Question

Let N denote the set of all natural numbers and R 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 $a+d=b+c$
• C. $(a,b)R(c,d)$ if $ad=bc$
• D. All the above

Answer 1

The correct option is D All the above

Given: N denote the set of natural numbers and R a reation on $N\times N$

A) $(a,b)R(c,d)$ id $ad(b+c)=bc(a+d)$

B) $(a,b)R(c,d)$ if $(a+d)=(b+c)$

C) $(a,b)R(c,d)$ if $ad=bc$.

Consider A $(a,b)R(c,d)$ if $ad(b+c)=bc(a+d)$

1) Reflexive

Let $(a+b)\in N\times N$

and $(a,b)R(a,b)$

$\Rightarrow ab(b+a)=ba(a+b)$

$\Rightarrow ab(b+a)=ab(b+a)$

$\Rightarrow R$ is reflective .....(I)

2) Symmetric

Let $(a,b)R(c,d)$

where $(a,v)\&(c,d)\in N$

$\Rightarrow ad(b+c)=bc(a+d)$

$=bc(a+d)=ad(b+c)$

$\Rightarrow cb(d+a)=da(c+b)$

$\Rightarrow (c,d)R(a,b)$

$\Rightarrow R$ is symmetric ...(II)

3) Transitive

Let $(a,b),(c,d)$ and $(c,f)\in N\times N$

consider,

$(a,b)R(c,d)$

$\Rightarrow ad(b+c)=bc(a+d)$

$\Rightarrow abd+adc=abc+bcd$

$\Rightarrow abd-abc=bcd-adc$

$\Rightarrow ab(d-c)=dc(b-a)$

$\Rightarrow \frac{ab}{(b-a)}=\frac{cd}{d-c}$ ...(1)

Now, consider,

$(c,d)R(e,f)$

$\Rightarrow cf(d+e)=dc(e+f)$

$\Rightarrow cfd+cef=ced+edf$

$\Rightarrow cfd-ced=edf+cef$

$\Rightarrow cd\left(fe\right)=ef(d-c)$

$\Rightarrow \frac{cd}{d-c}=\frac{ef}{f-e}$ ....(2)

from (1) and (2)

$\frac{ab}{b-c}=\frac{cf}{f-e}$

$\Rightarrow ab(f-e)=ef(b-a)$

$\Rightarrow abf-abe=efb-efa$

$\Rightarrow abf+efa=efb+abe$

$\Rightarrow af(b+e)=be(f+a)$

$\Rightarrow (a,b)R(e,f)$ (by definition of R)

$\Rightarrow R$ is trasitive .....(III)

from (I), (II) and (III)

$R$ is on equivalence Relation or $N\times N$.

Consider B $(a,b)R(c,d)$ if $(a+d)=(b+c)$

1) Reflexive

Let $(a+b)\in N\times N$

and $(a,b)R(a,b)$

$\Rightarrow (a+b)=(b+a)$

$\Rightarrow (a+b)=(a+b)$

$\Rightarrow R$ is reflective .....(IV)

2) Symmetric

Let $(a,b)R(c,d)$

where $(a,b)\&(c,d)\in N\times N$

$\Rightarrow (a+d)=(b+c)$

$\Rightarrow (b+c)=(a+d)$

$\Rightarrow (c+b)=(d+a)$

$\Rightarrow (c,d)R(a,b)$

$\Rightarrow R$ is symmetric ...(V)

3) Transitive

Let $(a,b),(c,d)$ and $(c,f)\in N\times N$

If $(a,b)\&(c,d)$ and $(c,d)\&(c,f)$

then $(a,b)R(c,f)$

consider,

$(a,b)R(c,d)$

$\Rightarrow (a+d)=(b+c)$ by definition by $R$

$\Rightarrow a-b=c-d$ ....(3)

Now,

$(c,d)R(e,f)$

$\Rightarrow (c+f)=(d+e)$ by definition of $x$

$\Rightarrow c-d+e-f$ ...(4)

from (3) and (4)

$a-b=e-f$

$\Rightarrow a+f=e+b$

$\Rightarrow (a,b)R(e,f)$

$\Rightarrow R$ is transitive ...(VI)

from (IV), (V), (VI) we have,

$R$ is an equivalence reaction.$\Rightarrow abf+efa=efb+abe$

$\Rightarrow af(b+e)=be(f+a)$

$\Rightarrow (a,b)R(e,f)$ (by definition of R)

$\Rightarrow R$ is trasitive .....(III)

from (I), (II) and (III)

$R$ is on equivalence Relation or $N\times N$.

Consider C $(a,b)R(c,d)$ if $ad=bc$

1) Reflexive

Let $(a,b)\in N\times N$

and $(a,b)R(a,b)$

$\Rightarrow ab=ba$

$\Rightarrow ab=ab$

$\Rightarrow R$ is reflective .....(VII)

2) Symmetric

Let $(a,b)R(c,d)$ where $(a,b),(c,d)\in N\times N$

$\Rightarrow ad=bc$

$\Rightarrow bc=ad$

$\Rightarrow cb=da$

$\Rightarrow (c,d)R(a,b)$

$\Rightarrow R$ is symmetric ...(VIII)

3) Transitive

Let $(a,b),(c,d)$ and (c,f) \in N \times N$

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

then $(a,b)R(c,f)$

consider,

$(a,b)R(c,d)$

$\Rightarrow ad=bc$ by definition by $R$

$\Rightarrow \frac{a}{b}=\frac{c}{d}$ ....(5)

Consider,

$(c,d)R(e,f)$

$\Rightarrow \left(cf\right)=\left(de\right)$ by definition of $R$

$\Rightarrow \frac{c}{d}=\frac{e}{f}$ ...(6)

from (5) and (6) we have,

$\frac{a}{b}=\frac{e}{f}$

$\Rightarrow R$ is transitive ...(IX)

from (VII), (VIII), (IX) we have,

$R$ is an equivalence reaction.

$\Rightarrow A,B,C$ all are equivalence relation.

$\therefore$ option all of the above is correct answer.