Question 1: If A = \{ 1,2,3 \} , B = \{ 4,5,6 \} , which of the following are relations from A to B ? Give reasons in support of your answer: (i) \{(1, 6), (3, 4), (5, 2) \} (ii) \{ (1,5), (2, 6), (3,4), (3,6) \} (iii) \{ (4, 2), (4,3), (5, 1) \} {iv} A \times B

Answer:

Given A = \{ 1,2,3 \} , B = \{ 4,5,6 \}

A \times B = \{ 1,2,3 \} \times \{ 4,5,6 \} \\ \hspace*{1.2cm} = \{ (1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6) \} 

i) \{(1, 6), (3, 4), (5, 2) \} is not a relation from A to B as it is not a subset of A \times B

ii) \{ (1,5), (2, 6), (3,4), (3,6) \} is a subset of A \times B . So it is a relation from A to B .

iii) \{ (4, 2), (4,3), (5, 1) \} is not a relation from A to B as it is not a subset of A \times B

iv) A \times B is a relation from A to B .

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Question 2: A relation R is defined from a set A = \{ 2, 3, 4, 5 \} to a set B = \{ 3, 6, 7, 10 \} as follows: (x, y) \in R \Leftrightarrow x is relatively prime to y . Express R as a set of ordered pairs and determine its domain and range.

Answer:

Given A = \{ 2, 3, 4, 5 \} and B = \{ 3, 6, 7, 10 \}

Also (x, y) \in R \Leftrightarrow x is relatively prime to y

Therefore 2 is a co-prime to 3 and 7

3 is a co-prime to 7 and 10

4 is a co-prime to 3 and 7

5 is a co-prime to 3, 6 and 7

\therefore R = \{ (2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7) \}

Hence Domain of R = \{ 2, 3, 4, 5 \}

and Range of R = \{ 3, 6,7, 10 \}

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Question 3: Let A be the set of first five natural numbers and let R be a relation on A defined as follows: (x, y) \in R \Leftrightarrow  x \leq y . Express R and R^{-1} as sets of ordered pairs. Determine also (i) the domain of R^{-1} (ii) the range of R .

Answer:

Given A = \{ 1, 2, 3, 4, 5 \} [ \because A is the set of first five natural numbers ]

Also it is given that R is a relation on A defined as (x, y) \in R \Leftrightarrow  x \leq y

\therefore R = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), \\ \hspace*{1.2cm} (3, 5), (4, 4), (4, 5) , (5,5) \}

R^{-1} = \{ (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (2, 2), (3, 2), (4, 2), (5, 2), (3, 3), (4, 3), (5, 3), \\ \hspace*{1.2cm} (4, 4), (5, 4), (5, 5) \}

\therefore i) Domain (R^{-1}) = \{ 1, 2, 3, 4, 5 \}

ii) Range (R) = \{ 1, 2, 3, 4, 5 \}

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Question 4: Find the inverse relation R^{ - 1} in each of the following cases:  (i) R = \{ (1, 2), (1, 3), (2, 3), (3, 2), (5, 6) \}     (ii) R = \{ (x,y) : x , y \in N, x+2y =8 \}   (iii) R is a relation from \{ 11, 12,13 \} to \{8, 10, 12 \} defined by y = x - 3 .

Answer:

i) Given R = \{ (1, 2), (1, 3), (2, 3), (3, 2), (5, 6) \}

\Rightarrow R^{-1} = \{ (2, 1), (3, 1), (3, 2), (2, 3), (6, 5) \}

ii) Given R = \{ (x,y) : x , y \in N, x+2y =8 \}

Now x + 2y = 8 \Rightarrow x = 8 - 2y

Putting y = 1, 2, 3   we get x = 6, 4, 2 respectively

For y = 4 , we get x = 0 \notin N and for y > 4 , x \notin N

\therefore R = \{ (6, 1), (4, 2), (2, 3) \}

Hence R^{-1} = \{ (1, 6), (2, 4), (3, 2) \}

iii) Given R is a relation from \{ 11, 12,13 \} to \{8, 10, 12 \} defined by y = x - 3

Now, y = x - 3

Putting x = 11, 12, 13 we get y = 8, 9, 10 respectively.

