Question

R and S are two relations on A set.

Prove That:
i) R⊂S then R-¹⊂S-¹
ii) (R∪T)-¹= R-¹∪T-¹

PLEASE SOLVE THIS,​

Answer 1

✪PROOF✪

1) Now,

Let $\:\:\: (x, y) \in R^{-1}$

$\implies (y, x) \in R$

$\implies (y, x) \in S$ [Since $R \subset S$ ]

$\implies (x, y) \in S^{-1}$

$(x, y) \in R^{-1}\implies (x, y) \in S^{-1}$

Thus, $\bf{ R^{-1} \subset S^{-1}}$

Hence Proved.

2)

Let $\:\:\: (x, y) \in (R\cup S)^{-1}$

$\implies (y, x) \in (R\cup S)$

$\implies (y, x) \in R\:or(y, x) \in S$

$\implies (x, y) \in R^{-1}\:or(x, y) \in S^{-1}$

$\implies (x, y) \in R^{-1}\cup S^{-1}$

$So, (R\cup S)^{-1}\subseteq R^{-1}\cup S^{-1}.. (1)$

And

Let $\:\:\: (x, y) \in R^{-1}\cup S^{-1}$

$\implies (x, y) \in R^{-1}\:or(x, y) \in S^{-1}$

$\implies (y, x) \in R\:or(y, x) \in S$

$\implies (y, x) \in (R\cup S)$

$\implies (x, y) \in (R\cup S)^{-1}$

$So, R^{-1}\cup S^{-1} \subseteq (R\cup S)^{-1}.. (2)$

From (1) and (2), we can conclude that

$\bf{(R\cup S)^{-1} = R^{-1}\cup S^{-1}}$

Hence Proved

Answer 2

(i)R and S are symmetric.

Let (a,b)∈R⇒(b,a)∈R

Let (a,b)∈S⇒(b,a)∈S

R is a subset of A×A and S is a subset of A×A

R∩S is a subset of A×A

Let (a,b)∈R⇒(a,b)∈R∩S ......(1)

(a,b)∈R and (a,b)∈S

(b,a)∈R and (b,a)∈S

(b,a)∈R∩S ......(2)

From (1) and (2) we have R∩S is symmetric.

Also,let (a,b)∈A such that (a,b)∈R∪S .......(3)

⇒(a,b)∈R or (a,b)∈S

(b,a)∈R∪S .......(4)

From (3) and (4) we have R∪S is symmetric.

(ii)R is reflexive and S is any relation.

We have (a,a)∈R since R is reflexive.

⇒(a,a)∈A since R is a subset of A×A

⇒(a,a)∈R∪S

∴R∪S is reflexive.