Recognising incomplete or invalid induction arguments | Year 12

01 / LEARN

The idea, step by step

Check both obligations

A true base does not prove the implication, and a correct implication does not give a true base. Check them separately, then check that the step reaches the entire claimed domain.

If P(k) implies P(k+2), starting at 1 reaches only odd indices. Supply an even starting case too, or establish a one-step implication.

Find the unjustified line

Correct algebra may be written in a circular order. Ask which hypothesis justifies each line and whether the desired next case has already been assumed. A verified graph or example is not a general implication.

One allowed counterexample refutes a universal statement. To repair an incomplete proof of a true claim, supply the missing base, bridge or implication rather than adding numerical checks.

A common trap

Believing more examples fix circular reasoning.

Instead: Examples diagnose problems but do not supply the missing general implication.

02 / WORKED EXAMPLES

See the thinking

Example 1

A false starting case

A student claims $1 + 2 + \dots + n = n (n + 1) / 2 + 1$ and obtains plausible successor algebra. Why does the proof fail?

At $n = 1$ , the left side is 1 and the right side is 2.
The conditional step can preserve an error.
There is no valid base to start the chain.

The base case is false.

Example 2

Missing even indices

P(1) is checked and P(k) implies P(k+2) is proved. What is missing?

The chain reaches 1,3,5,... only.
Even indices have no established starting case.
Check P(2) as well, or give a one-step implication.

A valid even base or a one-step implication.

Check your understanding

How many allowed counterexamples refute a universal claim?

Show answer

One.

Every claimed case must hold, so one false case is enough.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

A student assumes the result at k+1 and rearranges it to the hypothesis at k. Explain the logical gap.
3 marks
Why do correct checks for the first one hundred positive integers not constitute induction?
2 marks
The step is proved only for k at least 5 and the student checks P(3). What is missing to establish every n at least 3?
3 marks
Refute the claim that $n^{2} + n + 41$ is prime for every positive integer n.
3 marks
A sum of odd numbers proof adds $2 k - 1$ to the assumed total $k^{2}$ . Correct the next term.
2 marks
A divisibility proof concludes $A (k + 1) = d (q + 1 / 2)$ with integer q. Explain the problem.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A student assumes the result at k+1 and rearranges it to the hypothesis at k. Explain the logical gap.
3 marks
Derive P(k+1) from P(k).
If reversing algebra, explicitly justify that each step is reversible.
The direction written is the reverse of the required implication.
Why do correct checks for the first one hundred positive integers not constitute induction?
2 marks
An untested case could still fail.
A starting case and a general successor implication are both required.
There is no implication proved for an arbitrary integer.
The step is proved only for k at least 5 and the student checks P(3). What is missing to establish every n at least 3?
3 marks
The implication cannot be used at k=3 or k=4.
P(5) is needed to start the valid chain.
P(4) and P(5).
Refute the claim that $n^{2} + n + 41$ is prime for every positive integer n.
3 marks
This allowed case is composite.
One counterexample refutes a universal claim.
$n = 41$ gives $41^{2} + 41 + 41 = 41 (43)$ .
A sum of odd numbers proof adds $2 k - 1$ to the assumed total $k^{2}$ . Correct the next term.
2 marks
The assumed sum already contains $2 k - 1$ .
The next odd number is $2 (k + 1) - 1$ .
It must add $2 k + 1$ .
A divisibility proof concludes $A (k + 1) = d (q + 1 / 2)$ with integer q. Explain the problem.
2 marks
Integer divisibility requires an integer multiplier.
An arbitrary rational multiplier does not suffice.
The proposed multiplier is not an integer.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist mathematical induction. Explicit summation, power and divisibility identities; induction of inequalities or recursive formulae is not required.

Official curriculum source ↗

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