Mathematics/Lessons/Proof/Proof by Exhaustion
Pure · Proof · Year 1

Proof by Exhaustion

When a claim only has to hold for a finite, manageable set of cases, you can prove it by simply checking every single one — leaving no case unexamined.

9 min read 3 worked examples

Most proofs have to cover infinitely many cases at once — that's why we use a general $n$. But some claims only need to be true for a finite number of possibilities. When that's the case, there's a blunt but perfectly valid strategy: check every possibility, one by one. If the claim holds in every single case, and you've genuinely covered them all, then it's proven. This is proof by exhaustion — you exhaust the possibilities.

The name is literal. To "exhaust" a set of cases means to use them all up, leaving none behind. The entire validity of the method rests on one condition: you must be certain you have covered every case. Miss one, and the proof collapses — because that one unchecked case might be the exception.

Why this actually counts as proof

It feels almost too simple — "I just checked them all?" But think about what proof means: showing a statement is true in every situation it claims to cover. If a claim only covers, say, the integers from 1 to 5, then checking those five cases isn't a sample — it is the whole truth. There are no other cases hiding. That's the crucial difference from testing examples in an infinite claim, where you can never check them all.

When you're allowed to use it

Exhaustion only works when the number of cases is finite and small enough to list. You can use it when:

• the claim is about a specific, limited range (e.g. "for all integers $n$ with $1 \leq n \leq 5$"), or
• the possibilities naturally split into a few distinct types that together cover everything (e.g. "every integer is either even or odd" — just two cases to check).

That second kind is the clever one. Even when there are infinitely many numbers, they might fall into just two or three categories, and you can check each category. We'll see that in example 3.

The danger

Exhaustion is only as good as your case list. If you claim to have covered everything but secretly missed a case, the "proof" proves nothing. Before writing the final line, always ask: "Have I genuinely accounted for every possibility?"

Worked example 1

Prove a claim over a small range
Prove that every integer $n$ with $2 \leq n \leq 6$ can be written as a sum of at most two prime numbers.
ProofChecking each integer from 2 to 6
1
There are exactly five integers to consider: $2, 3, 4, 5, 6$.
The range $2 \leq n \leq 6$ is finite, so exhaustion applies.
2
$2 = 2$ (prime);   $3 = 3$ (prime).
Each is itself prime — a sum of one prime.
3
$4 = 2 + 2$;   $5 = 2 + 3$;   $6 = 3 + 3$.
Each is a sum of two primes.
4
Every integer in the range has been shown to work, and no case remains.
All five cases checked — the set is exhausted.
5
Therefore every integer $n$ with $2 \leq n \leq 6$ is a sum of at most two primes.
The claim is proven for the whole range.
▪ proven
Examiner's eye

The final two lines matter as much as the checking. Step 4 explicitly states that every case is covered — that's what turns a list of calculations into a proof. Never just show the arithmetic and stop.

Worked example 2

A claim about squares · finite range
Prove that for all integers $n$ with $1 \leq n \leq 4$, the value of $n^2 + n$ is even.
ProofTesting each value of n
1
The possible values are $n = 1, 2, 3, 4$ — four cases.
Finite range, so we can check each.
2
$n=1:\ 1^2 + 1 = 2$, which is even.
3
$n=2:\ 2^2 + 2 = 6$, which is even.
4
$n=3:\ 3^2 + 3 = 12$, which is even.
5
$n=4:\ 4^2 + 4 = 20$, which is even.
6
All four cases give an even result, and no case is left unchecked.
Therefore $n^2+n$ is even for every $n$ in the range.
▪ proven
Look deeper

Notice this only proves it for $n = 1$ to $4$. The result is actually true for all integers (because $n^2+n = n(n+1)$ is a product of consecutive integers, one of which must be even) — but exhaustion can't show that, since there are infinitely many integers. This is a perfect illustration of exhaustion's limit: it's watertight over a finite range, and useless beyond it.

Worked example 3

The powerful kind · splitting into cases that cover everything
Prove that the square of any integer is either a multiple of 4 or one more than a multiple of 4.

Here there are infinitely many integers — so we can't check them one by one. But every integer is either even or odd. Those two categories cover all integers, so if the claim holds for both, it holds for everything. Two cases to exhaust.

ProofBy cases: even and odd
1
Case 1 — even. Let the integer be $2n$. Then $(2n)^2 = 4n^2$.
$4n^2$ is a multiple of 4.
2
Case 2 — odd. Let the integer be $2n+1$. Then $(2n+1)^2 = 4n^2 + 4n + 1 = 4(n^2+n) + 1$.
This is one more than a multiple of 4.
3
Every integer is even or odd, so these two cases cover all integers.
The cases are exhaustive — nothing is left out.
4
Therefore the square of any integer is a multiple of 4, or one more than a multiple of 4.
Proven for all integers via two exhaustive cases.
▪ proven
The key insight

This is exhaustion at its most powerful. We couldn't check infinitely many integers — but we could check the two types every integer belongs to. The whole method hinges on step 3: proving the cases genuinely cover everything. Choosing the right way to split into cases is the real skill.

Check yourself

Prove that every integer $n$ with $1 \leq n \leq 5$ satisfies $n^3 \geq n^2$.

1
Five cases to check: $n = 1, 2, 3, 4, 5$.
2
$n=1:\ 1 \geq 1$ ✓    $n=2:\ 8 \geq 4$ ✓    $n=3:\ 27 \geq 9$ ✓
3
$n=4:\ 64 \geq 16$ ✓    $n=5:\ 125 \geq 25$ ✓
4
All five cases hold, and none is left unchecked — so $n^3 \geq n^2$ for every $n$ in the range.
▪ proven