Question 10

# Given an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

Answer

**(i)** Let *A* = {5, 6, 7}.

Define a relation R on* A* as R = {(5, 6), (6, 5)}.

Relation R is not reflexive as (5, 5), (6, 6), (7, 7) ∉ R.

Now, as (5, 6) ∈ R and also (6, 5) ∈ R, R is symmetric.

=> (5, 6), (6, 5) ∈ R, but (5, 5) ∉ R

∴R is not transitive.

Hence, relation R is symmetric but not reflexive or transitive.

**(ii) **Consider a relation R in **R **defined as:

R = {(*a*, *b*): *a* < *b*}

For any *a *∈ R, we have (*a*, *a*) ∉ R since *a* cannot be strictly less than *a* itself. In fact, *a* = *a*.

**∴ **R is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 2)

But, 2 is not less than 1.

**∴ **(2, 1) ∉ R

**∴ **R is not symmetric.

Now, let (*a*, *b*), (*b*, *c*) ∈ R.

⇒ *a* < *b* and *b* < *c*

⇒ *a* < *c*

⇒ (*a*, *c*) ∈ R

**∴ **R is transitive.

Hence, relation R is transitive but not reflexive and symmetric.

**(iii)** Let *A* = {4, 6, 8}.

Define a relation R on A as:

*A* = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}

Relation R is reflexive since for every *a* ∈ *A*, (*a*, *a*) ∈R i.e., (4, 4), (6, 6), (8, 8)} ∈ R.

Relation R is symmetric since (*a*, *b*) ∈ R ⇒ (*b*, *a*) ∈ R for all *a*, *b* ∈ R.

Relation R is not transitive since (4, 6), (6, 8) ∈ R, but (4, 8) ∉ R.

Hence, relation R is reflexive and symmetric but not transitive.

**(iv)** Define a relation R in **R** as:

R = {*a*, *b*): *a*^{3} ≥ *b*^{3}}

Clearly (*a*, *a*) ∈ R as *a*^{3} = *a*^{3}.

**∴ **R is reflexive.

Now,

(2, 1) ∈ R (as 2^{3} ≥ 1^{3})

But,

(1, 2) ∉ R (as 1^{3} < 2^{3})

**∴** R is not symmetric.

Now,

Let (*a*, *b*), (*b*, *c*) ∈ R.

⇒ *a*^{3} ≥ *b*^{3} and *b*^{3} ≥ *c*^{3}

⇒ *a*^{3} ≥ *c*^{3}

⇒ (*a*, *c*) ∈ R

**∴ **R is transitive.

Hence, relation R is reflexive and transitive but not symmetric.

**(v) ** Let *A* = {−5, −6}.

Define a relation R on *A* as:

R = {(−5, −6), (−6, −5), (−5, −5)}

Relation R is not reflexive as (−6, −6) ∉ R.

Relation R is symmetric as (−5, −6) ∈ R and (−6, −5}∈R.

It is seen that (−5, −6), (−6, −5) ∈ R. Also, (−5, −5) ∈ R.

**∴ **The relation R is transitive.

Hence, relation R is symmetric and transitive but not reflexive.

callboy
2019-08-23 02:10:20

in part v set is trans. then (-6,-5) & (-5,-6) both are in relation

angshika
2019-08-21 09:28:45

Thanks for the help

Kajol
2018-12-23 12:01:37

In v. If -6,-6 belongs to R then it will be reflexive (a,a) belongs to R therefore v answer is correct

Sunny
2018-07-15 21:04:10

Try to improve much more

Sachin
2015-04-17 13:33:47

I think, it is correct because (-6,-6) does not belongs to relation set R. Properties of Relation is A realtion R on set A is reflexive if aRa for all a belongs to A i.e. is (a,a) belongs to R for all a belongs to R => each element a of A is related to itself. Ex: Let A = {a,b} and R = {(a,a),(a,b),(b,a)} then R is reflexive as aRa belongs to R but it is not reflexive for pair (b,b) does not belongs to R.

imer
2015-04-16 11:55:15

plz check part v it does not seems correct as -6,-6 doesnot belongs to R

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