Unit 10 Review and Checkpoint
A free Logic lesson from the “Counterexamples, Sets, and Diagrams” unit, with a worked example and practice problems including step-by-step solutions.
All A are B means inclusion A⊆B; some A are B means a nonempty intersection; no A are B means an empty intersection. Do not reverse inclusion. For a doubtful inference, construct small finite sets satisfying its premises and test the conclusion. This review checks performance on mathematical tasks. For extended written work, use the separate rubric with a tutor or teacher; the automatic quiz does not certify proof-writing proficiency.
What you'll learn
- Review counterexamples, sets, diagrams, and classification reasoning
- Choose the reasoning tool that matches the statement
- Explain why an answer is valid, invalid, true, false, or unsupported
Worked example
Problem. Let A={46, 47} and B={46, 47, 48}. Which is true?
- Each member of A is in B.
- B contains one additional element.
Answer: A is a proper subset of B
Lesson preview: this page shows 5 practice items. The interactive activity contains 8 practice items and 8 quiz items (multiple choice). Counts include repeated prompts and variants.
Practice problems
1. Which integer disproves “Every multiple of 7 is a multiple of 14”?
Choices: 14 · 28 · 42 · 7
Show solution
- 7 satisfies the hypothesis but is not divisible by 14.
Answer: 7
2. Which value disproves “Every odd positive integer is prime”?
Choices: 5 · 7 · 9 · 3
Show solution
- 9 is odd, positive and composite because 9=3×3.
Answer: 9
3. Which real x disproves x²=x for every real x?
Choices: Both 0 and 1 · 7 · 0 · 1
Show solution
- 7²=49, which is not 7.
- Both 0 and 1 actually satisfy the equation.
Answer: 7
4. A claim concerns all integers greater than 10. A proposed counterexample is 8. Why is it invalid?
Choices: It does not satisfy the hypothesis · A counterexample must be prime · A counterexample must prove the conclusion · One example can never disprove a universal claim
Show solution
- A counterexample must belong to the stated domain and satisfy the setup.
Answer: It does not satisfy the hypothesis
5. Let A={10, 11} and B={10, 11, 12}. Which is true?
Choices: B is a subset of A · A and B are disjoint · A equals B · A is a proper subset of B
Show solution
- Each member of A is in B.
- B contains one additional element.
Answer: A is a proper subset of B
Practice this interactively with instant feedback and an AI tutor.
Practice Unit 10 Review and Checkpoint Take the free placement check