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properties of certain monotone functions
In the definitions of some partially ordered algebraic systems such as pogroups and porings, the multiplication is set to be compatible with the partial ordering on the universe in the following sense:
$ab\leq ac\quad\mbox{iff}\quad b\leq c\qquad\mbox{ and }\qquad ab\leq cb\quad% \mbox{iff}\quad a\leq c$ 
This is no coincidence. In fact, these “definitions” are actually consequences of properties concerning monotone functions satisfying certain algebraic rules, which is the focus of this entry.
Recall that an $n$ary function $f$ on a set $A$ is said to be monotone if it is monotone in each of its variables. In other words, for every $i=1,2,\ldots,n$, the function $f(a_{1},\ldots,a_{{i1}},x,a_{{i+1}},\ldots,a_{n})$ is monotone in $x$, where each of the $a_{j}$ is a fixed but arbitrary element of $A$. We use the notation $\uparrow,\downarrow,\updownarrow$ to denote the monotonicity of each variable in $f$. For example, $(\uparrow,\uparrow,\uparrow)$ denotes a ternary isotone function, whereas $(\downarrow,\updownarrow)$ denotes a binary function which is antitone with respect to its first variable, and both isotone/antitone with respect to the second.
Proposition 1.
Let $f$ be an $n$ary commutative monotone operation on a set $A$. Then $f$ is either isotone or antitone.
Proof.
Suppose $f$ is isotone (or antitone) in its first variable. Since $f(x,a_{1},\ldots,a_{{n1}})=f(a_{1},x,\ldots,a_{{n1}})=\cdots=f(a_{1},\ldots,% a_{{n1}},x)$, $f$ is isotone (or antitone) in each of its remaining variables. ∎
Proposition 2.
Let $f$ be an $n$ary monotone operation on a set $A$ with an identity element $e\in A$. In other words, $f(x,e,\ldots,e)=f(e,x,\ldots,e)=\cdots=f(e,e,\ldots,x)=x$. Then $f$ is either strictly isotone or strictly antitone.
Proof.
Proposition 3.
Let $f$ be a binary monotone operation on a set $A$ such that it is isotone (antitone) with respect to its first variable. Suppose $g$ is a unary operation on $A$ such that $f(x,g(x))$ is a fixed element of $A$. Then $g$ is antitone (isotone).
Proposition 4.
Let $f$ be an $n$ary associative monotone operation on a set $A$. Then

$f$ is isotone if $n$ is even

$f$ is either isotone, or is $(\underbrace{\uparrow,\ldots,\uparrow}_{{m}},\downarrow,\underbrace{\uparrow,% \ldots,\uparrow}_{{m}})$, if $n$ is odd, say $n=2m+1$.
Proof.
Suppose first that $n=2m$, $i\leq m$, and $g(x)=f(a_{1},\ldots,a_{{i1}},x,\ldots,a_{{m+1}},\ldots,a_{{2m}})$ is antitone. Then $g(g(x))$ is isotone. By the associativity of $f$, $g(g(x))$ is
$\displaystyle f(a_{1},\ldots,a_{{i1}},f(a_{1},\ldots,a_{{i1}},x,a_{{i+1}},% \ldots,a_{m},\ldots,a_{{2m}}),\ldots,a_{{m+1}},\ldots,a_{{2m}})$  
$\displaystyle=$  $\displaystyle f(a_{1},\ldots,a_{{i1}},a_{1},\ldots,a_{{i1}},x,a_{{i+1}},% \ldots,f(a_{{2mi+1}},\ldots,a_{{2m}},a_{{i+1}},\ldots,a_{{m+1}},\ldots,a_{{2m% }})).$ 
In the second expression, the position of $x$ is $2i2\leq 2m1<2m$, therefore implying that $g(g(x))$ is antitone, which is a contradiction! Therefore, $g(x)$ is isotone. Now, if $i>m$, and $h(x)=f(b_{1},\ldots,b_{m},\ldots,b_{{i1}},x,b_{{i+1}},\ldots,b_{{2m}})$ is antitone, then $h(h(x))$ is isotone. But
$\displaystyle f(b_{1},\ldots,b_{m},\ldots,b_{{i1}},f(b_{1},\ldots,b_{m},% \ldots,b_{{i1}},x,b_{{i+1}},\ldots,b_{{2m}}),b_{{i+1}},\ldots,b_{{2m}})$  
$\displaystyle=$  $\displaystyle f(f(b_{1},\ldots,b_{m},\ldots,b_{{i1}},b_{1},\ldots,b_{{2mi+1}% }),\ldots,b_{{i1}},x,b_{{i+1}},\ldots,b_{{2m}},b_{{i+1}},\ldots,b_{{2m}})),$ 
and the position of $x$ is the second expression is $(i1)(2mi+1)+2=2i2m>1$, therefore implying that $h(h(x))$ is antitone, again a contradiction. As a result, $f$ is isotone for all $i=1,\ldots,n$.
The argument above also works when $n$ is odd, say $n=2m+1$ and $i\neq m+1$. Finally, since $f$ is monotone, it is monotone with respect to the $i$th variable when $i=m+1$, so $f$ is one of the following three forms:
$(\underbrace{\uparrow,\ldots,\uparrow}_{{2m+1}}),\qquad(\underbrace{\uparrow,% \ldots,\uparrow}_{{m}},\updownarrow,\underbrace{\uparrow,\ldots,\uparrow}_{{m}% }),\qquad(\underbrace{\uparrow,\ldots,\uparrow}_{{m}},\downarrow,\underbrace{% \uparrow,\ldots,\uparrow}_{{m}}),$ 
the first two of which imply that $f$ is isotone. ∎
An example of an associative function that is, say $(\uparrow,\downarrow,\uparrow)$, is given by
$f:\mathbb{Z}^{3}\to\mathbb{Z}\qquad\mbox{where}\qquad f(x,y,z)=xy+z.$ 
$f$ is associative since $f(f(r,s,t),u,v)=f(r,f(s,t,u),v)=f(r,s,f(t,u,v))=rs+tu+v$.
Mathematics Subject Classification
06F99 no label found08C99 no label found08A99 no label found Forums
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