Question
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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

Answer Verified Verified
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.

Complete step-by-step answer:
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.