An induction proof does not prove a statement "for all positive integers" in a floating, context-free way. It proves the statement from the base-case index onward. If you verify P(1) and run the inductive step from there, you have proved P(n) for n \geq 1. If you verify only P(5) and run the inductive step from there, you have proved P(n) for n \geq 5 — and P(1), P(2), P(3), P(4) are still unverified.

This is a detail that trips up many students. The inductive step is the propagation; the base case sets the starting index. Move the starting index, and the region of proved integers moves with it.

The slider below lets you drag the base case from n = 1 up to n = 5 and watch the proved region shrink accordingly.

The shrinking-region visualisation

Number line from $n = 1$ to $n = 10$. The orange marker is the base-case index $n_0$. Everything to its right is the proved region; everything to its left is unproven by this particular induction proof. Toggle to the $2^n > n^2$ statement and watch how setting $n_0 = 5$ lines up exactly with where the statement starts being true (plus the lucky $n=1$ outside the proof's reach).

What the picture is saying

An induction proof has two inputs: the base-case index n_0 and a proof of the inductive step valid for k \geq n_0. What it produces is the statement:

P(n) is true for all integers n \geq n_0.

If n_0 = 1, the conclusion is "P(n) for all positive integers." If n_0 = 5, the conclusion is "P(n) for all integers n \geq 5" — nothing said about n = 1, 2, 3, 4.

This matters when the statement is genuinely false or problematic at small n. Many inequalities and combinatorial identities only hold from some threshold onward. For those, you must start the base case at the threshold, and you must not claim the statement for smaller n — because it is not proved, and often not true.

Why students sometimes misread "proved from n = 5" as "proved for all n": the sentence "by induction, P(n) holds" is ambiguous without a specified starting index. When a student writes just that line without saying "n \geq 5," the reader can default to thinking n \geq 1 is implied. The fix is to always explicitly name the base-case index in the conclusion: "by induction, P(n) holds for all n \geq 5." This is not pedantry; it is what the proof actually established.

A real statement whose base case is not n = 1

Consider the inequality: "2^n > n^2 for all n \geq 5."

The inequality is true at n = 1 by accident (the pattern has not yet started), false at n = 2, 3, 4, and true again from n = 5 onward. A proof by induction starting at n = 5 covers exactly the region where the statement is true (plus the accidental n = 1).

If you wrote "by induction, 2^n > n^2 for all positive integers n," your claim would be false — refuted by n = 2. The correct claim is "2^n > n^2 for n \geq 5." The base-case index is load-bearing in the conclusion.

The inductive step changes with the base-case index

A detail that is easy to miss: when you move the base case from n_0 = 1 to n_0 = 5, the inductive step you must prove changes too. You now need "for all k \geq 5, P(k) \Rightarrow P(k+1)." You do not need the step to work for k = 1, 2, 3, 4 — those indices are outside the region the proof covers.

This flexibility is sometimes useful: if your inductive step only goes through for large k, shifting the base case to where the step starts working is a standard move. The cost is that you give up the small-n cases, but you can often verify those separately by direct checking if they happen to be true.

When to extend the base case block

A related pattern: some proofs need multiple base cases — not just P(5), but P(5), P(6), P(7) — typically because the inductive step looks back more than one step (strong-induction-style). In such a proof, the first c + 1 values must be verified directly, and the claim is "for all n \geq n_0," where n_0 is the smallest index you verified. The slider analogy still applies: move the block of base cases, and the proved region moves with it.

The one-line takeaway

Induction proves the statement from the base-case index onward — no earlier. If your base case is at n_0 = 5, your conclusion is "for all n \geq 5," and you must not make any claim about n < 5. Always write the starting index in the conclusion.

Related: Mathematical Induction · Induction Ladder — Climb From P(k) to P(k+1), One Rung at a Time · Missing Base Case — The Domino Chain That Never Starts · Strong Induction as a Jenga Tower — Each Block Rests on Many Below · Missing Inductive Step — The Domino Chain With a Gap at Tile 7