Question

Let $\mathbb{N}$ be the set of all natural numbers and let R be a relation on $\mathbb{N}\times \mathbb{N}$, defined by
$\left( a,b \right)R\left( c,d \right)\Leftrightarrow ad=bc$.
Show that R is an equivalence relation in $\mathbb{N}\times \mathbb{N}$.

Answer 1

Hint: Recall the definition of an equivalence relation. A relation is said to be an equivalence relation if it is reflexive, symmetric and transitive. Prove that the given relation is reflexive, symmetric as well as transitive. Hence prove that the given relation is an equivalence relation.

Complete step-by-step answer:
Before solving the question, we need to understand what is an equivalence relation.
Reflexive relation: A relation R on a set “A” is said to be reflexive if $\forall a\in A$ we have $aRa$.
Symmetric relation: A relation R on a set “A” is said to be symmetric if $aRb\Rightarrow bRa$
Transitive relation: A relation R on a set “A” is said to be transitive if $aRb,bRc\Rightarrow aRc$.
Equivalence relation: A relation R on a set “A” is said to be an equivalence relation if the relation is reflexive, symmetric and transitive.
Reflexivity: We have $\forall \left( a,b \right)\in \mathbb{Z},a\times b=b\times a$ we have $\left( a,b \right)R\left( a,b \right)$ and hence the relation is reflexive.
Symmetricity: We have $\left( a,b \right)R\left( c,d \right)\Rightarrow ad=bc$. Hence, we have $cb=da\Rightarrow \left( c,d \right)R\left( a,b \right)$. Hence the relation is symmetric.
Transitivity: We have $\left( a,b \right)R\left( c,d \right),\left( c,d \right)R\left( e,f \right)\Rightarrow ad=bc,cf=ed$
Hence, we have
$\begin{aligned} & ad=bc\text{ (i)} \\\ & ed=cf\text{ (ii)} \\\ \end{aligned}$
Dividing equation (i) by equation (ii), we get
$\dfrac{ad}{ed}=\dfrac{bc}{cf}\Rightarrow \dfrac{a}{e}=\dfrac{b}{f}$
Cross multiplying, we get
$af=be\Rightarrow \left( a,b \right)R\left( e,f \right)$.
Hence, the relation is transitive.
Since the relation is reflexive, symmetric and transitive, the relation is an equivalence relation.

Note: Alternative solution:
Observe that if $\left( a,b \right)R\left( c,d \right)$ then $a:b::c:d$
Since every ratio is proportional to itself, we have $a:b::a:b\Rightarrow \left( a,b \right)R\left( a,b \right)$
Hence the relation is reflexive.
We know that if a:b::c:d then c:d::a:b
Hence, we have $\left( a,b \right)R\left( c,d \right)\Rightarrow \left( c,d \right)R\left( a,b \right)$
Hence the relation is symmetric.
Also, we know that if a:b::c:d and c:d::e:f then a:b::e:f
Hence, we have $\left( a,b \right)R\left( c,d \right),\left( c,d \right)R\left( e,f \right)\Rightarrow \left( a,b \right)R\left( e,f \right)$.
Hence the relation is transitive, and hence the relation is an equivalence relation.