\Rightarrow (11, 8) \in R, (12, 9) \notin R \ and \  (13, 10) \in R

\therefore R = \{ (11, 8), (13, 10) \}

and R^{-1} = \{ (8, 11), (10, 13) \}

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Question 5: Write the following relations as the sets of ordered pairs:
(i) A relation R from the set \{ 2, 3, 4 , 5, 6 \}   to the set \{1, 2, 3\} defined by x =2y .
(ii) A relation R on the set \{1,2,3,4,5,6,7\} defined by (x,y) \in R \Leftrightarrow x is relatively prime to y .
(iii) A relation R on the set \{ 0, 1, 2, ...,10 \} defined by 2x + 3y =12 .
(iv) A relation R from a set A = \{ 5,6,7,8 \} to the set B= \{ 10,12,15,16,18 \} defined by (x,y) \in R \Leftrightarrow x divides y .

Answer:

i) Given x = 2y ,

Putting y = 1, 2, 3 we get x = 2, 4, 6 respectively

\therefore R = \{ (2, 1), (4, 2), (6, 1) \}

ii) Given relation R on the set \{1,2,3,4,5,6,7\} defined by (x,y) \in R \Leftrightarrow x is relatively prime to y

Therefore 2 is  relatively prime to 3, 5 and 7

3 is  relatively prime to 2, 4, 5 and 7

4 is  relatively prime to 3, 5 and 7

5 is  relatively prime to 2, 3, 4, 6 and 7

6 is  relatively prime to  5 and 7

\therefore R = \{ (2, 3), (2, 5), (2, 7), (3, 2), (3, 4), (3, 5), (3, 7), (4, 3), \\ \hspace{2.0cm} (4, 5), (4, 7), (5, 2), (5, 3),(5, 4), (5, 6), (5, 7), (6, 5), (6, 7) \}

iii) Given 2x+ 3y = 12 \Rightarrow 2x = 12 - 3y \Rightarrow x = \frac{1}{2} (12-3y) \}

Putting y = 0, 2, 4 we get x = 6, 3, 0 respectively

\therefore R = \{ (6, 0), (3, 2), (0, 4) \}

iv) Given a relation R from a set A = \{ 5,6,7,8 \} to the set B= \{ 10,12,15,16,18 \} defined by (x,y) \in R \Leftrightarrow x divides y

Here 5 divides 10 and 15

6 divides 12 and 18

8 divides 16

\therefore R = \{ (5, 10), (5,15), (6, 12), (6, 18), (8, 16) \}

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Question 6: Let R be a relation in N defined by (x,y) \in R \Leftrightarrow x+2y=8 . Express R and R^{-1} assets of ordered pairs.

Answer:

Given  (x,y) : x , y \in N, x+2y =8 

Now x + 2y = 8 \Rightarrow x = 8 - 2y

Putting y = 1, 2, 3   we get x = 6, 4, 2 respectively

For y = 4 , we get x = 0 \notin N and for y > 4 , x \notin N

\therefore R = \{ (6, 1), (4, 2), (2, 3) \}

Hence R^{-1} = \{ (1, 6), (2, 4), (3, 2) \}

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Question 7: Let A= \{ 3,5 \} and B = \{ 7,11 \} . Let R = \{a,b \}:  a \in A , b \in B, a-b \ is \  odd \} . Show that R is an empty relation from A  into B .

Answer:

Given A= \{ 3,5 \} and B = \{ 7,11 \}

Also R = \{a,b \}:  a \in A , b \in B, a-b \ is \  odd \}

For elements of A and B, 3-7 = -4, 3-11  = -8, 5-7=-2, 5-11=-6

They are all even

Therefore R is an empty  relation from A to B .

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Question 8: Let A = \{1, 2\} and B = \{3, 4 \} . Find the total number of relations from A into B .

Answer:

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

\Rightarrow n(A) = 2 \ \ \ \ \ \& \ \ \ \ \ n(B) = 2

\Rightarrow n(A) \times n(B) = 2 \times 2 = 4

\Rightarrow n(A \times B) = 4

[ \because n(A \times B) = n(A) \times n(B) ]

Therefore there are 2^4 = 16 relations from A to B

[ \because n(x) = a , n(y) = b \Rightarrow total number of relations = 2^{ab} ]

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Question 9: Determine the domain and range of the relation R defined by:

(i) R = \{ (x,x +5) : x \in \{0, 1,2, 3, 4,5 \} \}

(ii) R = \{ ( x, x^3 ) :  x \ is \ a \ prime \ number \ less \ than \ 10 \}

Answer:

i) Given R = \{ (x,x +5) : x \in  \{0, 1,2, 3, 4,5 \} \}

Therefore for elements given we get

R = \{ (0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10) \}

Therefore Domain (R) = \{ 0, 1, 2, 3, 4, 5 \}

Range (R) = \{5, 6, 7, 8, 9, 10 \}

ii) Given R = \{ ( x, x^3 ) :  x \ is \ a \ prime \ number \ less \ than \ 10 \}

\therefore x = 2, 3, 5, 7

\Rightarrow R = \{ (2, 8), (3, 27), (5, 125), (7, 343) \}

Therefore Domain (R) = \{ 2, 3, 5, 7 \}

Range (R) = \{ 8, 27, 125, 343 \}

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Question 10: Determine the domain and range of the following relations

(i) R = \{ (a, b): a \in N, a < 5, b = 4 \}    

ii) S = \{ (a, b) : b = |a-1|, a \in Z \ and \ |a| \leq 3 \}

Answer:

i) Given R = \{ (a, b): a \in N, a < 5, b = 4 \}

\Rightarrow a = 1, 2,3 ,4      \&   \ \ \ b = 4

\therefore R = \{ (1, 4), (2, 4), (3 ,4), (4,4) \}

Therefore Domain (R) = \{ 1, 2, 3, 4 \}

Range (R) = \{ 4 \}

ii) Given S = \{ (a, b) : b = |a-1|, a \in Z \ and \ |a| \leq 3 \}

\Rightarrow a = -3, -2, -1, 0, 1, 2, 3  \therefore b = 4, 3, 2, 1, 0, 1, 2 respectively

\therefore S = \{(-3, 4), (-2, 3), (-1, 2), ), (0, 1), (1, 0), (2, 1), (3, 2) \}

Therefore Domain (S) = \{  -3, -2, -1, 0, 1, 2, 3 \}

Range (S) = \{ 0, 1, 2, 3, 4 \}

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Question 11: Let A = \{ a,b \} . List all relations on A and find their number.

Answer:

Given A = \{ a,b \}

We know that number of relations = 2^{mn} = 2^4 = 16

Ordered pairs = (a, a), (a, b), (b, a), (b, b)

\therefore R =  \Big\{ \{ (a, a), (a, a) \}, \{ (a, a), (a, b) \}, \{ (a, a), (b, a) \}, \{ (a, a), (b, b) \}, \{ (a, b), (a, a) \}, \\ \hspace*{1.5cm}  \{ (a, b), (a, b) \},  \{ (a, b), (b, a) \}, \{ (a, b), (b, b) \}, \{ (b, a), (a, a) \}, \{ (b, a), (a, b) \}, \\ \hspace*{1.5cm}  \{ (b, a), (b, a) \},  \{ (b, a), (b, b) \}, \{ (b, b), (a, a) \}, \{ (b, b), (a, b) \}, \{ (b, b), (b, a) \}, \\ \hspace*{1.5cm}  \{ (b, b), (b, b) \} \Big\}

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Question 12: Let A = \{ x, y, z \} and B = \{ a, b \} . Find the total number of relations from A into B .

Answer:

Given A = \{ x, y, z \} and B = \{ a, b \}

\Rightarrow n(A) = 3 and n(B) = 2

\Rightarrow n(A) \times n(B) = 3 \times 2 = 6

\Rightarrow n(A \times B) = 6

Therefore there are 2^6 = 64 relations from A to B

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Question 13: Let R be a relation from N to N defined by R = \{ (a,b):a,b \in N and a=b^2 . Are the following statements true?

(i) (a,a) \in R for all a \in N

(ii) (a,b) \in R \Rightarrow (b,a) \in R

(iii) (a,b) \in R and (b,c) \in R \Rightarrow  (a,c) \in R

Answer:

Given R = \{ (a,b):a,b \in N and a=b^2 \}

i) (a,a) \in R for all a \in N

False: Statement is not true because (2, 2) \notin R 

ii) (a, b) \in R \Rightarrow (b, a) \in R

False: The statement is not true because (4, 2) \in R but (2, 4) \notin R

iii) (a, b) \in R and (b, c) \in R \Rightarrow (a, c) \in R

False: Statement is not true because (36, 6) \in R and (25, 5) \in R but (36, 5) \notin R

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Question 14: Let A= \{ 1,2,3,...,14 \} . Define a relation on a set A by R = \{ (x,y):3x-y=0, \ where \ x,y \in A \} . Depict this relationship using an arrow diagram. Write down its domain, co-domain and range.

