# Proof by Cases

We often have to break up a proof into several cases and prove each case separately. In a proof by cases, the premise is that the cases listed cover all the possibilities. Since each case leads to the desired consequence, we have established the claim.

Example 1: Let x denote the ceiling of x, that is, the smallest integer larger than x. (In other words, x is the integer you get by rounding up from x.)

Claim: For all x ≥ 0, x/2 x /2 .

Proof: We break up the proof into three cases:
1. x is an even integer.
2. x is an even integer plus some ε, 0 < ε < 1.
3. x is an odd integer.
4. x is an odd integer plus some ε, 0 < ε < 1.

Case 1: x = 2n for some integer n ≥ 0.
Then,

x/2 = 2n/2 = n = n.
Also,
x /2 = 2n /2 = 2n/2 = n.
Thus, x/2 x /2 .

Case 2: x = 2n + ε for some integer n ≥ 0 and some real number ε where 0 < ε < 1.
Then,

x/2 = (2n + ε)/2 = n + ε/2 = n + ε/2 = n + 1.
We know ε/2 = 1 because 0 < ε/2 < 1/2. Also,
x /2 = 2n + ε /2 = (2n + ε ) /2 = (2n + 1) /2 = n + 1/2.
Again, we have x/2 x /2 .

Case 3: x = 2n+1 for some integer n ≥ 0.
Then,

x/2 = (2n + 1)/2 = n + 1/2 = n + 1/2 = n + 1.
Also,
x /2 = 2n + 1 /2 = (2n + 1)/2 = n + 1/2.
Thus, x/2 x /2 .

Case 4: x = 2n + 1 + ε for some integer n ≥ 0 and some real number ε where 0 ≤ ε < 1.
Then,

x/2 = (2n + 1 + ε)/2 = n + 1/2 + ε/2 = n + 1/2 + ε/2 = n + 1.
We know 1/2 + ε/2 = 1 because ε < 1. Also,
x /2 = 2n + 1 + ε /2 = (2n + 1 + ε ) /2 = (2n + 2) /2 = n + 1.
Thus, we have x/2 x /2 in the final case.
QED

Example 2:

Claim: The graph K3,3 shown below is not a planar graph.

Proof: Consider the vertices a, f, c and d. They form a cycle in the graph K3,3. Any planar drawing K3,3 cannot have the edges cross, so the cycle divides the plane into two regions: inside and outside the cycle. Now, the vertex b can either be placed inside or outside the a-f-c-d-a cycle:

Case 1: Suppose that the vertex b is placed inside the cycle. Now, consider where the vertex e can be placed. There are 3 possible regions: inside the region bounded by a-f-b-d-a, inside the region bounded by c-d-b-f-c or outside the a-f-c-d-a cycle.

In each of these possibilities, there is an edge that cannot be drawn without edges crossing.

Case 2: If the vertex b is place outside the a-f-c-d-a cycle, we make the same argument about the placement of vertex e, except the roles of vertices a and b are interchanged.

QED