From: mutation@dorsai.com (Florian Lengyel)
Subject: Mathematical Monthly problem
Organization: The Dorsai Embassy +1.718.729.5018
X-Newsreader: Tin 1.1 PL4
Date: Fri, 6 Nov 1992 08:19:31 GMT

I would appreciate someone giving me a trivial reason why I misunderstood
the following trivial problem from the American Mathematical Monthly:

E3427. {\sl Proposed by R. Padmanabhan, N.S. Mendelson, and B. Wolk,
University of Manitoba, Winnipeg, Canada}
\medskip
 
My trivial solution follows:

Let $S$ denote the set obtained by formally adjoining an element $e$
to the set {\bf Q} of rational numbers. On $S$ define a binary operation
$\circ$ as follows:
$$
\eqalign{  % first eqalign in this math display because only one & is 
% allowed per \eqalign and this display requires two of them!
p \circ q &= (3 + pq)/(p + q)   \cr 
p \circ q &= e                        \cr
x \circ e &= x                        \cr                        
} % end of first eqalign
\qquad  % leave some room (which is needed)
\eqalign{ % second eqalign in this math display
&{{\sl if\ }} p \in {\bf Q}, q \in {\bf Q}, p \ne q,          \cr
&{{\sl if\ }} p \in {\bf Q}, q \in {\bf Q} , p = -q,         \cr
&{\sl for\ all\ } x \in S.                \cr
} % end of second eqalign
$$
(a) Prove that $(S, \circ)$ is an abelian group isomorphic to a subgroup
of the real numbers.
\smallskip\noindent
(b) If $p$ is a positive rational number, put $p_1 = p, p_2 = p\circ p,
p_3 = p\circ p\circ p, \ldots$. Show that $\lim_{n \to \infty} p_n$ exists
and find the limit. For which values of p is the sequence 
$\{p_n\}_{n=1}^\infty$ monotonic?
 
\medskip
{\baselineskip = 2\baselineskip
To show (a), it suffices to exhibit a bijection $f: S \to {\bf R}^{\ne 0}$
such that the image of $f$ is a subgroup of $({\bf R}^{\ne 0}, *)$ and
$\forall x, y \in S, f(x \circ y) = f(x)f(y)$. In that case the axioms
for an abelian group will be satisfied by $(S, \circ)$ by virtue of
$x \circ y = f^{-1}(f(x)f(y))$ and because these axioms hold for
$({\bf R}^{\ne 0}, *)$.
 
A mapping $f: S \to {\bf R}$ can be obtained by differentiating the
homomorphism equation (interpreted over {\bf R} with $p \ne -q$, 
for the time being) $f(p \circ q) = f(p) f(q)$ with respect to $p$:
$$
        f'\left(3 + pq \over p + q\right) 
        \left(q^2 - 3 \over (p + q)^2 \right) 
        = f'(p) f(q)
$$
For fixed $q$, $p = \pm\sqrt{3}$ is a fixed point of 
$p \mapsto p \circ q$:
$$
        {3 + pq \over p + q} = p \Leftrightarrow 3 + pq = p^2 + pq
                                 \Leftrightarrow p^2 = 3.
$$
Therefore,
$$
        f'(\sqrt{3}) 
        { (q - \sqrt{3})(q + \sqrt{3}) \over (q + \sqrt{3})(q + \sqrt{3}) }
        = f'(\sqrt{3})f(q) \Rightarrow f(q) =
        { q - \sqrt{3} \over q + \sqrt{3} }.
$$
} % \baselineskip
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\vfill\eject
\centerline{ Florian Lengyel, 35--37 West 64th Street, Apt. 2H, NY 10023. 
(212) 580--8188. }
\bigskip
 
{\baselineskip = 2\baselineskip
{\sl NOTE: The statement of the problem leaves open the value of 
$\circ : S \times S \to S$ on $\{e\}\times {\bf Q}$; however, 
if $(S, \circ)$ defined as above is a group then $e$ must be its 
identity element: 
$  \exists x\in {\bf Q}, \forall y \in {\bf Q}, y \circ x = x $
implies that $x = \pm\sqrt{3}$, a contradiction. The only possible way to
complete the definition of $\circ$ so that $(S, \circ)$ will be
a group is to define $e \circ x = x$ for each $x \in S$; any other
assignment of values to $e \circ x \in S$ for $x \in S$ will
not complete the definition of $\circ$ to a group operation.
From now on it will be assumed that $x \circ e = e \circ x = x$
for all $x \in S$. } % slanted type for emphasis.
\medskip 
 
