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Sets and commonly used logic: mixed assessment

Read the idea, work independently, then explain what changed.

TOPIC 01

Sets and commonly used logic: mixed assessment

A mixed assessment: identify the method, justify it and revise your reasoning.

What you will be able to explain

Connect the chapter skills without relying on the order of the exercises.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Your turn

Count the distinct elements.

A=\{7,8,7,9\}

Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Check the membership rule and count distinct objects.

Hint 2

Remove the repeated entry.

Worked solution

Check the membership rule and count distinct objects.
Calculate or simplify this relation.
$A=\{7,8,9\};\quad |A|=3$
Order and repetition do not change a set.

The requested value is 3.

Checks and common pitfalls: Order and repetition do not change a set.

Think first. Reveal a hint when the class is ready.

02 / Foundation#Your turn

How many proper subsets does A have?

|A|=8

Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Apply the element-by-element definition of inclusion.

Hint 2

Remove only the full set.

Worked solution

Apply the element-by-element definition of inclusion.
Calculate or simplify this relation.
$2^{8}-1=255$
For a nonempty set, the empty set remains a proper subset.

The requested value is 255.

Checks and common pitfalls: For a nonempty set, the empty set remains a proper subset.

Think first. Reveal a hint when the class is ready.

03 / Foundation#Your turn

Find the cardinality of the complement.

A\subseteq U,\ |U|=26,\ |A|=9

Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Translate membership into “and”, “or”, or “not”.

Hint 2

Partition the universal set into two disjoint parts.

Worked solution

Translate membership into “and”, “or”, or “not”.
Calculate or simplify this relation.
$|U\setminus A|=26-9=17$
The complement depends on the specified universal set.

The requested value is 17.

Checks and common pitfalls: The complement depends on the specified universal set.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Solve both parts and justify the conditions used. Part A: Repair this proposed implication. Part B: Disprove: a positive product forces both factors to be positive.

\begin{gathered}\text{A: }xy=xz\Rightarrow y=z\\\text{B: }ab>0\end{gathered}

Use the domain, units and sampling assumptions stated in the question.
Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

A: Test both directions and seek a counterexample when a direction fails.

Hint 2

B: Test both directions and seek a counterexample when a direction fails.

Worked solution

Part A reasoning
Test both directions and seek a counterexample when a direction fails.
Calculate or simplify this relation.
$x(y-z)=0\Rightarrow x=0\text{ or }y=z$
Cancellation requires a nonzero factor.
Part B reasoning
Test both directions and seek a counterexample when a direction fails.
Calculate or simplify this relation.
$(-8)(-9)=72>0$
A single admissible counterexample disproves the universal claim.

A: It holds if x≠0; without this condition it can fail. B: Choose a=-8, b=-9.

Checks and common pitfalls: Cancellation requires a nonzero factor. A single admissible counterexample disproves the universal claim.

Reasoning checklist · self / teacher assessment

Solve part A with its stated restrictions.
Solve part B using an appropriate representation.
Give the reasoning and check conditions in both parts.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

Count the possible integer witnesses.

-7<x<7

Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Keep the domain explicit and distinguish all values from one witness.

Hint 2

Exclude both endpoints.

Worked solution

Keep the domain explicit and distinguish all values from one witness.
Calculate or simplify this relation.
$(6)-(-6)+1=13$
Existence needs one witness; counting asks for all admissible witnesses.

The requested value is 13.

Checks and common pitfalls: Existence needs one witness; counting asks for all admissible witnesses.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

Are the two sets equal? Explain.

A=\{8,9\},\ B=\{9,8,8\}

Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Check the membership rule and count distinct objects.

Hint 2

Compare membership, not position.

Worked solution

Check the membership rule and count distinct objects.
Calculate or simplify this relation.
$A\subseteq B,\quad B\subseteq A$
Every element of either set occurs in the other.

Yes; the members coincide.

Checks and common pitfalls: Every element of either set occurs in the other.

Reasoning checklist · self / teacher assessment

State the relevant definition, condition or model.
Show a valid calculation, proof or counterexample.
Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

07 / Standard#Your turn

Solve both parts and justify the conditions used. Part A: Prove that the empty set is a subset of every set. Part B: How many subsets does A have?

\begin{gathered}\text{A: }\varnothing\subseteq A\\\text{B: }|A|=10\end{gathered}

Use the domain, units and sampling assumptions stated in the question.
Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

A: Apply the element-by-element definition of inclusion.

