Sets - Exercise 6(C)

Let A = {2, 4, 6, 8}, B = {6, 8, 10, 12} and C = {7, 8, 9, 10}. Find :

(i) A ∪ B

(ii) A ∪ C

(iii) B ∪ C

(iv) A ∩ B

(v) A ∩ C

(vi) B ∩ C

Answer

(i) A ∪ B

List all elements from both A and B, avoiding duplicates.

We have:

A = {2, 4, 6, 8}

B = {6, 8, 10, 12}

A ∪ B = {2, 4, 6, 8} ∪ {6, 8, 10, 12}

∴ A ∪ B = {2, 4, 6, 8, 10, 12}

(ii) A ∪ C

List all elements from both A and C, avoiding duplicates.

We have:

A = {2, 4, 6, 8}

C = {7, 8, 9, 10}

A ∪ C = {2, 4, 6, 8} ∪ {7, 8, 9, 10}

∴ A ∪ C = {2, 4, 6, 7, 8, 9, 10}

(iii) B ∪ C

List all elements from both B and C, avoiding duplicates.

We have:

B = {6, 8, 10, 12}

C = {7, 8, 9, 10}

B ∪ C = {6, 8, 10, 12} ∪ {7, 8, 9, 10}

∴ B ∪ C = {6, 7, 8, 9, 10, 12}

(iv) A ∩ B

List all elements common to both A and B.

We have:

A = {2, 4, 6, 8}

B = {6, 8, 10, 12}

A ∩ B = {2, 4, 6, 8} ∩ {6, 8, 10, 12}

∴ A ∩ B = {6, 8}

(v) A ∩ C

List all elements common to both A and C.

We have:

A = {2, 4, 6, 8}

C = {7, 8, 9, 10}

A ∩ C = {2, 4, 6, 8} ∩ {7, 8, 9, 10}

∴ A ∩ C = {8}

(vi) B ∩ C

List all elements common to both B and C.

We have:

B = {6, 8, 10, 12}

C = {7, 8, 9, 10}

B ∩ C = {6, 8, 10, 12} ∩ {7, 8, 9, 10}

∴ B ∩ C = {8, 10}

Let P = {x : x is a factor of 18} and Q = {x : x is a factor of 24}.

(i) Write each one of P and Q in Roster form.

(ii) Find:

(a) P ∪ Q

(b) P ∩ Q

Answer

(i) Write each one of P and Q in Roster form.

P = {x : x is a factor of 18}

Factors of 18 are numbers that divide 18 exactly: 1, 2, 3, 6, 9, 18.

P = {1, 2, 3, 6, 9, 18}

And

Q = {x : x is a factor of 24}

Factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24.

Q = {1, 2, 3, 4, 6, 8, 12, 24}

∴ P = {1, 2, 3, 6, 9, 18}, Q = {1, 2, 3, 4, 6, 8, 12, 24}

(ii) Find:

(a) P ∪ Q

P = {1, 2, 3, 6, 9, 18}

Q = {1, 2, 3, 4, 6, 8, 12, 24}

P ∪ Q = {1, 2, 3, 6, 9, 18} ∪ {1, 2, 3, 4, 6, 8, 12, 24}

∴ P ∪ Q = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24}

(b) P ∩ Q

P = {1, 2, 3, 6, 9, 18}

Q = {1, 2, 3, 4, 6, 8, 12, 24}

P ∩ Q = {1, 2, 3, 6, 9, 18} ∩ {1, 2, 3, 4, 6, 8, 12, 24}

∴ P ∩ Q = {1, 2, 3, 6}

Let A = {a, b, c}, B = {b, d, e} and C = {e, f, g}, verify that :

(i) A ∪ B = B ∪ A

(ii) (A ∪ B) ∪ C = A ∪ (B ∪ C)

(iii) A ∩ B = B ∩ A

(iv) (A ∩ B) ∩ C = A ∩ (B ∩ C)

