Case Analysis Quiz: Break Problems into Logical Scenarios

These two cases cover all integers without overlap, ensuring every n falls into exactly one case. A proper case split must be both exhaustive and mutually exclusive. By dividing integers into two non-overlapping categories that together cover every possibility, the argument guarantees that every integer n will be handled exactly once, preventing logical gaps or double counting.

Prime/composite misses integers like 1 and negatives, so it is not a full partition of ℤ. A valid partition must include all integers. Since 1 is neither prime nor composite, and negative integers also fall into neither group, the split fails to cover the entire domain, making it unsuitable for case-based reasoning over ℤ.

Leaving out x=0 makes the analysis incomplete, since 0 is a real number requiring its own case. If a proof claims to analyze all real numbers but only considers x>0 and x

Every real number is either rational or irrational; the two sets form a complete partition of ℝ. Because rational and irrational numbers are disjoint and together account for every real number, splitting ℝ into these two categories ensures full coverage and avoids overlap, making this a model example of a clean binary partition.

Even and odd are mutually exclusive and cover all integers, making them ideal for parity proofs. Each integer is exactly one of the two—never both—and no integer fails to fit into the classification, which is why parity arguments rely so heavily on the even/odd dichotomy.

Positive/negative/zero fully covers ℤ without overlap and is the standard sign‑based partition. These three cases are disjoint and collectively exhaustive: every integer is either above zero, below zero, or equal to zero. This neat separation allows clear logical branching depending on sign.

x≥0 includes x>0, so the cases overlap and fail the mutually exclusive requirement. The conditions x≥0 and x>0 cannot form separate cases because one contains the other. Case analysis requires non-overlapping categories so that each value is counted once, which this pair fails to achieve.

A complete modulo‑7 split must list all 7 residue classes from 0 to 6. Working mod 7 partitions the integers into exactly seven disjoint categories, and omitting any one residue class leaves the partition incomplete, blocking the validity of any proof relying on full modular coverage.

Even/odd (mod 2) directly controls parity, making it the simplest and most efficient split. Since parity arguments depend entirely on whether a number leaves remainder 0 or 1 when divided by 2, the mod-2 partition is perfectly aligned with the structure of proofs involving divisibility, sums, or products of integers.

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Missing a legitimate case leaves the argument incomplete, breaking the logic of case analysis. If a value in the domain does not fall into any listed case, the proof fails to address all possibilities. Without complete coverage, the conclusion cannot be justified for the entire domain.

This split is not exhaustive because many integers (e.g., 5, 7, 11) are divisible by neither 2 nor 3.

It is also not mutually exclusive, since some integers (e.g., 6, 12) are divisible by both 2 and 3.

Thus it fails both requirements for a proper case split, and B is the only correct choice.

Parity arguments rely on the dichotomy n even or n odd, which corresponds to n mod 2. Because every integer leaves remainder 0 or 1 when divided by 2, splitting by parity creates two clean, non-overlapping categories. This makes properties involving sums, products, or divisibility especially transparent, since parity behaves predictably under arithmetic operations.

Absolute value behavior is clearest when treating positive, zero, and negative x separately. The definition |x| = x for x≥0 and |x| = −x for x

Case analysis requires covering all possibilities; otherwise the proof addresses only part of the domain. To argue that a statement holds universally, each possible situation for the variable must be included. If even one scenario is omitted, the proof becomes incomplete and the conclusion cannot be guaranteed for every element in the domain.

Adjacent integers differ by 1, so one is always even; splitting on even/odd exposes this directly. If n and n+1 are considered, one of them is divisible by 2 while the other is not. Using parity as the basis for the case split highlights this precise structure and often simplifies proofs about consecutive integers.

A full modulo‑4 partition must include all four residue classes. Working mod 4 means integers fall into classes congruent to 0, 1, 2, or 3 modulo 4. Leaving out any one of these classes results in a partial, invalid partition, since every integer must belong to exactly one of the listed categories.

Zero is neither positive nor negative, so it requires its own case to ensure completeness. Splitting ℤ or ℝ only into positive and negative values ignores 0 entirely, which breaks the completeness requirement for case-based proofs. Including zero as a separate case ensures every number is accounted for.

Each case must independently demonstrate the conclusion, guaranteeing the statement holds universally. The logic of case analysis demands that no matter which case the variable falls into, the argument must reach the same final conclusion. Once each branch proves the statement separately, the universal statement follows automatically.

Modulo‑3 arithmetic requires the full set of residues {0,1,2} to cover all integers. Every integer divided by 3 gives a remainder of 0, 1, or 2, so these three cases represent a complete and disjoint partition. Modular proofs rely on this completeness to ensure no integer escapes the analysis.