Question

Show that the relation $ R\,\,on\,\,R $ , the set of real no’s defined as $ R = \{ (a,b);a \leqslant b\} $ is reflexive transitive but not symmetric.

Answer 1

Hint: Reflexive relation: A relation $ R $ is called reflexive on a set $ A $ if $ (a,a) \in R $ holds for every element $ \forall a \in A $ ,
i.e. if set $ A = \{ a,b\} $ , then is reflexive relation.
Symmetric relation;
A relation $ R $ on a set $ A $ is called symmetric of $ (b,a) \in R $ holds true when $ (a,b) \in R $
Ex: $ R = \{ (4,5),(5,4),(5,6),(6,5)\} $ on set $ A = \{ 4,5,6\} $ is symmetric
Transitive:
Relation: A relation $ R $ on set $ A $ is called transitive of $ (a,b) \in R\,\,and\,\,(b,c) \in R $
then $ (a,c) \in R $ for every $ a,b,c \in R $ .
Ex. Relation $ R = \{ (1,2);(2,4),(1,4)\} $
on set $ A = \{ 1,2,3\} $ is transitive relation.
So, here we learned the basic definition of reflexive, symmetric and transitive relation.

Complete step-by-step answer:
Here, given relation is: $ R\{ (a,b)\,\,\therefore a \leqslant b\} $
Checking for reflexive
If the relation is reflexive then $ (a,a) \in R $
Here, $ a \leqslant a $ $ (i.e.\,\,a $ is equal to $ a) $
\[\therefore f(a)\] is true
So, relation is reflexive.
Checking for symmetric
If the relation is symmetric,
Then if $ (a,b) \in R, $ then $ (b,a) $ must $ \in R $
Here, if $ a \leqslant b\,\,\,\, \Rightarrow b \leqslant a $ ,
Taking $ a = 2,b = 4 $
i.e. $ 2 < 4\,\, \Rightarrow 2 > 4 $ (false)
So, relation is not symmetric
Checking for transitive
Let $ (a,b) \in R $ and $ (b,c) \in R $
If $ (a \leqslant b)\,and\,(b \leqslant c) $
 $ \Rightarrow (a \leqslant c) $ for relation to be transitive
Taking $ a = 2 $
 $ b = 4 $
 $
  c = 6 \\
  i.e\,\,2 < 4\,\,and\,\,4 < 6\,\, \Rightarrow 2 < 6(true) \\
  \,\,\,\,\,\,(a < b)\,\,\,\,\,\,\,\,\,\,\,(b < c)\,\,\,\,\,\,\,(a < c) \\
  $
 $ \Rightarrow $ Relation is transitive
Hence, the above given relation is reflexive and transitive but not symmetric

Note: If any relation is reflexive, symmetric and transitive, then the relation is called equivalence relation.
So, the above relation is not equivalence because it is not symmetric.