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Equal Sets = Two sets A and B are said to be equal if they have exactly the same elements & we write A = B

We have, 

J = {2, 3} 

and K = {x : x ϵ Z, (x² + 5x + 6) = 0} 

Here, x ∈ Z and x² + 5x + 6 = 0 

The given equation can be solved as: 

x² + 5x + 6 = 0 

⇒ x² + 2x + 3x + 6 = 0 

⇒ x(x + 2) + 3(x + 2) = 0 

⇒ (x + 2)(x + 3) = 0 

⇒ x = -2 and -3 

∴ K = {-2, -3} 

∴ J ≠ K because elements of both the sets are not equal.

A total number of People n(P) = ?. 

People who play Basketball n(B) =21. 

People who play Football n(F) = 29. 

People who play Hockey n(H) = 26. 

People who play Basketball and Hockey n(B ∩ H) = 14 

People who play Football and Hockey n(H ∩ F) = 15 

People who play Basketball and Football n(B ∩ F) = 12 

People who play all games. n(H ∩B ∩ F) = 8 

Total number of people would be n(H or B or F) = n(H∪B∪F) 

We know,

 n(H∪B∪F) = n(H)+n(B)+n(F) – n (H ∩ B)– n (H ∩ F)– n (B ∩ F)+ n (H ∩B ∩ F) 

n(H∪B∪F) = 26+21+29–14–15–12+8 

n(H∪B∪F) = 43. 

Hence, 

there are 43 members in all.

A ∪ B = {x: x ϵ A or x ϵ B} 

= {1, 2, 3, 4, 6, 8} 

(A∪B)’ means Complement of (A∪B) with respect to universal set U. 

So, 

(A∪B)’ = U – (A∪B)’ 

U – ( A∪B)’ is defined as {x ϵ U : x ∉ (A∪B)’} 

U = {1, 2, 3, 4, 5, 6, 7, 8, 9} 

(A∪B)’ = {1, 2, 3, 4, 6, 8} 

U – ( A∪B)’ = {5, 9}

A = {x : x is a natural number} = {1,2,3, , } 

B = {x : x is an even number} = {2,4,6, } 

C = {x : x is an odd number} . = {1,3,5, …} 

D = {x : x is a prime number = {2,3,5, } 

A ∪ B = {1, 2, 3 …………}{2, 4, 6……………. } = {1,2,3……………} 

A ∩ C = {1, 2, 3…………. } ∩ {1, 3, 5………….. } = {1, 3, 5………….. } 

B ∩ C = {2,4,6……. }∩{1,3,5……….. } = {} = ø 

B ∩ D = {2,4,6,…} ∩ {2,3,5,…} = {2} 

Noticed that A∪ B = A 

A ∩ C = C

(i) LHS = A ∪ B 

= {a, b, c, d, e}∪ {a, c, e, g} 

= { a, b, c, d, e, g} 

= {a, c, e, g}∪{a, b, c, d, e} 

= B ∪ A 

= RHS 

Hence proved. 

(ii) To prove: A ∪ C = C ∪ A 

Since the element of set C is not provided, 

let x be any element of C. 

LHS = A ∪ C 

= {a, b, c, d, e} ∪ { x |x∈C} 

= { a, b, c, d, e, x} 

= {x, a, b, c, d, e }

= { x |x∈C} ∪ { a, b, c, d, e} 

= C ∪ A 

= RHS

Hence proved. 

(iii) To prove: B ∪ C = C ∪ B 

Since the element of set C is not provided, 

let x be any element of C. 

LHS = B ∪ C 

= { a, c, e, g } ∪ { x |x∈C} 

= { a, c, e, g, x} 

= {x, a, c, e, g } 

= {x |x∈C} ∪ { a, c, e, g }

= C∪B 

= RHS 

Hence proved. 

