Why isn't checking examples a proof by induction?

Checking n = 1, 2, 3 and so on is only finitely many cases. It never covers every n, and a single later value can be a counterexample. The inductive step is what extends the result to all n.

Why isn't checking examples a proof by induction?

Checking n = 1, 2, 3 and so on is only finitely many cases. It never covers every n, and a single later value can be a counterexample. The inductive step is what extends the result to all n.

Year 12 Maths Extension 1 (2027) Proof

Identify errors in false ‘proofs by induction’

Q: Can a proof by induction have a valid step but still be false?

Yes. A false statement can have a perfectly valid inductive step. Without a true base case nothing ever makes the statement start being true, so the whole claim can still be false. Both parts are required.

Q: What is wrong with assuming the result for all n?

In the inductive step you may only assume the statement for the one value n = k, not for all n. Assuming it for all n assumes what you are trying to prove, which is circular.

Q: How do you spot an error in a proof by induction?

Check that the base case is actually true, that the step genuinely derives the k + 1 case from the k case rather than asserting it, that the correct k + 1 term and starting value are used, and that the step size matches the claim, for example k to k + 2 for odd n.

Q: What is the false proof example in the NESA syllabus?

The syllabus shows that the inductive step can be proven for the false proposition that 1 + 2 + 3 + ... + n equals one half of n minus 1 times n plus 2, but the initial case does not hold, so the statement is false.

Q: Is identifying errors in false proofs in Extension 1?

Yes. It is part of the Proof by mathematical induction focus area, outcome ME1-12-01, which asks students to identify errors in false proofs by induction, such as cases where only one of the two required steps is true.

20 practice questions 2 video lessons Theory + worked examples

Sharpen your reasoning by learning to identify errors in false proofs by induction for NSW Year 12 Mathematics Extension 1. A convincing-looking argument can still fail, most often when the inductive step works but the base case is never actually true.

You will learn to test the base case, check that each step genuinely follows from the inductive hypothesis, and pinpoint where a flawed proof breaks down — a critical-thinking induction skill that deepens the rigour examiners reward in Extension 1.

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Practice questions

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Why we need proof by mathematical induction Watch
HSC Mathematics Extension 1 | False Proof By Induction Watch

Theory

A proof by induction needs both parts: a true base case and a valid inductive step. This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-01) is about spotting the error when one part is missing or wrong — and seeing why the statement can then be false.

A valid proof by mathematical induction has two parts, and it needs both: a base case (the statement is true at the start) and an inductive step (truth at $n = k$ forces truth at $n = k + 1$ ). Break either link and the argument fails.

Step without a base case. A false statement can have a perfectly valid inductive step. The step only says "if it is true at $k$ , then it is true at $k + 1$ " — but if nothing ever makes it start being true, the statement can be false for every $n$ .

Base case without a step. Checking $n = 1, 2, 3, \dots$ is only finitely many cases. It never covers all $n$ , and a single later value can be a counterexample that disproves the claim.

NESA link. Part of the Year 12 Proof by mathematical induction focus area, outcome ME1-12-01 (with MAO-WM-01). The syllabus asks students to identify errors in false "proofs by induction", such as cases where only one of the two required steps is true, and gives the example in Example 1 below.

A valid step with no base case: nothing starts, so every domino stays standing.

Checking a few cases is not a proof — the chain can break further along.

A valid induction proof requires both of these to be true:

(1) Base case: S (n_{0}) is true.

(2) Inductive step: S (k) \Rightarrow S (k + 1) .

S (n_{0}) and S (k) \Rightarrow S (k + 1)

If either fails, the "proof" is invalid — and to show the statement itself is false you only need a single counterexample.

Two failure modes. A valid step with a false base case proves nothing (the statement may be false). A true base case with no step proves nothing about all $n$ (checking finitely many cases is never a proof).

Errors to watch for

Error	Fix
Asserting $S (k + 1)$ instead of deriving it	Use $S (k)$ to derive $S (k + 1)$
Adding the wrong $(k + 1)$ th term	The next term is the one at $n = k + 1$
Miscounting or skipping the base case	Check the smallest $n$ actually works
Starting at the wrong value $n_{0}$	Begin at the correct $n_{0}$
Claim only for odd/even $n$	Step $k \to k + 2$ ; prove the right base case
Assuming the result for all $n$	Assume it for $n = k$ only

How to audit a claimed proof by induction

Check the base case is actually true. Substitute the smallest $n$ into both sides. If they differ, the base case fails.
Check the step derives $S (k + 1)$ . The working must use the assumption $S (k)$ to reach $S (k + 1)$ — not simply assert it, and not assume the result for all $n$ .
Check the mechanics. Confirm the correct $(k + 1)$ th term, the correct starting value $n_{0}$ , and a step size that matches the claim ( $k \to k + 2$ for an odd/even claim).
Decide. If the base fails or the step fails, the proof is invalid. To show the statement is false, produce one counterexample.

