GATE CSE 1994 | Question: 2.3

Question

Amongst the properties $\left\{\text{reflexivity, symmetry, anti-symmetry, transitivity}\right\}$ the relation $R=\{(x, y) \in N^2|x \neq y\}$ satisfies _________

Comments

alternate approach:

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@Prateek Raghuvanshi $R=(x,y)$ ∈ $N^{2}$ here means $R=(x,y)$ ∈ $N\times N$ not $R=(x,y)$ ∈ $N^{2} \times N^{2}$ as suggested by your solution.

Answer 1

• It is not reflexive as $xRx$ is not possible.
• It is symmetric as if $xRy$ then $yRx$.
• It not antisymmetric as $xRy$ and $yRx$ are possible and we can have $x \neq y$.
• It is not transitive as if $xRy$ and $yRz$ then $xRz$ need not be true. This is violated when $z = x$.

So, symmetry is the answer.

Comments

Isn't R={ (2,4),(3,9),(4,16)....} what does belong to N square mean.

How is it not anti-symmetric and transitive can you please give counter examples for the Relation

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@sripo (4,16) and (16,4) both belong to R, since these are symmetric pairs and 4 !=16 which violates antisymmetric property therefore R is not antisymmetric.

And For Transitive : Both (4,16) and (16,4) belongs to R therefore (4,4) must also belong to R but since here

"x !=y" is the condition therefore (4,4) is not allowed and it is not Transitive as well. 

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Let’s say N={1,2,3}

NxN ={ (1,2), (1,3)  , (2,3)}

and in ques it is saying (x,y) belongs to NxN and x!=y

so in NxN every element satisfies X!=y condition and it is following transitive property also as  (1,2), (2,3)  there so (1,3)  also there so transitive possible?

Please correct me

Answer 2

Is this a valid answer

 

as x!=y Reflexive never holds true

For symmetry R={(1,2),(2,1)...} is possible

Having Symmetry invalidates anti-symmetry

Transitive is not possible as (1,2),(2,1) should imply (1,1) which we cannot get as it is not reflexive hence transitivity fails

Comments

@sripo 

Relation should be {(1,4)(4,1)...}.  (According to the given question)

$\because (x,y) \in N^2$

 

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@sripo

Having Symmetry invalidates anti-symmetry

This is not always true

Counter Example:

Let $A=\{1,2,3,4\}$ and $B=\{1,2,3\}$ and $R$ is a relation from $A$ to $B$

$R=\{ (1,1)(3,3)\}$

$R$ is symmetric as well as antisymmetric. 

Answer 3

$\therefore$ The given relation  satifies only $\large symmetry$