(iv) LHS = A ∩ B 

= {a, b, c, d, e}∪ {a, c, e, g} 

= {a, c, e}

RHS = B ∩ A 

= {a, c, e, g}∩{a, b, c, d, e} 

= {a, c, e} 

∴ A ∩ B = B ∩ A 

(v) Let x be an element of B ∩ C 

x∈B ∩ C 

x∈B and x∈C

x∈C and x∈B [by definition of intersection] 

x∈C∩B 

B∩C \(\subset\)C ∩ B ….(i) 

Now let x be an element of C∩B 

Then, x∈C∩B 

x∈C and x∈B

x∈B and x∈C [by definition of intersection]

x∈B∩C

C∩B \(\subset\)B ∩ C ….(ii) 

From (i) and (ii) we have, 

B∩C = C∩B [ every set is a subset of itself] 

Hence proved. 

(vi) Let x be an element of A ∩ C 

x∈A∩C 

x∈A and x∈C

x∈C and x∈A [by definition of intersection]

x∈C∩A 

A∩C \(\subset\)C∩A ….(i) 

Now let x be an element of C ∩ A

 Then, x∈C∩A

x∈C and x∈A

x∈A and x∈C [by definition of intersection] 

x∈A ∩ C

C ∩ A\(\subset\) A ∩ C ….(ii) 

From (i) and (ii) we have, 

A ∩ C = C ∩ A [ every set is a subset of itself] 

Hence proved. 

(vii) Let x be any element of (A ∪ B) ∪ C 

x∈(A ∪ B) or x∈C

x∈A or x∈B or x∈C 

x∈A or x∈(B ∪ C) 

x∈A ∪ (B ∪ C) 

(A ∪ B) ∪ C\(\subset\) A ∪ (B ∪ C) …..(i)

Now, let x be an element of A ∪ (B ∪ C) 

Then, x∈ A or (B ∪ C)

x∈A or x∈B or x∈C

x∈(A ∪ B) or x∈C

x∈(A ∪ B) ∪ C 

A ∪ (B ∪ C) \(\subset\)(A ∪ B) ∪ C …..(ii) 

From i and ii, (A ∪ B) ∪ C = A ∪ (B ∪ C)

[ every set is a subset of itself]

Hence , proved.

(viii) Let x be any element of (A ∩ B) ∩ C 

x∈(A ∩ B) and x∈C

x∈A and x∈B and x∈C

x∈A and x (B ∩ C)

x∈A ∩ (B ∩ C) 

(A ∩ B) ∩ C \(\subset\)A ∩ (B ∩ C) …..(i) 

Now, let x be an element of A ∩ (B ∩ C) 

Then, x∈A and (B ∩ C)

x∈A and x∈B and x∈C

x∈(A ∩ B) and x∈C

x∈(A ∩ B) ∩ C

A ∩ (B ∩ C) \(\subset\)(A ∩ B) ∩ C …..(ii) 

From i and ii, (A ∩ B) ∩ C = A ∩ (B ∩ C)

[every set is a subset of itself] 

Hence, proved.

i) A = B 

ii) A ≠ B

iii) A ≠ B 

iv) A ≠ B

i) Subsets of ‘B’ are {p}, {q}, {p, q}, φ 

ii) Subsets of ‘C’ are {x}, {y} {z}, {x, y}, {y, z}, {z, x}, {x, y, z} and φ (2 = 8) 

iii) Subsets of ‘D’ are {a}, {b}, {c}, {d}, {a,b}, {b,c}, {c, d}, {a, c}, {a, d}, {b, d}, {a, b, c}, {b, c, d}, {a, b, d}, {a, c, d}, {a, b, c, d} and φ 

iv) Subsets of ‘E’ are φ, {1}, {4}, {9}, {16}, {1,4}, {1,9}, {1, 16}, {4, 9}, {4, 16}, (9, 16}, {1, 4, 9}, {1, 9, 16}, {4, 9, 16}, {1, 4, 16}, {1, 4, 9, 16} 

v) Subsets of ‘F’ are φ, {10}, {100}, {1000}, {10, 100}, {100, 1000}, {10, 1000}, {10, 100, 1000}.

i) A = B 

ii) A ≠ E 

iii) C = D 

iv) D ≠ F 

v) F ≠ A 

vi) D ≠ E 

vii) F ≠ B

i) In R.H.S ‘x’ is greater than 1 and less than 10 but L.H.S is having both 1 and 10. 

ii) L.H.S ≠ R.H.S R.H.S: x = 2n + 1 is definition of odd numbers. L.H.S: Given set is even numbers set. 

iii) x is a multiple of 15. So 5 does not exist.

iv) x is a prime number but 9 is not a prime number.

