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

Hint: 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
\[\begin{align}
  & =x-x+\sqrt{2} \\
 & =\sqrt{2} \\
\end{align}\]
\[\sqrt{2}\] 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
\[\begin{align}
  & =\sqrt{2}-2+\sqrt{2} \\
 & =2\sqrt{2}-2 \\
\end{align}\]
\[2\sqrt{2}-2\] 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
\[\begin{align}
  & =2-\sqrt{2}+\sqrt{2} \\
 & =2 \\
\end{align}\]
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
\[\begin{align}
  & =1-\sqrt{2}+\sqrt{2} \\
 & =1 \\
\end{align}\]
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.