Question

Q: Determine whether the relation is reflexive, symmetric and transitive:
Relation R is the set N of natural numbers defined as,
ㅤㅤㅤ R = {(x, y) : y = x + 5 and x < 4}​

Answer 1

EXPLANATION.

R is set of N natural number defined as,

⇒ R = {(x, y) : y = x + 5 and x < 4}.

As we know that,

x and y ∈ N.

⇒ x < 4.

Values of x = {1, 2, 3}.

⇒ y = x + 5.

Put the value of x = 1 in the equation, we get.

⇒ y = 1 + 5.

⇒ y = 6.

Put the value of x = 2 in the equation, we get.

⇒ y = 2 + 5.

⇒ y = 7.

Put the value of x = 3 in the equation, we get.

⇒ y = 3 + 5.

⇒ y = 8.

⇒ R = {(1,6), (2,7), (3,8)}.

For reflexive.

⇒ x ∈ (a, a) for every a ∈ N.

⇒ (1,1), (2,2), (3,3) ∉ R.

So, it is not a reflexive.

For transitive.

If (a, b) ∈ R then (b, c) ∈ R and (a, c) ∈ R.

In this equation no one pair is matched.

So, it is not a transitive.

For symmetric.

If (a, b) ∈ R then (b, a) ∈ R.

In this equation no one pair is like this.

So, it is not a symmetric.

Answer 2

☀️  Given that, R is set of N natural number defined as R = {(x, y) : y = x + 5 and x < 4}.

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━

❍ Here,

• x and y are natural numbers and x is less than 4. So, we can take it's value as :

   x = [ 1, 2, 3 ]

Taking x as 1 -

y = x + 5

y = 1 + 5

• y = 6

Taking x as 2 -

y = x + 5

y = 2 + 5

• y = 7

Taking x as 3 -

y = x + 5

y = 3 + 5

• y = 8

• Therefore, R = {( 1, 6 ), ( 2,7 ), ( 3,8 )}

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━

Checking for Reflexive :-

• When the relation is reflexive, then ( a,a ) ∈ R for every a ∈ N.

→ ( 1, 1 ), ( 2,2 ), ( 3,3 )  ∉ R

∴ It is not reflexive.

Checking for Symmetric :-

• When the relation is symmetric then If (a, b) ∈ R then (b, c) ∈ R.

→ ( 1,6 ) ∈  R

→ ( 6,1 )  ∉ R

∴ It is not symmetric.

Checking for Transitive :-

• When the relation is transitive, then If (a, b) ∈ R then (b, c) ∈ R and (a, c) ∈ R.

There's no such pair.

∴ It is not symmetric.

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━