# Another alternative proof for rationals are dense in the reals

I recently asked Alternative proof and hints for: "rationals are dense in the reals" from Rudin's real analysis book and got a very satisfying answer. But then for the fun of I read the following question Proving the rationals are dense in R that follows Rudin's proof on page 1.20. In that question the top answer notes that an alternative proof to Rudin's proof would be to follow by cases. The goal of this question is to attempt to do that.

The cases are:

1. $x>0$
2. $x=0$
3. $x<0$

When $x>0,x=0$ we can just re-use Rudin's after line $m-1 \leq nx < m$ and using the fact $m > nx$.

For $x<0$ then we can't really use Rudin's proof directly because $m > nx$ will probably yield a integer that isn't close enough to $nx$ (since $nx$ could be really negative and $m$ will be as small as 1). Thus instead of showing that $m$ is in the interval $nx,ny$ I will proceed that we can construct a integer in that range. My proof is as follows:

The idea is still very similar, we will use the Archimedean property/principle to yield an integer that instead bounds the interval $nx$ to $ny$ from bellow (and is as close as one can to the interval, this part is by the WOP) and we will argue that by adding 1 to it, we guarantee that the incremented integer is the the interval $nx$,$ny$ i.e. $nx < \hat k+1 < ny$ and that completes the proof since $x<\frac{\hat k+1}{n}<y$.

First use the Archimedean principle to get an integer $n$ s.t.:

$$n(y-x) > 1 \iff ny > nx+1$$

then we yield an integer $k$ that bounds the interval from bellow using the Archimedean principle: $k > -nx \iff -k < nx$ and we choose the smallest by the WOP. Call it $-k = \hat k$. Consider the incremented integer $k' = \hat k + 1$. Thus:

$$\hat k < nx \leq \hat k +1 \iff k'-1 < nx \leq k'$$

thus, we also notice that $k' < nx + 1$. Thus we have so far:

$$k'-1 < nx \leq k' < nx+1$$

and we note that out initial $n$ satifies $ny > nx+1$ so we get:

$$k'-1 < nx \leq k' < nx+1 < ny$$

which is equivalent to:

$$nx \leq k' < ny \iff x \leq \frac{k'}{n} < y$$

My question is that this seems to be awfully close to what we wanted except that we have a less than or equal instead of a strictly less than on the LHS of the inequality. Is that an issue? Did I make a mistake in the proof or is that ok? Is there a way to fix it if its wrong?

I suspect that its ok because $x < \frac{k'}{n} < y$ is a stronger statement and thus, is still true. In other words, $k'$ has either of the two properties, but the case $x>0,x=0$ yielded a $k'$ that is strictly greater, thus using that one makes this second statement still true since we need it to be less than OR equal to and since one of the two is satisfied by the other two cases, it thus, conclues the proof correct. Is that right?