(a) Suppose $A\subset\mathbb{Z}$ and $A\neq\emptyset$ . If all elements of $A$ are negative, then the set $-A=\{n|-n\in A\}\subset\mathbb{Z}_{+}$ has the smallest element, which is the largest element of $A$ . If $0\in A$ and there are no positive numbers in $A$ , then $0$ is the largest element of $A$ . Finally, if $A$ has positive integers, then, using Exercise 4(a) and the fact that $A$ is bounded above by some real number, and, hence, by a positive integer number $m$ , i.e. $A\cap\mathbb{Z}_{+}\subset\{1,\ldots,m\}$ , we conclude that $A$ has a largest element.

(b) Consider the set $A\subset\mathbb{Z}$ of integers less than $x$ . It is nonempty ($-n\in A$ where $n>-x$ ), bounded from above by $x$ , has a largest element (by (a)), which is $n$ such that $n<x$ and $n+1\ge x$ . But $x\notin\mathbb{Z}$ , so that $x<n+1$ .

(c) Use (b) to find $n$ such that $n<x\le n+1$ (i.e., unlike (b), $x$ can be an integer number, in which case we let $n=x-1$ ), and use other proved properties to verify that this $y<x-1\le n$ .

(d) Consider $x-y>0$ , then find $n\in\mathbb{Z}_{+}$ such that $n>1/(x-y)>0$ . We have $nx-ny>1$ , so we can use (c) to find $m\in\mathbb{Z}$ such that $ny<m<nx$ . It follows that $m/n$ is a rational number between $x$ and $y$ .