Relations and Functions — Exercise 2.1

Find $a$ and $b$ if

(i) $(a + 1, b - 2) = (3, 1)$

(ii) $\left(\dfrac{a}{3} + 1,\ b - \dfrac{2}{3}\right) = \left(\dfrac{5}{3},\ \dfrac{1}{3}\right)$

(iii) $(2a, a + b) = (6, 2)$

(iv) $(a + b, 3b - 2) = (7, -5)$.

Answer

Two ordered pairs are equal if and only if their corresponding components are equal i.e. $(a, b) = (c, d)$ ⇔ $a = c$ and $b = d$.

(i) $(a + 1, b - 2) = (3, 1)$

Equating the corresponding components,

⇒ $a + 1 = 3$ and $b - 2 = 1$

⇒ $a = 3 - 1$ and $b = 1 + 2$

⇒ $a = 2$ and $b = 3$

Hence, $\bm{a = 2}$ and $\bm{b = 3}$.

(ii) $\left(\dfrac{a}{3} + 1,\ b - \dfrac{2}{3}\right) = \left(\dfrac{5}{3},\ \dfrac{1}{3}\right)$

Equating the corresponding components,

⇒ $\dfrac{a}{3} + 1 = \dfrac{5}{3}$ and $b - \dfrac{2}{3} = \dfrac{1}{3}$

⇒ $\dfrac{a}{3} = \dfrac{5}{3} - 1$ and $b = \dfrac{1}{3} + \dfrac{2}{3}$

⇒ $\dfrac{a}{3} = \dfrac{2}{3}$ and $b = \dfrac{3}{3}$

⇒ $a = 2$ and $b = 1$

Hence, $\bm{a = 2}$ and $\bm{b = 1}$.

(iii) $(2a, a + b) = (6, 2)$

Equating the corresponding components,

⇒ $2a = 6$ and $a + b = 2$

⇒ $a = 3$

Substituting $a = 3$ in $a + b = 2$,

⇒ $3 + b = 2$

⇒ $b = 2 - 3 = -1$

Hence, $\bm{a = 3}$ and $\bm{b = -1}$.

(iv) $(a + b, 3b - 2) = (7, -5)$

Equating the corresponding components,

⇒ $a + b = 7$ and $3b - 2 = -5$

⇒ $3b = -5 + 2$

⇒ $3b = -3$

⇒ $b = -1$

Substituting $b = -1$ in $a + b = 7$,

⇒ $a - 1 = 7$

⇒ $a = 8$

Hence, $\bm{a = 8}$ and $\bm{b = -1}$.

Find x and y if

(i) (4x + 3, y) = (3x + 5, −2)

(ii) (x − y, x + y) = (6, 10)

Answer

Two ordered pairs are equal if and only if their corresponding components are equal.

(i) (4x + 3, y) = (3x + 5, −2)

Equating the corresponding components,

⇒ 4x + 3 = 3x + 5 and y = −2

⇒ 4x − 3x = 5 − 3

⇒ x = 2

Hence, x = 2 and y = −2.

(ii) (x − y, x + y) = (6, 10)

Equating the corresponding components,

x − y = 6 ...........(1)

x + y = 10 ...........(2)

Adding equations (1) and (2), we get :

⇒ 2x = 16

⇒ x = 8

Substituting x = 8 in equation (2), we get :

⇒ 8 + y = 10

⇒ y = 2

Hence, x = 8 and y = 2.

If the ordered pairs (a, −1) and (5, b) belong to {(x, y) : y = 2x − 3}, find the values of a and b.

Answer

Given,

The ordered pairs (a, −1) and (5, b) belong to the set {(x, y) : y = 2x − 3}, so each of them satisfies y = 2x − 3.

For the ordered pair (a, −1), we have x = a and y = −1.

⇒ −1 = 2a − 3

⇒ 2a = −1 + 3

⇒ 2a = 2

⇒ a = 1

For the ordered pair (5, b), we have x = 5 and y = b.

⇒ b = 2(5) − 3

⇒ b = 10 − 3

⇒ b = 7

Hence, a = 1 and b = 7.