Define a mapping $f : S \to {\bf R}^{\ne 0}$ by
$$
p  \mapsto 
\cases{ 
  {\displaystyle p - \sqrt{3} \over\displaystyle p + \sqrt{3}} 
                          &if $p \in {\bf Q}$;\cr
   &\cr
   1   &if $p = e \in S$.\cr
}  % \cases        
$$
The inverse of $f$ on $f(S)$ is given by
$$
f^{-1}(x) = 
\cases{
        {\displaystyle x + 1 \over\displaystyle 1 - x}\sqrt{3} 
        &if $ x \in f(S) \wedge x \ne 1$;\cr
        &\cr
        e        &if $x = 1$\cr
} % \cases
$$
Claim: $\forall x, y \in S, f(x \circ y) = f(x)f(y)$  (for the 
corrected definition of $\circ$ in the statement of the problem).
For $x, y \in {\bf Q}, x \ne -y$: 
$$
  f(x \circ y) =
  f\left( { 3 + xy \over x + y }\right) =
  { \displaystyle{ 3 + xy \over x + y } - \sqrt{3}
    \over
    \displaystyle{ 3 + xy \over x + y } + \sqrt{3}
  } =
  { xy - (x+y)\sqrt{3} + 3
    \over
    xy + (x+y)\sqrt{3} + 3
  } =
  {x-\sqrt{3} \over x+\sqrt{3}}  {y-\sqrt{3} \over y+\sqrt{3}}  
= f(x)f(y)
$$
For $x, y \in {\bf Q}, x = -y, f(x \circ y) = f(e) = 1 = f(x)f(x)^{-1}
= f(x)f(y)$. For $x \in S, f(x \circ e) = f(x) = f(x)f(e) = f(e)f(x) =
f(e \circ x).$
 
Note that $f(S)$ is a multiplicative subgroup of ${\bf R^{\ne 0}}$, so that 
the relation $\forall x, y \in S, x \circ y = f^{-1}(f(x)f(y))$ 
makes $(S, \circ)$ into abelian group, 
isomorphic to $(f(S), *)$ under $f : S \to f(S)$.  
 
\medskip
 
For (b), the limit of the sequence $\{p_n\}_{n = 1}^\infty$ is $\sqrt{3}$;
the sequence is monotonic for  $p > \sqrt{3}$. 
 
For the first assertion, notice that
$   f(p_n) = f(p \circ {\ldots\atop n} \circ p) = f(p)^n =
    \displaystyle{\left( { p - \sqrt{3}} \over {p + \sqrt{3}} \right)^n}.
$
} % \baselineskip
 
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\vfill\eject
\centerline{ Florian Lengyel, 35--37 West 64th Street, Apt. 2H, NY 10023. 
(212) 580--8188. }
\bigskip
 
{\baselineskip = 2\baselineskip
 
For $p > 0$ we have
$$
    \left|{ {p - \sqrt{3}} \over {p + \sqrt{3}}}\right| < 1 \Leftrightarrow
    -1 < { p - \sqrt{3} \over p + \sqrt{3} } < 1 \Leftrightarrow
    -p - \sqrt{3} < p - \sqrt{3} < p + \sqrt{3} \Leftrightarrow
    -p < p < p + 2 \sqrt{3}
$$
which is true since $p > 0$; this implies
$\displaystyle{
\lim_{n \to \infty} \left({{p - \sqrt{3}} \over {p + \sqrt{3}}}\right)^n = 0
}$.
\smallskip\noindent
By continuity of the inverse,
$$
\lim_{n \to \infty} p_n =
\lim_{n \to \infty} f^{-1}(f(p_n)) =
f^{-1}(\lim_{n \to \infty} f(p_n)) =
f^{-1}(0) = \sqrt{3}.
$$
For the second assertion, notice that $f$ is order preserving 
on ${\bf Q}^{> 0}$ (and on ${\bf R}^{> 0}$ for that matter):
$$
{{p - \sqrt{3}} \over {p + \sqrt{3}}} < {{q - \sqrt{3}} \over {q + \sqrt{3}}}
\Leftrightarrow
pq + (p-q)\sqrt{3} + 3 < pq + (q-p)\sqrt{3} + 3
\Leftrightarrow
p - q < q - p
\Leftrightarrow
2p < 2q 
\Leftrightarrow p < q.
$$
For $ 0 < p < \sqrt{3}, 
\displaystyle{\left(p - \sqrt{3} \over p + \sqrt{3}\right)^n} $
oscillates around 0: since $f$ (and so $f^{-1}$) is order preserving,
$\{p_n\}_{n=1}^{\infty}$ 
is not monotonic. 
On the other hand,
if $p > \sqrt{3}$, the sequence 
$ \displaystyle{\left(p - \sqrt{3} \over p + \sqrt{3}\right)^n} $
strictly decreases to zero; since $f^{-1}$ is order preserving,
$\{p_n\}_{n=1}^{\infty}$ is  monotonic. 
} % \baselineskip
\bye
-- 
MUTATION@DORSAI.COM
FLORIAN@NEOTERIC.COM
Newsgroups: sci.math