Burton Archimedean property proof Can anyone help me understand the following proof of Archimedean property in Burton's Elementary Number Theory book:

Theorem 1.1  Archimedean  property.  If  $a$  and  $b$  are  any 
  positive  integers,  then  there  exists  a positive  integer  $n$ 
  such  that  $na  \ge    b$.  Proof.  Assume  that  the  statement  of 
  the  theorem  is not  true,  so  that  for  some  $a$  and  $b$,  $na <  b$  for  every  positive  integer  n.  Then  the  set  $S =  \{b- na  |  n$  a positive  integer$\}$  consists  entirely  of  positive  integers.  By  the  Well-Ordering  Principle,  $S$  will 
  possess  a  least  element,  say,  $b- ma$.  Notice  that  $b- (m +  1)a$  also  lies  in $S$,  because  $S$  contains  all  integers  of  this  form.  Furthermore,  we  have  $b- (m +  1)a =  (b- ma)- a<  b- ma$  contrary  to  the  choice  of  $b - ma$  as  the  smallest  integer  in  $S$.  This  contradiction  arose  out  of  our  original  assumption  that  the  Archimedean 
  property  did  not  hold;  hence,  this  property  is proven  true.

If he defined $b-ma$ as the least element of the set then how can he define another even lesser element $b-(m+1)a$? Wouldn't that mean that would be a negative number?

Eventually it means that $b-(m+1)a$ is negative, but the fact that you "know" this is jumping ahead a bit.  The assumption is that $na<b$ for every positive integer $n$.  That has to include $n=m+1$, so you're assuming $(m+1)a<b$ which is to say $b-(m+1)a >0.$  This leads to a contradiction and then you can conclude that $b-(m+1)a$ is negative.  You're letting your intuition get aheas of your logic.
