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Struggling a bit on this question regarding relations. Would appreciate your help immensely.

Determine if a relation is reflexive, symmetric, anti-symmetric & transitive, for:

$$R = \{(x,y) : xy = 4\} \text{ defined by } A = {0,1,2,3}$$

I've determined that it is not reflexive, as $x^2 = 4$ is only true for $(2,2)$.

I think it is not symmetric as $yx = 4$ is only true for $(2,2)$

I don't know how to prove anti-symmetry or transitive in this case, either.

The third question is the same, but the set was $\{0,1,2,3,4\}$ so I proved it was anti-symmetric by using $(1,4)\in \mathbb{R}$ and $(4,1)\in \mathbb{R}$, where $1\ne4$.

Thank you.

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    $\begingroup$ check the definition of symmetric $\endgroup$ – J. W. Tanner May 10 at 19:53
  • $\begingroup$ Would it be 4 = x/y? @J.W.Tanner $\endgroup$ – Ian May 10 at 20:25
  • $\begingroup$ It would be $(x,y)\in R \iff (y,x)\in R$ $\endgroup$ – J. W. Tanner May 10 at 20:29
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Check the definitions.

Symmetric means $(x,y)\in R\iff (y,x)\in R $ ; i.e., $xy=4 \iff yx=4,$

which is true because multiplication is commutative, so $R$ is symmetric.

Anti-symmetric means if $(x,y)\in R$ and $(y,x)\in R$ then $x=y.$

When the set is {$0,1,2,3$} and $(x,y)\in R$ and $(y,x)\in R$, then $x=y=2,$ so $R$ is anti-symmetric,

but when the set is {$0,1,2,3,4$} then $(1,4),(4,1)\in R$ so $R$ is not anti-symmetric because $1\ne4$.

Transitive means if $(x,y)\in R$ and $(y,z)\in R$ then $(x,z)\in R.$

$R$ is not transitive for {$0,1,2,3,4$} because $(1,4)\in R $ and $(4,1)\in R$ but $(1,1)\not\in R$,

but $R$ is transitive for {$0,1,2,3$} because in that case only $(2,2)\in R$.

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