Question

Let N denote the set of all natural numbers and R be the relation on $N \times N$ defined by $(a,b)R(c,d)$ if $ad(b + c) = bc(a + d)$, then R is
1. Symmetric only
2. Reflexive only
3. Transitive only
4. An equivalence relation

Answer 1

A relation R on a set A is said to be an equivalence relation if and only if the relation R is reflexive, symmetric and transitive. For solving such type of problem we check that the given relation satisfies all three types of relation by their definition as:
Reflexive: A relation is said to be reflexive, if $(a,a) \in R$ for every $a \in R$ .
Symmetric: A relation is said to be symmetric, if $(a,b) \in R$ then $(b,a) \in R$ .
Transitive: A relation is said to be transitive if $(a,b) \in R$ and $(b,c) \in R$ then $(a,c) \in R$ .

Complete step-by-step solution:
Each equivalence relation provides a partition of the underlying set into disjoint equivalence classes. Two elements of the given set are equivalent to each other, if and only if they belong to the same equivalence class.
We are given that R is a relation defined as$(a,b)R(c,d)$ if $ad(b + c) = bc(a + d)$
Reflexivity:
Since sum and product of natural numbers obeys commutative property. Hence We can write $ab(b + a) = ba(a + b)$ for all $a,b \in N$
Therefore we have $(a,b)R(a,b)$for all $a,b \in N \times N$
Hence R is Reflexive.
Symmetry:
Let $(a,b)R(c,d)$
Therefore we have $ad(b + c) = bc(a + d)$
Since sum and product of natural numbers obeys commutative property. Hence We have
$da(c + b) = cb(d + a)$
Or we can write it as $cb(d + a) = da(c + b)$
Hence $(c,d)R(a,b)$
Hence R is Symmetric.
Transitivity:
Let $(a,b)R(c,d)$ and $(c,d)R(e,f)$
Therefore we have , $ad(b + c) = bc(a + d)$
$adb + adc = abc + bcd$ …(1)
Which gives $cd(a - b) = ab(c - d)$
Also we have $cf(d + e) = de(c + f)$
$cfd + cfe = dec + def$
Which gives $cd(f - e) = ef(d - c)$ …(2)
From (1) and (2) we have
$\dfrac{{a - b}}{{f - e}} = - \dfrac{{ab}}{{ef}}$
On cross multiplication we get ,
$efa - efb = - abf + abe$
On arranging the terms we get ,
$efa + abf = abe + efb$
$af\left( {b + e} \right) = be(a + f)$
Therefore $(a,b)R(e,f)$
Hence R is Transitive.
Thus R being Reflexive, Symmetric and Transitive is an equivalence relation on $N \times N$.
Therefore option (4) is the correct answer.

Note: A relation R on a set A is said to be an equivalence relation if and only if the relation R is reflexive, symmetric and transitive. If any of these properties does not hold true then the relation R is never an equivalence relation . Each equivalence relation provides a partition of the underlying set into disjoint equivalence classes.