Question

Let $N$ denote the set of all natural numbers, and $R$ is a relation on $N\times N$. Which of the following is an equivalence relation?

• A. $(a,b)R(c,d)$ if $ad(b+c)=bc(a+d)$
• B. $(a,b)R(c,d)$ if $a+b=b+c$
• C. $(a,b)R(c,d)$ if $ab=bc$
• D. All the above

Answer 1

Answer: (D)

All the above

The relation in $\left(A\right)$ is reflexive because $ab=ba$
and $a+b=b+a$ , so that
$ab(b+a)=ba(a+b)$.i.e.$(a,b)R(a,b)$.
It is also symmetric because
$ad(b+c)=bc(a+d)$, i.e. $(a,b)R(c,d)$, implies
$cb(d+a)=da(c+b)$, i.e. $(c,d)R(a,b)$.

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

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

$⟹adb+adc=bca+bcd$ and $cfd+cfe=dec+def$

$⟹ab(d-c)=cd(b-a)$ and $cd(f-e)=ef(d-c)$

$⟹\frac{(d-c)}{cd}=\frac{(b-a)}{\left(ab\right)}$ and $\frac{(f-e)}{ef}=\frac{(d-c)}{cd}$

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

$⟹abf-abe=efb-efa$

$⟹af(e+b)=eb(a+f)$

$⟹(a,b)R(e,f)$

Therefore, relation $\left(A\right)$ is transitive too

And similarly, relations in $\left(B\right)$ and $\left(C\right)$ are also reflexive, symmetric and transitive.

So all relations are equivalence relations.