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

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we can also check like this- 0

@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.

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 $z = x$.

by Boss (33.9k points)
edited by
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Hi,

Can some one confirm if the below relations are symmetric or anti symmetric over {1,2}

1. {(1,1),(2,2), (1,2),(2,1)}

2. {(1,1),(1,2),(2,1)}

3.{(1,1),(2,2),(1,2)}
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1.  {(1,1),(2,2), (1,2),(2,1)} = Symmetric
2.  {(1,1),(1,2),(2,1)} = Symmetric
3. {(1,1),(2,2),(1,2)} = AntiSymmetric
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Hey sachin can you please explain this question in detail?
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Since, $x \not=y$ reflexive pairs are not allowed.Not Reflexive.

The matrix of this relation would be like

$\begin{bmatrix} 0 & & & \\ & 0& & \\ & & ...& \\ & & & 0 \end{bmatrix}$

only diagonals would be zero,and all other would have an entry 1.

Surely this matrix would be symmetric hence relation also symmetric.

Anti-Symmetry says that you can take the diagonal pairs and either one of the lower triangular or upper triangular pairs but not both at the same time, and in this matrix, we have both upper as well as lower triangular matrix pairs in R, so this is not Anti-Symmetric.

Transitive, surely not!!. Because since, reflexivity fails, this failure may create trouble

like (2,1) and (1,2) both belong to R but (2,2) won't. So not transitive.
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@Ayush You are correct about reflexivity and symmetry. The main diagonal is all 1’s, so the relation is reflexive, and the matrix is symmetric about the main diagonal, so the relation is symmetric.

As for

• A={1,2,3,}
• $\begin{bmatrix} a11 & a12 & a13 \\ a21& a22 &a23 \\ a31& a32& a33 \end{bmatrix}$
• please explain it more or with example
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For Anti-Symmetry you can consider below implication

$\forall x \forall y ((xRy \land yRx) \rightarrow x=y)$

this implication allows reflexive pairs but not symmetric pairs with different elements in pairs like

if 1R2 and 2R1 then $1 \not= 2$ makes the implication false. Hence, this causes Anti-symmetry to be broken.
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@ayush what you have explained I understood this

But in matrix m not understanding what u have explained
"Anti-Symmetry says that you can take the diagonal pairs and either one of the lower triangular or upper triangular pairs but not both at the same time, and in this matrix, we have both upper as well as lower triangular matrix pairs in R, so this is not Anti-Symmetric."

In lower triangular matrix ,

suppose if I take diagonal element a11 then 0!=0 which is false that's why antisymmetry doesn't hold because 0should equal to 0 and if I take only lower triangular matrix element a21 then then 2!=1 then antisymmetry holds but if I take upper triangular elements as well then 1!=2 then({ 2!=1 and1!=2 }->2!=2 )which is false so antisymmetry doesn't hold

Correct me if m wrong
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@Prince-No, if your reflexivity fails, that won't affect Anti-Symmetry. But for Even one pair say (1,2) and(2,1) exists, Your anti-symmetry is broken.

More clarity can come if you take a finite set, under the same relation in this question, try to draw the digraph of the ordered pairs of R. Now, for any two nodes, if you have bi-directional paths between them, say from node x to  y and from y to x, your anti-symmetry is broken. So same here, if your digraph has self-loops or not then that won't affect anti-symmetry.
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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.

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
by Active (2.4k points)
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Relation should be {(1,4)(4,1)...}.  (According to the given question)

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