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Class 11 | CBSE | MATHEMATICS
SETS | Miscellaneous Exercise on Chapter 1
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Exercise : Miscellaneous Exercise on Chapter 1
Question 1. Decide, among the following sets, which sets are subsets of one another:
A = { x : x ∈ ℝ and x satisfies x² − 8x + 12 = 0 }
B = { 2, 4, 6 }
C = { 2, 4, 6, 8, … }
D = { 6 }
Question 1. Decide, among the following sets, which sets are subsets of one another:
A = { x : x ∈ ℝ and x satisfies x² − 8x + 12 = 0 }
B = { 2, 4, 6 }
C = { 2, 4, 6, 8, … }
D = { 6 }
Step 1: Find the elements of Set A
Solve the quadratic equation:
x² − 8x + 12 = 0
Factorising:
(x − 6)(x − 2) = 0
So,
x = 2 or x = 6
Hence,
A = { 2, 6 }
Solve the quadratic equation:
x² − 8x + 12 = 0
Factorising:
(x − 6)(x − 2) = 0
So,
x = 2 or x = 6
Hence,
A = { 2, 6 }
Step 2: Compare elements of all sets
Set A: { 2, 6 }
Set B: { 2, 4, 6 }
Set C: { 2, 4, 6, 8, … } (all positive even integers)
Set D: { 6 }
Set A: { 2, 6 }
Set B: { 2, 4, 6 }
Set C: { 2, 4, 6, 8, … } (all positive even integers)
Set D: { 6 }
Step 3: Check subset relationships
• Every element of A is present in B
⇒ A ⊂ B
• Every element of A is also present in C
⇒ A ⊂ C
• Every element of D (which is 6) is present in A
⇒ D ⊂ A
• Every element of D is present in B
⇒ D ⊂ B
• Every element of D is present in C
⇒ D ⊂ C
• Every element of A is present in B
⇒ A ⊂ B
• Every element of A is also present in C
⇒ A ⊂ C
• Every element of D (which is 6) is present in A
⇒ D ⊂ A
• Every element of D is present in B
⇒ D ⊂ B
• Every element of D is present in C
⇒ D ⊂ C
✔ Subset relations are:
A ⊂ B, A ⊂ C
D ⊂ A, D ⊂ B, D ⊂ C
A ⊂ B, A ⊂ C
D ⊂ A, D ⊂ B, D ⊂ C
🧠 Neurological Insight:
The brain understands sets faster when they are broken into small comparisons. Solving Set A first reduces cognitive load, allowing the brain to treat it as a visual object. Comparing one set at a time activates pattern-matching neurons, which improves recall and reduces confusion.
👉 Memory trick: “Small sets fit inside bigger ones.” If your eyes can “see” it, your brain will remember it.
The brain understands sets faster when they are broken into small comparisons. Solving Set A first reduces cognitive load, allowing the brain to treat it as a visual object. Comparing one set at a time activates pattern-matching neurons, which improves recall and reduces confusion.
👉 Memory trick: “Small sets fit inside bigger ones.” If your eyes can “see” it, your brain will remember it.
Exercise : Miscellaneous Exercise on Chapter 1
Question 2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B, then x ∈ B
(ii) If A ⊂ B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
(iv) If A ⊄ B and B ⊄ C, then A ⊄ C
(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A
Question 2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B, then x ∈ B
(ii) If A ⊂ B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
(iv) If A ⊄ B and B ⊄ C, then A ⊄ C
(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A
(i) Analysis
Given: x ∈ A and A ∈ B
Here, A is an element of B, not a subset of B. So elements of A need not be elements of B.
Counter Example:
Let A = {1, 2}
Let B = {A, 3}
Then 1 ∈ A but 1 ∉ B
Given: x ∈ A and A ∈ B
Here, A is an element of B, not a subset of B. So elements of A need not be elements of B.
Counter Example:
Let A = {1, 2}
Let B = {A, 3}
Then 1 ∈ A but 1 ∉ B
(i) ❌ False
(ii) Analysis
Given: A ⊂ B and B ∈ C
B being an element of C does not mean its subsets are also elements of C.
