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CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 3 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Relations
Quiz 3
Let $$R$$ be a relation such that $$R = \{(1,4), (3, 7), (4, 5), (4, 6), (7, 6) \}$$ then $$(R^{-1} oR)^{-1} =$$
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$$\{(1, 1), (3, 3), (4, 4), (7, 7), (4, 7), (7, 4), (4, 3)\}$$
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$$\{(1, 1), (3, 3), (4, 4), (7, 7), (4, 7), (7, 4) \}$$
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$$\{(1, 1), (3, 3), (4, 4) \}$$
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$$\phi$$
If $$A$$ and $$B$$ are two sets, then $$A \times B = B \times A$$ if
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$$A \subset B$$
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$$B \subset A$$
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$$A = B$$
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None of these
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 $$R$$.
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$$z, z$$
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$$z^+, z$$
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$$z, z^-$$
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None of these
Explanation
Given:
$$R=\{ (a, b) : a, b, \in z, a-b \text{ is an integer}\}$$
As difference of integers are also integers so,
Domain of $$R = z$$
Range of $$R = z$$, as
$$a-b$$ spans the whole integer values.
Let $$R$$ be a relation on the set $$N$$ given by $$R=\left\{ \left( a,b \right) :a=b-2,b>6 \right\}$$. Then
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$$(2,4)\in R$$
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$$(3,8)\in R$$
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$$(6,8)\in R$$
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$$(8,7)\in R$$
Explanation
$$a=b-2,b>6$$
Above relation also imply $$a>6-2\rightarrow a>4$$
Among the options
$$C$$ is correct answer as $$6>4$$ and aslo $$6=8-2$$ .
Which of the following is not an equivalence relation on $$Z$$?
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$$aRb\Leftrightarrow a+b$$ is an even integer
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$$aRb\Leftrightarrow a-b$$ is an even integer
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$$aRb\Leftrightarrow a< b$$
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$$aRb\Leftrightarrow a= b$$
Explanation
Among the Options
$$Opt:[C]$$
$$aRb\Leftrightarrow a< b$$
As, $$a<b$$ doesn't imply $$b<a$$ ,which make the relation asymmetric and thus it is not an equivalence relation .
A relation $$R$$ is defined from $$\left\{ 2,3,4,5 \right\} $$ to $$\left\{ 3,6,7,10 \right\} $$ by:
$$xRy\Leftrightarrow x$$ is relatively prime to $$y$$. Then, domain of $$R$$ is
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$$\left\{ 2,3,5 \right\} $$
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$$\left\{ 3,5 \right\} $$
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$$\left\{ 2,3,4 \right\} $$
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$$\left\{ 2,3,4,5 \right\} $$
Explanation
Let's first write R as a set of ordered pairs,
$$R=\{(2,3),(2,7),(3,7),(3,10),(4,3),(4,7),(5,3),(5,6),(5,7)\}$$
So, Domain of R is $$\{2,3,4,5\}$$
The relation $$R$$ in $$N\times N$$ such that $$(a,b)R(c,d)\Leftrightarrow a+d=b+c$$ is
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reflexive but not symmetric
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reflexive and transitive but not symmetric
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an equivalence relation
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none of these
Explanation
If (a,b)R(c,d) then (c,d)R(a,b) as if a+d=b+c then b+d=a+c .
Thus relation R is symmetric .
For some a,b in N , (a,b)R(a,b) as a+b=b+a.
Thus every element of $$N\times N$$ is related to itself,
So relation R is also reflexive .
For some a,b,c,d,e,f in N, if (a,b)R(c,d) and (c,d)R(e,f) then (a,b)R(e,f),
As if a+d=b+c...1 and c+f=d+e….2 then a+f=b+e [since eq.1- eq.2 gives a-e=b-f$$\rightarrow$$a+f=b+e ]
So, R is also symmetric relation .
Thus R is a equivalence relation.So, C is correct answer .
