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A relation $R$ is defined on $\mathbb{N}\times\mathbb{N}$ by $(a,b)R(c,d)$ iff $ad = bc$. Show that $R$ is an equivalence relation.

I know that in order to find the equivalence relation we need to determine if $R$ is reflexive, symmetric, and transitive.

But I'm having some struggle in understanding this $\mathbb{N}\times\mathbb{N}$.

I'll be grateful for any help.

Thank you.

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  • $\begingroup$ Its just the set of natural numbers $\endgroup$ – Archis Welankar Apr 7 '17 at 13:06
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    $\begingroup$ Might be good to specify that (apparently) you intend $0\notin \mathbb N$, or else everything would be related to $(0,0)$. In that case, the equivalence classes are just a copy of the positive rational numbers ($(a,b)=\frac{a}{b}$) $\endgroup$ – rschwieb Apr 7 '17 at 13:16
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    $\begingroup$ @rschwieb Good point. If we consider $0 \in \mathbb{N}$, then the relation can still work on $\mathbb{N} \times( \mathbb{N} \setminus \{0\})$ though. $\endgroup$ – benguin Apr 7 '17 at 13:20
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    $\begingroup$ @benguin and then the set of classes is the canonical construction of the rational numbers. $\endgroup$ – lhf Apr 7 '17 at 13:24
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    $\begingroup$ @benguin Equally effective, although necessitating change in notation :) $\endgroup$ – rschwieb Apr 7 '17 at 13:25
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$\mathbb N$ is the set of natural numbers, and $\mathbb N\times\mathbb N$ is the set of pairs of natural numbers. So, each element of $\mathbb N\times \mathbb N$ is a pair $(n_1, n_2)$ such that $n_1\in\mathbb N$ and $n_2\in\mathbb N$.


Your job then, is to show that:

  1. $R$ is symmetric, that is, for every pair of natural number pairs $(n_1, n_2)$ and $(m_1, m_2)$, you know that if $(n_1, n_2)R(m_1, m_2)$, then $(m_1, m_2)R(n_1,n_2)$
  2. $R$ is reflexive, so for each pair $(n_1, n_2)\in\mathbb N\times \mathbb N$, you know that $(n_1, n_2)R(n_1, n_2)$
  3. $R$ is transitive, so for each three pairs $(n_1, n_2), (m_1, m_2), (k_1, k_2)\in\mathbb N\times\mathbb N$, you know that if $(n_1, n_2)R(m_1, m_2)$ and $(m_1,m_2)R(k_1, k_2)$, then $(n_1, n_2)R(k_1, k_2)$

Sidenote:

Since relations on a set $A$ are by definition subsets of the set $A\times A$, this means that your particular relation is, strictly speaking, a subset of $(\mathbb N\times \mathbb N)\times(\mathbb N\times \mathbb N)$

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  • $\begingroup$ Might be better to write "pair of natural number pairs" instead of "pair of integer pairs". $\endgroup$ – rschwieb Apr 7 '17 at 13:14

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