Counter Example:
Let A = {1}
Let B = {1, 2}
Let C = {B}
Then A ⊂ B and B ∈ C but A ∉ C
Given: A ⊂ B and B ∈ C
B being an element of C does not mean its subsets are also elements of C.
Counter Example:
Let A = {1}
Let B = {1, 2}
Let C = {B}
Then A ⊂ B and B ∈ C but A ∉ C
(ii) ❌ False
(iii) Analysis
Given: A ⊂ B and B ⊂ C
Every element of A is in B, and every element of B is in C.
Therefore, every element of A is also in C.
Given: A ⊂ B and B ⊂ C
Every element of A is in B, and every element of B is in C.
Therefore, every element of A is also in C.
(iii) ✔ True ⇒ A ⊂ C
(iv) Analysis
Given: A ⊄ B and B ⊄ C
This does not guarantee that A is not a subset of C.
Counter Example:
Let A = {1}
Let B = {2}
Let C = {1, 3}
Then A ⊄ B, B ⊄ C, but A ⊂ C
Given: A ⊄ B and B ⊄ C
This does not guarantee that A is not a subset of C.
Counter Example:
Let A = {1}
Let B = {2}
Let C = {1, 3}
Then A ⊄ B, B ⊄ C, but A ⊂ C
(iv) ❌ False
(v) Analysis
Given: x ∈ A and A ⊄ B
Just because A is not a subset of B does not mean every element of A is outside B.
Counter Example:
Let A = {1, 2}
Let B = {2}
Then A ⊄ B, but 2 ∈ A and 2 ∈ B
Given: x ∈ A and A ⊄ B
Just because A is not a subset of B does not mean every element of A is outside B.
Counter Example:
Let A = {1, 2}
Let B = {2}
Then A ⊄ B, but 2 ∈ A and 2 ∈ B
(v) ❌ False
(vi) Analysis
Given: A ⊂ B and x ∉ B
Since all elements of A are inside B, if x is not in B, it cannot be in A.
Given: A ⊂ B and x ∉ B
Since all elements of A are inside B, if x is not in B, it cannot be in A.
(vi) ✔ True
🧠 Neurological Insight:
The human brain often confuses ∈ (element) and ⊂ (subset). Separating each statement activates logical sequencing neurons, reducing overload.
👉 Brain rule:
Elements live inside sets, subsets live inside set boundaries.
This visual hierarchy helps the brain store concepts as mental containers, improving long-term recall.
The human brain often confuses ∈ (element) and ⊂ (subset). Separating each statement activates logical sequencing neurons, reducing overload.
👉 Brain rule:
Elements live inside sets, subsets live inside set boundaries.
This visual hierarchy helps the brain store concepts as mental containers, improving long-term recall.
Exercise : Miscellaneous Exercise on Chapter 1
Question 3. Let A, B, and C be the sets such that
A ∪ B = A ∪ C and A ∩ B = A ∩ C.
Show that B = C.
Question 3. Let A, B, and C be the sets such that
A ∪ B = A ∪ C and A ∩ B = A ∩ C.
Show that B = C.
Step 1: Understand what must be proved
To prove B = C, we must show:
• Every element of B is in C
• Every element of C is in B
To prove B = C, we must show:
• Every element of B is in C
• Every element of C is in B
Step 2: Take an arbitrary element of B
Let x ∈ B.
Then two cases are possible:
Case 1: x ∈ A
Then x ∈ A ∩ B
Since A ∩ B = A ∩ C, we get x ∈ A ∩ C
Hence x ∈ C
Case 2: x ∉ A
Then x ∈ A ∪ B
Since A ∪ B = A ∪ C, we get x ∈ A ∪ C
As x ∉ A, it must be that x ∈ C
Let x ∈ B.
Then two cases are possible:
Case 1: x ∈ A
Then x ∈ A ∩ B
Since A ∩ B = A ∩ C, we get x ∈ A ∩ C
Hence x ∈ C
Case 2: x ∉ A
Then x ∈ A ∪ B
Since A ∪ B = A ∪ C, we get x ∈ A ∪ C
As x ∉ A, it must be that x ∈ C
Step 3: Conclude B ⊂ C
From both cases, every element of B belongs to C.
