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?
Best Answer
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.