Question
Let $$R$$ be the set of real numbers.
Statement - 1: $$A$$ = {($$x, y$$ ) $$ \in R \times R:$$ $$y - x$$ is an integer} is an equivalence relation on $$R.$$
Statement - 2 : $$B$$ = {($$x, y$$ ) $$ \in R \times R:$$ $$x = \alpha y$$ for some rational number $$\alpha $$ } is an equivalence relation on $$R.$$
A.
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.
B.
Statement - 1 is true, Statement - 2 is false.
C.
Statement - 1 is false, Statement - 2 is true.
D.
Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1.
Answer :
Statement - 1 is true, Statement - 2 is false.
Solution :
$$x - y$$ is an integer.
$$\because \,\,x - x = 0$$ is an integer
⇒ $$A$$ is reflexive.
Let $$x - y$$ is an integer
⇒ $$y - x$$ is an integer
⇒ $$A$$ is symmetric
Let $$x - y, y - z$$ is also an integer
⇒ $$x - y + y - z$$ is also an integer
⇒ $$x - z$$ is an integer
⇒ $$A$$ is transitive
∴ $$A$$ is an equivalence relation.
Hence statement - 1 is true.
Also $$B$$ can be considered as
$$xBy$$ if $$\frac{x}{y} = \alpha ,$$ a rational number
$$\because \,\,\frac{x}{x} = 1$$ is a rational number
⇒ $$B$$ is reflexive
But $$\frac{x}{y} = \alpha ,$$ a rational number need not imply $$\frac{y}{x} = \frac{1}{\alpha },$$ a rational number because
$$\frac{0}{1}$$ is rational
⇒ $$\frac{1}{0}$$ is not rational
∴ $$B$$ is not an equivalence relation.