Therefore,
B ⊂ C
From both cases, every element of B belongs to C.
Therefore,
B ⊂ C
Step 4: By symmetry, prove C ⊂ B
Interchanging the roles of B and C in the given conditions, we similarly obtain:
C ⊂ B
Interchanging the roles of B and C in the given conditions, we similarly obtain:
C ⊂ B
✔ Since B ⊂ C and C ⊂ B,
therefore B = C.
therefore B = C.
🧠 Neurological Insight:
The brain understands this proof faster when it is divided into binary cases (inside A / outside A). This mirrors how the prefrontal cortex processes decisions—by splitting possibilities into minimal branches.
👉 Memory anchor:
“Same union + same intersection ⇒ same set.”
Visualizing A as a fixed container and B, C as movable pieces locks this concept into long-term memory.
The brain understands this proof faster when it is divided into binary cases (inside A / outside A). This mirrors how the prefrontal cortex processes decisions—by splitting possibilities into minimal branches.
👉 Memory anchor:
“Same union + same intersection ⇒ same set.”
Visualizing A as a fixed container and B, C as movable pieces locks this concept into long-term memory.
Exercise : Miscellaneous Exercise on Chapter 1
Question 4. Show that the following four conditions are equivalent:
(i) A ⊂ B
(ii) A − B = φ
(iii) A ∪ B = B
(iv) A ∩ B = A
Question 4. Show that the following four conditions are equivalent:
(i) A ⊂ B
(ii) A − B = φ
(iii) A ∪ B = B
(iv) A ∩ B = A
Step 1: Meaning of equivalence
The conditions are said to be equivalent if each condition implies the other.
We will prove this in a circular manner:
(i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i)
The conditions are said to be equivalent if each condition implies the other.
We will prove this in a circular manner:
(i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i)
Step 2: Prove (i) ⇒ (ii)
Given: A ⊂ B
This means every element of A belongs to B.
Hence, there is no element in A that is not in B.
Therefore,
A − B = φ
Given: A ⊂ B
This means every element of A belongs to B.
Hence, there is no element in A that is not in B.
Therefore,
A − B = φ
Step 3: Prove (ii) ⇒ (iii)
Given: A − B = φ
This means A has no element outside B.
So adding A to B does not introduce any new element.
Hence,
A ∪ B = B
Given: A − B = φ
This means A has no element outside B.
So adding A to B does not introduce any new element.
Hence,
A ∪ B = B
Step 4: Prove (iii) ⇒ (iv)
Given: A ∪ B = B
This means all elements of A are already contained in B.
Hence, the common part of A and B is exactly A itself.
Therefore,
A ∩ B = A
Given: A ∪ B = B
This means all elements of A are already contained in B.
Hence, the common part of A and B is exactly A itself.
Therefore,
A ∩ B = A
Step 5: Prove (iv) ⇒ (i)
Given: A ∩ B = A
This implies every element of A is also an element of B.
Hence,
A ⊂ B
Given: A ∩ B = A
This implies every element of A is also an element of B.
Hence,
A ⊂ B
✔ Conditions (i), (ii), (iii), and (iv) are
all equivalent.
🧠 Neurological Insight:
The brain learns equivalence faster when concepts are arranged in a logical loop. Circular proofs activate pattern completion neurons, helping the brain recognize that different expressions represent the same idea.
👉 Memory cue:
“Subset = nothing extra = no change in union = same intersection.”
This compressed mental chain improves long-term recall and exam performance.
The brain learns equivalence faster when concepts are arranged in a logical loop. Circular proofs activate pattern completion neurons, helping the brain recognize that different expressions represent the same idea.
👉 Memory cue:
“Subset = nothing extra = no change in union = same intersection.”
This compressed mental chain improves long-term recall and exam performance.
Exercise : Miscellaneous Exercise on Chapter 1
Question 5. Show that if A ⊂ B, then C − B ⊂ C − A.
