Option A , $$aRb$$ if $$a+b$$ is an even integer. Let $$a, b,c \in I$$ Clearly, $$a+a =2a$$ is an even integer. Hence, $$aRa$$ for all $$a\in I$$ Now, let $$aRb$$ $$\Rightarrow a+b$$ is an even integer. or $$b+a$$ is an even integer. $$\Rightarrow bRa$$ Next, let $$aRb, bRc$$ $$\Rightarrow a+b $$ in an even integer and $$ b+c$$ is an even integer $$\Rightarrow a$$ and $$b$$ both are even or both are odd . Also, $$b$$ and $$c$$ both are even or both are odd. Hence, if $$a$$ and $$b$$ are even .So $$c$$ is also even . Hence, $$a+c$$ is even. And if $$a$$ and $$b$$ are odd.So, $$c$$ is also odd . Again $$a+c$$ is even. Hence, $$aRc$$ Hence, option A is equivalence relation.Option B , $$aRb$$ if $$a-b$$ is an even integer. Let $$a, b,c \in I$$ Clearly, $$a-a =0$$ is an even integer. Hence, $$aRa$$ for all $$a\in I$$ Now, let $$aRb$$ $$\Rightarrow a-b$$ is an even integer. or $$-(a-b)$$ is also an even integer. $$\Rightarrow \ bRa$$ Next, let $$aRb, bRc$$ $$\Rightarrow a-b $$ in an even integer and $$ b-c$$ is an even integer $$a+c=a-b+b-c=$$ even+even Hence, $$a+c$$ is even. Hence, $$aRc$$ Hence, option $$B$$ is equivalence relation.Option C $$aRb$$ if $$a<b$$ Let $$a, b,c \in I$$ Clearly,symmetry does not hold here Now, let $$aRb$$ $$\Rightarrow a<b$$ But it does not implies $$b<a$$ Hence, option C is not an equivalence relation.Option D , $$aRb$$ if $$a=b$$ Let $$a, b,c \in I$$ Clearly, $$a=a$$ Hence, $$aRa$$ for all $$a\in I$$ Now, let $$aRb$$ $$\Rightarrow a=b$$ or $$b=a$$ is an even integer. $$\Rightarrow bRa$$ Next, let $$aRb, bRc$$ $$\Rightarrow a=b $$and $$ b=c$$ $$\Rightarrow a=b=c$$ Hence, $$aRc$$ Hence, option D is equivalence relation.