
The relation $ S $ defined on the set of real numbers by the rule $ aSb $ , iff $ a \geqslant b $ is an:
a) An equivalence relation
b) Reflexive, transitive but not symmetric
c) Symmetric, transitive but not reflexive
d) Neither transitive nor reflexive but symmetric.
Answer
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Hint: A relation can be categorized as:
a) Reflexive: if $ aRa $
b) Symmetric: if $ aRb \Rightarrow bRa $
c) Transitive: if $ aRb,bRc \Rightarrow aRc $
If any relation satisfies all of the above mentioned criteria then it is said to be an equivalence relation , failure of any one condition it cannot be an equivalence relation.
Complete step-by-step answer:
The given relation is defined as $ aSb \Rightarrow a \geqslant b $ for all real numbers.
We have to check the reflective relation.
Let ‘a’ be any real number that is $ a \in \mathbb{R} $
So we can say $ a = a $
So the inequality $ a \geqslant a $ holds good.
Thus, $ aSa $ is true. So, the relation is reflexive.
Now we have to check the symmetric relation.
Let ‘a’ and ‘b’ be any two distinct real number that is $ a,b \in \mathbb{R} $
So let’s say that $ a \ne b $
Now we have two possibilities that is $ a > b $ or $ a < b $
Let $ a > b $ then the inequality $ a \geqslant b $ holds good so we can say ‘a’ is related to ‘b’ with the relation “ $ aSb $ “
Now $ a > b \Rightarrow b < a $ then the inequality $ a \geqslant b $ does not holds good so we cannot say ‘a’ is related to ‘b’ with the relation “ $ aSb $ “
Same will happen in the other case which is $ a < b $ .
Thus, the relation is not symmetric.
Now we have to check the transitive relation.
Let ‘a’, ‘b’ and ‘c’ be any three distinct real number that is $ a,b,c \in \mathbb{R} $
So let’s say that $ aSb $ and $ bSc $
Now we have $ a \geqslant b $ and $ b \geqslant c $
Adding both the inequalities we have,
$ a + b \geqslant b + c \Rightarrow a \geqslant c $
So we have $ aSc $
Thus, the relation is transitive as $ aRb,bRc \Rightarrow aRc $
Therefore the relation “S” defined on real numbers is Reflexive, transitive but not symmetric. So, option ‘b’ is correct.
So, the correct answer is “Option B”.
Note: You need not show that $ a \geqslant b,b \geqslant c \Rightarrow a \geqslant c $ as it is obvious in case of real numbers. This is a known fact which can also be verified taking three distinct numbers. If a relation is reflexive, symmetric and transitive then the relation is called an equivalence relation.
a) Reflexive: if $ aRa $
b) Symmetric: if $ aRb \Rightarrow bRa $
c) Transitive: if $ aRb,bRc \Rightarrow aRc $
If any relation satisfies all of the above mentioned criteria then it is said to be an equivalence relation , failure of any one condition it cannot be an equivalence relation.
Complete step-by-step answer:
The given relation is defined as $ aSb \Rightarrow a \geqslant b $ for all real numbers.
We have to check the reflective relation.
Let ‘a’ be any real number that is $ a \in \mathbb{R} $
So we can say $ a = a $
So the inequality $ a \geqslant a $ holds good.
Thus, $ aSa $ is true. So, the relation is reflexive.
Now we have to check the symmetric relation.
Let ‘a’ and ‘b’ be any two distinct real number that is $ a,b \in \mathbb{R} $
So let’s say that $ a \ne b $
Now we have two possibilities that is $ a > b $ or $ a < b $
Let $ a > b $ then the inequality $ a \geqslant b $ holds good so we can say ‘a’ is related to ‘b’ with the relation “ $ aSb $ “
Now $ a > b \Rightarrow b < a $ then the inequality $ a \geqslant b $ does not holds good so we cannot say ‘a’ is related to ‘b’ with the relation “ $ aSb $ “
Same will happen in the other case which is $ a < b $ .
Thus, the relation is not symmetric.
Now we have to check the transitive relation.
Let ‘a’, ‘b’ and ‘c’ be any three distinct real number that is $ a,b,c \in \mathbb{R} $
So let’s say that $ aSb $ and $ bSc $
Now we have $ a \geqslant b $ and $ b \geqslant c $
Adding both the inequalities we have,
$ a + b \geqslant b + c \Rightarrow a \geqslant c $
So we have $ aSc $
Thus, the relation is transitive as $ aRb,bRc \Rightarrow aRc $
Therefore the relation “S” defined on real numbers is Reflexive, transitive but not symmetric. So, option ‘b’ is correct.
So, the correct answer is “Option B”.
Note: You need not show that $ a \geqslant b,b \geqslant c \Rightarrow a \geqslant c $ as it is obvious in case of real numbers. This is a known fact which can also be verified taking three distinct numbers. If a relation is reflexive, symmetric and transitive then the relation is called an equivalence relation.
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