If A = {−1, 0, 1} and B = {3, 5}, write the following:

(i) A × B

(ii) B × A

(iii) B × B.

Answer

Given,

A = {−1, 0, 1} and B = {3, 5}

We know that,

A × B = {(a, b) : for all a ∈ A, b ∈ B}

(i) A × B

Pairing each element of A with each element of B,

Hence, A × B = {(−1, 3), (−1, 5), (0, 3), (0, 5), (1, 3), (1, 5)}.

(ii) B × A

Pairing each element of B with each element of A,

Hence, B × A = {(3, −1), (3, 0), (3, 1), (5, −1), (5, 0), (5, 1)}.

(iii) B × B

Pairing each element of B with each element of B,

Hence, B × B = {(3, 3), (3, 5), (5, 3), (5, 5)}.

If A is a set such that n(A) = 3 and B = {3, 4, 5}, then what is the number of elements in A × B?

Answer

We know that,

If A and B are finite sets, then n(A × B) = n(A) × n(B).

Given,

n(A) = 3 and B = {3, 4, 5}, so n(B) = 3.

∴ n(A × B) = n(A) × n(B) = 3 × 3 = 9

Hence, the number of elements in A × B is 9.

If A = {−3, −1, 0, 4} and B = {−1, 0, 1, 2, 3}, then write the number of elements in each of the following cartesian products:

(i) A × B

(ii) B × A

(iii) A × A

(iv) B × B.

Answer

Given,

A = {−3, −1, 0, 4} and B = {−1, 0, 1, 2, 3}

∴ n(A) = 4 and n(B) = 5

We know that, n(X × Y) = n(X) × n(Y).

(i) A × B

n(A × B) = n(A) × n(B) = 4 × 5 = 20

Hence, the number of elements in A × B is 20.

(ii) B × A

n(B × A) = n(B) × n(A) = 5 × 4 = 20

Hence, the number of elements in B × A is 20.

(iii) A × A

n(A × A) = n(A) × n(A) = 4 × 4 = 16

Hence, the number of elements in A × A is 16.

(iv) B × B

n(B × B) = n(B) × n(B) = 5 × 5 = 25

Hence, the number of elements in B × B is 25.

If A = {1, 2} and B = {3, 4}, then how many subsets will A × B have?

Answer

Given,

A = {1, 2} and B = {3, 4}

∴ A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

⇒ n(A × B) = 4

We know that,

A set containing n elements has 2ⁿ subsets.

∴ Number of subsets of A × B = 2⁴ = 16

Hence, A × B will have 16 subsets.

If x ∈ {2, 3, 5} and y ∈ {2, 4, 6}, form the set of all ordered pairs (x, y) such that x < y.

Answer

Given,

x ∈ {2, 3, 5} and y ∈ {2, 4, 6}

Taking each value of x in turn and choosing those values of y which are greater than x,

when x = 2, y = 4, 6 (as 2 < 4 and 2 < 6),

when x = 3, y = 4, 6 (as 3 < 4 and 3 < 6),

when x = 5, y = 6 (as 5 < 6).

Thus, the corresponding ordered pairs are (2, 4), (2, 6), (3, 4), (3, 6) and (5, 6).

Hence, the required set of ordered pairs is {(2, 4), (2, 6), (3, 4), (3, 6), (5, 6)}.

If x ∈ {−1, 2, 3, 4, 5} and y ∈ {0, 3, 6}, form the set of all ordered pairs (x, y) such that x + y = 5.

Answer

Given,

x ∈ {−1, 2, 3, 4, 5} and y ∈ {0, 3, 6} with x + y = 5 i.e. y = 5 − x.

Taking each value of x in turn,

when x = −1, y = 5 − (−1) = 6 ∈ {0, 3, 6},

when x = 2, y = 5 − 2 = 3 ∈ {0, 3, 6},

when x = 3, y = 5 − 3 = 2 ∉ {0, 3, 6},

when x = 4, y = 5 − 4 = 1 ∉ {0, 3, 6},

when x = 5, y = 5 − 5 = 0 ∈ {0, 3, 6}.

