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Hint:For solving this question we will consider some suitable examples for proving that relation $R=\left\{ \left( a,b \right):\left( 1+ab \right)>0 \right\}$ is reflexive and symmetric but not transitive separately.
Complete step-by-step answer:
Given:
It is given that there is a $S$ of all real numbers and let $R$ be a relation in $S$ defined by $R=\left\{ \left( a,b \right):\left( 1+ab \right)>0 \right\}$ . And we have to prove that $R$ is reflexive and symmetric but not transitive.
Now, we will prove that the given relation $R$ is reflexive and symmetric but not transitive one by one.
1. If $a=b$ where $a$ is any real number then $ab={{a}^{2}}$ and always $1+ab=1+{{a}^{2}}>0$ as ${{a}^{2}}$ will be always positive. Now, we can conclude that $\left( a,a \right)\in R$ where $a$ is any real number from the set $S$ . Thus, the given relation is reflexive.
2. If $a$ and $b$ are any two different real numbers such that $1+ab>0$ which means that $\left( a,b \right)\in R$ then, $\left( b,a \right)\in R$ always as $1+ba>0$ . Now, we can conclude that if $\left( a,b \right)\in R$ then we can say that $\left( b,a \right)\in R$ always. Thus, if $\left( a,b \right)\in R$ then $\left( b,a \right)\in R$ so, the given relation is symmetric.
3. If $a$ , $b$ and $c$ are three different real numbers such that $1+ab>0$ and $1+bc>0$ which means $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then, it not always necessary that $1+ac>0$ and we can understand this easily with the help of an example: take $a=-2$ , $b=0.25$ and $c=3$ then, $1+ab=1-0.5=0.5>0$ and $1+bc=1+0.75=1.75>0$ but $1+ac=1-6=-5<0$ so, $1+ac>0$ 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 $R$ is reflexive and symmetric but not transitive.
Hence, proved.
Note: Here, the student must try to understand the relation given in the problem then prove that the relation is reflexive and symmetric but not 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.
Complete step-by-step answer:
Given:
It is given that there is a $S$ of all real numbers and let $R$ be a relation in $S$ defined by $R=\left\{ \left( a,b \right):\left( 1+ab \right)>0 \right\}$ . And we have to prove that $R$ is reflexive and symmetric but not transitive.
Now, we will prove that the given relation $R$ is reflexive and symmetric but not transitive one by one.
1. If $a=b$ where $a$ is any real number then $ab={{a}^{2}}$ and always $1+ab=1+{{a}^{2}}>0$ as ${{a}^{2}}$ will be always positive. Now, we can conclude that $\left( a,a \right)\in R$ where $a$ is any real number from the set $S$ . Thus, the given relation is reflexive.
2. If $a$ and $b$ are any two different real numbers such that $1+ab>0$ which means that $\left( a,b \right)\in R$ then, $\left( b,a \right)\in R$ always as $1+ba>0$ . Now, we can conclude that if $\left( a,b \right)\in R$ then we can say that $\left( b,a \right)\in R$ always. Thus, if $\left( a,b \right)\in R$ then $\left( b,a \right)\in R$ so, the given relation is symmetric.
3. If $a$ , $b$ and $c$ are three different real numbers such that $1+ab>0$ and $1+bc>0$ which means $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then, it not always necessary that $1+ac>0$ and we can understand this easily with the help of an example: take $a=-2$ , $b=0.25$ and $c=3$ then, $1+ab=1-0.5=0.5>0$ and $1+bc=1+0.75=1.75>0$ but $1+ac=1-6=-5<0$ so, $1+ac>0$ 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 $R$ is reflexive and symmetric but not transitive.
Hence, proved.
Note: Here, the student must try to understand the relation given in the problem then prove that the relation is reflexive and symmetric but not 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.
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