Let $x$ be a real number. Prove the existence of a unique integer $a$ such that $a \leq x < a+1$

Question

Let $x\in \mathbb{R}$ , Using the Well-Ordering Property of $\mathbb{N}$ and the Archimedean Property of $\mathbb{R}$, show that there exist a unique $a \in \mathbb{Z}$ such that $a \leq x < a+1$

My approach so far: Suppose $x$ greater than some Integer $a$.

$x\geq a$

By Archimedean Property of $\mathbb{R}$, there exist $n_{x} \in \mathbb{N}, x<n_{x}$

Combining those 2 inequality I have $a\leq x < n_{x}$.

But I do not know how to proceed from here, any help or insights is deeply appreciated.

Thank you for reading my post.

Answer 1

I'd break this up into 3 cases

Case 1: $x=0$

This case is really simple so I'll leave it for you to do.

Case 2: $x>0$

Let $S=\{n\in \mathbb{N}:n>x-1\}$, which exists by the Archimedean postulate. By the well-ordering property $S$ contains a least element $a$. Obviously $a>x-1$ Now assume by way of contradiction that $a-1>x-1$. Thus $a-1\in S$ contradicting $a$ being the least element of the set. So then $a-1\leq x-1$. We can therefore conclude that there exists a $a$ such that $a \leq x <a+1$ (by using our properties of the order axioms.)

Case 3: $x<0$

This argument is very similar to the one used in case 2. Note that $\mathbb{Z^-}=\{-n:n \in \mathbb{N}\}$. Thus we can construct any subset in the negative integers by mapping the corresponding subset in the positive integers under $f:\mathbb{N} \to \mathbb{Z^-}$ where $f(n)=-n$. We can see that $f$ is bijective. Finally remember that for positive $x,y\in \mathbb{Z}$ if $x<y$ then $-x>-y$. With the above information you can now use the well-ordering property of the positive integers to see that every subset of the negative integers has a greatest element.

Answer 2

I'll post my answer for discussion (please don't be hard):

Let $x\in \mathbb{R}$. Let $A=\{n\in\mathbb{Z}\ |\ n\leq x \}$. It is possible to show that this set is non-empty. Clearly $A$ is upper bounded by $x$ in $\mathbb{R}$ so it must have a maximum. Call this maximum $m$. Notice that $m+1\notin A$, otherwise we would have a contradiction.

Let $A'=\{n\in \mathbb{Z} \ | \ x<n \}$. By a consequence of the Archimedean Principle, this set is non-empty. Clearly $A'$ is lower bounded by $x$ in $\mathbb{R}$. Hence it has a minimum. Call this minimum $n'$. Notice that $A\cap A' = \emptyset$ and that $A\cup A' = \mathbb{Z}$. Since $m+1\notin A$ then $m+1\in A'$. Since $m+1$ is the consecutive integer of $m$ then $m+1=n'$. We thus have have $m\leq x< m+1$.

Moreover because the minimum and maximum are unique, thus $m$ is unique.