Answer

(i) We have:

A = {a, b, c}

B = {b, d, e}

A ∪ B = {a, b, c} ∪ {b, d, e} = {a, b, c, d, e} $\quad$.....(1)

B ∪ A = {b, d, e} ∪ {a, b, c} = {a, b, c, d, e} $\quad$......(2)

Since, (1) and (2) are equal,

∴ A ∪ B = B ∪ A

(ii) (A ∪ B) ∪ C = A ∪ (B ∪ C)

We have:

A = {a, b, c}

B = {b, d, e}

C = {e, f, g}

A ∪ B = {a, b, c} ∪ {b, d, e} = {a, b, c, d, e}

(A ∪ B) ∪ C = {a, b, c, d, e} ∪ {e, f, g} = {a, b, c, d, e, f, g} $\quad$.....(1)

Again, B ∪ C = {b, d, e} ∪ {e, f, g} = {b, d, e, f, g}

A ∪ (B ∪ C) = {a, b, c} ∪ {b, d, e, f, g} = {a, b, c, d, e, f, g} $\quad$......(2)

Since, (1) and (2) are equal,

∴ (A ∪ B) ∪ C = A ∪ (B ∪ C)

(iii) A ∩ B = B ∩ A

We have:

A = {a, b, c}

B = {b, d, e}

A ∩ B = {a, b, c} ∩ {b, d, e} = {b} $\quad$.....(1)

B ∩ A = {b, d, e} ∩ {a, b, c} = {b} $\quad$......(2)

Since, (1) and (2) are equal,

∴ A ∩ B = B ∩ A

(iv) (A ∩ B) ∩ C = A ∩ (B ∩ C)

We have:

A = {a, b, c}

B = {b, d, e}

C = {e, f, g}

A ∩ B = {a, b, c} ∩ {b, d, e} = {b}

(A ∩ B) ∩ C = {b} ∩ {e, f, g} = { } or ϕ $\quad$.....(1)

B ∩ C = {b, d, e} ∩ {e, f, g} = {e}

A ∩ (B ∩ C) = {a, b, c} ∩ {e} = { } or ϕ $\quad$......(2)

Since, (1) and (2) are equal,

∴ (A ∩ B) ∩ C = A ∩ (B ∩ C)

Let A = {x : x is a multiple of 2, x < 15}, B = {x : x is a multiple of 3, x < 20}, C = {x : x is a prime, x < 20}.

(i) Write each one of the sets A, B, C in Roster form.

(ii) Find:

(a) A ∪ B

(b) A ∪ C

(c) B ∪ C

(d) A ∩ B

(e) A ∩ C

(f) B ∩ C

Answer

(i) Write each one of the sets A, B, C in Roster form.

Given:

A = {x : x is a multiple of 2, x < 15}

Multiples of 2 less than 15 are: 2, 4, 6, 8, 10, 12, 14.

A = {2, 4, 6, 8, 10, 12, 14}

B = {x : x is a multiple of 3, x < 20}

Multiples of 3 less than 20 are: 3, 6, 9, 12, 15, 18.

B = {3, 6, 9, 12, 15, 18}

C = {x : x is a prime, x < 20}

Prime numbers between 1 and 20 are: 2, 3, 5, 7, 11, 13, 17, 19.

C = {2, 3, 5, 7, 11, 13, 17, 19}

∴ A = {2, 4, 6, 8, 10, 12, 14}, B = {3, 6, 9, 12, 15, 18} and C = {2, 3, 5, 7, 11, 13, 17, 19}

(ii) Find:

(a) A ∪ B

Given:

A = {2, 4, 6, 8, 10, 12, 14}

B = {3, 6, 9, 12, 15, 18}

A ∪ B = {2, 4, 6, 8, 10, 12, 14} ∪ {3, 6, 9, 12, 15, 18} = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 18}

∴ A ∪ B = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 18}

(b) A ∪ C

Given:

