Relations Test 23

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Relations Test 23

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Question 1
1 / -0

$x^2=xy$ is a relation which is:

A
Symmetric

B
Reflexive

C
Transitive

D
All of these

Solution

$xRx \Rightarrow x^2=x.x$ Hence reflexive

$xRy \Rightarrow x^2=xy. \Rightarrow x=y$ $yRx \Rightarrow y^2=xy \Rightarrow y=x$ Hence symmetric

$xRy$ and $yRz$ $\Rightarrow x^2=xy\Rightarrow x=y.$ Also $y^2=yz\Rightarrow y=z$

Since $x = z, x.x=z.x \Rightarrow x^2=xz \Rightarrow xRz$ . Hence transitive.
Question 2
1 / -0

The relation $R=\{(1, 1), (2, 2), (3, 3)\}$ on the set $\{1, 2, 3\}$ is

A
Symmetric only

B
Reflexive only

C
An equivalence relation

D
transitive only

Solution

Given Relation
$R = \{(1,1),(2,2), (3,3)\}$

Reflexive: If a relation has $\{(a,b)\}$ as its element, then it should also have $\{(a,a), (b,b)\}$ as its elements too.

Symmetric: If a relation has $(a,b)$ as its element, then it should also have $\{(b,a)\}$ as its element too.

Transitive: If a relation has $\{(a,b), (b,c)\}$ as its elements, then it should also have $\{(a,c)\}$ as its element too.

Now, the given relation satisfies all these three properties.
Therefore, its an equivalence relation.
Question 3
1 / -0

The minimum number of elements that must be added to the relation $R =\{(1,2)(2,3)\}$ on the set of natural numbers so that it is an equivalence is

A
$4$

B
$7$

C
$6$

D
$5$

Solution

For $R$ to be equivalence relation, it should be reflexive, symmetric and transitive.

Now, for $R$ reflexive:
$\Rightarrow (x, x) \in R$ for all $x \in \{1, 2, 3\}$
$\Rightarrow (1,1), (2,2), (3,3) \in R$

For $R$ is symmetric:
$\Rightarrow (1,2), (2,3) \in R$
$\Rightarrow (2,1), (3,2) \in R$

For $R$ is transitive:
$\Rightarrow (1,2), (2,3) \in R$
$\Rightarrow (1,3) \in R$
Also $(3,1) \in R$ as $R$ is symmetric.
So, the total number of elements is $9$ .
Hence, minimum $7$ elements must be added.
Question 4
1 / -0

Determine all ordered pairs that satisfy $(x - y)^{2} + x^{2} = 25$ , where $x$ and $y$ are integers and $x \geq 0$ . Find the number of different values of $y$ that occur

A
$3$

B
$4$

C
$5$

D
$6$

Solution

${ \left( x-y \right) }^{ 2 }+{ x }^{ 2 }=25$
Now, ${ 3 }^{ 2 }+{ 4 }^{ 2 }={ 5 }^{ 2 }$
$9+16=25$
$\therefore$ There are 2 possibilities:
$I.{ \left( x-y \right) }^{ 2 }=9$ and ${ x }^{ 2 }=16$
$\therefore x=\pm 4$ and $x-y=\pm 3$
$\left( i \right) .x-y=3\Rightarrow \left( 4,1 \right)$ and $\left( -4,-7 \right)$
$\left( ii \right) .x-y=-3\Rightarrow \left( 4,7 \right)$ and $\left( -4,-1 \right)$
$II.{ \left( x-y \right) }^{ 2 }=16$ and ${ x }^{ 2 }=9$
$\therefore x=\pm 3$ and $x-y=\pm 4$
$\left( i \right) .x-y=4\Rightarrow \left( 3,-1 \right)$ and $\left( -3,-7 \right)$
$\left( ii \right) .x-y=-4\Rightarrow \left( 3,7 \right)$ and $\left( -3,-1 \right)$
$\therefore$ Different values of y are $1,-1,7,-7$
$\therefore 4$ different values of y occur.
Question 5
1 / -0

Let $R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}$ be a relation on the set $A = \{1, 2, 3, 4\}$ . The relation $R$ is

