Question

Let N denotes the set of natural numbers and R is a relation in N $\times$ N. which of the following is not an equivalence relation in N $\times$ N?

• A. (a, b) R (c, d) $\iff$ a + d = b + c
• B. (a, b) R (c, d) $\iff$ ad = bc
• C. (a, b) R (c, d) $\iff$ ad (b + c) = ad (a + d)
• D. (a, b) R (c, d) $\iff$ (b + c) = ad (a + d)

Answer 1

The correct option is D (a, b) R (c, d) $\iff$ (b + c) = ad (a + d)

Let (a, b), (c, d) $ϵ$ N $\times$ N.
The relation R defined by (a, b) R (c, d)
$\iff$ a + d = b + c is an equivalence relation.
The relation R defined by (a, b) R (c, d)
$\iff$ ad = bc is equivalence relation.
Let R = {((a, b), (c, d)) : ad (b + c)= bc (a + b), (a, b), (c, d) $ϵ$ N $\times$ N}
Ad (b+c) = bc (a+b) \(\Leftrightarrow \frac{b+c}{bc}=\frac{a+d}{ad}\)
$\iff \frac{1}{c}+\frac{1}{b}=\frac{1}{d}+\iff \frac{1}{a}-\frac{1}{b}=\frac{1}{c}-\frac{1}{d}$
Let (a,b) $ϵ$ N $\times$ N
We have, $\frac{1}{a}-\frac{1}{b}=\frac{1}{a}-\frac{1}{b}$
$\therefore$ (a,b)R (a,b) $\therefore$ R is reflexive
Let (a,b) R(c, d). $\therefore \frac{1}{a}-\frac{1}{b}=\frac{1}{c}-\frac{1}{d}$
$\Rightarrow \frac{1}{c}-\frac{1}{d}=\frac{1}{a}-\frac{1}{b}\Rightarrow$ (c,d) R(a, b)
$\Rightarrow$ R is symmetric
Let (a, b) R (c, d) and (c, d) R (e, f).
$\therefore \frac{1}{a}-\frac{1}{b}=\frac{1}{c}-\frac{1}{d}$ and $\frac{1}{c}-\frac{1}{d}=\frac{1}{e}-\frac{1}{f}$
$\Rightarrow \frac{1}{a}-\frac{1}{b}=\frac{1}{e}-\frac{1}{f}\Rightarrow (a,b)R(e,f)$
$\Rightarrow$ R is transitive.
$\therefore$ R is equivalence relation.
Let R = {((a, b), (c, d)) :bc (b + c) = ad (a + d), (a, b), (c, d) $ϵ$ N $\times$ N}
Let (a, b) $ϵ$ N $\times$ N
We have, ba (b + a) = ab (a + b). $\Rightarrow$ R is reflexive.
Let (a, b) R (c, d)
$\therefore$ bc (b+c) = ad (a+b)
$\Rightarrow$ ad (a + b) = bc (b + c)
$\Rightarrow$ ad (d + a) = cb (c + b)
$\Rightarrow$ (c, d) R (a, b)
$\Rightarrow$ R is symmetric
Let (a, b) R (c, d) and (c, d) R (e, f).
$\therefore$ bc (b+c) = ad (a+d) and de (d+e)=cf (c+f)
These equations need no imply be (b + e) = af(a + f).
$\therefore$ (a,b) may not be R – related to (c, f).
$\therefore$ R is not transitive
$\therefore$ R is not an equivalence relation.
$\therefore$ The correct answer is (d).