Question

For real number x and y, define a relation ${{R}_{x}},{{R}_{y}}$ , if and only if $x-y+\sqrt{2}$ is an irrational number. Then, the relation R is
(A) reflexive
(B) symmetric
(C) transitive
(D) an equivalence relation

Answer 1

First of all, check for reflexive. Put $x=y$ in the expression $x-y+\sqrt{2}$ and check whether the result is rational and irrational. Now, put $x=\sqrt{2}$ and $y=2$ , and check whether it is symmetric or not. Now, let us assume that $x-y+\sqrt{2}$ is irrational and $y-z+\sqrt{2}$ is irrational. Take $x=1$ , $y=2\sqrt{2}$ and $z=\sqrt{2}$ , and check whether the expression, $x-z+\sqrt{2}$ is irrational or not.

Complete step-by-step solution:
According to the question, we have the expression $x-y+\sqrt{2}$ . We have to define a relation ${{R}_{x}},{{R}_{y}}$ such that the expression $x-y+\sqrt{2}$ is an irrational number.
First of all, let us check for reflexive.
Now, on taking values of x such that x belongs to R, we get
Now, on putting $x=y$ in the expression $x-y+\sqrt{2}$ , we get

& =x-x+\sqrt{2} \\\ & =\sqrt{2} \\\ \end{aligned}$$ $$\sqrt{2}$$ is an irrational number. Therefore, it is reflexive ……………………………………..(1) Now, let us check for symmetric. Let us assume $$x=\sqrt{2}$$ and $$y=2$$ . Now, on putting $$x=\sqrt{2}$$ and $$y=2$$ in the expression $$x-y+\sqrt{2}$$ , we get $$\begin{aligned} & =\sqrt{2}-2+\sqrt{2} \\\ & =2\sqrt{2}-2 \\\ \end{aligned}$$ $$2\sqrt{2}-2$$ is an irrational number ……………………………………(2) Now, on swapping the values of x and y, we get $$x=2$$ and $$y=\sqrt{2}$$ . Now, on putting $$x=2$$ and $$y=\sqrt{2}$$ in the expression $$x-y+\sqrt{2}$$ , we get $$\begin{aligned} & =2-\sqrt{2}+\sqrt{2} \\\ & =2 \\\ \end{aligned}$$ But, 2 is a rational number ………………………………………(3) From equation (1) and equation (2), we can see that on swapping, the values of x and y we don’t get the same thing. Therefore, it is not symmetric ……………………………………….(4) Now, let us check for transitive. Let us assume that $$x-y+\sqrt{2}$$ is irrational and $$y-z+\sqrt{2}$$ is irrational. Now, take $$x=1$$ , $$y=2\sqrt{2}$$ and $$z=\sqrt{2}$$ . These value of x, y, and z satisfies the condition that $$x-y+\sqrt{2}$$ is irrational and $$y-z+\sqrt{2}$$ is irrational. Now, on putting $$x=1$$ and $$z=\sqrt{2}$$ in the expression $$x-z+\sqrt{2}$$ , we get $$\begin{aligned} & =1-\sqrt{2}+\sqrt{2} \\\ & =1 \\\ \end{aligned}$$ But, 1 is a rational number. It is not irrational. Therefore, it is not transitive ……………………………………..(5) Now, from equation (1), equation (4), and equation (5), we can say that the given relation is reflexive. **Hence, the correct option is (A).** **Note:** In this question, one might do a silly mistake while checking for the relationship whether it is symmetric or not. Here, one might put $$x=\sqrt{3}$$ and $$y=2$$ in the expression $$x-y+\sqrt{2}$$ . If we do so, then after solving the expression we get an irrational number. After swapping the value of x and y, we again get an irrational number which states that the relation is symmetric. But, this is wrong. Put, $$x=\sqrt{2}$$ and $$y=2$$ , and then check for symmetry.