Proof by Cases

Proof by cases divides the hypotheses into alternatives that cover every possibility. The proof then establishes the conclusion separately in each case. Each case may itself use a direct proof, an indirect proof, or another suitable method.

The essential requirement is coverage: every allowed input must belong to at least one case. If the cases overlap, that is harmless; if they omit a possibility, the argument is incomplete.

ExampleThree consecutive integers

Prove that the sum of any three consecutive integers is divisible by $3$.

Proof

Write the integers as $n,n+1,n+2$. Every integer has one of the forms $3k$, $3k+1$, or $3k+2$.

• If $n=3k$, the sum is $9k+3=3(3k+1)$.
• If $n=3k+1$, the sum is $9k+6=3(3k+2)$.
• If $n=3k+2$, the sum is $9k+9=3(3k+3)$.

All three cases give a multiple of $3$, so the claim follows.

ExerciseA product of consecutive integers

Prove that $n(n+1)$ is even for every integer $n$.

Solution

If $n$ is even, the product has an even factor. If $n$ is odd, then $n+1$ is even. These cases exhaust the integers, so $n(n+1)$ is always even.

RemarkChecking a proposed theorem

The source draft also claimed that $n^2+4$ is never prime for $n>1$. This statement is false: for $n=3$, $n^2+4=13$, which is prime. The proposed factorisation

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

does not hold. This is a useful example of why a proof must be checked algebraically and against small cases before publication.

Case analysis is most useful when the cases follow naturally from a definition, such as parity or residue classes modulo an integer.