Hint 2

B: Apply the element-by-element definition of inclusion.

Worked solution

Part A reasoning
Apply the element-by-element definition of inclusion.
Calculate or simplify this relation.
$\neg(\varnothing\subseteq A)\Rightarrow\exists x\in\varnothing:x\notin A$
The proposed counterexample cannot exist.
Part B reasoning
Apply the element-by-element definition of inclusion.
Calculate or simplify this relation.
$2^{10}=1024$
The empty set and the full set both count.

A: No element of the empty set can violate inclusion. B: The requested value is 1024.

Checks and common pitfalls: The proposed counterexample cannot exist. The empty set and the full set both count.

Reasoning checklist · self / teacher assessment

Solve part A with its stated restrictions.
Solve part B using an appropriate representation.
Give the reasoning and check conditions in both parts.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

Find the interval intersection.

[-8,11)\cap(8,20]

Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Translate membership into “and”, “or”, or “not”.

Hint 2

Use the tighter bound at each end.

Worked solution

Translate membership into “and”, “or”, or “not”.
Calculate or simplify this relation.
$8<x<11$
Each endpoint fails at least one original condition.

The intersection is (8,11).

Checks and common pitfalls: Each endpoint fails at least one original condition.

Reasoning checklist · self / teacher assessment

State the relevant definition, condition or model.
Show a valid calculation, proof or counterexample.
Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

Solve both parts and justify the conditions used. Part A: Disprove: a positive product forces both factors to be positive. Part B: Give a necessary and sufficient condition for the product to vanish.

\begin{gathered}\text{A: }ab>0\\\text{B: }ab=0\end{gathered}

Use the domain, units and sampling assumptions stated in the question.
Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

A: Test both directions and seek a counterexample when a direction fails.

Hint 2

B: Test both directions and seek a counterexample when a direction fails.

Worked solution

Part A reasoning
Test both directions and seek a counterexample when a direction fails.
Calculate or simplify this relation.
$(-8)(-9)=72>0$
A single admissible counterexample disproves the universal claim.
Part B reasoning
Test both directions and seek a counterexample when a direction fails.
Calculate or simplify this relation.
$ab=0\iff(a=0)\lor(b=0)$
The inclusive “or” allows both factors to be zero.

A: Choose a=-8, b=-9. B: At least one factor is zero.

Checks and common pitfalls: A single admissible counterexample disproves the universal claim. The inclusive “or” allows both factors to be zero.

Reasoning checklist · self / teacher assessment

Solve part A with its stated restrictions.
Solve part B using an appropriate representation.
Give the reasoning and check conditions in both parts.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

Negate this statement.

\forall x\in\mathbb R,\ x^2>8

Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Keep the domain explicit and distinguish all values from one witness.

Hint 2

Swap ∀ for ∃ and negate >.

Worked solution

Keep the domain explicit and distinguish all values from one witness.
Calculate or simplify this relation.
$\exists x\in\mathbb R,\ x^2\le8$
The equality case belongs in the negation.

Some real x satisfies x²≤8.

Checks and common pitfalls: The equality case belongs in the negation.

Reasoning checklist · self / teacher assessment

State the relevant definition, condition or model.
Show a valid calculation, proof or counterexample.
Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

Solve both parts and justify the conditions used. Part A: Does “interesting books” specify a well-defined set? Repair the description. Part B: Count the distinct elements.

\begin{gathered}\text{B: }A=\{7,8,7,9\}\end{gathered}

Use the domain, units and sampling assumptions stated in the question.
Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

A: Check the membership rule and count distinct objects.

Hint 2

B: Check the membership rule and count distinct objects.

Worked solution

Part A reasoning
Check the membership rule and count distinct objects.
Different readers may disagree about membership. Fix a catalogue and a publication year to make membership decidable.
Set membership must be definite rather than a personal preference.
Part B reasoning
Check the membership rule and count distinct objects.
Calculate or simplify this relation.
$A=\{7,8,9\};\quad |A|=3$
Order and repetition do not change a set.

A: No; use an objective criterion such as books published in a fixed year. B: The requested value is 3.

