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

Hint: 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.