Miscellaneous Exercise answers: Sets

Open NCERT p. 21Matches NCERT’s answer

1. Solve x² − 8x + 12 = 0: it factors as (x − 2)(x − 6) = 0, so x = 2 or x = 6. So A = {2, 6}.
2. B = {2, 4, 6}, C = {2, 4, 6, 8, ...} (all even natural numbers) and D = {6}.
3. D ⊂ A, D ⊂ B and D ⊂ C, because 6 is in each of A, B and C.
4. A ⊂ B, because 2 and 6 are both in B; A ⊂ C, because 2 and 6 are both even.
5. B ⊂ C, because 2, 4 and 6 are all even.
6. No other pair works: 2 ∈ A but 2 ∉ D, so A ⊄ D; 4 ∈ B but 4 ∉ A and 4 ∉ D, so B ⊄ A and B ⊄ D; 8 ∈ C but 8 is not in A, B or D, so C is not a subset of any of them.

AnswerD ⊂ A, D ⊂ B, D ⊂ C, A ⊂ B, A ⊂ C and B ⊂ C (in short, D ⊂ A ⊂ B ⊂ C)

Watch this explained “Four sets in a line”, 12:30 into Containment, proper containment, and why the empty set is inside everything

Open NCERT p. 21Matches NCERT’s answer

(i) If x ∈ A … then x ∈ B

1. Let A = {1,2,3} and B = {A, 4} — so B's two members are the set A itself, and the number 4.
2. x = 1 is in A. Also A is in B (A is one of B's two members).
3. But is 1 an actual member of B? B's members are only the set A and the number 4 — not 1 directly.
4. So x ∈ A and A ∈ B do not force x ∈ B.

AnswerFalse — being an element of an element does not make you an element

(ii) If A ⊂ B … then A ∈ C

1. Let A = {2}, B = {2, 4}, and C = {B, 5} — C's members are the set B and the number 5.
2. A ⊂ B, since 2 ∈ B. Also B ∈ C.
3. But A itself is not one of C's members (C only holds B and 5, not the smaller set A).
4. So A ⊂ B and B ∈ C do not force A ∈ C.

AnswerFalse — being a subset of a member does not make you a member

(iii) If A ⊂ B … then A ⊂ C

1. Take any x ∈ A. Since A ⊂ B, x ∈ B.
2. Since B ⊂ C, that same x ∈ C.
3. So every member of A is a member of C, which is exactly A ⊂ C.

AnswerTrue — containment carries straight through

(iv) If A ⊄ B … then A ⊄ C

1. Let A = {3}, B = {1, 2}, C = {3, 5}.
2. A ⊄ B, since 3 ∉ B. B ⊄ C, since 1 ∉ C.
3. But A ⊂ C, since 3 ∈ C — so A ⊄ C is false here.

AnswerFalse — the two non-containments do not stop A from sitting inside C anyway

(v) If x ∈ A and A ⊄ B, then x ∈ B

1. Let A = {3, 5, 7} and B = {3, 4, 6}, and take x = 5.
2. x = 5 is in A. A ⊄ B, because 5 and 7 are not in B.
3. But x = 5 is not in B either.
4. So x ∈ A and A ⊄ B do not force x ∈ B.

AnswerFalse — x can fail to be in B even though A is not fully inside B

(vi) If A ⊂ B … then x ∉ A

1. Suppose, for contradiction, that x ∈ A.
2. Since A ⊂ B, every member of A is in B, so x would have to be in B.
3. That contradicts x ∉ B. So x cannot be in A.

AnswerTrue — this is just the contrapositive of the definition of subset

Watch this explained “Which chain survives”, 8:16 into Containment, proper containment, and why the empty set is inside everything

Open NCERT p. 21One way to think about it

1. Take any b ∈ B. Either b ∈ A or b ∉ A.
2. If b ∈ A, then b ∈ A ∩ B, and since A ∩ B = A ∩ C, b ∈ A ∩ C, so b ∈ C.
3. If b ∉ A, then since b ∈ B, b ∈ A ∪ B, and since A ∪ B = A ∪ C, b ∈ A ∪ C. As b ∉ A, this forces b ∈ C.
4. Either way b ∈ C, so B ⊂ C.
5. Swap the roles of B and C in exactly the same argument (using A ∪ C = A ∪ B and A ∩ C = A ∩ B) to get C ⊂ B.
6. B ⊂ C and C ⊂ B together mean B = C.

In shortB = C, proved by showing B ⊂ C and C ⊂ B

Watch this explained “The check, and it has two halves”, 0:51 into Why order and repetition cannot make two sets different

Open NCERT p. 21One way to think about it

1. (i) ⟹ (ii): If A ⊂ B, every member of A is in B, so no member of A is left out of B — A − B has nothing in it, i.e. A − B = φ.
2. (ii) ⟹ (iii): If A − B = φ, every member of A is already in B, so A ⊂ B; then A adds nothing new to B, so A ∪ B = B.
3. (iii) ⟹ (iv): If A ∪ B = B, then A ⊂ A ∪ B = B, so A ⊂ B; then every member of A is in B, so A ∩ B keeps all of A, giving A ∩ B = A.
4. (iv) ⟹ (i): If A ∩ B = A, then since A ∩ B ⊂ B always, A ⊂ B.
5. The chain (i) ⟹ (ii) ⟹ (iii) ⟹ (iv) ⟹ (i) shows all four say the same thing.

