Answer to Question #225253 in Discrete Mathematics for Amir

Question

Let A= {0, 1, 2, 3} and define relations R, S and T on A as follows:
R= { (0,0),(0,1),(0,3),(1,0),(1,1),(2,2),(3,0),(3,3)}
S= {(0, 0),(2,2),(1,1), (0, 2), (0, 3), (2, 3),(2,2)}
T= {(0, 1), (2, 3),(0,0),(2,2),(1,0),(3,3),(3,2)}
i. Is R Reflexive? Symmetric? AntiSymmetric?
ii. Is S Reflexive? Symmetric? AntiSymmetric?
iii. Is T Reflexive? Symmetric? AntiSymmetric?

Answer 1

Solution:

Given, A= {0, 1, 2, 3}

R= { (0,0),(0,1),(0,3),(1,0),(1,1),(2,2),(3,0),(3,3)}

S= {(0, 0),(2,2),(1,1), (0, 2), (0, 3), (2, 3),(2,2)}

T= {(0, 1), (2, 3),(0,0),(2,2),(1,0),(3,3),(3,2)}

Reflexive: $(a,a)\in R\forall a\in A$

Symmetric: $(a,b)\in R\Rightarrow (b,a)\in R, \forall a,b\in A$

Anti-Symmetric: $(a,b)\in R,(b,a)\in R,\Rightarrow a=b, \forall a,b\in A$

(i) Using these definitions, R is reflexive, symmetric but not anti-symmetric as (0,3),(3,0)$\in R$ but $0\ne3$

(ii) S is not reflexive as (3,3) is not in S.

S is symmetric and anti-symmetric.

(iii) T is not reflexive as (1,1) is not in T.

T is symmetric.

But T is not anti-symmetric as (0,1),(1,0) $\in T$ but $1\ne 0$