B–C is defined as {x ϵ B : x ∉ C} 

B = {2, 3, 5, 6} 

C = {4, 5, 6, 7} 

B–C = {2, 3} 

(A ∩ (B – C)) = {x:x ϵ A and x ϵ (B – C)} 

= {2} 

R.H.S: 

(A ∩ B) = {x:x ϵ A and x ϵ B} 

= {2, 5} 

(A ∩ C) = {x:x ϵ A and x ϵ C} 

= {4, 5} 

(A ∩ B) – (A ∩ C) is defined as {x ϵ (A ∩ B) : x ∉ (A ∩ C)} 

= {2}. 

Hence Verified.

n(A) = 5 , n(B) = 6 

n(A ∪ B) = 9 , n(A ∩ B) = 2 

Now, n(A ∪ B) = 9 …(i)

n(A) + n(B) – n(A ∩ B) = 5 + 6 – 2 = 9 …(ii) 

∴ n(A ∪ B) = n(A) + n(B) – n(A ∩ B). …[From (i) and (ii)]

Here, 4 ∈ B but 4 ∉ A

∴ A and B are not equal sets,

i.e. A ≠ B

A = {1, 2, 3, 5, 7, 9, 11, 13}

B = {1, 2, 4, 6, 8, 12, 13}

A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13} 

A ∩ B= {1, 2, 13}

n(A) = 8, n(B) = 7,

n(A ∪ B) = 12, n(A ∩ B) = 3

n(A ∩ B) = 12 …(i) 

n(A) + n(B) – n(A ∩ B) = 8 + 7 – 3 = 12 …(ii)

∴ n(A ∪ B) = n(A) + n(B) – n(A ∩ B) … [From (i) and (ii)]

L.H.S. 

(B ∪ C) = {x: x ϵ B or x ϵ C } 

= {2, 3, 4, 5, 6, 7}. 

A–(B ∪ C) is defined as {x ϵ A : x ∉ (B ∪ C)} 

A = {1, 2, 4, 5} 

(B ∪ C) = {2, 3, 4, 5, 6, 7}. 

A–(B ∪ C) = {1} 

R.H.S 

(A – B) = 

A–B is defined as {x ϵ A : x ∉ B} 

A = {1, 2, 4, 5} 

B = {2, 3, 5, 6} 

A–B = {1, 4} 

(A–C) = 

A–C is defined as {x ϵ A : x ∉ C} 

A = {1, 2, 4, 5} 

C = {4, 5, 6, 7} 

A–C = {1, 2} 

(A – B) ∩ (A–C) = {x:x ϵ (A – B) and x ϵ (A – C) }. 

= {1, 2} 

Hence verified.

L.H.S. 

(B ∪ C) = {x: x ϵ B or x ϵ C } 

= {2, 3, 4, 5, 6, 7}. 

(A ∩ (B ∪ C)) = {x:x ϵ A and x ϵ (B ∪ C)} 

= {2, 4, 5} 

R.H.S: 

(A ∩ B) = {x:x ϵ A and x ϵ B} 

= {2, 5} 

(A ∩ C) = {x:x ϵ A and x ϵ C} 

= {4, 5} 

(A ∩ B) ∪ (A ∩ C) = {x:x ϵ (A∩B) and x ϵ (A∩C) }. 

= {2, 4, 5}. 

