Explain why set of integers is complete Given definition: An ordered set A is complete if it has the "least upper bound property" (completeness). 
We are given $ \mathbb{N}, \mathbb{Z}$ are complete.
and $\mathbb{Q}$ is not complete. 
I understand the inf of the naturals is 1 and has no sup. not sure what the inf or sup of the integers is. For the rationals, I would appreciate any info here as well.
 A: The important part of the least upper bound property for an ordered set $A$ is that if you start with any subset $X$ of $A$ (not just $A$ itself) that is bounded above, the supremum of $X$ must be an element of $A$. 
$\mathbb{Z}$ has neither and infimum nor a supremum, but that is OK because any bounded subset of the integers has a largest element which is the set's supremum.
The rationals, on the other hand, do not have the property because it is possible to find a bounded subset of $\mathbb{Q}$ which has an irrational supremum. As an example you can take the set obtained by writing the first $n$ decimal places of $\pi$ for each $n\in\mathbb{N}$
$$X= \{3.1, 3.14, 3.141,3.1415,\ldots\} $$
Since any number with a terminating decimal representation is rational, $X\subset\mathbb{Q}$. But $\sup X = \pi \notin \mathbb{Q}$ so $\mathbb{Q}$ is not complete.
A: Just for the completeness of the definition of least upper bound property, you can only start with any non-empty subset X of A that is bounded above, as anything could be an upper bound of an empty set which can be proven by contradiction.