Let $$R$$ be the relation over the set of all straight lines in a plane such that $${l}_{1}$$ $$R$$ $${l}_{2}\Leftrightarrow {l}_{1}\bot {l}_{2}$$. Then, $$R$$ is
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symmetric
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reflexive
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transitive
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an equivalence relation
Explanation
For some lines $$l_1,l_2$$, We know that $$l_1\perp l_2 \leftrightarrow l_2\perp l_1$$
So, given relation is symmetric .
For some lines $$l_1,l_2,l_3$$, We know that if $$l_1\perp l_2 \ and \ l_2\perp l_3$$, then $$l_1\parallel l_3$$
So, given relation is not transitive .
For some line $$l$$, We know that it is never possible that $$l\perp l$$.
So, given relation is not reflexive .
Thus $$A$$ is correct answer .
If $$A=\left\{ a,b,c \right\} $$, then the relation $$R=\left\{ \left( b,c \right) \right\} $$ on $$A$$ is
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reflexive only
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symmetric only
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transitive only
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reflexive and transitive only
Explanation
The relation R contains (b,c) but do not contain (c,b) , so it is not symmetric .
The relation R do not relates any element of set A with itself, so it is not reflexive .
The relation R contains only one pair (b,c), so it is transitive .
Thus option C is correct .
Let $$A=\left\{ 1,2,3 \right\} $$. Then, the number of equivalence relations containing $$(1,2)$$ over set A is
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$$1$$
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$$2$$
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$$3$$
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$$4$$
Explanation
Relation to be equivalent firstly must contain (1,1),(2,2),(3,3) to make it reflexive relation over given set .
Also since (1,2) is to be contained in set (2,1) must also be there to make relation symmetric .
So, one option of relation R is $$R=\{(1,1)(2,2)(3,3)(1,2)(2,1)\}$$
Now if we add (1,3) we have to also add (3,1) to keep relation symmetric .
So, second option of relation R is $$R=\{(1,1)(2,2)(3,3)(1,2)(2,1)(1,3)(3,1)\}$$
And, third option of relation R is $$R=\{(1,1)(2,2)(3,3)(1,2)(2,1)(1,3)(3,1)(2,3)(3,2)\}$$
Thus, there can be three equivalence relations containing (1,2).
Let $$A=\left\{ 1,2,3 \right\} $$ and $$R=\left\{ \left( 1,2 \right) ,\left( 2,3 \right) ,\left( 1,3 \right) \right\} $$ be a relation on $$A$$. Then $$R$$ is
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neither reflexive nor transitive
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neither symmetric nor transitive
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transitive
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none of these
Explanation
The given relation is not reflexive as $$(1,1),(2,2),(3,3)\not \in R$$.
Again the relation is not symmetric as $$(1,2)\in R$$ but $$(2,1)\not \in R$$.
But the relation is transitive as $$(1,2)\in R,(2,3)\in R\Rightarrow (1,2)\in R$$ for all $$1,2,3\in A$$.
If $$A=\left\{ 1,2,3 \right\} , B=\left\{ 1,4,6,9 \right\} $$ and $$R$$ is a relation from $$A$$ to $$B$$ defined by $$x$$ is greater than $$y$$. The range of $$R$$ is
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$$\left\{ 1,4,6,9 \right\} $$
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$$\left\{ 4,6,9 \right\} $$
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$$\left\{ 1 \right\} $$
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none of these
Explanation
First Let's write R as set of ordered pairs,
$$R=\{(2,1),(3,1)\}$$
We can simply see from set that range of $$R = \{1\}$$
A relation $$\phi$$ from $$C$$ to $$R$$ is defined by $$x\phi y\Leftrightarrow \left| x \right| =y$$. Which one is correct?
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$$(2+3i)\phi 13$$
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$$3\phi (-3)$$
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$$(1+i)\phi 2$$
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$$i\phi 1$$
Explanation
Given,
a relation $$\phi$$ from $$C$$ to $$R$$ is defined by $$x\phi y\Leftrightarrow \left| x \right| =y$$.
