Let us define $\eqclass {\tuple {a, b} } \ominus$ as in the Inverse Completion of Integral Domain Exists. That is, $\eqclass {\tuple {a, b} } \ominus$ is an equivalence class of elements of $D \times D^*$ under the congruence relation $\ominus$. $\ominus$ is the congruence relation defined on $D \times D^*$ by $\tuple {x_1, y_1} \ominus \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$. By the method of its construction, $\dfrac p q \equiv \eqclass {\tuple {p, q} } \ominus$. From Equality of Division Products, two elements $\dfrac a b, \dfrac c d$ of $K$ are equal {{iff}} $a \circ d = b \circ c$. This correlates with the fact that two elements $\eqclass {\tuple {a, b} } \ominus, \eqclass {\tuple {c, d} } \ominus$ of $K$ are equal iff $a \circ d = b \circ c$. Suppose $a = 0_D$. {{begin-eqn}} {{eqn | l = a       | r = 0_D       | c =  }} {{eqn | ll= \leadsto       | l = 0_D \circ d       | r = b \circ c       | c =  }} {{eqn | ll= \leadsto       | l = b \circ c       | r = 0_D       | c =  }} {{eqn | ll= \leadsto       | l = c       | r = 0_D       | c = as $b \in D^*$, so $b \ne 0$ }} {{end-eqn}} Hence: :$\eqclass {\tuple {0_D, b} } \ominus  = \eqclass {\tuple {0_D, d} } \ominus$ Thus all elements of $K$ of the form $\eqclass {\tuple {0_D, k} } \ominus$ are equal, for all $k \in D^*$. To emphasise the irrelevance of the $k$, we will abuse our notation and write: :$\eqclass {\tuple {0_D, k} } \ominus$ as :$\eqclass {0_D} \ominus$ Next, by Product of Division Products, we have that $\ds \frac a b \circ \frac c d = \frac {a \circ b} {c \circ d}$. Again abusing our notation, we will write: :$\eqclass {\tuple {a, b} } \ominus \circ \eqclass {\tuple {c, d} } \ominus$ to mean: :$\eqclass {\tuple {a \circ c, b \circ d} } \ominus$ So: {{begin-eqn}} {{eqn | l = \eqclass {0_D} \ominus \circ \eqclass {\tuple {a, b} } \ominus       | r = \eqclass {\tuple {0_D, k} } \ominus \circ \eqclass {\tuple {a, b} } \ominus       | c =  }} {{eqn | r = \eqclass {\tuple {0_D \circ a, k \circ b} } \ominus       | c =  }} {{eqn | r = \eqclass {\tuple {0_D, k \circ b} } \ominus       | c =  }} {{eqn | r = \eqclass {0_D} \ominus       | c =  }} {{eqn | r = \eqclass {\tuple {a \circ 0_D, b \circ k} } \ominus       | c =  }} {{eqn | r = \eqclass {\tuple {a, b} } \ominus \circ \eqclass {\tuple {0_D, k} } \ominus       | c =  }} {{eqn | r = \eqclass {\tuple {a, b} } \ominus \circ \eqclass {0_D} \ominus       | c =  }} {{end-eqn}} Hence: :$\eqclass {0_D} \ominus \circ \eqclass {\tuple {a, b} } \ominus = \eqclass {\tuple {a, b} } \ominus = \eqclass {\tuple {a, b} } \ominus \circ \eqclass {0_D} \ominus$ So $\eqclass {0_D} \ominus$ fulfils the role of a zero for $\tuple {K, \circ}$ as required. Also we have that: {{begin-eqn}} {{eqn | l = \eqclass {0_D} \ominus \circ \eqclass {0_D} \ominus       | r = \eqclass {\tuple {0_D, k} } \ominus \circ \eqclass {\tuple {0_D, k} } \ominus       | c =  }} {{eqn | r = \eqclass {\tuple {0_D \circ 0_D, k \circ k} } \ominus       | c =  }} {{eqn | r = \eqclass {\tuple {0_D, k \circ k} } \ominus       | c =  }} {{eqn | r = \eqclass {0_D} \ominus       | c =  }} {{end-eqn}} So $\eqclass {0_D} \ominus$ is idempotent. It follows that $\eqclass {0_D} \ominus$ can be identified with $0_D$ from the mapping $\psi$ as defined in Construction of Inverse Completion. {{qed}} Category:Integral Domains Category:Inverse Completions retg8czrb5r3plkykgpud4aram1uv1f