Q. 5 B5.0( 2 Votes )

The following relations are defined on the set of real numbers :

aRb if 1 + ab > 0

Find whether these relations are reflexive, symmetric or transitive.

Answer :

Let set of real numbers be .

So, 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.


We have


aRb if 1 + ab > 0


Check for Reflexivity:


For a


If aRa,


1 + aa > 0


1 + a2 > 0


If a is a real number.


[Positive or negative, large or small, whole numbers or decimal numbers are all Real Numbers. Real numbers are so called because they are ‘Real’ not ‘imaginary’.]


This means, even if ‘a’ was negative.


a2 = positive.


a2 + 1 = positive


And any positive number is greater than 0.


Hence, 1 + a2 > 0


aRa is true.


So, a , then aRa is true.


R is reflexive.


R is reflexive.


Check for Symmetry:


a, b


If aRb,


1 + ab > 0


Replace a by b and b by a, we get


1 + ba > 0


Whether we write ab or ba, it is equal.


ab = ba


So, 1 + ba > 0


bRa is true.


So, if aRb is true, then bRa is true.


a, b


R is symmetric.


R is symmetric.


Check for Transitivity:


a, b, c


If aRb and bRc.


1 + ab > 0 and 1 + bc > 0


1 + ac > 0 or not.


Let us check.


1 + ab > 0 means ab > –1.


1 + bc > 0 means bc > –1.


1 + ac > 0 means ac > –1.


If ab > –1 and bc > –1.


ac > –1 should be true.


Take a = –1, b = 0.9 and c = 1


ab > –1


(–1)(0.9) > –1


–0.9 > –1, is true on the number line.


bc > –1


(0.9)(1) > –1


0.9 > –1, is true on the number line.


ac > –1


(–1)(1) > –1


–1 > –1, is not true as –1 cannot be greater than itself.


ac > –1 is not true.


1 + ac > 0 is not true.


aRc is not true.


So, if aRb is true and bRc is true, then aRc is not true.


a, b, c


R is not transitive.


R is not transitive.


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