For an ordered groupoid  $S$  without zero, we 

denote by  $S^0$  the groupoid arising from  $S$  by the 

adjunction of a zero element  0.   If  $S$  is an ordered 

groupoid without zero and  $R$  a right ideal of  $S$,  then the 

set  $R \cup \{0\}$  is a right ideal of  $S^0$.  If  $S$  is an 

ordered groupoid and  $R$  a right ideal of  $S^0$  such that  $R 

\neq \{0\}$,  then the set  $R\setminus \{0\}$  is a right ideal 

of  $S$.  Moreover, the mapping  $R  \to R \cup \{0\}$  of the 

set  $\rrr (S)$ of right ideals of  $S$  onto the set of all 

non-zero right ideals of  $S^0$  is a homomorphism, reverse 

isotone -and so  ($1-1$), which means that the sets  $\rrr (S)$  

and  $\rrr (S^0)\setminus \{\{0\}\}$  are isomorphic.  As a 

consequence, any theorem concerning right ideals in an ordered 

groupoid  $S$  without zero implies an evident corollary 

concerning right ideals in  $S^0$.  Similar results for left or 

two-sided ideals bi- and quasi-ideals also hold.