excersie 1.1

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.

solutions =

 

R = {(a, b): a ≤ b²}

It can be observed that 

∴R is not reflexive.

Now, (1, 4) ∈ R as 1 < 4²

But, 4 is not less than 1².

∴(4, 1) ∉ R

∴R is not symmetric.

Now,

(3, 2), (2, 1.5) ∈ R

(as 3 < 2² = 4 and 2 < (1.5)² = 2.25)

But, 3 > (1.5)² = 2.25

∴(3, 1.5) ∉ R

∴ R is not transitive.

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

Question 3:

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as

R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

solution =

Let A = {1, 2, 3, 4, 5, 6}.

A relation R is defined on set A as:

R = {(a, b): b = a + 1}

∴R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

We can find (a, a) ∉ R, where a ∈ A.

For instance,

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R

∴R is not reflexive.

It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R.

∴R is not symmetric.

Now, (1, 2), (2, 3) ∈ R

But,

(1, 3) ∉ R

∴R is not transitive

Hence

Question 4:

Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.

solution=

 

R = {(a, b); a ≤ b}

Clearly (a, a) ∈ R as a = a.

∴R is reflexive.

Now,

(2, 4) ∈ R (as 2 < 4)

But, (4, 2) ∉ R as 4 is greater than 2.

∴ R is not symmetric.

Now, let (a, b), (b, c) ∈ R.

Then,

a ≤ b and b ≤ c

⇒ a ≤ c

⇒ (a, c) ∈ R

∴R is transitive.

Hence,R is reflexive and transitive but not symmetric.

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

Question 5:

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

solution =

R = {(a, b): a ≤ b³}

It is observed that

∴ R is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 2³ = 8)

But,

(2, 1) ∉ R (as 2 > 1³ = 1)

∴ R is not symmetric.

We have

But 

∴ R is not transitive.

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