In shortAll four conditions are equivalent, shown by the cycle (i) ⟹ (ii) ⟹ (iii) ⟹ (iv) ⟹ (i)

Watch this explained “One implication, about any member”, 0:53 into Containment, proper containment, and why the empty set is inside everything

Open NCERT p. 21One way to think about it

1. Take any x ∈ C − B. This means x ∈ C and x ∉ B.
2. Since A ⊂ B, if x were in A it would also be in B — but x ∉ B, so x ∉ A.
3. So x ∈ C and x ∉ A, which is exactly x ∈ C − A.
4. Every member of C − B is a member of C − A, so C − B ⊂ C − A.

In shortC − B ⊂ C − A, proved directly from A ⊂ B

Watch this explained “Any set will do, not just the universe”, 9:49 into Why complementing turns each of the two operations into the other

Open NCERT p. 21One way to think about it

1. First identity — take any x ∈ A. Either x ∈ B or x ∉ B.
2. If x ∈ B, then x ∈ A ∩ B. If x ∉ B, then x ∈ A − B. Either way x ∈ (A ∩ B) ∪ (A − B), so A ⊂ (A ∩ B) ∪ (A − B).
3. Both A ∩ B and A − B are made only of members of A, so (A ∩ B) ∪ (A − B) ⊂ A.
4. Both containments give A = (A ∩ B) ∪ (A − B).
5. Second identity — take any x ∈ A ∪ (B − A). If x ∈ A, then x ∈ A ∪ B. If x ∈ B − A, then x ∈ B, so x ∈ A ∪ B. So A ∪ (B − A) ⊂ A ∪ B.
6. Now take any x ∈ A ∪ B. If x ∈ A, it is in A ∪ (B − A) directly. If x ∈ B and x ∉ A, it is in B − A, hence in A ∪ (B − A). If x ∈ B and x ∈ A, it is already in A ∪ (B − A) through A.
7. So A ∪ B ⊂ A ∪ (B − A) too, giving equality.

In shortBoth identities hold, proved by showing containment in both directions

Watch this explained “Three pieces that cannot overlap”, 3:38 into Difference and complement: the same idea with and without a universal set

Open NCERT p. 21One way to think about it

(i) A ∪ (A ∩ B) = A

1. A ∩ B is always a subset of A, since it only keeps members already in A.
2. Taking the union of A with one of its own subsets adds nothing new — A ⊂ A ∪ (A ∩ B) and A ∪ (A ∩ B) ⊂ A ∪ A = A.
3. So A ∪ (A ∩ B) = A.

In shortA ∪ (A ∩ B) = A

(ii) A ∩ (A ∪ B) = A

1. A is always a subset of A ∪ B, since A ∪ B holds everything in A already.
2. Meeting A with a bigger set that already contains it changes nothing: A ∩ (A ∪ B) = A.

In shortA ∩ (A ∪ B) = A

Watch this explained “When one set holds the other”, 2:43 into Union and intersection, and what it means for two sets to miss each other entirely

Open NCERT p. 21Checked by computerAnswers can differ: one example

1. Pick A = {1, 2}, B = {2, 3}, C = {2, 4}.
2. A ∩ B = {2}, and A ∩ C = {2} as well — the two intersections agree.
3. But B = {2, 3} and C = {2, 4} are not the same set, since 3 ∈ B but 3 ∉ C.
4. So A ∩ B = A ∩ C does not force B = C.

AnswerA = {1, 2}, B = {2, 3}, C = {2, 4}: A ∩ B = A ∩ C = {2}, yet B ≠ C

Open NCERT p. 22One way to think about it

1. Start from the hint: A = A ∩ (A ∪ X), because A ⊂ A ∪ X always.
2. Since A ∪ X = B ∪ X (given), A = A ∩ (B ∪ X).
3. By the distributive law, A ∩ (B ∪ X) = (A ∩ B) ∪ (A ∩ X).
4. But A ∩ X = φ (given), so this is just (A ∩ B) ∪ φ = A ∩ B. So A = A ∩ B, which means A ⊂ B.
5. By the same steps with A and B swapped: B = B ∩ (B ∪ X) = B ∩ (A ∪ X) = (B ∩ A) ∪ (B ∩ X) = (A ∩ B) ∪ φ = A ∩ B. So B = A ∩ B, which means B ⊂ A.
6. A ⊂ B and B ⊂ A together give A = B.

In shortA = B, proved using A = A ∩ (A ∪ X) and the distributive law

Watch this explained “The law that mixes them”, 10:25 into Union and intersection, and what it means for two sets to miss each other entirely

Open NCERT p. 22Checked by computerAnswers can differ: one example

1. Pick A = {1, 2}, B = {2, 3}, C = {1, 3}.
2. A ∩ B = {2}, which is non-empty. B ∩ C = {3}, non-empty. A ∩ C = {1}, non-empty.
3. Is there a number in all three sets at once? 1 is in A and C but not B; 2 is in A and B but not C; 3 is in B and C but not A.
4. So no member is common to all three, meaning A ∩ B ∩ C = φ, even though every pair overlaps.

AnswerA = {1, 2}, B = {2, 3}, C = {1, 3}: every pair overlaps, but all three together share nothing

Watch this explained “Four regions, then eight”, 3:34 into Turning a claim about sets into a picture you can read off