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Let $R$ be a relation on the set of ordered pairs of positive integers such that $((p,q),(r,s))∈R$ if and only if $p−s=q−r$. Is this relation an equivalence relation ?

My Try :

  • It is not a reflexive relation as $((p,q),(p,q))∈R$ then, $p-q = q-p$ and it is not equal.

  • It is a symmetric relation as $((p,q),(r,s))∈R$ then, the set $((r,s),(p,q))∈R$ should also be present. So, $r-q = s- p$ which is equal to $p−s=q−r$

  • It is transitive relation too as, if $((p,q),(r,s))∈R$ and if $((r,s),(e,f))∈R$ then, $((p,q),(e,f))∈R$ which is true .

Hence, it is not a equivalence relation as out of reflexive, symmetric and transitive, reflexive property is not satisfied.

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    What don't you understand? A relation is subset of S x S for some S. In this case S=NxN, so R is the subset of (NxN)x (NxN). In this case R= {((p,q), (r,s))|p-s=q-r}. – 2017-01-18
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    @fleablood Thanks !! Got that . – 2017-01-18

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Hence, it is not a equivalence relation as out of reflexive, symmetric and transitive, reflexive property is not satisfied.

That is so.

But, what is the meaning of this line "Let $R$ be a relation on the set of ordered pairs of positive integers" ?

It means that $R\subseteq (\Bbb Z^+{\times} \Bbb Z^+){\times}(\Bbb Z^+{\times} \Bbb Z^+)$ .   A relation on a set is a subset of the Cartession square of that set, and the set of interest is the Cartessian square of the positive integers.   The positive integers are $\{1, 2, 3, \ldots\}$, that is the natural numbers excluding zero.

$$R\subseteq \Big\{\big((p,q),(r,s)\big): p\in\Bbb Z^+, q\in\Bbb Z^+,r\in\Bbb Z^+,s\in\Bbb Z^+\Big\}$$

Specifically, $$R = \Big\{\big((p,q),(r,s)\big): p\in\Bbb Z^+, q\in\Bbb Z^+,r\in\Bbb Z^+,s\in\Bbb Z^+, p+r=q+s\Big\}$$