A = {2, 4, 6, 8, 10, 12, 14}

C = {2, 3, 5, 7, 11, 13, 17, 19}

A ∪ C = {2, 4, 6, 8, 10, 12, 14} ∪ {2, 3, 5, 7, 11, 13, 17, 19} = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 17, 19}

∴ A ∪ C = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 17, 19}

(c) B ∪ C

Given:

B = {3, 6, 9, 12, 15, 18}

C = {2, 3, 5, 7, 11, 13, 17, 19}

B ∪ C = {3, 6, 9, 12, 15, 18} ∪ {2, 3, 5, 7, 11, 13, 17, 19} = {2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19}

∴ B ∪ C = {2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19}

(d) A ∩ B

Given:

A = {2, 4, 6, 8, 10, 12, 14}

B = {3, 6, 9, 12, 15, 18}

A ∩ B = {2, 4, 6, 8, 10, 12, 14} ∩ {3, 6, 9, 12, 15, 18} = {6, 12}

∴ A ∩ B = {6, 12}

(e) A ∩ C

Given:

A = {2, 4, 6, 8, 10, 12, 14}

C = {2, 3, 5, 7, 11, 13, 17, 19}

A ∩ C = {2, 4, 6, 8, 10, 12, 14} ∩ {2, 3, 5, 7, 11, 13, 17, 19} = {2}

∴ A ∩ C = {2}

(f) B ∩ C

Given:

B = {3, 6, 9, 12, 15, 18}

C = {2, 3, 5, 7, 11, 13, 17, 19}

B ∩ C = {3, 6, 9, 12, 15, 18} ∩ {2, 3, 5, 7, 11, 13, 17, 19} = {3}

∴ B ∩ C = {3}

Let A = {b, d, e, f}, B = {c, d, g, h} and C = {e, f, g, h}. Find :

(i) A - B

(ii) B - C

(iii) C - A

(iv) (A - B) ∪ (B - A)

(v) (B - C) ∪ (C - B)

Answer

(i) A - B

Given:

A = {b, d, e, f}

B = {c, d, g, h}

A - B = Elements of A which are not in B

A - B = {b, d, e, f} - {c, d, g, h} = {b, e, f}

∴ A - B = {b, e, f}

(ii) B - C

Given:

B = {c, d, g, h}

C = {e, f, g, h}

B - C = Elements of B which are not in C.

B - C = {c, d, g, h} - {e, f, g, h} = {c, d}

∴ B - C = {c, d}

(iii) C - A

Given:

C = {e, f, g, h}

A = {b, d, e, f}

C - A = Elements of C which are not in A.

C - A = {e, f, g, h} - {b, d, e, f} = {g, h}

∴ C - A = {g, h}

(iv) (A - B) ∪ (B - A)

Given:

A = {b, d, e, f}

B = {c, d, g, h}

(A - B) = {b, d, e, f} - {c, d, g, h} = {b, e, f}

(B - A) = {c, d, g, h} - {b, d, e, f} = {c, g, h}

(A - B) ∪ (B - A) = {b, e, f} ∪ {c, g, h} = {b, c, e, f, g, h}

∴ (A - B) ∪ (B - A) = {b, c, e, f, g, h}

(v) (B - C) ∪ (C - B)

Given:

B = {c, d, g, h}

C = {e, f, g, h}

B - C = {c, d, g, h} - {e, f, g, h} = {c, d}

(C - B) = {e, f, g, h} - {c, d, g, h} = {e, f}

(B - C) ∪ (C - B) = {c, d} ∪ {e, f} = {c, d, e, f}

∴ (B - C) ∪ (C - B) = {c, d, e, f}

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} be the universal set and let A = {2, 3, 4, 5, 6} and B = {3, 5, 7, 8} be its subsets.

