Sunday, 14 April 2013

Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive.


Question 2: Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a  b2} is neither reflexive nor symmetric nor transitive.
Answer: R = {(a, b): a  b2}
It can be observed that 
R is not reflexive.
Now, (1, 2) R as 1 < 22
But, 2 is not less than 12.
(2, 1) R
R is not symmetric.
Now,
(5, 3), (3, 2) R
(as 5 < 32 = 9 and 3 < (2) 2 = 4)
But, 5 > (2) 2= 4
(5, 2) R
R is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

3 comments:

  1. (1,1) => 1<=1^2
    (2,2) => 2<=2^2
    etc...

    Then, why it is not reflexive?

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    2. Let see
      (1\2,1\2) belongs to R
      But 1/2 is > than (1/2)^2
      Therefore 1/2 is ≠ or not greater then (1\2)^2
      Hence R is not reflexive

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