Question

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

Answer 1

In the above question they asked us whether the given relation is equivalent. To solve this problem we will use the concepts of relations and equivalence of relation. Then if the conditions for equivalency are satisfied we can say that the relation is equivalent on $R$ .

Complete step by step answer:
Before going into conditions for equivalency of a relation we shall define relation as is clear means an establishment of a rule, between two clusters or sets of any commodities or quantities. Domain and Range are the characteristic properties, which are extensions to the definition of relation.
Now for example let us consider two sets.
Set A contains the elements 1,2,3,4,5 and 6 such that it is written as:
And there exists another set B which contains the elements:
Let R be a relation defined from A to B. defined $R:A \to B$ such that the elements of the relation set R are
$\\{ ({x_1},f({x_1})),({x_2},f({x_2})),({x_3},f({x_3}))............\\} = \\{ (1,2),(2,3),(3,4),(4,5),(5,7),(6,9)\\}$
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.
Reflexive: A relation is said to be reflexive, if (a, a) ∈ R, for every a ∈ A.
Symmetric: A relation is said to be symmetric, if (a, b) ∈ R, then (b, a) ∈ R.
Transitive: A relation is said to be transitive if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
In the above question they gave $\left( {a,b} \right)R(c,d)$ as $ad(b + c) = bc(a + d)$
$\Rightarrow$As $\forall {\text{ a,b }} \in {\text{ N,}}$ we can write $ab(b + a) = ba(a + b)$ or $(a,b)R(a,b)$ …….. $(1)$
$\Rightarrow$Using $(1)$ we can say $R$ is reflexive (using above theorems).
Let $\left( {a,b} \right)R(c,d)$ for $(a,b),(c,d) \in N \times N$
We write $ad(b + c) = bc(a + d)$ ……… (2) and also we can find $(c,d)R(a,b)$ which can be given by
$cb(d + a) = da(c + b)$ [By commutation of addition and multiplication on N]……….. (3)
$\Rightarrow$As both 2 and 3 the same we can say $R$ is symmetric.
Let $\left( {a,b} \right)R(c,d)$ and $(c,d)R(e,f)$ for $a,b,c,d,e,f \in N$
As given in question we can write $ad(b + c) = bc(a + d)$ ……. $(4)$
And $cf(e + d) = ed(c + f)$ ……… $(5)$
We now divide equations $(4)$ by $abcd$ and $(5)$ by $cdef$ we get
$\Rightarrow$ $\dfrac{1}{c} + \dfrac{1}{b} = \dfrac{1}{d} + \dfrac{1}{a}$ and $\dfrac{1}{e} + \dfrac{1}{d} = \dfrac{1}{f} + \dfrac{1}{c}$ now we add these both and get as follows
\Rightarrow $$$$\dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{e} + \dfrac{1}{d} = \dfrac{1}{d} + \dfrac{1}{a} + \dfrac{1}{f} + \dfrac{1}{c}, we can also write this as $af\left( {b + e} \right) = be\left( {a + f} \right)$ which means $(a,b)R(e,f)$ .
$\Rightarrow$This implies $R$ is transitive.
Hence we say $R$ is an equivalence relation.

Note:
It is necessary to verify all the definitions of reflexivity, symmetry, and transitivity for every relation to find whether it is an equivalence relation or not and if all these three conditions are satisfied, then only is a equivalence relation.It is important that we are familiar with the concept of relations .