First question:

Second question answer:

Let A= {0, 1, 2} so we can determined A×A as under

{(0,0), (0,1), (0, 2), (1,0), (1,1), (1, 2), A× A = (2,0), (2,1), (2,2)}

Here R = A×A so R = {(0,0), (0,1), (0,2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}

For Reflexive:

R is reflexive if and only if, for all aÎ A , (a,a)ÎR

A= {0, 1, 2}

A× A = {(0,0), (0,1), (0, 2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}

R = {(0,0), (0,1), (0,2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}

We can see that {(0,0), (1,1), (2, 2)}ÎR so R is Reflexive

For Anti-symmetric:

R is anti-symmetric if and only if, for all bÎ A if

(a,b)ÎR and (b, a)ÎR then a = b

aRb and bRa then a = b

A= {0, 1, 2}

A× A = {(0,0), (0,1), (0, 2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}

R = {(0,0), (0,1), (0,2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}

We can see that {(0,0), (0,1), (0,2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}

As we can see that,

(0,1)ÎR and (1,0)ÎR then 0=0 [As a = b]

(2,1)ÎR and (1, 2)ÎR then 2=2 [As a = b]

all pairs are available such that (a,b)ÎR and (b, a)ÎR so R is Anti-symmetric

For Transitive:

R is transitive if and only if, for all a,b, cÎ A if

(a,b)ÎR and (b, c)ÎR then (a, c)ÎR

That is aRb and bRc then aRc

A= {0, 1, 2}

A× A = {(0,0), (0,1), (0, 2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}

R = {(0,0), (0,1), (0,2), (1,0), (1,1), (1, 2), (2,0), (2,1), (2,2)}

As we can see

(0, 2)ÎR and (2,1)ÎR then (1,0)ÎR

(0,1)ÎR and (1, 2)ÎR then (0, 2)ÎR and so on…….

all pairs are available such that (a,b)ÎR and (b, c)ÎR then (a, c)ÎR so R is Transitive