Equivalence relation

Question

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.

Answer 1

$\mathbb N$ 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 $n_2\in\mathbb N$.

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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 $(n_1, n_2)R(k_1, k_2)$

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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 $(\mathbb N\times \mathbb N)\times(\mathbb N\times \mathbb N)$