For all $p, q, m, n\in N$, we define $R$ as follows:$$(p,q)R(m,n)\Leftrightarrow pn=qm$$
Prove that $R$ is an equivalence relation on $N$.
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It should be $\mathbb{N}\times\mathbb{N}$ – Adam Martens Dec 09 '19 at 01:33
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The claim in the title is false. – Dec 09 '19 at 01:33
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$R$ is not an equivalence relation because $(1,1)R(0,0)$ and $(0,0)R(2,1)$ and $\neg ((1,1)R(2,1))$.
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@NitinUniyal That’s dependent on your definition of $\mathbb N$. Really, this should’ve been clarified by the OP. – ViHdzP Dec 09 '19 at 09:03
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View the pair $(p,q)$ as a fraction $\frac{p}{q}$. Proving $R$ is reflexive and symmetric are clear. What does transitivity mean in this new interpretation?
Adam Martens
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$(p,q)R(p,q)$ holds as $pq=qp$ so $R$ is reflexive.
$(p,q)R(m,n)\implies pn=qm\implies mq=np\implies (m,n)R(p,q)$ so $R$ is symmetric.
$(p,q)R(m,n)$ and $(m,n)R(r,s)\implies pn=qm$ & $ms=nr$
On multiplication, this gives $pnms=qmnr\implies ps=qr$ on cancellation. This implies $(p,q)R(r,s)$ so $R$ is transitive.
Addendum- I assumed that the set of natural numbers $N$ doesn't include $0$.
Nitin Uniyal
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