# Thread: Proof involving the Archimedean property.

1. ## Proof involving the Archimedean property.

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

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

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

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

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.

Proof: Let x $\displaystyle \in$ R. By the Archimedean property, there exists some n > x, n $\displaystyle \in$ Z. Define S = { y $\displaystyle \in$ 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 $\displaystyle \in$ S, then n > x. However, n - 1 $\displaystyle \notin$ S and so n - 1 $\displaystyle \leq$ x < n.

I haven't proved the uniqueness component.

4. 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?

5. Allow me to give you a different proof.
You need the completeness axiom.
We need to have proved that the set of integers, $\displaystyle \mathbb{Z}$, is not bounded.

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