MTH202 complete Solution

Second question answerLet 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

First Question: