Q. 134.2( 5 Votes )

Show that the rel

Answer :

We have

The relation “≥” on the set R of all real numbers.


Recall that for any binary relation R on set A. We have,


R is reflexive if for all x A, xRx.


R is symmetric if for all x, y A, if xRy, then yRx.


R is transitive if for all x, y, z A, if xRy and yRz, then xRz.


So, let the relation having “≥” be P.


We can write


P = {(a, b): a ≥ b, a, b R}


Check for Reflexivity:


a R


If (a, a) P


a ≥ a, which is true.


Since, every real number is equal to itself.


So, a R, then (a, a) P.


P is reflexive.


P is reflexive.


Check for Symmetry:


a, b R


If (a, b) P


a ≥ b


Now, replace a by b and b by a. We get


b ≥ a, might or might not be true.


Let us check:


Take a = 7 and b = 5.


a ≥ b


7 ≥ 5, holds.


b ≥ a


5 ≥ 7, is not true as 5 < 7.


b ≥ a, is not true.


(b, a) P


So, if (a, b) P, then (b, a) P


a, b R


P is not symmetric.


P is not symmetric.


Check for Transitivity:


a, b, c R


If (a, b) P and (b, c) P


a ≥ b and b ≥ c


a ≥ b ≥ c


a ≥ c


(a, c) P


So, if (a, b) P and (b, c) P, then (a, c) P


a, b, c R


P is transitive.


P is transitive.


Thus, shown that the relation “≥” on the set R of all the real numbers are reflexive and transitive but not symmetric.


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