Question 5. Show that if A ⊂ B, then C − B ⊂ C − A.
Step 1: Understand the statement
We are given that every element of A is also an element of B.
We must prove that every element of C − B is also an element of C − A.
We are given that every element of A is also an element of B.
We must prove that every element of C − B is also an element of C − A.
Step 2: Take an arbitrary element of C − B
Let x ∈ C − B.
By definition of set difference:
x ∈ C and x ∉ B
Let x ∈ C − B.
By definition of set difference:
x ∈ C and x ∉ B
Step 3: Use the given condition A ⊂ B
Since A ⊂ B, every element of A is contained in B.
If x ∉ B, then it is impossible for x to be in A.
Hence,
x ∉ A
Since A ⊂ B, every element of A is contained in B.
If x ∉ B, then it is impossible for x to be in A.
Hence,
x ∉ A
Step 4: Conclude membership in C − A
We already have:
x ∈ C and x ∉ A
Therefore,
x ∈ C − A
We already have:
x ∈ C and x ∉ A
Therefore,
x ∈ C − A
✔ Since every element of C − B is also an element of
C − A,
therefore C − B ⊂ C − A.
therefore C − B ⊂ C − A.
🧠 Neurological Insight:
The brain processes set subtraction by exclusion mapping. When A is inside B, removing B from C automatically removes A as well.
👉 Visual memory cue:
“B is a bigger filter than A.”
Anything that escapes the bigger filter (B) will surely escape the smaller one (A), making this relation easy to remember during exams.
The brain processes set subtraction by exclusion mapping. When A is inside B, removing B from C automatically removes A as well.
👉 Visual memory cue:
“B is a bigger filter than A.”
Anything that escapes the bigger filter (B) will surely escape the smaller one (A), making this relation easy to remember during exams.
Exercise : Miscellaneous Exercise on Chapter 1
Question 6. Show that for any sets A and B,
(i) A = ( A ∩ B ) ∪ ( A − B )
(ii) A ∪ ( B − A ) = ( A ∪ B )
Question 6. Show that for any sets A and B,
(i) A = ( A ∩ B ) ∪ ( A − B )
(ii) A ∪ ( B − A ) = ( A ∪ B )
Part (i): Prove
A = ( A ∩ B ) ∪ ( A − B )
Step 1: Take an arbitrary element of A
Let x ∈ A.
Two situations are possible:
Case 1: x ∈ B
Then x ∈ A ∩ B
Case 2: x ∉ B
Then x ∈ A − B
Let x ∈ A.
Two situations are possible:
Case 1: x ∈ B
Then x ∈ A ∩ B
Case 2: x ∉ B
Then x ∈ A − B
Step 2: Conclude inclusion
In both cases,
x ∈ ( A ∩ B ) ∪ ( A − B )
Hence,
A ⊂ ( A ∩ B ) ∪ ( A − B )
In both cases,
x ∈ ( A ∩ B ) ∪ ( A − B )
Hence,
A ⊂ ( A ∩ B ) ∪ ( A − B )
Step 3: Reverse inclusion
Let x ∈ ( A ∩ B ) ∪ ( A − B ).
Then either x ∈ A ∩ B or x ∈ A − B.
In both cases, x ∈ A.
Therefore,
( A ∩ B ) ∪ ( A − B ) ⊂ A
Let x ∈ ( A ∩ B ) ∪ ( A − B ).
Then either x ∈ A ∩ B or x ∈ A − B.
In both cases, x ∈ A.
Therefore,
( A ∩ B ) ∪ ( A − B ) ⊂ A
✔ Hence, A = ( A ∩ B ) ∪ ( A − B )
Part (ii): Prove
A ∪ ( B − A ) = ( A ∪ B )
Step 4: Take an arbitrary element of A ∪ ( B − A )
Let x ∈ A ∪ ( B − A ).
Then either:
• x ∈ A, or
• x ∈ B − A ⇒ x ∈ B
In both cases,
x ∈ A ∪ B
Let x ∈ A ∪ ( B − A ).
