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A relation $R$ is defined on $N \times N$, such that $(a,b) R (c,d)$ iff $a+d = b+c.$ The relation $R$ is

1. reflexive but not transitive
2. reflexive and transitive but not symmetric
3. an equivalence relation
4. a partial order

(1,2)R(6,5) then (6,5) not in relation (1,2) so not symmetric. and (6,5)R(2,1) then (1,2) not in relation (2,1) so not transitive. then how is the answer equivalence since it has to satisfy reflexive, symmetric and transitive property.

$(a,b) R (c,d)$ iff $a+d = b+c$.

So,

$(c,d) R (a,b)$ iff $c+b = d+a$.

From property of addition, $d+a = a + d, c+b = b+c$.

So, $(a,b) R (c,d)$ iff $(c,d) R (a,b)$. Hence, $R$ is symmetric.

$a+b = b+a$, so $(a,b) R (a,b)$. Hence, $R$ is reflexive.

If, $(a,b) R (c,d)$ and $(c,d) R (e,f)$, we have $a+d = b+c$ and $c+f = d+e$. So, $a + d + c + f = b + c + d + e \implies a + f = b + e \implies (a,b) R (e,f).$

Hence, $R$ is transitive.

Since $R$ is reflexive, transitive and symmetric, it is an equivalence relation.
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edited by
thank you sir. makes sense now.

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