Relations Test 12

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Relations Test 12

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Weekly Quiz Competition

Question 1
1 / -0

A $$\times$$ (B - C) =

A
$$(A \times B) - (A \times C)$$

B
$$(A \times C) - (A \times B)$$

C
$$(A \times B) \bigcup (A \times C)$$

D
$$(A \times B) \bigcap (A \times C)$$

Solution
Question 2
1 / -0

If A $$=$$ {1, 2}, B $$=$$ {3, 4}, then A$$\times$$B $$=$$

A
{(1, 3), (1, 4), (2, 3), (2, 4)}

B
{(1, 1), (2, 2), (3, 3), (4, 4)}

C
{(4, 1), (3, 1), (4, 2), (3, 2)}

D
All the above

Solution

$${\textbf{Step -1: Define the Cartesian product.}}$$

$${\text{Cartesian product: If }}A{\text{ and }}B{\text{ are two non empty sets, then }}$$
$${\text{Cartesian product }}A \times B{\text{ is set of all ordered pairs }}\left( {a,b} \right)$$ $${\text{such that }}a \in A{\text{ and }}b \in B.$$

$${\textbf{Step -2: Find the Cartesian product of given sets.}}$$

$${\text{We have given,}}$$
$$A = \left\{ {1,2} \right\}$$ $${\text{and}}$$ $$B = \left\{ {3,4} \right\}$$

$${\text{So,}}$$ $$A \times B = \left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}$$

$${\textbf{Hence, option A. }}\left\{ \mathbf{\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}$$ $${\textbf{is correct answer.}}$$
Question 3
1 / -0

If $$ABC\sim PQR$$, then AB:PQ =

A
$$AC:PB$$

B
$$AC:PR$$

C
$$AB:PR$$

D
$$AC:RQ$$

Solution
Question 4
1 / -0

Let $$A=\left \{ 1, 2, 3 \right \}$$. Then number of equivalence relations containing (1, 2) is:

A
1

B
2

C
3

D
4

Solution

Observe that 1 is related to 2. So, we have two possible cases.

Case 1: When 1 is not related to 3, then the relation. $$R_{1}=\left \{ (1, 1),(1, 2),(2, 1),(2, 2),(3, 3) \right \}$$ is the only equivalence relation containing $$(1, 2)$$.

Case 2: When 1 is related to 3, then $$A \times A$$ = $$\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)\}$$ is the only equivalence relation containing $$ (1,2) $$

$$\therefore $$ There are two required equivalence relations.
Question 5
1 / -0

Let A and B be finite sets containing m and n elements respectively. The number of relations that can be defined from A and B is:

A
mn

B
$$2^{mn}$$

C
$$2^{m+n}$$

D
$$n^{m}$$

Solution

Here, $$O(A)=m$$ and $$O(B)=n$$.
Hence $$O(A×B)=mn$$
Since every subset of $$A×B$$ is a relation from $$A$$ to $$B$$, therefore, number of relations from A to B is equal to the number of the subsets of $$A×B$$, i.e., $$2^{m{n}}$$
Question 6
1 / -0

Which of the following are not equivalence relations on I?

A
a R b if $$a+b$$ is an even integer

B
a R b if $$a-b$$ is an even integer

C
a R b if $$a< b$$

D
a R b if $$a= b$$

Solution

For Option $$[A]$$
aRb if $$a+b$$ is an even integer
$$Reflexive:$$ $$a+a=2a\Rightarrow$$ is $$even$$ $$number$$
$$Symmetric$$ $$a+b=b+a\Rightarrow$$ is $$even$$ $$number$$
$$Transitive$$ $$a+b=2k;b+c=2p$$ then $$a+c=2k+2p-2b=2(k+p-b)\Rightarrow$$ is an $$even$$ $$number$$
$$\therefore$$ this is a equivalence relation

For option $$[B]$$
aRb if $$a-b$$ is an even number
$$Reflexive:$$ $$a-a=0\Rightarrow$$  is $$even$$ $$number$$
$$Symmetric$$ $$a-b=2k;b-a=-2k\Rightarrow$$ is $$even$$ $$number$$
$$Transitive$$ $$a-b=2k;b-c=2p$$
$$a-b+b-c=a-c=2k+2p=2(k+p)$$  is $$even$$ $$number$$
$$\therefore$$ this is a equivalence relation