Then either:
• x ∈ A, or
• x ∈ B − A ⇒ x ∈ B
In both cases,
x ∈ A ∪ B
Step 5: Reverse inclusion
Let x ∈ A ∪ B.
If x ∈ A, then x ∈ A ∪ ( B − A )
If x ∈ B and x ∉ A, then x ∈ B − A
Hence in both cases,
x ∈ A ∪ ( B − A )
Let x ∈ A ∪ B.
If x ∈ A, then x ∈ A ∪ ( B − A )
If x ∈ B and x ∉ A, then x ∈ B − A
Hence in both cases,
x ∈ A ∪ ( B − A )
✔ Therefore, A ∪ ( B − A ) = A ∪ B
🧠 Neurological Insight:
The brain understands sets best when they are broken into mutually exclusive parts. Here, A is mentally split into “part common with B” and “part outside B”.
👉 Memory model:
“Whole = common part + leftover part.”
This mirrors how the brain organizes objects spatially, making both identities easy to recall under exam pressure.
The brain understands sets best when they are broken into mutually exclusive parts. Here, A is mentally split into “part common with B” and “part outside B”.
👉 Memory model:
“Whole = common part + leftover part.”
This mirrors how the brain organizes objects spatially, making both identities easy to recall under exam pressure.
Exercise : Miscellaneous Exercise on Chapter 1
Question 7. Using properties of sets, show that:
(i) A ∪ ( A ∩ B ) = A
(ii) A ∩ ( A ∪ B ) = A
Question 7. Using properties of sets, show that:
(i) A ∪ ( A ∩ B ) = A
(ii) A ∩ ( A ∪ B ) = A
Part (i): Prove
A ∪ ( A ∩ B ) = A
Step 1: Use distributive law
Using the distributive property of union over intersection:
A ∪ ( A ∩ B ) = ( A ∪ A ) ∩ ( A ∪ B )
Using the distributive property of union over intersection:
A ∪ ( A ∩ B ) = ( A ∪ A ) ∩ ( A ∪ B )
Step 2: Apply identity laws
Since A ∪ A = A, we get:
A ∪ ( A ∩ B ) = A ∩ ( A ∪ B )
Since A ∪ A = A, we get:
A ∪ ( A ∩ B ) = A ∩ ( A ∪ B )
Step 3: Apply absorption law
Using the absorption property:
A ∩ ( A ∪ B ) = A
Using the absorption property:
A ∩ ( A ∪ B ) = A
✔ Hence, A ∪ ( A ∩ B ) = A
Part (ii): Prove
A ∩ ( A ∪ B ) = A
Step 4: Use distributive law
Using distributive property of intersection over union:
A ∩ ( A ∪ B ) = ( A ∩ A ) ∪ ( A ∩ B )
Using distributive property of intersection over union:
A ∩ ( A ∪ B ) = ( A ∩ A ) ∪ ( A ∩ B )
Step 5: Apply identity laws
Since A ∩ A = A, we get:
A ∩ ( A ∪ B ) = A ∪ ( A ∩ B )
Since A ∩ A = A, we get:
A ∩ ( A ∪ B ) = A ∪ ( A ∩ B )
Step 6: Apply absorption law
Using the absorption property:
A ∪ ( A ∩ B ) = A
Using the absorption property:
A ∪ ( A ∩ B ) = A
✔ Hence, A ∩ ( A ∪ B ) = A
🧠 Neurological Insight:
The brain remembers algebraic identities faster when they are reversible. These two results are mirror images, activating symmetry-detection neurons.
👉 One-line memory rule:
“A absorbs anything made from A.”
Visualizing A as a dominant region that swallows overlaps helps lock these identities into long-term memory for exams.
The brain remembers algebraic identities faster when they are reversible. These two results are mirror images, activating symmetry-detection neurons.
👉 One-line memory rule:
“A absorbs anything made from A.”
Visualizing A as a dominant region that swallows overlaps helps lock these identities into long-term memory for exams.
Exercise : Miscellaneous Exercise on Chapter 1
Question 8. Show that A ∩ B = A ∩ C need not imply B = C.
