Reasoning by Cases
A free Logic lesson from the “Proof Readiness” unit, with a worked example and practice problems including step-by-step solutions.
Choose cases whose union includes every allowed input. Prove the claim in each case and state why the cases are exhaustive. Even/odd covers every integer; positive/negative does not cover zero. The conclusion must hold in every case, not merely most cases.
What you'll learn
- Split a problem into complete cases and handle each one
- Explain the idea in plain English before using symbols
- Use examples, non-examples, or counterexamples to check the reasoning
Worked example
Problem. To prove n(n+1) is even for every integer n, which exhaustive cases work?
- Every integer is exactly even or odd.
- Positive/negative alone would omit zero.
Answer: n is even; n is odd
Lesson preview: this page shows 5 practice items. The interactive activity contains 6 practice items and 6 quiz items (multiple choice). Counts include repeated prompts and variants.
Practice problems
1. To prove n(n+1) is even for every integer n, which exhaustive cases work?
Choices: n is even; n is odd · n=1; n=2 · n is positive; n is negative · n is prime; n is greater than 10
Show solution
- Every integer is exactly even or odd.
- Positive/negative alone would omit zero.
Answer: n is even; n is odd
2. In the even case n=2k, rewrite n(n+1) to show evenness.
Choices: 2k+1 · k(k+1) · 2k²+1 · 2[k(2k+1)]
Show solution
- Substitute n=2k and factor out 2.
- The bracketed expression is an integer.
Answer: 2[k(2k+1)]
3. In the odd case n=2k+1, why is n(n+1) even?
Choices: n is already even · Only n=1 matters · n+1=2(k+1), so the product has a factor of 2 · Odd times odd is even
Show solution
- The next integer after an odd integer is even.
- Multiplying by that even factor preserves divisibility by 2.
Answer: n+1=2(k+1), so the product has a factor of 2
4. A proof for all real x handles x>0 and x<0. Which case is missing?
Choices: x=2 · x=0 · x=1 · x=-1
Show solution
- Zero lies in neither of the two strict-inequality cases.
Answer: x=0
5. To evaluate |x| by cases, which rules are correct?
Choices: |x|=x for x>=0; |x|=-x for x<0 · |x|=x for every real x · |x|=-x for every real x · |x|=0 for x<0
Show solution
- Absolute value is the nonnegative distance to zero.
- Negative inputs therefore need their signs reversed.
Answer: |x|=x for x>=0; |x|=-x for x<0
Explain and apply
Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.
Write a proof by cases
Prove n(n+1) is even for every integer n, including zero and negative integers.
Review rubric and solution
- State the exhaustive even/odd cases.
- Show a factor of 2 in the even case.
- Show a factor of 2 in the odd case.
- Conclude the assertion for every integer.
- If n=2k, then n(n+1)=2[k(2k+1)].
- If n=2k+1, then n(n+1)=2[(2k+1)(k+1)].
- The bracketed expressions are integers, and every integer is even or odd.
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