Question

Let $N$ denote the set of all natural numbers and $R$ be the relation on $N\times N$ defined by $(a,b)R(c,d)⟺ad(b+c)=bc(a+d)$. Check weather $R$ is an equivalence relation.

Answer 1

Let R be defined on $N\times N$ as

$(a,b)R(c,d)⟺ad(b+c)=bc(a+d)$. ....(1)

Reflexivity:

We can write $ab(b+a)=ba(a+b)$ for all $a,b\in N$

Since, sum and product of natural numbers obeys commutative property

Hence, by def (1), we can write

$(a,b)R(a,b)$ for all $(a,b)\in N\times N$

Hence, $R$ is reflexive.

Symmmetry :

Let $(a,b)R(c,d)$

$\Rightarrow ad(b+c)=bc(a+d)$

$\Rightarrow da(c+b)=cb(d+a)$ (Since, sum and product of natural numbers obeys commutative property)

or $cb(d+a)=da(c+b)$

$\Rightarrow (c,d)R(a,b)$

Hence, $R$ is symmetric

Transitivity :

Let $(a,b),(c,d),(e,f)\in N\times N$

Let $(a,b)R(c,d)$ and $(c,d)R(e,f)$

$ad(b+c)=bc(a+d)$ and $cf(d+e)=de(c+f)$

$\Rightarrow \frac{ab}{a-b}=\frac{cd}{c-d}$ and $\frac{cd}{c-d}=\frac{ef}{e-f}$

$\Rightarrow \frac{ab}{a-b}=\frac{ef}{e-f}$

$\Rightarrow af(b+e)=be(a+f)$

$\Rightarrow (a,b)R(e,f)$

Hence ,$R$ is transitive

$\therefore R$ is Equivalence Relation.