A
A function

B
Transitive

C
Not symmetric

D
Reflexive

Solution

Let $R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}$ be a relation on the set $A = \{1, 2, 3, 4\}$ , then
(a) Since $(2, 4)$ $\in$ $R$ and $(2, 3)$ $\in$ $R$ , so $R$ is not a function.
(b) Since $(1, 3)$ $\in$ $R$ and $(3, 1)$ $\notin$ $R$ but $(1, 1)$ $\notin$ $R$ , so $R$ is not transitive:
(c) Since $(2, 3)$ $\in$ $R$ but $(3, 2)$ $\notin$ $'R$ , so $R$ is not symmetric.
(d) Since $(1, 1)$ $\notin$ $R$ , so $R$ is not reflexive.
Hence, option C is correct.
Question 6
1 / -0

Let $X$ be the set of all persons living in a city. Persons $x, y$ in $X$ are said to be related as $x < y$ if $y$ is at least $5$ years older than $x$ . Which one of the following is correct?

A
The relation is an equivalence relation on $X$

B
The relation is transitive but neither reflexive nor symmetric

C
The relation is reflexive but neither nor symmetric

D
The relation is symmetric but neither transitive nor reflexive

Solution

Given, $X$ is the set of persons living ins a city.
Relation(R) = {(x,y): x,y $\in$ $R$ $\&$ x<y if y is atleast $5$ years older than x}

For Reflexive: $(x,x)$ should $\in R$ for all $x \in X$
now, $x$ cannot be be $5$ years older than himself. So the relation is not reflexive.

For Symmetric: If $(x,y) \in R \Rightarrow (y,x) \in R$
$(x,y) \in R \Rightarrow y$ is at least $5$ years older than $x$ .
$(y,x) \in R \Rightarrow x$ is at least $5$ years older than $y$ . This contradicts the above statement. Hence the relation is not symmetric

For Transitive: If $(x,y) \in R$ and $(y,z) \in R \Rightarrow (x,z) \in R$ .
$(x,y) \in R \Rightarrow y$ is at least $5$ years older than $x$ .
$(y,z) \in R \Rightarrow z$ is at least $5$ years older than $y$ .
Then, $(x,z) \in R \Rightarrow z$ is at least $5$ years older than $x$ .
Since, $z$ is at least $10$ years older than $x$ . The relation is transitive.
Question 7
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For any two real numbers $\theta$ $\phi$ , we define $\theta R \phi$ if and only if $\sec ^{ 2 }{ \theta } -\tan ^{ 2 }{ \phi } =1$ . The relation $R$ is

A
Reflexive but not transitive

B
Symmetric but not reflexive

C
Both reflexive and symmetric but not transitive

D
An equivalence relation

Solution

Given , $\theta$ R $\phi$ where $\theta$ and $\phi$ are two real numbers.
$\sec^{2} { \theta } -\tan ^{2} { \phi } =1$
$\theta=\phi$
R is reflexive and symmetric.
R is not transitive since there are only two numbers.
Question 8
1 / -0

The relation $R$ defined on set $A = \left \{x :|x| < 3, x\epsilon I\right \}$ by $R = \left \{(x, y) : y = |x|\right \}$ is

A
$\left \{(-2, 2), (-1, 1), (0, 0), (1, 1), (2, 2)\right \}$

B
$\left \{(-2, 2), (-2, 2), (-1, 1), (0, 0), (1, -2), (2, -1), (2, -2)\right \}$

C
$\left \{(0, 0), (1, 1), (2, 2)\right \}$

D
None of the above

Solution

$R$ defined on set $A = \left \{x :|x| < 3, x\epsilon I\right \}$ by $R = \left \{(x, y) : y = |x|\right \}$
$A = \left \{x :|x| < 3, x\epsilon I\right \}$
$=\left\{x:-3<x<3, x\epsilon I\right\}$
$\therefore$ $R = \left \{(x, y) : y = |x|\right \}$
$=\left\{(-2,2),(-1,1),(0,0),(1,1),(2,2)\right\}$
Hence, option A is correct.
Question 9
1 / -0

Let the number of elements of the sets $A$ and $B$ be $p$ and $q$ respectively. Then the number of relations from the set $A$ to the set $B$ is

A
${ 2 }^{ p+q }$

B
${ 2 }^{ pq }$

C
$p+q$

D
$pq$

Solution

Given $A$ and $B$ are two sets with number of elements $p$ and $q$ respectively.

