2.5.tex

\documentclass[a4paper,12pt]{article}
\usepackage{listings}
\lstset{language=Lisp}

\begin{document}
Let $a$, $b$, $c$ and $b$ be non negative integers such that,
$ 2^a 3^b = 2^c 3^d$.

\medskip
We have $\gcd(2, 3) = 1$, so $2^a$ divide $2^c$.  Thus we deduce that
$2^a \le 2^c$ thus $a \le c$.  By symmetry, we have too $c \le a$, so
$a = c$.  We then conclude too that $b = d$.

The uniqueness of the couple $a$ and $b$ permits the representation of
a pair of non negative integers as $2^a 3^b$.
\begin{lstlisting}
(define (cons a b)
  (* (expt 2 a)
     (expt 3 b)))
(define (car n)
  (if (odd? n)
      0
      (+ 1 (car (/ n 2)))))
(define (cdr n)
  (if (even? n)
      0
      (+ 1 (cdr (/ n 3)))))
\end{lstlisting}
\end{document}