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Let $R$ be a set containing at least two distinct elements $0$ and $1$ and $*$ a ternary operation on $R$ Write $a*b*c$ the image of $(a,b,c)$ under $*$ We call $(R,*,0,1)$ or simply just $R$ a ternary ring if
- $a*0*b=0*a*b=b$ for any $a,b\in R$
- $1*a*0=a*1*0=a$ for any $a\in R$
- given $a,b,c,d\in R$ with $a\ne b$ the equation $x*a*b=x*c*d$ has a unique solution for $x$
- given $a,b,c\in R$ the equation $a*b*x=c$ has a unique solution for $x$
- given $a,b,c,d\in R$ with $a\ne c$ the system of equations
has a unique solution for $(x,y)$
Given a ternary ring $R$ we may form two binary operations on $R$ one called the addition $+$ and the other the multiplication $\cdot$ on $R$ \begin{eqnarray*} a+b &:=& a*1*b, \\ a\cdot b &:=& a*b*0. \end{eqnarray*}
Proposition 1 $(R,+,0)$ and $(R-\lbrace 0\rbrace,\cdot,1)$ are loops, and $0$ is the zero element in $R$ under the multiplication $\cdot$
Proof. We first show that $(R,+,0)$ is a loop. Given $a,b\in R$ there is a unique $c\in R$ such that $a*1*c = b$ but this is exactly $a+c=b$ In addition, there is a unique $d\in R$ such that $d*1*a = b = d*0*b$ But $d*1*a=b$ means $d+a=b$ This shows that $(R,+)$ is a quasigroup. Now, $a+0=a*1*0=a$ and $0+a=0*1*a=a$ so $0$ is the identity with respect to $+$ Therefore, $(R,+,0)$ is
a loop.
Next we show that $(S,\cdot,1)$ is a loop, where $S=R-\lbrace 0\rbrace$ Given $a,b\in S$ there is a unique $c\in R$ such that $c*a*0=b=c*0*b=b$ since $a\ne 0$ From $c*a*0=b$ we get $c\cdot a=b$ Furthermore, $c\ne 0$ for otherwise $b=c*a*0=0*a*0=0$ contradicting the fact that $b\in S$ In addition, there is a unique $d\in R$ such that we have a system of equations $a*d*0=b$ and $0*d*0=0$ From $a*d*0=b$ we get $a\cdot d=b$ Furthermore $d\ne 0$ for otherwise $b=a*d*0=a*0*0=0$ contradicting the fact that
$b\in S$ Thus, $(S,\cdot,1)$ is a quasigroup. Now, $a\cdot 1 = a*1*0=a$ and $1\cdot a=1*a*0=a$ showing that $(S,\cdot,1)$ is a loop.
Finally, for any $a\in R$ $a\cdot 0 = a*0*0=0$ and $0\cdot a= 0*a*0=0$ 
Let $R$ be a ternary ring, and $a,b,c$ are arbitrary elements of $R$ $R$ is said to be linear if $a*b*c = a\cdot b+c$ for all $a,b,c\in R$ left distributive if $a\cdot (b+c)=a\cdot b+a\cdot c$ right distributive if $(a+b)\cdot c = a\cdot c +b\cdot c$ and distributive if it is both left and right distributive.
For example, any division ring $D$ associative or not, is a linear ternary ring if we define the ternary operation $*$ on $D$ by $a*b*c:=a\cdot b+c$ Any associative division ring $D$ is a distributive ternary ring. This easy verification is left to the reader.
Remark. Ternary rings were invented by Marshall Hall in his studies of axiomatic projective and affine planes. It can be shown that in every affine plane, one can set up a coordinate system, and from this coordinate system, one can construct a ternary ring. Conversely, given any ternary ring, one can define an affine plane so that its coordinate system corresponds to this ternary ring.
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