# Proof involving the Archimedean property.

• Sep 5th 2010, 04:38 PM
mjlaz
Proof involving the Archimedean property.
Given any x $\in$ R prove that there exists a unique n $\in$ Z such that n - 1 $\leq$ x < n.

Proof: Suppose x $\in$ R. By the Archimedean property, there is an n $\in$ Z such that n > x. Since N is well-defined, we have that n - 1 $\in$ N and n - 1 < n. Then it follows that n - 1 $\leq$ x < n for x $\in$ R.

I feel like I've jumped to conclusions, but to my non-math-genius brain, I feel like this is correct. Can someone prod me along?

(Clapping)
• Sep 5th 2010, 05:37 PM
HallsofIvy
Quote:

Originally Posted by mjlaz
Given any x $\in$ R prove that there exists a unique n $\in$ Z such that n - 1 $\leq$ x < n.

Proof: Suppose x $\in$ R. By the Archimedean property, there is an n $\in$ Z such that n > x. Since N is well-defined, we have that n - 1 $\in$ N and n - 1 < n. Then it follows that n - 1 $\leq$ x < n for x $\in$ R.

I feel like I've jumped to conclusions, but to my non-math-genius brain, I feel like this is correct. Can someone prod me along?

(Clapping)

Your error is in asserting that n-1< x. The Archimedean property only tells you that there is an integer larger than x. it does not tell you that that integer is anywhere near x. For example, if the x= 1.5, the Archimedean property might be telling you that the number 1,000,000 is larger than 1.5.

You will also need to use the "well ordering property" of the positive integers: every non-empty set of positive integers contains a smallest member. Apply that to the set of all integers larger than x which the Archimedean property tells you is non-empty.
• Sep 6th 2010, 12:15 PM
mjlaz

Proof: Let x $\in$ R. By the Archimedean property, there exists some n > x, n $\in$ Z. Define S = { y $\in$ Z : y > x}. Since S is bounded below by x, then by the well-ordering principle of integers, there is a least element in S, namely n = min(S). Since n $\in$ S, then n > x. However, n - 1 $\notin$ S and so n - 1 $\leq$ x < n.

I haven't proved the uniqueness component.
• Sep 6th 2010, 01:17 PM
Plato
Quote:

Originally Posted by mjlaz
I haven't proved the uniqueness component.

Is it not the case that a least element (from well ordering) is unique?
• Sep 6th 2010, 04:49 PM
Plato
Allow me to give you a different proof.
You need the completeness axiom.
We need to have proved that the set of integers, $\mathbb{Z}$, is not bounded.

If $a\in \mathbb{R}$ the define $\mathcal{S}=\{n\in \mathbb{Z}:n\le a\}$.
Clearly we can say that $b=\sup(\mathcal{S})$.
Also $\left( {\exists c \in \mathcal{S}} \right)\left[ {b - 1 < c \leqslant b \leqslant a} \right]$.
From which we get $b < c + 1 \leqslant b + 1$ which means that $c + 1 \notin \mathcal{S}\, \Rightarrow \,a < c + 1\, \Rightarrow \,c \leqslant a < c + 1$.
That completes the proof.