Some claims are easy once you know a little more about the number: whether it is even or odd, or what remainder it leaves. A proof by cases splits the problem so that each piece is a short direct proof.
Key idea
To prove $Q$ by cases, list cases $C_{1}, C_{2}, \ldots$ that together cover every possibility, then prove that each case leads to $Q$. Standard splits: $n$ even or odd ($n = 2k$ or $2k + 1$); remainders when dividing by $3$ ($n = 3k$, $3k + 1$ or $3k + 2$); signs ($x > 0$, $x = 0$, $x < 0$).
Why it works
Logically, $(C_{1} \vee C_{2})$, $C_{1} \to Q$ and $C_{2} \to Q$ together give $Q$: whichever case actually happens, $Q$ follows. The argument is only as good as the first fact, that the cases cover everything. Every integer leaves a remainder of $0$, $1$ or $2$ when divided by $3$, so the three forms $3k$, $3k + 1$, $3k + 2$ really do cover all integers.
Worked examples
Worked example: $n^{2} + n$ is always even
Prove: for every integer $n$, $n^{2} + n$ is even.
- Case 1, $n$ even: $n = 2k$, so $n^{2} + n = 4k^{2} + 2k = 2(2k^{2} + k)$, even.
- Case 2, $n$ odd: $n = 2k + 1$, so $$n^{2} + n = (4k^{2} + 4k + 1) + (2k + 1) = 2(2k^{2} + 3k + 1),$$ which is even.
- Every integer is even or odd, so the cases cover everything.
Answer: Even in both cases.
Worked example: Squares and the number $3$
Prove: for every integer $n$, $n^{2}$ leaves remainder $0$ or $1$ when divided by $3$.
- Case $n = 3k$: $n^{2} = 9k^{2} = 3(3k^{2})$, remainder $0$.
- Case $n = 3k + 1$: $n^{2} = 9k^{2} + 6k + 1 = 3(3k^{2} + 2k) + 1$, remainder $1$.
- Case $n = 3k + 2$: $n^{2} = 9k^{2} + 12k + 4 = 3(3k^{2} + 4k + 1) + 1$, remainder $1$.
- So a perfect square never leaves remainder $2$. (That is why, for example, $3m + 2$ is never a perfect square.)
Answer: Remainder $0$ or $1$ in every case.
Worked example: Absolute values
Prove: for all real numbers $x$ and $y$, $|xy| = |x|\,|y|$.
- Case $x \ge 0, y \ge 0$: $xy \ge 0$, so $|xy| = xy = |x|\,|y|$.
- Case $x \ge 0, y < 0$: $xy \le 0$, so $|xy| = -xy = x(-y) = |x|\,|y|$. The case $x < 0, y \ge 0$ is the same with the roles swapped.
- Case $x < 0, y < 0$: $xy > 0$, so $|xy| = xy = (-x)(-y) = |x|\,|y|$.
Answer: True in all four sign cases.
Common mistake
Leaving out a case. A proof that checks "$x > 0$" and "$x < 0$" has skipped $x = 0$; a proof about integers that only handles positive $n$ has skipped zero and the negatives. Check: before writing any case, ask whether every possible value lands in at least one case. For remainder cases, list one remainder for each value from $0$ to the divisor minus $1$.