Answer:

Given 3x-y = 0 \Rightarrow y = 3x 2019-08-23_21-45-06

Putting x = 1, 2, 3, 4 then y = 3, 6, 9, 12 respectively

For x > 4, y > 14 which does not belong to A

\therefore R = \{ (1, 3), (2, 6), (3, 9), (4, 12) \}

Therefore Domain (R) = \{ 1, 2, 3, 4 \}

Range (R) = \{ 3, 6, 9, 12\}

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Question 15: Define a relation R on the set N of natural numbers by R=\{(x,y):y=x+5, \ x \ is \ a \ natural \ number \ less \ than \ 4,x,y \in N \} Depict this relationship using (i) roster form (ii) an arrow diagram. Write down the domain and range or R.

Answer:

Given R=\{(x,y):y=x+5, \ x \ is \ a \ natural \ number \ less \ than \ 4,x,y \in N \}

i) Putting x = 1, 2,3 we get y = 6, 7, 8 respectively.2019-08-23_21-47-50

\therefore R = \{ (1, 6), (2, 7), (3, 8) \}

ii) Domain (R) = \{ 1, 2, 3  \}

Range (R) = \{ 6, 7, 8 \}

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Question 16: A = \{1,2,3,5 \} and B= \{4,6,9 \} . Define a relation R  from A to B by R=\{(x,y): \ the \ difference \ between x \ and \ y \ is \ odd, \ x \in A, \ y \in B \} . Write R in Roster form.

Answer:

Given A = \{1,2,3,5 \} and B= \{4,6,9 \}

Also R=\{(x,y): \ the \ difference \ between x \ and \ y \ is \ odd, \ x \in A, \ y \in B \}

For x = 1, y = 4, 6

For x =2, y = 9

For x = 3, y = 4, 3

For x = 5, y = 4, 6

\therefore R = \{ (1, 4), (1, 6), (2, 9), (3, 4), (3, 3), (5, 4), (5, 6) \}

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Question 17: Write the relation R=\{(x,x^3):x \ is \ a \ prime \ number \ less \ than 10 \} in roster form.

Answer:

Given R=\{(x,x^3):x \ is \ a \ prime \ number \ less \ than 10 \}

\therefore x = 2, 3, 5, 7

Hence R = \{ (2, 8), (3, 27), (5, 125), (7, 343) \}

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Question 18: Let A = \{ 1,2, 3, 4,5, 6 \} . Let R be a relation on A defined by \{(a,b):a,b \in A,b \ is \ exactly \ divisible \ by a \} (i) Write R  in roster form (ii) Find the domain of R (iii) Find the range of R

Answer:

Given A = \{ 1,2, 3, 4,5, 6 \} and R be a relation on A defined by \{(a,b):a,b \in A, b \ is \ exactly \ divisible \ by \ a \}

For a = 1, b = 1, 2, 3, 4, 5, 6

For a = 2, b = 2, 4, 6

For a = 3, b = 3, 6

For a = 4, b = 4

For a = 5, b = 5

For a = 6, b = 6

i) R = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), \\ \hspace*{1.0cm} (3, 6), (4, 4), (5, 5) , (6, 6) \}

ii) Domain (R) = \{ 1, 2, 3, 4, 5, 6 \}

iii) Range (R) =  \{ 1, 2, 3, 4, 5, 6 \}

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2019-08-24_6-27-40.jpgQuestion 19: Figure below shows a relationship between the sets P and Q . Write this relation in (i) set builder form (ii) roster form. What is its domain and range?

Answer:

Given R = \{ (x, y) : y = x-2, x \in P , y \in Q \}

P = \{ 5, 6, 7 \} and Q = \{ 3, 4, 5 \}

When x = 5 \Rightarrow y = 3

x = 6 \Rightarrow y = 4

x = 7 \Rightarrow y = 5

\therefore R = \{ (5, 3), (6, 4), (7, 5) \}

Therefore Domain (R) = \{ 5, 6, 7 \}

and Range (R) = \{ 3, 4, 5 \}

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Question 20: Let R be the relation on Z defined by R= \{(a,b): a,b \in Z, a-b \ is \ an \ integer \} . Find the domain and range of R .

Answer:

Given R= \{(a,b): a,b \in Z, a-b \ is \ an \ integer \}

Difference between two integers would also be an integer.

Therefore Domain (R) = Z

And Range (R) = Z

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Question 21: For the relation R_1 defined on R by the rule (a,b) \in R_1 \Leftrightarrow 1 + ab>0 . Prove that (a,b) \in R_1 and (b , c) \in R_1 \Rightarrow (a, c) \in R_1 is not true for all a, b, c \in R .

Answer:

To prove: (a,b) \in R_1 and (b , c) \in R_1 \Rightarrow (a, c) \in R_1 is not true for all a, b, c \in R .

