute k = 1 or k = 2 into the step and watch for a set that turns out empty or an index that goes out of range.
  4. Name the failing transitionSay exactly which k → k+1 fails and why. "The proof is wrong" earns nothing.

Worked example

Where does the "all horses are the same colour" induction fail?

  1. Claim: in any set of n horses, all have the same colour. Base n = 1 is fine.
  2. Step: given k+1 horses, remove the last to get k horses — all the same colour by hypothesis. Remove the first to get another k, also uniform.
  3. The argument concludes the two groups share a horse, so all k+1 match.
  4. At k+1 = 2, the two groups are {horse 1} and {horse 2}, which have no horse in common.

Answer. The step fails exactly at k = 1, i.e. the transition from n = 1 to n = 2, because the overlap the argument relies on is empty.

Where marks get dropped

These are the specific errors that cost credit on spotting the flaw in a bogus induction proof questions — QED's rubric penalises each of them separately.

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Spotting the flaw in a bogus induction proof — frequently asked questions

What are the common flaw types?

Missing or false base case; a step relying on k ≥ 2 without saying so; applying the hypothesis to a set the hypothesis does not cover; and hidden division by an expression that can be zero.

Why is the horses proof so instructive?

Because every individual sentence is valid and only the overlap assumption fails — it shows that a plausible-looking step needs checking at its smallest instance.

How do I avoid these in my own proofs?

Always substitute the smallest k into the step and check it literally. If the step needs k ≥ 2, add a second base case.

The rest of Induction & Recursion

Mathematical, strong & structural induction; recursion. Each subtopic below has its own method, worked example and mark-losing traps.

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