Now, $$|(2+3i)|=\sqrt{2^2+3^2}=\sqrt{13}$$.
So $$(2+3i)$$ and $$13$$ are not related.
Again, since $$y>0$$ this gives $$3$$ not related to $$-3$$.
$$|1+i|=\sqrt{1+1}=\sqrt 2$$
So $$1+i$$
and $$2$$ are not related.
Also $$|i|=1$$ so $$i\phi 1$$ holds.
Let $$R$$ be a relation on $$N$$ defined by $$x+2y=8$$. The domain of $$R$$ is
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$$\left\{ 2,4,8 \right\} $$
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$$\left\{ 2,4,6,8 \right\} $$
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$$\left\{ 2,4,6 \right\} $$
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$$\left\{ 1,2,3,4 \right\} $$
Explanation
$$x+2y=8$$
$$x=8-2y$$
Since y and x both must be natural,
Possible values of y are 1,2,3 which corresponds to possible value of x to be 6,4,2 .
Thus range of R = $$\{2,4,6\}$$
If $$A=\left\{ 1,2,3 \right\} $$, then a relation $$R=\left\{ \left( 2,3 \right) \right\} $$ on $$A$$ is
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symmetric and transitive only
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symmetric only
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transitive only
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none of these
Explanation
A binary relation R over Set X is transitive if for all elements a,b,c in X , if a is related to b and b is related to c, then a is related to c.
A binary relation R over Set X is symmetric iff $$\forall a,b \in X (aRb \leftrightarrow bRa)$$
But Given relation is not symmetric as it contains (2,3) but do not contain (3,2).
Since Given relation R contains only one pair (2,3), it is transitive .
Let $$A=\left\{ 1,2,3 \right\} $$ and consider the relation $$R=\left\{ \left( 1,1 \right) ,\left( 2,2 \right) \left( 3,3 \right) ,\left( 1,2 \right) ,\left( 2,3 \right) ,\left( 1,3 \right) \right\} $$, then $$R$$ is
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reflexive but not symmetric
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reflexive but not transitive
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symmetric and transitive
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neither symmetric nor transitive
Explanation
Relation $$R$$ is reflexive relation over set $$A$$ as every element of $$A$$ is related to itself in $$R$$.
Relation $$R$$ is not symmetric as $$(1,2)$$ is in $$R$$ but $$(2,1)$$ is not in $$R$$.
Relation $$R$$ is also transitive as for $$a,b,c$$ in $$A$$, if $$(a,b)$$ is in $$R$$ and $$(b,c)$$ is in $$R$$, then $$(a,c)$$ is also in $$R$$.
Thus $$A$$ is correct answer .
If $$n(A) = 4$$ and $$n(B) = 5$$, then $$n(A \times B) = $$
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$$20$$
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$$25$$
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$$4$$
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$$15$$
Explanation
Given, $$n\left( A \right) =4$$ and $$n\left( B \right) =5$$
$$ \Rightarrow n\left( A\times B \right) =n\left( A \right) \cdot n\left( B \right) $$
$$\Rightarrow n(A\times B)=4\cdot 5=20$$
So, answer is $$20$$.
In the set $$Z$$ of all integers, which of the following relation $$R$$ is an equivalence relation?
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$$xRy:$$ if $$x\le y$$
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$$xRy:$$ if $$x-= y$$
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$$xRy:$$ if $$x- y$$ is an even integer
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$$xRy:$$ if $$x\equiv y$$ (mod 3)
Explanation
For option C, xRy:if x-y is an even integer .
Let a be any number in Z,
Then, a-a=0 is an even integer .
So, Every number is related to itself .
Thus the relation is reflexive .
Let a,b be two numbers in Z,
Also let $$ a-b=k$$ be an even integer (which implies aRb)
Then, $$ b-a=-k $$ is also an even integer (which implies bRa).
