5. Proof by Exhaustion

Video coming soon2 worked examples

Theory

Proof by exhaustion works when the statement need only be checked over a finite number of cases. You split the possibilities into a complete set of cases and verify the claim in every one. If it holds in all of them, it holds in general — because there is nothing left to check.

The method is only available when the cases really are finite and genuinely cover every possibility. A common device is to split the integers by their remainder on division by some number: every integer is of the form $3k$ , $3k+1$ or $3k+2$ , for instance, which is three cases covering all integers.

The Golden Rule: the cases must be exhaustive — together they must account for every possibility, with none omitted. Before concluding, say in words that the cases are complete; a proof that quietly skips a case is not a proof.

⚠️ Common Examiner Traps

Missing a case: if you split by remainder mod $3$ , you need all three of $3k$ , $3k+1$ , $3k+2$ — leaving one out breaks the proof entirely.
Using it on infinite cases: exhaustion cannot prove a statement about “all integers” by testing a few examples. The cases must be finite, even though the numbers they represent need not be.
Checking examples instead of cases: verifying $n = 1, 2, 3$ is not a proof; verifying every form an integer can take is.

Worked examples

Example 1

Prove that $n^2 + n$ is even for every integer $n$ , by considering the two cases for the parity of $n$ .

Step 1: Every integer is either even or odd — two cases that together exhaust all possibilities.

Case 1 — $n$ even. Write $n = 2k$ :

n^2 + n = 4k^2 + 2k = 2(2k^2 + k)

This is a multiple of $2$ , so even.

Case 2 — $n$ odd. Write $n = 2k + 1$ :

n^2 + n = (2k+1)^2 + (2k+1) = 4k^2 + 4k + 1 + 2k + 1 = 2(2k^2 + 3k + 1)

This is also a multiple of $2$ , so even.

Step 2: The claim holds in both cases, and the two cases cover every integer, so $n^2 + n$ is even for all integers $n$ .

(A neater direct proof notes that $n^2 + n = n(n+1)$ is a product of consecutive integers, one of which is always even — but exhaustion makes the two cases explicit.)

Example 2

Prove that $n^3 - n$ is divisible by $3$ for every integer $n$ , by splitting into cases according to the remainder when $n$ is divided by $3$ .

Step 1: Every integer leaves remainder $0$ , $1$ or $2$ on division by $3$ , so it has one of the forms $3k$ , $3k+1$ or $3k+2$ . These three cases are exhaustive. First factorise:

n^3 - n = n(n^2 - 1) = (n-1)\,n\,(n+1)

Case 1 — $n = 3k$ . Then $n$ itself is a multiple of $3$ , so the product is divisible by $3$ .

Case 2 — $n = 3k+1$ . Then $n - 1 = 3k$ is a multiple of $3$ , so the product is divisible by $3$ .

Case 3 — $n = 3k+2$ . Then $n + 1 = 3k + 3 = 3(k+1)$ is a multiple of $3$ , so the product is divisible by $3$ .

Step 2: In every case the product $(n-1)n(n+1)$ contains a factor of $3$ , and the three cases cover all integers. Therefore $n^3 - n$ is divisible by $3$ for every integer $n$ .