Let $$R$$ be the set of real numbers. This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-1: $$A = \{(x, y) \in R \times R : y - x \text{ is an integer}\}$$ is an equivalence relation on $$R$$. Statement-2: $$B = \{(x, y) \in R \times R : x = \alpha y \text{ for some rational number } \alpha\}$$ is an equivalence relation on $$R$$.

For any relation on a set to be an equivalence relation it must be reflexive, symmetric and transitive. We verify these three properties for the two relations given.

Relation A : $$A=\{(x,y)\in\mathbb{R}\times\mathbb{R}:y-x\text{ is an integer}\}$$

• Reflexive  -  For every $$x\in\mathbb{R}$$ we have $$x-x=0$$, and $$0$$ is an integer. Hence $$(x,x)\in A$$ for all $$x$$.
• Symmetric  -  If $$(x,y)\in A$$, then $$y-x=n$$ for some $$n\in\mathbb{Z}$$. Therefore $$x-y=-n$$, which is also an integer, so $$(y,x)\in A$$.
• Transitive  -  If $$(x,y)\in A$$ and $$(y,z)\in A$$, then $$y-x=m,\;z-y=n$$ with $$m,n\in\mathbb{Z}$$. Adding, $$z-x=m+n\in\mathbb{Z}$$, hence $$(x,z)\in A$$.

Since A is reflexive, symmetric and transitive, it is an equivalence relation on $$\mathbb{R}$$. Thus Statement-1 is true.

Relation B : $$B=\{(x,y)\in\mathbb{R}\times\mathbb{R}:x=\alpha y\text{ for some rational }\alpha\}$$

• Reflexive  -  For every $$x\in\mathbb{R}$$ we can choose $$\alpha=1$$ (a rational number) so that $$x=1\cdot x$$. Hence $$(x,x)\in B$$; reflexivity holds.
• Symmetric  -  Consider the pair $$(0,1)$$. We have $$0=0\cdot1$$ with $$0\in\mathbb{Q}$$, so $$(0,1)\in B$$. For symmetry we would need $$(1,0)\in B$$, i.e. $$1=\beta\cdot0$$ for some rational $$\beta$$, which is impossible. Therefore symmetry fails.
• Since symmetry already fails, transitivity need not be tested further.

Thus B is not an equivalence relation. Statement-2 is false.

Both statements are not simultaneously true, so Statement-2 cannot explain Statement-1. The correct choice is:

Option B which is: Statement-1 is true, Statement-2 is false.