Thus, It is evident $$aRb\leftrightarrow bRA$$
Thus the relation is also symmetric .
Let a,b,c be three numbers in Z,
Also let aRb and bRc,
then $$a-b=2k_1$$ and $$b-c=2k_2$$ for $$k_1,k_2$$ belonging to Z
which gives $$a-b+b-c=2k_1+2k_2\Rightarrow a-c=2(k_1+k_2)$$
which implies $$aRc$$ .
Thus relation is also transitive .
Thus we finally arrive at a conclusion that relation is Equivalence relation .
Let $$L$$ denote the set of all straight lines in a plane, Let a relation $$R$$ be defined by $$lRm$$, iff $$l$$ is perpendicular to $$m$$ for all $$l \in L$$. Then, $$R$$ is
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reflexive
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symmetric
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transitive
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none of these
Explanation
We know that if line l is perpendicular to line m, line m must also be perpendicular to line l,
So, $$lRm \Leftrightarrow mRl$$ and thus Relation R is symmetric .
Let $$T$$ be the set of all triangles in the Euclidean plane, and let a relation $$R$$ on $$T$$ be defined as $$aRb$$, if $$a$$ is congruent to $$b$$ for all $$a,b\in T$$. Then, $$R$$ is
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reflexive but not symmetric
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transitive but not symmetric
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equivalence
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none of these
Explanation
Let there be three triangles a,b,c in Euclidean plane such that aRb and bRc.
We know that if triangle a is congruent to triangle b and triangle b is congruent to triangle c, triangle a must also be congruent to triangle c .
So, aRc and thus relation R is transitive .
Also we know that any triangle (say t) in Euclidean plane is always congruent to itself .
So, tRt and thus relation R is reflexive .
Let there be two triangles n and m in Euclidean plane.
We know that if n is congruent to m, m must also be congruent to n.
So, $$nRm \Leftrightarrow mRn$$ and thus relation R is symmetric .
Thus, finally we arrive at conclusion that realation R is equivalence relation .
Let $$R$$ be a relation on the set $$N$$ of natural numbers defined by $$nRm$$, iff $$n$$ divides $$m$$. Then, $$R$$ is
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Reflexive and symmetric
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Transitive and symmetric
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Equivalence
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Reflexive, transitive but not symmetric
Explanation
Let there be a natural number $$n$$,
We know that $$n$$ divides $$n$$, which implies $$nRn$$.
So, Every natural number is related to itself in relation $$R$$.
Thus relation $$R$$ is reflexive .
Let there be three natural numbers $$a,b,c$$ and let $$aRb,bRc$$
$$aRb$$ implies $$a$$ divides $$b$$ and $$bRc$$ implies $$b$$ divides $$c$$, which combinedly implies that $$a$$ divides $$c$$ i.e. $$aRc$$.
So, Relation R is also transitive .
Let there be two natural numbers $$a,b$$ and let $$aRb$$,
$$aRb$$ implies $$a$$ divides $$b$$ but it can't be assured that $$b$$ necessarily divides $$a$$.
For ex, $$2R4$$ as 2 divides 4 but 4 does not divide 2 .
Thus Relation R is not symmetric .
The relation $$R=\left\{ \left( 1,1 \right) ,\left( 2,2 \right) \left( 3,3 \right) \right\} $$ on the set $$A=\left\{ 1,2,3 \right\} $$ is
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symmetric only
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reflexive only
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an equivalence relation
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transitive only
Explanation
The relation R is reflexive relation over set A as every element of set A is related to itself in relation R.
The relation R is also symmetric as for a,b in A, if (a,b) is in R, then (b,a) is also in R .
The relation R is also transitive as for a,b,c in A, if (a,b) is in R and (b,c) is in R, then (a,c) is also in R.
Thus relation R is an equivalence relation, So C is correct answer .
Which one of the following relations on Z is equivalence relation?
