Proof there is a rational between any two reals

This is a problem from Rudin, but I wanted to add my own intuition to it. It uses Rudin's definition of Archimedean property. I'd just like to know if my version holds

If $x \in \mathbb R$, $y\in \mathbb R$ and $x<y$, then $\exists p \in \mathbb Q$ such that $x < p < y$

Since $x < y$, then $y-x>0$. Applying Archimedean property, we see $\exists n \in \mathbb Z^+$ such that $n(y-x)>1$.

Allow $A=\{i\in \mathbb Z\mid i > nx\}$. Obviously $A$ is bounded by $nx$, so take $m = \inf A$. So we have $m > nx$ and $m-1 \leq nx$ by set and $\inf$ definition.

This gives us $m-1 \leq nx < m$. Combining and rearranging with $n(y-x) > 1$ gives us $nx < m \leq nx +1 < ny \implies nx < m < ny$.

Thus $x < \frac{m}{n} < y$

Solution 1:

This proof works, mostly.

For completeness:

1. You should mention why $A$ is non-empty. The proof of this uses the Archimedean property.

2. You have shown that $A$ has an infimum, but you haven't shown that the infimum is in $A.$ For lots of sets of real numbers, the infimum is not in the set, so why is $A$ different?

3. Worth being explicit with $nx+1<nx+n(y-x)=ny$.

4. You could re-iterate that $n>0$ which is why you can can divide by $n$ in the last step.

But the second thing is the big problem. Why is $m\in A?$

The key is the result:

If $A\subset \mathbb Z$ is non-empty with an integer lower bound, then $A$ has a minimum element. That is, $\inf A \in A.$

This follows from the well-ordering principal of the positive integers:

If $A\subset \mathbb Z^+$ is non-empty, then it has a minimal element.

We know that our $A$ has a real lower bound, $xi$. So we need to show that there is an integer $k$ such that $k<xi,$ which is then an integer lower bound for $A.$

The integer $k$ can be shown to exist again by the Archimedean property. Find positive integer $K$ so that $-xi<K\cdot 1$ and then $k=-K.$

So what this proof is showing is how much our intuitions about the real numbers and integers are related to the Archimedean property. It is used here to:

1. Show that $n$ exists.
2. Show that $A$ is non-empty.
3. Show that $A$ has a least element.

Solution 2:

I think it's fine. It would be nice if you say that $\inf A$ exists because lower bounded subsets of $\Bbb Z$ are well-ordered.