The cartesian product of $A$ and $B = A \times B = \{(a,b): (a\in A)$ and $(b\in B)\}$
Number of elements in $A \times B = |A\times B| =|A|.|B| = pq$

Any relation from $A$ to $B$ is a subset of $A\times B$ .
Hence number of relations from $A$ to $B$ is the number of subsets of $A\times B$
$= 2^{|A\times B|} = 2^{pq}$
Question 10
1 / -0

If $N$ denote the set of all natural numbers and $R$ be the relation on $N\times N$ defined by $(a, b)R(c, d)$ . if $ad(b + c) = bc (a + d)$ , then $R$ is

A
Symmetric only

B
Reflexive only

C
Transitive only

D
An equivalence relation

Solution

For $(a, b), (c, d)\epsilon N\times N$
$(a, b)R(c, d)$
$\Rightarrow ad(b + c) = bc(a + d)$
Reflexive: $ab(b + a) = ba(a + b), \forall ab\epsilon N$
$\therefore (a, b)R(a, b)$
So, $R$ is reflexive,
Symmetric: $ad(b + c) = bc(a + d)$
$\Rightarrow bc(a + d) = ad(b + c)$
$\Rightarrow cd(d + a) = da(c + b)$
$\Rightarrow (c, d)R(a, b)$
So, $R$ is symmetric.
Transitive: For $(a, b), (c, d), (e, f)\epsilon N\times N$
Let $(a, b)R(c, d),\ (c, d) R(e, f)$
$\therefore ad(b + c) = bc(a+ d)\ and\ cf(d + e) = de(c + f)$
$\Rightarrow adb + adc = bca + bcd .... (i)$
and $cfd + cfe = dec + def ... (ii)$
On multiplying eq. (i) by $ef$ and eq. (ii) by $ab$ and then adding, we have
$adbef + adcef + cfdab + cfeab$
$= bcaef + bcdef + decab + defab$
$\Rightarrow adcf (b + e) = bcde (a + f)$
$\Rightarrow af (b + e) = be (a + f)$
$\Rightarrow (a, b)R(e, f)$
So, $R$ is transitive.
Hence, $R$ is an equivalence relation.

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Relations Test 23

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Relations Test 23

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SHARING IS CARING

Question 1

Symmetric

Reflexive

Transitive

All of these

Solution

Question 2

Symmetric only

Reflexive only

An equivalence relation

transitive only

Solution

Question 3

444

777

666

555

Solution

Question 4

333

444

555

666

Solution

Question 5

A function

Transitive

Not symmetric

Reflexive

Solution

Question 6

The relation is an equivalence relation on XXX

The relation is transitive but neither reflexive nor symmetric

The relation is reflexive but neither nor symmetric

The relation is symmetric but neither transitive nor reflexive

Solution

Question 7

Reflexive but not transitive

Symmetric but not reflexive

Both reflexive and symmetric but not transitive

An equivalence relation

Solution

Question 8

{(−2,2),(−1,1),(0,0),(1,1),(2,2)}\left \{(-2, 2), (-1, 1), (0, 0), (1, 1), (2, 2)\right \}{(−2,2),(−1,1),(0,0),(1,1),(2,2)}

{(−2,2),(−2,2),(−1,1),(0,0),(1,−2),(2,−1),(2,−2)}\left \{(-2, 2), (-2, 2), (-1, 1), (0, 0), (1, -2), (2, -1), (2, -2)\right \}{(−2,2),(−2,2),(−1,1),(0,0),(1,−2),(2,−1),(2,−2)}

{(0,0),(1,1),(2,2)}\left \{(0, 0), (1, 1), (2, 2)\right \}{(0,0),(1,1),(2,2)}

None of the above

Solution

Question 9

2p+q{ 2 }^{ p+q }2p+q

2pq{ 2 }^{ pq }2pq

p+qp+qp+q

pqpqpq

Solution

Question 10

Symmetric only

Reflexive only

Transitive only

An equivalence relation

Solution

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$4$

$7$

$6$

$5$

$3$

$4$

$5$

$6$

The relation is an equivalence relation on $X$

$\left \{(-2, 2), (-1, 1), (0, 0), (1, 1), (2, 2)\right \}$

$\left \{(-2, 2), (-2, 2), (-1, 1), (0, 0), (1, -2), (2, -1), (2, -2)\right \}$

$\left \{(0, 0), (1, 1), (2, 2)\right \}$

${ 2 }^{ p+q }$

${ 2 }^{ pq }$

$p+q$

$pq$