For option $$[C]$$
$$Reflexive:$$ $$a\not <a\Rightarrow$$ $$not$$ $$Reflexive:$$
$$Symmetric:$$ $$a<b$$ but $$b\not< a$$ $$\Rightarrow$$ $$not$$ $$Symmetric$$
$$Transitive$$ $$a<b;b<c\Rightarrow a<c$$ $$\Rightarrow$$ $$Transitive$$
$$\therefore$$ not a Equivalence relation

For option $$[D]$$
aRb if a=b $$\Rightarrow$$ $$Reflexive,Symmetric,Transitive$$
$$\therefore$$ this is a equivalence relation
Question 7
1 / -0

Let R be a reflexive relation on a finite set A having n elements, and let there be m ordered pairs in R. Then:

A
$$m\geq n$$

B
$$m\leq n$$

C
$$m=n$$

D
$$m=-n$$

Solution

The set consists of n elements and for relation to be reflexive it must have at least n ordered pairs. It has m ordered pairs therefore $$ m≥n$$.
Question 8
1 / -0

If $$X=\left \{ 1, 2, 3, 4, 5 \right \}$$ and $$Y=\left \{ 1, 3, 5, 7, 9 \right \}$$, determine which of the following sets represent a relation and also a mapping.

A
$$R_{1}=\left \{ (x, y):y=x+2, x\in X, y\in Y \right \}$$

B
$$R_{2}=\left \{ (1, 1),(2, 1),(3, 3),(4, 3),(5, 5) \right \}$$

C
$$R_{3}=\left \{ (1, 1),(1, 3),(3, 5),(3, 7),(5, 7) \right \}$$

D
$$R_{4}=\left \{ (1, 3),(2, 5),(4, 7),(5, 9),(3, 1) \right \}$$

Solution

$$R_1={(1,3),(2,4),(3,5),(4,6),(5,7)}$$
Since $$4$$ and $$6$$ do not belong to $$Y$$
$$(2,4),(4,6)∉R_1$$
$$R_1={(1,3),(3,5),(5,7)}⊂A×B$$
Hence $$R_1$$ is a relation but not a mapping as the elements $$2$$ and $$4$$ do not have any image.
$$R_2$$ : It is certainly a mapping and since every mapping is a relation, it is a relation as well.
$$R_3$$: It is a relation being a subset of $$A×B$$ but the elements $$1$$ and $$3$$ do not have a unique image and hence it is not mapping.
$$R_4$$ : It is both a mapping and a relation. Each element in A has a unique image. It is also one-one and onto mapping and hence a bijection
Question 9
1 / -0

Let N denote the set of all natural numbers and R a relation on $$N\times N$$. Which of the following is an equivalence relation?

A
$$(a, b) R(c, d) $$ if $$ad(b+c)=bc(a+d)$$

B
$$(a, b) R(c, d) $$ if $$ a+d=b+c$$

C
$$(a, b) R(c, d)$$ if $$ ad=bc$$

D
All the above

Solution

Given: N denote the set of natural numbers and R a reation on $$N\times N$$
A) $$(a, b) R (c, d)$$ id $$ad(b+c)=bc(a+d)$$
B) $$(a, b) R (c, d)$$ if $$(a+d) = (b+c)$$
C) $$(a, b) R (c, d)$$ if $$ad = bc$$.

