Question

Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b 2 } is neither reflexive nor symmetric nor transitive.

Answer 1

$\text{\bf{reflexive relation}}$ A relation is said to be reflexive relation, if every element of A is related to itself.
thus,(a,a) ∈ R , for all a∈ A => R is reflexive

in given relation if we take (1/2,1/2)
we observed , 1/2 $\nleq$ 1/2²
hence, it is not relfexive relation.

$\text{\bf{symmetric relation}}$ A relation is said to be symmetric relation, if
(a, b) ∈ R => (b, a) ∈ R , for all a, b ∈ A
e.g., aRb => bRa, for all a, b ∈ A => R is symmetric.
if we take (a, b) = (1,4)
then, (b , a) = (4, 1) doesn't belong to relation
because 4² $\nleq$1²

$\text{\bf{transitive relation}}$ A relation is said to be transitive relation if
(a, b) ∈ R and (b, c) ∈ R => (a,c) ∈ R for all a,b,c ∈ A
Now, (3, 2), (2, 1.5) ϵ R
But, 3 > (1.5)2 = 2.25.
Then, (3, 1.5) ∉ R


Therefore, R is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive

Answer 2

Step-by-step explanation:

Let a relation

on the set

of real numbers be defined as

for all

. Show that

is reflexive and symmetric but not transitive.