Final Course Assessment
A free Logic lesson from the “Logic Applications and Final Review” unit, with a worked example and practice problems including step-by-step solutions.
This untimed final has 36 independent questions, three from each course unit. Show truth tables, countermodels and algebraic reasoning on paper as needed. The automatic result summarizes the sampled skills. Complete the optional written proof tasks separately and ask a teacher or tutor to review the rubric; the automatic score does not establish independent proof-writing proficiency.
What you'll learn
- Apply logical tools across all twelve course units
- Evaluate truth values, quantifier order, counterexamples and argument validity
- Distinguish automatic proof-interpretation results from separately reviewed written proofs
Worked example
Problem. A program permits access when registered AND (member OR guest). For registered=true, member=false, guest=true, what happens?
- member OR guest is true.
- registered AND true is true.
Answer: Access is permitted
Lesson preview: this page shows 5 practice items. The interactive activity contains 30 practice items and 36 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.
Practice problems
1. Which sentence makes a claim with a truth value?
Choices: Calculate 15 + 2. · Is 15 even? · What an interesting number! · 15 is an even integer.
Show solution
- A false declarative claim is still a statement.
- 15 is odd, so the selected statement is false rather than not a statement.
Answer: 15 is an even integer.
2. The sentence “The sealed envelope contains a blue card” refers to one particular envelope. How should it be classified before opening it?
Choices: Automatically false · A command · A statement whose truth is not yet known to us · Not a statement because it is unobserved
Show solution
- Whether we know the truth is different from whether the sentence makes a definite claim.
Answer: A statement whose truth is not yet known to us
3. Evaluate “9 > 7 and 9 < 8.”
Choices: True · False
Show solution
- The first part is true and the second is false.
- A conjunction needs both parts true.
Answer: False
4. Rewrite “x is at most 7” exactly.
Choices: x <= 7 · x < 7 · x >= 7 · x = 7
Show solution
- At most includes the boundary and every smaller value.
Answer: x <= 7
5. All members are registered. Ada is a member. Which premise is needed to conclude Ada is registered?
Choices: All registered people are members · No members are registered · Ada is registered · All members are registered
Show solution
- The universal premise connects membership to registration.
- Do not assume the desired conclusion as the missing reason.
Answer: All members are registered
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 direct proof
Prove that the sum of an even integer and an odd integer is odd. Write a proof for arbitrary integers rather than a list of examples.
Review rubric and solution
- Represent the integers as 2r and 2s+1 with r,s integers.
- Derive 2(r+s)+1.
- State that r+s is an integer and invoke the definition of odd.
- Let a=2r and b=2s+1 for integers r,s.
- Then a+b=2r+2s+1=2(r+s)+1.
- Because r+s is an integer, the sum is odd.
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.
Write a contradiction proof
Prove that there is no smallest positive real number. Explain why zero is not a valid substitute for your constructed number.
Review rubric and solution
- Assume a smallest positive real r exists.
- Construct r/2 and verify 0<r/2<r.
- Identify the contradiction and reject the assumption.
- Explain that zero is outside the positive domain.
- Assume r is the smallest positive real.
- Since r>0, 0<r/2<r, contradicting the minimality of r.
- Thus no smallest positive real exists. Zero could not contradict minimality because it is not positive.
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