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\title{Solution to Questions  8.2 and 8.3}
\author{Geoff Smith \copyright 1998}
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We address question 8.2. The solution to question 8.3 is very similar. 

Let $r$ be a real number. We have to decide what that means; either
$r$ is a Dedekind cut or it is an equivalence class of Cauchy sequences.
We won't chase everything all the way back to these two starting points,
but we will come close. 

Before we get going, we remind ourselves of the floor function 
(integer part function) from 
${\bf R}$ to ${\bf Z}$ defined by $\lfloor x \rfloor
= \max \{ z \mid z \in Z,\ z \leq x \}.$

Now for the argument itself. 
By adding an integer $z$ to $r$ we can ensure that
$r' = r + z \in [0,1).$ If we can show $r'$ has a unique decimal expansion
then so to must $r = r' - z$ (since rival decimal representations
of $r'$ would give rise to rival decimal representations
of $r'$ and {\it vice versa}.

Thus we may assume that $r \in [0,1)$ (because we are really thinking about 
$r'$). Now if the decimal representation of $r$ is
\[0 \cdot a_1 a_2 a_3 \ldots\] it seems that 
we have $a_i = \left \lfloor 10^i r \right \rfloor \mbox{ mod } 10$ for $i \in {
\bf N}.$
In order for this to be true, we need that every decimal representation
\[[0 \cdot b_1 b_2 b_3 \ldots\] is less that 1 (so that the integer part
of $10^i r$
is what we think it is) provided the string of $b$'s does not end in 
infinitely many $9$'s.

To see this, 
choose $j$ such that $b_j \not = 9,$ then 
\[0 \cdot b_1 b_2 b_3 \ldots \leq  b_1 b_2 b_3 \ldots b_{j-1} 9 < 1 \] 
and we are done. 

{\it Well, almost. This is where we should  
start worrying about Dedekind cuts and Cauchy sequences. In fact we have 
disguised earlier reliance on Dedekind cuts and Cauchy sequences -- 
from the moment we added an integer $z$ to $r$ to pop it in $[0,1),$ we
really needed to know how to add a real number to an integer, so 
we ought to have gone {\it back to basics} even at that stage. However,
life is too short. Mine is anyway.}  
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