Question

# For real numbers $x$ and $y$, we define $xRy$ if $x - y + \sqrt 5$ is irrational. Then, which of the following is true about R?A. R is reflexive.B. R is symmetric.C. R is transitive.D. None of these

Hint: We will first write the meanings of R being reflexive, symmetric and transitive. Then, we will try to see if any of the properties are followed by the given relation and then mark that as correct.

Let us first understand the meanings of all the options:
Let R be a binary relation on a set A. Then,
R is reflexive if for all $x\; \in A,xRx$.
R is symmetric if for all $x,y \in A$, if $xRy$, then $yRx$.
R is transitive if for all $x,y,z\; \in A$, if $xRy$ and $yRx$, then $xRz$.
R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.
Now, let us come back to our question. We define $xRy$ if $x - y + \sqrt 5$ is irrational.
Let us see if this relation is reflexive or not.
So, if this is reflexive for all $x\; \in A,xRx$.
Consider $x - x + \sqrt 5 = \sqrt 5$, which is irrational.
Hence, R is reflexive.
Let us see if this relation is symmetric or not.
Now, for all $x,y \in A$, if $xRy$ we need $yRx$.
Consider $(\sqrt 5 ,1)$, we have $x - y + \sqrt 5 = \sqrt 5 - 1 + \sqrt 5 = 2\sqrt 5 - 1$, which is irrational but $(1,\sqrt 5 )$, we get: $x - y + \sqrt 5 = 1 - \sqrt 5 + \sqrt 5 = 1$, which is rational.
Hence, R is not symmetric.
Let us see if this relation is transitive or not.
Now, for all $x,y,z\; \in A$, if $xRy$ and $yRx$, then we need $xRz$.
Consider $(\sqrt 5 ,1)$, we have: $x - y + \sqrt 5 = \sqrt 5 - 1 + \sqrt 5 = 2\sqrt 5 - 1$ , which is irrational.
And see $(1,2\sqrt 5 )$, we have $x - y + \sqrt 5 = 1 - 2\sqrt 5 + \sqrt 5 = 1 - \sqrt 5$ , which is irrational.
But in $(\sqrt 5 ,2\sqrt 5 )$, we get: $x - y + \sqrt 5 = \sqrt 5 - 2\sqrt 5 + \sqrt 5 = 0$ , which is rational.
Hence, it is not transitive as well.

So, the correct answer is “Option A”.

Note: The students might be tempted to leave the question after proving the A part to be correct. They may do that, if we have the same options and the same kind of question, but there is a possibility that you have an equivalence relation in the option. So, to prove that incorrect or correct, you will have to check every property.