A student proves 'for all real numbers x, |x| ≥ 0' with two cases: Case 1 (x > 0) and Case 2 (x < 0). Is this a valid proof?
AYes — positive and negative numbers cover all real numbers
BNo — the case x = 0 is missing, leaving a gap in the proof
CYes — since the claim holds in both cases, it holds for all reals
DNo — absolute value proofs must use the definition directly, not cases
The case x = 0 is neither positive nor negative, and the two cases {x > 0} and {x < 0} do not cover it — their union is all nonzero reals, not all reals. The proof has a gap. For x = 0, we need to verify that |0| = 0 ≥ 0 separately. Exhaustiveness is the non-negotiable requirement: every element of the domain must fall into at least one case, and 0 falls into none of the student's cases.
Question 2 Multiple Choice
What is the key structural requirement that makes a proof by cases logically valid?
AEach case must use a different proof technique so the argument is varied
BThe cases must be non-overlapping and together cover every element of the domain
CThere must be exactly two cases representing a binary division of the domain
DThe cases must be ordered from simplest to most complex to aid readability
Exhaustiveness is the requirement that cannot be relaxed: the union of all cases must equal the entire domain. Every element must be covered by at least one case. Non-overlapping (disjointness) is convenient but not strictly required — an element in two overlapping cases is fine as long as no element falls through a gap. The key is coverage, not separation.
Question 3 True / False
In a proof by cases, the cases are required to be non-overlapping — if any two cases share even one element, the proof is invalid.
TTrue
FFalse
Answer: False
Overlap is permitted. An element that falls into two cases is simply proved twice, which is harmless. What is never permitted is an element that falls into zero cases — that element is uncovered, and the proof makes no claim about it. The formal requirement is that the union of all cases equals the domain. Disjointness (non-overlapping) is a common organizational choice but is not a logical necessity.
Question 4 True / False
A proof by cases establishes a universal claim by verifying the claim separately within each region of an exhaustive partition of the domain.
TTrue
FFalse
Answer: True
This captures the logic exactly. 'For all x in D, P(x)' is proved by: (1) partition D into cases C₁, C₂, ..., Cₙ with C₁ ∪ ... ∪ Cₙ = D; (2) prove P(x) holds for every x in each Cᵢ. Since every element of D is in at least one case, and P holds within every case, P holds for all elements of D. The partition is the bridge between local arguments and the universal claim.
Question 5 Short Answer
Why is exhaustiveness the non-negotiable requirement in a proof by cases? What goes wrong logically if even one element of the domain falls through the cracks?
Think about your answer, then reveal below.
Model answer: If an element x₀ falls into none of the cases, the proof says nothing about x₀. The universal claim 'for all x, P(x)' would remain unverified for x₀. The conclusion could be false at x₀ — and in fact this is often exactly where counterexamples hide. A proof with a missing case has a logical gap that cannot be patched by the quality of the argument within the other cases.
The entire force of a proof by cases depends on the claim 'every element is somewhere in my case structure.' Without exhaustiveness, the proof is not a proof but a collection of conditional arguments: 'IF x is in case 1, THEN P(x).' The universal quantifier requires unconditional coverage. This is why checking exhaustiveness is the first thing to verify when reviewing a cases-based proof.