Checks and common pitfalls: Set membership must be definite rather than a personal preference. Order and repetition do not change a set.

Reasoning checklist · self / teacher assessment

Solve part A with its stated restrictions.
Solve part B using an appropriate representation.
Give the reasoning and check conditions in both parts.

Think first. Reveal a hint when the class is ready.

12 / Standard#Your turn

How many subsets does A have?

|A|=10

Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Apply the element-by-element definition of inclusion.

Hint 2

Each element may be included or omitted.

Worked solution

Apply the element-by-element definition of inclusion.
Calculate or simplify this relation.
$2^{10}=1024$
The empty set and the full set both count.

The requested value is 1024.

Checks and common pitfalls: The empty set and the full set both count.

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

Find the cardinality of the union.

|A|=13,\ |B|=12,\ |A\cap B|=2

Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Translate membership into “and”, “or”, or “not”.

Hint 2

Remove the overlap counted twice.

Worked solution

Translate membership into “and”, “or”, or “not”.
Calculate or simplify this relation.
$|A\cup B|=13+12-2=23$
Every union member must be counted once.

The requested value is 23.

Checks and common pitfalls: Every union member must be counted once.

Think first. Reveal a hint when the class is ready.

14 / Transfer#Your turn

Solve both parts and justify the conditions used. Part A: Give a necessary and sufficient condition for the product to vanish. Part B: Repair this proposed implication.

\begin{gathered}\text{A: }ab=0\\\text{B: }xy=xz\Rightarrow y=z\end{gathered}

Use the domain, units and sampling assumptions stated in the question.
Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

A: Test both directions and seek a counterexample when a direction fails.

Hint 2

B: Test both directions and seek a counterexample when a direction fails.

Worked solution

Part A reasoning
Test both directions and seek a counterexample when a direction fails.
Calculate or simplify this relation.
$ab=0\iff(a=0)\lor(b=0)$
The inclusive “or” allows both factors to be zero.
Part B reasoning
Test both directions and seek a counterexample when a direction fails.
Calculate or simplify this relation.
$x(y-z)=0\Rightarrow x=0\text{ or }y=z$
Cancellation requires a nonzero factor.

A: At least one factor is zero. B: It holds if x≠0; without this condition it can fail.

Checks and common pitfalls: The inclusive “or” allows both factors to be zero. Cancellation requires a nonzero factor.

Reasoning checklist · self / teacher assessment

Solve part A with its stated restrictions.
Solve part B using an appropriate representation.
Give the reasoning and check conditions in both parts.

Think first. Reveal a hint when the class is ready.

15 / Transfer#Your turn

Solve both parts and justify the conditions used. Part A: Give an integer counterexample. Part B: Count the possible integer witnesses.

\begin{gathered}\text{A: }\forall n\in\mathbb Z,\ n^2>n\\\text{B: }-7<x<7\end{gathered}

Use the domain, units and sampling assumptions stated in the question.
Use the domain, units and sampling assumptions stated in the question.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

A: Keep the domain explicit and distinguish all values from one witness.

Hint 2

B: Keep the domain explicit and distinguish all values from one witness.

Worked solution

Part A reasoning
Keep the domain explicit and distinguish all values from one witness.
Calculate or simplify this relation.
$0^2=0\not>0$
One witness of failure is sufficient.
Part B reasoning
Keep the domain explicit and distinguish all values from one witness.
Calculate or simplify this relation.
$(6)-(-6)+1=13$
Existence needs one witness; counting asks for all admissible witnesses.

A: n=0 is a counterexample. B: The requested value is 13.

Checks and common pitfalls: One witness of failure is sufficient. Existence needs one witness; counting asks for all admissible witnesses.

Reasoning checklist · self / teacher assessment

Solve part A with its stated restrictions.
Solve part B using an appropriate representation.
Give the reasoning and check conditions in both parts.

Think first. Reveal a hint when the class is ready.

End-of-lesson check and correction

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Teacher preparation and assessment

Question sequence

Ask students to name the relevant condition before calculating.

Board plan

Compare valid methods and annotate their conditions.

Anticipated thinking

A correct final value may still hide a missing assumption.

Assessment checklist

Check the method, conditions, reasoning and interpretation separately.