high school Sets & Propositions

Sets & Propositions

A set collects clearly defined objects; a proposition is a sentence you can call true or false. Drawing two sets' intersection and union as a Venn diagram makes their relationship clear at a glance. Flipping or negating the parts of 'if p, then q' gives the converse, inverse, and contrapositive, and only the contrapositive always matches the original in truth value. Move union counts and the contrapositive table and they line up. Complements and De Morgan continue in the formula boxes below.

Sets — via Venn Diagrams

n(A)7

n(B)5

n(A∩B)3

💡 Why subtract?

Add 7 in A and 5 in B and you'd get 12, but the overlap is counted twice. Subtracting the intersection once gives the correct union count. That's the heart of the union formula.

Cardinality of the Union

Union Formula

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

Subtract the overlap once to remove duplicates

📐 Visual Check

①A-only region = n(A) − n(A∩B)

②B-only region = n(B) − n(A∩B)

③Intersection = n(A∩B)

④Sum of the three = n(A) + n(B) − n(A∩B)

⑤Verify with the sliders above

Set Operations & De Morgan

De Morgan's Law

(A ∪ B)^c = A^c ∩ B^c

Complement of union = intersection of complements

De Morgan (2)

(A ∩ B)^c = A^c ∪ B^c

Complement of intersection = union of complements

💡 Memorization Tip

①Taking the complement flips ∪ ↔ ∩

②Intuition: 'not in A or B' = 'not in A AND not in B'

③Difference: A − B = A ∩ Bᶜ (subtract the overlap)

Propositions & Contrapositive

p (0=false, 1=true)1

q (0=false, 1=true)1

Proposition p → q

p → q is false only when p is true and q is false

A true premise with a false conclusion makes the implication false

🔍 Converse, Inverse, Contrapositive

①Converse: q → p (truth value unrelated)

②Inverse: ~p → ~q (truth value unrelated)

③Contrapositive: ~q → ~p (always equivalent to original!)

④Proof by contrapositive: prove ~q → ~p when p → q is hard

⑤If p→q is true, p is sufficient for q and q is necessary for p

Work It Out

Example 1

For subsets A, B of the universal set, n(A)=20, n(B)=15, n(A∩B)=8. Find n(A∪B).

Write the formula for the size of a union.

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

Substitute the values.

n(A ∪ B) = 20 + 15 − 8 = 27

▸ n(A ∪ B) = 27

The key is subtracting the overlapping intersection exactly once.

Example 2

For the universal set U, n(U)=30, n(A)=18, n(B)=14, n(A∩B)=7. Find n(Aᶜ ∩ Bᶜ).

First find the size of the union.

n(A ∪ B) = 18 + 14 − 7 = 25

By De Morgan, Aᶜ ∩ Bᶜ = (A ∪ B)ᶜ, so subtract from the universal set.

n(Aᶜ ∩ Bᶜ) = n(U) − n(A ∪ B) = 30 − 25 = 5

▸ n(Aᶜ ∩ Bᶜ) = 5

Turn an intersection of complements into the complement of a union via De Morgan.

Wrap-up

Grade-10 education-office assessment type

For real x, two conditions are p: 1 ≤ x ≤ 5 and q: x ≥ k. Find the largest real k for which p is sufficient for q.

①-1

②0

③1

④2

⑤3

▸ ③ 1

If p is sufficient for q, the truth sets satisfy P ⊆ Q.

P = {x | 1 ≤ x ≤ 5}, Q = {x | x ≥ k}, P ⊆ Q

For every element of P to satisfy x ≥ k, k must not exceed the minimum of P, which is 1.

k ≤ 1, so the maximum is k = 1

🎯 Exam Points

①n(A∪B) = n(A) + n(B) − n(A∩B)

②De Morgan: complement flips ∪ ↔ ∩

③p→q is false ⟺ p true, q false

④Original ⟺ contrapositive (always)

⑤Sufficiency: if p→q is true, p is sufficient for q, q is necessary for p

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