Higher Mathematics · Classes 9-10 · Chapter 1
Cartesian Product and Propositions of Set Operations.
Pick one element from A = {1, 2} and one from B = {a, b, c} and pair them: how many pairs can you make? And when you move the brackets around a union or an intersection, or break the bar of a complement, when does the answer change? This note takes the rules one at a time, with their proofs.
1. The Cartesian product
The Cartesian product of two sets A and B is the set of all ordered pairs whose first entry comes from A and whose second entry comes from B: A × B = {(x, y) : x ∈ A and y ∈ B}.
Example 12: with A = {1, 2} and B = {a, b, c}, A × B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)}. There are 2 × 3 = 6 ordered pairs. The first entry always comes from A, so (a, 1) is not in this set.
See A × B as a grid: A down the side, B along the top, one ordered pair in each cell. Add or remove elements and watch the count stay n(A) × n(B).
Counting the cells of A × B
3 and d are not in the book's example; they are there so you can change the sets yourself.
2. Commutative, associative and distributive
Commutative law: A ∪ B = B ∪ A and A ∩ B = B ∩ A. Check: with A = {1, 2, 4} and B = {2, 3, 5}, the union is {1, 2, 3, 4, 5} and the intersection {2} in either order.
Associative law: (A ∪ B) ∪ C = A ∪ (B ∪ C) and (A ∩ B) ∩ C = A ∩ (B ∩ C). Check: with A = {a, b, c, d}, B = {b, c, f}, C = {c, d, g}, both groupings give the union {a, b, c, d, f, g} and the intersection {c}.
Distributive law: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) and A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). Each of union and intersection is distributive with respect to the other.
The book's activity check: with A = {1, 2, 3, 6}, B = {2, 3, 4, 5}, C = {3, 5, 6, 7}, B ∩ C = {3, 5} and A ∪ (B ∩ C) = {1, 2, 3, 5, 6}. On the other side A ∪ B = {1, 2, 3, 4, 5, 6} and A ∪ C = {1, 2, 3, 5, 6, 7}, whose intersection is {1, 2, 3, 5, 6}. Both sides agree.
Now see it. Press one side's button, then the other's. If the same regions get shaded, the two sets are the same.
See the law
A hand-drawn Venn diagram, not to scale.
3. De Morgan's laws
Take the complement and ∪ and ∩ swap. De Morgan's laws are two: (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′.
Here are eight equations from the four laws. Say which law each one is.
Order, brackets or complement?
The proof of Proposition 1 chases one element. If x ∈ (A ∪ B)′ then x ∉ A ∪ B, so x ∉ A and x ∉ B, that is x ∈ A′ and x ∈ B′, that is x ∈ A′ ∩ B′. So (A ∪ B)′ ⊆ A′ ∩ B′. Run the same steps backwards and you get A′ ∩ B′ ⊆ (A ∪ B)′. Together they give equality. Put the steps in order.
The De Morgan proof
4. More formulas
A ∪ A = A, A ∩ A = A. A ∪ ∅ = A, A ∩ ∅ = ∅. A ∪ U = U, A ∩ U = A.
If A ⊆ B, then A ∪ B = B, A ∩ B = A and B′ ⊆ A′.
For any A and B, A ⊆ A ∪ B, B ⊆ A ∪ B, A ∩ B ⊆ A and A ∩ B ⊆ B. The union contains every element of A, and every element of the intersection is in A.
5. Propositions 2 and 3
Proposition 2: A \ B = A ∩ B′. Proof: x ∈ A \ B ⟺ x ∈ A and x ∉ B ⟺ x ∈ A and x ∈ B′ ⟺ x ∈ A ∩ B′.
Proposition 3: A × (B ∩ C) = (A × B) ∩ (A × C) and A × (B ∪ C) = (A × B) ∪ (A × C). It is proved from the definition by taking (x, y) and showing inclusion both ways.
A numerical check: with A = {a, b}, B = {2, 3}, C = {2, 4, 5}, B ∪ C = {2, 3, 4, 5}, so A × (B ∪ C) has 2 × 4 = 8 ordered pairs. A × B has 4 and A × C has 6; count the shared (a, 2) and (b, 2) once and the union has exactly 8.
Every proof has the same shape. To prove two sets equal, take a general element x and first show left side ⊆ right side, then right side ⊆ left side. Together they give equality.
6. Identities from the activities and exercises
A \ A = ∅ and A \ (A \ A) = A \ ∅ = A.
A′ \ B′ = B \ A, because by Proposition 2, A′ \ B′ = A′ ∩ (B′)′ = A′ ∩ B = B ∩ A′ = B \ A.
De Morgan works for three sets too: (A ∪ B ∪ C)′ = A′ ∩ B′ ∩ C′ and (A ∩ B ∩ C)′ = A′ ∪ B′ ∪ C′.
A ∩ (B ∩ C) = (A ∩ B) ∩ (A ∩ C), because the right side is A ∩ A ∩ B ∩ C = A ∩ B ∩ C.
Five conditions are equivalent to A ⊆ B: A ∩ B = A, A ∪ B = B, B′ ⊆ A′, A ∩ B′ = ∅, B ∪ A′ = U. If any one holds, A ⊆ B.
If A ⊂ B and C ⊂ D, then A × C ⊂ B × D, because (x, y) ∈ A × C gives x ∈ B and y ∈ D.
7. Check yourself
Start with the first question.
Question 1
For any sets A, B and C, which of these is a distributive law?
The second is a little different in style.
Question 2
If A = {1, 2} and B = {a, b, c}, how many ordered pairs are in A × B?
Use all the laws on the last one.
Question 3
If A ⊂ B, which of these statements are correct?
i. A ∪ B = B
ii. B \ A = ∅
iii. A ∩ B = A
The whole note on seven cards.
Seven cards
Source: Higher Mathematics, Classes 9-10 (NCTB), Chapter 1 “Set and Function”, the “Cartesian Product and Propositions of Set Operations” part.