Find:

(i) A'

(ii) B'

(iii) A' ∩ B'

(iv) A' ∪ B'

Verify that:

(v) (A ∪ B)' = (A' ∩ B')

(vi) (A ∩ B)' = (A' ∪ B')

Answer

Given:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {2, 3, 4, 5, 6}

B = {3, 5, 7, 8}

(i) A'

A' = Elements in U which are not in A.

A' = U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {2, 3, 4, 5, 6} = {1, 7, 8, 9, 10}

∴ A' = {1, 7, 8, 9, 10}

(ii) B'

B' = Elements in U which are not in B.

B' = U - B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 5, 7, 8} = {1, 2, 4, 6, 9, 10}

∴ B' = {1, 2, 4, 6, 9, 10}

(iii) A' ∩ B'

We have:

A' = {1, 7, 8, 9, 10}

B' = {1, 2, 4, 6, 9, 10}

A' ∩ B' = {1, 7, 8, 9, 10} ∩ {1, 2, 4, 6, 9, 10} = {1, 9, 10}

∴ A' ∩ B' = {1, 9, 10}

(iv) A' ∪ B'

We have:

A' = {1, 7, 8, 9, 10}

B' = {1, 2, 4, 6, 9, 10}

A' ∪ B' = {1, 7, 8, 9, 10} ∪ {1, 2, 4, 6, 9, 10} = {1, 2, 4, 6, 7, 8, 9, 10}

∴ A' ∪ B' = {1, 2, 4, 6, 7, 8, 9, 10}

Verify

(v) (A ∪ B)' = (A' ∩ B')

We have:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {2, 3, 4, 5, 6}

B = {3, 5, 7, 8}

A' = {1, 7, 8, 9, 10}

B' = {1, 2, 4, 6, 9, 10}

First let us find A ∪ B:

A ∪ B = {2, 3, 4, 5, 6} ∪ {3, 5, 7, 8} = {2, 3, 4, 5, 6, 7, 8}

(A ∪ B)' = Elements in U which are not in (A ∪ B).

LHS = (A ∪ B)' = U - (A ∪ B)

LHS = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {2, 3, 4, 5, 6, 7, 8}

LHS = {1, 9, 10}

RHS = (A' ∩ B') = {1, 7, 8, 9, 10} ∩ {1, 2, 4, 6, 9, 10}

RHS = {1, 9, 10}

Since LHS = RHS,

∴ The statement (A ∪ B)' = (A' ∩ B') is verified.

(vi) (A ∩ B)' = (A' ∪ B')

Given:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {2, 3, 4, 5, 6}

B = {3, 5, 7, 8}

A' = {1, 7, 8, 9, 10}

B' = {1, 2, 4, 6, 9, 10}

First let us find A ∩ B:

A ∩ B = {2, 3, 4, 5, 6} ∩ {3, 5, 7, 8} = {3, 5}

(A ∩ B)' = Elements in U which are not in (A ∩ B).

LHS = (A ∩ B)' = U - (A ∩ B)

LHS = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 5}

LHS = {1, 2, 4, 6, 7, 8, 9, 10}

RHS = (A' ∪ B') = {1, 7, 8, 9, 10} ∪ {1, 2, 4, 6, 9, 10}

RHS = {1, 2, 4, 6, 7, 8, 9, 10}

Since LHS = RHS,

∴ The statement (A ∩ B)' = (A' ∪ B') is verified.

Let U = {a, b, c, d, e, f, g} be the universal set and let its subsets be A = {a, b, d, e} and B = {b, e, g}.

Verify that:

(i) (A ∪ B)' = (A' ∩ B')

(ii) (A ∩ B)' = (A' ∪ B')

Answer

Given:

Universal set U = {a, b, c, d, e, f, g}

Subset A = {a, b, d, e}

Subset B = {b, e, g}

(i) (A ∪ B)' = (A' ∩ B')

First let us find A ∪ B:

A ∪ B = {a, b, d, e} ∪ {b, e, g} = {a, b, d, e, g}

(A ∪ B)' = Elements in U which are not in (A ∪ B).