Thus, the corresponding ordered pairs are (−1, 6), (2, 3) and (5, 0).

Hence, the required set of ordered pairs is {(−1, 6), (2, 3), (5, 0)}.

If x ∈ {2, 3, 4} and y ∈ {4, 6, 9, 10}, form the set of all ordered pairs (x, y) such that x is a factor of y.

Answer

Given,

x ∈ {2, 3, 4} and y ∈ {4, 6, 9, 10}

Taking each value of x in turn and choosing those values of y of which x is a factor,

when x = 2, y = 4, 6, 10 (as 2 divides 4, 6 and 10),

when x = 3, y = 6, 9 (as 3 divides 6 and 9),

when x = 4, y = 4 (as 4 divides 4).

Thus, the corresponding ordered pairs are (2, 4), (2, 6), (2, 10), (3, 6), (3, 9) and (4, 4).

Hence, the required set of ordered pairs is {(2, 4), (2, 6), (2, 10), (3, 6), (3, 9), (4, 4)}.

State whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly.

(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ B and y ∈ A.

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.

Answer

(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}

Given,

P = {m, n} and Q = {n, m}

Pairing each element of P with each element of Q,

P × Q = {(m, n), (m, m), (n, n), (n, m)}

which has four ordered pairs and not two.

Hence, the given statement is false and the correct statement is P × Q = {(m, n), (m, m), (n, n), (n, m)}.

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ B and y ∈ A

By definition, A × B = {(x, y) : for all x ∈ A, y ∈ B}, so the first component is taken from A and the second component from B.

Hence, the given statement is false and the correct statement is: if A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ

Given,

A = {1, 2} and B = {3, 4}

The empty set φ has no element, so B and φ have no element in common.

⇒ B ∩ φ = φ

∴ A × (B ∩ φ) = A × φ = φ, as there is no second component available to form an ordered pair.

Hence, the given statement is true.

If A = {−1, 0, 1, 2, 3}, write the subset S of A × A such that the second component of the elements of S is 0.

Answer

Given,

A = {−1, 0, 1, 2, 3}

S is the set of those ordered pairs of A × A whose second component is 0, so the first component can be any element of A while the second component is fixed as 0.

Thus, the corresponding ordered pairs are (−1, 0), (0, 0), (1, 0), (2, 0) and (3, 0).

Hence, S = {(−1, 0), (0, 0), (1, 0), (2, 0), (3, 0)}.

If A × B = {(−1, 1), (−1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}, find A and B.

Answer

Given,

A × B = {(−1, 1), (−1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}

We know that,

A is the set of all first components of the ordered pairs of A × B and B is the set of all second components.

The first components are −1, −1, 2, 2, 3, 3.

∴ A = {−1, 2, 3}

The second components are 1, 2, 1, 2, 1, 2.

∴ B = {1, 2}

Hence, A = {−1, 2, 3} and B = {1, 2}.

If $A = {x, y, z}$ and some elements of $A \times B$ are $(x, 1)$, $(y, 2)$, $(z, 1)$, then write the set $B$ such that $n(A \times B) = 6$.

Answer

Given,

$A = {x, y, z}$, so $n(A) = 3$.

Also, $n(A \times B) = 6$

We know that,

$n(A \times B) = n(A) \times n(B)$.

⇒ $6 = 3 \times n(B)$

⇒ $n(B) = \dfrac{6}{3} = 2$

The ordered pairs $(x, 1)$, $(y, 2)$ and $(z, 1)$ belong to $A \times B$, so their second components 1 and 2 belong to $B$.

As $B$ contains only 2 elements, $B$ can have no element besides 1 and 2.

Hence, $\bm{B = {1, 2}}$.

If A × B = {(x, 1), (y, 2), (x, 3), (y, 3), (y, 1), (x, 2)}, then find B × A.

Answer

Given,

A × B = {(x, 1), (y, 2), (x, 3), (y, 3), (y, 1), (x, 2)}

B × A is obtained from A × B by interchanging the components of each ordered pair.

