Determining if a relation is reflexive, symmetric, or transitive Let $A = \{0,1,2,3\}$
Define a relation $T$ on $A$ as follows:
$T = \{(0,1),(2,3)\}$
Is $T$ reflexive? symmetric? transitive?
 A: To be Reflexive  you should have  (0,0) (1,1) (2,2) (3,3)   
To be symmetric  if you have     (0,1), then you should have (1,0) 
To be transitive if you have  (1,2) (2,4), then you should have (1,4) 
To answer your question
 NO for Symmetric and Reflexive 
 Yes for transitive because no counter example
A: Are you familiar with the definitions of reflexive, symmetric and transitive relations?

A reflexive relation is a binary relation on a set for which every element is related to itself.

As you can clearly see $(0,0),(1,1)$ etc. are not contained in your relation, so it is not reflexive.

A relation is symmetric if $aRb \implies bRa$.

Once again, $(0,1)$ is there in your relation but $(1,0)$ isn't. So it is not symmetric.

Relation is transitive if $(aRb \wedge bRc) \implies aRc$.

There is no counter-example to this in your relation, so it is transitive.
A: A relation is reflexive $\iff$ $\forall_{\alpha}(\alpha \in A)$$(\langle\alpha,\alpha\rangle \in T)$. This is clearly not the case for T. 
A relation is symmetric $\iff$ $\forall_{\alpha, \beta}(\alpha \in A)([\langle\alpha,\beta\rangle\in T] \rightarrow [\langle\beta,\alpha\rangle \in T])$. This isn't the case for T either.
A relation is transitive $\iff$ $\forall_{\alpha,\beta,\gamma}([\langle\alpha,\beta\rangle \in T \land \langle\beta,\gamma\rangle \in T] \rightarrow [\langle\alpha,\gamma\rangle \in T])$. Yes, this statement is vacuously true, I believe.  
