Q. 5 C5.0( 3 Votes )

The following rel

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 |a| ≤ b


Check for Reflexivity:


For a


If aRa,


|a| ≤ a, is true


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.


|a| = positive.


& |a| ≤ a


Hence, |a| ≤ a.


aRa is true.


So, a , then aRa is true.


R is reflexive.


R is reflexive.


Check for Symmetry:


a, b


If aRb,


|a| ≤ b


Replace a by b and b by a, we get


|b| ≤ a, which might be true or not.


Let a = 2 and b = 3.


|a| ≤ b


|2| ≤ 3, is true


|b| ≤ a


|3| ≤ 2, is not true


bRa is not true.


So, if aRb is true, then bRa is not true.


a, b


R is not symmetric.


R is not symmetric.


Check for Transitivity:


a, b, c


If aRb and bRc.


|a| ≤ b and |b| ≤ c


|a| ≤ c or not.


Let us check.


If |a| ≤ b and |b| ≤ c


b ≠ |b|


Say, if b = –2


–2 ≠ |–2|


–2 ≠ 2


But, from |a| ≤ b


b ≥ 0 in every case otherwise the statement would not hold true.


b can only accept positive values including 0.


b is a whole number.


if |a| ≤ b and |b| ≤ c


|a| ≤ b, b ≤ c


|a| ≤ b ≤ c


|a| ≤ c


aRc is true.


So, if aRb is true and bRc is true, then aRc is true.


a, b, c


R is transitive.


R is transitive.


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