Question

For real numbers $x$ and$y$, define a relation$R$, $xRy$ 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

In this problem, we need to check whether the given relation is reflexive, symmetric or transitive. If every element of the given relation is mapping itself, it will be a reflexive relation. The given relation is said to be symmetric if $xRy \Rightarrow yRx$.

Complete step by step answer:
The given relation $R$ for the real numbers $x$ and $y$ is shown below.
$xRy \Rightarrow x - y + \sqrt 2$
(i) For every value of $x \in R$,

$$\,\,\,\,\,x - x + \sqrt 2 \\\ \Rightarrow \sqrt 2 \\\$$

Here, $\sqrt 2$ is an irrational number, therefore, the given relation $R$ is reflexive.
(ii) Now, consider $x = 2$ and $y = \sqrt 2$, then,

$$xRy \Rightarrow 2 - \sqrt 2 + \sqrt 2 \\\ xRy \Rightarrow 2\left( {{\text{not irrational}}} \right) \\\$$

Again, consider $x = \sqrt 2$ and $y = 2$, then,

$$xRy \Rightarrow \sqrt 2 - 2 + \sqrt 2 \\\ xRy \Rightarrow 2\sqrt 2 - 2\left( {{\text{irrational}}} \right) \\\$$

Therefore, the given relation is not symmetric.
(iii) Now, consider the relation $xRy \Rightarrow x - y + \sqrt 2$ and $yRz \Rightarrow y - z + \sqrt 2$ is irrational.
Now, let $x = 1,y = 2\sqrt 2$ and $z = \sqrt 2$ then the relation $xRz \Rightarrow x - z + \sqrt 2$ is shown below.

$$xRz \Rightarrow 1 - \sqrt 2 + \sqrt 2 \\\ xRz \Rightarrow 1\left( {{\text{not irrational}}} \right) \\\$$

Therefore, the given relation is not transitive.

So, the correct answer is “Option A”.

Note: If the relation between the given elements is reflexive, symmetric and transitive, it will be an equivalence relation. A relation is said to be reflexive, if a=a is true for all values of a. A relation is said to be symmetric a=b, is also true for b=a. A relation is said to be transitive if a=b, b=c such that a=c.