Consider A $$(a, b) R (c, d)$$ if $$ad(b+c) = bc (a+d)$$
1) Reflexive
Let $$(a+b) \in N\times N$$
and $$(a,b)R(a,b)$$
$$\Rightarrow ab(b+a) = ba(a+b)$$
$$\Rightarrow ab(b+a) = ab(b+a)$$
$$\Rightarrow R$$ is reflective .....(I)

2) Symmetric
Let $$(a, b) R ( c,d)$$
where $$(a, v) \& (c,d) \in N$$
$$\Rightarrow ad(b+c) = bc(a+d)$$
$$=bc(a+d)=ad(b+c)$$
$$\Rightarrow cb(d+a)=da(c+b)$$
$$\Rightarrow (c,d)R(a,b)$$
$$\Rightarrow R$$ is symmetric ...(II)

3) Transitive
Let $$(a,b), (c,d)$$ and $$(c,f) \in N \times N$$
consider,
$$(a, b)R (c,d)$$
$$\Rightarrow ad (b+c) = bc(a+d)$$
$$\Rightarrow abd + adc = abc + bcd$$
$$\Rightarrow abd - abc = bcd - adc$$
$$\Rightarrow ab(d-c) = dc(b-a)$$
$$\Rightarrow \dfrac{ab}{(b-a)} = \dfrac{cd}{d-c}$$ ...(1)
Now, consider,
$$(c,d)R(e,f)$$
$$\Rightarrow cf(d+e) = dc(e+f)$$
$$\Rightarrow cfd + cef = ced + edf$$
$$\Rightarrow cfd - ced = edf + cef$$
$$\Rightarrow cd(fe) = ef(d-c)$$
$$\Rightarrow \dfrac{cd}{d-c} = \dfrac{ef}{f-e} $$ ....(2)
from (1) and (2)
$$\dfrac{ab}{b-c} = \dfrac{cf}{f-e}$$
$$\Rightarrow ab (f-e) = ef(b-a)$$
$$\Rightarrow abf -abe = efb - efa$$
$$\Rightarrow abf + efa = efb + abe$$
$$\Rightarrow af(b+e) = be (f+a)$$
$$\Rightarrow (a, b) R (e,f)$$ (by definition of R)
$$\Rightarrow R$$ is trasitive .....(III)
from (I), (II) and (III)
$$R$$ is on equivalence Relation or $$N \times N$$.

Consider B $$(a, b) R (c, d)$$ if $$(a+d) = (b+c)$$
1) Reflexive
Let $$(a+b) \in N\times N$$
and $$(a,b)R(a,b)$$
$$\Rightarrow (a+b) = (b+a)$$
$$\Rightarrow (a+b) = (a+b)$$
$$\Rightarrow R$$ is reflective .....(IV)

2) Symmetric
Let $$(a, b) R ( c,d)$$
where $$(a, b) \& (c,d) \in N\times N$$
$$\Rightarrow (a+d) = (b+c)$$
$$\Rightarrow (b+c)=(a+d)$$
$$\Rightarrow (c+b)=(d+a)$$
$$\Rightarrow (c,d)R(a,b)$$
$$\Rightarrow R$$ is symmetric ...(V)

3) Transitive
Let $$(a,b), (c,d)$$ and $$(c,f) \in N \times N$$
If $$(a,b)\&(c,d)$$ and $$(c,d)\& (c,f)$$
then $$(a,b)R(c,f)$$
consider,
$$(a, b)R (c,d)$$
$$\Rightarrow (a+d) = (b+c)$$ by definition by $$R$$
$$\Rightarrow a-b = c-d$$ ....(3)
Now,
$$(c,d)R(e,f)$$
$$\Rightarrow (c+f) = (d+e)$$ by definition of $$x$$
$$\Rightarrow c-d + e-f $$ ...(4)
from (3) and (4)
$$a-b=e-f$$
$$\Rightarrow a+f = e+b$$
$$\Rightarrow (a,b) R(e,f)$$
$$\Rightarrow R$$ is transitive ...(VI)
from (IV), (V), (VI) we have,
$$R$$ is an equivalence reaction.$$\Rightarrow abf + efa = efb + abe$$
$$\Rightarrow af(b+e) = be (f+a)$$
$$\Rightarrow (a, b) R (e,f)$$ (by definition of R)
$$\Rightarrow R$$ is trasitive .....(III)
from (I), (II) and (III)
$$R$$ is on equivalence Relation or $$N \times N$$.

