Question

Let $S$ be the set of all real numbers and let $R$ be a relation in $S$ , defined by R=\left\\{ \left( a,b \right):a\le {{b}^{2}} \right\\} . Show that $R$ satisfies none of reflexivity, symmetry, and transitivity.

Answer 1

Hint:For solving this question we will consider some suitable examples for proving that relation R=\left\\{ \left( a,b \right):a\le {{b}^{2}} \right\\} satisfies none of reflexivity, symmetry, and transitivity separately.

Complete step-by-step answer:
Given:
It is given that there is a set $S$ of all real numbers and let $R$ be a relation in $S$ , defined by R=\left\\{ \left( a,b \right):a\le {{b}^{2}} \right\\} and we have to prove that $R$ satisfies none of reflexivity, symmetry, and transitivity.
Now, we will prove that the given relation is neither reflexive, symmetric, and nor transitive one by one.
1. If $a=b$ where $a$ is any real number then it is not necessary that always $a\le {{b}^{2}}$ or $a\le {{a}^{2}}$ and we can see this with the help of an example: take $a=0.5$ then, ${{a}^{2}}=0.25$ and it is evident that here $a\le {{a}^{2}}$ doesn’t hold. Now, we can conclude that $\left( a,a \right)\notin R$ where $a$ is any real number from the set $S$ . Thus, the given relation is not reflexive.

2. If $a$ and $b$ are two different real numbers such that $a\le {{b}^{2}}$ so, $\left( a,b \right)\in R$ . Then, it not always necessary that $b\le {{a}^{2}}$ and we can see this with the help of an example: take $a=2$ , $b=5$ then, ${{b}^{2}}=25$ and it is evident that $a\le {{b}^{2}}$ but, ${{a}^{2}}=4$ and $b\le {{a}^{2}}$ doesn’t hold. Now, we can conclude that if $\left( a,b \right)\in R$ then we cannot always say that $\left( b,a \right)\in R$ always. Thus, if $\left( a,b \right)\in R$ then $\left( b,a \right)\notin R$ so, the given relation is not symmetric.

3. If $a$ , $b$ and $c$ are three different real numbers such that $a\le {{b}^{2}}$ and $b\le {{c}^{2}}$ . Then, it is not always necessary that $a\le {{c}^{2}}$ and we can see this with the help of an example: take $a=4$ , $b=3$ and $c=1.8$ then, ${{b}^{2}}=9$ and it is evident that $a\le {{b}^{2}}$ moreover, ${{c}^{2}}=3.24$ and it is evident that $b\le {{c}^{2}}$ but as $a=4$ so, $a\le {{c}^{2}}$ doesn’t hold. Now, we conclude that if $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then, it is not necessary that always $\left( a,c \right)\in R$ . Thus, if $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then $\left( a,c \right)\notin R$ so, the given relation is not transitive.
Now, as we have proved that the given relation is neither reflexive, symmetric and nor transitive.
Hence, proved.

Note: Here, the student must try to understand the relation given in the problem then prove that the relation is neither reflexive, symmetric, and nor transitive separately and don’t mix up their conditions with each other to avoid the confusion. Moreover, we should understand with the help of suitable examples for better clarity of the concept.