Thus, the corresponding ordered pairs are (1, x), (2, y), (3, x), (3, y), (1, y) and (2, x).

Hence, B × A = {(1, x), (2, y), (3, x), (3, y), (1, y), (2, x)}.

If A = {1, 2, 3, 4} and B = {5, 7, 9}, find

(i) A × B

(ii) B × A

(iii) Is A × B = B × A?

(iv) Is n(A × B) = n(B × A)?

Answer

Given,

A = {1, 2, 3, 4} and B = {5, 7, 9}

(i) Pairing each element of A with each element of B,

Hence, A × B = {(1, 5), (1, 7), (1, 9), (2, 5), (2, 7), (2, 9), (3, 5), (3, 7), (3, 9), (4, 5), (4, 7), (4, 9)}.

(ii) Pairing each element of B with each element of A,

Hence, B × A = {(5, 1), (5, 2), (5, 3), (5, 4), (7, 1), (7, 2), (7, 3), (7, 4), (9, 1), (9, 2), (9, 3), (9, 4)}.

(iii) The ordered pair (1, 5) ∈ A × B, but (1, 5) ∉ B × A, since the pair (1, 5) is not equal to the pair (5, 1).

Hence, A × B ≠ B × A.

(iv) Here n(A) = 4 and n(B) = 3.

∴ n(A × B) = n(A) × n(B) = 4 × 3 = 12

and n(B × A) = n(B) × n(A) = 3 × 4 = 12

Hence, n(A × B) = n(B × A).

Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}.

Verify that

(i) A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) A × C is a subset of B × D.

Answer

Given,

A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) A × (B ∩ C) = (A × B) ∩ (A × C)

The sets B and C have no element in common.

⇒ B ∩ C = φ

∴ A × (B ∩ C) = A × φ

⇒ A × (B ∩ C) = φ ...........(1)

Now,

A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)} and

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

These two sets have no ordered pair in common.

∴ (A × B) ∩ (A × C) = φ ...........(2)

From equations (1) and (2), we get

Hence, A × (B ∩ C) = (A × B) ∩ (A × C).

(ii) A × C is a subset of B × D

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

Each of the ordered pairs (1, 5), (1, 6), (2, 5) and (2, 6) of A × C is also an element of B × D.

Hence, A × C is a subset of B × D.

If A = {x : x ∈ W, x < 2}, B = {x : x ∈ N, 1 < x < 5} and C = {3, 5}, find

(i) A × (B ∩ C)

(ii) A × (B ∪ C).

Answer

Given,

A = {x : x ∈ W, x < 2}

The whole numbers less than 2 are 0 and 1.

∴ A = {0, 1}

B = {x : x ∈ N, 1 < x < 5}

The natural numbers lying between 1 and 5 are 2, 3 and 4.

∴ B = {2, 3, 4}

and C = {3, 5}

(i) A × (B ∩ C)

The only element common to B and C is 3.

⇒ B ∩ C = {3}

∴ A × (B ∩ C) = {0, 1} × {3}

Hence, A × (B ∩ C) = {(0, 3), (1, 3)}.

(ii) A × (B ∪ C)

Collecting all the elements which belong to B or C or both,

⇒ B ∪ C = {2, 3, 4, 5}

∴ A × (B ∪ C) = {0, 1} × {2, 3, 4, 5}

= {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}

Hence, A × (B ∪ C) = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}.

If A = {x : x ∈ N and x ≤ 3}, B = {x : x ∈ I, −1 ≤ x ≤ 1} and C = {1, 2}, verify that

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

(iii) (A − B) × C = A × C − B × C.

Answer

Given,

A = {x : x ∈ N and x ≤ 3}

The natural numbers not exceeding 3 are 1, 2 and 3.

∴ A = {1, 2, 3}

B = {x : x ∈ I, −1 ≤ x ≤ 1}

The integers from −1 to 1 are −1, 0 and 1.

