Let $A = \struct {A_R, \oplus}$ be a unitary division algebra whose zero is $\mathbf 0_A$. From the unitary nature of $A$ we have that $\oplus$ has a unit $1$ such that: :$\forall a \in A_R, a \ne \mathbf 0_A: a \oplus 1 = a = 1 \oplus a$ From the division algebra nature of $A$ we have that $R$ is a field and that: : $\forall a, b \in A_R, b \ne \mathbf 0_A: \exists_1 x \in A_R, y \in A_R: b \oplus x = a = y \oplus b$ As $1 \in A_R$ it follows that the above holds if $1$ is substituted for $a$: :$\exists_1 x \in A_R, y \in A_R: b \oplus x = 1 = y \oplus b$ Hence: {{begin-eqn}} {{eqn | l = 1       | r = b \oplus x       | c = for some $x \in A_R$ }} {{eqn | ll= \leadsto       | l = y \oplus 1       | r = y \oplus \paren {b \oplus x}       | c =  }} {{eqn | ll= \leadsto       | l = y \oplus 1       | r = \paren {y \oplus b} \oplus x       | c = Associativity of $\oplus$ }} {{eqn | ll= \leadsto       | l = y \oplus 1       | r = 1 \oplus x       | c = as $1 = y \oplus b$ }} {{eqn | ll= \leadsto       | l = y       | r = x       | c = Definition of $1$ }} {{end-eqn}} So if we let $b^{-1} := x = y$ in the above, we have: :$\exists_1 b^{-1} \in A_R: 1 = b \oplus b^{-1} = 1 = b^{-1} \oplus b$ That is, $b^{-1}$ is a unique multiplicative inverse of $b$. So every non-zero element of $A_R$ has a unique multiplicative inverse in $A_R$. {{qed|lemma}} Now suppose that $A_R$ has a unique multiplicative inverse in $A_R$. It follows directly that, by definition, $\struct {A_R, \oplus}$ has to be a unitary algebra or there is no $1$ for $a \oplus a^{-1}$ to equal. First we prove existence, i.e we show that: : $\forall a, b \in A_R, b \ne \mathbf 0_A: \exists x \in A_R, y \in A_R: b \oplus x = a = y \oplus b$ Take $a \oplus b^{-1}$ where $b^{-1}$ is the unique multiplicative inverse of $b$. Then: {{begin-eqn}} {{eqn | l = a \oplus b^{-1}       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = \paren {a \oplus b^{-1} } \oplus b       | r = x \circ b       | c =  }} {{eqn | ll= \leadsto       | l = a \oplus \paren {b^{-1} \oplus b}       | r = x \oplus b       | c = Associativity of $\oplus$ }} {{eqn | ll= \leadsto       | l = a \oplus 1       | r = x \oplus b       | c = as $b^{-1} \oplus b = 1$ }} {{eqn | l = a       | r = x \oplus b       | c =  }} {{end-eqn}} Similarly, we show that $\exists y \in A_R: a = y \oplus b$. Now we prove uniqueness. Now suppose $\exists x_1, x_2 \in A_R: a = x_1 \oplus b, a = x_2 \oplus b$. Then: {{begin-eqn}} {{eqn | l = a       | r = x_1 \oplus b       | c =  }} {{eqn | ll= \leadsto       | l = a \oplus b^{-1}       | r = \paren {x_1 \oplus b} \oplus b^{-1}       | c = $b^{-1}$ being the unique multiplicative inverse of $b$ }} {{eqn | ll= \leadsto       | l = a \oplus b^{-1}       | r = x_1 \oplus \paren {b \oplus b^{-1} }       | c = Associativity of $\oplus$ }} {{eqn | ll= \leadsto       | l = a \oplus b^{-1}       | r = x_1 \circ 1       | c = as $b \oplus b^{-1} = 1$ }} {{eqn | ll= \leadsto       | l = a \oplus b^{-1}       | r = x_1       | c =  }} {{end-eqn}} In exactly the same way, $a = x_2 \oplus b \implies a \oplus b^{-1} = x_2$. And so $x_1 = x_2$ thus proving uniqueness. In a similar way we prove that  :$\forall a, b \in A_R, b \ne \mathbf 0_A: \exists_1 y \in A_R: a = y \oplus b$ {{qed}}