Relations Test 23

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Relations Test 23

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Weekly Quiz Competition

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
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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
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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
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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

$$4$$

$$7$$

$$6$$

$$5$$

Solution

Question 4

$$3$$

$$4$$

$$5$$

$$6$$

Solution

Question 5

A function

Transitive

Not symmetric

Reflexive

Solution

Question 6

The relation is an equivalence relation on $$X$$

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

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

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

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

None of the above

Solution

Question 9

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

$${ 2 }^{ pq }$$

$$p+q$$

$$pq$$

Solution

Question 10

Symmetric only

Reflexive only

Transitive only

An equivalence relation

Solution

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