Example 1 — NESA: step holds, base fails

The statement

S (n) : 1 + 2 + \dots + n = \frac{(n - 1) (n + 2)}{2}

has a valid inductive step. Show the step holds, yet the statement is false. (This is the NESA syllabus example.)

Solution

Step: assume $S (k)$ and add the next term $(k + 1)$ :

$\frac{(k - 1) (k + 2)}{2} + (k + 1)$	$=$	$\frac{(k - 1) (k + 2) + 2 (k + 1)}{2}$
	$=$	$\frac{k^{2} + 3 k}{2} = \frac{k (k + 3)}{2}$
	$=$	$\frac{((k + 1) - 1) ((k + 1) + 2)}{2}$

\frac{(k - 1) (k + 2)}{2} + (k + 1) = \frac{k (k + 3)}{2}

This is $S (k + 1)$ , so the step is valid. Base $(n = 1)$ : LHS $= 1$ , RHS $= \frac{0 \cdot 3}{2} = 0$ . Since $1 \neq 0$ , the base case fails, so $S (n)$ is false (the correct sum is $\frac{n (n + 1)}{2}$ ).

Example 2 — The wrong term

To prove

2 + 4 + 6 + \dots + 2 n = n (n + 1)

, a student assumes the

n = k

case and adds

2 k

to both sides. Identify the error and give the correct step.

Solution

The $(k + 1)$ th term is $2 (k + 1) = 2 k + 2$ , not $2 k$ . Adding $2 k$ uses the wrong term, so the step is never actually proven. Correctly:

$k (k + 1) + 2 (k + 1)$	$=$	$(k + 1) (k + 2)$
	$=$	$(k + 1) ((k + 1) + 1)$

k (k + 1) + 2 (k + 1) = (k + 1) (k + 2)

This is the formula at $n = k + 1$ , completing the corrected step.

Example 3 — Wrong step size

A student "proves" that

3^{n} + 7^{n}

is divisible by

10

for all

n

, stepping

k \to k + 1

. Find the error.

Solution

Base $(n = 1)$ : $3 + 7 = 10$ — true. Step $k \to k + 1$ , assuming $3^{k} + 7^{k} = 10 M$ :

$3^{k + 1} + 7^{k + 1}$	$=$	$3 (3^{k} + 7^{k}) + 4 \cdot 7^{k}$
	$=$	$3 (10 M) + 4 \cdot 7^{k}$

3^{k + 1} + 7^{k + 1} = 3 (10 M) + 4 \cdot 7^{k}

The term $4 \cdot 7^{k}$ is not divisible by $10$ , so the step fails. In fact the claim is false for even $n$ : at $n = 2$ , $9 + 49 = 58$ is not a multiple of $10$ . The result holds only for odd $n$ , which needs the step $k \to k + 2$ .

Example 4 — Cases are not a proof

A student notices

n^{2} + n + 11

is prime for

n = 0, 1, 2, \dots, 9

and concludes it is prime for all

n

. What is wrong?

Solution

Checking cases is not a proof — there is no inductive step extending the pattern to all $n$ . A single counterexample settles it:

$n = 10 : 10^{2} + 10 + 11$	$=$	$121$
	$=$	$11 \times 11$

10^{2} + 10 + 11 = 121 = 11 \times 11

$121$ is not prime, so the claim is false. Any number of confirming cases can still hide a counterexample.

Common pitfalls

Trusting a valid step alone. A valid inductive step with a false base case proves nothing — the statement may be false. Always verify the base case really holds.

Treating cases as a proof. Checking finitely many values is never a proof, and a single counterexample can disprove the claim outright.

Over-assuming. In the step you may assume the result for

n = k

only, not for all

n

; assuming all

n

is circular.

Wrong mechanics. Watch the

(k + 1)

th term, the starting value, and the step size (

k \to k + 2

for odd or even claims).

Frequently asked questions

Can a proof by induction have a valid step but still be false?

Yes. A false statement can have a valid inductive step. Without a true base case nothing makes it start being true, so the claim can be false for every $n$ . Both parts are required.

Why isn't checking examples a proof?

Checking $n = 1, 2, 3, \dots$ is only finitely many cases; it never covers every $n$ , and a later value can be a counterexample. The inductive step is what reaches all $n$ .

What is wrong with assuming the result for all n?

In the step you may assume $S (k)$ for the single value $k$ only. Assuming it for all $n$ assumes what you are trying to prove — that is circular.

How do you spot an error in a proof by induction?

Check the base case is true, the step genuinely derives $S (k + 1)$ from $S (k)$ , and the $(k + 1)$ th term, starting value and step size are all correct.

What is the false proof example in the NESA syllabus?

It shows the inductive step can be proven for the false proposition $1 + 2 + \dots + n = \frac{1}{2} (n - 1) (n + 2)$ , but the initial case does not hold, so the statement is false (Example 1).

Is identifying errors in false proofs in Extension 1?

Yes — it is part of the Proof by mathematical induction focus area, outcome ME1-12-01, which asks students to identify errors in false proofs where only one of the two required steps is true.

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Errors to watch for

How to audit a claimed proof by induction

Common pitfalls

Frequently asked questions

Theory