TITLE: Every real number lies between 2 consecutive integers
QUESTION [10 upvotes]: Is it possible to prove that every real number must between two integers without using the completeness property?

REPLY [9 votes]: Yes. It's a consequence of the Archimedean Property of $\mathbb{R}$.
Fact: Archimedean Property: If $a,b$ are positive real numbers, then $\exists n \in \mathbb{N}$ s.t. $na > b$.
Proposition: if $a \in \mathbb{R}$, then $\exists N \in \mathbb{Z}$ s.t. $N-1 \leq a < N$.
Proof: Let $S = \{n \in \mathbb{Z} | n > a \}$. Then by the Archimedean property, $S \neq \emptyset$ and S is bounded below. By the Well-Ordering Principle, S has a least element. Then $N-1 \notin S$, so $N-1 \leq a < N$.

REPLY [3 votes]: Assume there was a number that wasn't between two integers, than it must be greater than every integer or less than every integer which is a contradiction to the Archimedean property. While this holds, I think it implicitly relies on completeness.