Question 8. Show that A ∩ B = A ∩ C need not imply B = C.
Step 1: Understand what is being claimed
The statement says that even if B and C have the same common part with A, they may still be different sets.
To disprove an implication, it is enough to give one counterexample.
The statement says that even if B and C have the same common part with A, they may still be different sets.
To disprove an implication, it is enough to give one counterexample.
Step 2: Choose suitable sets
Let A = {1}
B = {1, 2}
C = {1, 3}
Let A = {1}
B = {1, 2}
C = {1, 3}
Step 3: Compute intersections
A ∩ B = {1}
A ∩ C = {1}
Hence,
A ∩ B = A ∩ C
A ∩ B = {1}
A ∩ C = {1}
Hence,
A ∩ B = A ∩ C
Step 4: Compare B and C
B = {1, 2}
C = {1, 3}
Clearly,
B ≠ C
B = {1, 2}
C = {1, 3}
Clearly,
B ≠ C
✔ Therefore,
A ∩ B = A ∩ C does not imply
B = C.
🧠 Neurological Insight:
The brain often commits a pattern-overgeneralization error: seeing the same overlap and assuming complete equality.
Here, A acts like a filter. Different sets can look identical after passing through the same filter.
👉 Memory image:
“Same shadow does not mean same object.”
This analogy helps the brain separate partial equality from total equality in set theory.
The brain often commits a pattern-overgeneralization error: seeing the same overlap and assuming complete equality.
Here, A acts like a filter. Different sets can look identical after passing through the same filter.
👉 Memory image:
“Same shadow does not mean same object.”
This analogy helps the brain separate partial equality from total equality in set theory.
Exercise : Miscellaneous Exercise on Chapter 1
Question 9. Let A and B be sets. If
A ∩ X = φ, B ∩ X = φ
and
A ∪ X = B ∪ X for some set X, show that A = B.
(Hint: A = A ∩ ( A ∪ X ), B = B ∩ ( B ∪ X ) and use Distributive law)
Question 9. Let A and B be sets. If
A ∩ X = φ, B ∩ X = φ
and
A ∪ X = B ∪ X for some set X, show that A = B.
(Hint: A = A ∩ ( A ∪ X ), B = B ∩ ( B ∪ X ) and use Distributive law)
Step 1: Express A using the given hint
Using the identity law of sets:
A = A ∩ ( A ∪ X )
Using the identity law of sets:
A = A ∩ ( A ∪ X )
Step 2: Apply distributive law
Using distributive property:
A ∩ ( A ∪ X ) = ( A ∩ A ) ∪ ( A ∩ X )
Since A ∩ A = A and A ∩ X = φ, we get:
A = A ∪ φ = A
Using distributive property:
A ∩ ( A ∪ X ) = ( A ∩ A ) ∪ ( A ∩ X )
Since A ∩ A = A and A ∩ X = φ, we get:
A = A ∪ φ = A
Step 3: Express B similarly
Using the same identity:
B = B ∩ ( B ∪ X )
Using the same identity:
B = B ∩ ( B ∪ X )
Step 4: Apply distributive law to B
B ∩ ( B ∪ X ) = ( B ∩ B ) ∪ ( B ∩ X )
Since B ∩ B = B and B ∩ X = φ, we get:
B = B ∪ φ = B
B ∩ ( B ∪ X ) = ( B ∩ B ) ∪ ( B ∩ X )
Since B ∩ B = B and B ∩ X = φ, we get:
B = B ∪ φ = B
Step 5: Use the given condition A ∪ X = B ∪ X
Given:
A ∪ X = B ∪ X
Intersect both sides with A:
A ∩ ( A ∪ X ) = A ∩ ( B ∪ X )
Given:
A ∪ X = B ∪ X
Intersect both sides with A:
A ∩ ( A ∪ X ) = A ∩ ( B ∪ X )
Step 6: Simplify both sides
Left side:
A ∩ ( A ∪ X ) = A
Right side using distributive law:
A ∩ ( B ∪ X ) = ( A ∩ B ) ∪ ( A ∩ X )
Since A ∩ X = φ, we get:
A = A ∩ B
Left side:
A ∩ ( A ∪ X ) = A
Right side using distributive law:
A ∩ ( B ∪ X ) = ( A ∩ B ) ∪ ( A ∩ X )
Since A ∩ X = φ, we get:
A = A ∩ B
Step 7: Conclude A ⊂ B
From A = A ∩ B, every element of A belongs to B.
