Question

N denotes the set of all natural numbers and R be the relation on $N \times N$ defined by (a, b) R(c, 0), if $ad(b + c) = bc(a + d)$, then show that R is a equivalence relation.

Answer 1

Hint:
A relation is said to be equivalence relations if it satisfies the condition of transitivity, reflexivity and symmetry. Remember all these three conditions should be satisfied by the same relation.

Stepwise solution:
Given:
R is the relation on $N \times N$ defined by (a, b) (c, d) $ \Leftrightarrow $
$ab(b + c) = bs(a + d)$

To show that the given relation is an equivalence relation we have to prove reflexivity, symmetry and transitivity of a relation. Thus, starting with
Symmetry:
Given,
\[\left( {a,{\text{ }}b} \right){\text{ }}R{\text{ }}\left( {c,{\text{ }}d} \right)\]
$ \Rightarrow ad(b + c) = bc(a + d)$
The above relation can be written as
$dd(c + b) = cb(d + a)$
Now,
This can be further written as
$cb(d + a) = da(c + b)$ [Since, if a=b is correct, b=a is also correct]
Thus,
$(c,d)\,R\,(a,b)$ can be written from the above relation. Thus, it is a symmetric relation.
Transitivity
Let,
$(a,b)\,,\,(c,d)\,,\,(e,f)\,\, \in \,\,N \times N$
Thus, we can also assume that
$(a,\,b)\,R\,(c,d)\,and\,(c,d)\,R\,(e,f)$
Therefore, we can write the above equation as
$ad(b + c) = bc(a + d)\,and\,cf(d + e) = de(c + f)$
$\therefore \,adb + adc = bca + bcd$
$ \Rightarrow \,\,abc - abd = \,acd - bcd$
$ \Rightarrow ab(c - d)\, = cd(a - b)$
$ \Rightarrow \dfrac{{ab}}{{a - b}} = \dfrac{{cd}}{{c - d}}$ eq. (2)
Similarly, $cf(d + e) = de(c + f)$
$ \Rightarrow cfd + cfe = dec + def$
$ \Rightarrow cef + cdf = cde + def$
$ \Rightarrow cef - def = cde - cdf$
$ \Rightarrow ef(c - d) = cd(e - f)$
$ \Rightarrow \dfrac{{ef}}{{ef}} = \dfrac{{cd}}{{c - d}}$ eq. (3)
Equating equation (2) and (3), we get
$ \Rightarrow \,\dfrac{{ab}}{{a - b}} = \dfrac{{ef}}{{e - f}}$
$ \Rightarrow ab(e - f) = ef(a - b)$
$ \Rightarrow abe - abf = efa - efb$
$ \Rightarrow abf + afe = abe + efb$
$ \Rightarrow af(b + e) = be(a + f)$
$\therefore \,(a,b)\,R\,(e,f)$
Thus, we can say that the relation is a transitive relation.
Now, for reflexivity
Let, $ab(b + a) = ba(a + b)$ for all $a,b\, \in \,N$
Since, sum and product obey continuous rule.
Hence, by equation (1) we can write
$(a,b)\,R\,(a,b)\,for\,all\,(a,b)\, \in \,N \times N$
Thus, R is a reflexive relation.
Hence, we have seen that the relation R on $N \times N$ defined by $(a,b)\,R\,(c,d)\,\, \Leftrightarrow $ $ad(b + c) = bc(a + d)$
is symmetry, transitive and reflexive. So, it is a type of relation which can be said to be an equivalence relation.

Note:
Prove all the three conditions, not solving any one will not satisfy the equivalency of the relation.