Thread: Proof involving the Archimedean property.

Given any x R prove that there exists a unique n Z such that n - 1 x < n.

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

Originally Posted by mjlaz

Given any x R prove that there exists a unique n Z such that n - 1 x < n.

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

After visiting the professor today, I made a little ground. How about this variation?

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

I haven't proved the uniqueness component.

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?

Allow me to give you a different proof.
You need the completeness axiom.
We need to have proved that the set of integers, , is not bounded.

If the define .
Clearly we can say that .
Also .
From which we get which means that .
That completes the proof.