UGC NET CSE | July 2018 | Part 2 | Question: 89

Question

Which of the following is an equivalence relation on the set of all functions from Z to Z?

• A. $\{ f, \:g) \mid f(x) - g(x) =1 \: \forall \: x \in \: Z \}$
• B. $\{ f, \:g) \mid f(0) = g(0) \text{ or } f(1) = g(1) \}$
• C. $\{ f, \:g) \mid f(0) = g(1) \text{ and } f(1) = g(0) \}$
• D. $\{ f, \:g) \mid f(x) - g(x) =k  \text{ for some } k \in Z \}$

Answer 1

Answer : D

Let $A$ be Set of All functions from $Z$ to $Z$.

Option A :

$R_1$ = $\left \{ (f,g) | f(x) =  g(x) = 1, \forall x \in Z \right \}$ .. It is Not Reflexive because $A$ is Set of All the functions from $Z$ to $Z$. And the given condition $f(x) =  g(x) = 1, \forall x \in Z $ will only be satisfied by an Unique pair $(f,f)$. So, $R_1$ is Symmetric, Transitive But Not Reflexive. 

$R_1$ = Not Reflexive, Symmetric, Transitive, Not Irreflexive, Antisymmetric, Not Asymmetric, Not an Equivalence Relation, Not a Partial Order relation.

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Option B : 

$R_2$ = $\left \{ (f,g) | f(0) =  g(0)  \,\,or\,\,f(1) = g(1) \right \}$

It is Reflexive, Symmetric But Not Transitive.

for e.g. : Let $(f,g), (g,h) \in R_2$ such that $f(0) = g(0) = 1$ and $g(1) = h(1) = 2$ and $h(0) = 4, f(1) = 5$ 

So, We can see that $(f,h) \notin R_2 $. Hence, Not Transitive.

$R_2$ = Reflexive, Symmetric, Not Transitive, Not Irreflexive, Not Antisymmetric, Not Asymmetric, Not an Equivalence Relation, Not a Partial Order relation.

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Option 3 : 

$R_3$ = $\left \{ (f,g) | f(0) =  g(1)  \,\,and\,\,f(1) = g(0) \right \}$

It is Not Reflexive because $A$ is Set of All the functions from $Z$ to $Z$. And the given condition $f(0) =  g(1) $ and $f(1) = g(0)$ will Not be satisfied by all the Pairs $(f,f)$. 

for e.g. : Let $h$ be a function such that $h(0) = 1$ and $h(1) = 2$ , So, we can see that $(h,h) \notin R_3$

$R_3$ = Not Reflexive, Symmetric, Not Transitive, Not Irreflexive, Not Antisymmetric, Not Asymmetric, Not an Equivalence Relation, Not a Partial Order relation.

See, $R_3$ is Not Transitive.

for e.g. : Let $(f,g), (g,h) \in R_3$ such that $f(0) = g(1) = 5$ and $f(1) = g(0) = 6$..So,  $g(0) = h(1) = 6$ and $g(1) = h(0) = 5$ 

So, We can see that $(f,h) \notin R_3 $ because $f(0) = 5 $ But $h(1) = 6$ Hence, Not Transitive.

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Option D :

 $R_4$ = $\left \{ (f,g) | f(x) -  g(x) = K,  \,\,for\,\,some\,\,K \in Z \right \}$

$R_4$ is Reflexive, Symmetric, Transitive. Hence, An Equivalence Relation.

NOTE that $R_4$ is Reflexive, Symmetric, Transitive because Range of All the functions is $Z$ and We know that $Z - Z = Z$

$R_4$ = Reflexive, Symmetric, Transitive, Not Irreflexive, Not Antisymmetric, Not Asymmetric, An Equivalence Relation, Not a Partial Order relation.

Comments

check the option A again from this edited question where f(x)-g(x)=1

and option 4 should be true only for value K=0