Question

Let S be the set of all real numbers and let R be a binary relation on S defined by (a, b) belongs to R = 1 + ab > O for all a, b s belongs to S. Show that R is reflexive as well as symmetric. Give an example to show that R is not transitive​

Answer 1

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For reflexive:-

if the relation is reflexive

then,(a,a) € R

i.e, 1 + a²>0

Since, square number is always positive

Hence, 1+a²>0 is true for all value of a.

So, the given relation is reflexive

For symmetric:-

To check whether symmetric or not,

if, (a,b)€ R, then (b,a)€ R

i.e, if 1+ab<0

then, 1+ba>0

since, if 1+ab>0, then 1 + ba>0

is always true for all value of a and b.

Hence, the given relation is symmetric.

For transitive:-

To check whether symmetric or not,

if, (a,b)€R and (b,c)€R, then (a,c)€R

i.e, if 1+ab>0 and 1+bc>0 then 1+ac>0

Lets, take an example

a= -8, b= -2 , c= 1/4

1+ab= 1+(-8)(-2)=1+16=17, { 17>0}

1+bc= 1+(-2)(1/4)=1-1/2=1/2, {1/2>0}

1+ac= 1+(-8)(1/4)=1-2= -1 ,{-1<0}

Since, 1+ac < 0

when, 1+ab>0 and 1+bc>0

•°• The condition is not true for all value of a,b and c

Hence, the given relation is not transitive.

Therefore, the relation is reflexive, symmetric but not transitive.

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Answer 2

Answer:

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For reflexive:-

if the relation is reflexive

then,(a,a) € R

i.e, 1 + a²>0

Since, square number is always positive

Hence, 1+a²>0 is true for all value of a.

So, the given relation is reflexive

For symmetric:-

To check whether symmetric or not,

if, (a,b)€ R, then (b,a)€ R

i.e, if 1+ab<0

then, 1+ba>0

since, if 1+ab>0, then 1 + ba>0

is always true for all value of a and b.

Hence, the given relation is symmetric.

For transitive:-

To check whether symmetric or not,

if, (a,b)€R and (b,c)€R, then (a,c)€R

i.e, if 1+ab>0 and 1+bc>0 then 1+ac>0

Lets, take an example

a= -8, b= -2 , c= 1/4

1+ab= 1+(-8)(-2)=1+16=17, { 17>0}

1+bc= 1+(-2)(1/4)=1-1/2=1/2, {1/2>0}

1+ac= 1+(-8)(1/4)=1-2= -1 ,{-1<0}

Since, 1+ac < 0

when, 1+ab>0 and 1+bc>0

•°• The condition is not true for all value of a,b and c

Hence, the given relation is not transitive.

Therefore, the relation is reflexive, symmetric but not transitive.

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