Hence verified

A ∪ {1, 2} = {1, 2, 3, 5, 9} 

Elements of A and {1, 2} together give us the result 

So smallest se of A can be A 

= {1, 2, 3, 5, 9} – {1, 2} 

A = {3, 5, 9}.

(i) B - C represents all elements in B that are not in C

B - C = {a, c}

A∩(B - C) = {a, c}

A∩B = {a, c, e}

A∩C = {b, e} 

(A∩B) - (A∩C) = {a, c}

A∩(B - C) = (A∩B) - (A∩C) 

Hence proved

(ii) B∩C = {e, g} 

A - (B∩C) = {a, b, c, d}

(A - B) = {b, d}

(A - C) = {a, c, d}

(A - B)∪(A - C) = {a, b, c, d}

A - (B∩C) = (A - B)∪(A - C) 

Hence proved 

L.H.S = 

(B ∩ C) = {x:x ϵ B and x ϵ C} 

= {5, 6} 

A ∪ (B ∩ C) = {x: x ϵ A or x ϵ (B ∩ C) } 

= {1, 2, 4, 5, 6}. 

R.H.S = 

(A ∪ B) = {x: x ϵ A or x ϵ B } 

= {1, 2, 4, 5, 6}. 

(A ∪ C) = {x: x ϵ A or x ϵ C } 

= {1, 2, 4, 5, 6, 7}. 

(A ∪ B) ∩ (A ∪ C) = {x:x ϵ (A ∪ B) and x ϵ (A ∪ C) }. 

= {1, 2, 4, 5, 6}. 

Hence Verified.

(B ∩ C) = {x:x ϵ B and x ϵ C} 

= {5, 6} 

(A – (B ∩ C)) = 

A–(B ∩ C) is defined as {x ϵ A : x ∉(B ∩ C)} 

A = {1, 2, 4, 5} 

(B ∩ C) = {5, 6} 

(A – (B ∩ C)) = {1, 2, 4} 

R.H.S: 

(A – B) = 

A–B is defined as {x ϵ A : x ∉ B} 

A = {1, 2, 4, 5} 

B = {2, 3, 5, 6} 

A–B = {1, 4} 

(A–C) = 

A–C is defined as {x ϵ A : x ∉ C} 

A = {1, 2, 4, 5} 

C = {4, 5, 6, 7} 

A–C = {1, 2} 

(A – B) ∪ (A–C) = {x:x ϵ (A – B) OR x ϵ (A – C) }. 

= {1, 2, 4} 

Hence verified

(i) A∪B = {2, 3, 4, 5, 6, 7, 8} 

(A∪B)’ = {1, 9} 

A’ = {1, 3, 5, 7, 9} 

B’ = {1, 4, 6, 8, 9} 

A’∩B’ = {1, 9} 

(A∪B)’ = A’∩B’ 

Hence proved 

(ii) A∩B = {2} 

(A∩B)’ = {1, 3, 4, 5, 6, 7, 8, 9}

A’∪B’ = {1, 3, 4, 5, 6, 7, 8, 9}

(A∩B)’ = A’UB’ 

Hence proved 

These are also known as De Morgan’s theorem

(i) A’ = {d, e, f} 

(A’)’ = {a, b, c} = A 

Hence proved 

(ii) A∪B = {a, b, c, d, e} 

(A∪B)’ = {f} 

A’ = {d, e, f} 

B’ = {a, f} 

A’∩B’ = {f} 

(A∪B)’ = (A’∩B’) 

Hence proved 

(iii) A’∪B’ = {a, d, e, f} 

A∩B = (b, c} 

(A∩B)’ = {a, d, e, f} 

(A∩B)’ = A’∪B’ 

Hence proved

1. A = {1, 3, 4, 8} ; B = {2, 3, 5}

2. A – B = {1, 4, 8}; A ∩ B = {3}

⇒ (A – B) ∪ (A ∩ B) = {1, 3, 4, 8}

Hence; (A – B) ∪ (A ∩ B) = A

3. (A ∩ B)’ = {1, 2, 4, 5, 6, 7, 8, 9}