Check whether the relation R on R defined by R = {(a, b) : a ≤ b3} is reflexive, symmetric or transitive.

We have the set of real numbers, R.

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

R = {(a, b): a ≤ b3}

Check for Reflexivity:

For a R

If (a, a) R,

a ≤ a3, which is not true.

Say, if a = – 2.

a ≤ a3

– 2 ≤ – 8

–2 ≤ –8, which is not true as – 2 > – 8.

Hence, (a, a) R

So, a R, then (a, a) R.

R is not reflexive.

R is not reflexive.

Check for Symmetry:

a, b R

If (a, b) R

a ≤ b3

Replace a by b and b by a, we get

b ≤ a3

[Take a = –2 and b = 3.

a ≤ b3

–2 ≤ 33

–2 ≤ 27, which is a true statement.

Now, b ≤ a3

3 ≤ (–2)3

3 ≤ –8, which is not a true statement as 3 > –8]

(b, a) R

So, if (a, b) R, then (b, a) R

a, b R

R is not symmetric.

R is not symmetric.

Check for Transitivity:

a, b, c R

If (a, b) R and (b, c) R

a ≤ b3 and b ≤ c3

a ≤ c3 or not.

Let us check.

Take a = 3, and .

a ≤ b3

3 ≤ 3.37, which is true.

b ≤ c3

1.5 ≤ 1.728

a ≤ c3

3 ≤ 1.728, which is not true as 3 > 1.728.

Hence, (a, c) R.

So, if (a, b) R and (b, c) R, then (a, c) R.

a, b, c

R is not transitive.

R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Functions - 0152 mins
Some standard real functions61 mins
Quick Revision of Types of Relations59 mins
Range of Functions58 mins
Battle of Graphs | various functions & their Graphs48 mins
Functions - 0947 mins
Quick Recap lecture of important graphs & functions58 mins