Given R_1 = \{ (a, b): 1 + ab > 0 \}

Let a = 1, b = -0.5, c = -4

Here, (1, -0.5) \in R_1 \ \ \ [ \because 1+(1 \times -0.5) = 0.5 > 0 ]

And, (-0.5, -4) \in R_1 \ \ \ [ \because 1+(-0.5 \times -4) = 3 > 0]

But, (1, -4) \notin R_1 \ \ \ [ \because  1+(1\times -4) = -3 < 0]

\therefore (a, b) \in R_1  and (b,c) \in R_1 \Rightarrow (a, c) \in R_1  is not true for all a, b, c \in R

Hence Proved.

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Question 22: Let R be a relation on N \times N defined by (a,b) R (c,d) \Leftrightarrow a+d=b+c for all (a,b),(c,d) \in N \times N . Show that:

(i)  (a,b) \ R \ (a,b) for all (a, b) \in N \times N

(ii)  (a, b) \ R \ (c, d) \Rightarrow (c, d) \ R \ (a, b) for all (a, b), (c, d) \in N \times N

(iii)  (a,b) \ R \ (c,d) and (c,d) \ R \ ( e,f) \Rightarrow (a,b) \ R \ (e,f) for all (a,b),(c,d),(e,f) \in N \times N

Answer:

Given (a,b) R (c,d) \Leftrightarrow a+d=b+c for all (a,b),(c,d) \in N \times N

i) (a,b) \ R \ (a,b) for all (a, b) \in N \times N

\because (a+b) = (b + a) for all a, b \in N

\therefore (a, b) \ R \ (a, b) for all a, b \in N

ii) (a, b) \ R \ (c, d) \Rightarrow (c, d) \ R \ (a, b) for all (a, b), (c, d) \in N \times N

(a, b) R (c, d) \Rightarrow a + d = b + c

\Rightarrow c + d = d + a

\Rightarrow (c, d) \ R \ (a, b)

iii) (a,b) \ R \ (c,d) and (c,d) \ R \ ( e,f) \Rightarrow (a,b) \ R \ (e,f) for all (a,b),(c,d),(e,f) \in N \times N

\therefore (a+b) \ R \ (c, d) and (c, d) \ R \ ( e, f)

\Rightarrow a + d = b + c and c + f = d + e

\Rightarrow a + d + c + f = b + c + d + e

\Rightarrow a + f = b + e

\Rightarrow (a, b) \ R \ (e, f)

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Question 23: Find the linear relation between the components of the ordered pairs of the relation R , where (i) R = \{ (0, 2), (- 1,5), (2, -4), \ldots \} (ii) R = \{ (- 1, -1), (0,2), (1, 5), \ldots \}

Answer:

i) Given R = \{ (0, 2), (- 1,5), (2, -4), \ldots \}

x and y are related to each other by linear equation

Let the relation be defined as y = mx + c

For (0, 2) we get c = 2

For (-1, 5) we get 5 = -m + 2 \Rightarrow m = -3

Therefore y = -3x + 2 or y = 2 - 3x

ii) Given R = \{ (- 1, -1), (0,2), (1, 5), \ldots \}

x and y are related to each other by linear equation

Let the relation be defined as y = mx + c

For ( 0, 2) we get 2 = 0 + c \Rightarrow c = 2

For (-1, -1) we get -1 = -m + 2 \Rightarrow m = 3

Therefore y = 3x + 2

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Question 24: If A = \{ 1, 3,5, 6 \} and B = \{ 3,4,5 \} , write the relation R as a set of ordered pairs, if
(i) R = \{(x,y) : (x, y) \in A \times B, x+y \ is \ even \}

(ii) R = \{ (x,y) : (x,y) \in A \times B, xy \ is \ odd \}

Answer:

Given A = \{ 1, 3,5, 6 \} and B = \{ 3,4,5 \}

i) R = \{(x,y) : (x, y) \in A \times B, x+y \ is \ even \}

A \times B = \{ (1, 3), (1, 4), (1, 5), (3, 3), (3, 4), (3, 5), (5, 3), (5, 4), (5, 5), (6, 3), (6, 4), (6, 5) \}

If x + y is even then R = \{ (1, 3),  (1, 5), (3, 3),  (3, 5), (5, 3),  (5, 5),  (6, 4) \}

ii) if xy is odd, then R = \{ (1, 3), (1, 5), (3, 3), (3, 5), (5, 3), (5, 5) \}

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