Is antisymmetric the same as reflexive? Note: The following definitions from my book, Discrete Mathematics and Its Applications [7th ed, 598].
This is my book's definition for a reflexive relation

This is my book's definition for a anti symmetric relation

Is a reflexive relation just the same as a anti symmetric relation?  From what I've, the only way to meet that antisymmetric requirement is to have the same ordered pair, say an element a from Set A, (a,a). If you have anything other than the same ordered pair, (1,2) and (2,1), it will not meet the antisymmetric requirement. But the overall definition of reflexive relation is that it's the same ordered pair. Are they just two ways of saying the same thing? Is it possible to have one and not the other?

Reflexivity requires all elements to be both way related with themselves.  However, this does not prohibit non-equal elements from being both way related with each other.
That is:  $\qquad\forall a\in A: (a,a)\in R$ from which we can prove: $$\forall a\in A\;\forall b\in A:\Big(a=b \;\to\; (a,b)\in R \wedge (b,a)\in R\Big)$$
Antisymetricality requires any two elements which are both way related with each other to be equal.  Which means it prohibits non-equal elements from being both-way related to each other; however, this does not require all elements to be both way related with themselves.
$$\forall a\in A, \forall b\in A:\Big((a,b)\in R \wedge (b,a)\in R \;\to\; a=b\Big)$$
Or via contraposition: $\qquad\forall a\in A, \forall b\in A:\Big(a\neq b \;\to\; (a,b)\notin R \vee (b,a)\notin R\Big)$
Thus they are not the same thing.

Examples are easy to generate.  From the set $A=\{0, 1\}$ The following relations are:
$$\begin{array}{c|c|c}
R & \text{Reflexive} & \text{Antisymmetric}
\\ \hline
\{(0,0), (1,1)\} & \text{Yes} & \text{Yes}
\\
\{(0,0), (0,1), (1,0), (1,1)\} & \text{Yes} & \text{No}
\\
\{(0,0), (0,1), (1,0)\} & \text{No} & \text{No}
\\
\{(0,0), (0,1)\} & \text{No} & \text{Yes}
\end{array}$$