Consider C $$(a, b) R (c, d)$$ if $$ad = bc$$
1) Reflexive
Let $$(a,b) \in N\times N$$
and $$(a,b)R(a,b)$$
$$\Rightarrow ab = ba$$
$$\Rightarrow ab = ab$$
$$\Rightarrow R$$ is reflective .....(VII)

2) Symmetric
Let $$(a, b) R ( c,d)$$ where $$(a, b), (c,d) \in N\times N$$
$$\Rightarrow ad=bc$$
$$\Rightarrow bc=ad$$
$$\Rightarrow cb=da$$
$$\Rightarrow (c,d)R(a,b)$$
$$\Rightarrow R$$ is symmetric ...(VIII)

3) Transitive
Let $$(a,b), (c,d)$$ and (c,f) \in N \times N$$
If $$(a,b)R(c,d)$$ and $$(c,d)R (c,f)$$
then $$(a,b)R(c,f)$$
consider,
$$(a, b)R (c,d)$$
$$\Rightarrow ad = bc$$ by definition by $$R$$
$$\Rightarrow \dfrac{a}{b} = \dfrac{c}{d}$$ ....(5)
Consider,
$$(c,d)R(e,f)$$
$$\Rightarrow (cf) = (de)$$ by definition of $$R$$
$$\Rightarrow \dfrac{c}{d} = \dfrac{e}{f} $$ ...(6)
from (5) and (6) we have,
$$\dfrac{a}{b}=\dfrac{e}{f}$$
$$\Rightarrow R$$ is transitive ...(IX)
from (VII), (VIII), (IX) we have,
$$R$$ is an equivalence reaction.
$$\Rightarrow A, B, C$$ all are equivalence relation.
$$\therefore$$ option all of the above is correct answer.
Question 10
1 / -0

Let A be a finite set containing n distinct elements. The number of relations that can be defined on A is:

A
$$2^{n}$$

B
$$n^{2}$$

C
$$2^{n^{2}}$$

D
2n

Solution

Since $$A$$ contains $$n$$ distinct elements, therefore, $$A×A$$ contains $$n×n=n^2$$ distinct elements. since every subset of $$A×A$$ is a relation on $$A$$., therefore, number of relations on $$A$$ is equal to the order of the power set of $$A×A$$, is equal to the order of the power set of $$A×A$$, i.e., $$2^{n^{2}}$$

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Re-Attempt Weekly Quiz Competition

Relations Test 12

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Relations Test 12

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SHARING IS CARING

Question 1

$$(A \times B) - (A \times C)$$

$$(A \times C) - (A \times B)$$

$$(A \times B) \bigcup (A \times C)$$

$$(A \times B) \bigcap (A \times C)$$

Solution

Question 2

{(1, 3), (1, 4), (2, 3), (2, 4)}

{(1, 1), (2, 2), (3, 3), (4, 4)}

{(4, 1), (3, 1), (4, 2), (3, 2)}

All the above

Solution

Question 3

$$AC:PB$$

$$AC:PR$$

$$AB:PR$$

$$AC:RQ$$

Solution

Question 4

1

2

3

4

Solution

Question 5

mn

$$2^{mn}$$

$$2^{m+n}$$

$$n^{m}$$

Solution

Question 6

a R b if $$a+b$$ is an even integer

a R b if $$a-b$$ is an even integer

a R b if $$a< b$$

a R b if $$a= b$$

Solution

Question 7

$$m\geq n$$

$$m\leq n$$

$$m=n$$

$$m=-n$$

Solution

Question 8

$$R_{1}=\left \{ (x, y):y=x+2, x\in X, y\in Y \right \}$$

$$R_{2}=\left \{ (1, 1),(2, 1),(3, 3),(4, 3),(5, 5) \right \}$$

$$R_{3}=\left \{ (1, 1),(1, 3),(3, 5),(3, 7),(5, 7) \right \}$$

$$R_{4}=\left \{ (1, 3),(2, 5),(4, 7),(5, 9),(3, 1) \right \}$$

Solution

Question 9

$$(a, b) R(c, d) $$ if $$ad(b+c)=bc(a+d)$$

$$(a, b) R(c, d) $$ if $$ a+d=b+c$$

$$(a, b) R(c, d)$$ if $$ ad=bc$$

All the above

Solution

Question 10

$$2^{n}$$

$$n^{2}$$

$$2^{n^{2}}$$

2n

Solution

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