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$$xR_1 y\Leftrightarrow |x| = |y|$$
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$$xR_2 y\Leftrightarrow x \geq y$$
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$$xR_3 y\Leftrightarrow \dfrac{x}{y}$$
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$$xR_4 y\Leftrightarrow x < y$$
Explanation
A relation is equivalence if it is symmetric, transitive and reflexive
Consider, $$xR_1y\Leftrightarrow \left | x \right |=\left | y \right |$$
For reflexive:
$$xR_1x\Leftrightarrow \left | x \right |=\left | x \right |$$, clearly it satisfies $$R_{1}$$
Thus, $$R_{1}$$ is reflexive.
For symmetric:
$$xR_{1}y\Leftrightarrow \left | x \right |=\left | y \right |$$
$$\implies yR_{1}x\Rightarrow \left | y \right |=\left | x \right |$$ as equality communicative
So, $$ R_{1}$$ is symmetric.
For transitive:
Let $$(x,y)$$ and $$(y,z)$$ satisfy $$R_{1}$$
$$\implies xR_{1}y\Leftrightarrow \left | x \right |=\left | y \right |$$ and $$yR_{1}z\Leftrightarrow \left | y \right |=\left | z \right |$$
$$\implies |x|=|y|$$ and $$|y |=|z|$$
$$\implies |x |=|z|$$
$$\therefore xR_{1}z\Leftrightarrow \left | x \right |=\left | z \right |$$
$$\therefore \left ( x,z \right )$$ satisfies $$R_{1}$$
Thus, $$R_1$$ is a transitive relation.
Hence, $$ R_{1}$$ is equivalent.
Also, relations $$R_{2},\ R_{3}$$ and $$R_{4}$$ are not symmeteric on $$Z,$$ hence not an equivalence relations.
Let $$R$$ be a reflexive relation on a finite set $$A$$ having $$n$$ elements, and let there be $$m$$ ordered pairs in $$R,$$ then:
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$$m\geq n$$
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$$m\leq n$$
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$$m=n$$
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$$m< n$$
Explanation
By the definition for a relation with $$n$$ elements to be reflexive relation it must have at least $$n$$ ordered pairs.
it has $$m$$ ordered pairs therefore $$\mathrm{m}\geq \mathrm{n}$$.
Total number of equivalence relations defined in the set $$S =\{a,b,c\}$$ is
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$$5$$
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$$3$$!
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$$2^{3}$$
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$$3^{3}$$
Explanation
The smallest equivalence relation is the identity
relation $$R_1=\{(a,a),(b,b),(c,c)\}$$
Then two ordered pairs of two distinct elements
can be added to give three more equivalence
relations.
$$R_2=\{(a,a),(b,b),( c,c),(a,b),(b,a)\}$$
Similarly $$R_3$$and $$R_4$$ can be made by taking $$(b,c),(c,b)$$ and $$(a,c),(c,a)$$ respectively.
Finally the largest equivalence relation i.e., the universal relation
$$R_5=\{(a,a),(b, b),(c, c),(a, b),(b, a),(a, c),(c,a ),(b, c),(c, b) \}$$
Hence, total $$5$$ equivalence relations can be created.