LHS = (A ∪ B)' = U - (A ∪ B)

LHS = {a, b, c, d, e, f, g} - {a, b, d, e, g}

LHS = {c, f}

Now, find A' and B':

A' = U - A

A' = {a, b, c, d, e, f, g} - {a, b, d, e} = {c, f, g}

B' = U - B

B' = {a, b, c, d, e, f, g} - {b, e, g} = {a, c, d, f}

RHS = (A' ∩ B') = {c, f, g} ∩ {a, c, d, f}

RHS = {c, f}

Since LHS = RHS,

∴ The statement (A ∪ B)' = (A' ∩ B') is verified.

(ii) (A ∩ B)' = (A' ∪ B')

First let us find A ∩ B:

A ∩ B = {a, b, d, e} ∩ {b, e, g} = {b, e}

LHS = (A ∩ B)' = U - (A ∩ B).

LHS = (A ∩ B)' = {a, b, c, d, e, f, g} - {b, e}

LHS = {a, c, d, f, g}

Now, find A' and B':

A' = U - A

A' = {a, b, c, d, e, f, g} - {a, b, d, e} = {c, f, g}

B' = U - B

B' = {a, b, c, d, e, f, g} - {b, e, g} = {a, c, d, f}

RHS = (A' ∪ B') = {c, f, g} ∪ {a, c, d, f}

RHS = {a, c, d, f, g}

Since LHS = RHS,

∴ The statement (A ∩ B)' = (A' ∪ B') is verified.

Let U = {3, 6, 9, 12, 15, 18, 21, 24} be the universal set and let A = {6, 12, 18, 24} be its subset.

Verify that:

(i) A ∪ A = A

(ii) A ∩ A = A

(iii) A ∩ A' = Φ

(iv) A ∪ A' = U

(v) (A')' = A

Answer

Given:

Universal set U = {3, 6, 9, 12, 15, 18, 21, 24}

Subset A = {6, 12, 18, 24}

(i) A ∪ A = A

LHS = A ∪ A = {6, 12, 18, 24} ∪ {6, 12, 18, 24}

LHS = {6, 12, 18, 24} $\quad$[since repeated elements are written once]

RHS = A = {6, 12, 18, 24}

Since LHS = RHS,

∴ The statement A ∪ A = A is verified.

(ii) A ∩ A = A

LHS = A ∩ A = {6, 12, 18, 24} ∩ {6, 12, 18, 24}

LHS = {6, 12, 18, 24}

RHS = A = {6, 12, 18, 24}

Since LHS = RHS,

∴ The statement A ∩ A = A is verified.

(iii) A ∩ A' = Φ

A' = U - A = {3, 6, 9, 12, 15, 18, 21, 24} - {6, 12, 18, 24} = {3, 9, 15, 21}

LHS = A ∩ A' = {6, 12, 18, 24} ∩ {3, 9, 15, 21}

LHS = Φ

RHS = Φ

Since LHS = RHS,

∴ The statement A ∩ A' = Φ is verified.

(iv) A ∪ A' = U

A' = {3, 9, 15, 21} $\quad$[From previous step]

LHS = A ∪ A' = {6, 12, 18, 24} ∪ {3, 9, 15, 21}

LHS = {3, 6, 9, 12, 15, 18, 21, 24}

RHS = U = {3, 6, 9, 12, 15, 18, 21, 24}

Since LHS = RHS,

∴ The statement A ∪ A' = U is verified.

(v) (A')' = A

We know A' = {3, 9, 15, 21}

LHS = (A')' = U - A' = {3, 6, 9, 12, 15, 18, 21, 24} - {3, 9, 15, 21}

LHS = {6, 12, 18, 24}

RHS = A = {6, 12, 18, 24}

Since LHS = RHS,

∴ The statement (A')' = A is verified.