Missing Inductive Step — The Domino Chain With a Gap at Tile 7

The beauty of induction is the chain reaction. The danger of induction is that a chain is only as strong as its weakest link. If your inductive step fails at even one value of k, the fall stops at that point — and every statement beyond it is, from the proof's perspective, completely unjustified.

The animation below makes this visible. A row of dominoes is set up correctly for positions 1 through 6, but between tile 6 and tile 7 the spacing is too wide. Push the first tile, watch the chain reaction march through tiles 1, 2, 3, 4, 5, 6 — and then stop. Tiles 7 through 20 stand untouched, no matter how hard you push the first one.

This is exactly what happens to an induction proof when the inductive step fails somewhere. You do not notice the failure by looking at the base case. You do not notice it by checking the first few cases by hand. You notice it only when you examine the step P(k) \Rightarrow P(k+1) for an arbitrary k, and realise there is some k for which the implication breaks.

The gap visualisation

Gap position k (step P(k)⇒P(k+1) fails):

Twenty dominoes with a gap at the slider's chosen position. Click Play and watch the chain fire from tile 1 forward — until it reaches the gap, where the inductive step $P(k) \Rightarrow P(k+1)$ fails and the fall stops dead. Move the slider to change where the gap sits; the number of unproven tiles changes with it. The picture says: a single broken step kills everything beyond it.

What the gap represents in a real induction proof

In a real proof, the "gap" is a value of k for which the argument P(k) \Rightarrow P(k+1) does not go through. This can happen in several ways:

The algebra silently breaks at a specific k. You wrote an expression that divides by k - 6, say, and the step fails when k = 6 because the denominator vanishes. Every step for k \neq 6 works; the k = 6 step is the gap.
The argument relied on an implicit assumption that fails for small k. You cited "P(k-1) and P(k) both hold" but the inductive hypothesis only gave you P(k), not P(k-1). The step works from k = 2 onward but not for k = 1 (or wherever the first hypothesis is missing).
A case split in the inductive step misses a case. You proved P(k) \Rightarrow P(k+1) when k is even, and proved it when k is odd and k > 3, but forgot the odd k \leq 3. The chain has small gaps near the start.

In all three failure modes, the visible behaviour is exactly what the animation shows: the chain reaction fires for a while and then stalls, leaving a trailing region where the statement is unverified. Often a student testing P(1), P(2), P(3) by hand will see all three come out true and assume the proof is correct — because the base case and the first few steps happen to work, and the gap is further up the chain.

Why checking P(1), P(2), P(3) by hand is not enough: those individual checks verify the statement at those specific values, but they say nothing about whether the inductive mechanism continues beyond them. The proof needs the inductive step to work at every k, not just at small k. A chain of tested values followed by an unverified inductive step at k = 7 is exactly the gap pattern — and exactly the kind of bug that slips past undergraduate homework.

A concrete statement where the gap appears

Here is a made-up but honest illustration. Consider the false claim "for every positive integer n, we have n < 7."

Base case. P(1): is 1 < 7? Yes.
Inductive step attempt. Assume P(k): k < 7. We want P(k+1): k + 1 < 7.

The inductive step does not work in general. For most k (say k = 1, 2, 3, 4, 5), the step produces k + 1 < 7, which is true — the inductive step "fires." But at k = 6, the inductive step says: if 6 < 7 then 7 < 7, which is false. The implication P(6) \Rightarrow P(7) is the gap.

If you had a naive induction proof of this claim and only checked the base case plus k = 1, 2, 3, everything would look fine. You would conclude P(n) for all n, which is false for n = 7, 8, 9, \dots The proof broke at the gap, and the gap was invisible until you interrogated the inductive step at every k.

What to do when you find a gap in your own proof

When you are writing an induction proof and you find a k for which the step fails, there are three recovery strategies:

Strengthen the inductive hypothesis. Instead of proving P(n), prove a stronger statement Q(n) for which the step does work at every k, and observe that Q(n) implies P(n).
Case-split the gap. Handle the problematic k as a separate direct calculation, then show the inductive step works for all other k. The base case may now need to extend — verify P(1), P(2), \dots, P(k_0) directly, and run induction from k_0 onward.
Accept that induction may be the wrong tool. If no repair works, the statement may not be provable by induction. Try a different technique — direct proof, contradiction, or a closed-form algebraic argument.

The gap diagnostic is almost always useful: even if you cannot fix the proof, understanding exactly where and why the chain broke tells you something about the structure of the statement you are trying to prove.

The one-line takeaway

If any inductive step fails at any single k, the chain stops there — and nothing beyond that k is proved. The base case and the first few correct steps are not enough. The inductive step must hold at every k, or the fall never reaches the right end of the chain.