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.
(1,1) => 1<=1^2
ReplyDelete(2,2) => 2<=2^2
etc...
Then, why it is not reflexive?
This comment has been removed by the author.
DeleteLet see
Delete(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