Proof by Cases

What is Proof by Cases?

Sometimes you can’t prove something in one go.

But you can split it into scenarios and prove each one separately.

The Idea

If there are only a few possibilities:

  1. Split into Case 1, Case 2, Case 3…
  2. Prove the result holds in each case
  3. Conclude — since all cases are covered, the result is true

Can’t do it all at once? Split into cases. Cover them all.

The Method

To prove P:

  1. Identify all possible cases (they must cover everything)
  2. Prove P is true in Case 1
  3. Prove P is true in Case 2
  4. (Continue for all cases)
  5. Done — P is always true

Example 1: n² + n is Even

Claim: For any integer nn, the expression n2+nn^2 + n is even.

The split:

Any integer is either even or odd. Two cases.

Case 1: nn is even

  • n=2kn = 2k for some integer kk

Substitute:

n2+n=(2k)2+2k=4k2+2k=2(2k2+k)\begin{aligned} n^2 + n &= (2k)^2 + 2k \\ &= 4k^2 + 2k \\ &= 2(2k^2 + k) \end{aligned}
  • That’s 2×(something)2 \times (\text{something}) — even ✓

Case 2: nn is odd

  • n=2k+1n = 2k + 1 for some integer kk

Substitute:

n2+n=(2k+1)2+(2k+1)=4k2+4k+1+2k+1=4k2+6k+2=2(2k2+3k+1)\begin{aligned} n^2 + n &= (2k+1)^2 + (2k+1) \\ &= 4k^2 + 4k + 1 + 2k + 1 \\ &= 4k^2 + 6k + 2 \\ &= 2(2k^2 + 3k + 1) \end{aligned}
  • That’s 2×(something)2 \times (\text{something}) — even ✓

Conclusion:

Both cases give even. So n2+nn^2 + n is always even.

We didn’t know if n was even or odd. So we checked both.

Example 2: Consecutive Product

Claim: The product of two consecutive integers is always even.

In other words: n(n+1)n(n+1) is even for any integer nn.

The split:

Any integer is either even or odd. Two cases.

Case 1: nn is even

  • n=2kn = 2k for some integer kk

Substitute:

n(n+1)=2k(n+1)=2×k(n+1)\begin{aligned} n(n+1) &= 2k(n+1) \\ &= 2 \times k(n+1) \end{aligned}
  • That’s 2×(something)2 \times (\text{something}) — even ✓

Case 2: nn is odd

  • If nn is odd, then n+1n+1 is even
  • So n+1=2mn+1 = 2m for some integer mm

Substitute:

n(n+1)=n×2m=2×nm\begin{aligned} n(n+1) &= n \times 2m \\ &= 2 \times nm \end{aligned}
  • That’s 2×(something)2 \times (\text{something}) — even ✓

Conclusion:

Both cases give even. So n(n+1)n(n+1) is always even.

Consecutive integers — one is always even. So the product is always even.

When to Use Proof by Cases

  • When the situation naturally splits (even/odd, positive/negative, etc.)
  • When a direct approach doesn’t work for all values at once
  • When you need to handle edge cases separately