Hence:
A ⊂ B
From A = A ∩ B, every element of A belongs to B.
Hence:
A ⊂ B
Step 8: Similarly prove B ⊂ A
Intersect B ∪ X with B:
B ∩ ( B ∪ X ) = B ∩ ( A ∪ X )
Simplifying similarly, we obtain:
B = B ∩ A
Hence:
B ⊂ A
Intersect B ∪ X with B:
B ∩ ( B ∪ X ) = B ∩ ( A ∪ X )
Simplifying similarly, we obtain:
B = B ∩ A
Hence:
B ⊂ A
✔ Since A ⊂ B and B ⊂ A,
therefore A = B.
🧠 Neurological Insight:
The brain grasps this proof by treating X as a neutral blocker. Since A and B never intersect X, adding X does not change their internal structure.
👉 Mental image:
“Same addition + no overlap ⇒ same original set.”
This mirrors how the brain cancels irrelevant noise, making the conclusion intuitive and exam-friendly.
The brain grasps this proof by treating X as a neutral blocker. Since A and B never intersect X, adding X does not change their internal structure.
👉 Mental image:
“Same addition + no overlap ⇒ same original set.”
This mirrors how the brain cancels irrelevant noise, making the conclusion intuitive and exam-friendly.
Exercise : Miscellaneous Exercise on Chapter 1
Question 10. Find sets A, B and C such that:
• A ∩ B, B ∩ C and A ∩ C are non-empty sets
• A ∩ B ∩ C = φ
Question 10. Find sets A, B and C such that:
• A ∩ B, B ∩ C and A ∩ C are non-empty sets
• A ∩ B ∩ C = φ
Step 1: Understand the requirement
We need:
✔ Every pair of sets must have at least one common element
✘ But no single element should be common to all three sets
We need:
✔ Every pair of sets must have at least one common element
✘ But no single element should be common to all three sets
Step 2: Choose sets carefully
Let us define:
A = {1, 2}
B = {2, 3}
C = {1, 3}
Let us define:
A = {1, 2}
B = {2, 3}
C = {1, 3}
Step 3: Check pairwise intersections
A ∩ B = {2} → non-empty ✔
B ∩ C = {3} → non-empty ✔
A ∩ C = {1} → non-empty ✔
A ∩ B = {2} → non-empty ✔
B ∩ C = {3} → non-empty ✔
A ∩ C = {1} → non-empty ✔
Step 4: Check intersection of all three sets
A ∩ B ∩ C
= {1,2} ∩ {2,3} ∩ {1,3}
No element is common to all three sets.
Hence:
A ∩ B ∩ C = φ
A ∩ B ∩ C
= {1,2} ∩ {2,3} ∩ {1,3}
No element is common to all three sets.
Hence:
A ∩ B ∩ C = φ
✔ One possible choice is:
A = {1, 2}, B = {2, 3}, C = {1, 3}
A = {1, 2}, B = {2, 3}, C = {1, 3}
🧠 Neurological Insight:
The brain finds this problem tricky because it must balance overlap and separation simultaneously. Pairwise overlaps activate association neurons, while excluding a common element activates inhibitory control.
👉 Visual memory trick:
“Three friends shaking hands pairwise, but no group hug.”
This mental image makes the condition intuitive and easy to recall in exams.
The brain finds this problem tricky because it must balance overlap and separation simultaneously. Pairwise overlaps activate association neurons, while excluding a common element activates inhibitory control.
👉 Visual memory trick:
“Three friends shaking hands pairwise, but no group hug.”
This mental image makes the condition intuitive and easy to recall in exams.