∴ B = {−1, 0, 1}

and C = {1, 2}

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

B ∪ C = {−1, 0, 1, 2}

∴ A × (B ∪ C) = {(1, −1), (1, 0), (1, 1), (1, 2), (2, −1), (2, 0), (2, 1), (2, 2), (3, −1), (3, 0), (3, 1), (3, 2)} ...........(1)

Now,

A × B = {(1, −1), (1, 0), (1, 1), (2, −1), (2, 0), (2, 1), (3, −1), (3, 0), (3, 1)} and

A × C = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}

Collecting all the ordered pairs which belong to A × B or A × C or both,

∴ (A × B) ∪ (A × C) = {(1, −1), (1, 0), (1, 1), (1, 2), (2, −1), (2, 0), (2, 1), (2, 2), (3, −1), (3, 0), (3, 1), (3, 2)} ...........(2)

From equations (1) and (2), we get

Hence, A × (B ∪ C) = (A × B) ∪ (A × C).

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

The only element common to B and C is 1.

⇒ B ∩ C = {1}

∴ A × (B ∩ C) = {(1, 1), (2, 1), (3, 1)} ...........(1)

The ordered pairs common to A × B and A × C are (1, 1), (2, 1) and (3, 1).

∴ (A × B) ∩ (A × C) = {(1, 1), (2, 1), (3, 1)} ...........(2)

From equations (1) and (2), we get

Hence, A × (B ∩ C) = (A × B) ∩ (A × C).

(iii) (A − B) × C = A × C − B × C

Removing from A the element 1 which also belongs to B,

⇒ A − B = {2, 3}

∴ (A − B) × C = {(2, 1), (2, 2), (3, 1), (3, 2)} ...........(1)

Now,

A × C = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)} and

B × C = {(−1, 1), (−1, 2), (0, 1), (0, 2), (1, 1), (1, 2)}

Removing from A × C the ordered pairs (1, 1) and (1, 2) which also belong to B × C,

∴ A × C − B × C = {(2, 1), (2, 2), (3, 1), (3, 2)} ...........(2)

From equations (1) and (2), we get

Hence, (A − B) × C = A × C − B × C.

If P = {1, 2}, form the set P × P × P.

Answer

Given,

P = {1, 2}

We know that,

P × P × P = {(a, b, c) : for all a ∈ P, b ∈ P, c ∈ P}

Forming all the triplets whose components are taken from P,

Hence, P × P × P = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}.

If A = {a, b, c} and some elements of A × B are (a, p), (b, q), (c, p), write the set B and find the remaining ordered pairs of A × B such that n(A × B) = 6.

Answer

Given,

A = {a, b, c}, so n(A) = 3.

Also, n(A × B) = 6

We know that,

n(A × B) = n(A) × n(B).

⇒ 6 = 3 × n(B)

⇒ n(B) = 2

The ordered pairs (a, p), (b, q) and (c, p) belong to A × B, so their second components p and q belong to B.

As n(B) = 2, B can have no element besides p and q.

∴ B = {p, q}

Now,

A × B = {(a, p), (a, q), (b, p), (b, q), (c, p), (c, q)}

Removing the three given ordered pairs (a, p), (b, q) and (c, p),

Hence, B = {p, q} and the remaining ordered pairs of A × B are (a, q), (b, p) and (c, q).

Given B = {2, 3, 5} and some elements of A × B are (a, 2), (b, 3), (c, 5). Find the set A and the remaining ordered pairs of A × B such that A × B is least.

Answer

Given,

B = {2, 3, 5}, so n(B) = 3.

The ordered pairs (a, 2), (b, 3) and (c, 5) belong to A × B, so their first components a, b and c belong to A.

For A × B to have the least number of elements, A must have the least number of elements, i.e. A can have no element besides a, b and c.

∴ A = {a, b, c}, thus n(A) = 3

n(A × B) = n(A) × n(B) = 3 × 3 = 9

Now,

A × B = {(a, 2), (a, 3), (a, 5), (b, 2), (b, 3), (b, 5), (c, 2), (c, 3), (c, 5)}

Removing the three given ordered pairs (a, 2), (b, 3) and (c, 5),

Hence, A = {a, b, c} and the remaining ordered pairs of A × B are (a, 3), (a, 5), (b, 2), (b, 5), (c, 2) and (c, 3).