Questions & Answers

Question

Answers

A. R is reflexive.

B. R is symmetric.

C. R is transitive.

D. None of these

Answer
Verified

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.