If relation $$R$$ is defined by $$\mathrm{R}=\{(\mathrm{x},\ \mathrm{y}):2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6\}$$, then the domain of $$\mathrm{R}$$ is
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$$[-3,3]$$
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$$[-\sqrt{3},\sqrt{3}]$$
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$$[-\sqrt{2},\sqrt{2}]$$
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$$[-2,2]$$
Explanation
Domain $$=$$ value of $$\mathrm{x}$$ in $$f=2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6$$ then
$$2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6$$ = $$\displaystyle \frac{\mathrm{x}^{2}}{3}+\frac{\mathrm{y}^{2}}{2}\leq 1$$
Since it is an equation of ellipse then value of $$x$$ $$ \Rightarrow \mathrm{x}\in[-\sqrt{3},\sqrt{3}]$$
If $$\displaystyle A=\left \{ 2, 3, 5 \right \}, B=\left \{ 2, 5, 6 \right \}$$ then $$\left ( A-B \right )\times \left ( A\cap B \right )$$ is
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$$\displaystyle \left \{ \left ( 3, 2 \right ), \left ( 3, 3 \right ), \left ( 3, 5 \right )\right \}$$
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$$\displaystyle \left \{ \left ( 3, 2 \right ), \left ( 3, 5 \right ), \left ( 3, 6 \right )\right \}$$
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$$\displaystyle \left \{ \left ( 3, 2 \right ), \left ( 3, 5 \right )\right \}$$
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None of these
Explanation
$$\displaystyle A-B=\left \{ 3 \right \},A\cap B=\left ( 2, 5 \right ).$$
$$\displaystyle \therefore \left ( A-B \right )\times \left ( A\cap B \right )=\left \{ \left ( 3, 2 \right ),\left ( 3, 5 \right ) \right \}.$$
The domain and range of relation $$R=\{(x,y) | x, y \in N$$, $$x+2y=5\} $$ is?
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$$\{1,3\}, \{2,1\}$$
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$$\{2,1\}, \{3,2\}$$
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$$\{1,3\}, \{1,1\}$$
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$$\{1,2\}, \{1,3\}$$
Explanation
The possible values of $$x$$ which satisfies the given relation is the domain of the relation
The possible values of $$y$$ which satisfies the given relation is the range of the relation
Given that $$x,y$$ are natural numbers and the relation is $$x+2y=5$$
For $$x=1$$ , the value of $$y$$ is $$2$$
For $$x=3$$ , the value of $$y$$ is $$1$$
If $$x\geq 5$$, then $$y$$ is negative which does not belong to naturals.
If $$x=2,4$$ then $$y$$ becomes rational number which is not in naturals.
Thus, only $$1$$ and $$3$$ gives $$y$$ as natural numbers $$2$$ and $$1$$.
Therefore, the domain is $$\{1,3\}$$ and the range is $$\{2,1\}$$
Let $$A$$ be a non-empty set such that $$A \times A$$ has $$9 $$ elements among which are found $$(-1, 0)$$ and $$(0, 1)$$, then
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$$A=\left\{ -1,0 \right\}$$
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$$A=\left\{ 0,1 \right\}$$
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$$A=\left\{ -1,0,1 \right\}$$
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$$A=\left\{ -1,1 \right\}$$
Explanation
Given $$A\times A$$ has $$9$$ elements $$\Rightarrow n(A)=3$$
Elements of $$A\times A $$ are $$(-1,0)$$ & $$(0,1)$$.
$$\therefore$$ $$A$$ needs to contain $$-1, 0$$ and $$1$$.
$$\therefore$$ Since $$n(A)=3 \Rightarrow A=\left \{ -1,0,1 \right \}$$
Given the relation $$R =\{(1,2) (2,3)\}$$ on the set $$A=\{1,2,3\}$$, the minimum number of ordered pairs which when added to $$R$$ to make it an equivalence relation is
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$$5$$
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$$6$$
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$$7$$
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$$8$$
Explanation
$$R$$ is reflexive if it contains $$(1, 1), (2, 2), (3, 3)$$.
$$\because (1, 2)\in R$$ and $$ (2,3)\in R$$
$$\therefore R$$ is symmetric if $$(2, 1), (3, 2) \in R.$$
As $$(1, 2), (2,3)\in R$$
Then, $$R$$ is transitive is $$(1,3)\in R$$
Also, $$(3,1)\in R$$ for symmetric property.
Now,
$$R=\{(1,1),(2,2),(3,3),(2,1),(3,2),(2,3),(1,2),(1,3),(3,1)\}$$
Thus, $$R$$ becomes an equivalence relation by
adding
$$(1,1),(2,2),(3,3),(2,1),(3,2),(1,3),(3,1)$$
Hence, the total number of ordered pairs
is $$7$$.
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